A LEVEL BIOLOGY NOTES

Father God let it not be to my praise; but that in everything glory and praise be given to you; and you
alone. Remember the learners who shall use these notes. I pray for their success; bless them according to
your riches in mercy and grace. In Jesus Christ’s name . AMEN.

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 FORM 5 TERM 1
TOPIC 1: CELL STRUCTURE AND FUNCTIONS
Patience and perseverance have a magical effect before which difficulties
disappear and obstacles vanish
KEY CONCEPT OBJECTIVES CONTENT SUGGESTED SUGGESTED
Learners should be (ATTITUDES, SKILLS LEARNING RESOURCES
able to: AND KNOWLEDGE) ACTIVITIES AND
NOTES
8.1.1 Microscopy  Calibrate eyepiece - calibration and  Calibrating  Relevant reference
graticule measurement eyepiece graticule. materials
 ICT tools
 Draw and - units of measurement  Observing cells  Braille
determine linear (millimetre, micrometre using light software/Jaws
dimensions of and nanometre) microscope.  Light Microscope
specimens  Measuring linear (X4, X10, X40
dimensions of objective lenses)
specimens.  Hand lenses
-magnification and  Graticules
 distinguish between resolution (refer to light  Discussing the  Stage micrometers
magnification and and electron concepts  Stains
resolution microscopes) magnification and  Prepared slides
resolution.
- wet mounts  Mounting
 prepare temporary - staining temporary slides.
slides  Staining wet
mounts with
appropriate stains.

MICROMETRY

Micrometry: The measurement of microscopic objects.

─ A micrometer is used.
─ In biology we often need to measure very small objects.
─ When measuring cells or parts of cells, the most useful unit is the micrometer (µm).
─ One micrometer is 1000 of a millimeter.
─ 1µm =1/1000mm.
─ Even smaller structures such as the organelles with such small sizes are measured using
even smaller units.
─ The Nanometers are used (nm).
─ 1nm=1/1000µm.

NB* Note that centimeters are not units in Biology, nm, µm, & mm are used.

Microscopy

─ The use of microscopes in Biology is called microscopy.

─ Microscopes used are light microscope and electron microscope.

Distinguish between resolution and magnification.

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 ─ Resolution is the ability of a microscope to distinguish two objects close together rather
than to see them as one object.
─ Magnification is the number of times an object is enlarged.

𝒊𝒎𝒂𝒈𝒆
𝑴𝒂𝒈𝒏𝒊𝒇𝒊𝒄𝒂𝒕𝒊𝒐𝒏 = 𝒔𝒊𝒛𝒆𝒐𝒇 𝒔𝒊𝒛𝒆𝒐𝒇𝒐𝒃𝒋𝒆𝒄𝒕.
𝒓𝒆𝒂𝒍

─ E.g. A person makes a drawing of an incisor tooth and the width of the actual tooth is 5mm
while the tooth drawing is 12mm .Calculate the magnification of the drawing?

Magnification = 12/5 =x2, 4

Calculation of magnification and the conversion of units.

─ Let’s say, the real diagram of a red blood cell is 7nm and asked to calculate the
magnification.

Step 1– measure the diameter of the cell in the diagram. You find that it is 30mm.

Step2-we have been given its real size so we need to convert 30mm to µm .there are
1000µm in a mm , so 30mm =30x1000µm

Step 3-we can now put the numbers in the equation.

Calculating magnification from a scale bar

─ When given a scale bar, there is no need to measure the leukocyte .We can simply use the
scale bar.
─ All you need to do is to measure the length of the scale bar and then substitute it’s measured
length and the length that it represents on the scale bar.
─ Remember to convert the measurements to µm.

Step 1– measure the scale bar.Here it is 24mm

Step2-substutute into the equation.

𝒊𝒎𝒂𝒈𝒆
𝑴𝒂𝒈𝒏𝒊𝒇𝒊𝒄𝒂𝒕𝒊𝒐𝒏 = 𝒔𝒊𝒛𝒆𝒐𝒇 𝒔𝒊𝒛𝒆𝒐𝒇𝒐𝒃𝒋𝒆𝒄𝒕.
𝒓𝒆𝒂𝒍

= length of scale Bar

Length of scale bar represents

Light microscope

─ A light microscope (compound microscope) uses the magnifying powers of the convex lens
to produce a magnified image of a small object.
─ Therefore the magnification of the light microscope I equal to the objective lens & the
magnifying lens.

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Common stains used in light microscopy

Stain Use Colours produced

METHYLENE Staining living cells Dark blue nucleus , light cytoplasm 9in bacteria the
BLUE whole cell takes up the stain)
Iodine Staining living plant cells Very dark blue starch grains
Acidified Staining lignin (the Bright red
phloroglucinal substance in the cell walls
of xylem vessels)
Acetin orcein Nuclei and chromosomes Red
Light green Staining plant cell walls Green

Electron microscope.

─ Is like an up-side-down light microscope.

─ The radiation enters @ the top and the specimen at the bottom is viewed.
─ The principle is the same as on the light microscope in that, a beam of radiation electrons is
focused by condenser lenses and the image is magnified by further lenses.
─ A photograph taken by the electron microscope is called an electron micrograph
─ The major advantage of an electron microscope is its high resolving power i.e., 0, 5 nm in
practice.

Comparison of a light microscope and an electron microscope

Electron microscope Light microscope

Radiation source Electrons light
Wavelength About 0,005nm 400-700nm
Max resolution 0,5NM 200nm
Max magnification X250000 X1500
Lenses Electromagnets glass

Qn.Describe what is meant by the term resolution. [2]

Qn. State the maximum magnification that can be achieved by a light microscope and a
transmission electron microscope.Select your answers from the list below.

10x 40x 100x 400x 1500x 25 000x 50 000x 500 000x

light microscope ................................... x

transmission electron microscope ................................... x [Total 2 marks]

Qn. Explain why it would not be possible to see the same detail of a cell ,as can be seen using a
electron microscope on a light microscope. [3marks]

 Electron microscope has high resolution

 Because of shorter wave length
 More detail can be seen/ much clearer
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  That are close together

Microscope techniques

1. Permanent preparation
(i) Fixation
 This is the preservation of material in a life like condition.
 Tissues must be killed rapidly
 The chemical used is called a Fixative.
 The original shape and structure is maintained.
 The tissue hardens so that thin sections can be cut.
(ii) Dehydration
 Refers to the removal of water
 Its done to prepare the material for infiltration with an embedding medium or
mounding medium.
 For the preservation of fine details, dehydration should be gradual eg. Acetone,
propanone, ethanol.
(iii) Clearing
 Alcohols do not mix with some of the common embedding and mounting media.
 When it’s the case, it is replaced with a medium (clearing agent) eg xylol.
(iv) Sectioning
 When material is too thick to allow sufficient light to pass through for
microscopic investigations it is usually cut into very thin slices for the material
(sections) to be seen.
 The razor is called Microtome.
 Sections should be 8-12micrometers thick.
(v) Mounting
 Usually done on slides

(2) Temporary preparations

 Stages involved are fixation, staining and mounting.

 70% alcohol can be used as fixative

PLANT AND ANIMAL CELLS

KEY CONCEPT OBJECTIVES CONTENT SUGGESTED SUGGESTED

Learners should (ATTITUDES, SKILLS AND LEARNING RESOURCES
be able to: KNOWLEDGE) ACTIVITIES AND
NOTES
8.1.2 Plant and  identify plant and - Ultrastructure of the  Observing plant  Photomicrographs
Animal Cells animal cells plant and animal cells and animal cells.  Print media
 Drawing plant  ICT tools
and animal cells.  Braille software/Jaws
 Microscope
 compare plant - Rough and smooth  Discussing the  Prepared slides
and animal cells endoplasmic reticula, similarities and
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KEY CONCEPT OBJECTIVES CONTENT SUGGESTED SUGGESTED
Learners should (ATTITUDES, SKILLS AND LEARNING RESOURCES
be able to: KNOWLEDGE) ACTIVITIES AND
NOTES
Golgi body, differences  Models
mitochondria,ribosomes, between plant
chloroplasts, cell surface and animal cells.
membrane,
nuclear envelope,
centrioles, nucleus and
nucleolus
8.1.3 Organelles  outline the - Functions of  Discussing  Photomicrographs
and their functions functions of organelles listed functions of cell  Print media
organelles above organelles.  ICT tools
 Braille software/Jaws

I am not rich enough to give you gold; here is what I have..it may not be the best gift.. but I know you
may get one or two..share with others …

Ultra-structure of a typical animal cell

─ A typical animal cell is surrounded by a membrane known as the cell surface or plasma
membrane.
─ Inside the membrane is a jelly-like fluid known as the cytoplasm.
─ It contains the nucleus and other organelles.
─ The cytoplasm and the nucleus together are known as the protoplasm.

Nucleus

(describe the structure and Function of the Nucleus. [6])

─ It is the largest organelle in a (plant;animal) cell

─ Spherical structure 10-20 micrometers in diameter
─ Surrounded by a double membrane called nuclear envelope
─ Contains nucleoplasm
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 ─ Nuclear envelope comportmentalises chemical reactions taking place in the nucleus
─ Nucleoplasm contains chromosomes and nucleoli
─ Chromosomes contain DNA attarched to proteins (histones)
─ Nucleoplasm also contains RNA (3 types of RNA)
─ Nucleus functions to control the synthesis of proteins
─ Controls the cell’s activities
─ Devide at the start of cell division ,ensuring that daughter cells have exact copies of the
cell’s genetic material
─ To assemble ribosomes

Endoplasmic reticulum

─ It is a system of flattened membrane bound sacs called cisternae, forming tubes and sheets.
─ It is continuous with the outer membrane of the nuclear envelope.
(a) Rough endoplasmic reticulum
─ Have ribosomes on them.
─ Their role is to manufacture proteins.
─ Rough endoplasmic reticulum is abundant in cell that either secrete proteins or that are
growing rapidly.
(b) Smooth endoplasmic reticulum
─ Lacks ribosomes on its surface.
─ Abundant in cells that secrete steroids or lipid substances
─ These are the site for lipid and steroid synthesis.

Functions of endoplasmic reticulum

─ To provide area for biochemical reactions.

─ To act as a pathway for the transport and exchange of material.
─ To manufacture proteins e.g. enzymes
─ To manufacture lipids and steroids.
─ To collect and store any synthesized material.
─ To form a structural skeleton for maintaining cellular shape

Ribosomes

─ Very small organelles consisting of a large subunit and a small subunit.

─ They are made of roughly equal amounts of protein and RNA.
─ There are two types i.e. 70s and 80s ribosomes.
─ They are responsible for protein synthesis .
─ They are either bound to ER or lie free in the cytoplasm.
─ They form polysomes i.e. collection of ribosomes strung along messenger RNA

Mitochondria

(Describe the structure of a mitochondrion and outline its function in a plant cell.[8])

─ 0.5–1.0 μm, diameter / width ;

─ double membrane ;
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 ─ inner membrane folded / cristae ;
─ hold, stalked particles / ATP synthase / ATP synthetase ;
─ site of ETC ;
─ ref. H+ and intermembrane space ;
─ ATP production ;
─ oxidative phosphorylation / chemiosmosis ;
─ matrix is site of, link reaction / Krebs cycle ;
─ enzymes in matrix ;
─ 70S ribosomes ;
─ (mitochondrial) DNA ;
─ They are envelope bound and the inner membrane folds to form cristae.
─ It consists of a matrix with few ribosomes, a circular DNA molecule and phosphate granules.
─ In aerobic respiration, cristae are the sites for oxidative phosphorylation and electron
transport chain.
─ Matrix is the site for Krebs’s cycle of enzymes.

Golgi apparatus.

─ Composed of a series of parallel membranes that are flattened fluid spaces .

─ The cristae are slightly curved the entire structure appear concave.
─ Functions of Golgi apparatus:
o Manufacture of glycoproteins which are required for secretions.
o Production of secretory enzymes.
o Production of carbohydrates e.g. those involved in the manufacture of new cell walls.
o Transport and storage of lipids.
o Formation of lysosomes.

Lysosomes.

─ A simple spherical sac bound by a single membrane and containing digestive and or
hydrolytic enzymes.
─ Functions of Lysosomes:
o To contain enzymes capable of digesting a wide variety of substances.
o To digest cytoplasmic organelles.
o To act as suicide bags which help to rapidly digest entire cells that are old.

Vacuoles
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 ─ They are temporary membrane bound pockets of cell sap.
─ White blood cells in higher animals form similar vacuoles around pathogens which are
engulfed.

Centrioles

─ Found as a pair near the nucleus.

─ They are made up of bundles of tubules.
─ They pull apart during cell division to produce a spindle made of microtubules which are
involved in chromosome movement.

Cilia and flagella

─ They are concerned with cell movement e.g. they are outgrowths from cells which can beat
either in one direction (cilia) or in a wavelike manner (flagella).
─ Flagella are larger than cilia.
─ Both cilia and flagella have a characteristic 9+2 arrangement of microtubules.

Plant cells

─ Plant cells tend to be uniform in their shape because the cell is bound by a rigid cell wall.
─ The cells give strength and support to plants due to insoluble cellulose fibres which are
meshed in a matrix of carbohydrates called pectates or hemicelluloses.
─ Plant Cells have have the following:

Vacuole

─ Is a fluid filled space in the cytoplasm.

─ In plants a vacuole is a permanent feature.
─ It is surrounded by a membrane called tonoplast.
─ Is filled with cell sap.
─ The vacuole determines the osmotic properties of a plant cell.
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Chloroplast.

(Relate the structure of the chloroplast to their roles in photosynthesis. [7])

─ Are large organelles containing their own DNA and have a double membrane.
─ Chloroplasts have a folded inner membrane which gives a greater surface area for
biochemical reactions to occur.
─ Thylakoid membranes contains pigments/ electron carriers/ enzymes
─ Used in cyclic and non – cyclic photophoshorilation/ light dependent reactions
─ Stroma (contain enzymes) for the calvin cycle/ dark reaction
─ Grana a network of proteins holding pigments into photosynthesis
─ Light reactions on thylakoid which contain ATP/ stalked particles
─ Membraine system separates the reactions of photosynthesis from other cell reactions
─ Stroma fluid which surrounds grana so that products light dependent stage can easily pass
into stroma.
─ Contain DNA /ribosomes for manufacture of proteins needed for protein synthesis
─ Granna are interconnected by membranes called lamella.

(List any 3 structural similarities between a mitochondrion and a chloroplast. [3])

─ Circular DNA
─ Double membranes
─ 70s ribosomes

Prokaryotes and eukaryotes

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Learners should (ATTITUDES, SKILLS AND LEARNING RESOURCES
be able to: KNOWLEDGE) ACTIVITIES AND
NOTES
8.1.4 Eukaryotic and  compare - Structure of eukaryotic  Observing and  Prepared slides
Prokaryotic Cells eukaryotic and and prokaryotic cells drawing  Microscope
prokaryotic cells eukaryotic and  Print media
prokaryotic cells.  ICT tools
 Discussing the  Braille software/Jaws
similarities and
differences
between the
cells.
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 ─ Prokaryotes are organisms which are single celled e.g. bacteria.
─ Eukaryotes are organisms usually complex and these include animal and plant cells.

Feature Prokaryotes Eukaryotes

Cell division Binary fission. Mitosis and meiosis.
No spindle Spindle in animal cells.
Genetic material. DNA is circular and lies free in DNA is linear , often associated
the cytoplasm . with proteins to form
No true nucleus. chromosomes .
It is contained in the nucleus.
Protein synthesis. 70s ribosomes. 80s ribosomes.
No ER present. ER present and ribosomes may
be attached to ER.
Organelles. Few organelles. Many organelles ,.
None envelope bound. Envelope bound organelles e.g.
nucleus.
Cell walls. Rigid, contain polysaccharides Rigid , contain polysaccharides
of amino acids , murein is the , lignin is the main
main strengthening compound. strengthening material.
Respiration Use mesosomes except blue Use mitochondria for aerobic
green bacteria cytoplasmic respiration.
membranes.
Photosynthesis No chloroplasts Contain chloroplasts
Nitrogen fixation Some have the ability None have the ability.

(Compare and contrast the structure of a typical animal and plant cell. [6])

Structure Plant cell Animal cell

Cell wall Cellulose No cell wall
Cell membrane Protein/ phospholipid or Protein/ phospholipid or
phospholipid bilayer (same phospholipid bilayer (same
as animal) as animal)
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 Cytoplasm Fluid with dissolved Fluid with dissolved
substances substances
Vacuole Large central vacuole If present, small and
temporary
Shape Regular Irregular
Chloroplast Present Absent
Mitochondrial Present Present
Plasmodesmata Present Absent

MOVEMENT OF SUBSTANCES INTO AND OUT OF CELLS

OBJECTIVES CONTENT SUGGESTED SUGGESTED
KEY CONCEPT Learners should (ATTITUDES, SKILLS AND LEARNING RESOURCES
be able to: KNOWLEDGE) ACTIVITIES AND
NOTES
8.1.5 Movement of  describe and - Fluid mosaic model  Drawing the cell  Print media
substances into and explain the cell including the roles of surface  Photomicrographs
out of cells surface phospholipids, membrane.  ICT tools
membrane cholesterol, glycolipids,  Identifying the  Braille software/Jaws
structure proteins and components.
glycoproteins  Discussing the
functions of parts
of the cell
surface
membrane.
 relate the - Diffusion  Designing and  Onion
structure of the - Facilitated diffusion carrying out  Potatoes
membrane to - Osmosis experiments to  Slides
movement of - Active uptake demonstrate  Microscope
substances into - Endocytosis osmosis (include  Egg membrane
and out of cells - Exocytosis serial dilutions).  Visking tubing

Membranes

─ Cell membranes separate their contents from their external environment controlling the
exchange of materials between the two.
─ Membranes also act as receptor sites for hormones and neurotransmitters and other
chemicals.
─ Membranes are described as selectively permeable, since other substances such as
glucose , amino acids, fatty acids , glycerol and ions can diffuse through slowly.
─ Membranes are made almost entirely of proteins and lipids.

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Phospholipids

─ Each phospholipid molecule consist of a polar head containing a phosphate and two fatty
acids.
─ The polar head is hydrophilic and the tails are hydrophobic.
─ In aqueous environments, the hydrophilic heads face the external environments and the
hydrophobic fatty acid tails come into close intact to exclude water.

Proteins

─ Freeze fracturing reveals the presence of proteins which penetrate into and often through
the phospholipids bilayer.
─ The more metabolically the membrane is, the more the protein cuticles are found.

Glycolipids and cholesterol

─ Glycolipids are lipids with carbohydrate residues, like phospholipids they have polar heads
and non-polar tails.
─ Cholesterol acts as a plug reducing even further exit and entry of molecules through the
membrane.

Fluid mosaic model membrane

(why is it called the fluid mosaic model membrane? [2marks])

─ This was so called because of the protein molecules which float about hap-hazard in the
phospholipid bilayer
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 ─ The membrane is 7nm thick.
─ Its basic structure is the phospholipid bilayer.
─ The phospholipids are fluid and move about rapidly by diffusion in their own layers.
─ Unsaturated fatty acids are bent.
─ Most protein molecules float about in the phospholipid bilayer forming a fluid mosaic
pattern.
─ The proteins stay in the membranes because they have regions of hydrophobic amino acids
which interact with fatty acid tails to exclude water.
─ The two sides of the membrane can differ in composition and function

(Explain how the phospholipid bilayer prevents the flow of protons)

─ The phospholipid bilayer is non polar whilst the hydrogen ions are polar because of the
charge they possesses so the polar substance cannot easily pass through the non-polar
unless there is a hydrophilic channel.
─ The cytoplasm is negatively charged whilst the hydrogen ions are positively charged such
that they attract each other , so the phospholipid bilayer prevents the flow because the
hydrogen ions will cause a change in electrochemical gradient of the cytoplasm.

Channel proteins and carrier proteins

─ These are involved in the selective transport of polar molecules and ions across the
membranes.
 Qn. Outline the process of cell signaling? [4marks]
 Qn. Outline the route by which proteins are exported from a cell. [3marks]

Transport across the cell surface membrane.

Outline four reasons why transport of substances across membranes is vital to a cell.

─ To generate ionic gradient for action potentials/ muscle contraction

─ To secrete useful substances;
─ To excrete waste substances;
─ To obtain nutrients;
─ To maintain suitable pH/ ionic concentration;

Qn. Small non-polar substances enter cells in different ways to large or polar substances.

Outline the ways in which substances, other than water, can enter a cell through the
plasma (cell surface) membrane.

In your answer, you should use appropriate technical terms, spelt correctly.

Small, non-polar substances

Large substances

Polar substances [Total 5 marks]

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State the 3 features/ properties of a substance which influence its ability to pass through a cell
surface membrane.

─ Polarity (polar or non-polar)

─ Size of the particle

─ Solubility (lipid solubility/ water solubility )

Qn. Outline the roles of the membrane? [4marks]

Qn. Explain why the cell surface membrane is impermeable to most biological molecules.
[4marks]

Endocytosis is one method by which substances enter cells

Describe the process of endocytosis.

─ membrane , folding in / engulfing / invaginates / AW ;

─ fuses with itself / pinches off ;
─ formation of , vesicle / vacuole ; A completely surrounded by membrane
─ fate of vesicle ; e.g. moves through cytoplasm / fate of contents
─ ref. fluid nature (of membrane) / requires energy ;
─ A active / ATP R active transport
─ triggered by binding of molecule (to receptor site) ;
─ ref. to uptake of solid and liquid (not name alone) ;

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TOPIC 2: Biological molecules and Water
Someone asked me, “Why do you insist on taking the hard road?” and I replied,
“Why do you assume that I see two roads?” So challenges are what make life
interesting and overcoming them is what makes life meaningful.

KEY CONCEPT OBJECTIVES CONTENT SUGGESTED SUGGESTED

Learners should be (ATTITUDES, SKILLS LEARNING ACTIVITIES RESOURCES
able to: AND KNOWLEDGE) AND NOTES
8.2.1Carbohydrates  carry out tests to - Reducing sugars  Performing the  Benedict’s solution
identify - Non- reducing reducing and non-  Reducing sugars
carbohydrates sugars reducing sugar tests.  Non-reducing
- Starch  Carrying out the sugars
- (Qualitative and starch test.  Potassium iodide
Quantitative tests) solution
 Colorimeters
 describe the - Glycosidic bond  Illustrating formation
formation and - Starch and breakage of
breakage of - Glycogen glycosidic bonds.  ICT tools
glycosidic bond - cellulose  Braille
software/Jaws
 describe the  Discussing the  models
synthesis and synthesis and
molecular molecular structure of
structure of starch, glycogen and
polysaccharides cellulose.
 Observing molecular
 relate structures structures of
of polysaccharides.
polysaccharides
to their functions  Discussing the link
in living between the structure
organisms and the function of
each polysaccharide.

Macromolecule

─ A macromolecule is a giant molecule made from many repeating molecule unit.

─ Poly-many, saccharides -carbohydrates.
─ Such molecules are called polymers.
─ The individual units are known as monomers.
─ The units are joined together by condensation reaction which is the removal ofwater
molecules.
─ The bonds can bebroken by hydrolysis .Example of macromolecules are polysaccharides,
proteins and nuclei acids e.g DNA and their constituent monomers are monosaccharides,
amino acidsand nucleotide respectively.

Carbohydrates

─ Carbohydrates aresubstances which contains the elements (carbon, hydrogen,and

oxygen)and hence a general fomular Cx(H2O)y where x and y are variable number .
─ They have the following general properties :
i. They area aldehyde or keto
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 ii. They all contain several hydroxyl group.

Monosaccharide

─ Monosaccharide are single sugar units.

─ Their general formula is (H2O)n
─ They are classified according to the number of carbon atom i.e triose, tenthose, pentose,
hexose are the most common.
─ Triose e.g glyceraldehyde and dihydroxyacetone are intermediates in respiration.
─ Pentose e.g ribose, deoxyribose and ribulose aid in the synthesis of nucleic acid and
coenzyme .

Aidose and Ketoses

─ In monosaccharaides all carbon atoms have a hydroxyl group attarched . The remain carbon
atom is either c-atom.

Open chain and closed chain

─ The rings structure of pentose and hexose are the usual forms with only a small portion of
the molecules in the open chain forms.
─ These ring structure is the form used to make disaccharides and polysaccharides.

Alpha and Beta isomers

─ Glucose can exists in two different ring forms called the Alpha and Beta forms.
─ The hydroxyl group on the carbon atom 1 projects below the ring (alpha glucose) or above
the ring (BETA GLUCOSE).
─ These two are said to be isomers of each other.
─ Alpha glucose is used to make starch (monomer of starch) and Beta glucose used to make
cellulose.

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Disaccharide

─ Are formed when two monosaccharides usually hexose combines by condensation reaction
forming a glyosidic bond.
─ It normally forms between C-atoms 1 & 4 of neighboring (1,4 linkages ).
─ Most common disaccharidesare maltose, lactose and sucrose.

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Polysaccharides

─ Function as store of food and energy e.g starch and glycogen and as structural material e.g
cellulose.
─ They are convenient storage molecules cause their large size make them more insoluble in
water, hence they do not exert on osmotic pressure on the cell.
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Starch.

(Describe the molecular structure of starch. [6])

─ Is a polymer of alpha glucose.

─ It is an energy store in plant.
─ It has two components which are amylase and amylopectin.
─ Both are polymers of alpha glucose
─ Amylase have a straight chain and the glucose residues are joined together by 1,4 glycosidic
bond.
─ Amylase forms 20% of starch
─ These bonds cause the chains to coil helically into more compact shape.
─ Amylopectin is also more compact as it have many branches formed by 1,4 and 1,6
glycosidic bonds .
─ Coiled chains (may) contain 1 500 monomers with braches every ten (10) units
─ Forms 80% of Starch
─ As suspension of amylase in water gives a blue black colour with iodine (potassium iodide)
andamylopectin give a red violet colour .

Glycogen

─ Is of animal equivalent of starch and act as a store of energy.

─ It is a polysaccharide made from alpha glucose.
─ In vertebrates, glycogen is stored in the liver and muscles where it work as an energy
reserve.
─ Its conversion back to insulin is controlled by insulin

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 Structure Function related to structure
Starch It consist branched and The large size of the molecule makes
unbranched chain of amylase starch insoluble and can be used for food
and amylopectin molecules storage in plant.
.these chain comprises large
alpha glucose units that’s are
folded.

Cellulose It consist of Beta GLUCOSE The cross bonds give cellulose its tough
units in a long unbranched and resistance properties to provide
chain linked by many cross structural strength to cell walls
bridges between the chains.

Glycogen It consist of alpha glucose units It is a mean of food storage in animals due
in more branched and compact to its compact shape that is not bulk.
chains.

Cellulose

─ Is a polymer of Beta glucose and has a structural role

─ One of the glucose molecule is rotated 180 for the bond to be formed. This rotation gives
cellulose a different structure to starch.
─ It consist of long chain of glucose residues with about 10 000 residues per chain.
─ The Beta 1, 4 linkages make the chain in contrast to starch where 1,4 linkages cause of chain
to be curve.
─ Hydroxyl project out wards from each chain in all direction and form hydrogen bonds with
neighboring chains .
─ The chain associate in group of appropriate 60-70 to form micro fibrils which are arranged
in larger bundle to form micro fibrils.
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 ─ These have tremendous tensile strength.
─ Several layers of cell wall are found in cell wall.
─ This prevents the cell from bursting when water enters the cell by osmosis and help to
determine the shape of plant cells.
─ The layers are freely permeable to water and food material.

Summary of differences between starch, glycogen and cellulose

Starch Glycogen cellulose

Name of monomer Glucose glucose Beta glucose
Bonds between I,4 glycosidic 1,4+1,6 glycosidic 1,4 glycosidic
monomers (amylase) 1,4+1,6
(amylopectin)
Characteristics of Coiled unbranched Short branched chains Straight long
chains + unbranched chains
long branched

Lipids
KEY CONCEPT OBJECTIVES CONTENT SUGGESTED SUGGESTED
Learners should be (ATTITUDES, SKILLS LEARNING ACTIVITIES RESOURCES
able to: AND KNOWLEDGE) AND NOTES
8.2.2 Lipids  identify lipids in - emulsion test - Carrying out the  Lipids
different emulsion test.  Alcohol
substances
- triglycerides - Illustrating the  ICT tools
 describe the - phospholipids molecular
molecular structures of a  Braille
structures of a triglyceride and a software/Jaws
triglyceride and a phospholipid.  Models
phospholipid
- Observing the
 relate the molecular
Structures of structures
triglycerides and - Discussing the
phospholipids to relationship
their functions in between structures
living organisms and functions.

─ Lipids are classified in general as water insoluble substances extracted from cells by solvent
e.g benzene,chloroform,ether. Lipids are made of glycerol and fatty acids.

Fatty acids

─ Contain the acidic group (-COOH).

─ Have the general formula R-COOH .R is the hydrogen or group such as –CH3-C2H3 etc.
increasing CH3 IN each subsequent member of the series.
─ Most naturally occurring fatty acids have an even number of carbon atom between 14 and
22.

Steric acids

─ The tails are hydrophobic.

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 ─ Fatty acids with one or more double bonds etc. oleic acid are said to be unsaturated while
those lacking double bond are saturated. Unsaturated bonds melt at a lower temperature
than saturated bonds.

Triglyceride

─ Glycerol has 3 hydroxyl all of which can bond with 3 fatty acids to form a triglyceride
─ Condensation reaction takes place resulting in an ester link and is known as esterification.

Function of triglycerides

─ They are non-polar and no even distribution charge which mean they do not form hydrogen
bonds with water molecules and do not dissolve in water (hydrophobic).
─ They are less dense than water, they float on the top and they act as energy store.
─ Have high energy value than carbohydrates cause of higher proportion of hydrogen as
compared to carbohydrates.
─ Animal excess energy to fats and in vertebrates the fats act as the insulator.
─ When fats are oxidized H2O is a product to desert animals e.g. kangaroo rat.

Explain how the molecular structure of triglycerides is related to their functions.[8]

─ possess hydrophobic tails of fatty acids;

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 ─ which cause the molecule to be insoluble in water;
─ they are not so easily dissolved out of the cell;
─ this functions to provide the properties of the phospholipid bilayer in cell
membranes;
─ acts as energy store for the cell;
─ due to their higher proportion of hydrogen compared to carbohydrates;
─ as a result the breakdown of triglycerides yields ore energy;
─ due to the lower proportion of oxygen to carbon that requires more oxygen for
complete oxidation to occur;
─ triglycerides also float in water due to their lighter density;
─ this enables them to aid in the buoyancy of aquatic animals;

(Describe with examples the roles of lipids in an organism. [6])

─ Used as high energy stores e.g triglycerides (fat or oil)

─ Used in waterproofing coverings e.g. in wax cuticles found in insects and leaves to prevent
water loss while sea birds prune their feathers with oil
─ Form an insulating layer against heat loss
─ Phospholipids are major components of cell membranes
─ Steroids such as cholesterol is an important component of cell membranes and nerve fibres.
Steroids form the basis of many hormones
─ Lipids form the basis of scents that attract insects for pollination

Phospholipids

─ These are the containing phosphate, one of the OH (hydroxyl group) of glycerol, combines
with phosphoric acid (H3PO4) and the other combine with fatty acids.

Describe the structure of a phospholipid [3marks]

─ Glycerol
─ 2 fatty acids/hydrocarbon(s) (tails /chains)
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 ─ Phosphate

Suggest 2 possible functions of carbohydrate chains attached to the membrane [2 marks]

─ Cell recognition
─ Carbohydrates cement/ stick cells together
─ Receptor sites for chemical signals

Glycolipid

─ Is the association of lipids and carbohydrates.

─ The carbohydrates chain forms at the polar heard of the molecules.
─ Glycolipids are found in membrane.

Structure Role in cell

Triglyceride  Possess hydrophobic tails of fatty  Alternative energy for the
acids which cause the molecule to cell
be insoluble in water
Phospholipids  Possessing a hydrophilic  Major component of the cell
phosphoric head and hydrophobic membrane bilayer.
tail of fatty acid.
Glycolipid  Possessing a carbohydrate  Found as one of the
component in the lipid structure component in the cell
membrane bilayer and act as
recognition site .it also make
the cell membrane more
stable.

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PROTEINS

KEY CONCEPT OBJECTIVES CONTENT SUGGESTED SUGGESTED

Learners should be (ATTITUDES, SKILLS LEARNING ACTIVITIES RESOURCES
able to: AND KNOWLEDGE) AND NOTES
8.2.3 Proteins  Identify proteins in - Biuret test - Carrying out the  Biuret reagents
different food Biuret test for
substances proteins.

- Amino acid - Observing the  ICT tools

 describe the molecular structure  Braille
structure of an of amino acid. software/Jaws
amino acid - Demonstrating  Print media
 outline the - Peptide Bond peptide bond  Models
formation and - Dipeptides formation and (buttons/beads
breakage of a - Polypeptides breakage. threads)
peptide bond
- Illustrating
 explain the structures of
meaning of the - Primary, proteins.
terms primary, Secondary,
secondary, Tertiary,
tertiary and Quaternary
quaternary structures
structures of - Discussing the
proteins various bonds in
 describe the types proteins.
of bonds which - Hydrogen, ionic,
hold the protein disulphide and
molecules in hydrophobic - Making models of
shape interactions haemoglobin and
 describe the collagen.
molecular - Haemoglobin
structures of - Collagen
haemoglobin and - Discussing the
collagen relationship
 relate the between structure
structures of and function.
haemoglobin and
collagen to
theirfunctions in
living organisms
- Constructing
- Lock and key models to
 explain the mode
hypothesis demonstrate the
of action of
- Induced fit mode of action of  Catalase
enzymes
hypothesis enzymes.  Amylase
 Substrates
 follow the  Buffers
- Enzyme catalyzed - Measuring the rate
progress of an
reactions of formation of  Acids and bases
enzyme catalyzed
products or rates of  Inhibitors
reaction
disappearance of  Models of
substrates. enzymes

 explain factors - Effects of - Carrying out

affecting rate of temperature, pH, experiments to
enzyme catalysed enzyme show effects of the
reactions concentration and factors on the rate
substrate of reactions.
concentration
 explain the effect - Reversible and - Demonstrating
of competitive and non- reversible effects of inhibitors
non – competitive inhibition on enzyme
inhibitors on Inhibitors such as catalysed
enzyme activity heavy metals reactions.
(cyanide, mercury),
insecticides

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Amino acids

─ These are the basic units from which the proteins are made.
─ Plants are able to make all the amino acids they need in their diet, these are called essential
amino acids they require fromthese.

─ There is a central carbon atom attached to an acidic carboxyl group, a basic amino group i.e.
NH2 and a hydrogen atom.
─ The 4 position is occupied by the R group. This group give each amino acids have its
uniqueness.
─ Amino acids are amphoteric (are acidic and basic) because they contain an acid and basic
part.Such ions are dipolar and are known as zwitterions.
─ Amino acids maintain the pH cause in acid conditions it absorbs protons while in basic
condition it donates proteins.

Bonds used in protein structure

─ Amino acids combine to form protein and are joined together to form peptides bonds
─ The protein folds into a particular shape as a result of 4 types of bonds. These are
ionic,disulphide, hydrogen and hydrophobic interaction.

Peptide bonds

─ Is formed when condensation reaction occurs between the amino group of one amino acids
and the carboxyl group of another.
─ A polypeptide is formed when amino acids are joined this way.

Qn. State 2 sites in a eukaryotic cell were peptide bonds are formed. [2]

─ cytoplasm and Rough endoplasmic reticulum

Qn. Explain the significance of the R groups of different amino –acids to protein structure. [3]

─ it gives amino acid its uniqueness, by differentiating one amino acid from another.
─ the R groups determine the chemical characteristics of the whole amino acid

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Ionic bonds

─ Acidic R groups are negatively charged and basic R groups are positively charged.
─ They can be attracted to each other, formic ionic bonds.
─ In aqueous environment, this bond is weaker than a covalent bond and can be broken by
changing the pH.
─ As a result pH changing have a destructive effects on protein structure.

Disulphide bonds

─ The amino acid cistern have contain a sulphide group –SH as its R group.
─ If two molecules of cistern lie up alongside each other neighboring sulphidryl groups can be
oxidized and form disulphide bonds
─ Disulphide can be formed between different parts of chain of amino acids e.g.
insulin,causing the molecule to fold into a particular shape.

Hydrogen bonds

─ When hydrogen is part of an OH group /NH group it become slightly positively charged
(electropositive) cause the electron which are shared and which are negatively charged are
attracted more towards the oxygen and nitrogen atoms .
─ The hydrogen may then be attracted towards a neighboring electronegative oxygen /
nitrogen atom.

Hydrophobic interaction

─ If a polypeptide contain non-polar R groups i.e. hydrophobic and is in aqueous environment,

the chain will tend to fold so that hydrophobic groups come into close contact and exclude
water.
─ This how many globular protein folds up.
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Protein

─ Are made from amino acid therefore contain the elements C, H, O and N & in some cases
sulphur.
─ Proteins are macromolecules of high Mr, consisting of amino acids.
─ The potential variety of protein is limited because the sequence of amino acids in each
protein is specific for that protein and is generally controlled by the DNA of the cell on
which it is made.

Structure of protein

─ Each protein contain a characteristic of 3 dimension shape, its conformation.

─ There are four levels of structure which are as follows :

Primary structure

─ It is the number of sequence of amino acids.

─ The sequence of amino acids of a protein detects its biological function.
─ In turn this sequence is controlled by the sequence of bases on DNA.
─ Substitution of a single amino acid can cause major alteration in the protein function, as in
the sickle cell anemia.

Secondary structure

─ The most common secondary structure is Alpha helix whose structure is maintained by
hydrogen bonds which are formed between neighboring CO and NH groups.
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 ─ The hydrogen atom of the NH group is bonded to oxygen of carbon dioxide group, 4 amino
acids away.
─ Amino acids would be bonded to 5 &2 to 6 etc.
─ Alpha helix make one complete turn for every 36 amino acids.

Keratin
─ Is a protein which is alpha helical and is the structural protein of hair protein, nail.
─ Its hardness and stretchability may vary with the degree of cross linkages ofdisulphide
bonds between neighboring chains.
─ Another type of secondary structure is B pleated sheet.
─ The proteins that make silk, is fibrin is entirely in this B pleated form.
─ It is made up of a number of adjacent chains which are more extended than the alpha helix.
─ They are arranged in parallel fashion ,either running in the same direction, or in opposite
direction they are joined by hydrogen bond , formed between the CO and NH groups ,of one
chain and groups are involved in hydrogen bonding to the structure is rigid and very stable .
─ The whole structure is known as the B pleated sheet.
─ In globular protein a single polypeptide chain commonly fold back to itself several times to
form B pleated sheet.
─ Another arrangements is seen in the protein collagen.
─ Three polypeptides chain are found wound around each other.
─ They are like strands of a rope to form a triple helix.
─ There are about 1 000 amino acids in each chain and the complete triple helix structure is
called tropollogen.
─ The 3 strands are held together by hydrogen bonds.
─ Many triple helix can be parallel to each other to form fibrils.
─ They are joined by covalent bonds.
─ Fibrils in turn unite to form fibers.

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 Collagen
─ Collagen is the main protein that form teuckans,ligaments, connective, tissues and skin.

Tertiary structure

─ Usually the polypeptide chain bends and fall extensively, forming a compact globular
structure.
─ This is the tertiary structure and is maintained by ionic, disulphide and hydrogen bonds and
hydrophobic interaction (all the bonds).
─ Myoglobin is formed found in the muscles where its function is rto store oxygen.
─ Oxygen combines with heam group contained in myoglobin just like in heamoglobin.

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Quaternary structure

─ Highly complex protein consists of more than one polypeptide chain.

─ The chains are held together by hydrogen bonds ionic bonds hydrophobic interaction.
─ Their precise arrangement is known as quaternary structure.
─ It consists of four separate chains of two types which is Alpha chains and Beta CHAINS.
─ Theseresembles myoglobin in structure to Alpha CHAINS, EACH CONTAIN 146.
─ A mutation which causes one of the hydrophilic amino acids to be replaced by a
hydrophobic amino acids there by reducing its solubility is responsible for sickle cell
anemia.

An enzyme, such as amylase, has a specific 3-dimensional shape.’

Explain how DNA structure determines the specific shape of enzymes.

─ DNA codes for , protein / polypeptide ;

─ transcription and translation (or described) ;
─ enzyme is globular (protein) ;
─ 3 bases 1 amino acid ;

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 ─ sequence of , bases / triplets , determines , sequence of amino acids / primary
─ structure ;
─ coiling / helix / pleated sheet / particular secondary structure ;
─ determines projecting side groups ;
─ folding / bonding , for tertiary structure ;
─ 3-D structure is tertiary structure ;
─ AVP ; e.g. ref. active site related to shape
─ 2 or more genes produce quaternary structure

Denaturation of protein

─ Denaturation the loss of three dimensional shape of a protein molecule.

─ The amino acids sequence of protein remains unaffected (the primary structure remains the
same )
─ If denaturation occurs, the molecules unfolds and can no longer perform its normal
function.

Qn (a) In globular proteins, the polypeptide chain bends and folds to give a more compact shape.
This is the tertiary structure of the protein. Name three types of bond that help to maintain the
tertiary structure.
[3]
(b) Monosaccharide can also be linked together to form long chain molecules
calledPolysaccharides. State two ways, other than the names of the monomers present, in which the
structure of a Polysaccharide chain differs from that of a polypeptide chain.
[2]
(c) The fibrous protein collagen and the polysaccharide cellulose both possess considerable tensile
strength. List two features that contribute to the strength of
(i) Collagen; [2]
(ii) Cellulose. [2]

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Key points
Primary structure
 Number + sequence of amino acids
 By peptide bonds, involving condensation reaction;
 Sequence determined by DNA
Secondary Structure
 Folding/twisting;
 Resulting from hydrogen bonding;
 Produce alpha helix;
 And beta pleated sheet;
 Carboxyl group of each amino acid hydrogen bonded to amide hydrogen of amino acid
4 residues away in chain, eg myosin in muscles
 Protein chains lie side by side held in position by Hydrogen bonds between amide
carboxyl oxygens of one chain and amide hydrogen of adjacent chains;
 Eg silk
Tertiary structure
 Bending and folding of proteins
 To form a specific three-dimensional shape;
 Results from interactions between R side chains of amino acids residues;
 R-group interactions (a) Disulphide bridges between two cysteine residues that are
close to each other in chain/in different chains
(b)ionic bonds between ionized side chain of acidic amino acid
(-COO-) and side chain of basic amino acid (-NH3+)
(c)Hydrogen bonds between a variety of R groups /side chains
especially those that possess the functional group
-OH
-NH2
-CONH2
(d)Hydrophobic reactions: result when non polar groups are attracted to one another of forced
together by their mutual repulsion of aqueous solvent. Interactions of this type common
between nonpolar R groups

Quaternary Structure
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  Many functional proteins contain two or more polypeptide chains held together by ionic
interactions, disulphide bridges, hydrogen bonds + Hydrophobic interactions;
 Each of the polypeptide chains has its own primary structure, secondary structure , and
tertiary structure;
 Eg. Haemoglobin made of four chains- 2 alpha chains each with 141 amino acids
-2 Beta chains each with 146 amino acids
 Each of the subunits contains a haem group with iron in its center;

Qn. Monosaccharide can be linked together to form long chain molecules called
polysaccharides. State two ways, other than the names of the monomers present , in which the
structure of a polysaccharide chain differs from that of a polypeptide chain. [2]

Difference between globular proteins and Fibrous proteins

GLOBULAR PROTEIN FIBROUS PROTEINS
It has polypeptide chains with irregular It has polypeptide chains with regular repetitive
sequence of amino acids sequences of amino acids
Its shape is a compact globule of polypeptides It has long chins running parallel
It is chemically less stable and stable its activity It is chemically stable and relatively unaffected
if affected by factors such as its by temperature, concentration and pH
concentration,pH and temperature
Each molecule of the same type of globular Each molecule of the same type of Fibrous
proteins has a specific sequence proteins may vary in length with slightly
different sequences of amino acids
It is water soluble It is insoluble in water
It is involved in various body systems such as Its roles is mainly in helping to maintain
the digestive system, the endocrine system and structure and providing support.
the immune system

ENZYMES
─ Enzymes are protein molecules synthesized by living cells that speed up the rate of chemical
reactions. That is they are biological catalyst, globular proteins.
─ They are used to catalyze a vast number of reactions at temperatures suitable for living
organisms, between 5-40 degrees Celsius.

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─ The chemical which an enzyme works on is called a substrate.
─ An enzyme, combined with its substrate to form a short lived enzyme / substrate complex.
─ Once catalysis has occurred, the complex breaks up into product and enzyme.
─ The enzyme remain unchanged at the end of the reaction
─ A number of enzymes can be used in sequence to convert one substrate into one / several
products via a series of intermediate compounds.
─ The chain of such reaction is known as a metabolic cell owing to the specific enzymes.
─ This enzyme serves to control the chemical reaction that occurs within cells.

PROPERTIES OF ENZYMES

Enzymes possess the following properties

─ They are all globular proteins

─ Being proteins, they are coded for by the DNA
─ They are catalyst
─ They are very efficient, they have a high turnover rate .A typical enzyme undergoes 100
reactions.
─ They are highly specific; an enzyme will only catalyze one reaction.
─ Their catalyzed reaction is reversible
─ They are affected by pH, temperature, substrate concentration & enzyme concentration.
─ Enzyme lower Activation energy
─ Enzymes possess Active sites where the reaction take place
o Mechanism of enzymes
─ Activation energy is the amount of energy required for a reaction to occur.
─ Enzymes by function as catalyst serve to reduce the activation energy (E.a) required for a
chemical reaction to occur, thus the overall rate of reaction to greatly increased.

─ Enzymes are very specific. This is because enzymes have a particular shape into which the
substrate fit exactly. This is referred to as the lock and key hypothesis.
─ The substrate is thought to be like a key whose shape is complementary to the lock.
─ The site where substrate bite in the enzyme is called the active site and it is this which has a
specific shape.

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 ─ Active sites are usually a very small position of enzyme between 3 & 12 amino acids long.
These amino acids are brought together to for a particular shape of INDUCED FIT
HYPOTHESIS.
─ The active site, due to the interacting of hydrogen, ionic, disulphide linkages & hydrophobic
interactions or bonds – the tertiary structure of proteins.
─ Sometimes when the substrate do not fit exactly into the active site maybe induced or
moulded into a more precise shape as to fit the substrate for catalysis to occur more
effectively.
─ This is known as the Induced fit hypothesis
─ Enzymes change shape slightly as substrate enters active site making the fit more precise.

Factor affecting the rate of enzyme reaction

─ The rate of an enzyme controlled reaction is measured by the amount of substance changed
(e.g. use of amylase/ amount of product formed(e.g. the use catalase )formed during a
period of time
─ The rate is determined by measuring the shape of the tangent to the curve in the initial
stage of the reaction.

(with reference to a named example, describe one model of enzyme action. [6])

─ Enzyme are very specific

─ They have a particular shape into which the substrate or substrate fit e.g. the enzyme
amylase will only act on starch converting it to maltose
─ Although the enzyme molecule is large, overall, only a small region of it is functional.
─ This is known as the active site.
─ Active site-small hollow depression
─ Substrate –molecule on which enzyme acts e.g starch for the enzyme salvary amylase
─ Substrate fits into depression to form an enzyme- substrate complex
─ Substrate molecule is held within the active site by bonds that form between certain amino
acids and the active site
─ One model, the lock and key model proposes that enzymes work in the same way as a key
operates a lock
─ Substrate only fit the active site of one particular enzyme e.g. starch on amylase only
─ Shape of substrate exactly fits the active site.

(Describe the role of the “active site”. [2])

─ The site were enzyme binds

─ to substrate/to bacterial cell wall
─ by lock and key mechanism/ induced fit theory
─ substrate / bacterial cell wall broken down into subunits/products/ smaller units

(Explain why enzymes are effective in very small quantities. [2])

─ not used up in reaction/ can be used up again and again

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 ─ often work fast/high turn over

(1)Enzyme concentration

─ Assuming that the substrate concentration is maintained at a level , ceteris paribus , the rate
of reaction is proportional to the enzyme concentration increases, the rate also increases

(2)Substrate concentration

─ For a given enzyme concentration , the rate of an enzyme controlled reaction increased with
increasing substrate concentration but there comes a time when any further increase in
substrate concentration does not result in significant increase in the rate of reaction.
─ This is because at high substrate concentration , virtually all the active sites will be
saturated & any extra substrate have to wait until the active sites are free, that means that
V-max is reached.

(3)Temperature

─ As temperature increases, the rate of enzyme controlled reaction increases.

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 ─ This is because the E.K of molecules of both the enzyme & the substrate such that they start
to vibrate quickly & their chances of fruitful collision are greatly increased.
─ The temperature beyond the optimum temperature will result in the decrease of the rate of
reaction, as the secondary & tertiary structure of the enzyme is disrupted thus the enzyme
becomes denatured & catalysis declines.
─ Low temperature inactivates enzymes.
─ Catalysis becomes very low or almost ceases.
─ This is basic of refrigeration in preserving food for a long time catalysis gradually resumes
as the temperature gradually increases.
─ The effect of temperature on enzyme activity can also be expressed as the temperature co-
efficient Q10.

Q10 = Rate of reaction (a) (x+10) degrees Celsius divided by the rate of reaction (a) x degrees
Celsius.

─ Over a range of 0 – 40 degrees Celsius, Q10 for an enzyme controlled reaction is 2 e.g. there
rate of an enzyme controlled reaction doubles for every 10 degrees Celsius rise.

(4) pH

─ Enzymes are very sensitive to a slight change in pH changes & as such operate in very
narrow pH ranges. The optimum pH is that at which the maximum rate of reaction occur.
─ When the pH is altered, above or below, this optimum value, the rate of enzyme activity
diminishes significantly.
─ Pepsin has an optimum pH of 2, while Arginine has 9,7 for instance. Changes in pH alter the
ionic charge of acidic & basic group & therefore disrupt the ionic bonds that maintain the
specific shape of the enzyme.
─ Extreme pH denatures the enzyme. The peptide bond can be hydrolyzed.

