2.

1 Molecules to Metabolism

Metabolic Processes

Define metabolism
The totality of all enzyme-catalysed reactions in a cell or organism
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Compare anabolism and catabolism (including an example of each)

Anabolism Catabolism

• The build up of complex molecules from more

………………………………………………………………………… • The break down of complex molecules into more
…………………………………………………………………………
simple subunits (i.e. monomers)
………………………………………………………………………… simple subunits (i.e. monomers)
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• Requires condensation reactions to proceed
………………………………………………………………………… • Requires hydrolysis reactions to proceed
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(water is produced as a by-product) (water is consumed as part of the reaction)
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• Example: Photosynthesis • Example: Cell respiration
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Differentiate between organic and inorganic compounds

Organic molecules contain carbon and are synthesised by living organisms (anything else is inorganic)
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Exceptions include carbides, carbonates, oxides of carbon and cyanides
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Explain how the structure of the carbon atom contributes to the formation of organic life
Carbon forms the basis of organic life due to its capacity to form large molecules
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Carbon has four valence electrons and can form four covalent bonds
…………………………………………………………………………………………………………………… +
6p
6n
This allows it to function as a stable backbone in a wide variety of compounds
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Explain the theory of vitalism and how it was falsified

Vitalism proposed that organic molecules could ONLY be synthesised by living organisms
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It was falsified by Frederick Woehler in 1828
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He was able to synthesise urea (an organic molecule) from an inorganic salt under laboratory conditions
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Biomacromolecules

Define monomer
A recurring subunit within a more complex polymer
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Compare the four different types of biomacromolecules

Macromolecule Monomer / Subunit Polymer Bond Involved

Carbohydrate Monosaccharide Polysaccharide Glycosidic Linkage

Lipid Glycerol + fatty acids (x3) Triglyceride Ester linkage

Protein Amino acid Polypeptide Peptide bond

Nucleic Acid Nucleotide DNA or RNA Phosphodiester bond

Write the chemical formula for an unsaturated fatty acid

CH3 – (CH2)n – COOH
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Draw molecular diagrams of glucose and ribose

Glucose (α-D) Ribose

Explain how the following may be identified based on their chemical formula
Have C,H,O in a common ratio according to formula (CH2O)n
Carbohydrate: ………………………………………………………………………………………………………………………………..
May contain sulphur (some amino acids include sulphur)
Protein: …………………………………………………………………………………………………………………………………………
Will contain phosphorus in relatively large amounts (nucleotides include a phosphate group)
Nucleic Acid: ………………………………………………………………………………………………………………………………….
 2.2 Water

Water Structure

Explain the polarity of water

• Water is made up of two hydrogen atoms covalently bonded to an oxygen atom
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• The oxygen atom has a high electronegativity and attracts the shared electrons more strongly
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• This results in polarity (the O atom is slightly negative, while the H atoms are slightly positive)
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• Due to this polarity, water can form hydrogen bonds between the O of one molecule and the H of another
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Draw a diagram of two water molecules, showing the intermolecular bonding between them

+ +
–
δ δ H H δ
+
O +
δ δ O
H H –
Hydrogen bond δ

Water Properties

Distinguish between cohesion and adhesion with relation to water

Cohesion: Cohesion is when two identical molecules ‘stick’ together (via intermolecular bonding)
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Water molecules are cohesive (they can ‘stick’ together via hydrogen bonding)
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Adhesion: Adhesion is when two different molecules ‘stick’ together (via intermolecular bonding)
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Water molecules are adhesive with polar or charged substances

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Describe the biological significance of the cohesive and adhesive properties of water

• The cohesive properties of water results in water having a high surface tension (due to extensive H bonding)
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• This is biologically significant as it allows small insects to move along the surface of water (e.g. water striders)
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• The adhesive properties of water result in capillary action when in contact with charged or polar surfaces
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• This is biologically significant as it allows for a transpiration stream in plants (flow of water against gravity)
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Explain the thermal properties of water
• Water molecules can form extensive hydrogen bonding between molecules, which require energy to break
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• This means it takes a lot of thermal energy (heat) to change the temperature (or state) of water
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• Hence, water has a high specific heat capacity (and a high heat of vaporisation / fusion)
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Describe the biological significance of the thermal properties of water

• Because water has a high specific heat capacity, it functions as an excellent biological coolant
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• Sweating results in evaporative cooling, as ambient heat is absorbed to evaporate water (break H bonds)
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• This cools the air surrounding the skin and also directly draws heat from the skin
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Compare water (H2O) and methane (CH4)

Similarities:
• They have a similar size and molecular weight (water = 18 daltons, methane = 16 daltons)
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• They have comparable valence structures (tetrahedral orbitals– although water is bent due to unbonded pairs)
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Differences:
• Water has a significantly higher boiling point and melting point
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• Water has a higher specific heat capacity (plus a higher heat of vaporisation / fusion)
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• This is because water is polar and can form hydrogen bonds (methane is non-polar and doesn’t form H bonds)
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Distinguish between hydrophilic and hydrophobic

Hydrophilic substances are soluble in water (hydrophilic = water ‘loving’)
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Hydrophobic substances are insoluble in water (hydrophobic = water ‘hating