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Enzyme inhibition

─ Molecules/ substance that reduce the rate of an enzyme catalyzed reaction are referred to
as enzyme inhibitors.
─ Inhibition is a normal part of the regulation of enzyme activity & many drugs & poisons act
as enzyme inhibition to achieve their effects.
─ Inhibition can be divided into competitive & non competitive

Competitive inhibition
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 ─ This occurs when a compound (inhibitor) have a shape similar to the actual substrate such
that they both competes for the active sites of the enzyme.
─ Normally the molecule / inhibitor does not take part in the reaction, but while it occupies
the active site, it prevent the actual substrate to be catalyzed hence the rate of the reaction
decreases.
─ However, a characteristic feature of competitive inhibitor is that of the substrate is
increased the rate of reaction also increases

Non competitive inhibitor- reversible

─ This type of inhibitor has no structural similarity to the substrate & combines with the
enzyme at a point other than the active site.
─ It does not affect the ability of the substrate to bind with the enzyme, but makes it
impossible for catalysis for catalysis to occur.
─ The rate of reaction decreases with inhibitor concentration to almost nil, when inhibitor
saturation is reached.
─ However, increasing substrate concentration does not increase the rate of reaction.
─ When the inhibitor is removed, the enzyme regains its catalytic activity hence reversible

Non competitive inhibitor/irreversible

─ Some chemicals can cause irreversible inhibition of enzymes.

─ Some concentration of chemicals reagents e.g. heave metals or certain iodine containing
compounds, complete inhibit enzymes.
─ They may be in active site or elsewhere.
─ Once such reactions have occurred the enzyme loses its catalytic activity effectually /
permanently.
─ The change may cause the enzyme protein to precipitate.

Allosteric enzymes

─ These are enzymes which can change their shape.

─ They are regulated by compound bind to the enzyme at specific site well away from the
active site.
─ While there, they cause the active site to change, there by affecting the ability of the
substrate to bind the enzyme.
─ Compounds of this nature are called allosteric inhibitor.

Enzyme Co- factor

─ Enzymes require non proteins components called co- factors for their effective activity.
─ Co- factors vary from simple inorganic ions to complex organs molecules &may either
remain unchanged at the end of a reaction or be generated by a later process.
─ There are 3 types of co-factors & these are inorganic ions prosthetic group & coenzymes.

Prosthetic group (e.g. Haem, FAD, hydrogen carrier molecules)

─ There are organic molecules that are tightly bound on a permanent basis to the enzyme.
WRITTEN BY MR. R. GWANZURA WHATSAPP: 0773266377/CALL 0717558917 HOLY CROSS HIGH
 ─ They assist in catalytic activity of the enzyme e.g. fluorine adenine dinucleotide (FAD) which
contains fluorine.
─ Its function is to accept hydrogen. Haem is found in the catalase & peroxides which
catalyzes the breakdown of hydrogen peroxide in to water & oxygen.

Inorganic ions (enzyme activators (CL)

─ These are thought to mold either the enzyme or substrate in a shape that easily allows an
enzyme / substrate complex to be formed.
─ Hence, they greatly increase the chance of the reaction occurring, salivary amylase is
activated by chloride ions.

Coenzyme (NAD, NADP, Coenzyme A & ATP)

─ Like prosthetic group, coenzymes are organic molecules which act as co- factors, but they
do not remain attached to the enzyme between reactions.
─ Coenzymes are derived from victims. NAD (Nicotinamide Adenine Dinucleotide) is derived
from vitamins nicotinic & can exist in both reduced & oxidized form.
─ It function as a hydrogen acceptor.

(Explain the effect of inhibitors on enzyme action. [6])

─ A variety of small molecules exist which can reduce the rate of an enzyme controlled
reaction
─ These are inhibitors
─ Inhibition can be
1. Competitive
2. Non-competitive
─ They can also be classified as reversive and non-reversive (permanent)
─ Reversive inhibitors-can be competitive or non competitive
─ Competitive- have molecular shape similar to substrate
─ Fit into active site, substrate prevented from occupying it- no of enzyme substrate
molecules reduced-active site directed inhibitors
─ Compete for active site with substrate
─ Non- competitive inhibitors bind to other parts of enzyme
─ Overall shape of the enzyme molecules including the active site
─ Substrate-enzyme complex not formed

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Water molecules
KEY CONCEPT OBJECTIVES CONTENT SUGGESTED SUGGESTED
Learners should be (ATTITUDES, SKILLS LEARNING ACTIVITIES RESOURCES
able to: AND KNOWLEDGE) AND NOTES
8.2.4 Water  describe the - Structure of a water - Constructing a  ICT tools
structure and molecule water molecule  Braille
properties of - Physical and model. software/Jaws
water chemical properties - Performing  Models
of water experiments
illustrating various
properties of water.

- Roles of water in - Visiting water

 explain the roles living organisms bodies.
of water in living - Discussing the
organisms and as roles of water in
an environment living organisms.

─ Is a vital chemical constituent of living organisms

THE STRUCTURE AND PROPETIES OF WATER

BIOLOGICAL SIGNIFINCE OF WATER

─ WATER IS IMPOTANT FOR TWO MAIN REASONS :

1) IT IS VITAL FOR CHEMICAL CONSTITUENT OF LIVING CELLS.
2) It provides environment needed by living organisms that leave in water.
─ The chemical of water are due to its small size, its polarity and hydrogen bonding between
its molecules.
─ Polarity is an even distribution of a charge in a molecule.
─ The electronegative oxygen atom draws electrons from the hydrogen atom making the
oxygen atom more negative, relative to the hydrogen atom.
─ Thus what is dipole (contain both negative and positive poles )
─ Water molecules therefore have weak attraction for each other, with positive charges
coming together and cause them to stick to each other like links in a chain.
─ These bonds are relatively weaker than the ionic and the covalent bond bonds and are
called hydrogen bonds

The solvent properties of water

─ To dissolve chemicals inside the cells/ metabolic reaction + place in aqeous solutions
─ Uptake of mineral salts from soil

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 ─ Excretion of waste products dissolved in water e.g urine
─ Water has high solvent of polar molecules e.g. ionic compounds like NaCL and non ionic
substances like sugar that contain polar group.
─ On contact with water the ions and the polar group are surrounded by water molecules
which separate the molecules from each other.
─ Biochemical reaction takes place in aqueous conditions.
─ Water as a solvent acts as transport medium e.g. in blood, lymphatic system and xylem and
the phloem vessel.

High heat capacity

─ Constant environment for aquatic plants

─ Constant temperature in cell hence no denaturation of enzymes
─ Cooling by evaporation
─ Contains of the cell do not freeze easily
o Water has high heat capacity i.e. (the total amount of water required to raise water
temperature of 1Kg of water by 1 degree.
o A large amount of heat energy results in a small rise in temperature.
o Temperature changes within water or aqueous unitary therefore minimized.
o Biochemical processes consequently operate over a small temperature range and are less
likely to be affected extremities of temperatures.
o Water also provides a very constant external environment for many cell and organisms.

High heat of evaporation

─ Latent of evaporation is the measure of vaporization is a measure of heat energy required to

vaporize a liquid.
─ A large amount of water is required to make water vapour.
─ This is due to hydrogen bonding of water molecules.
─ Therefore water have usually high boiling point.
─ Evaporating water as a result takes a lot of heat energy with them from the surrounding
thus cooling takes place.
─ This is made use of in the transpiration, sweating and panting of mammals.
─ High heat of evaporation also means that a large amount of heat can be lost with minimal
loss the body, plant etc.

High heat of fusion

─ Latent heat of fusion is the measure of the heat energy required to melt a solid i.e. (ice)
─ Ice requires a relative large amount of heat energy to thaw it.
─ Conversely, liquid must loss a relative large amount of heat to freeze.
─ Content of a cell and their environment are less likely to freeze.

Density and freezing properties

─ Water has highest density at 4 degrees

─ Its density increases as the temperature decreases

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 ─ Ice therefore tends to float.
─ Ice insulates the water below it thus increasing survival chance of organisms below it.
─ Since water below 4 degrees tend to rise, this also tends to maintain circulation in lentx
ecosystem.
─ This may result in nutrient cycling and colonization of water to greater depth.

High surface tension and cohesion

─ Cohesion is the force where individual molecules stick together at the surface, a force
called surface tension between the molecules as they try to occupy the least possible
space (ideally a sphere).
─ Water has a higher surface tension which makes it possible for small organism to skate
over its surface.
─ The high cohesion of water is important in cell and in the translocation of water in the
xylem vessel.
─ Movement of water in the xylem
─ Habitat for less dense aquatic organism

Water as a reagent

─ Water is biological significant as an essential metabolite that is , it takes part in chemical

reaction of metabolism .
─ For example it is used as a source of hydrogen and electron.

Functions of water

─ Turgidity of exoskeletons
─ Preventing friction of joints /lubricant
─ Fertilization of swimming gametes
─ Component of blood
─ Raw material for photosynthesis
─ Part of fluid produced e.g. mucus and tears to clean the body

Some small insects are able to stand on the surface of water.

State the property of water that enable insects to stand on water – high surface tension

Outline the importance of high heat of vaporization of water in living organisms

─ Enable cooling/ energy transferred to water molecules

─ Which allows them to vaporize resulting in loss of heat from the surrounding

Explain why cell contents and their environment are less likely to freeze under extremely cold
temperatures

─ Due to high heat capacity

─ Were high amount of heat is needed to break hydrogen bonds (in water)
─ Which restricts movement of molecules / (boiling)
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 ─ Water must loose large amounts of heat to freeze

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 FORM 5 TERM 2
TOPIC 3: Cell and Nuclear division
KEY CONCEPT OBJECTIVES CONTENT SUGGESTED LEARNING SUGGESTED
Learners should be (ATTITUDES, SKILLS AND ACTIVITIES AND NOTES RESOURCES
able to: KNOWLEDGE)
8.3.1 The Cell  outline the cell - Interphase  Illustrating the cell  ICT tools
Cycle cycle - Mitosis cycle.  Braille
- Cytokinesis software/Jaws
 Print media
- Growth  Outlining DNA
 describe - DNA replication replication.
interphase

Life is like riding a bicycle. You can‘t fall off unless you stop pedaling. A bend in the
road is not the ends of the road unless if you fail to make the turn.

Chromosomes

─ A composed of deoxyribonucleic acid [DNA] and protein with small amount of chromosomal RNA
DNA has a negative charge distributed along it’s length and positively charged protein molecule
called histones and bond to it .
─ Between division of the nucleus each chromosome contain one DNA molecule .
─ Before the nucleus divides the DNA replicates such that at nuclear division the chromosomes is a
double structure, containing two identical DNA molecules.
─ The two parts of the chromosomes are referred to as chromatids.
─ Each chromatids one of the two identical molecules.
─ Species in which there are two sets of chromosomes are referred to as diploid given symbol 2n
. A few simple organisms have only one set of chromosomes and are referred to as hyploid
symbol n
─ Garments either sex are haploid.

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 ─ There are two types of nuclear division: meiosis and mitosis.
─ Mitosis is a process by which a cell nucleus divides to produce daughter nuclei containing
identical sets of chromosomes of parent cell.
─ Usually it is followed by division of whole cell to form two daughter cells a process known
as cell division.
─ Mitosis with cell division results in increase in cell number and is the method by which
growth replacement and repair of cell occurs in eukaryotes.
─ In unicellular eukaryotes, mitosis results in asexual reproduction.

Meiosis

─ Is the process by which a cell nuclear divide to form four daughter cells containing half
the number of chromosomes of the original cell .
─ It is also called reduction division since it reduces the number of chromosomes in the cell
from the diploid 2n to hypoid n.
─ Meiosis occurs during gametogenesis in animals and during spore formation in plants .

THE CELL CYCLE

PHASE EVENTS WITHIN THE CELL

GROWTH PHASE Intensive cellular synthesis mitochondrion,chloroplast divide .energy store
G1/G2 increases mitotic spindle begins to form.
M Nuclear division occurs in four phases.
C Equal distribution of organelles and cytoplasm into each daughter cells.

─ The sequence of events which occur between one cell division and next is called cell cycle. It
has three main stages.

INTERPHASE MITOSIS CELL DIVISION

This is a period of synthesis The actual process of nuclear This is the process of division
and growth. The cell produces division. of the cytoplasm into the
the material required or it’s daughter cells.
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own growth ,DNA replication
also occurs .

PHASE EVENTS IN THE CELL

G1 Intensive cellular synthesis including new organelle cell growth occurs.
S DNA replication occurs. Protein molecules [histones] produced .At this stage the cell
is 4n.

MITOSIS

8.3.2 Mitosis  describe the - Prophase  Observing behavior of  Onion root tips
behaviour of - Metaphase chromosomes in a root  Microscope
chromosomes, - Anaphase tip squash  Stains
nuclear - Telophase  Drawing of diagrams  Prepared slides
envelope, cell showing phases of  ICT tools
membrane, mitosis.  Braille
centrioles and software/Jaws
spindles during  Print media
mitosis
- Cytokinesis  Discussing
 distinguish cytokinesis in plant
between and animal cells.
cytokinesis in
plants and - Growth  Discussing the
animals - Repair importance of mitosis.
- Asexual
 explain the reproduction
importance of - Production of
mitosis genetically
identical cells

 identify factors
that increase
chances of  Discussing factors
cancerous - Carcinogens associated with
- Mutations
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 growth - Radiation cancerous growth.

 outline the
stages involved - Uncontrolled cell  Watching video clips.
in the division  Analysing video clips.
development of
cancer

─ Has 5 stages / phases

1. INTERPHASE
─ Variable according to the function of the cell. .
─ DNA replicates
─ Each chromosome exits as a pair of chromatids joined at the centromere.
─ The cell now has 4n.
─ Chromosomes now visible as loosely coiled threads called chromatin.
─ Centrioles replicate

2. PROPHASE
─ Formation of spindle.
─ It is the longest phase.
─ Chromosomes shorten and thicken by coiling.
─ Chromosomes now available as a double structure.
─ In animal cells centrioles move to opposite poles.
─ A star from short microtubules radiating from centrioles.
─ The spindle is formed.
─ Chromatids form chromosomes.
─ Nuclear envelope disappear.

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 3. METAPHASE
 Chromosomes line up at the equator of the spindle attached by the centromeres to the
spindle.
 Chromosomes move to metaphase plate.

4. ANAPHASE

 A very rapid stage.

 The centromers split into two and the spindle pull the daughter chromosomes to opposite
poles.
 The separated are pulled behind the centromeres.
 The shortening of the spindle fiber by the removal of the subunits account to moving of
chromatids during anaphase.

5. TELOPHASE
 The chromatids reach the opposite poles.
 They uncoil and lengthen to from chromatin again.
 The spindle fiber disintegrate and centrioles replicate.
 Nuclear envelope reforms.

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CYTOKINESISE

─ Is the actual division of the cytoplasm to form two daughter cells.

─ The cell become evenly distributed towards the two poles.
─ In animal cells the cell surface membrane invigilates around the equator forming a farrow
which eventually meet up and completely separate the cells.
─ In plant cells, spindle fibre remain around the equatorial plane and increase in number to
form a barrel shaped region called the phagmoplast. Cell organelles line at the equator
─ Forming a cell plate. The cell plate spreads across the equator meeting the cell wall, there by
separating the two daughter cells.
─ The new cell wall is called a primary cell wall, it later develops to a secondary cell wall by
the deposition of cellulose andlignin.

SIGNIFICANCE OF MITOSIS

1. GENETIC STABILITY
 Mitosis produces two cells with the same chromosomal number as the mother cell .No
variation is introduced during mitosis.

2. GROWTH
 The number of cells in organism increase with mitosis .This is the basic of growth in a
multicellular organism.

3. CELL REPLACEMENT

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  Replacement of cell tissue involve mitosis.

4. REGENARATION
 Some animals are also regenerated some parts of their body by mitosis.

5. ASEXUAL REPRODUCTION
─ Mitosis is the basic of asexual reproduction of new individuals of a spice by one parent
organism.

Difference between cell division in plants and animals

PLANT ANIMAL
Centrioles Absent Present
Asters Absent Asters form
Cell division Involves formation of cell Involves furrowing and
plate cleavage of cytoplasm
Site of occurrence Occurs mainly in meristems Occurs in tissues
throughout body

MEOSIS [REPRODUCTION DIVISION]

KEY CONCEPT OBJECTIVES CONTENT SUGGESTED LEARNING SUGGESTED

Learners should be (ATTITUDES, SKILLS AND ACTIVITIES AND NOTES RESOURCES
able to: KNOWLEDGE)
8.3.3 Meiosis  explain the - Haploid  Illustrating haploid cells,  ICT tools
meanings of - Diploid diploid cells and  Braille
the terms - Homologous homologous software/Jaws
haploid, diploid - Chromosomes chromosomes.
and
homologous
chromosomes  Print media
- Interphase  Observing behaviour of
 Describe the - Meiosis I chromosomes during
behaviour of - Meiosis II pollen grain formation
chromosomes, - Cytokinesis  Drawing of diagrams
nuclear showing phases of
envelope, cell meiosis.
membrane,  Microscope
centrioles and  Prepared slides
spindles during  Discussing the importance  Flowers
meiosis - Gamete formation of meiosis.
- Genetic variation
 discuss the
importance of
meiosis  Discussing the
- Similarities and similarities and
differences between differences.
 compare and mitosis and meiosis
contrast mitosis
and meiosis

─ Like mitosis involves DNA replication during interphase in the parent cell but this is
followed by two cycles of nuclear division which are meiosis one and mitosis two.
─ Meiosis occurs during gametogenesis and spore formation in plants.

Meiosis I
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PROPHASE 1

─ Longest phase.
─ Crossing over may occurs.
─ Chromosomes shorten and become visible as a single structure.
─ Homologous pair up in a process called synapses.
─ Each pair is called a bivalent.
─ One of the homologues pair comes from the father paternal and the other from the mother
maternal.
─ The bivalents shorten and thicken by coiling.
─ The homologous chromosomes partially separate some for a fill points along the length
.These points are called chiasmata.
─ Genes from one chromosome may swap with genes from the other chromosome leading to
new gene combinations in the resulting chromatids.

METAPHASE 1

 The bivalents become arranged around the equator attached to their centromers .

ANAPHASE 1

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 ─ Spindle fibres pull homologues chromosomes apart centromeres face opposite poles of the
spindle.
─ This separates the chromosomes into two haploid one set at each of the opposite spindle.

TELOPHASE 1

─ Chromatides usually uncoil and nuclear envelope reforms.

─ Cytokinesis now occurs as in mitosis.
─ In many plants there is no telophase I.
─ Cell formation and interphase II and then the cell passes from anaphase IIto prophaseII.
─ Reduction ofchromosomes has occurred but sister chromatids are genetically different.

INTREPHASE II

─ Present only in animal cells and varies in length.

─ No synthesis occurs and no further and replication occurs.
─ The second meiotic division is similar to mitosis.

PROPHASE II

─ Only present if interphase two is present.

─ The nucleoli and the nuclear envelope disappears, the chromatids, shorten and thicken.
─ Centrioles if present move to opposite poles and the spindle axis of the first meiotic
division.

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METAPHASE II

─ Chromosomes arrange theme self at the centre of the equator and centromers appear as
double structure .The orientation or assortment of chromosomes is at the equator is
random.
─ Independent assortment occurs at metaphase II .

ANAPHASE II

─ The centromers divide and the spindle pulls to opposite poles and to double centromeres
─ The separated chromatids now called chromosomes are pulled along the centromeres.

TELOPHASE II

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 ─ The stage is similar to that found at mitosis .The chromosomes uncoil, lengthen and become
very indistinct .The spindle fiber disappear and the centrioles replicate.
─ Nuclear envelope reforms around the nucleus .Cell wall forms in plants and four daughter
cells are produced.

SIGNIFICANTS OF MEIOSIS

1. SEXUAL REPRODUCTION
─ All organisms reproduce sexually use meiosis.Garment production involves meiosis.
─ The number 2n is restored during fertilization.

2. GENETIC VARIATION
─ Meiosis provides a platform for new gene combinations of genes. This results in genetic
variation in offspring when garments fuse.

a) Independent assortment

─ The orientation of bivalents at the equator of the spindle in metaphase one is random. The
bivalents separate independently of those in other bivalents during anaphase.

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b) CROSSING OVER

─ Is the result when chiasmata crossover of segments of chromatids occurs between

homologues during prophase 1

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(Outline the difference between mitosis and meiosis. [8])

STAGE MITOSIS MEIOSIS

Homologues chromosomes Homologues chromosomes
remain separated Pair up in a process called
PROPHASE synapsis.
Crossing over occurs No crossing over
No Formation of chiasmata Formation of chiasmata
Pairs of chormatids line up on Pairs of chromosomes line up
METAPHASE
spindle equator on spindle equator
Centriomeres divide Centromeres don’t divide
Chromatids separate Chromosomes separate
ANAPHASE
Chromatids identical Chromatids may not be
identical
Cell contain the same number Cells contain half the number
of chromosomes as the mother of the chromosomes .
cell
May occur in haploid and Only occurs in diploid cells
TELOPHASE diploid or polyploidy cells
2 daughter cells identical 4 daughter cells
Genetically identical to parents Genetically different from
parents

Similarities between the prophase stage of mitosis and prophase I

─ Chromosomes shorten and thicken;

─ Spindle fibres form;
─ Nucleoli disappears.

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NATURALSELECTION AND ARTIFICIAL SELECTION
KEY CONCEPT OBJECTIVES CONTENT SUGGESTED SUGGESTED
Learners should be (ATTITUDES, SKILLS LEARNING RESOURCES
able to: AND KNOWLEDGE) ACTIVITIES AND
NOTES
8.10.1 Natural  explain, with - Natural Selection  Discussing how  Print media
selection examples, how - Mutations mutations and  ICT tools
mutations and environment may  Braille software/Jaws
environment may affect phenotype.
affect phenotype

 explain, with - Natural selection  Discussing with

examples, how - Environmental factors examples how
environmental environmental
factors act as factors act as forces
forces of natural of natural selection.
selection - Evolution
 Researching and
 explain how natural presenting on how
selection may bring natural selection
about evolution may bring about
evolution.
8.10.2 Artificial  describe one - Artificial selection  Outlining the  ICT tools
selection example of artificial examples of artificial  Print media
selection selection.  Braille software/Jaws

"Until one is committed, there is hesitancy, the chance to draw back, always
ineffectiveness. Concerning all acts of initiative (and creation) there is one elementary
truth, the ignorance of which kills countless ideas and splendid plans: that the moment
one definitely commits oneself, and then Providence moves too. All sorts of things occur
to help one that would never otherwise have occurred. A whole stream of events issues
from the decision, raising in one's favour all manner of unforeseen incidents and
meetings and material assistance, which no man could have dreamed would have come
his way. Whatever you can do, or dream you can, begin it. Boldness has genius, power,
and magic in it. "

Explain the role of natural selection

─ In an ecosystem individual within a particular population (of an organism) have a great

reproduction potential.

─ The population members of the individuals remain roughly constant in that community. This

occurs while other individuals may fail to survive or may fail to reproduce.

─ What determinewhether an individual survives and is able to reproduce are the environmental

factors.

─ The environment factors tend to support the survival of other members compared to others.

─ This is referred to as natural selection.

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 ─ Members that survive and reproduce are said to be selected for, while those that do not survive

or die or later fail to reproduce are said to be selected against.

─ What determine the selection is the variations that different individuals possesses.

─ some variations in characteristics make other members best adapted to survive and these ones

tend to reproduce, passing their alleles make it different for the members to survive thus they

are selected against.

─ It is this genetic variation which leads to a change in the phenotype thus overtime some alleles

can be completely lost from the genetic pool.

─ Overtime this produces an evolutionary change as new species which are well adapted tend to

rise.

─ A good illustration or a good example is a British moth during the industrialisation era. Where

lightly coloured moth (white) where selected against because of the environment that had

turned darker as a result of soot from industries.

─ Predators such as birds could easily identify the moth reducing its survival chances while the

dark coloured moth was selected for as they were difficult to identify.

─ these members tended to survive and reproduce passing on their alleles of dark colour

Outline how artificial selection differs from natural selection. [6]

artificial selection natural selection

selection (pressure by) humans or environmental selection pressure ;
genetic diversity lowered or genetic diversity remains high ;
inbreeding common or outbreeding common ;
loss of vigour / inbreeding
depression
or increased vigour / less chance of
inbreeding depression ;
increased homozygosity / decreased
heterozygosity
or decreased homozygosity / increased
heterozygosity ;
no isolation mechanisms operating or isolation mechanisms do operate ;
(usually) faster or (usually) slower ;
selected feature for human benefit or selected feature for organism’s benefit ;
not for, survival / evolution or promotes, survival / evolution ;

1. Describe the role of natural selection in evolution. [8]

 individuals in population have great reproductive potential / AW ;

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  numbers in population remain roughly constant ;
 many fail to survive / die ;
 do not reproduce ;
 due to environmental factors / named factor ;
 variation in members of population ;
 those best adapted survive ;
 reproduce / pass on alleles ; R genes
 genetic variation leads to change in phenotype ;
 ref: changes in gene pool ;
 over time produces evolutionary change ;
 new species arise from existing ones

2. Explain how natural selection may bring about evolution. [8]

 individuals in population have great reproductive potential / AW ;

 numbers in population remain roughly constant ;
 variation in members of population ;
 environmental factors / named factor (biotic or abiotic) ; linked to 17 and 18
 (cause) many, fail to survive / die / do not reproduce ;
 those best adapted survive / survival of the fittest ;
 (reproduce to) pass on alleles ; R genes
 genetic variation leads to change in phenotype ;
 ref: changes in, gene pool / allele frequency ;
 over time produces evolutionary change ;
 new species arise from existing ones / speciation ;
 directional / stabilising, selection ; [8 max]

3(a) Describe why variation is important in natural selection. [6]

 ref. continuous / discontinuous variation ;

 genetic / inherited variation ;
 variation in phenotype / characteristics / AW ;
 (can be due to) interaction of genotype and environment ;
 e.g. of characteristic that influences survival ;
 ref. intraspecific competition / struggle for existence ;
 those with favourable characteristics survive / AW ;
 pass on favourable characteristics to offspring ;
 those with disadvantageous characteristics die ; 6 max

(b) Explain the role of isolating mechanisms in the evolution of new species. [9]

 ref. to definition of species ;

 ref. allopatric ;
 geographical isolation ;
 ref. to examples e.g. islands / lakes / mountain chains / idea of barrier ;
 ref. to example organism ;
 ref. to populations prevented from interbreeding ;

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  isolated populations subjected to different selection pressures / conditions ;
 over time sufficient differences to prevent interbreeding ;
 ref. sympatric ;
 ref. to reproductive isolation ;
 ref. behavioural barriers (within a population) ;
 e.g. day active / night active ;
 correct ref. to gene pool ;
 change in allele frequencies ; 9 max

(c) Explain, using named examples, how mutation can affect phenotype. [7]

 gene) example ; (sickle cell / PKU )

 change in gene / DNA / base change ;
 different amino acid ;
 different polypeptide / different protein / non-functional protein ;
 AVP ; details
 AVP ; details
 (chromosome) example ; (Down’s, Turner’s syndromes)
 structural changes in chromosomes ;
 change in number of chromosomes ;
 change in sets of chromosomes / ref. polyploidy ;
 AVP ; details
 AVP ; details

(b) Explain, using examples, how the environment may affect the phenotype of an
organism. [8]

 phenotypic variation results from interaction of genotype and environment /

VP = VG+ VE ;
 environment may limit expression of gene(s) / AW ;
 e.g. for size / mass / height ;
 because, food / nutrients / ion, missing / malnutrition ;
 named, nutrient / ion / mineral, missing ;
 environment may, trigger / switch on, gene ;
 ref. low temperature and change in animal colour ;
 ref. high temperature and, curled wing in Drosophila / gender in crocodiles ;
 ref. UV light and melanin production ;
 ref. wavelength of light and, flowering / germination / fruit colour ;
 other named trigger plus example ;
 environment effect usually greater on polygenes / ora ;
 environment may induce mutation affecting phenotype ;

(a) Explain how meiosis and fertilisation can result in genetic variation amongst
offspring. [7]
 chiasma / crossing over ;
 between non-sister chromatids ;
 of, homologous chromosomes / bivalent ;

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  in prophase 1 ; linked to 1
 exchange of genetic material / AW ; R genes unqualified
 linkage groups broken ;
 new combination of alleles ;
 independent assortment ; R random assortment
 metaphase 1 ; linked to 8
 detail of independent assortment ;
 possible mutation ;
 random mating ;
 random fusion of gametes ;

(a) Explain the role of isolating mechanisms in the evolution of new species. [8]
 allopatric speciation ;
 geographical isolation / spatial separation ;
 e.g. of barrier ;
 e.g. of organism ; must relate to 3
 sympatric speciation ;
 example ;
 meiosis problems ;
 polyploidy ;
 behavioural / temporal / ecological / structural, isolation ;
 (isolated) populations, prevented from interbreeding / can only breed
 amongst themselves ;
 no, gene flow / gene mixing, (between populations) ;
 different selection pressures operate ;
 natural selection ;
 change in allele frequencies ;
 different gene pool ;
 over time (differences prevent interbreeding) ;
 reproductively isolated ;

(b) Describe and explain, using an example, the process of artificial selection. [7]

 humans ; must be linked to, choosing / selecting / mating etc

 parents with desirable feature ;
 e.g. organism and feature ;
 bred / crossed ;
 select offspring with desirable feature ;
 repeat over many generations ;
 increase in frequency of desired allele(s) / decrease in frequency of
 undesired allele(s) ;
 background genes ;
 loss of hybrid vigour / increase in homozygosity / ref. inbreeding depression ;
AVP ; e.g. detail of breeding techniques [7 max]

Explain the benefits of maintaining biodiversity. [4]

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  cultural / aesthetic / leisure, reasons ;
 moral / ethical, reasons ; e.g. right to exist / prevent extinction
 resource material ; e.g. wood for building / fibres for clothes / food for
 humans
 ecotourism ;
 economic benefits ;
 ref. resource / species, may have use in future / AW ; e.g. medical use
 maintains, food webs / food chains ; A description
 nutrient cycling / protection against erosion ;
 climate stability ;
 maintains, large gene pool / genetic variation ;

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CANCER

─ Are a group of diseases that are caused by uncontrolled cell division which involve mitosis.
─ The problem is caused by mutations /abnormal activation of genes which control cell
division. Such abnormal genes are called oncogenes.
─ An abnormal cell divides by mitosis to form an irregular mass of undifferentiated cells
called tumor.
─ Tumor cell can break away and spread to all parts of the body especially in blood stream or
lymphatic system causing secondary tumor called metastasis .
─ The process is known as metastasis.
─ Tumors that spread and eventually cause ill health and death are described as malignant
.The rest of tumors that do not spread such as the warts are described as benign .

CAUSES OF CANCER

─ Change in the genetic constitution of a cell / organism are called mutations . And any factor
that brings about a mutation is called a mutagen .
─ An agent that causes cancer is referred to as carcinogens are not always mutagens .
─ Development of malignant cancer cell involves several steps and is usually caused by more
than one factor operating over a considerate period of time .
1. Retroviruses
─ Retroviruses are RNA viruses which when they invade animal cells, use the enzyme
reverse transcriptase to make DNA copies of the viral RNA .
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 ─ The DNA contain a gene which alters host cell division genes , switching them on and
causing the cell

Describe the first division of meiosis (meiosis I) in animal cells. [6]

─ reduction division / (to) halve number of chromosomes / diploid to haploid / AW ;

─ homologous chromosomes pair up / bivalents form ;
─ ref. chiasmata / ref. crossing over ;
─ homologous chromosome pairs / bivalents, line up on equator ;
─ independent assortment ;
─ spindle / microtubules, attached to centromeres ;
─ chromosomes of each pair pulled to opposite poles ;
─ by shortening of, spindle / microtubules ;
─ nuclear envelopes re-form ;
─ cytokinesis / AW ;

(Describe how crossing over and independent assortment can lead to genetic variation.[9])
─ occur during meiosis I ;
─ crossing over
─ between non-sister chromatids ;
─ of, (a pair of) homologous chromosomes / a bivalent ;
─ in prophase 1 ;
─ at chiasma(ta) ;
─ exchange of genetic material / AW ;R genes unqualified
─ linkage groups broken / AW ;
─ new combination of alleles (within each chromosome) ;
─ independent assortment
─ of homologous chromosomes pairs / bivalents ;
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 ─ each pair lines up independently of others ;
─ line up on equator ;
─ (during) metaphase 1 ;
─ results in gametes that are genetically unique / AW ;

Describe how genetic variation in secondary oocytes arises.

─ During / prophase 1;
─ Crossing over/ chiasmata formation occurs
─ Leads to new combination of alleles;
─ During metaphase 1;
─ Homologous chromosomes position themselves either way up/ down on equator of spindles/ AW
─ Independent assortment
─ Segregation occurs;

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TOPIC4: GENE CONTROL
KEY CONCEPT OBJECTIVES CONTENT SUGGESTED SUGGESTED
Learners should be (ATTITUDES, SKILLS LEARNING ACTIVITIES RESOURCES
able to: AND KNOWLEDGE) AND NOTES
8.4.1 Nucleic  describe the - Nucleoside  Illustrating the  Models
Acids structure of a - Nucleotide structure of a  ICT tools
nucleotide nucleotide.  Braille
software/Jaws
 describe formation - Dinucleotide  Demonstrating the
of a dinucleotide - Phosphodiester formation of a
bonds phosphodiester
 distinguish bond.
between - RNA nucleotides
Ribonucleic acid  Discussing the
(RNA) and - DNA nucleotides differences.
Deoxyribonucleic between RNA and
acid (DNA) DNA nucleotides.
nucleotides
8.4.2 Structure  describe the - DNA structure  Constructing  ICT tools
and replication of structure of DNA models of DNA.  Braille
DNA software/Jaws
 explain how DNA - semi - conservative  Making DNA
replicates replication of DNA models illustrating  Print media
replication.  Models (zips,
- Messelson and Stahl beads, soft
experiment wires)

8.4.3 Protein  outline the - Transcription  Viewing simulations  ICT tools

synthesis process of protein - Translation including and videos of  Braille
synthesis role of messenger protein synthesis. software/Jaws
RNA, transfer RNA
and ribosomes

In life as in football, you won‘t go far unless you know where the goalposts are.
Have a vision in life. The vision must be followed by venture. It is not enough
to stair up the steps but you must step up the stairs.

Nucleotides

Nucleotides have three parts to them:

 a phosphate group , which is negatively charged, and gives nucleic acids their acidic
properties.

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  a pentose sugar, which has 5 carbon atoms in it. By convention the carbon atoms are numbered
as shown (1', 2', etc, read as "one prime", "two prime", etc), to distinguish them from the carbon
atoms in the base. If carbon 2' has a hydroxyl group attached (as shown), then the sugar is ribose,
found in RNA. If the carbon 2' just has a hydrogen atom attached instead, then the sugar is
deoxyribose, found in DNA.
 a nitrogenous base. There are five different bases (and you don't need to know their structures),
but they all contain the elements carbon, hydrogen, oxygen and nitrogen. Since there are five
bases, there are five different nucleotides:

Base: Adenine (A) Cytosine (C) Guanine (G) Thymine (T) Uracil (U)
Nucleotide: Adenosine Cytidine Guanosine Thymidine Uridine

-The base thymine is found in DNA only and the base uracil is found in RNA only, so there are
only four different bases present at a time in one nucleic acid molecule.

-The nucleotide above is shown with a single phosphate group, but in fact nucleotides can have
one, two or three phosphate groups.
-So for instance you can have adenosine monophosphate (AMP), adenosine diphosphate (ADP)
and adenosine triphosphate (ATP).
-These nucleotides are very common in cells and have many roles other than just part of DNA.
ATP is used as an energy store This shows a nucleotide triphosphate molecule.

Nucleotide Polymerisation
-Nucleotides polymerise by forming phosphodiester bonds between carbon 3' of the sugar and
an oxygen atom of the phosphate.This is a condensation polymerisation reaction.
-The bases do not take part in the polymerisation, so there is a sugar-phosphate backbone with
the bases extending off it. This means that the nucleotides can join together in any order along
the chain. Two nucleotides form a dinucleotide, three form a trinucleotide, a few form an
oligonucleotide, and many form a polynucleotide.

A polynucleotide has a free phosphate group at one end, called the 5' end because the phosphate
is attached to carbon 5' of the sugar, and a free OH group at the other end, called the 3' end
because it's on carbon 3' of the sugar. The terms 3' and 5' are often used to denote the different
ends of a DNA molecule.

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Structure of DNA
KEY CONCEPT OBJECTIVES CONTENT SUGGESTED SUGGESTED
Learners should be (ATTITUDES, SKILLS LEARNING ACTIVITIES RESOURCES
able to: AND KNOWLEDGE) AND NOTES
8.4.2 Structure  describe the - DNA structure  Constructing  ICT tools
and replication of structure of DNA models of DNA.  Braille
DNA software/Jaws
 explain how DNA - semi - conservative  Making DNA
replicates replication of DNA models illustrating  Print media
replication.  Models (zips,
- Messelson and Stahl beads, soft
experiment wires)

The main features of the structure are:

 DNA is double-stranded, so there are two polynucleotide stands alongside each other. The
strands are antiparallel, i.e. they run in opposite directions.
 The two strands are wound round each other to form a double helix (not a spiral, despite what
some textbooks say).
 The two strands are joined together by hydrogen bonds between the bases. The bases therefore
form base pairs, which are like rungs of a ladder.

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  The base pairs are specific. A only binds to T (and T with A), and C only binds to G (and G with
C). These are called complementary base pairs (or sometimes Watson-Crick base pairs). This
means that whatever the sequence of bases along one strand, the sequence of bases on the other
stand must be complementary to it.

FORM 5 TERM 3
TOPIC 6: INHERITED CHANGE AND EVOLUTION
KEY CONCEPT OBJECTIVES CONTENT SUGGESTED LEARNING SUGGESTED
Learners should be (ATTITUDES, SKILLS ACTIVITIES AND NOTES RESOURCES
able to: AND KNOWLEDGE)

8.5.1 Nature of  discuss the gene - Gene as unit of  Discussing the gene  ICT tools
Gene concept inheritance concept.  Braille
software/Jaws

 Print media
8.5.2 Monohybrid  use genetic - Co-dominance  Performing genetic  Print media
and Dihybridcrosses diagrams to - Sex linkage crosses.  Seeds
solve problems - Multiple alleles  Demonstrating  Pebbles
involving - Test crosses genetic crosses  Beads
monohybrid using beads, seeds  Scientific calculator
and dihybrid or pebbles.  Statistical tables
crosses
- Chi – squared  Applying the chi-
test squared test to
 use chi – results obtained
squared test to from the
test the demonstrations.
significance of
difference
between
observed and
expected
results

Someone asked me, “Why do you insist on taking the hard road?” and I replied, “Why
do you assume that I see two roads?” So challenges are what make life interesting and
overcoming them is what makes life meaningful
-Inheritance is a process in which the material is passed from the parents to the off-springs.

-In sexual reproduction the fusion of male and female gametes results in transfer of DNA from
parents

o A gene is a sequence of nucleotide on the DNA strand which codes for a certain peptide
chain. It’s also referred as a unit of inheritance.
o An allele is an alternative form of the same gene responsible for determining, construction
of characteristics. e.g. an allele for black skin color and an allele for white skin color.

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 o A dominant allele is an allele which influences the appearance of the phenotype presence
of an alternative allele. It suppose of expression of a recessive allele. It is represented by
capital letters.
o A recessive allele which influence the appearance of the phenotype only in the presence of
the other identical allele. It will not express itself in the presence of the alternative allele of
the same gene.
o A phenotype is the outward appearance of an organism or the external expression of a
gene or genotype e.g. black white etc.
o A genotype is the genetic constitution of an organism with respect to alleles under
consideration e.g. BB, Bb, bb.
o A locus is the position of an allele in the DNA molecule.
o Homozygous is a diploid condition in which the alleles at a given locus are identical e.g. BB
or bb.
o Heterozygous is a diploid condition in which the alleles at a given locus are different e.g.
Bb.
o A first filial generation (F1) these are off-springs produced by crossing parental genotypes
of organism.
o A second filial generation (F2) these are produced by crossing the parents from the F1
generation.

PRINCIPLE OF MANDALIAN INHERITANCE

1. MANDALIAN FIRST LAW OF SEGREGATION

─ In one of Gregor Mandel’s experiments he crossed pure breeding pea plants which are tall and
short. Cross pollination was presented and the F1 were all tall and no short plants were
produced. He intercrossed the F1 ones to produce the F2 generation. On counting these 787
were tall plants, 277 were short.
─ This is a ratio of 3:1 Mandel made the following conclusions:
─ There were no intermediate plants in F1 and F2 generation, this indicated that inheritance is not
a process in which parental features are blended. It is a process in which definite features which
may show themselves in F2 generation only, this implied that the F1 were carrying the allele for
shortness were being suppressed. All the F1 were tall and this indicated that the allele for
tallness was dominant and the allele for shortness was recessive.

THE GENETIC DIAGRAM

Let T represent the allele for dominant tallness. And t represents the recessive allele for shortness.

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Back cross test or test cross is then done by intercrossing the parents from F1 generation.

TEST 1 BACK CROSS TECHNIQUE

─ An organism showing dominant characteristics can have two possible genotypes, A tall
plant can have either homozygous or heterozygous tall. The phenotype will be the same but
the genotype is different and is determined by crossing the plant with the recessive
organism.
─ By crossing this organism with the unknown genotype with homozygous recessive it is
possible to determine the unknown or dominant characteristics genotype e.g. tall pea plant.

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 ─ If half the off-springs are tall and the other half is short this implies that the unknown
genotype is heterozygous tall.

MANDEL’S LAW OF SEGREGATION

─ It states that the characteristics of organisms are determined by internal factors which
occur in pairs and only one of the pair of such a pair is represented in a single gamete.
The internal factors are the genes.
─ This law’s explanation is obtained from meiosis, the alleles are located on one of the two
homologous chromosomes and during meiosis the homologous chromosomes come
together and they segregate into different gametes. Each gamete only receives one of each
type of chromosome and it also receives one of the pair of alleles.

ALBINISM

─ It is a condition in which the external segregation fails to develop due to lack of the skin
pigment melanin.
─ The individuals have light skin, white hair and pink eyes. It is an example of monohybrid
inheritance in humans caused by a recessive allele.

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 ─ This implies that it will only exert its effects in the homozygous condition. The genotype of
a normal person AA and of a carrier is Aa and the sufferer is aa. It is a result of gene
mutation.

CODOMINANCE

─ It is a condition in which two or more alleles fail to show complete dominance or

recessiveness to each other. This causes the alleles to show equally their effects on the
phenotype.
─ This is due to the failure of one of the alleles to be dominant in the heterozygous condition.
Co dominance is found in both plants and animals.
─ The heterozygous has a phenotype which is intermediate between homozygous dominant
and recessive condition produced by crossing pure breeding black and splashed white
fowls.
─ A cross between pure breeds of red and white cows produces an intermediate color called
ROAN.
─ The presence of the black plumage is the result of possession of an allele for black pigment
melanin. Splashed white fowls also lack this melanin. Heterozygous show a partial
development of this melanin which produces a blue sheen in the plumage.

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 ─ If the F1 generation intercrossed, the F2 generation shows the modification of the Normal
Mendelian F2 phenotypic monohybrid ratios of 3:1
─ A phenotypic ratio of 1:2:1 is product of alleles which are codominant.

SEX DETERMINATION

─ Sex chromosomes carry genes that determine an individual’s sex. In females the two sex
chromosomes are identical and are called X chromosomes.
─ The female genotype (autozome) is XX. In males the two sex chromosomes differ. They are
heterozomes X chromosome and one Y chromosome. Their genotype is XY.
─ The genotype XX is described as homogametic since it produces gamete cells with X
chromosomes. The XY genotype is described as heterogametic since half the gametes
contain X chromosomes and the other half contains the Y chromosomes.
─ The sex genotypes differ in other organisms e.g. in female butterflies have XY genotype and
male XX.

SEX LINKAGE-GENES ON SEX CHROMOSOMES

─ Genes carried on the sex chromosomes are said to be sex linked. In the case of
heterogametic sex there is a portion of X chromosome for which there is a non-homologous
region of the Y chromosomes.
─ Characteristics determined by genes carried on the non-homologous portion of the X
chromosome appear in males if they are recessive e.g. hemophilia and color blindness e.g.
hemophilia male is married to a carrier woman.

DIHYBRID INHERITANCE

─ Is concerned with the inheritance of two pairs of alleles consider the following example;
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 ─ Pea plants can produce the seeds which are round and wrinkled, Also they can be green and
yellow. One pure breed produce seeds that are round and yellow seeds while one pure
breed produces round and yellow seeds while the other pure breed winkled and green
seeds.

Let R the dominant allele for round seeds and r for winkled.

Let Y be the dominant allele for yellow seeds and y for green.

Parental phenotype: Round and Yellow x Winkled and green

Parental genotype: RRYY rryy

Meiosis:

Gametes : RY RY RY RY ry ry ry ry

Using PANNET’S square

RY RY RY RY
ry RrYy RrYy RrYy RrYy
ry RrYy RrYy RrYy RrYy
ry RrYy RrYy RrYy RrYy
ry RrYy RrYy RrYy RrYy

Genotypic ratio:1: 2 : 2 : 4

Phenotypic ratio:9:3:3:1

─ Basing on the results of the dihybrid cross therefore the presence of new combinations of
characteristics. Mandel postulated his second law known as “the principle of
independent assortment”.On one pair of characteristics may combine with either one. The

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 typical dihybrid ratio of 9:3:3:1 only apply to characteristics controlled by genes and
different chromosome are said to be linked, they form a linkage group.

Activity

In fruit Drosophila the gene for long wing length and for eye colour are sex linked. Normal wing and
red eye are dominant to miniature wing and white eye.

(i) Define the term sex-linked. [1]

(ii) A cross between a miniature wing, red eyed male and homozygous normal wing, white-
eyed female was carried out.
Using symbols N for normal wing R for red eye, r for white eye and n for miniature
wing, draw a genetic diagram to show the genotypes and phenotypes of the F1
generation in the space below. [5]

(source: June 2015 Zimsec)

HUMAN BLOOD GROUPS

─ They are controlled by an autozomal gene. The gene locus is represented by a single (I)
isohaemoglobin.
─ There are three alleles for the blood groups A, B, and O. A, AB, and B are codominant
(equally dominant) while O is recessive to both.
─ The presence of the single dominant allele, A or B results in the blood producing a substrate
agglutinin which acts as an antibody e.g. the genotype IAIOwould give rise to agglutionogen
A on the red blood cell membrane and plasma would contain agglutinin anti-B. There are
six possible genotypes and only four exist.

─ The A, B, O blood groups in humans are controlled by multiple alleles of a single autozomal gene.
The gene locus is usually represented by the symbols (IA, IB, and IO). There are three alleles
represented by the symbols IA, IB, IO. Allele IA and allele IBare equally dominant to IO which is
recessive to both.
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 THE SUMMARY OF MENDEL’S 2nd LAW.

─ Orientations on the equatorial spindle of bivalents during metaphase 1 and of chromosome

in metaphase 2 are random. The subsequent separation of these chromosomes during
anaphase 1 and 2 respectively produces new alleles recombination in gamete cells. This is
called independent assortment and results in the random assortment of maternal and paternal
chromosomes between daughter nuclei. This is the basis of Mendel’s second law.

A PEDIGREE CHART

─ Taking an example of albinism.

─ A pedigree chart is a diagram of a family tree over several generations, showing the
descendants. The relationships and the presence or absence of a specific trait in all the
members.
─ In the chart the males are represented by squares and females by circles.
─ Shading indicates the incidence of the particular genotype.
─ Under investigation, analysis of the pedigree chart enables us to detest the difference
between a dominant and a recessive trait.
─ A dominant trait occurs in members of every generation. A recessive trait is seen in
frequently often skipping one or more generations.

HOW MUTATION CAN AFFECT THE PHENOTYPE

─ Chromosome mutation – is a change in the structure and amount of the chromosomes, and
gene sequence on the chromosome. The changes are most likely to occur when chromatids
break and rejoin during cross over in prophase of meiosis. Sections of the chromosome may
be lost or incorporated into other chromosome i.e.
 detection,
 inversion,
 Translocation
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  duplication (reshuffling of genes).
─ CHANGE IN CHROMOSOME NUMBER
 An individual’s failure of chromosome segregation during anaphase in meiosis leads
to the production of 24 chromosomes in a gamete cells instead of the normal 23 in a
human being.
 Usually it is the 21𝑠𝑡 chromosome (which is very small) which fails to separate to
two separate chromosomes.
 An extra chromosome is left in the gamete to leave 24 chromosomes (2n+1).
 The outer gamete is left ladening another chromosome it is left with 22
chromosomes (2n-1). The process of a failure in chromosome segregation is called
non-disjunction.

EFFECTS OF MUTATIONS

(A) DOWN’S SYNDROME/ MONGOLISM

─ It is caused by non-disjunction. When a gamete with an extra chromosome fuses with a normal
gamete with 23 chromosomes the total number of chromosome will be 47 (2n+1) instead of 46
(2n).
─ DOWN’S SYNDROME CHILDREN HAVE THE FOLLOWING DISABILITIES
 Flat broad face.
 Squint eyes, with a skin fold in the inner corner.
 Furrowed and protruding tongue.
 Short stacky body thick neck.
 Thick neck.
 Short life expectancy.
 Low IQ (intelligence quotient)

Activity
1. Coat colour in cats is determined by a sex-linked gene with two alleles, black and orange. When
black cats are mated with orange cats, the female offspring are always tortoiseshell, their coats
show black and orange patches of various sizes, while the male offspring have the same coat colour
as their
(a) Using the symbols XBforXBfor black and XOforXOfor orange, draw genetic diagrams to
account forbothforboth these crosses.
(i) black female X orange male

(ii) orange female X black male

(b) List the genotypes and their phenotypes of the offspring that may result from mating
atortoiseshell female with a black male.
[4]
(c) Suggest an explanation for the tortoiseshell coat in terms of the activity of the Xchromosomes.
[1]
2. Fig. 4.1 shows four generations of a family in which some members of the family suffer from
sickle cell anaemia.

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(a) Using the symbols HNfor the allele for normal haemoglobin and HSfor the allele forSickle cell
haemoglobin, state the genotypes of the individuals A and C [1]
(b) Draw a genetic diagram to show the probability of the parents A and B producinganother child
with sickle cell anaemia. [5]

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WRITTEN BY MR. R. GWANZURA WHATSAPP: 0773266377/CALL 0717558917 HOLY CROSS HIGH
TOPIC 7: ENERGETICS
KEY CONCEPT OBJECTIVES CONTENT SUGGESTED SUGGESTED
Learners should be (ATTITUDES, SKILLS LEARNING RESOURCES
able to: AND KNOWLEDGE) ACTIVITIES AND
NOTES
8.6.1 ATP  outline the need for - Anabolic reactions  Discussing uses of  Print media
Structure and energy in living - Active transport energy.  ICT tools
Synthesis organisms - Movement  Braille software/Jaws
- Maintenance of body
temperature

 describe ATP - Structure of ATP  Illustrating the  Model

structure as a structure of ATP.
phosphorylated
nucleotide

 describe synthesis of - Chemiosmosis  Illustrating the

ATP by chemiosmosis
chemiosmosis coupling of ATP
synthesis.
I long to die knowing that I have lived well… I have taught well…. Helped them that needed
help n did come my way….

─ Need to use energy in living organisms.

─ Energy is defined as the capacity \ ability to do work.
─ Living organisms require a constant supply of energy for them to keep working and alive.
─ Energy required by living organisms for :
o Anabolic reactions e.g growth.
o Active transport of sustains against concentration gradient.
o Phagocytosis, pinocytosis , exocytosis, endocytosis.
o Electrical transmission of nerve impulse.
o Mechanical, contraction of muscles.
o Maintainace of temperature heat from respiration.
o Bioluminescence.
o Electrical discharge.

Structure of ATP

─ ATP is a energy carrier molecule made up of organic base adenine ,a pentose sugar ribose
and three phosphate groups.
─ The third phosphate can be debouched from ATP by the release producing ADP plus a
inorganic phosphate .

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 Hydrolyses

ATP + H2O ADP + Pi + energy [30 .6KJ/mol ]

Condensation

─ Adding Pi to ADP is known as phosphorylation. The enzyme ATPASE has catalyzed the
reaction.
─ All cells in every living organism use ATP as their energy source
─ ATP is known as the universal energy carrier molecule.

Explain the different energy values of carbohydrate, lipid and protein as respiratory
substrates. [6]

 idea of lipid > protein > carbohydrate / AW ; A lipid has more energy than either protein or
carbohydrate
 comparative figures ; e.g. 39.4, 17.0 and 15.8 accept any two
 kJ g-1 / per unit mass ;
 more hydrogen atoms in molecule, more energy ;
 lipid have more, hydrogen atoms / C-H bonds ;
 (most) energy comes from oxidation of hydrogen to water ;
 using reduced, NAD / FAD ;
 in ETC ;
 detail of ETC ;
 ATP production

Describe the structure and synthesis of ATP and its universal role as the energy currency in all living
organisms.[8]
─ nucleotide ;
─ adenine + ribose / pentose + three phosphates ;
─ loss of phosphate leads to energy release / hydrolysis releases
─ 30.5 kJ ;
─ ADP + Pi ↔ ATP (reversible reaction) ;
─ synthesised during, glycolysis / Krebs cycle / substrate level
─ phosphorylation ;
─ synthesised, using electron carriers / oxidative phosphorylation /
─ photophosphorylation ;
─ in, mitochondria / chloroplasts ;
─ ATP synthase / ATP synthetase ;
─ chemiosmosis / description;
─ used by cells as immediate energy donor ;
─ link between energy yielding and energy requiring reactions / AW ;
─ active transport / muscle contraction / Calvin cycle / protein synthesis

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PHOTOSYNTHESIS
KEY CONCEPT OBJECTIVES CONTENT SUGGESTED SUGGESTED
Learners should be (ATTITUDES, SKILLS LEARNING RESOURCES
able to: AND KNOWLEDGE) ACTIVITIES AND
NOTES
8.6.2 Photosynthesis  draw detailed - chloroplast structure  Drawing and labeling  Print media
structure of chloroplast.  Filter paper
chloroplast  Acetone
- Chloroplast pigments  Separating pigments  Different coloured
 identify chloroplast by paper leaves
pigments chromatography.  Leaf extracts
 Collecting different
- Absorption and coloured leaves.  ICT tools
Action spectra  Finding out other  Braille software/Jaws
 discuss the role of uses of pigments in
chloroplast pigments life.
in absorption and  Analysing absorption
action spectra and action spectra.
- Light dependent
reactions  Outlining the light
 describe the photo - (cyclic and non-cyclic dependent reactions
activation of photo of photosynthesis.
chlorophyll phosphorylation)

- Light – independent
reactions (Calvin  Illustrating the Calvin
 outline the Calvin Cycle) Cycle.
Cycle
- Carbon fixation in C4
plants
 Discussing carbon
 discuss fixation in C4 plants.
photosynthesis in C4
plants
- Light intensity and
wavelength
 Investigating the
 discuss the concept - Carbon dioxide
effects of limiting
of limiting factors concentration
factors on rate of
- Temperature
photosynthesis.

Photosynthetic pigments

─ Photosynthetic pigments of higher plants fall into two classes the chlorophylls and
carotenoides .
─ The role of the pigments is to absorb light .
─ There are located in the thylakoids .Chloroplasts absorb mainly red blue violet light
reflecting green.

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 ─ The chlorophyll has flat light absorbing and it has a magnesium at the centre .
─ It also as along hydrophobic tail .Chlorophyll is the most common pigment and it exists in
various forms each differ slightly according to it’s light absorbing peck.

CARTENOIDES

─ Cartenoides are yellow , orange and red carbon pigments that absorb strongly in the blue
violet range .
─ They are accessory pigments because they pass their energy they absorb to the chlorophyll.
─ Carotenoide have two types carotenes and xanthophylls.

PHOTOSYNTHESIS
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 ─ The chlorophyll and accessory pigments molecules are are located in two types of
photosystems known as photosystem 1 and photosystem 2 .
─ Each contain an antenna complex of pigments collect light of different wave length making
the process more efficient.
─ All the energy harvesting transferred to chrolophylls are known as P700 in PS1 and P680
in PS2.
─ The biochem of photosynthesis.
─ The commonly used equation for photosynthesis is

SUNLIGHT

6H2O +6CO2 C6 H12 O6 + 6O2 .

CHLOROPYLL

─ It implies that carbon dioxide reacts with H2O to give carbohydrates + oxygen in a one of
process yet the CO2 and H2O do not react together perse .
─ Photosynthesis is divided into stages. The first stage is the light depended stage and the
dark stage.

LIGHT DEPENDED STAGE: THE Z SCHEME / PHITOPHOSPHORYLATION

Describe the transfer of energy to ATP during photosynthesis [6]

─ light absorbed by chlorophyll

─ An electron from P700 orP680 is boosted to a higher energy level when light strikes the
photosystem;
─ the electron which acquires excitation is is accepted by electron acceptor x or y .
─ The electrone acceptor x or y become reduced and chlorophylls become oxidized with
positive change.

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 ─ The electrons with the excess energy of oxidation are very unstable tend to fall down to
their ground state .
─ The electrons then in terms of energy via a series of electrons acceptors.
─ The excess energy lose when they fall back to the ground state is coupled is coupled in the
production of ATP.
─ The positive change left in the p680 contribute to to the photochemical lysis of water which
releases electrons which lost from electron acceptor X .
─ Electrons flow X along a chain of electron carries loosing energy [used to phospholyate
ADP to ATP ] FILLING THE HOLE LEFT IN p700 ,electrons also pass down from Y to NADP
along a chain a chain of electron carries to and combine with hydrogen ions from water to
form reduced NADP.
─ This is called non cyclic photophosphorylation .
─ In cycle photophosphorylation electrons from Y are recycled to P700 via a electron chain
carries results in ATP being formed

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Describe the roles of the following substances in the light-independent stage of photosynthesis:
(i) RuBP [2]
(ii) (ii) reduced NADP [2]
(iii) (iii) ATP. [2]
(i) carbon dioxide fixation ;
production of GP ;
ref. to rubisco ;

(ii) reduction (of GP) / donates hydrogen ;

GP to TP ;

(iii) supplies, energy / phosphate ;

(to convert) GP to TP ;
(to) regenerate of RuBP ;

Outline the process of the photolysis of water and describe what happens to theproducts of
photolysis. [10]
─ PII absorbs light ;
─ enzyme (in PII) involved ;
─ to break down water / AW ;
─ 2H2O 4H+ + 4e– + O2 ;
─ oxygen is produced ;
─ used by cells for (aerobic) respiration ;
─ or released (out of plant) through stomata ;
─ protons used to reduce NADP ;
─ with electrons from PI ;
─ reduced NADP used in, light independent stage / Calvin cycle ;
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 ─ to convert GP to TP ;
─ electrons also used in ETC ;
─ to release energy for photophosphorylation ;
─ to produce ATP ;
─ electrons (from PII) go to PI ;
─ ref. re-stabilise PI ;

Light independent stage; The Calvin cycle

─ The light independent ( or dark ) reactions takes place in stroma and do not require
light .
─ they make use of the ATP ( energy ) and NADPH2 from from the light dependent stage of
photosynthesis .
─ The biochemistry of these reactions take place where promulgated by Calvin ,Benson
and Basham , hence they are known as Calvin –Benson cycle .
1. Carbon dioxide fixation
─ The first stage is carbon dioxide fixation or acceptance.
─ The carbon dioxide acceptor RuBP a5C compound combines with CO2 to form a highly
unstable 6C compound which quickly breaks down to form to 3C compound to glycerate
phosphate.
─ The enzyme Ribulose biphoshate carboxylase oxyfenal ( Rubisco) catalyse the reaction .

2. Reduction phase
─ The glycerate phosphate a 3C acid contains the carboxylic group (COOH) which is reduced
to an aldehyde group (-CHO).
─ Energy from ATP and hydrogen from NADPH2 are used to remove oxgen from GP.
─ TP is produced a sugar phosphate, the first carbohydrate made in photosynthesis.

3. Regeration of RuBP – the CO2 Acceptor

─ Some of the TP is used to generate the RuBP consumed in the first reaction.
─ The regeneration of RuBP involves a complex cycle containing 4,5,6. And 7C sugar
phosphates.

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(a) Outline/ describe the main features/reactions of the Calvin Cycle. [9]
─ RuBP 5C ;
─ combines with carbon dioxide ;
─ rubisco ;
─ to form an unstable 6C compound ;
─ which forms 2 X GP (PGA) ;
─ ATP;
─ energy source
─ and reduced NADP ;
─ forms TP (GALP) ;
─ TP used to form glucose / carbohydrates 1 lipids / amino acids ;
─ TP used in regeneration of RuBP
─ requires ATP ;
─ as source of phosphate ;
─ light independent ;
─ Grana increase surface area for light absorption

Qn. Fig. 1.1 shows the arrangement of photosystems, protein complexes containing
chlorophyll molecules, on the thylakoid membrane of a plant chloroplast.

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(a) Describe the photoactivation of chlorophyll. [3]
(b) Explain how the photolysis of water occurs. [3]
(c) Outline how ATP is formed in the chloroplast [3]
(d) Suggest an advantage of having photosystems, the electron transport chain and
ATPsynthase as part of the thylakoid membrane.
[1]
[Total: 10]

Solution

1 (a) 1 chlorophyll absorbs mainly red and blue light;

2 light absorbed by antenna complex;
3 energy transferred;
4 reaction centres/P700/P680;
5 light energy excites electron(s)/reference passing to higher energy level;
6 electron lost from chlorophyll 3 max
(c) 1 water is split into H+ and OH-;
2 electron removed from OH-;
3 to replace electron from photosystem/chlorophyll;
4 OH breaks down into O2 and water;
5 H+ used to form reduced NADP;
6 reference correct, balanced equation; 3 max
(d) 1 reference flow of electrons along ETC;
2 reference to pumping H+ across membrane;
3 reference to H+/proton gradient across the thylakoid membrane;
4 flow of protons down gradient;
5 via ATPase/stalked particles;
6 formation of ATP from ADP and Pi;
7 cyclic, electron returns to original photosystem;
8 non-cyclic, electrons from PSII to PSI; 3 max
(e) reference increased efficiency/short diffusion distance/close together;

Describe the arrangement and location of chloroplast pigments and discuss their effecton
absorption spectra. [8]

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 ─ chlorophyll a is primary pigment ;
─ carotenoids / chlorophyll b, is accessory pigment ;
─ arranged in, light harvesting clusters / photosystems ; A antenna complex
─ on, grana / thylakoids ;
─ ref. PI and PII ; A P700 and P680
─ primary pigment / chlorophyll a, in reaction centre ;
─ accessory pigments / carotenoids / chlorophyll b, surround primary pigment ;
─ light energy absorbed by, accessory pigments / carotenoids / chlorophyll b ;
─ (energy) passed on to, primary pigment / chlorophyll a / reaction centre ;
─ chlorophyll a and b absorb light in red and blue/violet region ;
─ carotenoids absorb light in blue/violet region ;
─ ref. absorption spectrum peaks ;
─ diagram of absorption spectrum ;
─ different combinations of pigments (in different plants) give different spectra

The C4 pathway

─ Certain plants (C4 plants) possess a characteristic leaf anatomy in which two rings of cells
are found around each of vascularbundles.
─ The inner ring the bundle sheath cells contain chloroplast which differ in form from those
in mesophyll cells thee are referred to as the kranz anatomy

Hatch slack pathway

─ The hatch slack pathway is a pathway for transporting CO2 from the mesophyll , it’s a
pumping mechanism for CO2 .
─ The carbon dioxide in acceptor is PEP and the reaction is catalysed by PEP carboxlase an
efficient enzyme with high affinity for CO2 and not inhibited by O2.

PEP caboxylase

PEP + CO2 oxaloacetate

─ Oxaloacetate is then reduced by NADPH 2 to MALATE .

─ Malate is then shunted through plasmodesm into the bundle sheath cells where they
accepted by RUBP in the normal C3 pathway .

C3 and C4 plants

─ C3 plants have GP and c4 have a four carbon compound oxaloacetate.

─ The enzyme Rubisco is also able to catalyes the combination of Rubp with oxygen .
─ This results in the lose of Rubp which can be used only in the presents of CO2frome the
plant.As this process has the overall effect of taking in O2and giving out CO2 and this is
called photorespiration.
─ However C4 prevent photorespiration by keeping Rubp and Rubisco away from O2.The
cells containing these two compoundes are arranged around the vascular bundle called
bundle sheath.
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 ─ They have nodirect contact with the air CO2 is absorbed by the mesophyl cells which are in
contact with air space .
─ The mesoptle cells contain a enzyme called PEP CARBOXYLASE which catalyses the
combination of CO2 with the three carbon molecule phophopyruvate.
─ The opd made frome the reaction is oxaloasetate .
─ Stil inside the the mesopyle cell oxaloacetate is converted to malate and this malate is
shunted into the bundle sheath.
─ Here CO2 is removed from the malate and delivered to Rubp by Rubisco in the usual way
.The calvin cycle then procceds normally.
─ Some of the most productive crop are C4 plants .Sugar cane is is especially efficient
converting around 8percent sunlight to energy in carbohydrates.

Difference between C4 AND C3 Plants

C3 C4
GP present oxaloacetate
High photorespiration Minimum photorespiration
Sheath bundle cells absent Bundle sheath cells present.
Have Rubisco PEP and rubisco are present .
No malate is fromed Malate is formed .
Combination of rubisco with CO2 Combination of CO2 with PEP.

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Explain how the palisade mesophyll cells of a leaf are adapted for photosynthesis. [6]

─ closely packed -- to absorb more incident light / AW ;

─ palisade mesophyll near upper surface of leaf -- to maximize light interception ;
─ . arranged at right angles to leaf surface -- to reduce number of light absorbing walls ;
─ cylindrical cells -- producing air spaces between cells ;
─ air spaces -- act as reservoir of carbon dioxide ;
─ large surface area -- for gas exchange ;
─ cell walls thin -- so short diffusion pathway ;
─ large vacuole -- pushes chloroplasts to edge of cell ;
─ chloroplasts on periphery -- to absorb light more efficiently ;
─ large number of chloroplasts -- to maximise light absorption ;
─ chloroplasts can move within cells -- towards light ;
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 ─ chloroplasts can move away from high light intensity -- to avoid damage ;
─ AVP ;

FACTORS AFFECTING PHOTORESPIRATION

1. TEMPERATURE
2. CARBON DIOXIDE
3. WATER
4. SUNLIGHT
5. CHLOROPHYLL

TEMPERATURE

─ AS little affect o the rate of photosynthesis since energy required is from sunlight not heat.
─ Calvin cycle is enzyme controlled hence needs optimum temperature to operate a maximum
rate.

LIGHT INTERNSITY

─ As for CO2 concentration light tends to be the rate limiting at low intensities but not at high
intensities.
─ Rate of photosynthesis is measured by the rate of O2 production.
─ Plants respire as well as photosynthesis .At low light intensities the plants tend to respire
.As light intensities increase the rate of photosynthesis.
─ This is called the light compensation point .Above this light intensity the rate of
photosynthesis exceeds the rate of respiration and so O2 is released from the plant.

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CARBON DIOXIDE

─ When CO2 is the rate limiting increase in the CO2 increases the rate of photosynthesis .
─ This makes a rising straight line on the graph on the rate against CO2 concentration.
─ Concentration point is the point where rate of concentration of CO2 is equal to O2 thus
photosynthesis equal to respiration.
─ When CO2 concentration is not a factor limiting increase in the concentration of CO2 will
not change the rate of photosynthesis .The graph is horizontal.

RESPIRATION
KEY CONCEPT OBJECTIVES CONTENT SUGGESTED SUGGESTED
Learners should be (ATTITUDES, SKILLS LEARNING RESOURCES
able to: AND KNOWLEDGE) ACTIVITIES AND
NOTES
8.6.3 Respiration  Draw detailed - Mitochondrion  Drawing and  ICT tools
structure of annotating  Print media
mitochondrion mitochondrion.  Braille software/Jaws

 outline the process - Glycolysis  Outlining the process

of glycolysis of glycolysis.

 describe the - Link reaction  Discussing the

formation of acetyl conversion of
Coenzyme A (CoA) pyruvate to acetyl
from pyruvate CoA.
- Krebs Cycle
 outline the Krebs  Illustrating the steps
Cycle in the conversion of
citrate to
- Decarboxylation oxaloacetate.
 explain - Dehydrogenation
decarboxylation and  Discussing the
dehydrogenation in processes of
relation to the link decarboxylation and
reaction and the dehydrogenation
Krebs cycle
- Election transport
 describe the process chain
of oxidative - Role of oxygen  Discussing oxidative
phosphorylation - Role of Nicotinamide phosphorylation.
in the mitochondrion Adenine Dinucleotide
(NAD)

 outline the process - Anaerobic respiration

of anaerobic  Discussing
respiration in plant/ anaerobic  Yeast
yeast and animal - fermentation respiration.  Sucrose/Glucose
cells
 design experiments
to compare rates of  Designing and
fermentation carrying out  Food samples
- Carbohydrates
experiments to
- Proteins
compare rates of
- Lipids
 explain the relative fermentation.
energy values of
carbohydrates, lipids  Performing
and proteins as experiments to  Respirometer
- RQ
respiratory determine energy  Small animals such
- Effect of temperature
substrates values. as beetles, harurwa
on respiration rates
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 caterpillars, amacimbi
define the term  Water bath
Respiratory Quotient  Designing and  Incubator
(RQ) carrying out
experiments using
simple respirometers
 calculate RQ to measure RQ.

 Calculating RQ.

─ The which makes ATP using in organic molecules and is subdivided into glycolysis and the
Krebs cycle and oxidative phosphorylation.
─ The glucose is dismantled streadility in a series’ of reactions known as the metabolic
pathway.
─ The metabolic path ways of respiration are divided into three main stages.
I. Firstly in the cytoplasm of the cell glucose is converted into pyruvate. {glycolysis}
II. . The next inside the mitochondrion is in cycle of reactions called Krebs cycle .
III. Finally in the mitochondrion the electrons produced in the Krebs cycle are passed to
the electron transport chain producing ATP in aprocess called oxidative
phosphorylation.
─ Glucose are uncreative so they are activated before glycolysis occurs.
─ This is done by addition of a phosphate to the glucose forming glucose phosphate.
─ The atoms in this molecule are then arranged to form fructose phosphate and another
group added to form fructose biphosphate.
─ Each of these additions of a phosphate group is done by transferring a phosphate group to
the sugar from ATP.
─ Theis split into 3 carbon molecule triose phosphate .Each of these converted into GP then
pyruvate in a series of steps.
─ These steps release energy which is used to ATP from GP .Four molecules of ATP are made
directly there are then in the cytoplasm using energy released as ATP are gradually
changed to pyruvate .
─ The conversion of TP to GP releases hydrogen ions to electrons which are transferred to
the coenzyme NAD to form reduced NAD.
─ These hydrogen and high energy electrons are passed into mitochondria where they can
be used to produce 5ATP molecules in oxidative phosphorylation only in the presents of
oxygen.

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The Krebs cycle

─ If oxygen is available, THE PYRUVATE FROMED IN GLYCOLYSIS IS passed into the

mitochondrion through the inner and outer membrane.
─ Once in the matrix of the mitochondrion pyruvate is converted into Acetylcoenzyme A .
─ One hydrogen ion, two electrons and one CO2 molecule is released.
─ The acetyl coenzyme A which conteins two carborn molecules is added to oxaloacetate
[4CO2].
─ The result Compound a six carbon compound citrate is gradually required than
oxaloacetate. At two stages in the Krebs cycle CO2 plus that was removed from the
compound involved. This is called decarboxylation.
─ This CO2 plus that which was produced when pyruvate was converted into acetyl coenzyme
A diffuses out of the mitochondrion out of the cell then out of the organism.
─ Electrones and hydrogen ion are both picked up by the oxidized form of the coenzyme NAD
and some by FAD.
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 ─ These coenzymes can hold electrons which will which will then be fed into the ETC.
─ When one glucose molecule is respired two pyruvate are produced and they result in the
formation of six reduced NAD and two reduced NAD are produced when pyruvate is
converted to acetyl coenzyme A

Describe what happens to the pyruvate so that the krebs cycle can continue.

─ enters mitochondrion;
─ active uptake / ATP used;
─ into matrix;
─ link reaction;
─ decarboxylation / carbon dioxide released / AW;
─ dehydrogenation / AW;
─ reduced NAD formed;
─ forms acetyl coenzyme A / combines with coenzyme A; A Co A
─ combines with oxaloacetate / forms citrate;

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OXIDATIVE PHOSPHORYLATION [ETC]

─ ATP is made by addition of inorganic phosphate to ADP are phosyphorylation reaction.

─ In the respiration process it requires oxygen and is known as oxidative phosphorylation.
─ The Krebs cycle takes place in the matrix of the mitochondrion.
─ Reduced NAD and FAD provide energy for ATP synthesis as they are passed in the ETC .
─ The electrons are paced along ATP is made.
─ At the end of the ETC oxygen is used to combine with electrons as they off the chain with
hydrogen ions to form water.
─ If there‘s no enough O2 the ETC wont work and hence resulting in the occurrence of
anaerobic respiration .

ANAROBIC RESPIRATION
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─ In anaerobic respiration glycolysis takes place as ussul pyruvate a small amount of ATP is formed
.If pyruvate was allowed to form it will inhibit glycolysis so it is converted to something else.
─ The reduced NAD which is produced in glycolysis must be oxidized back to NAD or the cell will
run out of ATP.

Alcohol fermentation

─ Yeast convents pyruvate to ethanol .CO2 is removed from pyruvate to produce ethanol.

CH3 COCOOH CH3 CHO + CO2

{ETHANOL}

─ The enzyme alcohol dehydrogenate convets the ethanol to ethenil . This requires hydrogen
which is taken from reduced NAD .
─ The overall reaction equation
─ Conversion of pyruvate acetyle CoA [link ring]

Lactase fermentation

─ In the lactase fermentation the pyruvate from glycolysis accepts the hydrogen ions from
reduced NAD directly .

CH3 COCOOH CH3 CHO HCOOH

NADH2 NAD {lactase}

─ If oxygen be made unavailable again lactase can be further broken down to to release it’s
remaining energy or alternatively be resynthesised to carbohydrates or extricated.
─ Lactase fermentation is useful to animals living in in flacuating levels of oxygen , a baby in
the period just after birth .

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LACTASE FERMATAION ALCOHOL FERMANTAION
Yield 2ATP Yield 2 ATP
NO CO2 PRODUCED CO2 PRODUCEDTION INVOLVED
Pyruvate, decarboxylase ,alcohol used Lactase ,dehydrogenase ,is used .
Yield 180kj Yield 210kj

Under anaerobic conditions, the reduced NAD cannot be oxidised using oxygen.
However. Without it being oxidised glycolysis will stop and no ATP will be formed.

Explain how the reduced NAD is oxidised under anaerobic conditions in mammalian
muscle tissue and in yeast.

Muscle

─ pyruvate converted to lactate; A lactic acid

─ hydrogen combines with pyruvate;
─ lactate dehydrogenase;

Yeast

─ pyruvate converted to ethanal;

─ release of carbon dioxide / decarboxylated;
─ hydrogen combines with ethanal;
─ ethanal converted to ethanol;
─ alcohol dehydrogenase;

Explain the role of NAD in aerobic respiration. [6]

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 ─ coenzyme;
─ for dehyrogenase;
─ reduced;
─ carries electrons;
─ and protons/H+/H/hydrogen; R H2/hydrogen molecules
─ from Krebs cycle;
─ and from glycolysis;
─ to cytochromes/electron transfer chain;
─ reoxidised/regenerated;
─ ATP produced;
─ 3/2.5 (molecules of ATP) per reduced NAD;

Describe the main features of the Krebs Cycle. [9]

─ matrix;
─ of mitochondrion;
─ acetyl CoA combines with oxaloacetate;
─ to form citrate;
─ 4C to 6C;
─ decarboxylation/produce CO2;
─ dehydrogenation/oxidation;
─ 2CO2 released;
─ reduced NAD produced; accept reduced coenzyme for one mark - annotate 9/10
─ reduced FAD produced;
─ ATP produced;
─ series of steps/intermediates;
─ enzyme catalysed reactions;
─ oxaloacetate regenerated;
─ AVP;

ALTERNATIVE RESPIRATORY SUBSTRATES

RESPIRATION OF FATS

─ The oxidation of fats is proceeded by it’s hydrolysis to glycerol and fatty acids .The glycerol
may then be phosphorylated and converted to TP.
─ This can be incorporated into the glycolysis pathway and subsequently the Krebs cycle.
─ The fatty acid compound is progressively broken down in the matrix of the mitochondrion
into two CO2 compounds which are converted to Actyle coenzyme A.
─ This then entres the krebs cycle with the consequent release of energy .The oxidation of fats
has the advantage of producing large amounts of H+ IONS.
─ These can be transported by hydrogen carries and used to produce ATP in the electron
transported by hydrogen carries for this region fats librerate move and double the the
energy for the same amount of carbohydrates.

RESPIRATION OF PROTEINS

─ Proteins must be hydrolyzed to constituent amino acid which then have the amino
[NH2]group removed in the liver by deamination.
─ The remaining portion of the amino acid then entre the respiratory pathway at a number of
points depending on their carbon content.
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 ─ 5 carbon amino acid and 4 C amino acid are converted to pyruvate ready to be converted to
acetyl coenzyme A
─ Other a, a with longer quantities of carbon undergo transmination reaction to convert
them to 3,4 or 5 amino acid .

REPIRATION QUNTIENTS

─ The respiratory [RQ] is the measure of the CO2 involved by the organisisms to the O2
consumed over a certain period.

─ In the fats ratio of O2 to CO2 is far smaller than on carbohydrate. A fat requires quantity of
O2 for as complete oxidation and thus RQ less than one.

GLYCEROL

─ Is first phosphorylated by ATP into GP and then dehydrogenated by NAD to the sugar
dehydrogenatephosphates.
─ This is next converted into isomer glycerddehyde 3 phosphate
─ The process consume ATP and yield ATP when hydrogen is to along the respiratory chain
─ Glycerated 3 phosphate is subsequently incorporated into the glycolysis pathway and Krebs
cycle liberating a further 17ATP.
─ Therefore a yield per one molecule of glycerol aerobically respires to 20/1 =19 ATP.
─ Respiring Q measures ofrotating ATP of CO2 evolved by one organism consumed over a
long period of time.

FATTY ACIDS

─ Each fatty acid molecule is oxidized by a process called oxidation which involves 2C
fragments of acetyl COEZYME A being split off from the acid molecule.
─ Each Acetyl coenzyme formed can entre Krebs cycle as usual to be oxidized to CO2 and
H2O

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 ─ A lot of energy is released per one molecule of glycerol. A lot of energy is released when a
fatty acid is oxidized.
─ Fatty acids therefore contribute more than half the normal energy required of heart, resting
skeletal muscle liver and kidney.

PROTEIN

─ Proteins are first hydrolyzed into their constituent aa then delaminated and transmission.

OXDATIVE DEAMINATION

─ Occurs in vertebrates liver cells on ammonium molecules is removed from the amino acids
by dehydrogenation and hydrolysis.
─ The deamination amino acids is an alpha keto acid .It may be respired like a carbohydrate
or via the fatty acid pathway.

TRANSMINATION

─ Is the transfer of an amino acid group

─ An amino acid to a keto acid .One amino acid can be converted to another .
─ This process also produce an alpha keto acid able to enter the normal respiratory pathway.

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TOPIC 8: TRANSPORT SYSTEMS
Transport in plants

KEY CONCEPT OBJECTIVES CONTENT SUGGESTED SUGGESTED

Learners should be (ATTITUDES, SKILLS AND LEARNING ACTIVITIES RESOURCES
able to: KNOWLEDGE) AND NOTES
8.7.1 Structure  describe the - Structure and - Examining fresh  microscope
and mechanisms of structures of the adaptations of xylem monocotyledonous  Slides
transport systems xylem vessels, vessels, sieve tube and dicotyledonous  Prepared slides
in plants sieve tube elements and plant roots and  Staining dyes
elements and companion cells stems.  Microtome
companion cells - Drawing cross  ICT tools
sectional diagrams  Braille
of monocot and software/Jaws
dicot plant roots
 Print media
and stems.
 Scalpel blades
- Discussing  Visking tubing
adaptations of  Live plants
 explain how the xylem and phloem.
xylem vessels and
phloem tubes are - Osmosis
adapted to their - Apoplast
functions - Symplast - Discussing the
- Vacuolar pathways.
 describe, clearly - Role of the
stating the Casparian strip
pathways, how
water is transported - Osmosis
from the soil to the - Root pressure - Observing effect of
xylem - Transpirationpull root pressure by
- Capillary effect cutting a stem of a
live plant.
 explain the
mechanisms by
which water is - Mass flow hypothesis - Demonstrating
transported from mass flow
soil to xylem and hypothesis.
from roots to leaves

 explain the
translocation of
sucrose
Father God; remember those I teach; not for my sake but for their sake. That in everything I do it may
be for their gain. That praises may belong only to thee , not me…amen

Xylem

─ The xylem has two major functions: the conductor of water, mineral salts and support.
─ It consists of both physiological and structural importance in plants.
─ It consists of four cell types :tracheids, vessel elements, parenchyma and fibres

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Tracheids

─ A single cell elongated and lignified and lapering end and wall that overlap with adjacent
tracheds in the same way as sclerenchyma fibres.
─ They have mechanical strength and give support to the plant.
─ They are dead with empty lumen when mature.
─ Water pass through the lumens without being obstructive by living organisms.
─ It pass from trenched to trached through the pits via the pit membrane or through
unlignified portions of the cell walls

Vessels

─ Are characteristic conducting units of angiosperms xylem.

─ They are very long tubular structures formed by fusion of several cell end to end in a row .
─ Each of the cell forming a xylem is equivalent to a trached and is called a vessel element.
─ A vessel is formed when a neighboring vessel element of a given raw fuse as a result of their
end walls break down.
─ A series of rims is left around the inner side of the vessel marking the remains of the end
wall.

Protoxylem

─ A mature protoxylem can be stretched as surrounding cells elongate because ligning is not
deposited over the entire cell wall but only in rings or spirals
─ There act as reinforcement for the tubes during elongation of stem nd roots
─ As growth proceed, more xylem vessel develop and these undergo more extensive
lignifiction completing their development in the mature reigns of the organs and forming
metaxylem
─ Mature xylem cannot grow since they are dead, rigid and fully lignified tubes
─ The long empty tubes of xylem provide an ideal system for translocation of large quantities
of water with minimal obstructions
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 ─ Water can pass through plants from vessel to vessel through unlignified portions of the cell
wall
─ High tensile strength due to lignifications preventing tube collapse when conducting water
under tension
─ In the primary body(plant) the distribution of xylem in the roots is central, helping to with
stand the turging strains of aerial plants as they bend or lean over

Xylem parenchyma

─ Also referred to as filling tissue

─ It has thin cellulose cell wall and living contents
─ Two systems of parenchyma exist in secondary xylem derived from meristems cell-ray initials
and fusiforms
─ Parenchyma is found in the medulcing rays. Its functions are food storage , deposition of
tannins ,crystals transport food ,water and gaseous exchange
─ Parenchyma is found in pith and cortex
─ Fusiforms initial give rise to xylem vessel or phloem sieve tubes and companion cells but
occasionally give rise to parenchyma cells-these form vertical rows of parenchyma in the
secondary xylem

Xylem fibres

─ They are shorter and narrow or tracheoids and have much thicker walls
─ They resemble sclerenchyma fibers ,having overlapping end walls
─ Have thicker walls and narrow and lumens hence conifer additional mechanical strength in xylem

The secondary wall of some of the cells is laid down in a variety of patterns. State the patterns that can
be found in cells

─ Ringed/annular
─ Scalariform
─ Reticulate

Describe ways in which the xylem is adapted for its function.

─ (Xylem vessels) are continuous this enables an unbroken column of water;

─ (Xylem vessels )lack contents/hollow/ dead this enables an unrestricted flow of water;
─ (Xylem vessels) walls are lignified for support and strength;
─ Lignin makes vessels water proof/ increases adhesion of water molecules
─ Vessels are narrow for capillarity; pits to allow lateral flow of water

Phloem

─ Sieve tubes & companion

─ Are tube like structure and translocate solution of organic solutes eg sucrose throughout the
plant
─ They are formed by end fusion of the sieve tube or sieve elements
─ The first phloem is to be produced to the photo phloem & is produced in the zone of elongation
of the growing root or stem
─ As the tissues around it grow and elongate it becomes stretched and much of it eventually
collapse & become non-functional
─ More phloem however has ceased to be called metaphloem
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─ Sieve tube have cell walls made up of cellulose and pectic substances with the cytoplasm being
confined to a thin layer around the periphery of the cell
─ Although they lack nuclei, sieve elements remain living but they depend on adjacent companion
cells which develops from the same original meristematic cells
─ Sieve plate is derived from two adjacent and walls of neighboring sieve elements
─ Originally plasmodesmata run along the wall but the canal enlarge to form pore allowing flow of
liquid from one element to the next
─ Secondary phloem which develops from the vascular cambium appears similar in structure to
primary phloem except it to be crossed by bound of lignified fibres and medullary rays of
parenchyma

Phloem parenchyma, fibres and sclereids

─ Are thin walls because they full up spaces

─ Companion cells accompanies the sieve plate hence they provide energy to more substance
─ They are living and have dense active cytoplasm. These have a nuclear and a lot of
mitochondrion .

Uptake of water by roots

─ Water moves across the root by pathways similar to these in the leaf i.e apoplast, symplast
& vacuolar pathway
─ The water potential gradient is maintained in two ways ;

1) By water moving up the xylem ,setting up tension in the xylem & lowering the water
potential in its sap

2)the xylem sap has a lower (more –ve )solute potential than the dilute solution

Symplast and vacuolar pathway

─ Symplast from cytoplasm to cytoplasm, vascular pathway from vacuole to vacuole

─ As water moves up the xylem in the root ,it is replaced by water from neighboring cells
(parenchyma) e.g cell A
─ As water leaves cell A the water potential of a cell A lowers & the water enters from cell B by
osmosis

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 ─ The water potential of a cell B then dorses & water enters it from the cell C & soon across
the root of the pulterous layer
─ The soil solution has a higher water potential cells of the pulferous layer which include the
root hair, waters the root to the pulferous layer

Water movement up xylem vessels

─ The removal of water from the top of the xylem vessels reduces hydrostatic
pressure(pressure exerted by liquids)
─ At the top of xylem hydrostatic pressure is lower than pressure at the bottom
─ Pressure differences causes water to move up xylem vessels
─ Water in xylem is under tension , hence xylem walls are lignified to stop them from
collapsing

Mass flow

─ This is the movement of water up xylem

─ All water molecules move together(due to cohesion forces) that attract water molecule
together + adhesive between them +xylem walls ) as a body of liquid

NB the cohesion & adhesion help to keep water in a xylem vessel moving as a continuous column

Set backs

o Air bubbles (air lock) which form in the column

o Wercome by the use of pits present in the lignified side walls of the xylem vessel
o Air backs + other blockages break water column
o Difference in presence between water at top of water at bottom can not be transmitted
through air lock hence water steps moving upwards …* small diameter of xylem vessels
helps to prevent such from occurring

Root pressure

1. water pressure at the top of the xylem vessel to reduced by transpiration; this causes water
to flow up the vessel
2. this increases pressure differences from top to bottom by raising water pressure at the base
of vessel
─ How; active secretion of solutes eg mineral ions in the water the xylem vessels in the roots ,this
requires energy
─ Pressure of solutes lowers the water from the cells. This in flow of water increase the water
pressure at the base of xylem vessel

Role of root pressure

─ Help water to move the xylem vessel

─ Water transport in plants in largely passive process tilled by transpiration from leaves
─ Water simply moves a continuous water potential gradient from air

Apoplast pathway

─ Most water travel from cell to cell via cell wall which is made up of cellulose fibres between
which are filled spaces. As the water evaporates into the substomal air space from the wall of
one cell ,it creates tension which pulls in water from spaces in the walls of surrounding cells. The

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 pull is transmitted through the plant by the cohesion forces between the water molecules which
due to hydrogen bonding are particularly strong

Symplast pathway

─ Some water is lost to the sub stomal air spaces from the cytoplasm of the cell surrounding it .
The major potential of this cytoplasm is thereby made more –ve. The plasmodesmata which link
the cytoplasm of one to that of the next. Water may pass along plasmodesma from adjacent cells
with a higher (less –ve)water potential. This loss of water makes the water potential of this
second cell which I have a higher water potential
─ In this water potential gradient is established between the substomatal space & the space & the
xylem vessels of the leaf
─ The symplast pathway carriers less water than the apoplast pathway.

Vacuole pathway

─ A little water passes by osmosis from the vacuole of one cell to the next , through the cell wall
,membrane & cytoplasm of adjacent cells.In the same way s symplast pathway ,a water
potential gradient exist between the xylem & substomal air space .It along gradient that the
water passes
─ NB The apoplast pathway is due to cohesion and adhesion tension & is independent of a water
potential gradient
─ The vacuole and symplast pathways are independent on a water potential gradient

TRANSPIRATION

─ is the loss water by the stomata of leaves into the atmosphere through cuticle
─ stomata –transpiration
─ lenti cells crack of buck by evaporation

Factors affecting transpiration

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(i)External(environmental factors)

 HUMIDITY
─ The humidity/ vapour pressure of the air affect the water potential gradient between the
atmosphere within the leaf of that out .When the external air has high humidity , the
gradient is reduced or less water is transpired .Low humidity high the rate of transpiration
 LIGHT
─ The stoma of most plants open in and close in the dark .The mechanisms are an increase in
volume of a guide causes increased bowing of the cell owing to the greatest expansion
occurring in the outer wall. When this occurs in two guard cells of stoma .The stomal
aperture enlarges .An increase in light intensity increase in transpiration rate and vice versa
 TEMPERATURE
─ A change in temp affects both the kinetic movement of water molecules and relative
humidity of air. A rise in temp increase the kinetic E* of water molecules and so increase
rate of evaporation of water. At the same time it lowers the relative humidity of yhe air
.Both changes increase the rate of transpiration .A fall in temp has the reverse effect of
reducing the amount of water transpired.
 WIND SPEED
─ In the absence of any movement , the water vapour which diffuses from stomata
accumulates near the leaf surface . This reduces the water potential gradient between the
most atmosphere in the stomata and the drier air at outside .The transpiration rate is
reduced .Any movement of tends to dispose the humid layer at the surface this increase the
transpiration rate .The faster the wind speed the more rapidly the moist air is removed and
the greater the rate of transpiration.
 WATER AVAILABILITY
─ A reductionin the availability of water to the plant as a result of a dry soil means there is a
reduced water potential gradient between the soil and leaf

(i) INTERNAL FACTORS AFFECTING TRANSPIRATION(PLANT FACTORS)

─ Leaf area as a proportion of water loss occurs through the cuticle , the greater the total area
of a plant the greater the rate of transpiration regardless of the number of stomata present.
In addition any reduction in leaf area invetable involves a reduction in the total number of
the stomata e.g. thin prime leaves.
 Cuticle- the cuticle is the wax coloring over the leaf surface which reduces water loss .The
thicker the cuticle ,the lower the rate of cuticular transpiration
 Density of stomata-the greater the number of stomata for a given area the higher the rate of
transpiration rate stomal rate of aboxial epidermis of plant may vary
 Distribution of stomata-in most plants the leaves are positioned at the
adoxial(upper)surface towards the light .The upper surface are subjected to greater temps
rises than lower ones owing warming effect of the sun. Transpiration is there for potentially
greater from upper side

XEROPHITIC PLANTS

─ These are plants that grow in areas which have unfavorable water balance and adapted to
the conditions.
Adaptations of Xerophitic plants
1. Thick cuticle
2. Reduces cuticula transpiration by forming a wax barrier preventing water loss
3. Rolling of leaves
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 4. Moist air is trapped within the leaf preventing diffusion out through the stomata which
are confined to the inner surface
5. Protective hairs on the leaf(pubescence)
6. Moist air is trapped in the hair layer ,increasing the length of the diffusion path so
reducing transpiration
7. Depression of stomata
Lengthens the diffusion path by trapping still moist air above the stomata so reducing the
transpiration
─ Small and circular leaves
To reduce transpiration rate .The shape also gives structural turgidity to prevent wilting
─ Orientation of leaves
The positions are constantly change(of the leaves)so that the sun strikes them
obliquely.This reduces their temp and hence the transpiration rate
─ A more positive water potential
The cells accumulates salts which makes their water potential positive. This makes it
difficult for water to be drawn from them
─ Succulent leaves and stems
For water storage
─ Nocturnal opening of stomata
The more efficient use of CO2 by C4 plants allows them to keep stomata closed during much
of the day so reducing transpiration
─ Shallow and extensive root systems
Allow efficient absorption of water over a wide area when the uppers are moistened by rain

TRANSLOCATION

MASS FLOW

─ Photosynthesing cells in the leaf have a lower potential due to accumulation of the sucrose
synthesized
─ Water enters the cells froms the xylem increasing their pressure potential.in the roots,
sucrose is either being utilized as a respiratory substrate / is being converted to starch for
storage
─ The sucrose content of these cells is therefore low giving them a higher water potential and
low pressure potential
─ Therefore, the gradient of pressure potential btwn the cells .the source of sucrose(the
leaves) point of utilization the sink (the root /other tissue)
─ The two are linked by the phloem and as result liquid flows to other tissue along sieve tube
elements
─ Movement of phloem up and down is by directional.The mvmnt in xylem is upward i.e
unidirectional, in translocation there is organic phloem and inorganic xylem.Sucrose moves
to the growing zones and is dependent on concentration
─ Root –growing zone meristematic zone

Diagram of loading of sucrose into phloem

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─ H+ can move back in companion cells down their conc gradient thru a protein which act as
carrier for both H+ and sucrose at same tym
─ Sucrose carried thru this co-transporter molecule.

DIAGRAM FOR MASS FLOW IN PHLOEM AND THE LABELLING

(a) With reference to Fig. 4.1,

(i) name an example of a source and a sink? [1]
(ii) name cells C and D. [2]
(b) With reference to Fig. 4.1, explain how sucrose travels from,
(i) the source to cell C
(ii) cell C to the sink [4]
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(c) Explain why multicellular plants require transport systems for substances, such as water
and sucrose. [2]

Solutions
(a) (i) source = leaf/mesophyll/palisade/spongy qualified
sink = flower/fruit/seed/stem/bud/root/tuber/storageorgan/young
leaf/meristem/pollen/nectary/AW ; [1]
(ii) C sieve, (tube) element/cell,
D companion/transfer, cell ; [1]

(b) source to cell C

- correct ref (sucrose) loaded ;
-H+ pumped out, sucrose moves in through co-transporter ;
-role of companion cells in moving sucrose into sieve tube element ;
- -sucrose diffuses down concentration gradient (anywhere) ;
- -ref to plasmodesmata ; [max. 2]
cell C to sink
-water enters by osmosis/water moves down its Ψ gradient ;
-hydrostatic pressure builds up ;
-(idea that sucrose) unloaded/used at sink ;
-water follows by osmosis ;
-idea there is a difference in pressure/pressure gradient (between source and sink)
;mass flow ; [max. 2]

Qn. Various hypotheses for the mechanism of transport in phloem have been suggested.
One hypothesis proposes that movement between sources and sinks occurs entirely
passively by the process of mass flow.

The diagram below shows a physical model to illustrate the principle of mass flow.

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 tube water

source sink

sugar rigid partially permeable water

solution membranes

(i) Give an example in plants of:

a source ..........................................................................................................

a sink ...............................................................................................................
[2]

Q1

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 [3]

2 (a) (i) Name structures A to C. [3]

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(ii) State the name given to the region labelled D that separates the two sieve tube

elements. [1]

(iii) Name one assimilate that is transported in the phloem. [1]

(b) Explain how the structure of sieve tube elements helps the translocation of substances in the
phloem. [3]

(c) Describe the role of companion cells in translocation in the phloem. [2]

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Qn 4

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WRITTEN BY MR. R. GWANZURA WHATSAPP: 0773266377/CALL 0717558917 HOLY CROSS HIGH
TRANSPORT IN MAMMALS
KEY OBJECTIVES CONTENT SUGGESTED SUGGESTED
CONCEPT Learners should be able to: (ATTITUDES, SKILLS AND LEARNING RESOURCES
KNOWLEDGE) ACTIVITIES
AND NOTES
8.11.1  identify arteries, veins and - Arteries, veins and  Recognising  Microscope
Mammalian capillaries capillaries the vessels  Prepared slides
circulatory under the  Photomicrographs
system light  ICT tools
microscope.  Print media
 Drawing plan  Braille software/Jaws
diagrams of
 explain the role of - Transportation of blood
haemoglobin in the oxygen and carbon vessels.
transportation of oxygen and dioxide
carbon dioxide  Discussing
the
 explain the Bohr effect transportatio
- Oxygen dissociation n of oxygen
 explain the significance of the curves and carbon
difference in the affinity for dioxide.
oxygen between: - Difference in oxygen
i. Haemoglobin and affinity between:
myoglobin i. Haemoglobin and  Analysing
ii. Maternal and myoglobin oxygen
foetal ii. Maternal and dissociation
haemoglobin foetal curves.
haemoglobin
 describe the cardiac cycle  Discussing
the  Heart models
 explain how heart action is differences in
initiated and controlled oxygen
affinity.
- Cardiac cycle
- Pacemaker
 explain the meaning of the
terms systolic blood pressure,
diastolic blood pressure and
hypertension
- Myogenic control  Observing  Sphygmomanometer
- Systolic and diastolic
 discuss the long term cardiac cycle  Stethoscope
blood pressure
consequences of exercise on
- Hypertension
simulations.  Research tools
the cardiovascular system  Observing
heart
initiation
- Improved cardiac output simulations.
- Normal resting pulse
rate
 Measuring
- Efficient cardiovascular
blood
system
pressure.
 Analysing the
results.

 Discussing
the long term
consequence
s of exercise.

Father God; remember those I teach; not for my sake but for their sake. That in everything I do it
may be for their gain. That praises may belong only to thee , not me…amen

Arteries

-Arteries transport swiftly and at high pressure to the tissues. Arteries are made up of three layers which
are:-

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 (i) An inner endothelium
─ (Lining tissue) made up of a layer of flatting cells fitting together. This layer is very smooth,
minimizing friction with the moving blood.

(ii) Tunica media

─ Contain smooth muscle and elastic fibre.

(iii) Tunica extern

─ -Contain elastic fibre and collagen fibre.

─ -Arteries have narrow lumen maintaining
high pressure facilitate faster movement of
blood.
─ -Elastic walls to allow for expansion when
pressure increases.
─ -Blood moves in pulse, rapidly and in
pulses. Transport blood from the heart.
─ -It carries oxygenated blood except in
pulmonary artery.

Veins

─ Have thin muscular wall with little elastic

tissue.
─ Large lumen relative to diameter. Unable
to constrict and not permeable. Have
valves throughout to prevent backflow of
blood.
─ Transport blood to heart and is under low
pressure (1KPa). No pulses and blood
flows slowly. Deoxygenated blood except
pulmonary vein.
Capillary

─ Routes by which water, dissolved

substances and white blood cells can enter
or leave.
─ Have no muscles or elastic tissue.
Large lumen relative to diameter.
Permeable and unable to constrict. Link
arteries to vein. Blood flows slowly and no
pulses. Have gaps to allow leakage of blood
components. Pressure is 4-1KPa.

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 Haemoglobin

─ Most efficient respiratory pigment is a protein with 4 polypeptide chains, making the globin part
of the molecule.
─ Each of the chains, making the chain is associated with atoms that form a haem group with an
iron at its centre. Each of the iron bonds with oxygen molecule to form oxyhaemoglobin. When
oxygen combine with one haem group then haemoglobin changes shape making it easier for it
to bind with another oxygen molecules, and so on until 4 oxygen molecules have been bonded.

The haemoglobin dissociation curve

─ Haemoglobin is less likely to combine with oxygen if the oxygen concentration is low. As light
increases in oxygen concentration cause a huge increase in a % saturation of haemoglobin.
─ A high oxygen concentration most of the haemoglobin molecules are combined with 4 oxygen
molecules, as oxygen concentration decreases one oxygen molecule may leave hence making it
easier for the other oxygen molecules to leave since the haemoglobin changes shape.
─ Haemoglobin is more likely to bind oxygen at low oxygen concentration.

Carbon dioxide transport and the Bohr Effect

─ There are three ways in which carbon dioxide are carried from tissue to lungs.
(i) As hydrogen carbonate ions
─ Carbon dioxide produced by respiring cells diffuses into the plasma in capillaries and into red
blood cells.
─ Inside the red blood cells, carbonic anhydrate catalyses the reaction.

Carbon dioxide + water carbonic acid

─ Carbonic acid is weak and therefore dissociates to produce hydrogen ions.

Carbonic acid hydrogen ion + hydrogen carbonate ions

125
Mr Gwanzura .R. Whatsapp: 0773266377/ 0717558916 HOLY CROSS HIGH SCHOOL ‘’TINOFAMBA
NEVANOFAMBA’’
─ Haemoglobin react with hydrogen ion to form haemoglobin acid. Haemoglobin therefore acts a
buffer as it removes hydrogen ion from solution preventing an increase in acidity. As the
haemoglobin combines with hydrogen ion it releases some of the oxygen it carries. With carbon
dioxide present, the haemoglobin is less saturated with oxygen at any given concentration. This is
called the Bohr Effect.

The loading tension is the partial pressure of oxygen at which 95% of the pigment is saturated with
oxygen. The unloading tension is the partial pressure of oxygen at which 50% of the pigment is
saturated with oxygen.

(i) From the graph, determine the difference between loading and unloading tensions
of haemoglobin. Show your working. [2]
(ii) State the location in the human body where partial pressure, lower than the
unloading tension may be reached. [1]
─ Muscle cells which are actively respiring
b) State and explain the effect of increasing carbon dioxide concentration in blood on the
loading and unloading tensions of human haemoglobin.
─ Curve shifts to the right due to the bohr effect
Reasons
o With increase in carbon dioxide; more hydrogen ions are produced causing ore
oxygen to be released from hemoglobin;
o High carbon dioxide reduces the affinity of haemoglobin for oxygen;
o Both loading and unloading tension are higher.

Myoglobin

─ Is a dark red pigment found in muscle cells.

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─ Each myoglobin is made up of 1 polypeptide chain and can combine with one oxygen molecule to
form oxymyoglobin.
─ Once combine the oxymyoglobin is stable and will not release oxygen unless the partial pressure of
oxygen around is very low.
─ It therefore acts as storage device for oxygen. At the normal partial pressures of oxygen in a
respiring muscle cells, some of the oxygen is picked up by the myoglobin from the haemoglobin.
Only to release it when the oxygen concentration in the muscle drops very low.

Myoglobin Haemoglobin

Fetal haemoglobin

─ Fetal haemoglobin is more likely to combine with oxygen and therefore has a higher affinity
for oxygen than the adult haemoglobin.
─ A fetus obtains all its oxygen through the placenta from its mother blood where it is being
carried at oxyhaemoglobin.
─ The difference in affinity means that enough oxygen will leave the mothers haemoglobin
and combine with the fetus to supply the fetus with all its oxygen requirements.

Activity

1 (a) Oxygen is carried around the bodies of mammals, bound reversibly to the
pigment haemoglobin. The pigment is found in both adult and fetal red blood
cells.The graph below shows the dissociation curves for maternal and fetal
oxyhaemoglobin.

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 100

80
fetal

60
saturation of maternal
haemoglobin
with oxygen / %
40

20

0
0 2 4 6 8 10 12
partial pressure of oxygen / kPa

(i) State the difference in the percentage saturation of haemoglobin with

oxygen between the fetal and the maternal blood at an oxygen partial
pressure of 3 kPa. [1]

(ii) Explain why the difference between the two curves is essential for the
survival of the fetus . [4]

(b) After birth, the adult form of haemoglobin gradually replaces the fetal form of
haemoglobin.

Suggest why this is necessary. [2]

[Total 7 marks]
The cardiac cycle

─ Is the complete sequence of contraction and relaxation of the heart muscle.

─ The cycle can be divided into two stages, the systole and diastole.
─ During systole the muscle contract and diastole relaxes.
─ As the heart muscle contract it squeeze the blood in the chambers inward, decreasing the
volume and increase the pressure which forces the blood out towards region where
pressure is lower.
─ The muscle then relaxes allowing the volume of the chamber to increase again and pressure
drops. Blood then flows in from regions where pressure is higher.
─ There are four stage which are
 Arterial diastole
 Arterial systole

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  Ventricular systole
 Ventricular diastole.

Initiation and control of cardiac

─ Cardiac muscle is known to be myogenic meaning that they naturally contract and relax.
The heart has a Sino atria node (SAN) which stimulate the contraction of heart found in the
right atrium. The SAN initiates the heart beat but the rate at which it beats can be varied by
stimulation from the automatic nervous system.
─ The muscle cells of the SAN set the rhythm for all other cardiac muscles cells.
─ They have an inbuut rhythm of contraction which is slightly faster than the rest of the heart
muscle.
─ Each time they contract they set up an excitation wave and the wave of polarization which
speeds out rapidly over the whole of the atrial walls. The cardiac muscles in the arterial wall
respond to the excitation wave contracting at the same rhythm as SAN thus all the muscle in
both atria contracts almost simultaneously.
─ There is a delay before the excitation wave can pass from the atria to the ventricles the
delay is caused by a bond of fibres (non conducting fibre) which do not conduct the
excitation wane.
─ These transmit the excitation wave very rapidly down to the base of septum, from where it
spreads outwards and upwards through the ventricle walls. As it does so it causes the
cardiac muscle in these walls to contract from the bottom up, so squeezing blood upward
into the arteries.

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Regulation of cardiac output

─ If an increased volume of blood returns to the heart in the veins, the heart pumps fast and
harder to push out.
─ The incoming blood stretches the muscles of the heart cell and the muscles responding by
contracting harder than usual increasing the stroke volume.
─ The SAN is therefore directly faster than usual, slightly increasing the heart rate therefore
cardiac output is increased.
─ The heart has 2 nerves running into the VAGUS (Doras sympathetic and sympathetic).
─ The VAGUS nerves bring impulses from the brain to the SAN and AVN while the sympathetic
nerve brings impulses to many areas of the muscle wall in the heart.
─ If action arrives on a sympathetic, nerve, they speed up the heart rate and increase stroke
volume.
─ The parasympathetic (VEGUS) slows down the heart for decrease stroke volume.
─ Blood pressure inside the aorta and also in the walls of the carotid arteries are nerve
endings sensitive to stretching i.e. the baroreceptor of the stretch receptors. if blood
pressure rises, the artery walls are stretched stimulating the nerve ending, which send
impulses to the brain which sends impulses to the vagus nerve to the heart. This slows the
heart rate and stokes volume which can help to reduce pressure.
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 ─ Low blood pressure has the opposite effect. The baroreceptors are not stretched and do not
send impulses to the brain.
─ The cardiac vascular centre in the brain the sends massages along the sympathetic nerve
which increases cardiac output and thus blood pressure. Massages are also sending to
muscles in the atria walls which contrast a narrow the atrioles vasoconstriction so
increasing blood pressure.
─ Increased blood flow into heart stretches cardiac muscle fibers and they respond by
contracting more strongly during systole. Therefore increased volume of blood is pumped
out. This gives direct relationship between degree of stretching of cardiac muscle and
power of cardiac contraction.

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 Explain why the blood pressure in the left ventricles falls to zero in the cardiac cycle; but the
lowest pressure recorded in the arteries is about 10kpa?
─ Pressure falls to zero because all blood is expelled from ventricles;
─ Pressure falls to 10kpain arteries because of elastic recoil of the smooth muscles
─ And the narrow diameter of the capillaries and arterioles
─ This gives resistance to flow

Endocrine control
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─ Stress need for action. Adrenalin is produced, causes increase in heart rate and stroke volume
hence cardiac output increased.
─ Thyroxin increase metabolism and therefore, there is need to pump blood faster to the respiring
tissue to supply sufficient oxygen for the tissue metabolically.
─ Active hence there is an increase in cardiac output.

Change in blood composition

High pH decelerates heart rate

Low pH accelerates heart rate

(High carbon dioxide levels as in the case during active exercise)

Low temperature decelerates (because there is vasoconstriction)

High temperature accelerates due to vasodilatation.

Qn. Carbon dioxide is produced in tissues as a waste product of respiration.

The majority of carbon dioxide is carried as hydrogencarbonate ions (HCO3–) in the

plasma.
The figure below shows the chemical pathway in which carbon dioxide is converted into HCO3– in a
red blood cell.

red blood cell

capillary
wall
CO2 + H2O

CO2 in X
tissue
Y

HCO3– in
Z + HCO3–
plasma

Identify the following:

enzyme X ......................................................................................................
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 substance Y ......................................................................................................

ion Z .......................................................................................................
[Total 3 marks]

QN. Below is a simple diagram of a mammalian heart and associated blood vessels as seen
in front (ventral) view.

P Q

X
A B

D C
Y

right left

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 (a) (i) Draw arrows on the diagram to show the direction of blood flow through
the left side of the heart.
[1]

(ii) State the name of vessel X and valve Y.

vessel X .................................................................................................

valve Y ...................................................................................................
[2]

(iii) Explain why there are valves at P and Q. [2]

(b) The maximum thickness of the external wall of each of the four chambers was
measured. The measurements made are shown below.

2 mm 9 mm 16mm 2 mm

(i) From the list of measurements, select the one most likely to correspond to
each of the chambers, A, C and D. Write your answers in the table.

chamber thickness/mm

[3]

(ii) Explain the differences in the wall thickness of chambers A, C and D.

[3]

(c) In this question, one mark is available for the quality of written communication.

Describe how the heart beat is initiated and how the contractions of the four
chambers are coordinated.

(Allow one and a half lined pages).

[7]
[Total 18 marks]

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 FORM 6 TERM 1
TOPIC 9: NERVOUS CONTROL
KEY CONCEPT OBJECTIVES CONTENT SUGGESTED SUGGESTED
Learners should be (ATTITUDES, SKILLS LEARNING RESOURCES
able to: AND KNOWLEDGE) ACTIVITIES AND
NOTES
8.12.1 Need for  recognise the need - Neurones  Drawing neurones  Prepared slides
communication for communication - Need for from prepared slides
systems within communication  Discussing the need
living organisms for communication in
living organisms.

Homeostasis.

─ It is the maintenance of a constant body environment.

Control mechanism of feedback.

─ Reference point-the set level where the system operates.

─ Detector –signals the extent of any deviation from the reference point.
─ Controller-coordinates the information from various detectors and sends out instructions
which will correct the deviation.
─ Effector.- brings about the necessary change needed to return the system to the reference
point
─ Feedback loop- informs the detector of any in the system as a result of action by the effector.

Input detector effector output

Feedback loop

Negative feedback.

─ The detectors pick up the connected change and this causes them to send impulses to the
controller which bin turn sends an impulse to the effector instructing it to stop correcting the
changes. This causes the system to be turned off.
─ Explain the role of negative feedback in homeostasis in mammals. [4]
 maintains, constant / stable, internal environment ; R normal
 a change in, some parameter / example of parameter ; (like blood glucose or
 temperature)
 detected by a, sensor / receptor ;
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  brings about response via an effector / ref.corrective mechanism ;
 ref. return to, norm / set point ;
 named, receptor / effector ;

In mammals, changes in both the internal and external environment are detected by
receptors.

(a) State the general name given to changes in the environment that can be
detected by receptors.
─ stimuli; A stimulus
(b) Explain why it is important for mammals to be able to detect changes in their
internal environment.
─ need to keep internal conditions constant / homeostasis occurs / ora;
─ so enzymes / biochemical pathways / cells/ tissues / organs work (efficiently) /
ora;
─ corrective mechanism switched on / AW;
─ named mechanism;

Positive feedback.

─ It is when a disturbance occurs in a system which set in motion events which will increase the
disturbance even further.E.g. During labor, oxytocin is secreted causing contractions to
increase.

Transmission of action potential along a myelinated neuron

Active ion transport.

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 ─ A model of the sodium and potassium pump a special protein in the membrane activated by
reaction with ATP postulates flip-flop movements transferring sodium ions out across the
the membrane and potassium in.

Facilitated diffusion: diffusion occurs because of the concentration differences generated by active
transport of `ions and cause the membrane to be permeable to Na and K more, K diffuse out than Na
does in.

─ The inside of the cell is negative with respect to the exterior and the membrane is said to be
polarized
─ The potential difference s the resting potential.
─ The K+ are actively transported into the cell i.e. 2K+ into the cell to every 3Na+ transported
out.
─ The cytoplasm has a high concentration of potassium ions and a low concentration of
sodium ions.
─ These gradients are known as electrochemical gradients.
 Because of their electrochemical gradients, sodium ions tend to diffuse back and potassium
ions tend to diffuse out.
 The channel proteins through which diffusion occurs are much more permeable to
potassium ions than sodium ions.
 Due to this and that immobilized very charged protein ions retained inside the cell , the
outside of the cell, contains many more positive ions from inside compared to outside.
 Resting potential does not usually change but change when there is a stimuli and an
impulse is promoted
 The ATP used in setting up the resting potential provides the energy for the generation of
an impulse.

The impulse OR Action potential

KEY CONCEPT OBJECTIVES CONTENT SUGGESTED SUGGESTED
Learners should be (ATTITUDES, SKILLS LEARNING ACTIVITIES RESOURCES
able to: AND KNOWLEDGE) AND NOTES
8.12.2 Action  describe the - Action potential  Illustrating the  ICT tools
potential generation of an - Resting potential generation of an  Braille
action potential action potential. software/Jaws
 Watching simulations  Print media
on transmission of an
action potential.
 explain the - Myelinated neurone
transmission of an (importance of
action potential sodium and
along a myelinated potassium ions in

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KEY CONCEPT OBJECTIVES CONTENT SUGGESTED SUGGESTED
Learners should be (ATTITUDES, SKILLS LEARNING ACTIVITIES RESOURCES
able to: AND KNOWLEDGE) AND NOTES
neurone the impulse
transmission to be
emphasised).

 explain the  Demonstrating

importance of - Myelin sheath saltatory conduction
myelin sheath and - Saltatory conduction in myelinated
the refractory - Refractory period neurones
period in
determining speed
of impulse
transmission

─ An impulse or action potential is and local reversal of the resting potential, arising when an
axon is stimulated.
─ During an action potential, the membrane potential falls until the inside of the membrane
becomes positively charged with respect to the exterior.
─ It changes from about -70mV to + 40Mv at the point of stimulation.
─ At the first point the membrane is said to be depolarized.
─ The change in potential across a membrane come about causing one of the Na+ and K+
channels to have a voltage sensitive gate.
─ The channel open and close with respect to the change in the membrane potential
difference.
─ When the gates are closed, there is little ion movement but when open, ions flow through by
diffusion.
─ One type of gated channel protein is permeable to Na+ and the other to K= ions.
─ During resting potential all gates close.
─ As stimulus depolarizes a neuron’s plasma membrane by causing a local increase in
permeability of the membrane, to sodium ions.
─ This local depolarization opens the gated sodium channels allowing a large number of
sodium ions to flow down their electrochemical diffusion gradient.
─ This causes the interior to be progressively more positive with respect to the outside and
sodium gates are almost instantly closed.
─ The gated potassium channels open and potassium flow ions flow out from their
electrochemical diffusion gradient.
─ The interior of the membrane starts to become less positive again and the process i.e.
establishing the resting potential.
─ The impulse in form of this reversal of charges runs the length of the neuron fibre as a wave
of depolarization.
─ The impulse is propagated ( self-generated ) by the effect of sodium ions entering .they
create an area of positive charge causing a local current to be set up with the positively
charged region .the impulse lusts for 2milliseconds at each K+ along the fibre before the
resting potential is negatively established.

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1. During the resting potential, sodium channels and potassium channels are closed

2. Sodium channels open and Na+ rush in.

3. Interior of the membrane becomes increasingly more positively charged with respect to the
outside.

4. Equally suddenly, sodium channels close and K+ channels open causing K+ to flow out.

5. Interior of the membrane now starts to become less negative again.

6. Slight overshoot (more –ve) than the resting potential (hyper polarization)

7. Na+/K= pump working with facilitated diffusion and resting potential is reestablished.

Transmission of nerve impulse

─ Once action potential has been set up, it moves rapidly from one end of the neuron to the
other.
─ Two factors are important in determining the speed of conduction.
 Diameter of the axon: the greater the diameter the faster the speed of transmission.

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  The myelin sheath: Mylenated neurons conduct impulses faster than non
myelinated ones.

─ The myelinated sheath which is produced by the Schwann cells is not continuous along the
axon but is absent I points called nodes (salutatory conduction) increasing the speed with
which they are
transmitted.

KEY CONCEPT OBJECTIVES CONTENT SUGGESTED SUGGESTED

Learners should be (ATTITUDES, SKILLS LEARNING ACTIVITIES RESOURCES
able to: AND KNOWLEDGE) AND NOTES
8.12.3 Cholinergic  describe the - Cholinergic synapse  Demonstrating the  ICT tools
synapse structure and (Role of calcium function of  Braille
function of a ions to be cholinergic synapse software/Jaws
cholinergic emphasised) using animations.
synapse
 Discussing the effect
 explain the effects - Effects of of analgesics on the  Pain killers
of analgesic drugs analgesics on the nervous system.
on the nervous nervous system
system

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Transmission of action potential along a myelinated neuron

─ The arrival of an impulse at the synaptic knob opens the calcium channels in the
presynaptic membrane briefly and calcium ions flow in from the synaptic depth.
─ The calcium induces a few residues containing transmitter substance to fuse with the
presynaptic membrane and release their contents in the synaptic depth.
─ Once released, the transmitter diffuses across the synaptic depth where t binds with a
receptor protein on the membrane of the post synaptic neuron .The receptor protein
controls a channel in the membrane which when open allows more types of (Na+ & K+) to
pass.
─ When the transmitter substance binds to the receptor it causes the opening of Na+ and Na+
flows in,(an excitatory synapse).The entry of Na+ depolarizes the post synaptic membrane
.If depolarization reaches the threshold level < an action potential is generated in the post
synaptic neuron and travels down the axon to the next synapse or to an end plate.
─ The action of the neurotransmitter does not persist , renewal of neurotransmitter substance
from the synaptic debt (by enzyme action) prevents the continuous firing of the post
synaptic neuron e.g. enzyme choline sterase hydrolyses ACH to choline and ethanoic acid
which are:

Hormones; chemical messenger with following properties.

─ Travel in blood.
─ Has effect on site other than manufacture called target site / cell.
─ Fit precisely into receptor (protein) molecule sites in their target cells through lock & key
hypothesis mechanism.
 Hormones are specific to a particular target.
 Small soluble organic molecules (steroids).
 Effective I low concentration.

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Receptors are often described as biological transducers, structures which convert
energy from one form into another.

Explain how receptors in mammals convert energy into action potentials. Use named
examples of receptors in your answers.

─ rods / cones / retina / photoreceptors, detect light;

─ taste buds / olfactory cells / chemoreceptors, detect chemicals;
─ Pacinian / Meissner’s corpuscle / mechanoreceptors, detects pressure / touch;
─ Ruffini’s endings in skin / thermoreceptors, detect temperature changes;
─ proprioreceptors / stretch receptors in muscle, detect mechanical displacement /
AW;
─ hair cells / AW, in semicircular canals detect movement;
─ hair cells / stereocilia, in cochlea detect sound;
─ baroreceptors detect blood pressure changes;
─ osmoreceptors detect changes in blood water potential;
─ stimulus causes sodium channels to open;
─ sodium ions enter cell;
─ depolarisation;
─ receptor potential / generator potential;
─ greater than threshold / all or nothing principle;
─ increased stimulus leads to increased frequency of action potentials;
─ AVP; e.g. hyperpolarisation in rod cell
─ deformity of capsule in Pacinian corpuscle

Describe the structure of a myelin sheath and explain its role in the speed of transmission of a nerve
impulse. [8]
─ Schwann cells ;
─ wrap around axon ;
─ sheath mainly lipid ;
─ (sheath) insulates axon (membrane) ;
─ Na+ / K+, cannot pass through sheath / can only pass through
─ membrane at nodes ;
─ depolarisation (of axon membrane) cannot occur where there is
─ sheath / only at nodes of Ranvier ;
─ local circuits between nodes ;
─ action potentials ‘jump’ between nodes ;
─ saltatory conduction ;
─ increases speed / reduces time, of impulse transmission ;
─ up to 100 ms-1 ;
─ speed in non-myelinated neurones about 0.5 ms-1

Describe how a nerve impulse crosses a cholinergic synapse. [9]

─ action potential / depolarisation, reaches presynaptic membrane ;

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 ─ calcium (ion) channels open / presynaptic membrane becomes more permeable to
Ca2+ ;
─ Ca2+ flood into presynaptic neurone ; R membrane
─ this causes vesicles of (neuro)transmitter to move towards presynaptic membrane ;
─ ref. acetylcholine / ACh ;
─ vesicle fuses with presynaptic membrane / exocytosis ;
─ ACh released into synaptic cleft ;
─ ACh diffuses across (cleft) ;
─ ACh binds to receptor (proteins) / AW ;
─ on postsynaptic membrane ; R neurone
─ proteins change shape / channels open ;
─ sodium ions rush into postsynaptic neurone ; R membrane
─ postsynaptic membrane depolarised ;
─ action potential / nerve impulse ;
─ AVP ; e.g. action of acetylcholinesterase

Explain the roles of synapses in the nervous system. [6]

─ ensure one-way transmission ;

─ receptor (proteins) only in postsynaptic, membrane / neurone ; ora
─ vesicles only in presynaptic neurone ; ora
─ ref. adaptation ;
─ increased range of actions ;
─ due to interconnection of many nerve pathways ;
─ ref. inhibitory synapses ;
─ involved in memory / learning ;
─ due to new synapses being formed ;
─ AVP; e.g. summation / discrimination

Qn. Fig. 2.1 shows the changes in membrane potential in an axon during the passage of a single
impulse.

(a) Outline how the resting potential from A to B is maintained. [3]

(b) Describe how the changes in the membrane bring about depolarization from B to C. [3]
(c) Explain how the membrane is repolarised from C to D.
[3]
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 (d) State three differences between nervous and hormonal communication in mammals. [3]

Solution

(a) 1 reference to Na+/K+ pump;

2 active process/ATP used;
3 Na+ (pumped) out and K+ (pumped) in;
4 high Na+ outside and high K+ inside axon;
5 membrane slightly more leaky to K+/more K+ leaks out than Na+ leaks in/
reference to some K+ channels open;
6 inside more negative than outside; 3 max
(b) 1 reference stimulation;
2 opening of Na+ channels;
3 Na+ diffuses in (across axon membrane);
4 inside more positive than outside/outside more negative than inside;
potential across the membrane changes; 3 max
(c) 1 reference to closing Na+ channels;
2 opening of K+ channels;
3 K+ diffuses out (across axon membrane);
4 (charge on the K+) restores the membrane/resting potential;
5 reference to slight overshoot/hyperpolarisation;
6 reference K+ channels close; 3 max
(d) 1 electrical vs chemical;
2 (impulses) along nerve cells vs (hormones) through blood;
3 rapid vs slow;
4 response immediate vs relatively slow;
5 responses short lived vs long lived; 3 max

Describe the role of sodium ion channels in the transmission of a nerve impulse.[3]

─ ref. to voltage-gated sodium ion channels / ref. ligand gated channels ;

─ channels change shape (when, pd / voltage, changes) ;
─ open when, membrane depolarises / action potential arrives / neurotransmitter
─ binds to receptors ;
─ sodium ions flood in ;
─ diffuses / down concentration gradient ;
─ channels close when membrane, repolarises / potential reaches +30mV ;
─ ref. to sodium-potassium pump ;

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TOPIC 10 : Sexual Reproduction
KEY CONCEPT OBJECTIVES CONTENT SUGGESTED SUGGESTED RESOURCES
Learners should (ATTITUDES, SKILLS LEARNING
be able to: AND KNOWLEDGE) ACTIVITIES AND
NOTES
8.13.1 Sexual  describe - Anther structure  Discussing  ICT tools
Reproduction in anther - Pollen formation anther structure  Braille software/Jaws
Plants structure and and pollen  Flowers
pollen formation.  Microscope
formation  Observing and  Slides
drawing anther  Scalpels
structure and
- Ovule development pollen grains.
 describe ovule  Dissecting
development flowers.

 Observing and
drawing the cross
section of the
ovary.

 Discussing ovule
- Double fertilisation development.
 describe
double
fertilization  Discussing
double
 explain the fertilisation and
significance of its significance.
double  Conducting
fertilisation in educational tours
the embryo to plant breeders.
sac
8.13.2 Sexual  recognise the - Structure of the  Observing the  Mammalian specimens
Reproduction in microscopic ovary and testis microscopic  Models
Humans structure of the structures of  Microscope
ovary and ovary and testis  Prepared slides
testis from  Photomicrographs
photomicrograph  ICT
s and prepared  Braille software/Jaws
- Gametogenesis slides.  Print media
 describe
gametogenesis  Observing
gametogenesis
simulations.
 Outlining the
processes of
gametogenesis.
- Hormonal control of
 explain how gametogenesis
gametogenesis  Discussing
is controlled by homornal control
hormones of
gametogenesis.
 explain in detail - Menstrual cycle and
the role of hormones
hormones in  Interpreting
the menstrual graphical
cycle representation of
- Capacitation the menstrual
- Acrosome reaction cycle.
 describe - Cortical reaction
fertilisation - Fertilization  Observing
simulation of
- Structure of the fertilization.
placenta

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KEY CONCEPT OBJECTIVES CONTENT SUGGESTED SUGGESTED RESOURCES
Learners should (ATTITUDES, SKILLS LEARNING
be able to: AND KNOWLEDGE) ACTIVITIES AND
NOTES

 describe the - Transport  Observing and

structure of the - Hormonal drawing the
placenta production structure of the
placenta.
 explain the
roles of the  Observing
placenta - Contraception simulation of the
- Invitro fertilization mechanisms in
 discuss - Abortion placental
contraception transfer.
and abortion
from biological  Debating on
and ethical - Role of hormones biological and
view points ethical
viewpoints.
 outline the role
of hormones in
pre-menstrual
tension,  Discussing the
replacement role of hormones.
therapy and
menopause

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TOPIC 11: ECOLOGY
KEY CONCEPT OBJECTIVES CONTENT SUGGESTED SUGGESTED
Learners should be (ATTITUDES, SKILLS LEARNING RESOURCES
able to: AND KNOWLEDGE) ACTIVITIES AND
NOTES
8.14.1 Levels of  define the terms - Species  Explaining the  print media
Ecological used to describe - Habitat terms.
Organisation levels of ecological - Population  Stating examples of
organisation - Niche each of the terms.
- Community
- ecosystem

─ Is the study of organism within their environment, including the way in which organisms
interact with each other & the non living part of the environment

Terms used in ecology

1. Habitant
─ The place where organism live e.g. a pond.
─ The term habitant is often used to mean the kind of place in which a particular species of
organism can live, such as the range of pH of the water & the range of dissolved oxygen
concentration in which it is found.
2. Population
─ Refers to all the organism of one species that live in the same place at the same time, make
up a population on that species.
─ A population is a breeding group or it includes all the individuals of that species which can
interbreed with each other.
─ The population of dark weed is made up of the entire dark weed found in the pond.
─ A group of organisms of one species occupying defined area & usually isolated to some
degree from other similar to groups.
3. Community
─ All organisms of every species in a habitant. Communities may remain fairly stable over a
period of time or may be in a progress of gradual change (succession).
─ Eventually, succession may result in the formation of a stable community known as climax
community.
─ Any group of organism belonging to a number of different species that co- exist in the same
habitant/ area & interact through trophic & spatial relations.
4. Ecosystem.
─ The inter-relationship of the living (biotic) and non-living (abiotic) elemevts in any
biological system.
─ It is a self contained unit.
─ A community of organisms & their physical environment interact as an ecological unity.
5. Niche
─ Is a place which is occupied by particular organisms in an environment &its role in that
environment?
─ With in an ecosystem each species of organisms plays a particular role.
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 ─ The term niche is used to describe this role.
─ An organism niche has many aspects.
─ It includes what the organism eats, how it captures its food, what eat it, the secretory
material produces & so on behavior.
─ Within a community, each species has a niche which differs in at least some ways from the
niches of all the other species in the same community. They will be competition for
available resources.

Flow of energy

─ Living cells require energy for many purposes which are locomotion , growth& e.t.c
─ The immediate source of energy is almost ATP which is produced for respiration.
─ Respiration transfers energy from other organic molecules such as glucose to ATP
molecules.
─ The energy in these organic molecules can be thought as organic chemical energy.
─ There are 3 types of chains.
─ A food chain is a sequence of series of organisms feeding on one another.

Grazing food chain

-this is based on living plants

Grass cow lion decomposition.

─ Produces of the first level, they receive max energy which is able to sustain a large number
of organisms.
─ They are autotrophic organism, e.g. grass , leaves
─ Primary consumers : food directly on producers , these are herbivores
─ Secondary consumers: feed on herbivores usually carnivores.
─ Tertiary consumers: can be omnivores, herbivores or carnivores.
─ Decomposers: they feed on saprophytic organisms

Detrital food chain

This is based on dead plants materials

Grass earth worms shrew.

Food web

─ Is a group of interconnected food chains because there are sustaining organisms which do not
depend on one type after food?

Parasitic food chain

Rose bush aphids spider insect bird lawks.

─ At each stage in the food chin energy containing materials are transformed.
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─ The stage of the food chain is reorganized as feeding levels or trophic levels.
─ Most food chain with other chains, since most organisms are the prey of more than one predator.
─ Only a small portion of the energy containing materials obtained by a consumer as it feed
becomes built into the organism itself.
─ This is partly because most of the food is undigested & partly because most of the remainder is
used to provide energy for processes e.g. movement, digestion, excretion & reproduction.
─ More energy is lost & the last receive a small amount of energy.
─ Shorter food chain – more energy is required by the last feeding level.

(Explain how energy is lost from producers to secondary consumers?)

─ Respiration
─ Waste/urine/faeces/dead plants/exreta/excreation
─ Plant are swept/ away migrate
─ Not all parts of the plant / primary consumers are digestible
─ Energy losses to decomposers

(Outline how bacteria convert Nitrogen in the proteins to a form that may be taken up living
plants. [2])

─ protein amino acids

─ Deamination
─ Ammonification/ ammonia
─ Ammonia to nitrates
─ oxidation

Qn. Describe how energy flows in an ecosystem. [8]

-source of energy, the sun

-sun to plants/producers

-conversion of light to chemical energy

-from plants/ producers to primary consumers

-examples of a food chain

-secondary to tertiary consumers

-loss of energy at each tropic level/loss as heat/ respiration /excretion/egestion to the atmosphere

-dead and waste material to decomposers/destritus feeders to the atmosphere

-unidirectional flow of energy.

The Nitrogen cycle

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KEY CONCEPT OBJECTIVES CONTENT SUGGESTED SUGGESTED
Learners should be (ATTITUDES, SKILLS LEARNING RESOURCES
able to: AND KNOWLEDGE) ACTIVITIES AND
NOTES
8.14.2 Nitrogen Cycle  outline the nitrogen - Nitrogen cycle  Illustrating the  Print media
cycle nitrogen cycle.  ICT tools
- Roles of:  Observing  Braille software/Jaws
leguminous root  legumes
- nitrogen – fixing nodules.
bacteria
(Rhizobium)

- nitrifying bacteria
(Nitrosomonas and
Nitrobacter)

- denitrifying bacteria
(Pseudomonas and
Clostridium)

─ Nitrogen in the atmosphere is very insert & it takes place a lot of energy to split it, the bonds in
the molecule so that it can form other compounds such as nitrates & nitrites.
─ Nitrogen is an essential component of biological molecules such as protein & DNA
─ The only organism capable of splitting nitrogen are few bacteria & algae .they use it form nitrites
& nitrates, a process known s nitrification
─ This is the major way in which nitrogen enters the biotic component of the ecosystem.

Nitrogen fixation.

─ Is energy consuming process (endothermic reaction) because the two nitrogen atoms of the
nitrogen molecules must be separated?
─ Nitrogen fixers achieve this by using the enzyme nitrogenase & energy from ATP.
─ Non enzymatic separation requires the much greater energy of industrial process / of ionizing
radiation
─ There is no counter balancing removal mechanism taking industrial / fixed nitrogen back to
the atmospheric reserver pool.
─ A regulation n small amount of nitrogen ( 5- 10 %) is formed by ionizing events in the
atmosphere , the resulting nitrogen oxides dissolve in the rain forming nitrites
─ The legumes such as clover, soya beans & peas are probably the greatest natural sources of
fixed nitrogen.
─ Their roots possess character swellings called nodules which caused by colonies of nitrogen
fixing bacteria living within the soils.
─ The relationship is mutualistic because the plant gain fixed nitrogen in the form of ammonia
from the bacteria & in return the bacteria gains energy and certain nutrients such as
carbohydrates from the plant.
─ Legumes can contribute as much as 100 times more fixed nitrogen than free living bacteria.
─ All nitrogen fixers Inco-operate nitrogen into ammonia but this is immediately used to make
organic compounds mainly proteins.

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Qn. Describe the role played by microorganisms in the Nitrogen cycle [6]

Emphasis should have been placed on specific organisms and the part they play in the Nitrogen cycle.
The following points are expected:

─ saphrophytic bacteria/ fungi feed on dead organisms or waste/ decompose dead organisms or
waste
─ -releasing ammonium compounds /ammonia
─ -nitrifying bacteria/ nitrosomonas
─ -oxidize ammonia or ammonium compounds to nitrates.
─ -oxidation of nitrates to nitrates by free-living bacteria/ nitrobacter/ nitrococcus
─ -nitrigen fixing bacteria/ rhizobium/ mutualistic bluegreen bacteria/ nostoc/ free- living green
bacteria/ Azotobacter/ clostridium
─ -convert (gaseous) nitrogen into ammonia/nitrates
─ -denitrificans
─ -converts nitrates (in the soil)into gaseous nitrogen.
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Explain the role of nitrifying bacteria in the Nitrogen Cycle.

─ convert , ammonium / NH4+ , to , nitrate III / nitrite / NO2- ; A ammonia / NH3

─ nitrite , converted to , nitrate (V) / NO3- ;
─ A one mark for single step ‘ammonium to nitrate (V)’
─ requires , aerobic conditions / oxygen / aerated soil ;
─ (nitrate (V) ions) can be , taken up / used , by plants ;

Explain the role denitrifying bacteria in the Nitrogen cycle.

─ remove nitrate (V) (ions) / convert nitrate (V) (ions) to nitrogen (gas) ;
─ in , anaerobic conditions / oxygen poor soil / non-aerated soil ;
─ recycles nitrogen / further use of nitrogen (by fixing) ;
─ prevents nitrogen being trapped / AW ;

Pyramid of numbers

It presents the number of organisms in each trophic level.

Inverted pyramid of numbers

─ The producer is a single plant e.g. sycamore which is affected by parasites such as caterpillars
which are parasitized by protozoa.

Pyramids of biomass

─ Fresh mass of all the organism in that trophic level is called biomass.
─ We can have inverted pyramids in specific seasons.
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 ─ At certain times of the year the biomass of the timing herbivores that floats in lakes oceans
may exceed the biomass of tiny photosynthetic phytoplankton on which they feed.
─ Total mass of organisms (biomass) is estimated
─ For each tropic level
─ Estimates involve weighing of representatives individuals; recordingmembers labourous
and expensive in terms of time and equipment
─ Ideally dry masses should be compared
─ These can be estimated from wet masses or can be determined by destructive methods
─ Rectangles used in constructing the pyramids represent masses of organisms at each
tropic level per unit area or volume
─ Pyramid become a typical shape above this level

Pyramid of energy

─ Most ideal and fundamental way of representing relationship between organisms in

different tropic levels.
─ Takes into account the rate of production contast to pyramid of biomass
─ Each bar of pyramid represents the amount of energy per unit area
─ Allows different ecosystems to be compared , i.e the relative importance of populations
within one ecosystem can be compared and inverted pyramids can no be obtained
─ Input of solar energy can be added as an extra rectangle at the base of pyramid
─ There is energy loss at each feeding level such that there is no inverted pyramid of energy.
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 ─ Some of the energy is lost through respiration & excretion & egestion.

ANTHROPOGENIC IMPACT ON ECOSYSTEMS

KEY CONCEPT OBJECTIVES CONTENT SUGGESTED SUGGESTED
Learners should be (ATTITUDES, SKILLS LEARNING RESOURCES
able to: AND KNOWLEDGE) ACTIVITIES AND
NOTES
8.14.3 Anthropogenic  describe the - Human settlement  Discussing the  Ecosystems
Impact on effects of human - Deforestation human activities  ICT tools
Ecosystems activities on - Industrial activities that affect the  Braille software/Jaws
ecosystems - Agricultural activities ecosystems.
- Mining  Carrying out case
- Global warming studies.
- Invasive plant
species

How fossil fuels may affect the environment

─ Almost all air pollutants are gases from burning of fossil fuel.

Smoke

─ Is tiny particles of sooth (carbon) suspended in the air, which are produced from fossil fuels,
particularly coal & oil.

It has a number of harmful effects.

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 1. When breathed in, smoke may blacken the alveoli, causing damage to their delicate
epithelical lining, it also aggravates respiratory aliments e.g. bronchitis.
2. While it remains suspended in the air, it can reduce the light intensity at ground level. This
may lower the overall rate of photosynthesis.
3. Deposits of smoke / more particularly soot may coat plant leaves, reducing photosynthesis
by preventing the light penetrating, by blocking the stomata.
4. Smoke, soot & ash become deposited on clothes, cars & buildings. These are costly to clan.

Sulphur dioxide

─ It may increase soil fertility in areas where sulphates are deficient / even help to control
diseases such as black spot of roses by acting as a fungicides , it’s affect concentration are
largely harmful.
1. It causes irritation of the respiratory system & damage to the epithelical lining the alveoli; it
can also irritate the conjunctiva of the eye.
2. It reduces the growth of many plants e.g. barley, wheat, lettuce, while other such as lichens
may be killed, sulphur is soluble in water. The sulphurous & sulphuric acid. The rainfall
therefore has a low pH & is known as acid rain, lakes in the region affected by acid rain are
extremely acidic & many species with in them have been killed.

Carbon dioxide

─ Carbon dioxide is transparent to incoming short wave radiation from the sun but absorbs
strongly long wave radiation which the air re- radiates into space therefore traps going
radiation warming the lower atmosphere which in turn radiates energy back to the surface
of the earth .
─ The rise in temperature i.e. so called green house effect will cause the expansion of the
oceans & gradual melting of the polar ice caps, with consequent rise in sea level.
─ This would in turn cause flooding in low lying land, upon which it happens, many of the
world’s capital cities lie.

Carbon monoxide.

─ (Co) occurs in exhaust emission from cars & other vehicles.

─ It is poisonous on account of having an affinity for hemoglobin than oxygen.
─ Upon combining with hemoglobin it forms a stable compound which is prevents oxygen
combining with it.
─ Continued inhalation leads to death as all hemoglobin combine with carbon monoxide,
leaving non transport oxygen.
─ In small concentration it may cause headaches.

Nitrogen oxides

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 ─ Nitrogen oxides e.g. nitrogen dioxide are produced by the burning of fuels in car engines &
emitted as exhaust.
─ In themselves, they are poisonous, but importantly, they contribute to the formation of the
photochemical smog.
─ Under certain climatic conditions pollutants become trapped close to the ground.
─ The action of sunlight on the nitrogen oxides in these pollutants cause them to be converted
to peroxacyl nitrate (PAN).
─ The compounds are much more dangerous causing much more damage to vegetation & eye
lung irritation in man.

(Discuss the use and effects of nitrogen containing fertisers in agriculture. [8])

─ Nitrogen containing fertilizers increase yields

─ When a farmer rotate crops nitrogen containing fertilizers do not cause much harm
─ When a farmer applies the correct amount of fertilizer no harm occurs;
─ Problem arises when excess fertilizer are applied;
─ Excess application of fertilizer causes eutrophication of water bodies
─ Eutrophication of water bodies cause algal blooms;
─ Algal blooms decompose and reduce the amount of oxygen in water bodies;
─ Cause death of aquatic organisms;
─ Due to lack of oxygen;
─ Fertilizer apply no harm occurs when fert application crop is actively growing /correct time
─ Soil becomes acidic and this affects the crumb structure of the soil;

Lead

─ Most lead in the air is emitted from car exhaust.

─ Tetraethyl lead (TEL) is added to petrol as anti knock agent to help it to burn more evenly
in car engine.
─ The lead absorbed by the lung could have the following adverse effects.
1. Digestive problem e.g. intestinal colic.
2. Impering the functions of the kidney.
3. Nervous problem including convulsion.
4. Brain damage & mental retardation in children.
─ Anti knock agent which do not contain lead exist & in some countries legislation permits
only this time.
─ They are however, most expensive & cause an increase in the actual cost of petrol.

Methods of controlling air pollution.

1. Use of non lead & anti knock & the removal of pollution such as sulphur before smoke is
emitted from the chimney by passing the smoke through a spray of water at which much of
the sulphur dissolves.
2. Use of electric cars is a further means of limiting air pollution.
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 3. Using catalytic converters to make sure that there is complete combustion.
4. Using other forms of fuel which do produce the gases e.g. methane

How deforestation may affect the environment.

1. There is lost of traditionally harvested products such as timber, poles fire wood, twine,
fruits, honey game animals & herbs that at one time supplied local people with their need.
2. The demand for soft wood for building, pulp wood & tropical hard wood is rising globally
.long term suppliers are threatened.
3. If forest canopy is removed, the tropical soil surface backs hard in the intense heat;
4. More rapid run off
5. rainfall cannot easily penetrate the surface runoff.
6. More rapid runoff of water results in soil erosion. This can remove the topsoil & leave the
ground unsuitable for growing crops.
7. It can also lead to silting of reservoirs which reduces their useful life , whilst habbers
8. Deforestation increase global warming carbon dioxide which may have long term effect on
the global climate (green effect).
9. This may result in seasonal drift & flooding in some area.
10. Reduces transpiration
11. Leads to less rainfall, atmospheric moisture/ droughts/ desertification
12. Forest have species rich & diverse wild life communities , their destruction will lead to
enumerable extinctions of little known forms of life with the consequent loss of genetic
variety & potential resources.

Measures

1. Practicing re- forestation & afforestation.

2. Using other sources of fuel instead of wood e.g. nuclear fuel, wind & water.
3. Electrify rural areas.

Conservation
KEY CONCEPT OBJECTIVES CONTENT SUGGESTED SUGGESTED
Learners should be (ATTITUDES, SKILLS LEARNING RESOURCES
able to: AND KNOWLEDGE) ACTIVITIES AND
NOTES

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KEY CONCEPT OBJECTIVES CONTENT SUGGESTED SUGGESTED
Learners should be (ATTITUDES, SKILLS LEARNING RESOURCES
able to: AND KNOWLEDGE) ACTIVITIES AND
NOTES
8.14.4 Conservation  explain, using - conservation  Discussing the  ICT tools
specific examples, - role of concept of  Braille software/Jaws
how conservation Environmental conservation.  Environmental
may involve Management Management Act
preservation, Agency (EMA) and
management and CAMPFIRE
reclamation  Evaluating trends in
- The African the population
 discuss the Elephantand White numbers of the
conservation of the Rhinoceros African Elephant
African Elephant and White
(Loxodonta - Population numbers rhinoceros.
africana) and the - Reasons for
White Rhinoceros concern, measures  Researching on
(Ceratotherium introduced other endandered
simum) species.
- International co-
operation, conflict of  Discussing
interests economic
implications to
Zimbabwe.
 Conducting
Educational Tours.

Qn. Discuss the Conservation of the African elephant, L. Africana and African cyclotis, with
regard to population numbers, reasons of concern, measures introduced and international
co-operation. [8]

On population

-no serious predators except man

-poaching main problem

-dropped (from +1.0 million to +500 000)

Reasons for concern

-elephant complete with man for land used in agriculture/forestry/settlement/ destruction of

vegetation by elephants

-elephant killed for ivory

-mostly males for bigger tasks

Measures introduced

-ban elephant poaching

-sustainable management programme

-culling

-ref to camfire

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-ban on illegal trade of elephant products (by CITES)

International co-operation

-ref CITES difficult for allcountries to agree on total ban/ culling measures/ sale and marketing of
animal products

-ref to tourism and conservation agreements between countries.

(outline the reasons why conservationists are concerned about the population of the African elephant,
loxodonta Africana. [8])
─ low population of elephants/ref to extinction
─ population falling rapidly
─ loss of genetic diversity
─ conversion of suitable areas of habitat to agricultural use/AW/Human population rising
rapidly
─ high value of ivory/porverty makes poaching of ivory attractive
─ disruption of elephant families/ elephants have a matriachical family system; elephants
threaten human life
─ elephants destoy trees/crops/water installations
─ AVP

Endangered species – any species whose numbers have become so low that they are unlikely to be
maintained by normal rates of reproduction and are in danger of becoming extinct.

Biodiversity
─ The biodiversity of the planet is the result of evolution. In any ecosystem, there is a huge
interdependence between species and it is clear that biodiversity is essential to maintain
ecological balance and stability.
─ Another part of biodiversity is the extent of genetic diversity with species and populations. Such
genetic diversity is also essential for the stability and survival of a species.
The Need to maintain Biodiversity
─ Biodiversity is in decline – mostly as a result of a variety of man’s activities. It is now well
understood that it is important to try and halt this decline – indeed, conservation measures are
needed, not only to halt the decline, but to try and restore as much biodiversity as possible.
─ The need to maintain biodiversity may be considered in terms of biological reasons or reasons
from a human perspective :
Biological reasons
─ As mentioned above, it is essential that biodiversity is maintained if ecosystems (and the
whole planet) are to remain ecologically balanced and stable.
─ In addition, evolution has resulted in diverse gene pools within populations – the
maintenance of these gene pools and the genetic diversity of species is extremely important
if species are to be prevented from becoming extinct.

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Human reasons
─ Other species of animals and plants provide an important resource for humans.
─ These may be
i. For use in agriculture, either as potential food supplies or to be crossed with
ii. existing agricultural species to improve features, such as yield, hardiness or
iii. disease resistance.
iv. To provide possible medicines
v. To encourage tourism in some countries - ecotourism
vi. From an ethical point of view, if human activity has been largely responsible for the decline
in biodiversity, then humans have an obligation to reverse this decline. Equally, it is
important to try and maintain the current level of biodiversity for future generations.

Reasons why species have become endangered

─ In African ecology, the elephant is regarded as a keystone species. In 1930 there were
estimated to be 5 – 10 million African elephants. By 1979 the number was reduced to 1.3
million and when it was officially added to the endangered list in 1989, the numbers had
fallen to around 600,000 - less than 10% of its numbers earlier in the twentieth century.
─ A number of factors contributed to this dramatic decline in numbers:
Habitat loss – elephants eat a great deal and need a large amount of habitat.
─ During the twentieth century, the human population of Africa has increased massively and,
as a result, humans and elephants have become competitors for living space. The forest and
savannah habitats of the elephant have been reduced as humans have used timber for fuel
and building and land for growing crops and grazing livestock.
─ When humans and elephants live in close proximity, various problems arise – elephants
raid crops and, on occasion, will rampage through villages. Farmers and other residents
regard them as something of a pest and shoot them.
Hunting – this has been a major cause of the decline in elephant numbers.
─ Elephants became prized trophies for big-game hunters and, more recently, they have been
killed for their ivory tusks. Ivory is easily caved and regarded as a beautiful material
─ most of the ivory carving in the world takes place in Japan and other countries in Asia. At
one stage, ivory was more expensive than gold – indeed, it became known as ‘white gold’.
─ Hunting continues for the global ‘bushmeat’ trade
Poaching – it is no longer legal to hunt elephants in most African countries. However, the high
prices paid for ivory meant that elephants continued to be killed by poachers.
─ At its peak, the poachers became highly organised, using automatic weapons, vehicles and
even planes to herd and kill huge numbers at a time. The biggest elephants were usually
targeted (because they have the largest tusks) which meant that it was generally the adults
that were killed, leaving young elephants without any adults to learn from. As a result,
─ the social structure of the elephant populations broke down and many of the elephant
groups left were leaderless juveniles.
─ Habitat loss – competition between humans and elephants for space, treesand grazing
leading to loss and fragmentation of the elephant habitat
─ Hunting and poaching – for trophies, ivory, protection of villages and forBushmeat
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Methods of protecting endangered species
─ There are a variety of ways in which attempts are being made to protect endangered
species and prevent them becoming extinct.
─ The extent to which these attempts are successful is somewhat variable.
Zoos
─ One way of protecting endangered species of animals is to capture some from the wild and
place them in captivity.
─ In this way, it is possible to make sure that they are well fed, protected from predators and
disease and isolated from other potential problems which might be encountered in their
natural habitat.
─ If such animals are simply placed in zoos, the zoo is really acting as an ‘ark’ and little is
actually being achieved in terms of maintaining or increasing populations in the wild.
─ If the animals will breed in captivity, then it is possible to maintain or even increase
numbers.
─ If such captive-bred individuals can be returned to their natural habitat, then it might be
possible to increase numbers in the wild, thereby preventing the endangered from
becoming extinct.

Captive breeding has a number of advantages:

─ it is possible to monitor the health of the mother and the development of the
─ fetus during the pregnancy.
─ sperm and eggs can be obtained from the captive individuals
─ these can be stored in a frozen form
─ it allows the possibility of artificial insemination
─ also in-vitro fertilisation
─ fertilised embryos may be implanted in surrogate mothers (which might even
─ be of different species)
─ there is the possibility of international co-operation and the transfer of breeding individuals
between different zoos
─ it allows the keeping of breeding records and the genetic relatedness of captive individuals

Although some species of animals have been bred successfully in captivity and released back into
the wild, with other species this has not been straightforward and a number of problems have been
encountered. It has been found that some species simply do not breed successfully in captivity,
whilst, in some cases, there have been problems in releasing animals that have bred in captivity.

Captive Breeding
There are a number of reasons why animals do not always breed successfully when in captivity:
1 They are no longer living in their natural habitat

2 The conditions experienced in captivity can cause stress and behavioural changes

3 The stress can disrupt normal reproductive cycles and breeding behaviour

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4 They often have little choice of mate and may reject the chosen mate

Release of captive-bred individuals into the wild

Problems which reduce the success rate of releasing captive-bred individuals
include :
1. Habitat destruction (usually as a result of man’s activities) might mean that there is very
little suitable habitat available in which to release the animals
2. Having been in captivity, animals might not find it easy to move around in their natural
habitat
3. It may not be easy for them to find enough food – especially if they have been used to being
fed in captivity
4. They may not be able to communicate with other members of their species in the wild and
may not integrate into social groups
5. They may be susceptible to diseases in the wild Some of these problems are being overcome
by making sure that conditions within zoos are as close to the natural habitat of the species
as possible. Contact with humans is kept to an absolute minimum and individuals can be
‘acclimatised’ in cages before they are actually released into their natural habitat.

Botanic gardens
─ Endangered species of plants can be grown in botanic gardens. Clearly, it ispossible to
create ideal growing conditions – either outdoors or in glasshouses, when it is possible to
control very carefully the growing conditions.
─ This applies to the availability of light, nutrients, water and the atmospheric conditions.
Within such botanic gardens,
─ it is also possible to propagate endangered species
– either by growing from seed or by some means of vegetative propagation, such as cuttings.
Techniques of tissue culture also allow large numbers to be produced very quickly.
─ This allows the possibility of re-introducing endangered species of plants into their natural
habitat.

Seed banks
─ Many plants produce seeds which are very long-lived and large numbers can be stored in a
relatively small space. Such a collection of seeds is referred to as a seed bank. The life span
of such seeds can be extended if they are kept in carefully controlled conditions – especially
in an atmosphere of low oxygen levels, moisture and temperature.
─ Given that the seeds will contain all the genetic material of any given species, italso means
that the gene pool of that species is being maintained.
─ Clearly, if the seeds of endangered species are stored in this way, such seeds can be
germinated at any time and plants can be grown in Botanic gardens or restored to the wild.
─ Some species produce seeds which have a limited longevity (e.g. cocoa, rubber,coconut) –
keeping their seeds in seed banks is not possible. Such plants would need to be maintained
in botanic gardens.
National Parks (and other protected areas)
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 • Many countries have designated areas, such as National Parks, which are set up to conserve
rare / endangered species and maintain important habitats.
• Often, legislation is passed to ensure that such areas are protected under the law.
• The ways in which National Parks protect their resident species include :
1. Wardens, rangers and volunteers can be used to patrol the parks
2. Access by humans can be restricted – often footpaths are created and maintained to avoid
interference with wildlife habitats
3. Agricultural activities can be strictly controlled – traditional farming methods can be encouraged
4. Industrial activities and mining can be limited and controlled
5. The building of roads, dwellings and other developments can be strictly controlled
6. Visitor Centres can be established to educate the general public in the importance of
conservation within the Park – and elsewhere
7. Wildlife can be protected directly e.g. 24 hour surveillance of nests /breeding sites
In addition to National Parks (which usually occupy large areas of land), different countries can also
create other categories of conservation areas if they contain species or habitats which need some
form of protection

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 FORM 6 TERM 2
TOPIC 12: DIVERSITY OF ORGANISMS
KEY CONCEPT OBJECTIVES CONTENT SUGGESTED SUGGESTED
Learners should be (ATTITUDES, SKILLS AND LEARNING RESOURCES
able to: KNOWLEDGE) ACTIVITIES AND
NOTES
8.15.1  identify - Diagnostic features of the five  Observing  ICT tools
Classification organisms using Kingdoms organisms.  Braille
diagnostic  Classifying software/Jaws
features of the organisms into  Samples of
five Kingdoms the five organisms
Kingdoms.
- Diagnostic features of phyla  Dichotomous key
 use diagnostic  Collecting and
features to classifying
divide kingdoms - Kingdom organisms.
into phyla - Phyla
- Class  Outlining the
 state the - Order taxonomic
taxonomic - Family hierarchy.
hierarchy - Genus
- Species

- Binomial nomenclature
- Genus and species names  Discussing the
rules of binomial
 observe the nomenclature.
rules of binomial
nomenclature
8.15.2 Importance  describe the - socio-economic importance  Discussing the  ICT tools
of Biodiversity socio-economic of socio-economic  Brail software/Jaws
importance of I. Kingdom Prokaryotae importance of
the five o fermentation the five
Kingdoms o bio-technology kingdoms.
o food spoilage
o decomposition

II. Kingdom Protista

o Plasmodium sp -
malaria
o Schistosoma sp –
schistosomiasis
o Trypanosoma sp -
Trypanosomiasis

III. Kingdom Fungi

o Fermentation
o Penicillin production
o Decomposition
o Food spoilage
o Food

IV. Kingdom Plantae

o Producers
o Carbon sink
o Timber
o Medicinal use
o Tourism

V. Kingdom Animalia

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KEY CONCEPT OBJECTIVES CONTENT SUGGESTED SUGGESTED
Learners should be (ATTITUDES, SKILLS AND LEARNING RESOURCES
able to: KNOWLEDGE) ACTIVITIES AND
NOTES

o Tourism
o Food
o Hunting
o Leather
o Fishing

Introduction

─ Diversity is the variety of living organisms

─ Organisms have been grouped together for studies of their characteristics

─ The grouping of organisms is called classification

─ classification is based on agreed name of each organism

Hierarchy of classification

─ Systems of classification are hierchial i.e each successive group contains more and more different
kinds of organisms

─ Taxon is the general name given to each classification grouping

─ Taxonomy is the science of classification of organisms into groups called taxons

─ The longest taxon is the species and the most increasive or highest taxon is the kingdom

─ Phylogeny is the study of evolutionary traits

─ Natural classification of organisms is based on evolutionary relationships

DEFINITION OF TERMS

─ Kingdom is the largest grouping of organisms’ e.g animalia

─ Phylum consists of organisms with many similarities e.g bryophyte, cnidarians etc.

─ Class consists of organisms which are grouped into several orders with few similarities

─ Order is a group of apparently related families

─ Family is a group of apparently related genera

─ Genus is a group of similar and closely related species

─ Species is a group of organisms capable of interbreeding to produce fertile off springs

BINOMIAL CLASSIFICATION

─ In the 18th century a scientist called Carolus Linnaeus introduced a binomial system of classifying
organisms.
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─ it names each organism as a member of genus and species and is used on international agreement

─ This was in Latin a language that could be understood by all scientists around the world.

─ The scientific name is two parts the first denotes the genus which starts with a capital letter and the
second denotes species starts with a small letter.

─ e.g Homo sapiens

─ The generic name is shared with other related species considered to be sufficiently similar to be
grouped in the same genus e.g Homo erectus, Homo habilis

Kingdoms

─ Organisms were once classified into two kingdoms i.e Animalia and Plantae in which all organisms
that were not animals were placed in Plantae kingdom e.g Fungi, spirogyra/protoctists

─ This was known as two kingdom classification

─ This had an advantage that it was easy to classify the organisms into appropriate kingdom; it was also
consistent with the traditional literature

─ However it had drawbacks such as problems with protoctists e.g Euglena is photosynthetic-plant like
and motile-animal like

─ It was not very late that Margulis and Perutz introduced the five kingdom classification in which
organisms were classified as Plants, Fungi, Prokaryote, Protoctists and Animalia

(a)Kingdom Plantae

Diagnostic features of the Kingdom Plantae

─ Eukaryotic

─ Multicellular

─ Photosynthetic/autotrophic

─ Have cellulose cell walls

─ Non-motile

─ Have chloroplasts containing chlorophyll a and b

─ store carbohydrate as starch

─ reproduce sexually and asexually

─ Have vascular system or undeveloped vascular tissue

─ mainly terrestrial

─ some have true roots, leaves and roots

─ alteration of generations
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Kingdom Animaliae

Diagnostic features of the kingdom Animaliae

─ Eukaryotic

─ Multicellular

─ Non photosynthetic

─ Heterotrophic

─ no cellulose cell walls

─ store carbohydrate as glycogen

─ no chlorophyll

─ motile

─ have nervous system (C.N.S)

─ have endocrine system for homeostasis

─ reproduce sexually or asexually

─ body divided into head, abdomen and limbs

─ all have an alimentary canal

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 ─ have muscles and skeletal systems for maintaining body shape, protection and to provide support
for inner structures

─ have a transport system with a transport medium usually blood pumped around the body in
vessels by the heart

─ bilateral symmetry except cnidarians and echinoderms

─ triploblastic except cnidarians

─ some are segmented e.g annelids and arthropods

KINGDOM PROKARYOTAE

Diagnostic features of the kingdom Prokaryotae

─ lack true nucleus

─ circular D.N.A lies free in the cytoplasm

─ unicellular

─ no membrane bound organelles

─ mesosomes for respiration (instead of mitochondria)

─ have 70s ribosomes

─ cell walls of murein (peptidoglycan)

─ average diameter 0.5-5 micrometres

─ reproduce asexually by binary fission

Kingdom Fungi
Diagnostic features of the Kingdom Fungi

─ some are unicellular e.g yeast and some are multicellular e.g mushroom

─ non photosynthetic

─ heterotrophic/saprotrophic/parasitic/mutualistic

─ nutrition is absorptive-digestion takes place outside the body and nutrients are absorbed

─ cell walls made of chitin as the main fibrilar material

─ body is a mycelium a network of fine tubular filaments called hyphae growing from horizontal
hyphae the stolon

─ end of hyphae bears sporangia which are a reproductive organ for spore formation
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 ─ eukaryotic

─ store carbohydrate as glycogen

─ asexual reproduction by spore formation

─ non-motile

Kingdom Protoctista
-Made up of eukaryotes no longer classified as animals, plants or Fungi e.g algae and protozoa.

PROTOZOA ALGAE
Non-photosynthetic photosynthetic
Parasitic and some free living Free living/non parasitic
No cell walls Have cellulose cell walls
Small and temporary food vacuoles Large permanent vacuoles
Some motile and some non -Non motile
motile
Unicellular Multicellular or unicellular
No tissues formed No tissues
Some have differentiated anterior and posterior No distinct anterior and posterior
No transport system No transport/vascular system

-filamentous

-no leaf structure

-no roots

-no stems

-contain chlorophyll a and b

-unicellular algae

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IMPORTANCE OF BIODIVERSITY
KEY CONCEPT OBJECTIVES CONTENT SUGGESTED SUGGESTED
Learners should be (ATTITUDES, SKILLS AND LEARNING RESOURCES
able to: KNOWLEDGE) ACTIVITIES AND
NOTES
8.15.2 Importance  describe the - socio-economic importance  Discussing the  ICT tools
of Biodiversity socio-economic of socio-economic  Brail software/Jaws
importance of VI. Kingdom Prokaryotae importance of the
the five o fermentation five kingdoms.
Kingdoms o bio-technology
o food spoilage
o decomposition

VII. Kingdom Protista

o Plasmodium sp -
malaria
o Schistosoma sp –
schistosomiasis
o Trypanosoma sp -
Trypanosomiasis

VIII. Kingdom Fungi

o Fermentation
o Penicillin production
o Decomposition
o Food spoilage
o Food

IX. Kingdom Plantae

o Producers
o Carbon sink
o Timber
o Medicinal use
o Tourism

X. Kingdom Animalia

o Tourism
o Food
o Hunting
o Leather
o Fishing

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TOPIC 13: HEALTH AND DISEASES
KEY CONCEPT OBJECTIVES CONTENT SUGGESTED SUGGESTED
Learners should be (ATTITUDES, SKILLS LEARNING RESOURCES
able to: AND KNOWLEDGE) ACTIVITIES AND
NOTES
8.8.1 Drug and  explain the - Drug dependence  Discussing effects  Adverts
substance abuse meanings of the - Drug tolerance of drug abuse.  Resource persons
terms drug - Addiction  Print media
dependence and - (with reference to  ICT tools
drug tolerance alcohol, tobacco,  Braille software/Jaws
heroin, cough
mixtures, marijuana
{mbanje})

- Psychological  Visiting psychiatric

 distinguish dependence hospitals and
between - Physical dependence rehabilitation
psychological and centres.
physical  Observing video
dependence clips of evidence of
drug abuse.
 Carrying out
surveys on
- Atherosclerosis statistics on drug
- Coronary heart abuse.
disease  Resource persons
- Strokes
 describe the - Cancer
effects of tobacco
smoking in the
gaseous
exchange and - Social effects
cardiovascular
systems - Long and short term
effects of alcohol  Debating on effects
of tobacco smoking
 describe the and alcohol.  Fresh liver tissue
immediate and  Alcohol
long term  Illustrating the
consequences of effects of alcohol
alcohol on liver tissue in
consumption on the laboratory.
the brain, on the
peripheral nervous
system and on the
liver
8.8.2 Global  discuss the global - Malaria  Discussing and  Resource person
distribution of distribution of - Tuberculosis evaluating  Print media
diseases diseases - Ebola epidemiological  ICT tools
- HIV/AIDS evidence of  Braille software/Jaws
- Cholera diseases.
- Coronary heart  Visiting clinics.
disease
- Sickle cell anaemia
8.8.3 Immunity  recognise - Phagocytes,  Observing  Microscope
lymphocytes and lymphocytes phagocytes and  Photomicrographs
phagocytes lymphocytes.  Prepared slides

 describe the - Phagocytosis  Observing  ICT tools

origin, maturation phagocytosis
and mode of simulations.  Braille software/Jaws
action of
phagocytes  Print media
- B and T lymphocytes  Observing
 describe the - Mode of action simulations of
modes of action of modes of action of

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 B and T B and T
lymphocytes lymphocytes.
 Discussing modes
- Effects of HIV on T of action.
lymphocytes
 Discussing the
 describe the effects of HIV.
effects of HIV on T - Memory cells
lymphocytes
 Discussing the role
 explain the role of of memory cells.
memory cells in - Vaccination
long term
immunity  Researching on
eradication of
 discuss the smallpox and
reasons why reasons for failure
vaccination to eradicate
programmes have measles,
eradicated small tuberculosis,
pox but failed in malaria and
measles, cholera.
tuberculosis,
malaria and
cholera

Describe whether health is more than simply the absence of a disease

─ complete physical, mental and social being

─ linked to happiness/fulfilling life
─ having a positive outlook in life
─ socially well adjusted
─ ability to undertake physical/mental tasks without too much difficulty
─ feeling good physically/physical fitness
─ need for a balanced diet
─ need of regular exercises/lack of exercises likely to suffer certain diseases
─ both diet and exercise prevent obesity
─ access to medical care

Outline the aspects that contribute towards good health

─ good health is complete physical, mental and social well being

─ ref. to balanced diet
─ regular exercise/regular physical activity
─ which ensure body is in best condition
─ to combat disease
─ limit drug intake/alcohol intake/smoking
─ improving ability to cope with stress/other benefits of exercise
─ regular sleep/rest + reasons
Disease

─ difficult to define
─ is a disorder or malfunction of the body leading to departure from good health.
─ it is usually a disorder of a specific tissue or organ due to a single cause
─ diseases are characterised by signs and symptoms that are physical, mental or both
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 ─ symptoms give an indication of the nature of the disease
─ some diseases are acute and they last for a short time
─ some are chronic and the effects continue for months or years
─ many chronic diseases are extremely debilitating

Categories of diseases

─ there are different ways of classifying diseases

─ there are nine broad categories though some disease are classified into more than one
category

Using named examples for each, explain what is meant by degenerative and inherited
disease

degenerative

─ gradual decline in body functions

─ associated with characteristics of ageing
─ caused by deficiencies of nutrients during childhood
─ e.g. skeletal diseases/cardiovascular/cancers/Huntington‘s disease
inherited

─ an inherited genetic fault

─ mutation
─ ref to mechanism of mutation
─ pattern of inheritance
─ e.g. cancer, PKU, cystic fibrosis
physical

─ permanent or temporary damage to body parts

─ body parts damages
─ e.g. leprosy, multiple sclerosis, stroke
mental

─ disorder occurring in brain cells

─ no/a sign of physical damage may appear to the brain
─ e.g. schizophrenia, claustrophobia, anxiety
infectious

─ caused by pathogens which invade the body such as viruses, bacteria, fungi, worms,
protoctists and insects (e.g. lice)
─ are also called communicable diseases because the pathogens can be transferred from
person to person
─ e.g. TB, HIV, cholera, STIs
deficiency

─ nutritional diseases
─ due to poor nutrition/inadequate diet
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 ─ not passed to offspring
─ scurvy, kwashiorkor, obesity
self-inflicted

─ people‘s health is put at risk by their own decisions regarding their behaviour
─ due to actions of an individual
─ e.g. lung cancer/CHD/liver cirrhosis/anorexia nervosa, attempted suicide
social

─ due to social behaviour and living conditions

─ e.g. TB/cholera/infectious diseases , smoking related diseases

non-infectious

─ all diseases not caused by pathogens

─ no organism invasion
─ not transmitted from one person to another
─ e.g. deficiency disease, mental disease, night blindness

EPIDEMIOLOGY AND PATTERNS OF DISEASES

KEY CONCEPT OBJECTIVES CONTENT SUGGESTED SUGGESTED
Learners should be (ATTITUDES, SKILLS LEARNING RESOURCES
able to: AND KNOWLEDGE) ACTIVITIES AND
NOTES
8.8.2 Global  discuss the global - Malaria  Discussing and  Resource person
distribution of distribution of - Tuberculosis evaluating  Print media
diseases diseases - Ebola epidemiological  ICT tools
- HIV/AIDS evidence of  Braille software/Jaws
- Cholera diseases.
- Coronary heart  Visiting clinics.
disease
- Sickle cell anaemia

Epidemic

─ a disease that suddenly spreads to affect many people e.g. cholera

─ an outbreak of disease in a population
Endemic

─ an infectious disease which is always present in a population e.g. TB

─ this describes diseases that are always in a population
Pandemic

─ a disease that spreads over a large area e.g. continent/worldwide

─ an outbreak of disease that occurs across the world or across continents.
Prevalence

─ the number of people in a population with a disease within any given time

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Incidence

─ number of new cases within a population occurring for a given time e.g. week/month/year
Epidemiology

─ the study of patterns of disease and the various factors that affect the spread/distribution
of the disease
─ data collected on disease (morbidity) and death (mortality) reveal patterns that can
indicate how diseases are spread and their likely cause or causes

Discuss the possible reasons for the global distribution of coronary heart diseases

─ mainly confined to developed/affluent countries

─ mainly due to lifestyle/sedentary
─ fatty diets
─ high in saturated fats
─ cigarette smoking
─ alcohol intake
─ obesity
─ high blood pressure
─ lack of exercise
─ fast foods
Explain the possible reasons for the global distribution of TB

─ it is a pandemic disease (globally distributed)

─ it is an endemic disease (always present)

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 ─ most prevalent in developing countries
─ some of the TB strains becoming more resistant
─ AIDS pandemic
─ poor housing – overcrowding
─ breakdown of TB control programme
─ partial treatment of TB

─ poor sanitation
─ poor medical facilities
─ TB spread in meat and milk
─ high rate of transmission - droplet infection

Explain the possible reasons for the global distribution of measles

─ most commonly affects developing countries;

─ in places where conditions are overcrowded and insanitary;
─ measles requires several booster shots to develop full immunity;
─ in large cities with high birth rates, it can be difficult to give boosters or even follow up cases
of measles;
─ refugees from these areas can spread the disease around, which makes it much more difficult to
treat than smallpox;
─ measles is highly infectious;
─ poor response to vaccine (children do not respond well to one dose of vaccine);
─ deficiency immune system;
─ or protein energy malnutrition;
─ need several boosters which are expensive;
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─ high birth rates and flighting populations make it difficult for boosters;
─ follow up cases and trace contacts also impossible;

Explain the possible reasons for the global distribution of HIV/AIDS

─is a pandemic disease (globally distributed)

─an epidemic (always present)
─most prevalent in developing countries
─linked to TB
─some of the TB strains becoming more resistant
─AIDS pandemic
─partial treatment of due to inability to purchase ARVs
─poor medical facilities
─Highly confined in sub-Saharan Africa
─Rates of infection are lower in other parts of the world, but different subtypes of the virus
have spread to Europe, India, South and Southeast Asia, Latin America, and the Caribbean.
Rates of infection have leveled off somewhat in the United States and Europe.
─ In Asia the sharpest increases in HIV infections are found in China, Indonesia, and Vietnam.
─ Both the cost of these therapies and the poor health care delivery systems in many affected
countries need to be addressed before antiretrovirals can benefit the majority of people
living with HIV/AIDS.
─ Botswana, with the highest rate of infection, has experienced stable, democratic government
and a strong economy since independence in 1966.
─ Mozambique, with the lowest rate of infection, experienced sixteen years of devastating civil
war from which it only emerged in 1992.
─ While South Africa and Botswana are the two richest countries in Sub-Saharan Africa (as
measured in per capita gross domestic product),
─ They include poverty and economic marginalization, poor nutrition, opportunistic infection,
migration, sexual networking and patterns of sexual contact, armed conflict, and gender
inequality. Some of these will be discussed in more detail below.
─ HIV/AIDS, like all communicable diseases, is linked to poverty. The relationship is bi-
directional in that poverty is a key factor in transmission and HIV/AIDS can impoverish
people in such a way as to intensify the epidemic itself.
─ poverty does seem to be a crucial factor in the spread of HIV/AIDS. It should be emphasised
that poor people infected with HIV are considerably more likely to become sick and die
faster than the non-poor since they are likely to be malnourished, in poor health, and
lacking in health attention and medications.
─ In effect, all factors, which predispose people to HIV infection, are aggravated by poverty,
which ―creates an environment of risk‖.
─ Deep-rooted structural poverty, arising from such things as gender imbalance, land
ownership inequality, ethnic and geographical isolation, and lack of access to services.
─ Developmental poverty, created by unregulated socio-economic and demographic changes
such as rapid population growth, environmental degradation, rural-urban migration,
community dislocation, slums and marginal agriculture.
─ Poverty created by war, civil unrest, social disruption and refugees. High levels of rape and
the breakdown of traditional sexual mores are associated with military destabilisation,
refugee crisis and civil war (Walker, 2002: 7).
─ closely associated with patterns of human mobility

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 ─ Large-scale economic migration has been a feature particularly of the southern African
region
─ Massive migration of young, unmarried adults from presumably
―conservative‖ rural environments to more sexually permissive African cities in recent
years has been regarded as partly responsible for the much higher infection levels observed
in urban than in rural areas.

Values of Procreation:

─ In Africa fertility is seen as demonstrating the masculinity and manliness of men, as well
asproving the significance of women as good wives. Because procreation is highly valued in
Africansociety, both men and women are refusing to use condoms.
─ Even though condoms are successful in preventing the spread of AIDS, they also prevent
reproduction.
─ Thus, many individuals are willing to risk contracting AIDS and have unprotected sex because
fertility is so important to social status.
Myths:

Myths influence the spread of HIV/AIDS in many ways. One strong belief held by a number
of Africans is that the West wants to control the population growth of Africa, and that the
West is trying to do this by convincing Africans to use condoms. The West is encouraging
African nations to use condoms as protection against AIDS, but many Africans believe that
this is just a ploy to curb reproduction rates. Many Christians in Africa believe that God is
using AIDS as a weapon to punish sinners. Since AIDS is often associated with promiscuity,
many followers believe that God will protect the innocent spouse from contracting AIDS, but
use AIDS to punish the spouse that was involved in sexual practices outside of her/his
marriage. Two other popular myths are that some Africans believe that regular infusions of
sperm is required if a woman is to grow up to be beautiful, and that sleeping with a virgin
will rid an infected person from the disease.

Link between atherosclerosis and coronary heart disease

─ atherosclerosis is the main cause of CHD;

─ coronary arteries become narrower;
─ therefore blood pressure increases;

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 ─ can damage walls of arteries/cause aneurysm, causing wall to burst;
─ atheromas roughen lining of arteries causing blood clots/thromboses;
─ clot can block coronary artery;
─ clot may break away and lodge where artery narrows/embolism;
─ heart muscle is deprived of oxygen and diets;

Evaluate the epidemiology and experimental evidence linking smoking to lung cancer and
early

death

Epidemiology

─ more smokers died of cancer;

─ number of woman developing cancer increased with number of women smoking
─ number of cigarettes smoked per day increased per day linked with death rate;
Experimental

─ carcinogen identified in tar;

─ dogs exposed to cigarette smoke developed tumours;
─ rate of tumour development reduced when filter tipped brands used;

Discuss the epidemiological and experimental evidence which links smoking with disease.

─ more new diseases in smokers;

─ compared to non-smokers
─ the higher the number of smokers the higher the number of sufferers
─ experimental animals exposed to smoke developed diseases;
─ compared to those not exposed to smoke;
─ animals exposed to smoke have higher chances of developing disease
─ filtered vs. unfiltered

How cigarette smoking can lead to coronary heart disease

─ main cause of CHD is atherosclerosis;

─ carbon monoxide/nicotine in smoke responsible;
─ cholesterol deposited in, inner layers/.linings of artery walls;
─ form plaques which lead to restriction of blood flow/clotting which can block vessels;
─ smoking increases blood cholesterol/fat level;
─ nicotine causes constriction of coronary arteries/arterioles;
─ rise in blood pressure makes damage to walls more likely;
─ increases number of platelets stimulating formation of blood clots;
─ nicotine makes platelets more sticky;
─ smoking causes rise in ratio of VLDLs/LDLs, to HDLs in blood so more
atherosclerosis/cholesterol deposited;

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 ─ decrease concentration of antioxidants/Vitamin C/vitamin E, so increasing damage to
artery walls by free radicals;

The difficulty in achieving a balance between prevention and cure of coronary heart diseases

─ due to life style;

─ such as smoking/diet/lack of exercise most causes can be avoided;
─ government could take steps to encourage change of life;
─ a few patients are victims of their own genetic;
─ cure is expensive;
─ e.g. heart transplants/coronary by-pass/drug treatment;
─ ethical problems of who to treat, suitable example;
─ since donors are few;
─ problems associated with tissue rejection;
─ should patient change their life style before treatment is made available

Arguments for diverting funds from the treatment of coronary heart disease to its
prevention

─ cure is expensive;
─ e.g. heart transplant, coronary by-pass, drug treatment;
─ difficult to find enough donor hearts;
─ ethical problems of who to treat e.g. father with young family;
─ many of the risks are avoidable;
─ associated with life style - change will make people less susceptible;

Discuss the factors that should be taken into account when deciding how to share limited
resources

between prevention and treatment of coronary heart disease

─ treatment is expensive due to technology and professional expertise of surgeons;

─ after – care also expensive (immunosuppressant drugs, e.t.c.);
─ NHS working on tight/limited budget;
─ preventive measures cheaper;
─ not so dependent on expensive equipment/manpower;
─ very expensive to advertise/train/employ health educators;
─ difficulty in disseminating information;
─ prevention saves a lot of suffering for potential victims;
─ and families;
─ e.g. may cause financial difficulties if wage earner affected/fatherless family/e.t.c;
─ in terms of years of healthy life gained preventive measures may be better;
─ great demand for treatment because heart disease so common;
─ moral dimension – if a treatment is available should we not make resources available
to use it;
─ more lives can be saved by preventative measures;

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Global distribution of CHD

─ mainly confined to developed/affluent countries;

─ due to lifestyle + sedentary work;
─ fatty diets;
─ high saturated fats;
─ cigarette smoking;
─ alcohol intake;
─ obesity;
─ high blood pressure;
─ lack of exercise;
─ fast foods;

Reasons why coronary heart disease is so common in developed countries

─ caused by many factors most of which are common in developed countries;

─ smoking s common risk factor;
─ carbon monoxide and nicotine are the components responsible;
─ carbon monoxide reduces oxygen carrying capacity;
─ increasing strain on the cardiovascular system;
─ nicotine increases stickiness of platelets, raising risk of clotting;
─ diets tend to be rich in saturated fats;
─ increased blood cholesterol and hence atherosclerosis;
─ cause rise in ratio of LDLs to HHDLs;
─ hypertension/high blood pressure common which puts arteries under strain;
─ lack of exercise is a risk factor and life style/ occupations often sedentary

Common categories of diseases which could be applied to coronary heart disease

[Each category should be qualified with a suitable reason]

─ physical;
─ non-infectious;
─ self – inflicted;
─ degenerative;
─ social;
─ links with diet/lifestyle of developed countries;
─ inherited;

Why infectious diseases are the leading causes of death in developing countries

─ infectious disease are communicable/transmittable;

─ pathogens spread from infected to uninfected people;
─ some people have the disease but do not show an symptoms/are carriers;
─ makes tracing very difficult;
─ many developing countries struggle to pay health personnel;
─ and to carry out effective prevention and treatment programmes;
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 ─ problem of development of clean water supply;
─ good sanitation difficult to achieve;
─ natural disaster/wars/civil unrest;
─ growth of shanty towns;
─ rural areas not reached;
─ many developing countries in warmer regions where pathogens/parasites/insects/vectors
thrive;

CHOLERA

─ Cholera is a water-borne disease caused by a bacterium.

─ It often appears in a population following a natural disaster, such as a major earthquake or
flood.
─ Cholera is caused by a bacterium called Vibrio cholerae.

Signs and symptoms of cholera

─ Common symptoms of cholera and the dehydration it causes include:

─ watery, pale-colored diarrhea, often in large amounts.
─ nausea and vomiting.
─ cramps, particularly in the abdomen and legs.
─ irritability, lack of energy, or unusual sleepiness.
─ glassy or sunken eyes.
─ dry mouth and extreme thirst.
─ dry, shriveled skin.

Transmission of cholera

Describe how cholera can be spread from one person to another/ Explain how cholera is transmitted
from one person to another through water supply/

Describe and explain the ways through which cholera is transmitted

─ food and waterborne disease;

─ due to lack of proper sanitation/poor hygiene by infected person;
─ feaces from infected person;
─ contains Vibrio cholerae, the causative agent;
─ contaminate water supply/water borne;
─ water used to irrigate crops;
─ sewage and water supply not separated / human faeces (sewage)
─ contaminates water supply;
─ person drinks water / eats food / swims in contaminated water / AW
─ transmitted to uninfected person;
─ via contaminated water;
─ via contaminated food;
─ e.g. vegetables irrigated with raw sewage;
─ e.g. washing/bathing in contaminated water;
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 ─ pass out (of infected person) in feaces/(bacteria) leave person in faeces;
─ infected people handle food/cooking utensils without washing hands;

Global distribution of cholera

─ Cholera is now almost unknown in the developed world but can still cause large numbers
of deaths in less developed countries.

Outline the reasons why cholera is more likely to spread in less developed countries.

─ lack of education/knowledge of hygiene;

─ poor sanitation;
─ lack of sewage treatment;
─ raw sewage used to irrigate/fertilise crops;
─ lack of water treatment;
─ unable to control outbreaks due to lack of rehydration treatments;
─ natural disasters;
─ poor economy;
─ civil unrest/migrants;

Reasons why cholera does not show the same global distribution as malaria

─ malaria is transmitted by, mosquito / Anopheles / vector;

─ distribution is determined by mosquito which lives in, tropics / subtropics /described; ora
─ malaria not dependent on poor, hygiene / sanitation;
─ cholera transmitted via water or food;
─ sickle trait / genetic factors, influence distribution of malaria;
─ ref to natural disasters / manmade disasters;

Why infants who are breast fed rarely suffer from cholera

─ breast milk is sterile / idea;

─ infants do not drink (contaminated) water / drink milk;
passive immunity;

─ milk may contain antibodies (to cholera);

─ antibodies provide protection for infant;

Measures to control the spread of cholera

Sanitation should be qualified / explained

─ piped water;
─ ensure that water supply is separate from sewage;
─ hygienic, removal / disposal, of faeces; A ‗human waste‘ / sewage treatment

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─ latrines;
─ encourage breast feeding;
─ treat people with cholera / provide ORT / provide antibiotics / provide trained
─ medical personnel / medical facilities / access to medical facilities;
─ boiled drinking water / sterilized water / chlorinated water;
─ make sure people, eat cooked food / avoid raw food;
─ AVP;; e.g. ref to education, vaccination, contact tracing, cordon sanitaire, ref to flies

Role of economic factors in the prevention and control of cholera

─ waterborne disease;
─ caused when water is infected by feaces from carrier/sufferer;
─ important to purify water;
─ important to have proper sewage treatment;
─ developing countries often cannot afford the required measures;
─ partly because they have large debts;
─ education needed about importance of hygiene + economic link;
─ cost money to train teachers/run advertising campaigns/build schools;
─ locate and isolate carriers/sufferers;
─ cost money to build hospitals/isolation wards/trace contacts/train staff/pay nurses;
─ (economic) aid available in form of oral rehydration packs for treatment;
─ cheap and effective;
─ from international aid e.g. Red Cross;
─ antibiotics can cure but often too expensive;
─ little incentive for drug companies to develop cured because developing countries cannot afford
them;

Explain why it has been proved difficult to develop a vaccine to control the spread of cholera

─ V. cholerae in intestine;
─ out of reach of immune system;
─ antigenic concealment;
─ antibodies broken down in intestine;
─ antibodies are proteins;
─ ref to pH and effect on structure or shape; e.g. in the stomach
─ denaturation;
─ vaccine stimulates antibodies in, blood / lymph;
─ not in gut;
─ oral vaccine needed;
─ mutation;
─ different strain idea;
─ AVP; e.g. not required in developed countries
─ developing countries cannot afford to develop vaccines
─ no / limited, demand
─ cholera can be treated with ORT
─ can be treated with antibiotics

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MALARIA

─ malaria is caused by a single celled organism called Plasmodium.

─ the organism is transmitted from one person to another by female anopheles mosquito.
─ a mosquito takes up the gametes of a malaria parasite when it feeds on the blood of an
infected person.
─ fertilisation occurs in the stomach of the mosquito and the immature parasites reproduce.
─ infective stages of the malaria parasite migrate to the mosquito‘s salivary glands.
─ a new person becomes infected when the mosquito takes another meal of the infected
blood.
─ the parasites enter the liver of the new victim where further reproduction takes place
before migrating to the red blood cells.
─ when an organism such as mosquito is involved in transmission, it is called a vector.
─ the malarial parasite can also be transmitted by a vector

Transmission of malaria

Explain how an infected is likely to have acquired malaria

─ bitten by mosquito carrying malarial parasite;

─ genus Anopheles/female;
─ injects parasites with saliva/anticoagulant;
─ ref to vector;
─ mosquito fed on/bit/took a blood meal from an infected person;
─ transmission by needle;
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 ─ injected into blood;
─ after use by someone with malaria;
─ (needle) shared/reused/used but not sterile;
─ Transmission across placenta;
─ Blood transfusions;

Signs and symptoms of malaria

People with malaria have the following symptoms:

─ abdominal pain.
─ chills and sweats.
─ diarrhea, nausea, and vomiting (these symptoms only appear sometimes)
─ headache.
─ high fevers.
─ low blood pressure causing dizziness if moving from a lying or sitting position to a standing
position (also called orthostatic hypotension)

Ways of preventing the spread of malaria

Explain the ways in which the transmission of the malaria life cycle can be disruptedreduce mosquito
numbers

─ stock ponds with fish (Gambusia) to eat larvae ; R kill mosquitoes

─ oil on surface ;
─ spray bacteria (Bacillus thuringiensis) to kill mosquito larvae ;
─ DDT / pesticide spray ;
─ release of sterile male mosquitoes ;
─ draining, ponds / bodies of water ;
avoid being bitten by mosquitoes

─ wear insect repellant ;

─ long sleeved clothes ;
─ sleep under nets ;
─ nets soaked in, insecticide / repellant ;
─ sleep with, pigs / dogs ;

use drugs to prevent infection

─ use, prophylactic drug / quinine / chloroquine / larium / artimesinin / vibrimycin

─ tetracycline / antimalarial ;
─ use malaria vaccine ;

Social and biological factors in the prevention of malaria

Social

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 ─ poverty;
─ no access to treatment/access to anti-malaria drugs;
─ no access to mosquito nets/sleep under mosquito nets
─ cultural beliefs

Biological

─ Plasmodium strains resistant to some malaria drugs such as chloroquine

─ development of resistance in Plasmodium/breaking life cycle of vector;
─ development of resistance to pesticide;
─ use of natural predators of vectors;
─ some pesticides extend their effects to other innocent organisms;
─ vaccination failing;
─ Plasmodium an intracellular pathogen;
─ an enormous reserve in monkeys;
Plasmodium antigens change from time to time making vaccination difficulty;

Outline the problems associated with controlling the spread of malaria

─ resistance of, Plasmodium / pathogen, to drugs;

─ eukaryote / protoctist, has many genes;
─ many surface antigens / antigenic variation; A ref to mutation
─ inside red blood cells / in liver cells / antigen concealment;
─ difficult for immune system to operate / idea;
─ dormant / in body for a long time / symptomless carriers / long incubation;
─ different stages in life cycle in the body;
─ resistance of, vector / mosquito, to insecticides; A mutation / selection
─ mosquito, breeds in small areas of water; A implications
─ breeds quickly;
─ mosquitoes, spread over large area / widely distributed / fly a long way;
─ mosquito control programmes disrupted by war etc;
─ lack of infrastructure (for control programmes);
─ problems with sleeping nets, described;
─ more effective when soaked in insecticide;
- no vaccine;
─ people lose immunity if, malaria eradicated / move to non-endemic area;
─ poor primary health care / few doctors or other medical personnel;
─ ref to poor housing / slums / shanties;
─ ref to remote rural areas;
─ ref to cost of control programmes;
─ ref to travel / migration;
─ ref to change in climate;
─ ref to education;
─ ref to problems of biological control;
─ AVP; e.g. effects of insectides on, ecosystems / humans
─ AVP; side effects of drugs
─ impossible to isolate infected people
─ ref to sterilising male mosquitoes
─ opening new areas of tropics
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 ─ different, species / strains, of malaria
─ cost to individual
─ ref to detection in bloodstream
─ blood transfusions
─ mother to fetus across placenta

Outline the problems associated with the elimination of malaria

─ resistance of mosquitoes to insecticides;

─ such as DDT/dieldrin;
─ difficulty in controlling the breeding places;
─ resistance of some strains of Plasmodium/no effective vaccine against Plasmodium;
─ migration of both people (infected and uninfected);
─ expensive;
─ attitude of society to elimination programme by WHO;
─ environmental effects of insecticides;

Explain the link between malaria and sickle cell anaemia prevalence.

─ possession of sickle cell trait has selective advantage in malaria areas;

─ pathogens does not survive in red blood cells
─ do not suffer from malaria

Causes of sickle cell anaemia

─ a faulty occurs on the haemoglobin molecule;

─ the 6th amino acid of the beta chain (146 amino acids)
─ glutamic acid is replaced by valine;
─ glutamic acid carries a negative charge and its polar, valine is non-polar;
─ deoxygenated become less soluble;
─ haemoglobin crystalises into a rigid rod-shaped fibre;
─ it is a result of a base in substitution;
─ thymine of a DNA triplet code CTC is replaced by adenine to make a CAC;
─ the affected gene is on chromosome 11;
─ the faulty gene is codominant;
─ effect expressed only in homozygous condition

TUBERCULOSIS (TB)

Causes of TB

─ TB is caused by two bacteria; Mycobacteria tuberculosis and Mycobacteria bovis.

─ Tuberculosis is ultimately caused by the Mycobacterium tuberculosis, a bacterium that is spread
from person to person through airborne particles. Inhaling infected particles does not
necessarily mean that a person becomes infected.
─ One of three things may happen when Mycobacterium tuberculosis enters the human body:
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 ─ the bacterium is destroyed because the body has a strong immune system 9attacked by the
macrophages in the lungs);
─ the bacterium enters the body and remains as latent TB infection. The patient has no
symptoms and
cannot transmit it to other people;

(iii) the patient becomes ill with TB;

Signs and symptoms of TB

─ Most people who become infected with Mycobacterium tuberculosis do not present
symptoms of the disease.
─ However, when symptoms are present, they include:
─ unexplained weight loss
─ fatigue
─ shortness of breath
─ fever
─ night sweats
─ chills
─ loss of appetite.

Symptoms specific to the lungs include:

─ coughing that lasts for 3 or more weeks

─ coughing up blood
─ chest pain
─ painful breathing
─ pain when coughing.

Transmission of TB

─ TB is a communicable disease that is spread primarily by tiny airborne particles (droplet

nuclei);
─ the disease spreads when infected people with the active form of the illness cough or sneeze;
─ the bacteria spread through droplets in the saliva or sputum
─ it spreads most rapidly among people living in overcrowded conditions;
─ when a person with active TB coughs, sneezes, talks, or spits, tiny droplets containing the
bacteria
─ are released into the air and can be inhaled by people who are close by.
─ the bacteria can spread from the initial location in the lungs to other parts of the body
through the bloodstream.
─ only a small number of bacteria are needed to cause an infection.
─ persons with latent TB infection cannot transmit TB because bacteria are not present in their
saliva or sputum.
─ M. bovis causes TB in cattle and is spread to humans in meat and milk;
─ most people with TB are infected with M. tuberculosis.
─ people with active TB can transmit the bacteria through the air by coughing and sneezing.

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People at risk

Although anyone can be exposed to or get TB, some people are at higher risk for both exposure and
infection (though exposure does not necessarily result in infection). These higher risk groups
include, among others:

─ the close contacts of someone who is infectious

─ immigrants from areas where TB is common, such as Asia, Africa, and Latin America
─ the poor
─ the medically underserved
─ racial and ethnic minorities
─ persons living in congregate settings, such as correctional facilities;
─ alcoholics and persons who inject drugs
─ the homeless
─ persons with HIV infection
─ persons who are exposed to infectious TB on the job.
─ Of those infected with TB, the following run an especially high risk of developing active TB
disease:
─ persons with HIV
─ persons whose infection is relatively recent (within the previous 2 years)
─ injection drug users
─ those with a history of inadequately treated TB.
─ persons infected with both HIV and TB has the highest known risk factor for developing
active TB disease.
─ Whereas TB-infected persons who are not HIV-positive run a 10 percent lifetime risk of
developing active disease, those with both TB and HIV run a percent to 10 percent chance
per year of developing active disease
Global distribution of TB

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 Worldwide
Explain the possible reasons for the global distribution of TB

─ it is a pandemic disease (globally distributed)

─ it is an endemic disease (always present)
─ most prevalent in developing countries
─ some of the TB strains becoming more resistant
─ AIDS pandemic
─ poor housing – overcrowding
─ breakdown of TB control programme
─ partial treatment of TB
─ poor sanitation
─ poor medical facilities
─ TB spread in meat and milk
─ high rate of transmission - droplet infection

Problems associated in the prevention and control of TB

─ some strains of TB bacteria resistant to drugs;

─ the AIDS pandemic;
─ poor housing and rising homelessness in inner cities in the developed world;
─ the breakdown of TB control programmes particularly in the USA;
─ partial treatment for TB increases the chance of drug resistance in Mycobacterium;
─ attacks many of the poorest and socially disadvantaged because it is spread by airborne
droplets;
─ so people who are overcrowded are particularly at risk;
─ those with low immunity particularly because of malnutrition or being HVI+ are also
vulnerable;
─ transmission is easily achieved but the bacteria may remain in the lung, or in the lymphoid
tissue for years until they become active;

Link between TB and HIV/AIDS

─ TB is often the first opportunistic infection to strike HIV+ people;

─ HIV infection may reactivate dormant infections of Mycobacterium tuberculosis;
─ TB is now the leading cause of death of HIV+ people;
─ The HIV pandemic has been followed very closely by a TB pandemic;
─ There are high rates of incidence of all across the developed world and in the countries of the
former Soviet Union;
─ Very high rates are also found in areas of destitution in developed countries;
─ Social factors such as homelessness, neglect of primary health care and urban decay,
contribute to the spread of TB and these need to be addressed if the pandemic I to be
curbed

Prevention

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 ─ once someone appears with the symptoms of TB, the sputum (mucus and pus) from their lungs
is collected for analysis;
─ the identification of M. tuberculosis can be made very quickly by microscopy;
─ isolation of sufferers while they are in their most infectious stage;
─ this is particularly the case if they have an infection of a drug resistant strain;
─ the treatment involves use of several drugs to ensure that all bacteria are killed, not just a few;
otherwise drug resistant strains are left behind to continue the infection;
─ the WHO promotes a scheme to ensure that patients complete their course of drugs;
─ DOTS (Direct Observation Treatment Short Course) involves health workers or responsible
family members, making sure that patients take their medicine regularly for 6 – 8 months;
─ contact tracing and the subsequent testing of contacts for the bacterium is an essential part of
controlling TB;
though contacts are screened for TB, the diagnosis can take up to two week;
─ In children TB is prevented by vaccination;
─ the BCG vaccine is derived from M. bovis and protect up to 70% of teenagers and its
effectiveness decreases with age unless there is an exposure to TB;
─ the vaccine is effective in some parts of the world and less effective in others e.g. India;
an effective method of control is the dual approach of milk pasteurisation and TB testing of
cattle;
─ any cattle found to test positive are destroyed;
─ these measures have reduced the incidence of TB caused by M. bovis considerably it is hardly
a hazard to health in countries where these controls operate.

Treatment

─ the treatment is long (6 months to 1 year), but many people do not complete their course
of the drugas they think that when they feel better they are cured;
─ however it takes months to kill mycobacteria because they are slow growing;
─ they are intracellular parasites surviving inside cells of the immune system, where they are
metabolically inactive therefore they are difficult to treat with drugs;
─ strains of drug-resistant M. tuberculosis were identified when treatment with antibiotics, such
as streptomycin, began in the 1950s;
─ antibiotics act as selective agents killing drug-sensitive strains and leaving resistant ones
behind;
─ drug resistance happens as a result of mutation;
─ if three or four drugs are used in treatment, then the chance of resistance occurring is greatly
reduced,
─ if TB is not treated or the person stops treatment before the bacteria are completely eliminated,
bacteria spread throughout the body increasing the likelihood that mutation will arise;
─ prematurely stopping treatment means the M. tuberculosis develops resistance to all the drugs
being used;
─ patients under poorly managed treatment programmes return home to infect others
─ multiple drug resistant forms of TB (MDR-TB) now exist

Explain how an understanding of the disease tuberculosis (TB) can be used in its control and

prevention

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 ─ TB cause by Mycobacterium tuberculosis/bacillus;
─ bacteria can be treated/controlled by antibiotics;
─ sputum infected therefore dispose of sputum hygienically;
─ associated with poverty/poor housing/poor living conditions;
─ therefore social remedies important;
─ overcrowding facilitates spreading;
─ because droplet infection;
─ also spread by physical contact/can be contagious;
─ re-house if infected people cannot have own bedroom;
─ education needed – poor education associated with poor hygiene e.t.c.;
─ notifiable so can trace contacts;
─ infectious phase brief/spread easily/spread by droplets so can isolate patients;
─ bacteria can remain dormant for several years;
─ so people who recover should not work in the food industry /become teachers;
─ can be transmitted in milk, therefore pasteurize milk;
─ test cattle (TT herds) and slaughter affected cattle;
─ BCG vaccination given to children (aged 10 – 13 years);
─ poor diet reduces resistance therefore good nutrition needed;
─ all teachers X-rayed before starting job;
─ screening possible in areas of high risk because X-rays show up lung damage
─ high risk groups can have early vaccination (e.g. Asian immigrants I inner city poor housing);
─ ref. to MDR/cocktails of antibiotics used;
─ importance of finishing course of antibiotics;
─ AVP;

Reasons for classifying TB as a social disease and lung cancer as a self-inflicted disease

TB

─ social disease is due to social conditions/behaviour

─ social factors aid to the spread of TB;
─ poor housing;
─ poverty/social class;
─ poor diet, i.e. people with low immunity from malnutrition are also vulnerable;
─ overcrowding living conditions;
─ TB droplet infection/airborne infection;
─ people sleeping/living together are at increased risk;
─ prolonged exposure;
─ poor unlikely to seek/complete treatment;
─ TB is classified in other disease categories e.g. infectious;
Lung cancer

─ Self-inflicted disease is due to one‘s behaviour;

─ most lung cancers are caused by smoking;
─ tar in tobacco smoke is a carcinogen;
─ which alters DNA In epithelial cells;
─ mutation
─ development of malignant tumours;

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 ─ epidemiologists discovered a correlation between smoking and lung cancer, i.e. smokers are
18 times more likely to develop lung cancer than non-smokers;
─ risk of development of lung cancer decreases as soon as smoking decreases;
─ personal choice to smoke;
─ once started difficult to give up;
─ addiction to nicotine;
─ physical dependence;
─ psychological dependence;
─ lung cancer is classified in other categories e.g. degenerative, non-infectious;

ACQUIRED IMMUNODEFICIENCY SYNDROME

(AIDS)

─ caused by HIV;
─ the virus infects and destroys cells of the body‘s immune system (T-cells that control the body‘s
immune response to infection) so that their numbers gradually decrease;
─ when their numbers are low, the body is unable to defend itself against infection, so allowing a
range of parasites to cause a variety of different infections (known as opportunistic
infections);
─ AIDS is not a disease but it is a collection of rare opportunistic diseases associated with
immunodeficiency caused by HIV infection;
─ since HIV is an infective agent; AIDS is called an acquired immunodeficiency to distinguish it
from other types, for example an inherited form;

Structure of the HIV virus

─ the outer envelope contains two glycoproteins gp120 and gp 41;

─ the protein core contains the genetic material (RNA) and two enzymes, protease and
reverse transcriptase;
─ reverse transcriptase uses the RNA as a template to produce DNA once the virus is inside
a host cell;

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Transmission

─ in semen and vaginal fluid during sexual intercourse;

─ infected blood or blood products;
─ contaminated hypodermic syringes;
─ mother to fetus across placenta;
─ mother to infant in breast milk;
─ HIV is a virus that is spread by intimate human contact: there is no vector and the virus is
unable to survive outside the human body;
─ sexual intercourse is the main method of transmission;
─ the initial epidemic in North America and Europe was amongst male homosexuals who had
many sex partners and practised anal intercourse;
─ the mucus lining of the rectum is not as thick as that lining of the vagina;
─ it is often damaged during intercourse and the virus passes from the semen to the blood;
─ as many homosexuals were blood donors and also had heterosexual relationships, the virus
spread more widely;
─ at high risk of infection were haemophiliacs who were treated with clotting substance (factor 8)
isolated from blood pooled from many donors;
─ the transmission of HIV by heterosexual transmission is rising world wide

Viral Replication/Multiplication

─ the virus binds to receptors present in the surface of the T4 lymphocytes;

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 ─ from here it enters the lymphocytes by endocytosis or by fusing with the cell surface
membrane and injects its viral RNA directly into the cell;
─ the viral RNA is then copied into DNA by the activity of the enzyme reverse transcriptase;
─ the viral DNA enters the lymphocyte nucleus and becomes incorporated into the cell‘s own
DNA;
─ thus it becomes a permanent part of the cells of an infected individual;
─ every time the human cell divides, so does the viral DNA, and thus spread of the viral genes is
rapid;
─ the viral DNA may remain dormant for at least six years, the so-called latency period;
─ however suddenly, for some unknown reason, the lymphocytes begin to make some copies of
the viral genes in the form of mRNA;
─ these then migrate from the nucleus into the lymphocyte cytoplasm and direct the synthesis
of viral proteins and RNA;
─ these assemble to form the new HIV viruses which leave the lymphocyte by budding out from
underneath the cell surface membrane;
─ the viruses spread and infect many other lymphocytes and brain cells;
─ eventually the cells in which the virus has multiplied and killed;

HIV and Immunity

─ lymphocytes are very important white blood cells in the maintenance of normal
immunity
─ there are 2 types of lymphocytes in circulation- T and B lymphocytes
─ B-lymphocytes are responsible for cellular immunity
─ T- lymphocytes have a cluster of differentiation (CD) molecules receptors and co-
receptors
─ HIV attacks and destroys CD4 T- lymphocytes
─ HIV also attaches to CNS, gut, and lymph nodes
─ with the fall in CD4 lymphocyte count the individual becomes prone to opportunistic
infections
HIV INFECTION PROGRESSION

─ the phases of HIV infection include;

─ HIV infection
─ Window
period
(iii)Seroconv
ersion
─ Asymptomatic HIV
─ HIV /AIDS related illnesses
─ AIDS
1. HIV Infection

- initial infection with HIV

2. Window period

─ time lag between infection and detection of antibodies;

─ rapid multiplication of virus;
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 ─ person highly infectious;
─ no signs and symptoms of disease and no detectable HIV antibodies;
─ last 2-6 weeks or occasionally months;

3. Seroconversion (antibody – positive phase/ HIV-positive phase)

─ the development of antibodies to HIV;

─ part of the immune response;
─ when people develop antibodies to HIV, they "seroconvert" from antibody-negative to
antibody-positive;
─ a brief phase occurring 2-6 weeks up to a few months of exposure;
─ antibodies develop;
─ may be accompanied by flu-like illness, such as fever, head ache, muscle and joint aches,
sore throat, rash and diarrhea or rarely encephalitis with severe headache;
─ may be called ACUTE HIV SYNDROME;

4. Asymptomatic HIV

─ lasts from one year to 10-15 years or more;

─ antibodies present but no apparent symptoms;
─ this is the incubation period which may be accompanied by persistent generalized
lymphadenopathy (PGL) lasting for a long time without other disease symptoms;

5. HIV related illnesses (AIDS –related complex)

─ lasts months to years;

─ signs and symptoms increase because HIV is damaging the immune system;
─ the individual may contract a variety of conditions known as opportunistic infections;
─ symptoms are not life-threatening but become more serious and long lasting;
─ common bacterial, viral and fungal infections occur and are often noted for their
persistence and virulence;
─ oral and genital herpes or athlete‘s foot are common examples;
─ if a person goes into ARC the duration of this type of infection is lengthened compared with
that in a normal healthy person;
─ loss of weight may be seen at this stage;
─ a significant drop occurs in the number of T helper cells;
─ appropriate nursing is required since this stage is the first real onset of the disease
diagnosis;

6. AIDS

─ lasts usually less than one year unless treatment is available;

─ terminal stage;
─ life threatening infections and cancers occur because the immune system is severely
weakened and cannot cope;
─ life expectancy depend on:
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 ─ the condition that develops;
─ availability of treatment including ARVs, drugs for the treatment of other opportunistic
infections;
─ good nutrition;
─ availability of psychosocial support and holistic care;

Factors that affect progression

─ infection with different types of HIV;

─ natural genetic differences in individual immune system;
─ stress on the immune system through a general lack of fitness and exposure to repeated or
severe infections with different organisms;
─ repeated STIs that keep the immune system highly active and so appear to speed up HIV
replication;
─ state of mind;
─ other stressors such as overtiredness, poor diet, under nutrition and heavy drinking of
alcohol;

Global distribution

worldwide especially in Africa and South- East Asia;

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Explain the possible reasons for the global distribution of HIV/AIDS

─ is a pandemic disease (globally distributed)

─ an epidemic (always present)
─ most prevalent in developing countries
─ linked to TB
─ some of the TB strains becoming more resistant
─ AIDS pandemic
─ partial treatment of due to inability to purchase ARVs
─ poor medical facilities
─ Highly confined in sub-Saharan Africa
─ Rates of infection are lower in other parts of the world, but different subtypes of the virus
have spread to Europe, India, South and South-East Asia, Latin America, and the Caribbean.
Rates of infection have leveled off somewhat in the United States and Europe.
─ In Asia the sharpest increases in HIV infections are found in China, Indonesia, and Vietnam.
─ Both the cost of these therapies and the poor health care delivery systems in many affected
countries need to be addressed before antiretrovirals can benefit the majority of people
living with HIV/AIDS.

Prevention and treatment of HIV/AIDS

─ AIDS is caused by a virus and while bacteria can be controlled by antibiotics, these are not
effective against viruses;
─ most treatments are therefore limited to relieving symptoms;
─ present research on treatment and prevention is concentrating on three areas;
─ restoring or improving the damaged immune system;
─ developing drugs that will stop the growth of the virus and also treat the other infections
and symptoms that result from HIV infection;
─ developing a vaccine against the virus;
─ development of drugs
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The other obvious precautions which can be followed in trying to prevent the disease are:

─ use of a barrier during intercourse can prevent the virus from infecting through blood or
semen. Thus the use of sheath or condom is recommended.
─ restriction to one sexual partner and the absence of promiscuity will also clearly reduce the
risk of infection;
─ use of clean needles and syringes by drug addicts;
─ testing blood donated for the presence of antibodies to HIV which indicates whether or not the
donor is infected; blood containing these antibodies is not used;
─ educating people about the disease particularly in reassuring the public about the real risks.
─ contact tracing
─ needle exchange schemes operate in some places to exchange used needles for sterile needles
to reduce the chances of infection with HIV and other blood borne diseases;
─ in developed countries, blood collected from blood donors is routinely screened for HIV and
heat treated to kill any virus;
─ people who think they may have been exposed to the virus are not encouraged to donate blood;
─ HIV+ women are advised not to breast feed their children because of the risk of transmitting the
virus to their child because both viral particles and infected lymphocytes are found in breast
milk ;

Social and economic factors in the prevention of HIV AIDS

─ HIV transmitted by sexual intercourse with infected person/mother to child;

─ education on ways of preventing transmission is important;
─ education in work places on methods of preventing the spread of HIV;
─ education on use of condoms to reduce risk of infection during intercourse
─ sharing unsterilized needles among illicit drug users;
─ injecting drug users can be advised not to share needles;
─ sharp instruments like razors/shaving machines must not be shared;
─ HIV+ not willing to disclose status/unwilling to go for testing;
─ fear of stigmatization by society
─ ARVs available but not affordable to all;
─ prostitution due to poverty;
─ polygamy/promiscuity;
─ Government to inject more money in awareness campaigns;
─ HIV+ women (in developing countries) can be advised not to breastfeed;
─ HIV+ women to take antiretroviral drugs (navirapine) before delivery;
─ Blood collected from blood donors routinely screened for HIV and heat treated to kill any
viruses;

Describe why vaccination managed to eradicate small pox but not malaria

Small pox

─ varilosa virus stable

─ (harmless) strain of (live) vaccine effective
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 ─ vaccine could be kept for a long time (6 months)
─ infected people were easy to identify
─ ring vaccination was possible
─ political stability during that time
Malaria

─ no vaccine/no effective vaccine against the protozoan

─ resistance of Plasmodium to drugs
─ resistance of vector/mosquitoes to DDT/deldrin/insecticide
─ difficulty of mosquito control
─ expensive to expand the programme
─ civil wars disrupt the programmes
Measles

─ caused by an RNA virus

─ viruses are intracellular pathogens
─ antigenic concealment
─ have a short time in the blood
─ major changes in the epitate as a result of mutation
─ poor response to vaccine (children do not respond well to one dose of vaccine)
─ deficiency immune system
─ or protein energy malnutrition
─ need several boosters which are expensive
─ high birth rates and flighting populations make it difficult for boosters
─ follow up cases and trace contacts also impossible
─ refugees and immigrants from reservoirs of the infection
─ it is highly infectious resulting in the whole population requiring vaccination which is highly
expensive
─ the virus is of hiring attenuated virus can be virulent
Tuberculosis

─ some strains of TB bacteria resistant to drugs;

─ the AIDS pandemic;
─ poor housing and rising homelessness in inner cities in the developed world;
─ the breakdown of TB control programmes particularly in the USA;
─ partial treatment for TB increases the chance of drug resistance in Mycobacterium;
─ attacks many of the poorest and socially disadvantaged because it is spread by airborne
droplets;
─ so people who are overcrowded are particularly at risk;
─ those with low immunity particularly because of malnutrition or being HVI+ are also
vulnerable;
─ transmission is easily achieved but the bacteria may remain in the lung, or in the lymphoid
tissue for years until they become active;
Cholera

─ V. cholerae in intestine;
─ out of reach of immune system;
─ antigenic concealment;
─ antibodies broken down in intestine;
─ antibodies are proteins;
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 ─ ref to pH and effect on structure or shape; e.g. in the stomach
─ denaturation;
─ vaccine stimulates antibodies in, blood / lymph;
─ not in gut;
─ oral vaccine needed;
─ mutation;
─ different strain idea;
─ AVP; e.g. not required in developed countries
─ developing countries cannot afford to develop vaccines
─ no / limited, demand
─ cholera can be treated with ORT
─ can be treated with antibiotics

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TOPIC 5: GENE TECHNOLOGY
KEY CONCEPT OBJECTIVES CONTENT SUGGESTED SUGGESTED
Learners should be (ATTITUDES, SKILLS LEARNING RESOURCES
able to: AND KNOWLEDGE) ACTIVITIES AND
NOTES
8.9.1 Insulin  outline the - Steps involved in  Illustrating genetic Paper and scissors
Production synthesis of human the production of engineering using models
insulin by bacteria human insulin by paper and scissors.
bacteria  Conducting
educational tours to
Biotechnology  ICT tools
laboratories.  Braille
software/Jaws

 explain the  Discussing the

advantages of - Advantages of advantages of use
treating diabetics human insulin of insulin from gene
with human insulin produced by gene technology.
produced by gene technology in
technology treating diabetes

8.9.3 Gene Therapy  outline the basis of - Gene therapy  Discussing gene  ICT tools
gene therapy therapy.  Braille
software/Jaws

GENE TECHNOLOGY

Key Definitions
• Gene technology – this term really covers techniques such as genetic engineering, the creation
of genomic libraries of DNA and DNA fingerprinting.
• Genetic engineering – the transfer of a gene from one organism (the donor) to another (the
recipient) e.g. the genes coding for human insulin, growth hormone or the blood clotting factor,
Factor VIII may be removed from human cells and transferred to bacteria.
• Promoter – a length of DNA (usually about 40 bases long) situated next to genes and which
identify the point at which transcription should begin.
• Marker – a gene which is deliberately transferred along with the required gene during the
process of genetic engineering. It is easily recognised and used to identify those cells to which
the gene has been successfully transferred.
• Genetic fingerprinting – the analysis of DNA in order to identify the individual from which the
DNA was taken to establish the genetic relatedness of individuals. It is now commonly used in
forensic science (for example to identify someone from a blood sample) and to determine
whether individuals of endangered species in captivity have been bred or captured from the wild.
• DNA sequencing - the determination of the precise sequence of nucleotides in a sample of DNA
or even a whole genome e.g. the Human Genome Project.

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Steps involved in the genetic engineering of bacteria to synthesise human insulin

1. Human insulin gene must be identified. There are various ways in which this might have been
done, described at the end of this section. What was actually done, in the late 1970’s, was as
follows:
o Insulin-producing cells from human pancreas tissue synthesise large amounts of the
protein, insulin, for which they make large amounts of mRNA. This mRNA has a genetic
code complementary to the key portions (exons) of the human insulin gene. Some of this
mRNA was isolated from such cells.
o The mRNA was incubated with a mixture of free DNA nucleotides and reverse
transcriptase (an enzyme from viruses that use RNA as their genetic material). This
produced a single strand of DNA known as complementary DNA or cDNA, which is a
copy of the informational strand of the human insulin gene.
o The single strand of cDNA was then made double stranded using DNA polymerase, and
cloned to make many cDNA molecules using the polymerase chain reaction (PCR)
2. Additional, non-coding DNA was added to the ends of the cDNA insulin genes so that ‘sticky
ends’ could be produced using restriction enzymes (also called restriction endonucleases).
Restriction enzymes cut DNA at specific basesequences – their restriction site, for example,
EcoR1 cuts at the sequence GAATTC. Some restrictions enzymes leave ‘sticky ends’ (short
lengths of unpaired bases at each cut) as shown below.
GAATTC G AATTC
CTTAAG CTTAA G
The restriction enzyme was chosen so that it would not cut the insulin genes into pieces,
and would leave sticky ends at either end of the gene, shown below.

3. The gene is then transferred to a bacterial plasmid - a small, circular DNA molecule found in
bacteria and separate from the bacteria’s main DNA molecule. The gene was inserted into a
selected plasmid by cutting open the plasmid using the same restriction enzyme that was used to

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 make sticky ends at either end of the cDNA human insulin genes – again, leaving complementary
sticky ends. If the insulin genes and the cut plasmids are mixed, the complementary bases in the
sticky ends will pair up. This may join the gene into the plasmid. (Unfortunately some plasmids
rejoin without gaining the desired gene.) Ligase enzyme is used to re-join the breaks in the
sugarphosphate backbone of the DNA so that the gene is permanently added to the plasmid,
forming recombinant DNA.

4. The plasmids containing the human insulin gene are then transferred to bacterial cells. This is
brought about by mixing the plasmids with bacteria, some of which will take up the plasmids.
The bacteria which take up plasmids containing the human gene are said to have been
transformed. The transformed bacteria are then cloned to produce large numbers of genetically
identical offspring, each containing the recombinant plasmid, and grown on a large scale. Every
time a bacterium divides, it will replicate the human insulin gene. In each bacterium the gene will
be expressed, being transcribed and translated in the bacterium to produce human insulin.
5. The bacterium Escherichia coli has been transformed in this way and has been used since 1982 to
produce human insulin.
6. The steps above are a simplification of the process used to manufacture human insulin using
recombinant DNA. This is partly because it has been done several times, improving the process
each time it has been done as we understand more of the genetic mechanisms involved.Human
insulin is a small protein which does not contain the amino acid methionine, but does have quite a
complex structure, with two polypeptide cut site for same restriction chains, A and B, joined to
one another by covalent disulphide bonds. The presence of two chains means that it has a
quaternary structure, and also that two separate genes are used, one to make each polypeptide. In
order to produce each of these two polypeptide chains separately, the two genes were added into
the lac operon (see below, in the section on promoters) of the Bgalactosidase enzyme of E. coli.
Before the start of the cDNA code for each of the insulin genes was inserted an extra triplet,
ATG. Look in a DNA dictionary (e.g. at
http://users.rcn.com/jkimball.ma.ultranet/BiologyPages/C/Codons.html ) to confirm that this is
the DNA triplet code for methionine. To make sure that transcription stopped at the correct place,
two consecutive stop codes were added at the end of the cDNA for the A chain, and also the
cDNA for the B chain. In each case the triplet TAA was followed by the triplet TAG. Look these
up in a DNA genetic dictionary to confirm that they are stop codes. This had to be done
separately to different plasmids, so that some plasmids the genetically engineered E coli
containing both types of plasmid was grown in the presence of lactose, the lac operon genes were
turned on but instead of producing B-galactosidase, produced some proteins containing the first
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 part of the bacterial protein, followed by methionine and then either the insulin A chain or the
insulin B chain. When these proteins had been separated from the bacteria, they were treated with
cyanogen bromide, which cuts the amino acid sequence at methionine, separating the insulin
chains from the remains of the bacterial protein. When the mixture of A and B chains is treated to
promote formation of disulphide bonds, insulin forms.
7. The latest methods for manufacturing genetically engineered human insulin use eukaryotic yeast
cells rather than prokaryotic bacterial cells. The yeast cells can use eukaryotic promoter
sequences and have Golgi bodies, so that they produce insulin that is released already in the
correct 3-dimensional conformation to achieve maximum activity in humans.
http://www.littletree.com.au/dna.htm Other methods that could have been used to isolate the
insulin gene:
o The amino acid sequence of insulin is known, so a DNA dictionary could have been used,
and synthetic DNA with an appropriate base sequence synthesised.
o The DNA base sequence of the insulin gene has been found during the human genome
project, so a single-stranded DNA probe, radioactive or fluorescent in UV light, could
have been made, complementary to part of the insulin gene. If the DNA from human cells
was cut up into fragments using restriction enzymes, denatured into single stands by
heating, separated depending on mass using electrophoresis and then treated with the
probe, the probe would stick only to the DNA fragments containing the insulin gene,
allowing the insulin gene to be isolated from the rest of the DNA. Such methods have
been widely applied to isolate other genes for genetic engineering.

The advantages of treating diabetics with human insulin produced by gene technology

Until bacteria were used to produce human insulin, people with insulin-dependent diabetes were
injected with insulin derived from pigs or cattle. Although this type of insulin works in the
human body, pig or cow insulin does not have exactly the same primary structure as human
insulin, so its amino acids sequence, while similar to human insulin, is not identical.

There are a number of advantages of using the human insulin produced by genetically engineered
bacteria:
1. it is chemically identical to the insulin that would have been produced had they not been diabetic,
so there is little chance of an immune response
2. because it is an exact fit in the human insulin receptors in human cell surface membranes, it
brings about a much more rapid response than pig or cow insulin,
3. like natural human insulin, the duration of the response is much shorter than pig or cattle insulin,
4. it overcomes problems related to the development of a tolerance to insulin from pigs or cattle,
5. it avoids any ethical issues that might arise from the use pig or cattle insulin, for example,
religious objections to the use of pig insulin or objections from vegetarians to the use of animal
products.

Why promoters need to be transferred along with the desired genes

In the DNA of bacteria and other prokaryotes, base sequences called promoters are situated just
before (‘upstream’ of) each gene. These identify the point at which transcription should begin.
Usually, these consist of two short six base sequences,TATAAT, situated about 10 bases before
the gene and, TTGACA, situated about 35 bases before the gene. The presence of at least one of
these is usually necessary to initiate transcription of the gene in prokaryotes.
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In the case of insulin, the first successful recombinant DNA involved using the promoter of an
existing non-essential gene, for an enzyme involved in lactose metabolism (B-galactosidase).
The human insulin gene was inserted into the existing gene. The promoter for this gene remained
intact. There is also a lactosesensitive regulatory sequence that is designed to turn on the natural
B-galactosidase in the presence of lactose. The promoter, regulator and gene, are together called
an operon, in this case the lac-operon. The effect of all this is that when the genetically
engineered E coli, containing the human insulin gene in its lac-operon, was exposed to lactose, it
transcribed a polypeptide that contained the first part of the Bgalactosidase, followed by human
insulin.

Now that more is known about prokaryote promoters, synthetic DNA can be made, rather than
trying to make use of natural promoters in this way. In eukaryotes, the regulation of gene
expression is considerably more complex, and so eukaryote promoters may well not have the
intended effect in prokaryotic cells.
What this means in practice, is that if a gene, such as the human insulin gene, is transferred into
prokaryote DNA without adding a prokaryotic promoter, it will not be transcribed and hence will
not be expressed. When genes are transferred from eukaryotes to prokaryotes, it is therefore
essential that a suitable prokaryote promoter is added to the gene before it forms recombinant
DNA with the plasmid vector. The promoter initiates transcription of the gene so that the desired
product is expressed. If eukaryote promoters are to be transferred with eukaryotic genes, into
eukaryotic cells of a different species, then care must be taken to ensure that all of the relevant
code is included, which may include short base sequences close to the start of the gene (such as
TATAAA, [TATA box] within 50 bases of the start of the gene, promotes mRNA formation) or
other sequences further away from the gene (such as CACGTG [E box] which binds proteins
needed for transcription) some of which may cause the DNA to bend back on itself, so that the
promoter is several thousand bases before the gene. http://en.wikipedia.org/wiki/Promoter

Why fluorescent markers (or easily stained substances) are now used instead of antibiotic
resistance markers

When plasmids containing the human insulin gene are mixed with bacteria, only a small
proportion of the bacteria will actually take up the plasmids – this may be as low as 1%. There
needs to be some way of identifying those bacteria which have taken up the gene, so that they
can be separated from those that have not. The first methods used were based on antibiotic
resistance markers. Not all the bacteria in the culture will successfully take up the plasmid, and
not all the plasmids in the mixture will have successfully formed recombinant DNA containing a
viable copy of the cDNA insulin gene. The method used to identify the bacteria containing
the desired recombinant DNA is:
o The original selected plasmid has antibiotic resistance genes to two different antibiotics,
antibiotic A and antibiotic B. Any bacterium containing this plasmid will grow
successfully in the presence of these two antibiotics, but bacteria lacking the plasmid will
be killed by the antibiotics.
o The restriction enzyme is selected so that it cuts in the middle of one of these antibiotic
resistance genes, in this case the gene for resistance to antibiotic B. If a successful
recombinant is formed, this one antibiotic resistance gene will no longer work because it
is interrupted by the cDNA insulin gene.
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 o Bacteria that have taken up the plasmid all have a successfully working copy of antibiotic
resistance gene A. Many plasmids also have a working copy of antibiotic resistance gene
B, showing that the plasmids have failed to form recombinant DNA. However, those
bacteria that have taken up recombinant plasmids containing the cDNA insulin gene do
not contain a working copy of the antibiotic resistance gene B – which gives a way to
identify them as follows:
o The bacteria are spread out and cultured on an agar plate containing antibiotic A. Only bacteria
that have taken up the plasmid survive and reproduce to form colonies, each of which is a clone,
genetically identical to the original cell.
o A sponge is then touched briefly onto the agar, picking up some of the bacteria from each colony.
The original agar plate with the colonies is carefully refrigerated to preserve it. The sponge is
then touched briefly onto a sterile agar plate containing antibiotic B. Bacteria containing
recombinant DNA will be killed by this antibiotic, so that their location on the original plate is
now known.
o One potential problem with using antibiotic markers in this way is that they are present on
plasmids, which are commonly transferred between bacteria of the same species and also of
different species. This means that if the genetically engineered bacteria come into contact with
pathogenic bacteria (e.g. pathogenic strains of E. coli or even pathogens that cause TB or cholera)
the plasmid, with its antibiotic resistance genes, could be transferred into the pathogen, giving it
instant resistance to the antibiotics involved. If this did happen, it would then become much more
difficult to control the spread of such bacteria by using these antibiotics.

There is no evidence that such a transfer has ever happened so the risk is a hypothetical one.
This contrasts with the known damage caused by routine misuse of antibiotics which selects for
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naturally resistant bacteria very strongly. The potential risks led to development of alternative
methods of detecting successful genetic engineering. One method, used for example in genetic
manipulation of papaya, was to incorporate a markergene for a protein that fluoresces green
under ultra-violet light, along with the desired genes. The genes were added, as is now common
in plants, using a micro-projectile to shoot them into the plant cell nuclei. Compared to antibiotic
resistance markers, the process has been found to be both quicker and to produce a higher
proportion of transformed plants. A commonly used fluorescent protein gene comes from
jellyfish. Another approach is to incorporate alongside the desired gene, another marker gene
that produces a harmless product that is easily stained and is not normally produced by the cells.
An example of this is the gene for ß-glucuronidase (GUS) which produces a harmless product
that is easily stained blue. This can be made even safer by linking it to a promoter incorporating
a regulator requiring the presence of an unusual material to turn on the gene, which is thus only
expressed in the peculiar circumstances of the test. The DNA for the gene and the chemicals
required to detect the easily stained product are now widely available and have been used, for
example, in detecting successful transformation of fungi.

Activity
1 (a) Describe the use of recombinant DNA technology in the synthesis of human insulin
by bacteria [9]

 mRNA coding for insulin/isolate gene for human insulin;

 from beta cells of islets of Langerhans/pancreas;
 reference to reverse transcriptase;
 to cDNA;
 reference PCR/DNA polymerase/double strand;
 reference sticky ends/AW;
 use of vector/virus/plasmid;
 reference endonuclease/restriction enzymes;
 to cut plasmid;
 reference DNA ligase to join DNA;
 inserted into suitable host cell/E.coli/bacteria;
 reference method of insertion;
 identification of modified bacteria;
 reference growth/culture of engineered bacteria in fermenters; 9 max

(b) Explain the advantages of treating diabetics with human insulin produced by genetic
engineering [6]

 constant/reliable supply all year round/unlimited supply;

 less risk of contamination/infection;
 identical to insulin produced in the body;
 less/no risk of allergic reaction;
 does not stimulate the immune system;
 fewer side effects;
 can be produced without the killing of animals/ethical reason;
 cheaper/easier to extract and purify;
 more available/large amount;
 more rapid response; 6 max.
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2. Fig. 2.1 outlines the way in which the gene for human insulin is incorporated into
plasmid DNA and inserted into a bacterium.

(a) Describe how the plasmid DNA is cut. [3]

 restriction (endonuclease) enzyme ;
 named example e.g. EcoR1 ;
 specific sequence of bases ;
 ref. to sticky ends / exposed bases ;
(b) Explain how the human insulin gene is joined to the plasmid DNA. [3]
 ref. to complimentary base pairing ;
 of sticky ends ;
 ligase ;
 formation of phosphodiester bond ;

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 (c) List two advantages of treating diabetics with human insulin produced by genetic
engineering. [2]
 identical to human insulin (ref. to bovine / porcine insulin used previously) ;
 ref. to possible immune response ;
 easier to extract ;
 pure / uncontaminated ;
 regular production not dependent on livestock ;

Fig. 4.1 shows some of the steps involved in the production of bacteria capable of
synthesising human insulin.

State the role of each of the following enzymes in the production of bacteria capable of
synthesising human insulin,
reverse transcriptase
DNA polymerase
restriction enzymes (restriction endonucleases)
DNA ligase
[6]
reverse transcriptase
 makes, cDNA / single strand of DNA ;
 from (human) mRNA ;
DNA polymerase
 produces, second strand of DNA / double stranded DNA ;
 ref. links nucleotides (in context of backbone formation) ;
 ref. semiconservative replication / ref. complementary base pairing ; [max 2]
restriction enzymes
 cut DNA / cut plasmid ; R cuts gene A cuts out gene
 at specific sites / at palindromic sites ;
 to give sticky ends ; A blunt ends [max 2]
DNA ligase
 seals nicks in sugar-phosphate backbone ;
 forms rDNA ;
 by adding phosphate group ; [max 2]

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The secretion of insulin by the islets of Langerhans in the pancreas stimulates the liver
to reduce the blood glucose concentration.
(a) Describe how the liver reduces blood glucose concentration, when insulin issecreted.
(b) Almost all insulin used to treat type I diabetes is produced by genetically engineered
bacteria or yeast. A summary of this procedure is shown in Fig. 4.1.

(i) One way of carrying out step 1 is to collect mRNA from cells from the pancreas.
The relevant mRNA is then isolated and used to make DNA.
Suggest why isolating the mRNA coding for insulin in a cell is easier than isolating
the DNA for insulin in a cell. [2]
(ii) Outline the use of restriction enzymes in step 2. [2]

(a)
 binds to receptors (on liver cell membranes) ;
 conversion of glucose to glycogen / glycogenesis ;
 (because) insulin activates enzyme ; e.g. glucokinase / phosphofructokinase /
 glycogen synthase
 increased use of glucose in respiration ;
 increased uptake of glucose / increased permeability to glucose (of liver cells) ; [3 max]
(b) (i)
 mRNA (found in β cells) is only from gene coding for insulin / AW ;
 large numbers (of mRNA coding for insulin) ;
 (whereas) DNA has all genes ;
 (so) restriction enzymes needed ; [2 max]
(ii)

 cut plasmid (DNA) ;

 at specific, base sequence / site ;
 leaving sticky ends (that will join with insulin gene) ;
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The benefits and hazards of gene technology

KEY CONCEPT OBJECTIVES CONTENT SUGGESTED SUGGESTED

Learners should be (ATTITUDES, SKILLS LEARNING RESOURCES
able to: AND KNOWLEDGE) ACTIVITIES AND
NOTES
8.9.4 Benefits and  explain the benefits - Gene technology  Discussing benefits  ICT tool
hazards of Gene and hazards of - Its benefits and and hazards of gene  Braille
Technology gene technology hazards technology. software/Jaws

Benefits
Through gene technology, it is now possible to produce:
• genetically modified organisms for a specific purpose. Previously, such genetic change would
have to be brought about by selective breeding which requires organisms to be of the same
species (able to breed successfully together), takes many generations and involves transfer of
whole genomes, complete with undesirable background genes. Gene technology is much faster
and involves transferring one or few genes, which may come from completely unrelated
organisms, even from different kingdoms.
• specific products, such as human insulin and human growth hormone, thereby reducing the
dependence on products from other, less reliable sources, such as pig or cow insulin.
• reduce use of agrochemicals such as herbicides and pesticides since crops can be made resistant
to particular herbicides, or can be made to contain toxins that kill insects
• clean up specific pollutants and waste materials – bioremediation
• potential for use of gene technology to treat genetic diseases such as cystic fibrosis (see below)
and SCID (Severe Combined Immune Deficiency) as well as in cancer treatment.
Hazards
Genes inserted into bacteria could be transferred into other bacterial species, potentially
ncluding antibiotic resistance genes and those for other materials, which could result in antibiotic
resistance in pathogens, or in bacteria that can produce toxic materials or break down useful
materials. Regulation is designed to minimise the risks of escape of such genes. There is little
evidence that such genes have escaped into wild bacterial populations. Crop plants have, by their
nature, to be released into the environment to grow, and many millions of hectares of genetically
engineered crops, both experimental and commercial, are planted across the globe. So far, fears
that they might turn out to be ‘super-weeds’, resistant to herbicides and spreading uncontrollably,
or that their genes might transfer into other closely related wild species, forming a different kind
of ‘super-weed’, or that they might reduce biodiversity by genetic contamination of wild
relatives seem to have proved unfounded. A paper was published in Nature in 2001 showing that
Mexican wild maize populations were contaminated with genes from genetically manipulated
maize, but the methods used were flawed and subsequent studies have not confirmed this
contamination, suggesting that the wild maize is not genetically contaminated. There is some
evidence that Bt toxin, geneticially engineered into plants such as cotton and maize, whilst very
effective in killing the target species, may kill other, desirable, insects such as bees and
butterflies, and may also cause natural selection of Bt toxin resistant insects. Future events may
show that such environmental risks are greater than they look at present. Food that is derived
from genetically engineered organisms may prove to be unexpectedly toxic or to trigger allergic
reactions when consumed. There is little reliable evidence that this has been so, but the risk
remains. Food containing the expressed products of antibiotic resistance marker genes could be

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consumed at the same time as treatment with the antibiotic was occurring, which would
potentially reduce the effectiveness of the treatment. No examples of this are known.

The social and ethical implications of gene technology

KEY CONCEPT OBJECTIVES CONTENT SUGGESTED LEARNING

Learners should be able to: (ATTITUDES, SKILLS AND ACTIVITIES AND NOTES
KNOWLEDGE)
8.9.5 Ethical implications of  discuss the social and - Social and ethical  Researching and debating
Gene Technology ethical implications of implications of gene on the social and ethical
gene technology technology implications of gene
technology.

The social impact of gene technology is to do with its potential and actual impact of human
society and individuals. In terms of social impact, gene technology could:
o enhance crop yields and permit crops to grow outside their usual location or season so
that people have more food
o enhance the nutritional content of crops so that people are better fed
o permit better targeted clean-up of wastes and pollutants
o lead to production of more effective and cheaper medicines and treatments through
genetic manipulation of microorganisms and agricultural organisms to make medicines
and genetic manipulation of human cells and individuals (gene therapy)
o produce super-weeds or otherwise interfere with ecosystems in unexpected ways,
reducing crop yields so that people have less food
o increase costs of seed and prevent seed from being retained for sowing next year (by
inclusion of genes to kill any seed produced this way) reducing food production
o reduce crop biodiversity by out-competing natural crops so that people are less well fed
o damage useful materials such as oil or plastic in unexpected ways
o cause antibiotics to become less useful and cause allergic reactions or disease in other
unexpected ways
The ethical impact is about the application of moral frameworks concerning the principles of
conduct governing individuals and groups, including what might be thought to be right or wrong,
good or bad. So in the context of gene technology, it is to do with issues of whether is right or
wrong to conduct research and develop technologies, whether it is good or bad. Judgements may
be that
o It is good to conduct such research to develop technologies that might improve nutrition,
the environment or health
o It is good to use the results of such research to produce food, to enhance the environment
or improve health
o It is wrong to continue such research when the potential impact of the technology is
unknown and many aspects of it remain to be understood.
o It is wrong to use the results of such research even when the organisms are kept in
carefully regulated environments such as sterile fermenters as the risks of the organisms
or the genes they contain escaping are too great and unknown
o It is wrong to use the results of such research when this involves release of gene
technology into the environment as once it is released it cannot be taken back – the genes
are self-perpetuating, and the risks that they might cause in future are unknown

The social and ethical implications of gene technology are complex and relatively unfamiliar to
people who are not scientists, including those involved in the media and in government. This

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complexity and unfamiliarity is the cause of considerable concern and debate. In considering the
implications of gene technology the best approach is to avoid the general (e.g. avoid ‘it is bad to
play God’) and stick to the specific and balanced (e.g. it is possible to increase food crop yields
with gene technology so more people can be fed, but there is enough food already if it is properly
distributed, so people should not be forced to eat products with unknown risks).

The use of electrophoresis in genetic fingerprinting and DNA sequencing

KEY CONCEPT OBJECTIVES CONTENT SUGGESTED SUGGESTED

Learners should be (ATTITUDES, SKILLS LEARNING RESOURCES
able to: AND KNOWLEDGE) ACTIVITIES AND
NOTES
8.9.2 Genetic  describe how - Genetic screening  Discussing how  ICT
Screening and genetic screening genetic screening is  Braille software/Jaws
Fingerprinting is carried out carried out.  Ink pads
 Bond paper
 discuss the roles of - Roles of genetic  Discussing the  Hand lense
genetic screening screening roles of genetic
for genetic screening.
conditions and
need for genetic
counselling
- Genetic
 explain the fingerprinting  Observing
theoretical basis of simulations of
genetic electrophoresis
fingerprinting process.
 Discussing genetic
 outline how the fingerprinting.
process of genetic
fingerprinting is
carried out

Electrophoresis
Electrophoresis is a method of separating substances and analyzing molecular structure based on
the rate of movement of each component in a liquid medium while under the influence of an
electric field. In genetic fingerprinting and DNA sequencing, the components being separated are
fragments of DNA.

In this case, the type of electrophoresis used is gel electrophoresis – the gel appears solid but is
actually a colloid in which there are spaces between the molecules through which other
molecules can move. Electrodes are placed at either end of the gel, as a result of which the DNA
molecules move under the influence of an electric current. Usually the DNA is fragmented (cut
across) into a series of fragments using a restriction enzyme or mixture of restriction enzymes.
These enzymes cut the DNA at specific restriction sites (see above), but these sites are randomly
distributed along the length of the DNA so the fragments are of varied lengths.

The direction of movement depends on the fact that DNA molecules and fragments of DNA are
negatively charged and thus move towards the positive electrode (anode). The distance moved in
a given time will depend on the mass of the molecule of fragment. The smaller fragments move
further in a given time, and the larger fragments of DNA move less far. Taking humans as an
example, almost everyone has 46 chromosomes: 23 pairs if you are female and 22 pairs plus two
odd ones if you are male. The longest of these kinds of chromosomes has been numbered as
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chromosome 1 and the smallest as 22, the sex chromosomes being out of sequence and called X
and Y. The base sequence of every chromosome 1 in every human being is similar, but not
identical due to the existence of mutations and therefore of different alleles of genes. What this
means is that when the DNA is fragmented with a restriction enzyme, the fragments are similar
but not exactly the same in DNA from different people. The DNA is transparent and invisible, so
the fragments must be treated to make them visible. There are two key ways of doing this:
o One is based on staining all of the DNA fragments, for example using ethidium bromide
(toxic, fluoresces in short wave UV radiation), methylene blue (fades quickly and stains
gel as well as DNA)and nile blue A (does not stain gel and visible in ordinary light).
o The other is based on creating a gene probe that is complementary:
o either to a commonly repeated bit of DNA that will therefore be present on many of the
fragments,
o or to a base sequence that is specific to a particular gene or allele of a gene which will
therefore be present on no more than one of thefragments. The gene probe is a single
stranded piece of DNA with a base sequence complementary to the DNA that you wish to
identify.
In order to make it possible to locate which fragment or fragments the gene probe has attached itself to,
the gene probe must be labelled. The most common forms of labelling are:

o to make the probe radioactive and to detect it by its ability to expose the photographic
film used to make X-ray photographs
o to stain the probe with a fluorescent stain such as vital red, that will fluoresce with bright
visible light when placed in ultraviolet light, making the location of the probe and
therefore of the fragment or fragments visible.

Genetic fingerprinting
Once the DNA fragments have been separated by gel electrophoresis they can be compared with
other samples of DNA, thereby allowing determination of the source of the DNA (as in forensic
investigations) or whether the samples are derived from related individuals, as shown below:

DNA sequencing
The most publicised example of DNA sequencing is the Human Genome Project.
Electrophoresis is used to separate fragments of DNA to enable determination of the order of
bases within genes and chromosomes. The fragments vary in length by one base at a time and the
last base on each can be identified. Because the fragments are different lengths, they can be
separated by electrophoresis as shown below:

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The causes and symptoms of Cystic Fibrosis

Cystic Fibrosis (CF) is a genetic condition in humans. It is inherited and although it reduces
considerably the life expectancy of people with the condition, improved treatments have been
helping such people to live longer so that the average life-span is now about 35 years. There are
estimated to be around 50,000 people with CF worldwide.
Causes
Cystic fibrosis is caused by several different alleles of a key gene coding for a transmembrane
protein that transports chloride ions through cell surface membranes (cystic fibrosis
transmembrane regulator, CFTR). Its inheritance is autosomal (i.e. it is NOT sex-linked) and
recessive. The gene is located on chromosome 7. CF alleles originate by mutation of the CFTR
protein, but can then be inherited through many generations.

As CF alleles are recessive, individuals with a single copy of such an allele are heterozygous and
do not have the condition. There are about 10 million such carriers worldwide.

To have CF, it is necessary to be homozygous for CF alleles, most often by inheriting

one CF allele from each parent.

Effects of CF
Reduced chloride transport through cell membranes leads to production of thick,sticky mucus
that particularly affects the lungs, pancreas and reproductive organs.
o The mucus remains in the lungs rather than being swept out by the tracheal cilia, leading
to wheezing and repeated infections. The mucus may be removed by physiotherapy
o The mucus may block the pancreatic duct, preventing amylase and protease enzymes
from reaching the small intestine, compromising digestion and nutrition, and also causing
a build-up of protease in the pancreas, damaging the pancreatic tissue including the cells
that produce insulin, increasing the chance of diabetes.
o The mucus may block the sperm ducts, causing male infertility and may slow the
progress of eggs and sperm through the oviducts, reducing female fertility.

Progress towards treating Cystic Fibrosis with gene technology

Current treatments for CF deal with the symptoms rather than the causes, for example
physiotherapy to remove mucus from lungs, antibiotics to combat recurrent lung infections and
enzyme supplements to enhance digestion. These have been very successful in improving

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people’s quality of life and lifespan, but research continues to try and develop techniques for
adding functional copies of the CFTR gene to the cells of people with CF.

Since it is a recessive condition, such gene therapy does not need to remove or replace the
existing genes in the person’s cells – adding a working copy of the gene to a cell and having it
expressed would be sufficient to permit that cell to transport chloride ions normally. Since it is
the mucus in the lungs that generally limits lifespan in people with CF, it is these cells that have
been the focus of effort. It is thought that if even a proportion of lung cells could be given a
working copy of the gene, this would thin the mucus sufficiently to allow the cilia to operate
normally.
The approach that has been trialled with another recessive genetic condition, SCID, is to remove
cells from the body, add working copies of the gene and put the cells back. The working copies
of the gene integrate themselves into random positions in the genome of the treated cells. The
blood cells involved in this case only live for a few weeks so it has to be frequently repeated. Of
14 boys in one French trial, 3 have developed cancer, probably because the gene has been
inserted into a critical portion of one of the cells at some point. Clearly this approach cannot be
used with CF because the lung surface cells cannot be extracted from the body.

For CF, a vector must be used to deliver the DNA containing the functional CFTR
gene into the lung cells.
o Viral delivery systems – some viruses such as Adenoviruses can be used asthe vector.
Normally, viruses which infect lung cells are used – their virulence (ability to cause
disease) is removed and they are genetically engineered to carry the functional human
CFTR gene. Early trials have involved either injection with the genetically engineered
viruses or inhale them from an aerosol directly into the lungs. The intention is that the
lung surface cells are infected with the virus, which releases the genetic material into the
cells where it is expressed.
o Non-viral delivery systems – other systems are also being developed and have been
trialled for safety but have not been used therapeutically e.g.
1. Creation of a lipid sphere or liposome, containing the DNA. An aerosol is sprayed into the lungs
where the liposome will be able to pass through the target cell membrane and carry the DNA into
the cell.
2. DNA can be compressed into a very small volume which may directly enter cells.

Whether the DNA is introduced into the cells by viruses or some other system, the intention is
that the gene will be incorporated into the cell’s genome and will start to be expressed, to
produce CFTR protein to carry chloride ions through its membrane.

There is not yet a successful example of treatment of CF by gene therapy. This is because:
o current viral vectors have been found to stimulate allergic or other immune responses
o current liposome vectors have proved inefficient at delivering genes into cells
o the effect of the therapy on chloride ion transport has, so far, lasted only a few days
Research continues to solve these problems to develop a workable treatment for lung symptoms.
Further into the future, similar approaches may be possible for pancreatic symptoms. A cure
would require every one of the 50 x 1013 cells in the body to be altered, which is not currently
thought to be technically possible and would raise significant further ethical issues. To enable

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people with CF to have children would require germ-line gene therapy where changes are made
to human gamete cells that are inherited by the next generation. This would also raise very
significant further ethical issues and does not appear to be realistic at present.

Genetic screening and counseling

There are now many conditions known to be caused by varied alleles of varied genes and which
can therefore be inherited. The pattern of inheritance varies, according to whether the allele is
dominant, recessive or sex-linked.

Individuals may be tested for the presence of such alleles – such tests may be requested because
there is a history of a particular condition in the family of that person or because the person
belongs to an ethnic group which has a high percentage of individuals with a particular allele,
such as the alleles that cause Tay Sachs in people who are Ashkenazi Jews.

 Genetic screening: The testing of samples of DNA from a group of people to identify the
presence or absence of particular alleles and thus the risk of having or passing on particular
genetic conditions. Such screening may be:
 Carrier screening- all the individuals in a family may be screened if one family member develops
a particular condition that may be genetic. potential parents may be screened where there is the
possibility that one or both of them might carry a recessive allele for some particular condition
e.g. cystic fibrosis
o Prenatal screening – this is used to determine aspects of the genetic makeup of an unborn
child. Such testing can detect a number of genetic conditions:
o Chromosomal abnormalities, such as Downs Syndrome (of particular importance if the
mother is over 34), trisomy 13 and trisomy 18.
o Single gene disorders, such as haemophilia, sickle cell anaemia and cystic fibrosis
o Neural tube defects, such as spina bifida and anencephaly Pre-natal screening may be
carried out in different ways and at different stages of the pregnancy :
o Chorionic villus sampling – where the early placental tissue is sampled, usually done at
10 – 12 weeks of the pregnancy
o Amniocentesis – where fetal cells in amniotic fluid are sampled, usually done at 13 – 18
weeks of the pregnancy
o Intra-uterine blood test – where fetal blood is sampled, usually done at 16 – 18 weeks of
the pregnancy
o Newborn screening – in some countries, all newborn babies are screened for genetic
conditions such as phenylketonuria (pku) by a simple blood test. This test enables the
affected individual to be put onto a protective diet low in the amino acid phenylalanine,
for the rest of their life, to protect them from the damaging symptoms of the condition.

Once the results of a genetic test are known, it will be necessary for those involved to receive
Genetic Counselling. This will involve an explanation of the results and the implications in terms
of probabilities, dangers, diagnosis, and treatment.
o For the individual – depending on the nature of any detected allele (dominant or
recessive), it will be necessary to explain the possible future consequences in terms of the
health of the individual and whether this is likely to have repercussions on their education
or employment. In some cases, it might affect their prospects of obtaining insurance.
o For couples who want to have children – again, depending on the nature of the
inheritance, it will need to be explained what the probabilities are of any children
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 inheriting the defective allele – and the chances of any child actuallyhaving the disease
i.e. it showing in their phenotype. All of this will depend on whether the allele is
dominant, recessive or sex-linked.
In addition to the practical considerations of genetic screening and counselling, there are also some ethical
considerations :

o Who decides who should be screened or tested?

o Which specific disorders should be screened?
o Who should be providing the screening?
o Should we screen or test for disorders for which there is no known treatment or cure?
o What psychological impact might the results have on the individuals involved?
o Should the results be confidential?
o If not, who should be able to have access to the information?
o Should the results be made available to potential employers, insurers etc.?

(a) Explain how changes in the nucleotide sequence of DNA may affect the amino acid
sequence in a protein. [7]
 code is three, bases / nucleotides ; A triplet code
 (gene) mutation ; R chromosome mutation
 base, substitution / addition / deletion ;
 addition / deletion, large effect (on amino acid sequence) ;
 frame shift ;
 completely new code after mutation / alters every 3 base sequence which follows ;
 (substitution) often has no effect / silent mutation ;
 different triplet but same amino acid / new amino acid in non-functional part of protein ;
 (substitution) may have big effect (on amino acid sequence) ;
 could produce ‘stop’ codon ;
 sickle cell anaemia / PKU / cystic fibrosis ;
 reference to transcription or translation in correct context ; A description
AVP ; e.g. protein produced, is non-functional / not produced / incomplete [7 max]

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Discuss the use of genetic engineering in improving the quality and yield of
(ii) crop plants, [6]
(iii) animals. [6]

(a) (i) Describe the techniques used in embryo transplantation. [7]

(ii) Describe the process of cloning plants from tissue culture. [7]
(iii) Explain the advantages of using these two procedures in selective breeding. [6]
[Total : 20]

Describe the ways by which gene mutations can occur. [6]

 change in, base / nucleotide, sequence (in DNA) ;

 during DNA replication ;
 detail of change ; e.g. base, substitution / addition / deletion
 frame shifts / AW ;
 different / new, allele ;
 random / spontaneous ;
 mutagens ;
 ionising radiation ;
 UV radiation / mustard gas ;
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Outline the need for energy in living organisms using named examples. [9]

 ATP as universal energy currency ;

 light energy needed for photosynthesis ;
 ATP used conversion of GP to TP ;
 ATP used to regenerate RuBP ;
 (energy needed for) anabolic reactions ;
 protein synthesis / starch formation / triglyceride formation ;
 activation energy ;
 (activate) glucose in glycolysis ;
 active transport ;
 example ; e.g. sodium / potassium pump
 movement / locomotion ;
 example ; e.g. muscle contraction / cilia beating
 endocytosis / exocytosis / pinocytosis / bulk transport ;
 temperature regulation ;

Explain the advantages of treating diabetics with human insulin produced by genetic
engineering. [6]

─ constant/reliable supply all year round/unlimited supply;

─ less risk of contamination/infection;
─ identical to insulin produced in the body;
─ less/no risk of allergic reaction;
─ does not stimulate the immune system;
─ fewer side effects;
─ can be produced without the killing of animals/ethical reason;
─ cheaper/easier to extract and purify;
─ more available/large amount;
─ more rapid response;

Describe the use of recombinant DNA technology in the synthesis of human insulin by
bacteria [9]
mRNA coding for insulin/isolate gene for human insulin;
─ from beta cells of islets of Langerhans/pancreas;
─ reference to reverse transcriptase;
─ to cDNA;
─ reference PCR/DNA polymerase/double strand;
─ reference sticky ends/AW;
─ use of vector/virus/plasmid;
─ reference endonuclease/restriction enzymes;
─ to cut plasmid;
─ reference DNA ligase to join DNA;
─ inserted into suitable host cell/E.coli/bacteria;
─ reference method of insertion;
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 ─ identification of modified bacteria;
─ reference growth/culture of engineered bacteria in fermenters;

Explain why mammalian cells are used as host cells in recombinant DNA technology during
production of human factor VIII
─ human genes contain exons and introns
─ mammalian cells have enzymes to remove introns/ enzymes for methylation and splicing of
mRNA;
─ mammalian cells have enzymes / Golgi apparatus for post –translational modification;
─ presence of human regulator genes

State any biological methods other than the use of plasmids; of introducing genes into host cells.
─ Liposome transfer
─ Electroporation
─ Microinjection/ micropipette;
─ a DNA gun fires tungsten or gold particles coated with DNA
─ ballistic impregnation
─ use of vectors

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 A LEVEL BIOLOGY TEACHERS
PRACTICAL TRAINING MODULE

CONTENTS

Background 2

Practical 1: Introduction to Microscopy 2

Practical 2: Demonstration of plasmolysis using red onion skin cells 6

Practical 3: Measuring the water potential of potato tubers 9

Practical 4: Determination of the water potential of solutes 12

Practical 5: Identification of non-reducing sugars 17

Practical 6: Determination of relative quantities of reducing sugars in food items 18

Practical 7: Demonstration of the action of protease (pepsin) on enzymes 21

Practical 8: Demonstration of the effect of substrate concentration on enzyme action 23

Practical 9: Demonstration of regulation and control through dialysis 27

Practical 10: Plant Gross form and Tissues 32

Practical 11: The structure of flowers and adaptation to pollination 34

Practical 12: The basic leaf structure 36

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BACKGROUND

A practical assessment is undertaken for ZIMSEC Advanced level students in Paper 4 which is
written for 2 h, 30 min. The paper is allocated 60 marks and has a weighting of 20%. The paper
comprises of experiments and investigations based on the Core syllabus. The students are
expected to demonstrate planning and implementing skills in carrying out the practical exam.
After that they should demonstrate interpreting and concluding skills on the findings of the
investigations that they would have carried.

Good implementing skills can be demonstrated by carrying out experimental work in a

methodical and organised way with due regard for safety and living organisms. The students
should also use apparatus and materials in an appropriate way. They should be make accurate
and detailed observations of low power and high power drawings of a specimen. They should
also make measurements to the appropriate degree of precision allowed by apparatus.

Interpreting and concluding skills involve assessing the reliability and accuracy of experimental
data and techniques by identifying and assessing errors. Students should apply knowledge to
explain and interpret experimental results to reach valid conclusions. They are also expected to
communicate information, results and ideas in clear and appropriate ways using text, tabulation,
graphs and figures.

This A Level Practical Module has been developed to facilitate learning among students and to
guide teachers in the execution of practicals. The practicals were derived from the Cambridge
and ZIMSEC past examination papers.

The module provided the teacher with selected practicals that can be executed by the students.
For each practical a brief introduction of the concept being investigated is provided, the
objectives and a detailed procedure. A list of materials and equipment required for the practicals
are also provided.

PRACTICAL 1: INTRODUCTION TO MICROSCOPY

INTRODUCTION
Of all the techniques used in biology, microscopy is probably the most important. The vast
majority of living organisms are too small to be seen in any detail with the human eye, and cells
and their organelles can only be seen with the aid of a microscope. Cells were first seen in 1665
by Robert Hooke (who named them after monks' cells in a monastery), and were studied in more
detail by Leeuwehoek using a primitive microscope. The Light Microscope is the oldest,
simplest and most widely-used form of microscopy. Specimens are illuminated with light, which
is focused using glass lenses and viewed using the eye or photographic film. Figure 1 outlines
the parts of a light microscope. All light microscopes today are compound microscope, which
means they use several lenses to obtain high magnification. Light microscopy has a resolution of
about 200 nm, which is good enough to see cells, but not the details of cell organelles.
A light source, which is either external or an integral part of the microscope, is required to
illuminate the specimen. The condenser lens under the stage gathers the diffuse light rays from
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the light source and illuminates the specimen with an intense cone of light. This allows small
parts of the specimen to be seen after magnification. Two sets of light rays enter the objective
lenses. One set has not been altered by the specimen, coming directly from the many points on
the specimen and is brought into focus by the objective lens. Using the focusing controls the user
can change the relative distance between the specimen and the objective lens so that the final
image is in focus on the retina of the eye. The total magnification of the specimen is the product
of the magnification of the ocular lens (usually 10X) and that of the objective lens (10X, 40X
etc).
Specimens can be living or dead, but often need to be stained with a coloured dye to make them
visible. Various dyes/stains are used to enhance the contrast between the object and the
background medium. On the other hand, the most important property for seeing an unstained
object is scattering of light by diffusion. Absorption of light may also enhance contrast. A good
example is the wing of an insect. The thicker parts will absorb more light than the thinner bits,
which will appear brighter. Due to the difference in the transmitted light from the different parts
of the wing an image is seen

Figure 1: Diagram of a typical light microscope, showing the parts and the light path
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OBJECTIVES
The objective of this practical is to familiarize the student with the theory and practice of
operating the compound optical microscope specifically to (i) calibrate (ii) prepare and mount a
specimen for observation and, (iii) measure the size of cells under a light microscope.
MATERIALS AND METHODS/ REQUIREMENTS

1. Purple Onion
2. Light Microscope
3. Eye pierce graticule
4. Forceps
5. Fuse wire of known diameter
6. Slides
7. Cover slips
8. Blotting/filter paper
9. Distilled water
10. Iodine solution
Part A:Calibration of the Ocular Micrometer
As part of this practical you will need to calibrate a scale (an ocular micrometer or reticle) in
the right eyepiece of the microscope so that it can be used to measure the dimensions of different
types of cells. The ocular micrometer scale itself has no inherent units, and because different
objectives produce images with different degrees of magnification, the meaning of its intervals
varies from one objective to the next. The figure below shows a slide with some red blood cells
as seen through an ocular lens fitted with an ocular micrometer. In this case, the scale simply
goes from 1 to 10, and each interval is divided into 10 smaller units.

Figure 2: Shows the view through the objective lens of a microscope with an ocular
micrometer.
Because the ocular micrometer or reticle scale has no inherent units, it is necessary to calibrate
it using a stage micrometer. A stage micrometer is a special microscope slide with a ruler
etched on its surface, which has units of millimeters (mm) and micrometers (μm). To calibrate
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 the reticle, you will line up the stage micrometer with the ocular micrometer and count the
number of units or divisions on the ocular micrometer that corresponds to a particular distance in
millimeters or micrometers on the stage micrometer (Figure 2). The number of ocular units per
millimeter or micrometer will change as the magnification changes.

(a)

(b)
Figure 3: (a) The view through the microscope if the stage micrometer has been aligned
with the ocular micrometer and focused correctly. (b) Shows the scale on a sample stage
micrometer.

Calibration of an eye piece graticule

An eye pierce graticule is a special eye-piece containing an arbitrary scale – scale units which do
not mean anything until they have been calibrated. To calibrate, follow these steps:
1. Using the eye-piece graticule and a piece of fuse wire of known diameter calibrate your
microscope at 100X total magnification i.e. using 10X objective lens.
2. With the eye piece graticule in place, focus on the calibrations of a stage micrometer using the
low power objective.
3. Rotate the eyepiece until the ocular micrometer is aligned with the stage scale.
4. Determine the length of each division by dividing the total length sub-tendend on the stage scale.
Place the slide with the piece of wire of known diameter on the stage microscope. Measure how
many arbitrary eye unit equal the diameter of the wire. Perform a simple calculation and record
your answer which should be in the form:
1 eye piece unit = µm (using the 10X objective ONLY)

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5. Carefully change to 20X and 40X objective lenses and determine the lengths of each division as
in 4 above. Enter your values in a table with magnification and length in µm per division. Show
all your calculation on paper. Comment on the numerical relationship between the calibration
values for the 10X, 20X and 40X objective lenses.

NOTE:
(a) Not all the eye-piece graticules contain exactly the same size scales so you must calibrate the one
that you will be using to make your measurements
(b) The magnification used for doing the calibration must be the same as the one used for making
the measurements. Why?
(c) Now that you have calibrated your eye-piece scale you are now ready to measure the length of
epidermal cells. You will be preparing a onion epidermal cells for observation under the
microscope.

Part B: Preparing a slide of purple onion epidermal cells

The fleshy pieces inside an onion are leaves that store nutrients. Cut open an onion and separate
some of the leaves. Peel off one of the thin layers of epidermal tissue from the inner concave
surface of a leaf and transfer it to a drop of water on a microscope slide. Use forceps and
mounted needle to make sure that the tissue is not folded.
(a) Place two drops of iodine solution onto the tissue. Gently lower a coverslip onto the slide
using a mounted needle. Use a piece of filter paper to absorb excess stain. Place some filter paper
over the coverslip and gently press to flatten the specimen.
(b) Place the slide on the stage of the microscope and use the low-power objective to locate the
cells. Now use the high-power objective to select three adjacent cells that are clearly visible in
your field of view.
(c) Make a large, labelled drawing of these three epidermal cells.
Part C: Measuring the size of the cells
1. Use the calibrated eyepiece graticule to measure the length of one of the epidermal cells that you
have drawn. Now measure the same cell in your drawing.
2. Measure the length of 2-5 epidermal cells in eyepiece units at 100X magnification and calculate
the length in µm. Calculate the mean epidermal cell length.
3. Calculate the magnification of your drawing, using the formula:
𝑚𝑎𝑔𝑛𝑖𝑓𝑖𝑐𝑎𝑡𝑖𝑜𝑛 = 𝑙𝑒𝑛𝑔𝑡ℎ 𝑜𝑓 𝑑𝑟𝑎𝑤𝑖𝑛𝑔 𝑜𝑓 𝑐𝑒𝑙𝑙/ 𝑎𝑐𝑡𝑢𝑎𝑙 𝑙𝑒𝑛𝑔𝑡ℎ 𝑜𝑓 𝑐𝑒𝑙𝑙
NB. The lengths must be measured in the same units e.g. micrometers (µm)
Some useful definitions in Microscopy

Magnification, the degree of enlargement of the image of the object compared to its
real size. It is provided by a two lens system; ocular lens (8 or 10X) and objective lens
(4, 10, 40 or 100X). Total magnification is the product of these two magnifying power values.
Resolution, a measure of the clarity of an image. Resolving power is the ability of an optical
instrument to show two objects as separate. For example, what looks to your unaided eye like a
single star in the sky may be resolved as two stars with the help of a telescope. Any optical
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 device is limited by its resolving power. Contrast, is the ability to determine same particular
detail of specimen against its background. In a bright field light microscope, adjustable
condenser with aperture diaphragm control or adding dyes may increase contrast of a transparent
specimen.

The maximum magnification obtained through a light microscope is 400X, in other words the
closest two distinct points can be and still be resolved is 0.2 micrometer (µ m) about the size of
the smallest bacterium (1 µ m = 10-6 m). This limitation is the result of light being diffracted by
the object under observation and because diffracted light interferes with the image.
PRACTICAL 2: DEMONSTRATION OF PLASMOLYSIS USING RED
ONION SKIN CELLS
INTRODUCTION
One of the main functions of the cell membrane is to selectively regulate what passes into and
out of the cell. Molecules can cross the cell membrane by simple diffusion or osmosis.
Diffusion is the random movement of molecules from an area of higher concentration to an area
of lower concentration. This will occur until the two areas reach a dynamic equilibrium. When
this dynamic equilibrium is reached the concentration of molecules will be approximately equal
and there will be no net movement of molecules after this point. Osmosis is a special kind of
diffusion in which water moves through a selectively permeable membrane in the direction that
tends to equalize the osmotic concentrations on the two sides of the membrane.
Plant cells are composed of an outer cell wall and an inner cell membrane. The cell wall is a
relatively rigid structure composed of cellulose and several other polysaccharides. It is also
porous and allows most molecules to pass through it. When a plant cell is placed in a hypotonic
solution, water will enter the cell through osmosis. However, the rigid cell wall prevents the cell
from enlarging beyond a certain volume, leading to the development of hydrostatic pressure
called tugor pressure. Tugor pressure keeps a plant upright, and prevents more water from
entering the cell. When a plant cell is placed in a hypertonic solution, water will leave the cell
through osmosis, and the cell membrane pulls away from the cell wall. This phenomenon is
called plasmolysis. Plasmolysis is a reversible process: when the hypertonic solution is replaced
with a hypotonic solution, the cell regains its original volume and tugor pressure. When cells are
placed in isotonic solution, the movement of water out of the cell is exactly balanced by the
movement of water into the cell. An isotonic solution is when two solutions, separated by a
semipermeable membrane have equal concentrations of solutes and water. Since the solutions
have the same osmotic pressure this allows for the free movement of water across the membrane
without changing the concentration of solutes on either side.
OBJECTIVES
1. To observe osmotic responses typical of plant cells using purple onion skin cells under
hypotonic, hypertonic and isotonic conditions.

METHOD AND MATERIALS/REQUIREMENTS

1. Purple Onion
2. Light Microscope
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 3. Forceps/Tweezers
5. Slides
7. Cover slips
8. Blotting/filter paper
9. Distilled water
10. Isotonic Solution
11. Hypertonic Solution
12. Hyportonic solution
1. Using a razor blade cut out a 3 mm x 3 mm square on the outside skin (epidermis) of a red onion.
You only need to cut through the skin, which is consisting of a single layer of cells.
2. Using a pair of tweezers carefully peel off the skin and spread it onto a glass slide skin side up.

3. Add a drop of distilled water onto the onion skin and cover with a cover glass. When mounting a
cover glass, it is a good practice to contact the glass slide at a 45 degree angle then lower the
cover glass slowly. This way you will have less air bubbles trapped under the cover glass.
4. Observe the onion skin under 10X and 40X magnification. Start with the low magnification. Can
you see the cell wall, the cell membrane and the nucleus? Record your observations. Draw and
label 3 plant cells.
5. Take another strip of cells from onion, mount it on a slide and then add a drop of salt water
(hypertonic solution 10 NaCl) onto the onion skin and cover with glass slide. Observe the onion
skin under the microscope. Can you see the cell membrane now? Record your observation. Draw
and label 3 plant cells.
6. Take another strip of cells from onion, mount it on a slide and then add a drop of isotonic
solution onto the onion skin and cover with glass slide. Observe the onion skin under the
microscope. Can you see the cell membrane now? Record your observation. Draw and label 3
plant cells.

QUESTIONS
1. Describe the cells in distilled water. How are the cells in hypertonic solution (10%
sodium chloride) different from this?
2. Explain what happened to the cells in the hypertonic solution (10% sodium chloride
solution) using biological terms. Try to include these words. Cytoplasm, diffusion, water,

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 solvent, dissolved salts, solute, cell membrane, vacuole, cell wall, osmosis, plasmolysis,
turgid, flaccid, turgor.
3. What prevents the plant cells from bursting when they take in lots of water?
4. You’ve seen what happens to cells in epidermal tissue when they lose water. How does a
whole plant look when it is short of water? How does it change when you give it water?
5. Animal cells do not have the same structure as plant cells. What do you think could
happen to an animal cell in water?
6. For microscopic observation, why is it necessary to use only the thin skin of the onion
cells?
7. Why use purple onion cells and not white onions.

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PRACTICAL 3: MEASURING THE WATER POTENTIAL OF POTATO TUBERS

INTRODUCTION
Plants rely on physical forces to move water in and around their tissues, both short distances
from cell to cell and long distances through the xylem vessels. Water potential (), a measure of
the driving force that governs the movement of water from the soil into plants and finally into the
atmosphere, is the amount of energy per unit volume (or pressure) contained in a system like a
plant cell. Pure water in a free standing solution has a water potential of zero, while most plant
cells have a negative water potential. The atmosphere typically has a very negative water
potential, much more negative than a typical plant tissue, while the water potential of the soil
solution is typically less negative than a typical plant tissue. Water moves from a tissue or area of
higher water potential to a tissue or area of lower (more negative) water potential. It is
energetically favorable under normal conditions for water to move into a plant from the soil,
through the plant, and out to the atmosphere down a water potential gradient. The rate of
movement varies considerably depending upon soil moisture availability and relative humidity of
the atmosphere.

Water potential of a plant cell is made up of two important components. The relationship among
these components is expressed mathematically as: (Eq. 1)  = s + p
Where  is the overall water potential of a cell. s is the solute or osmotic potential and
represents the contribution made by dissolved solutes to . Adding dissolved solutes to a
system decreases the water potential. In a plant cell, mineral ions, sucrose, starch, amino acids,
proteins that can accumulate to high levels in the cytosol or vacuoles are iimportant contributors
to solute potential. p is the pressure potential and represents the contribution made by pressure
to . Fully turgid cells whose plasma membranes are pressing against the cell wall have a
positive p. Cells at incipient plasmolysis (the point at which the membrane is just barely
touching the cell wall) have a p of zero. Cells under tension, like those in the xylem during
active evapotranspiration, have a negative p.

When placed in a free standing sucrose solution, water will move into or out of a plant tissue
depending upon its water potential relative to the solution. Gain or loss of water can be detected
by weighing the plant tissue before and after immersion in the solution. By incubating tissues in
a series of sucrose solutions of different concentrations, the solution that causes no change in
tissue weight can be identified. The water potential of this isotonic solution is assumed to equal
the water potential of the tissue.

OBJECTIVES
The objective of this practical is to measure the water potential of potato tuber tissues.

METHODS AND MATERIALS/REQUIREMENTS

Requirements

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  Distilled water
 Sucrose solutions
 Potato tubers
 Scapels
 Balance
 Blotting paper

You are required to estimate the water potential of the plant material provided. Observe the
tubes carefully during the whole procedure.

1. Label six test-tubes as follows: DW for distilled water, 0.2, 0.4, 0.6, 0.8, and 1.0 molm3
and place them in a test tube rack.
2. Prepare the following sucrose solutions: Distilled water, 0.2, 0.4, 0.6, 0.8 and 1.0
molm3(mol/kg solution) sucrose.
3. Add 75 ml of each solution to the appropriate labeled test-tube using a separate graduated
cylinder for each solution.
4. Cut 6 cylinders from a potato and trim each cylinder to 4 cm in length with a scapel,
being careful to remove the peel. Work quickly to prevent the tissues from drying out as
you cut them. What problems would tissue drying cause in the experiment?
5. Place all 6 cylinders into a petri dish which must be lidded immediately.
6. Weigh one cylinder and record its initial mass in the Table below. Place this cylinder into
the test tube labeled DW.
7. Weigh a second cylinder and place it into the test tube labeled 0.2, taking care to record
its mass in the Table below.
8. Repeat the weighing for the remaining cylinders and put them into the test tubes labeled
0.4, 0.6, 0.8 and 1.0 molm3.
9. Immediately start a stop watch. Leave the cylinders in the test tubes for 25 minutes.
10. After 25 minutes, pour out the distilled water from the tube labeled DW. Place the
cylinder on a filter paper and blot any excess water quickly and gently. Do not squeeze
the cylinder. Reweigh this cylinder and record its final mass in the Table below.
11. Repeat the procedure above with the other cylinders and record their final masses.
12. Calculate the change in mass and percentage change in mass for each cylinder and record
in the Table below.

Table 1: Weight change in potato tissues in sucrose solutions of different concentration

Concentration Initial mass (g) Final mass (g) Change in % change in
of solution mass (g) mass
(moldm )3

D1
0.2
0.4
0.6
0.8
1.0

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 Calculations

First, subtract the initial tissue weights from the final weights. Second, divide the difference
by the initial weight and multiply by 100 to get the percent weight change. Record your
calculations in Table 1. Next, plot the percent change in weight (ordinate) vs. sucrose
concentration (abscissa) using a graph paper or in Microsoft Excel. Using this figure (and
the regression line determined from it), determine the exact concentration of sucrose that
would cause no change in weight in the potato tubers. The water potential of this solution
will equal the water potential of the potato tissue.

In an open solution where there is no turgor pressure, the p is equal to zero. Thus, the 
of such a solution is equal to the s of a solution. Calculate the s of the solution causing
no change in weight of the potato tissues using the following formula:

(Eq. 2) s = -miRT

m = molarity
i = ionization constant = 1 for sucrose
R = gas constant = 8.31 J K -1 mol -1
T = room temperature in K (ºC + 273 = K)

First, convert molality of the appropriate sucrose solution to mol m-3 (Note that 1 molal =
1 x 103 mol m-3), and then use Eq. 2 to calculate the s of your solution. The results of
this calculation will be in units of J m-3 (energy per unit volume), which is equal to a unit
of pressure in Pa. Convert your answer to Mpa by dividing by 106. Show your
calculations and include them in your report.

B. Relationship between molarity and water potential

The table below shows the relationship between molarity and water potential.

Molarity/ moldm3 Water potential

(kPa)
0.05 -130
0.10 -260
0.15 -410
0.20 -540
0.25 -680
0.30 -860
0.35 -970
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 0.40 -1120
0.45 -1280
0.50 -1450
0.55 -1620
0.60 -1800
0.65 -1980
0.70 -2180
0.75 -2390
0.80 -2580
0.85 -2790
0.90 -3000
0.95 -3250
1.00 -2500

Draw a graph to show the relationship between molarity and water potential. Account for
the shape of the graph.

QUESTIONS

a. Explain why the potato cylinders must not be squeezed during the blotting
process.

b. Why was sucrose used as the solute in the solutions? How might using another
solute influence the results?

c. Tissues in which treatments have a water potential equal to that of their solution
after the incubation period? How can you tell?

d. What influence would increasing temperature have on our calculation of water

potential?

e. The results of the calculations of water potential usually vary among years,
among laboratory sections, and even among groups within one laboratory. Why
might this be?

f. Suggest three ways in which you could improve this experiment to make your
results more reliable.

PRACTICAL 4: DETERMINATION OF WATER POTENTIAL OF SOLUTES

INTRODUCTION

Water Potential is the measurement of the tendency of water molecules to move from one place
to another. Water always moves down the water potential gradient, therefore moving from an
area of higher water potential to an area of lower water potential. Equilibrium is reached when
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the water potential in one region is equal to the water potential in another region. For example, if
a plant cell (like the potato tuber cells) is in equilibrium with an external solution of such a
concentration that there is no net gain or loss of water then the water potential of the external
solution will be equal to the water potential of the cell.

By convention, the water potential of pure water is set at zero. Knowing that solutes make the
water potential of solutions lower, solutes make solutions negative. Solute potential is the
amount that the solutes lower the water potential of a solution. Pressure potential is especially
important in plant cells. If a plant, for example the potato tuber cells, is placed in pure water (or
a dilute solution), the water (or solution) has a higher water potential than the plant cell. This
causes the movement of water to the cell due to the higher water potential in the cell. Water
enters a cell through the partially (semi) permeable membrane by osmosis.

Osmosis is the movement of water molecules from a region of higher water potential to a region
of lower water potential through a partially permeable membrane. A plant cell wall is extremely
inelastic. This property allows very little water to enter a plant cell - preventing the cell
from bursting. For plant cells water potential consists of a combination of solute potential and
pressure potential. Solute potential can be defined as the amount that the solute molecules lower
the water potential. It is evidently always negative. On the other hand, the pressure potential is
always positive. This is because it causes the water potential to be less negative.
Pressure potential can be defined as the contribution made by pressure to water potential.

OBJECTIVES

The aim of this practical is to determine the water potential of solutions, A and B.

MATERIALS AND METHODS

Prior knowledge :The water potential of a plant tissue can be found by immersing the plant
tissue in sucrose solutions of different water potential.

Sucrose solutions A and B are the solutions in which tissues from two different species of plant
did not change in mass after immersion for 30 minutes as shown in Figure1.

Figure 1: Plant tissue immersed in sucrose solution

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(a) (i) Use two of the following words to complete the sentences below.

gains less loses more

If the plant tissue ........................ water then the sucrose solution will become more dilute.
This will change the solution so that it becomes ........................ dense.

A blue dye is added to the two solutions, A and B, so that they can be seen. A drop of the
coloured solution is placed into a known concentration of sucrose solution. Fig. 2 shows how the
drop is released.

Figure 2: How the drop is released

Immediately the drop is released the syringe is removed. The drop may move up, move down or
remain at the same level.

(ii) Show clearly in diagrams how you would expect to see the drop move.

Preparations of different solutions (S, A, B, D and W).

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(i) S, at least 200 cm3 of 1mol dm¯3 sucrose solution, in a beaker or container, labelled S.
This is prepared by dissolving 68 g of sucrose in 100 cm3 of distilled water and making up to 200
cm3 with distilled water.

(ii) A, at least 10 cm3 of 0.7 mol dm¯3 sucrose solution, in a small beaker or container,
labelled A. This is prepared by dissolving 48 g of sucrose in 80 cm3 and making up to 200 cm3
with distilled water.

(iii) B, at least 10 cm3 of 0.25 mol dm¯3 sucrose solution, in a small beaker or container,
labelled B. This is prepared by dissolving 17 g of sucrose in 80 cm3 and making up to 200 cm3
with distilled water.

(iv) D, at least 10 cm3 of 0.01% methylene blue solution, in a small beaker or container,
labelled D. This is prepared by dissolving 1 g of methylene blue in a container and making up to
100 cm3 with distilled water. This forms a 1% methylene blue solution. Then take 1 cm3 of this
solution and make up to 100 cm3 with distilled water. Methylene blue is harmful and will stain
skin. Care should be taken when making up the stain.

(v) W, at least 200 cm3 of distilled water, in a beaker or container, labelled W.

It is advisable to wear safety glasses/goggles when handling chemicals.

Solutions and reagents provided to the students should be supplied in a suitable beaker, or
container, for removal of the solution using a syringe.

APPARATUS

1. Two 10 cm3 syringes or one 10 cm3 syringe and one 50 cm3 measuring cylinder.
2. At least two 5 cm3 syringes. If possible, up to six 5cm3 syringes can be provided.
3. Container with tap water, labelled “For washing”.
4. Container, labelled “For waste”.
5. Paper towels.
6. Eight small containers capable of holding at least 50 cm3.
7. One large test-tube, with the means to wash it out.
8. Small Petri dish or shallow container.
9. Teat pipette.
10. Test-tube rack or container to hold the large test-tube.
11. Glass marker pen.
12. Safety goggles/glasses.

The teacher should carry out the experiment and record his/ her results for comparison.

These solutions you have been prepared from the procedure above.
 200 cm3 of 1.0 mol dm-3 sucrose solution in a beaker, labelled S
 200 cm3 of distilled water in a beaker, labelled W
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  10 cm3 of sucrose solutions A and B
 10 cm3 of 0.01% methylene blue, labelled D.

If any methylene blue comes into contact with your skin wash off immediately with water.

It is recommended that you wear safety goggles/glasses.

1. To find the concentration of sucrose in samples A and B you will need to dilute the 1.0
mol dm-3 sucrose solution to provide a range of concentrations.
2. Decide on the concentrations of sucrose solution that you will prepare using the 1.0 mol
dm-3 sucrose solution and distilled water. You will need to make up 50 cm3 of each
sucrose solution. Fill in Table 1 to help you with your dilutions.

Table 1: Amounts of 1.0 mol dm-3 sucrose solution and distilled water needed to
make different sucrose concentrations
Concentrations of sucrose Volumes of 1.0 mol dm-3 Volumes of distilled water
solution sucrose solution

3. Place a 5 cm3 syringe on top of the large test-tube and use the glass marker to draw a line on the
test-tube at the same height as the end of the syringe nozzle
4. Use a 5 cm3 syringe to collect 4.0 cm3 of A and place it in a Petri dish. With a pipette, add
sufficient drops of D to turn the solution blue and stir.
5. Use the same syringe to collect 1.0 cm3 of the coloured solution A. Wipe the syringe with a
paper towel and label the syringe A.
6. Repeat steps 2 and 3 with sample B and label the second syringe B.
7. Put 35 cm3 of one of your sucrose solutions into the large test-tube.
8. Put syringe A into the large test-tube so the end of the nozzle is level with the mark. Hold the
syringe vertically and very gently push out a drop of the coloured solution.
9. Immediately observe the movement of the drop.
10. Record your observations.
11. Repeat steps 6 to 8 with sample B in syringe B.
12. Empty and wash the large test-tube.
13. Repeat steps 5 to 10 with each sucrose solution that you have made and record all your
observations.
14. Use your results to estimate the sucrose concentration of
sample A...............................mol dm-3
sample B............................... mol dm-3 .

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 In order to find the water potential of the solutions A and B a graph is required show the
relationship between sucrose concentration and water potential.

Table 1: The water potential of different sucrose concentrations.

15. Plot a graph of the data shown in Table 1.

16. Using your results and your graph estimate the water potential of sample A. Show clearly on
your graph how you obtained the water potential.
17. Describe how you would improve the investigation to obtain a more accurate estimate of the
water potential of sample A.

PRACTICAL 5: IDENTIFICATION OF NON-REDUCING SUGARS

INTRODUCTION
Non-reducing sugars are sugars which do not have an aldehyde functional group - the reducing
species. As non-reducing sugars do not have the aldehyde group, they cannot reduce copper (I)
(blue) to the copper(II) (red). Sucrose is the most common disaccharide non-reducing sugar. The
test for non-reducing sugars may either be qualitative, or it may be quantitative. The qualitative
tests produce a colour change from blue to green to yellow to orange to brick red (for Benedict’s
test and Fehling’s test) and a silver mirror (for the Tollens test) as highlighted in Table 1. The
principle of the Benedict's Test for non-reducing Sugars in that disaccharides are hydrolyzed to
their constituent monosaccharides when boiled in dilute hydrochloric acid. The
monosaccharide products of hydrolysis are reducing sugars i.e. have the aldehyde functional
group and can reduce copper in the presence of alkali producing the colour changes.

Table 1: Major tests for non-reducing sugars

Oxidising Benedict's Fehling's Tollen's

Reagent Solution Solution Reagent

copper sulfate copper sulfate silver nitrate

Composition in alkaline in alkaline in aqueous
citrate tartrate ammonia

Colour of Solution deep blue deep blue colourless

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 Colour After brick red brick red silver mirror
Reaction with a precipitate precipitate forms
Reducing Sugar Cu2O(s) Cu2O(s) Ag(s)

Species Being
Reduced Cu2+ Cu2+ Ag+
Cu2+ + e ---> Cu+ Cu2+ + e ---> Cu+ Ag+ + e ---> Ag(s)
(the oxidant)

Species Being reducing reducing

reducing sugar
Oxidised oxidised to sugar sugar
carboxylate oxidised to oxidised to
(the reductant) carboxylate carboxylate

OBJECTIVES

In this practical, we will identify the presence of non-reducing sugars. This is a qualitative test
hence it is important for the student to exercise their observation skills.

MATERIALS AND METHODS

1. 2% sucrose solution
2. Dilute hydrochloric acid
3. Dilute sodium hydroxide or dilute sodium hydrogen carbonate solution
4. pH paper
5. Benedict’s solution
6. Test tubes
7. Bunsen burner
8. Water bath.

1. Add 2cm³ of sucrose solution to a test tube.

2. Add equal amounts of the Benedict’s solution and observe any colour change. Explain
your observations.
3. In another test tube, add 2cm3 of sucrose.
4. Add 1cm³ of dilute hydrochloric acid.
5. Boil for one minute.
6. Cool the tube under running water.
7. Carefully neutralise with dilute sodium hydroxide or sodium hydrogen carbonate.
8. Add about 2 cm3 of the Benedict’s solution and observe any colour changes. Explain your
observations (N.B. Care should be taken as effervescent occurs.)

QUESTIONS

a) Suggest the role of hydrochloric acid in the reaction.

b) How else would you convert sucrose into reducing sugars?
c) What is the principle of the Benedict's Test for non-reducing sugars?

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PRACTICAL 6: DETERMINATION OF THE RELATIVE QUANTITIES OF
REDUCING SUGARS IN FOOD ITEMS

INTRODUCTION
Sugars are classified as reducing or non-reducing based on their ability to act as a reducing agent
during the Benedict's Test. A reducing agent donates electrons during a redox reaction and is
itself oxidized. The aldehyde functional group is the reducing agent in reducing sugars. Reducing
sugars have either an aldehyde functional group or have a ketone group - in an open chain form -
which can be converted into an aldehyde. Reducing sugars are simple sugars and include all
monosaccharides and most disaccharides. Some examples of monosaccharides are glucose,
fructose and galactose. Examples of reducing disaccharides are lactose and maltose.
Benedict’s solution is designed to detect the presence of reducing sugars. In hot alkaline
solutions, reducing sugars reduce the blue copper (II) ions to brick red copper (I) oxide
precipitate. As the reaction proceeds, the colour of the reaction mixture changes progressively
from blue to green, yellow, orange and red. When the conditions are carefully controlled, the
colouration developed and the amount of precipitate formed depends upon the amount of
reducing sugars present. Hence, in most conditions, a sufficiently good estimation of the
concentration of glucose-equivalent reducing sugars present in a sample can be obtained.

OBJECTIVE

The objective of this experiment is to ascertain relative quantities of reducing sugars in selected
food items.

MATERIALS AND METHODS/APPARATUS

 Pipette fillers and safety goggles should be used where necessary.
 Three test-tubes labelled X, Y and Z, containing 2 cm3 of the following solutions:
Solution Y consisting of 20 g of glucose, made up to 1 dm3 of solution.
Solution Z consisting of 20 cm3 of solution Y, made up to 1 dm3 of solution.
Solution X consisting of 20 cm3 of solution Z, made up to 1 dm3 of solution.
One dm3 of solution is sufficient for at least 450 candidates

1. 1 cm3 of freshly peeled potato, covered in water, in a dish, labelled P. Avoid using
sprouting potatoes if possible.
2. 1 cm3 of freshly peeled onion, covered in water, in a dish, labelled O.
3. Benedict’s solution, labelled as Benedict’s solution.
4. Syringe to measure 2 cm3.
5. Test-tube rack or similar with two empty test-tubes and bungs.
6. Means of holding hot test-tubes.
7. Access to a water bath. This could consist of a Bunsen burner, tripod, gauze and 250 cm3
or similar beaker or heatproof container, half full of hot water, with thermometer that can
read up to at least 100 °C. Alternatively, an electrically heated water bath can be set to at
least 80 °C.
8. Paper towel.
9. Tile.
10. Sharp knife or scalpel.
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 11. Thick glass or wood or metal rod to crush the tissue.
12. Sight of a clock.

You are provided with test-tubes, labelled X, Y and Z, each containing a different concentration
of reducing sugar.

1. Carry out the test for reducing sugars on samples X, Y and Z as follows
a. In another test tube, add 2cm3 of sucrose
b. Add 1cm³ of dilute hydrochloric acid.
c. Boil for one minute.
d. Cool the tube under running water.
e. Carefully neutralise with dilute sodium hydroxide or sodium hydrogen carbonate.
f. Add about 2 cm3 of the Benedict’s solution and observe any colour changes
g. Record your observations and conclusions for your observations in Table 1.0 below.

Table 1.0: Benedict’s test results

Retain the test-tubes for comparison with the results of your next step.

f. Determine the order of concentration of reducing sugar in the three solutions and

complete Table 1.1 below.

Table 1.1: The concentration of reducing sugars in three solutions

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 2. You are also provided with some potato tissue, labelled P, and some onion tissue,
labelled O.
a. Finely cut up tissue P on the tile and place the crushed tissue into one of the two
empty test-tubes provided.
b. Add 2 cm3 of water to the test-tube.
c. Place a bung in the open end of the test-tube and shake gently.
d. Repeat the process for tissue O using the other empty test-tube.
e. Carry out the test for reducing sugars on both samples.
f. Record your observations in Table 2.0.

Table 2.0 : Tests for reducing sugars

g. Compare your observations with the results obtained for (1) above.
h. Explain how you made sure that your tests produced a fair comparison.

Prior Knowledge of the Benedicts test

Table 2: Colour observations and their interpretation

Observations Interpretations

No Colour Change No non-reducing sugars present

(Blue)
Green Trace amounts of non-reducing sugars present
Yellow Low amounts of reducing sugars present
Orange Moderate amounts of reducing sugars present
Brick-red Large amounts of non-reducing sugars present

PRACTICAL 7: DEMONSTRATION OF THE ACTION OF PROTEASE (PEPSIN) ON

ENZYMES

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 An enzyme is a biological catalyst, a chemical agent that speeds up a reaction without being
consumed by the reaction. The initial investment of energy for starting a reaction so that bones
can break is known as the free energy or activation energy. Linus Pauling in 1948 stated that
enzymes are molecules that are complementary in structure to the activated complexes of the
reactions that they catalyse. The attractions of the enzymes molecule for the activated complex
would thus lead to a decrease in its energy, hence, to a decrease in the energy of activation of the
reaction and increase in the rate of reaction.

The large enzyme molecule presumably can have sufficient rigidity and sufficiently exclude
solvent to provide the necessary environment for sufficient catalysis. The large size further
provides sites for the binding of molecules that control enzyme activity. The large size may be
necessary to resist proteinases and other potentially dangerous/ damaging agents.

Factors affecting enzymically controlled driven reactions are: (i) Temperature -The rate
increases with increasing temperature especially between 35 and 40°C. Rate doubles for every
10°C rise in temperature between 30 to 40°C. (i)Enzyme inhibitors- Both naturally occurring
and synthetic compounds have the ability to bind reversible or irreversibly with/ to specific
enzymes and alter their activity. Inhibitors reduce/ eliminate the catalytic activity of the enzyme.
Such inhibitors are drugs, antibiotics, toxins and antimetabolites. (iii) Irreversible inhibitors
(non-competitive inhibitors)-They form covalent bonds with specific functional groups usually
an amino acid side chain, thereby blocking the active site and preventing enzyme action. They
cannot be reversed by dilution. They reduce velocity of the reaction to an extent that corresponds
to the fraction of enzyme molecules which have been inactivated. (iv) Reversible inhibition-
Chemicals resemble an enzyme’s normal substrate and compete with it for the active site. They
block the active site from the substrate.
(v) Co-factors-These include prosthetic groups and coenzymes. Prosthetics groups are organic
cofactors which permanently combine with the enzyme. Examples are FAD and Haem.
Coenzymes are not bonded to the enzyme molecules. Main examples are vitamin derivatives
such as NAD (respiratory coenzyme).

Proteases are enzymes that hydrolyse proteins into smaller peptide fragments. They work best on
denatured proteins. When eggs are cooked, protein is rapidly denatured by heat. Egg whites
contain several proteins, the most abundant being ovalbumin. When pepsin works on egg white
suspension the physical states in which it should be in the stomach, where pepsin is one of
several digestive enzymes, it rapidly clears or clarifies it. This is because the peptides formed by
enzyme action are water soluble, while the denatured proteins are not. The peptide bonds in
proteins can also be broken by boiling the protein in concentrated acid for several hours.

OBJECTIVES
The objective of this practical is to demonstrate the action of a protease (pepsin) on protein.

MATERIALS AND METHODS/ REQUIREMENTS

 Egg
 Pepsin (40 mg sample in tightly stoppered bottle)
 Stop watch
 Bunsen burner
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  500 ml beaker
 Test-tubes & rake
 Boiling tube & holder
 Water bath at 25°C
 1 cm ³ & 2 cm³ syringes
 100 cm³ conical flask
 10 cm³ measuring cylinder
 Dropping pipette
 Distilled water
 2M hydrochloric acid.

NB: Hydrochloric acid is corrosive. Use safety glasses when handling it and deal with spills
immediately by neutralising it with sodium hydrogen carbonate before wiping.

1. Break the egg and separate its white into a small beaker.
2. Put 25 cm³ distilled water into a boiling tube and add four drops of egg white, stirring the
suspension thoroughly after each drop.
3. Heat the test tube over a medium Bunsen flame until the suspension turns milky. The proteins
are denatured.
4. Cool the tube under a running cold water tap.
5. Make an acidified egg white suspension by adding 0.5 cm³ 2M HCl to the 10 cm³ measuring
cylinder using the 1 cm³ syringe. Then add egg white suspension to the 10 cm³ mark.
6. Add 20 cm³ distilled water to the bottle containing pepsin.
7. Transfer a 1 cm³ portion of this to a test tube and denature by heating in boiling water bath
(use the 500 cm³ beaker) for 5 minutes. Allow to cool.
8. Label four tubes A, B, C and D. Add 2cm³ egg white suspension to A, B and C. Add 3 cm³
distilled water to C. Incubate the tubes in the water for 5 minutes.
9. Now add 1cm³ pepsin solution to A and then 1cm³ portion of denatured pepsin to B and start
the stop watch.
10. Note and record the appearance of the tubes at suitable intervals for the next hour.
11. Finally, add another 2cm³ egg white suspension to test tube A.

(a) Record the time taken to clarify.

(b) Comment on the use of controls in this experiment.
(c) Which properties of enzymes are demonstrated by this experiment?
(d) Account for any faint cloudiness remaining after clarification.
(e) Giving reasons, would you expect pepsin to clarify a suspension of
(i) olive oil in water
(ii) cooked meat particles
(iii) Milk
12. How would you degrade the protein in egg white if no enzymes were available?
NB: In the presence of hydrochloric acid, pepsin turns a cloudy suspension of egg-white into a
clear solution

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 Tips for the Instructor
 The interpretation of results will be easier if the student knows already that egg-white suspension
consists of solid particles of albumen, suspended in water; enzymes are inactivated by boiling,
conditions in the stomach are acid.

 In the presence of hydrochloric acid, pepsin turns a cloudy suspension of egg-white into a clear
solution.

PRACTICAL 8: DEMONSTRATION OF THE EFFECT OF SUBSTRATE

CONCENTRATION ON ENZYME ACTION

INTRODUCTION

Enzymes such as Catalase are protein molecules which are found in living cells. They are used to
speed up specific reactions in the cells. They are all very specific as each enzyme just performs
one particular reaction. A catalase is an enzyme found in food such as potato and liver. It is used
for removing Hydrogen Peroxide from the cells. Hydrogen Peroxide is the poisonous by-product
of metabolism. A catalase speeds up the decomposition of Hydrogen Peroxide into water and
oxygen as shown in the equations below.

Formula:

Catalase
Hydrogen Peroxide---------------------->Water + Oxygen
Catalase
2H2O2------------------->2H2O+O2
It is able to speed up the decomposition of Hydrogen Peroxide because the shape of its active site
matches the shape of the Hydrogen Peroxide molecule. This type of reaction where a molecule is
broken down into smaller pieces is called an anabolic reaction.

In 1913, Leonor Michaelis and Maud Menten proposed a theory to explain the way in which the
rate of enzyme-catalysed reactions increases with substrate concentration. They noted that, at
low substrate concentrations, the rate is proportional to concentration, but as concentration
increases, the rate levels off until a maximum value is achieved. This rate can be only be
increased further if more enzyme is added. Michaelis and Monton defined two important
properties from their model, which are used today to compare the properties of different
enzymes: Km, the Michaelis constant, is the substrate concentration at which half the maximum
enzyme velocity (Vmax) is reached, (Figure 1).

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 Figure 1: Effect of substrate concentration on enzyme action

OBJECTIVES

The objective of this practical is to demonstrate the effect of substrate concentration on enzyme
action.

MATERIALS AND METHODS

1. Gas Syringe
2. Metal Stand
3. Yeast Catalase
4. Hydrogen Peroxide
5. Test Tubes
6. Beakers
7. Test Tube Rack
8. Stop Watch
9. Pipette
10. Pipette Filler
11. Tap Water

To test out how the concentration of hydrogen peroxide affects the rate of reaction, first set up
the apparatus below.

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 Figure 2: Apparatus to test the effect of substrate concentration on enzyme action.

1. Add 2cm3 of yeast to one test tube. Add 4cm3 of hydrogen peroxide solution at a
concentration of 20% to the other test tube. Use a pipette to measure out the volumes. It
is very important to accurately measure the amounts of Hydrogen Peroxide, Yeast and
water to ensure a fair test.

2. Pour the hydrogen peroxide solution into the test tube containing the yeast and
immediately put the gas syringe bung on the end of the test tube, at the same time start
the stopwatch.

3. Bubbles should start to rise up the tube and the gas syringe will move outwards, as soon
as the gas syringe passes the 30 cm3 mark stop the stopwatch and note the elapsed time
down to the nearest 1/10th of a second.

4. Repeat the experiment with hydrogen peroxide concentrations of 16%, 12%, 10%, 8%,
4% and 0%. The 0% concentration of hydrogen peroxide solution is done as a control
solution to show that at 0% concentration no reaction occurs. The different
concentrations of Hydrogen Peroxide are made by adding tap water to the 20%
Hydrogen Peroxide in the correct amounts. The table below shows what amounts of
Hydrogen Peroxide and water are needed to make the solutions.

Table 1: Amounts of Hydrogen Peroxide and water needed to make the solutions

Concentration Of Hydrogen Volume Of Hydrogen Volume Of Water

Peroxide Peroxide (cm3) (cm3)

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 20% 4 0

16% 3.2 0.8

12% 2.4 1.6

10% 2 2

8% 1.6 2.4

4% 0.8 3.2

0% 0 4

5. Repeat all the tests at least three times so that an average can be obtained. Repeating the
experiments several times will help to produce better and more accurate results as any
inaccuracies in one experiment should be compensated for by the other experiments.
Note all the results in a table such as the one below (Table 2).

Table 2: Effect of substrate concentration on enzyme action

Hydrogen Peroxide
0% 4% 8% 10% 12% 16% 20%
Concentration

Time Taken (Test 1)

Time Taken (Test 2)

Time Taken (Test 3)

Average of the Tests

Rate

The rate can then be worked out by

Rate=30/Average Time

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 This gives the rate in cm3 of oxygen produced per second, this is because we are timing how
long it takes to produce 30 cm3 of oxygen. From these results a graph can be plotted with
concentration on the x-axis and time taken on the y-axis.
6. List any 3 variables that should be kept constant to ensure a fair experimental procedure.
Tips for the Instructor.
 We are using yeast catalase as opposed to catalase from apples, potatoes or liver because it is
easier to get the desired amount of yeast catalase by simply measuring it off. To obtain catalase
from a substance such as potato would involve crushing it and with that method you would never
be sure of the concentration of the catalase. If the catalase was used up then another potato would
have to be crushed and this could produce catalase of a totally different concentration which
would lead to inaccuracies in the experiment making this an unfair test.
 To ensure this is a fair test all the variables except for the concentration of Hydrogen Peroxide
must be kept the same for all the experiments. Variables that must not be altered include:-
 Temperature,
 yeast concentration,
 type of yeast,
 batch of yeast,
 volume of yeast,
 volume of hydrogen peroxide,
 air pressure and
 humidity.
 When measuring the volumes of Hydrogen Peroxide, Yeast and Water the measurement should
be taken by looking at the scale at an angle of 90 degrees to it to avoid any parallax error.

PRACTICAL 9: DEMONSTRATION REGULATION AND CONTROL THROUGH

DIALYSIS

INTRODUCTION

During kidney dialysis an artificial kidney is used to remove solutes and toxins from the blood
by diffusion and osmosis. This shows that osmosis and diffusion are vital to the health of
organisms. Osmosis is useful for regulating solute concentrations in organisms in order to
maintain homeostasis. The process can be demonstrated by placing cells in solutions of varying
solute concentrations. In an isotonic solution there will be no net movement of water into or out
of the cell while in a hypertonic solution water would leave the cell in order to reach equilibrium.
In a hypotonic solution water would enter the cell in an effort to reach equilibrium.

Kidneys maintain homeostasis within the body by filtering the blood and producing urine. The
filtration system of the kidneys maintains the necessary ion levels in the blood. It removes the
waste products (e.g. urea). Urine is produced in order to maintain the internal body environment
through the regulation of certain solutes, such as potassium and sodium ions and other materials.
The kidney responds to the constantly changing internal conditions, adjusting and re-adjusting
itself to maintain homeostasis. In people with kidney malfunction the blood composition can be
regulated artificially by dialysis. The basic principle in dialysis involves removing a small
amount of blood from the body at a time and then filtering out urea, remove other waste products
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 and balance essential ion concentration from the blood through simple diffusion. This principle
will be demonstrated during this practical.

Hemodialysis involves temporary removal of blood from the body, flowing the blood through a
tube surrounded by a carefully selected permeable membrane which is surrounded by a fluid
called the dialysate, and then returning the filtered blood back into the body. Dialysis occurs so
that molecules can attain a state of equilibrium. Molecules diffuse through a membrane in order
to reach this state. If a hypertonic solution is surrounded by a hypotonic solution, the solute
particles will diffuse across the membrane. The dialysate is a solution that has been specially
formulated to remove specific materials from the blood before sending the blood back into the
body. Ideally, little to no urea should be present in the blood when it is sent back into the body.
As the blood passes through the tube, which would have pores large enough for urea to pass
through, the dialysate would contain no urea. This would cause urea to move through the
membrane into the dialysate, thereby reducing the concentration of urea in the blood. There are
multiple factors that will determine the rate of diffusion during dialysis. The rate of diffusion,
and therefore the success of the dialysis, is dependent on the concentration gradient between the
blood and the dialysate, the material used for the membrane and the size and properties of the
solute that is diffusing.

OBJECTIVES
The objective of this practical is to investigate how much glucose diffuses through selectively
permeable Visking (dialysis) tubing in 15 minutes.

MATERIALS AND METHODS/REQUIREMENTS

1. Fresh G
2. Benedict’s solution
3. W
4. Visking tubing
5. S1, S2 and S3

More of the solutions should be available if requested by the student.

Solutions and reagents provided to the students should be supplied in a suitable beaker, or
container, for removal of the solution using a syringe.

Preparations
(i) G: at least 25 cm3 of 10 % glucose solution in a small beaker or container, labelled G. This is
prepared by dissolving 10 g of glucose in 80 cm3 of distilled water and making up to 100 cm3
with distilled water.
(ii) At least 100 cm3 of Benedict’s solution, in a small beaker or container (so that a syringe can be
used), labelled Benedict’s solution.
(iii) W, at least 100 cm3 of distilled water, in a beaker or container, labelled W.
(iv) 20 cm length of Visking tubing (about 14 mm flat diameter) submerged in distilled water, in a
beaker or container, labelled V.

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(v) S1, at least 20 cm3 of 0.1 % glucose solution in a small beaker or container, labelled S1. This is
prepared by dissolving 1.0 g of glucose in 500 cm3 of distilled water and making up to 1 dm3.
(vi) S2, at least 20 cm3 of 0.2 % glucose solution in a small beaker or container, labelled S2. This is
prepared by dissolving 2.0 g of glucose in 500 cm3 of distilled water and making up to 1 dm3.
(vii) S3, at least 20 cm3 of 0.3 % glucose solution in a small beaker or container, labelled S3. This is
prepared by dissolving 3.0 g of glucose in 500 cm3 of distilled water and making up to 1 dm3.

These solutions can be made up the day before the practical and stored in a refrigerator.
However, these must be at room temperature for the practical.

It is advisable to wear safety glasses/goggles when handling chemicals.

Apparatus
1. Elastic band to fit around the top of a large test-tube.
2. One 10 cm3 syringe
3. Two 5 cm3 syringes, or four 2 cm3 syringes
4. Container with tap water, labelled “For Washing”
5. Container labelled “Waste”
6. One large test-tube
7. Four test-tubes suitable for heating
8. Test-tube rack or container to hold at least four test-tubes
9. Small beaker or container
10. Bunsen burner, tripod, gauze, bench mat
11. At least a 400 cm3 beaker suitable for a water-bath
12. Matches
13. Thermometer –10 °C to 110 °C
14. Stop clock, stop watch or sight or a clock with a second hand
15. Glass marker pen
16. Safety goggles/glasses

Method

You are required to investigate how much glucose diffuses through selectively permeable
Visking (dialysis) tubing in 15 minutes.

You are provided with

• 25 cm3 of 10% glucose solution, labelled G
• about 20 cm of Visking (dialysis) tubing in a container of distilled water, labelled V
• 100 cm3 of distilled water, labelled W
• 20 cm3 of 0.1 % glucose solution, labelled S1
• 20 cm3 of 0.2 % glucose solution, labelled S2
• 20 cm3 of 0.3 % glucose solution, labelled S3
• 100 cm3 of Benedict’s solution, labelled Benedict’s solution.

Proceed as follows:

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1. Tie a knot in the Visking tubing as close as possible to one end so that it seals the end.
2. To open the other end, wet the Visking tubing and rub the tubing gently between your fingers.
3. Use a syringe to put 10 cm3 of G into the open end of the Visking tubing.
4. Rinse the outside of the Visking tubing by dipping it into the water in the container labelled V.
5. Put the Visking tubing into a large test-tube in a test-tube rack.
6. Fold the open end of the Visking tubing over the top of the large test-tube as shown in
Figure 1.
7. Use an elastic band to hold the Visking tubing in place.
8. Use a syringe to put some of W into the large test-tube so that it surrounds the Visking tubing.
9. Immediately start a stop clock, stop watch or record the time on a clock to time for 15 minutes.
(a) Draw on Figure 1 a line to show the level of water in the large test-tube.

Figure 1

To find out how much glucose has diffused out of the Visking tubing after 15 minutes, you are
provided with solutions S1, S2 and S3.
In order to find how much glucose has diffused from inside the Visking tubing into the water you
will need to test a sample of the water with Benedict’s solution.

You should record the time taken for the first appearance of any green colour.
The result will be compared with the time taken for the first appearance of any green colour
obtained from testing solutions S1, S2 and S3 with Benedict’s solution.

To do this you need to use the same procedure.

(b) State the volume of Benedict’s solution and the volume of the solutions (S1, S2 and S3) and
the sample you are testing.
volume of Benedict’s solution ..................... cm3
volume of each solution (S1, S2 or S3) .................... cm3
volume of sample .................... cm3
(c) State one variable, other than volume, which needs to be kept constant when you do the tests
and describe how you will keep this variable constant.

10. After 15 minutes, pour the water from around the Visking tubing into a beaker or container
and label it sample.
11. Now test all four solutions, sample, S1, S2 and S3.
(d) (i) Record your results.

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(ii) Estimate the concentration of glucose in the sample.

(iii) Suggest how you might modify this investigation to find the effect of temperature on
the rate of diffusion of glucose through Visking tubing.

(e) A student investigated the rate of diffusion of a coloured solution through agar. A Petri
dish containing a layer of agar had a small well of 1 cm diameter, cut so that 10 drops
of the coloured solution could be placed in the well. The distance the coloured solution
diffused from the edge of the well was measured at 15 minute intervals.

Figure 2 below shows the surface view of the Petri dish after 75 minutes.

Figure 2: The surface view of the Petri dish after 75 minutes.

The results of the student’s investigation are shown in Table 1 below.

(i) Plot a graph to show the results in Table 1.

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 (ii) Use the graph to calculate the rate of diffusion of the solution between 10 minutes and 20
minutes. Show on your graph where you took the readings. Show all the steps in your
calculation.

(iii) Describe and explain the trend in the rate of diffusion shown in the graph you have
drawn in (e) (i).

(f) The ruler used to measure the distances in Table 1 is shown in Figure 3.

Figure 3
State the uncertainty of the measurements using this ruler.

PRACTICAL 10: PLANT GROSS FORM AND TISSUES

INTRODUCTION
An angiosperm plant consists of two basic parts: the root system and the shoot system. The root
system is below ground and the shoot system above ground (of course there are exceptions). The
root system constitutes a taproot and lateral roots. Sometimes there is no taproot, and the root
system is adventitious. The root system may be modified into various functional structures
(bulbs, nodules, etc.). The shoot system consists of a stem, together with leaves, flowers and
fruits. The points of attachment to the stem are located on nodes. In between any two nodes are
internodes. The stem may be modified into various functional structures. At the tip of the plant
is the terminal bud which contains the actively dividing meristems. The leaf may be attached to
the stem through a petiole (dicotyledons) or a leaf sheath (monocotyledons). Leaves are
variously arranged on the stem. The blade of the leaf may be divided into leaflets (compound
leaf), or it may be undivided (simple leaf). At the base of the petiole, small lateral outgrowths
(stipules) are found. Between the stem and the petiole of a leaf (leaf axial) lies the axillary bud.

OBJECTIVES

1. To study morphology of the shoot and root systems

2. To compare features of dicotyledons and monocotyledons

MATERIALS AND METHODS

PROCEDURE 1

You are provided with dicotyledon and monocotyledon plant material.

1. Study the shoot system from the provided material.
a) Make sketch diagrams of the stem, and (where present) label the following structures: stipule,

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 axillary bud, terminal bud, node, internode, leaf scar, vascular bundle scar, and lenticel.
b) Makes a sketch diagram of the leaf, and label the following structures: margin, lamina, tip,
midrib, and (where present) petiole, pulvinus, rachis, pinna, auricle and ligule.
c) Which major stem and leaf morphological characters distinguish dicotyledons from
monocotyledons?
d) Relate structure to function in the monocotyledon stem and leaf.
e) What is the functional significance of a compound leaf in a xerophytic environment?
f) Examine the various types of modified stems on display:
i. Potato tuber (a fleshy underground stem). The potato on display shows some “eyes”. What is the
functional significance of the “eye”?
ii. Onion bulb (a very short underground stem surrounded by fleshy leaf bases). Why is the bulb not
wholly a true stems?
iii. Corm (a short, thick stem that grows vertically underground). How does this structure differ from
a stem bulb? What structures are found in the corm, but lacking in the bulb? State the functional
significance of these structures.
iv. Rhizome (a horizontal underground stem) - State two functions associated with this structure.

v. Stolon (a horizontal above ground stem) - How does it differ from the rhizome? State its
functional significance.

2. Tissues of the plant body (an introduction)

When compared to gymnosperms, angiosperms have much more complex tissue structures. Also,
among the angiosperms, there are considerable differences in tissue organisation between
dicotyledons and monocotyledons. The following are some of the major tissues in angiosperms:
meristem, parenchyma, colenchyma, and sclerenchyma, dermal and vascular. Variation in structure
and organisation of tissues has taxonomic and evolutionary significance. Considerable variation in
tissue organisation is also found in different organs of the same plant. Variation in tissue structure
and organisation among plants is best studied at primary growth stage.
OBJECTIVES
1. To study structure and arrangement of tissues in stems and roots of angiosperms at primary growth
stage
2. To compare and contrast root anatomy in monocotyledons and dicotyledons at primary growth stage
3. To compare and contrast stem anatomy in dicotyledons and monocotyledons at primary growth
stage

PROCEDURE 2
Roots and stems arise from apical meristem. The meristem of the root, however, does not occur at
the extreme tip but just behind it. The tip of a root is covered by a thimble-like structure, the root
cap. For several millimetres behind the root cap, the root is smooth, representing the zone of
elongation - a region of undifferentiated tissue. Root hairs are found immediately behind the zone
of elongation.
A. The monocotyledon primary root
T/S of Zea mays (MAIZE)
a) Make a Low Power diagram (Plan diagram) of a whole section to show the distribution of tissues.
b) Make a High Power drawing of a representative section. Note and label the following:
i) epidermis ( a single layer of cells surrounding the entire root);
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 ii) cortex (the ground tissue, composed of several cell types whose primary function is
that of support, storage, secretion, and a variety of other functions):
EXODERMIS (several layers below the epidermis);
PARENCHYMA (bulk of cortex)
ENDODERMIS (with U-shaped thickening on radial walls, and at intervals
with some passage cells)
i) vascular system, comprising:
PERICYCLE
XYLEM (polyarch, note the protoxylem and metaxylem)
PHLOEM (alternating with xylem)
PITH parenchymatous cells (rest of the cells in between the vascular bundles)

B. The dicotyledon primary root

T/S of Phaseolus vulgaris (COMMON BEAN)
a) Make a Low Power diagram (Plan Diagram) of a whole section to show distribution of tissues.
b) Make a High Power drawing of a representative section.
Note and label the following:
i) Epidermis
ii) Cortex with EXODERMIS (single layer of cells beneath the epidermis), PARENCHYMA
(note intercellular spaces and starch grains) and ENDODERMIS (note the thickened
walls)
iii) Vascular system comprising:
PERICYCLE (a single layer of cells);
XYLEM (triarch or tetrarch, note the smaller, outer protoxylem vessels
and the larger, inner metaxylem vessels);
PHLOEM (patches of cells alternating with xylem);
PARENCHYMA (between xylem and phloem).

C. Comparing the T/S of a dicotyledon stem and that of a monocotyledon

1. Make a Low Power diagram and a High Power drawing of a representative section of the stem in
each case.
2. Draw up a comparison table between monocotyledon and dicotyledon roots and stems,
emphasising the diagnostic characters.

D. Transverse section of a plant organ

You are provided with slide Z1, which is a transverse section of a plant organ.
You are also provided with a microscope fitted with an eye piece graticle and a 30 cm
transparent ruler. You are required to examine Z1 carefully using both the low and high power
of your microscope.
(i) Draw an accurate plan drawing of Z1, to show distribution of the various tissues that make up
Z1.
Label your drawing.

(ii) Calculate the actual diameter of Z1, your answer should be in micrometers.
Actual diameter of Z1 …………………………………..𝜇𝑚
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(iii) Describe your method fully
………………………………………………………………………………………………………
………………………………………………………………………………………………………
………………………………………………………………………………………………………
………………………………………
(iv) Put a line on your diagram to show the size that you have measured.
(v) Drawing diameter ………………………………………..mm/ cm
(vi) Calculate the magnification of your drawing, showing working.
(vii) Make a detailed drawing of three adjacent storage cells.

PRACTICAL 11: THE STRUCTURE OF FLOWERS AND ADAPTATION TO

POLLINATION

INTRODUCTION
A typical flower is composed of four whorls of modified leaves (a) sepals (b) petals (c) stamens
and (c) carpel/carpels all attached to the modified stem end that supports these structure, the
receptacle. The sepals-enclose the other flower parts in the bud and are generally green. Sepals
taken collectively constitute the calyx. The petals are usually the conspicuous, coloured,
attractive flower parts. Petals together constitute the corolla. Stamens form a whorl, lying inside
the corolla. Each stamen has a slender stalk or filament at the top of which is an anther, the
pollen-bearing organ. The whorl or grouping of stamens is called androecium. The carpel/carpels
comprises the central whorl of the modified floral leaves, collectively carpels form the
gynoecium. Each individual structure within the gynoecium is referred to as a pistil. A pistil may
be composed of a single carpel or of several united carpels in the center of the flower. There are
generally 3 distinct parts to each pistil: (a) an expanded basal portion, the ovary, in which are
borne the ovules and (b) the style a slender stalk supporting (c) the stigma. Perianth is the term
applied to the calyx and corolla collectively.

Modification of the flower has occurred in relation to mode of pollination. In primitive flowers,
floral parts are usually large and of an indefinite number. Advanced flowers have fewer floral parts
and of a definite number.

Pollination of flowers may be brought about by either wind or insects and occasionally by water,
birds, bats and other small mammals. Flower structure is generally adapted to one or the other of
these pollinating vectors. Insect pollination adaptation may be very complex, indication a long
association of insect vectors and plants. Wind pollination is common in plants with
inconspicuous flowers, such as grasses. Such plants produce pollen in enormous quantities,
flowers lack odour and or nectar and hence are unattractive to insects, pollen is light and dry and
easily wind-borne and their stigmas are feathery and expose a large surface to catch flying pollen

Plants that are pollinated by living vectors usually possess a colourful perianth and produce a
sweet-tasting fluid (nectar) and volatile compounds having distinctive odours. Bees are able to
detect and distinguish between many odours and colours and degrees of sweetness. When the
insect attempts to reach nectar or collect pollen floral architecture ensures that the insect transfers
pollen to the stigma.
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OBJECTIVES
1. To study floral structure in a generalised flower
2. To study floral structures in highly modified flowers

METHODS AND MATERIALS/REQUIREMENTS

A. Dissecting and drawing flowers A, B, C

1. Dissect the provided flower (Flower A) and draw a half flower. Label all parts. Briefly comment
on the structure of the flower.

2. Dissect and draw a half flower from each of the provided flowers (Flowers B and C). Label all
parts. Briefly comment on the structure of the two flowers. Compare them with the flower provided
in 1, above.

B. Observing slides of flowers Z1 and Z2

You are provided with two slides of flowers, Z1 and Z2, from two different plant species.
(a) Examine Z1 using the light microscope under low and high power.
(i) Make a plan drawing of Z1.

(ii) Annotate the plan drawing of Z1.

(b) Examine Z2 using the light microscope under low and high power.

(i)Make a plan drawing of Z2.

(ii) Annotate the plan drawing of Z2.

(c) (i) List three structural differences between Z1 and Z2.

(ii) Suggest the type of pollination for Z1

PRACTICAL 12: THE BASIC LEAF STRUCTURE

INTRODUCTION

Leaves are the primary photosynthetic organs of the plant. Leaves are furnished with a large surface
area and an optimal orientation for maximum capture of light, numerous intercellular spaces for
efficient exchange of gases, and large vascular supply for the maintenance of optimal moisture
levels and efficient translocation of photosynthetic products. The upper and lower epidermis may be
covered with a waxy cuticle to minimise water loss and with stomata providing the only pathway
for gases.

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The leaf consists of dermal, vascular and ground tissue system. The epidermis comprises of a
compact arrangement of cells and the presence of cuticle and stomata. The stomata may occur on
both sides of the leaf (amphistomatic leaf) or only on one side either on the upper (epistomatic
leaf) or commonly on the lower side (hypostomatic leaf). The stomata may be on the same level
as other epidermal cells. The stomata may be located above the surface of the epidermis (raised
stomata) or below it (sunken stomata). Stomata may appear in a depression called a stomatal
crypt. Raised stomata are associated with a hydrophytic habitat providing a large supply of
water. Sunken stomata are associated with a xerophytic habitat characterized by a low supply of
available water. The mesophyll is the main part of the ground tissue of a leaf. The mesophyll
contain chloroplasts and a large volume of intercellular space. It is differentiated into palisade
parenchyma and spongy parenchyma. The palisade parenchyma consists of cells that are
elongated and perpendicular to the surface of the blade in one or more rows. The spongy
parenchyma consists of cells of various shapes. The vascular system of the leaf is distributed
throughout the blade.

The epidermis covers the leaf surface and is generally one cell thick. Its main function is
protection and gas exchange. There are special cells in leaf epidermis called guard cells which
form stomata. The mesophyll comprises of all internal cells of the leaf outside of vascular
bundles. The two main types of mesophyll cells are (i) palisade mesophyll- which are located
near upper epidermis with elongated cell that contain many chloroplasts and (ii) spongy
mesophyll- which are located near lower epidermis and contain many air spaces which are
important in transpiration and CO2 movement. The vascular tissues in leaves (veins) branch
extensively throughout the mesophyll of leaves. They provides support and are involved in water
transportation (xylem) and food transportation (phloem).

OBJECTIVES

To study the basic leaf structure and anatomy

To compare and contrast leaf anatomy between dicotyledons and monocotyledons
To draw a plan diagram of tissues of a transverse section of a dicotyledonous leaf
To calculate the linear magnification of drawings

METHODS AND MATERIALS/REQUIREMENTS

PROCEDURE

1. Make High Power drawings of representative sections of dicotyledon and monocotyledon

leaf specimens provided. Draw up a table to compare and contrast monocotyledon and
dicotyledon leaves, emphasising on diagnostic characters.

B. Experiment to test whether the lower epidermis of a plant has more stomata than the
upper epidermis

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 A student carried out an investigation using epidermal strips of a plant. These epidermal strips
were used to test the following hypothesis “The lower epidermis of the leaves of the plant has
more stomata than the upper epidermis”.

The student presented the results of the investigations as shown in the Table below.

Number of stomata/ mm2

Upper epidermis Lower epidermis
Leaf Leaf Leaf Leaf Leaf Mean Leaf Leaf Leaf Leaf Leaf Mean
1 2 3 4 5 1 2 3 4 5
Strip 1 30 27 35 32 29 32 37 39 33 36
Strip 2 33 29 38 30 32 36 31 40 35 38
Strip 3 31 32 30 31 27 36 34 37 32 35
Strip 4 34 29 33 36 30 39 30 32 38 31

(i) Describe a procedure by which the student could have obtained these results.
(ii) Calculate the mean number of stomata per mm2 on the lower epidermis and upper
epidermis.
(iii) Use the information and formula below to calculate the standard error for these
results.
S = Standard deviation
𝑆
SM = Standard error = 𝑛
√
Upper epidermis: S = 2.96
Lower epidermis: S = 3.04
Standard error, upper epidermis ______________________________
Standard error, lower epidermis ______________________________

Standard error is used to calculate confidence limits. These indicate how certain the student can
be that the true mean of a whole population lies within the range of the estimated sample mean.
The Table below shows some values of t.

Degrees of freedom 10 12 14 16 18 20 22 24 26 28 30 40 50 60
(v)
t Values when 2.23 2.18 2.14 2.12 2.10 2.09 2.07 2.06 2.06 2.05 2.04 2.02 2.01 2.00
probability = 0.05
t. values when 3.17 3.06 2.98 2.92 2.88 2.85 2.82 2.80 2.78 2.75 2.75 2.70 2.68 2.60
probability = 0.01

(ii) State the number of degrees of freedom for one epidermis for the data in the Table.

(iii) Use the information from the Table and the formula below to calculate the confidence intervals
at 95% certainty for the upper epidermis and for the lower epidermis of the leaves.
Confidence interval at 95% = t x SM
Express your answer in the form;
Mean + confidence interval
Show your working
Upper epidermis
Lower epidermis
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(iv) Draw an appropriate conclusion to the student’s experiment

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 FORM 6 TERM 3
REVISION

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BIOLOGY EXAM TIPS

General advice
• Use your syllabus all the time while you are revising and preparing for the examination papers. You
must know which topics you will be tested on.
• Make sure you have all the equipment you will need for the exam in a clear, plastic container. You need
two pens, pencils (preferably HB or B), a clean eraser, a ruler (which measures in mm), a pencil sharpener
and a calculator.

Answering questions
• The questions are designed to test your knowledge and understanding and your ability to apply the
skills you have gained during the course. When you are writing your answers remember that another
person has to be able to read them.
○ Do not waste time by writing out the question before you start to answer.
○ Keep your handwriting clear and legible.
○ Keep your answers on the lines on the question paper. Do not write in the left hand or right-hand
margins of the paper.
○ If you wish to change an answer, cross out your first answer and rewrite. Do not write over what you
have already written.
○ If you have to cross out something, put a line through it; do not scribble over it.
○ If you run out of space, use white space on another part of the exam paper for a continuation
answer; do not try to squeeze in your answer by using very small writing.
○ If you have to use a different space for a rewritten answer or to continue an answer, put a note to tell
the Examiner where it is, e.g. “see page 5” or “see back page”.
○ Always try to write accurately using the correct biological terms. This often helps you to gain marks.
○ If you want to use the word “it” or “they” – think “what is it?” or “what are they?” and then phrase your
answer more precisely.
○ If you want to use the word “affect” or “effect” – remember to write “how they affect” or “what
effect do they have?”

Example 1
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Question
Chronic obstructive pulmonary disease (COPD) is a progressive disease that develops in many
smokers. COPD refers to two conditions:
• chronic bronchitis
• emphysema.
(i) State two ways in which the lung tissue of someone with emphysema differs from the lung tissue
of someone with healthy lungs [2]
Correct answer for two marks
1 There are fewer alveoli than in a healthy lung.
2 The surface area for gas exchange is much smaller.
From the wording of the question it is clear that the answers refer to the lung tissue of emphysema.
Ambiguous answers for no marks
1 There are many air spaces.
2 There is less diffusion of oxygen and carbon dioxide.
3 There are fewer capillaries.
Both types of lung tissue have many air spaces. The technical term alveoli should be used as in the
correct answers. Even though the third answer is correct, it will not be marked as the question asks for two
ways.

Do not write the first answer that comes into your head. You are unlikely to think of exactly the
correct phraseology or have all the necessary detail to answer the question. Plan what you intend to write
before you start writing.
○ Remember to read the question carefully, plan an answer, write the answer clearly, re-read the
question, re-read your answer and then make any additions or corrections clearly. Always re-read your
answers to check them against the question.
○ During your course you will probably have seen many mark schemes from past papers. Do not learn
them. If you write out a mark scheme that you have learnt, it is unlikely to gain you many marks and often
none at all, as it is very unlikely to be relevant to the exact question you were supposed to be answering
○ Be prepared for questions on aspects of practical biology; they can appear on all the papers, not just
Papers 3

Terms
• These are the technical words used in biology. Many of them are given in the syllabus. These terms will
be used in questions. You will get more marks if you can use them correctly in your examination. Ask your
teacher if you are unsure of the meanings of the biological terms used in the syllabus and in any textbook
you are using. You will notice that many terms are defined in the syllabus, so that is a good place to start
when making your own dictionary. Many of the definitions in the 'Definitions' section of the syllabus are
quite long. It would be a good idea to write more concise definitions for yourself and use them to start
your own biological dictionary using your class notes, web sites and the glossaries from the back of text
books.

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○ Try to use the correct spelling. If you cannot remember how to spell a word, write it down as best you
can. The examiners will probably recognise what word you mean; if the spelling is too far out or
ambiguous, then they cannot allow you a mark.
○ Some biological terms have very similar spelling. Make sure you write clearly and always try to spell as
accurately as you can.
○ Do not try to mix the spellings of two words when you are not sure which of them is the correct answer.
For example, you might write “meitosis” when you are not sure whether the answer is mitosis or meiosis.
This answer will not get a mark.

Writing in your own words

• You often have to write two or more sentences to answer a question.
○ Use short sentences. If you write long sentences you can become confused and your meaning
is lost. You might also write something contradictory. It is hard for the examiner to find correct
statements in a muddled answer.
○ You are often asked to write down something you have learned. Make sure you have learnt the
meanings of the common terms used in biology, e.g. active transport, osmosis, photosynthesis and
respiration.
○ During your course take every opportunity to read and write as much as you can to improve the way
you express yourself.

What you should look for in a question

The number of marks
• Always look to see how many marks are available for each question.
° In Paper 1 there is one mark for each question.
° The number of marks is printed on the examination papers for Papers 2, 3, 4 and 5. The mark
available for each part question ((a), (b), (c)(i), etc.) is printed in square brackets, e.g. [2]. The number of
marks helps you decide how much to write. The total number of marks for each question is printed at the
end of the last part question, e.g. [Total: 12].
° The number of marks is a guide to how long to spend on each part of a question.
° Do not waste time and write a long answer for a question which has one or two marks. You will not get
any extra marks even if your answer is full of many correct and relevant statements.
° If there are two or more marks do not write the same thing in two different ways, e.g. “The leaf is very
large. The leaf has a large surface area”. Notice that the second sentence is more accurate and is
preferable to the first one.

The instructions
• These are called command words and tell you what to do.
○ You can find all the command words in the Glossary of terms used in science papers in the
'Appendix' section of the syllabus.
○ If a question asks you to 'name' or 'state' two things only the first two will be marked. Use the

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numbered lines for your answers if they are given on the question paper. If you write more than two and
the first is correct, the second one is wrong, and the third one correct, you will only get one mark (see
Example 1).
○ Some questions have two commands in the question, for example 'predict and explain'. This means that
you have to say what you think will happen AND then say why you think it will happen. Usually the word
and is printed in bold type to help you. See the section below for a tip about answering questions that
have two command terms and require an extended answer.
○ The table below has a list of terms used in biology papers to tell you what to do in an answer. Make sure
you know what you should do in response to each command word.

Example 2
Question
A learner investigated the effect of increasing the concentration of sucrose on the rate of activity of sucrase.
The results are shown in Fig. 4.1.
The graph in Fig. 4.1 shows that as the substrate concentration increases the rate of activity of sucrose
increases to a constant level.
Describe and explain the results shown in Fig. 4.1.
It is quite easy to forget that there are two parts to this question. Before writing your answer it is a good idea
to write description at the beginning of the first of the answer lines and then explanation about half way
down. You could write these in pencil and rub them out when you have finished your answer.
Alternatively, you may choose to write a description of the first part of the graph (activity increases) and
then explain it followed by a description and explanation of the plateau on the graph. That is also a perfectly
acceptable way to answer the question.

What the question is about

• Make sure you know which part of the syllabus is being tested.
○ Read the whole of a question carefully including all the stimulus material and parts (a), (b), (c) (i) and
(c) (ii), etc. before you begin to answer. Some of the parts may have similar answers so you need to work
out the differences between them. If you write exactly the same thing in different parts of the same
question, the answer cannot be correct for both parts.
○ There is often stimulus material for each question. This might be a photograph, diagram, drawing, flow
chart, table of data, graph or just some text. Read all of this information carefully and study any pictures,
tables or graphs that are included. All of it is relevant to the questions.
○ The stimulus material is often about something you have not studied. Do not panic. There will be
enough information in the question for you to work out an answer. You are being tested on your ability to
apply your knowledge to new information.
○ All the different parts of a question may be about the same topic, e.g. cells from section A or blood from
section G, but you should be prepared for questions that test different topics, e.g. the structure and
function of white blood cells (phagocytes and lymphocytes) involving sections of A, G and J.
○ Look for clues in the wording of the questions.

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○ If you are only given a Latin name or a name you do not recognise, e.g. impala, look to see if you are told
anything about it. If in a question on section K you are told that impala are herbivores, then you know
they eat plants.
○ Answer each question as far as you can. Do not spend a long time staring at a question.
○ If you do not know the answer or how to work it out, then leave it and come back to it later. It is best to
put a mark by the side of the question so you can find it easily. An asterisk (*) is a good idea or a large
question mark against the letter of the part question. Not all part questions have answer lines. You may
not realise that you have left out a part question when you check through your script towards the end of
the examination.
○ Try not to leave blanks. Always check through your script towards the end of the examination. When
you come back to a question you may remember what to write as an answer to a question that you left
out earlier in the exam.
○ Do not waste time by writing about things unrelated to the question.

Example 3
It helps to highlight the main features of a question. You cannot use a highlighter pen, so the best thing todo
is to underline or circle key words in the questions.

Command words
• You can find out more about command terms in the Glossary of terms towards the end of the syllabus.
These notes should help you respond to each of the command words.

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The style of questions
We use a great variety of different styles of questions. If you answer plenty of past papers during your
course you will gain lots of practice at these. Here are some:
• Putting ticks and crosses in a table to make comparisons. For example, comparing the properties of
different biological molecules.
• Completing tables of information by writing in single words, numbers or short phrases, e.g. what
happens to the four valves in the heart during different phases of the cardiac cycle.

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• Completing a passage of text with the missing terms.
• Writing definitions – make these as concise as you can; there is no need to use any examples unless
asked.
• Making a list – answers should also be concise; detail is not required.
• Matching pairs from two lists, e.g. matching the names for the stages of mitosis with descriptions of
what happens inside a cell during this type of nuclear division.
• Putting stages of a process into the correct sequence, e.g. the stages of protein synthesis.
• Labelling a diagram – label lines may already be on the diagram or you may have to add them yourself.
• Completing a genetic diagram (Paper 2).
• Describing and/or explaining data from a table or a graph.
• Explaining aspects of an investigation, e.g. a learner investigation that you might have carried out or a
piece of research that has been adapted from a scientific paper.
• Adding information to a flow chart.
• Writing a flow chart from information that you are given, e.g. drawing a food web from written
descriptions of the feeding relationships in a community.

Use information given in the question

• Questions may ask you to “Use examples from...” or “Use only the information in ....” or “With
reference to Fig. 6.2”. If you read instructions like these, find out what you are expected to use as
examples or take information from. You will not get any marks if you use examples from somewhere else.
The information can be given to you in different ways:
○ a diagram, such as a food web, a set of apparatus or a biological structure;
○ a graph, which could be a line graph, a bar chart or a histogram – always check the headings and units
carefully;
○ a table – always read the headings of the columns and/or rows carefully and look for any units.

Interpret tables and graphs

• The stimulus material may be in the form of a table, line graph, bar chart or histogram.
• Always read the introductory text very carefully before you study the table or graph. Underline key
points in the information that you are given. In Papers 2 and 3, there may be quite a bit of introductory
text explaining how the information was collected.

Tables
• Look at the column and row headings in a table and make sure you understand them. If you have read
the introduction carefully, then you will.
• Find the units that have been used. Make sure that you use the units if you give any figures in your
answer.
• Use a ruler to help read the table. Align the ruler with the first column. This should be the independent
variable and should increase in steps. Now put the ruler to the right of the next column and look at the
figures in this second column that should be the dependent variable. Look for a pattern or trend in the
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figures. Identify the pattern or trend first before thinking of an explanation. Move the ruler across to the
right of the third column if there is one and continue in the same way. It may help to sketch a little graph
on the exam paper to help you identify any pattern or trend.

Line graphs
• Look carefully at the x-axis which is the independent variable and make sure you understand what has
been changed. Look carefully at the y-axis which is the dependent variable. Both variables should be
described in the introduction to the question.
• Put your ruler against the y -axis and move it gradually across the graph from left to right. Follow the
pattern or trend of the line (or each line if there is more than one). Mark on the graph where something
significant happens. For example, the line might show that the dependent variable becomes constant
(gives a horizontal line).
• Use your ruler when taking figures from the graph. If the graph is plotted on a grid, then the examiners
may allow ± one small square or half a small square in taking your readings. If you use a ruler and rule
lines on the graph, you should take exact readings.

Bar charts and histograms

• Look carefully at the x-axis and the y-axis to see what has been plotted. Again, it is a good idea to move a
ruler across the bar graph or histogram from left to right to help you concentrate on one aspect at a time.
You can identify the highest and lowest figures and see if there is any pattern.
• You should make yourself some notes about the table, graph or histogram before answering the
questions.

Calculations
• If you are asked to do a calculation:
○ You may have to find the figures from a table or graph.
○ Write out all the working for your calculation. If you make a mistake and give the wrong answer, you
may well be awarded marks for showing how to do the calculation.
○ Make sure that you show the units in the calculation.
○ Make sure you include the units if they are not given on the answer line.
○ Always express your answer in the same way as other figures provided, e.g. in a table. If the other
figures are 5.6 and 4.6, then your answer should be given to one decimal place, e.g. 2.0 and 7.0, not 2 and
7.
○ Round up or down the result on your calculator – do not copy all the figures after the decimal point.

Make comparisons
• If you are asked to compare two things make sure you make it clear which thing you are writing about.
○ The question may ask you to compare two structures or two processes that you have learnt about.
Sometimes you may be expected to do this on answer lines in which case you must make clear the items
that you are comparing (see Example 4).

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○ You may be given a table to complete. This may be blank and you have to fill it in, or it may already have
some entries and you complete it.
○ If you are given lines to make the comparison, it is perfectly acceptable to draw a table for your answer.

Extended writing
• You are required to write longer answers to questions that have four or more marks. There are more of
these questions in Paper 2 than in the other papers. You do not have to write your whole answer in prose.
You can use labelled and annotated diagrams, flow charts, lists and bullet points. However you present
your material, you should write enough to make your meaning clear.

The rest of these tips concern the individual papers

Paper 1
• You have about one minute to read and answer each question. Each question may test one topic or
several topics from different parts of the syllabus.
• Some questions test what you know and understand.
• Some questions test if you can apply what you have learnt to understand new data. These questions will
often have a diagram, graph or table to use.
• Some of the choices can be very similar; read carefully and underline words that make each choice
distinct from the other three.
• Try to decide what the question is testing as you are reading it. The sequence of questions usually
follows the sequence of topics in the syllabus. Therefore you can expect the early questions to ask about
Section A on cells and those at the end will be on Sections J and K about immunity and ecology.
• Do not try to find a pattern in the order of your answers (e.g. A, B, C, D, A, B....)
○ The same letter could be the correct answer for several questions in a row.
○ Letter A might be the correct answers for more questions than B, C or D. Or there could be fewer
correct answers shown by letter D than any of the others.
○ Do not let what you have chosen for the previous questions influence which letter you choose.
• Some questions may ask about aspects of practical work, for example about different variables:
independent, dependent and controlled.
• It is important to understand how to use terminology, e.g. how to apply water potential terminology to
problems on cells and osmosis.

Paper 2
• This paper has a mix of short answers questions and those requiring slightly longer answers. There is no
essay.
• Longer answers will need four or five sentences with two or three different ideas. Always look at the
number of marks for each part question to help you decide how much to write.
• Look at the number of command words: ask yourself ‘do you have to do one or two things?’. See
Example 2.
• Use the lines given. Stick to the point and do not write too much.
• Only give the number of answers that are asked. Use the numbered lines and give one answer per line.

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• There will only be a few parts of questions that need extended writing. These will have four [4] or five
[5] marks. These questions will often be related to some information you are given. You will need to write
four or five sentences in a sequence that makes sense. You can think of it as “telling a story with a
beginning, a middle and an end”. Remember to refer to any information you are given.

Paper 4
General tips
Success at Paper 4 requires you to do plenty of practical work during your course and have several
attempts at past paper questions to find out how to complete everything in the time available. During the
practical exam you will have to make some decisions; if you practise plenty of past questions you will find
out what sort of decisions to expect. As you revise, make sure you know exactly how to carry out the
practical procedures described in the syllabus. You will be assessed on your skills at:

• manipulating apparatus to collect results and make observations

• data presentation
• analysis of results and observations
• evaluation of procedures and data.
You should make decisions, such as:
• identifying variables
• standardising the control variables
• how to change the independent variable
• choose the number of measurements to take
• decide the intervals between the values of the independent variable
• choosing a control experiment
• identifying any risks and stating appropriate precautions.

During the examination

• Read through the questions carefully, looking to see how many marks are given for each question.
• Read the instructions to the end; do not start a practical procedure without reading carefully all the
steps involved.
• As you read, check that you have the apparatus and materials described. If not alert the supervisor.
• Think about the apparatus that you will use for each step and imagine using it in your mind.
• Make sure that you have a sharp pencil to use for making drawings and for drawing graphs and charts.
Do not draw in ink because you cannot make changes as you can when using a pencil.
• Make sure you have a good, clean eraser for rubbing out your pencil lines if necessary. Do not press too
hard when using a pencil for making drawings, graphs or charts. Sometimes it is hard for an examiner to
tell which is your final line.

Following the instructions

• Follow the instructions for practical methods exactly. If you make a change in the method you can alter
the results.
• Do not take short cuts.
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• Always label test-tubes and other containers to help you remember which is which.
• If you are told to “Wash the apparatus thoroughly after each use” make sure you do. If there is anything
left in the apparatus the next stage may not work.

It is a good idea to put a tick by the side of each instruction when you have completed it. This helps you to
find the right place in the instructions, so that you do not leave out a step or repeat a step when it is not
required.
• Keep your exam paper on a part of the bench which you can keep dry. Do not pour liquids or use
syringes or pipettes over your exam paper. If you keep your exam paper away from the ‘wet’ part of your
bench you are unlikely to spill anything on it.

Recording your measurements and observations

You are expected to make observations and record them.
• You can record your observations:
○ as statements in writing
○ in tables
○ by using drawings
○ by constructing tally charts.
You will take readings from different apparatus. You must make the measurements as accurately and
reliably as you can. Numerical readings will normally be collected and presented in a table.
• Follow the instructions below about drawing tables.
• Make clear descriptions of colours and colour changes; refer to ‘blue’, ‘orange’ and ‘purple’ when
describing reagents used in biochemical tests. You may want to refer to slight differences, so use words
like ‘pale’ and ‘dark’.
• Make your measurements as accurate and reliable as possible.
• Accurate results are close to the actual or ‘true’ values; reliable results are those that are repeatable.
• If you can take repeat readings, then do so. There is not always enough time to do this.
• You can process your observations by:
○ carrying out calculations, e.g. percentages and percentage changes
○ plotting graphs – line graphs, bar charts and histograms.
• Use all the space available on the paper for your observations.
• Do not write an explanation until the question asks for one.
• Use a sharp HB or B pencil. It can be rubbed out easily if you need to correct a mistake. Use a good
eraser so that is clear to the examiner which is your final line.
• Do not forget to include headings for the columns and the rows in tables.

Drawings
These will be from microscope slides or photographs.
• Read the question carefully, the drawing may have to be an accurate size e.g. twice the original.
• Make each drawing as big as the space allows without writing over the text of the question and making
sure that you leave enough space for labels and annotations, if asked for.
• Use a ruler for labelling lines.
• Draw and label in pencil.
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• Use one clear continuous outline not an artistic drawing. Do not shade.
• Observe details carefully, such as the relative number of chloroplasts in different cells and the thickness
of cell walls in different cells in a vascular bundle. Show these accurately on your drawing.

A plan diagram shows the distribution of tissues in a section. It also shows the proportions of the
different tissues. Although called a low power plan diagram you may use high power to identify the
different tissues and to be sure you are putting the boundaries of those tissues in the right place. You
should not draw any cells in a lower power plan diagram.
When you make a plan diagram, follow these simple rules:

• make the drawing fill most of the space provided; leave space around the drawing for labels and
annotations (if required by the question)
• use a sharp HB or B pencil (never use a pen)
• use thin, single, unbroken lines (often called ‘clear and continuous lines’)
• show the outlines of the tissues
• make the proportions of tissues in the diagram the same as in the section
• do not include drawings of cells
• do not use any shading or colouring.
Add labels and annotations (notes) to your drawing only if you are asked for these in the question. Use a
pencil and a ruler to draw straight lines from the drawing to your labels and notes. Write labels and notes
in pencil in case you make a mistake and need to change them. You may leave your labels and notes in
pencil –do not write over them in ink.

High power drawings should show a small number of cells and they should be drawn a reasonable size
so you can show any detail inside them. When you make a high power drawing, follow these simple rules:
• make the drawing fill most of the space provided; leave space around the drawing for labels and
annotations (if required by the question)
• use a sharp HB or B pencil (never use a pen)
• use clear, continuous lines (see above)
• draw only what is asked in the question, e.g. three cell types or one named cell and all cells adjoining it
• show the outlines of the cells
• the proportions of the cells in the drawing must be the same as in the section you are drawing
• plant cell walls should be shown as double lines with a middle lamella between the cells; the
proportions of cell walls should be drawn carefully.
• show any details of the contents of cells – draw what you see not what you know should be present; for
example, in plant cells you may see nuclei, chloroplasts and vacuoles
• do not use any shading or colouring.

Taking measurements of specimens and photographs

Using an eyepiece graticule
An eyepiece graticule is a scale that fits inside the eyepiece on your microscope. It allows you to take
measurements of the specimens you view with the microscope. You can measure simply in graticule units,

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but you may be asked to make an actual measurement which involves calibrating the graticule using a
stage micrometer. This is done by lining up the graticule with the divisions on the micrometer.
• Make your measurements as accurate as you can. You will probably be able to measure to the nearest
division on the graticule.
• You may be asked to take several measurements and then calculate a mean.

Taking measurements from photographs

You may have to measure an object on a photograph and calculate the actual size of a structure or the
magnification of an image.
• Always measure photographs in millimetres, not centimetres.
• If you have to use your measurements in a calculation, write neatly and show your working. The person
marking your paper might be able to give you marks for knowing what to do even if you make a mistake
or do not finish the calculation.

Presenting data and observations

Tables
Before you start to draw a table, decide what you wish to record. Decide on how many columns and how
many rows you will need. Make sure you have read all the instructions before you draw the table outline.
Follow these rules:
• use the space provided, do not make the table too small
• leave some space to the right of the table in case you decide you need to add one or more columns
• make the table ready to take observations or readings so that you can write them directly into the table
rather than on another page and then copy them into the table (tables need to show all the raw data you
collect)
• draw the table outlines in pencil
• rule lines between the columns and rows
• rule lines around the whole table
• write brief, but informative headings for each column
• columns headed with physical quantities should have appropriate SI units
• when two or more columns are used to present data, the first column should be the independent
variable; the second and subsequent columns should contain the dependent variables
• entries in the body of the table should be brief – they should be single words, short descriptive phrases
or numbers
• data should be recorded in the table in the order in which it is collected – this is because the table is
prepared before the data collection. For example, if the instructions state that results from the highest
temperature or highest pH is to be recorded first then these go at the top of their respective columns. It is
more usual to arrange the values of the independent variable in ascending order (e.g. from 0 to 100) so
that patterns are easier to follow and that is how data in tables for Papers 1, 2, 3 and 4 is usually
presented
• numbers written into the body of the table do not have units (units only appear in the column
headings).

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You may have to process your results by calculating rates of reaction, changes in length, percentage
changes or means of repeat readings. These processed results can appear in the same table with the raw
data that you have collected or can be in a separate table with the independent variable. The solidus or
slash (/) meaning ‘per’ should not be used in units. For example, if you have to include concentrations as
in a table you do not write g per 100 cm3 as g/100 cm3. It should always be written out in full using ‘per’
or, better, as g 100 cm–3. The negative exponent, cm–3, means ‘per’.
Note that the solidus is used to separate what is measured from the unit in which it is measured. You may
notice that text books and examination papers use brackets around the units in tables. This is also an
accepted convention, but the solidus is the convention used in A LEVEL Biology.
Correct and incorrect ways of showing units in tables and graphs

A note on the uses of ticks and crosses in tables:

Do not use ticks and crosses in tables of results which should show observations, such as the colours
obtained in biochemical tests. Ticks and crosses may be used in tables of comparison if there is a key to
explain what they mean,
e.g. = present; x= absent.
You may want to show anomalous results in tables. If so circle them and put a note underneath the table
to explain that they are anomalous results. You may be asked to compare specimens viewed in the
microscope and/or in photographs. These comparisons must be organised into a table. Draw your table
so that it has a first column for the features that you have observed. You can then present both
similarities and differences:

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Line graphs
Line graphs are used to show relationships in data which are not immediately apparent from tables. The
term graph applies to the whole representation. The term curve should be used to describe both curves
and straight lines which are used to show trends.
Follow these guidelines:
• use at least half the grid provided, do not make the graph too small
• draw the graph in pencil
• the independent variable should be plotted on the x-axis
• the dependent variable should be plotted on the y-axis
• each axis should be marked with an appropriate scale. The origin should be indicated with a 0. The data
should be examined critically to establish whether it is necessary to start the scale(s) at zero. If not, you
may have a displaced origin for one or both axes, but this must be made obvious by labelling the displaced
origin very clearly
• each axis should be scaled using multiples of 1, 2, 5 or 10 for each 20 mm square on the grid. This makes
it easy for you to plot and extract data. Never use multiples of 3
• each axis should be labelled clearly with the quantity and SI unit(s) or derived (calculated) units as
appropriate, e.g. time/s and concentration/g dm–3; the axes labels and units must be the same as those in
the table
• plotted points must be clearly marked and easily distinguishable from the grid lines on the graph. Dots
in circles () or small, neatly drawn crosses (x) should be used; dots on their own should not. If you need to
plot three lines, vertical crosses (+) can also be used
• label each line carefully or use a key. Use a pencil for both lines; do not use a blue or black pen or
different colours
• in Paper 4 there are usually five or six results to plot.

After plotting the points you need to decide if any of them are anomalous. Ask yourself the question ‘do
they fit the trend?’. But what is the trend? You should know something about the theory behind the
investigation so you should be aware of the likely trend. If you think one or more of the results are
anomalous, then it is a good idea to ring them. Put a circle on the graph away from the line and put a key

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to state that the circled point(s) represent anomalous result(s). The next thing to decide is how to present
the curve.
• It may be obvious that the points lie on a straight line; for example, the effect of enzyme concentration
on the rate of an enzyme-catalysed reaction. If you have a result for the origin (0, 0) then that must be
included and you can place a clear plastic ruler on the grid and draw a straight line from the origin
making sure that there is an even number of points on either side of the line. If you do not have a result
for the origin, then start the line at the first plotted point. Do not continue the line past the last plotted
point.
• You should only draw a smooth curve if you know that the intermediate values fall on the curve. You
may be expecting the relationship to be a smooth curve and if the points seem to fit on a curve then draw
one. Again decide whether the origin is a point and, if not, start at the first plotted point. The curve should
go through as many points as possible, but try to make sure there is an even number of points on either
side of the line. Do not continue past the last plotted point.
• In the practical examination you may only have five or six results. These are likely to be single results
rather than means of replicate results. Therefore you cannot be sure of the relationship and should not
draw a straight line or a curve as described above. You should draw straight lines between the points.
This indicates uncertainty about the results for values of the independent variable between those plotted.
• If a graph shows more than one line or curve, then each should be labelled to show what it represents

If you have times in minutes and seconds, never use minutes as the unit on a graph. It is very difficult to
use a scale with each small square representing 3 or 6 seconds. Always plot results in seconds unless the
unit for time is whole minutes.

Analysis, conclusions and evaluation

As part of analysis you should be able to:
• identify anomalous results. Anomalous results are those that do not fit the trend
• process your results to calculate means, percentages, changes in mass or length, calculate percentage
changes and rates of reactions
• find unknown quantities by using axis intercepts or estimating from colour standards using known
concentrations
• describe the pattern or trend in data
• make conclusions to consider whether experimental data supports hypotheses or not.

Processing results
You should be prepared to calculate:
• means
• percentages
• percentage changes
• rates of reaction by calculating 1/t or 1000/t; the unit used is s–1.
You should know how to use line graphs to:
• find an intercept – where a line you have drawn crosses a key value on the x-axis; for example, finding
the water potential of a tissue using percentage change in length of plant tissues
• find the rate of a reaction by calculating the gradient of a line you have drawn.
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As part of evaluation you should be able to:
• identify systematic and random errors
• systematic errors are those that affect all the results in the same way
• random errors do not affect all the results in the same way
• identify the significant errors in your investigation
• estimate the uncertainty in measurements. The actual error is half the smallest division on the
apparatus you are using
• assess how effective you have been at standardising variables
• suggest improvements to the procedure you have followed
• suggest ways in which the investigation might be extended to answer a new question.

Estimating uncertainty in your results

You may have to estimate the uncertainty or error in your results. For particular apparatus, the error is
half the smallest graduation on the apparatus, e.g. if the smallest division is 1.0 cm3 then the uncertainty
would be ±0.5 cm3. So if you start your measuring at 0 the uncertainty applies where you take your
measurement –say at 6.3 cm3. So the result is expressed as 6.3 ± 0.5 cm3. BUT if you have to start at a
measurement other than 0 (for example when taking readings from a burette) the uncertainty applies at
both ends, so it is multiplied by two as there is an error at each end, e.g. 7.5 ± 1.0 cm3. Similarly, if using a
ruler then there would be an error at each end unless you start at 0. The same applies to measuring a
quantity in a syringe by sucking up from empty. The error would be half the minimum measurement. But
when you take two readings from the syringe (say delivering 2.0 cm3 by moving the plunger from 6.5
cm3 to 4.5 cm3) the
uncertainty is multiplied by two.
Percentage error is calculated as the error expressed as a percentage of the actual reading. For example if
the reading is 7.5 ± 1.0 cm3, then the percentage error is 1.0/7.5 × 100 = 13.3%.

Conclusions
• Conclusions are brief statements supported with explanations using your knowledge from the
syllabus
• Use your own results for your conclusions.

Before planning what to write for a conclusion, turn back to the beginning of the question and read the
introduction. You may have forgotten what you were told about the investigation you have just carried
out. Think about the theory and apply it to the results you have obtained.
• Sometimes you are expected to make conclusions about some other data, not the data you have
collected.
• Do not write the conclusion you have learned from a class experiment or from theory.
• You should also consider the confidence that you have in your conclusions. For this it is a good idea to
consider whether:
○ the standardised variables have been kept constant
○ there were any other variables that were not standardised
○ there were any anomalous results
○ any replicate results were similar or not.
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• If you are unsure about any aspect of the practical you have carried out, then you can say that you do
not have confidence in your conclusions and give a reason or reasons.

Suggesting improvements
You may be asked to suggest modifications or improvements that will increase the accuracy and
reliability of the results. As you carry out the practical procedure you should think critically about it and
make some notes. If asked to suggest improvements, then look back to these notes for ideas. You can
suggest:
• ways to improve the standardisation of variables, for example by using a thermostatically-controlled
water bath
• taking repeat readings (replicates) to assess the reliability of the data
• calculate mean results
• use a different way to measure the dependent variable so the results are more accurate
• use a different piece of apparatus to measure the dependent variable and reduce the percentage error
(see above)
You may also have to justify your suggested improvements. When you do this, make sure you explain how
they will improve the confidence you have in the data and therefore in the conclusion.

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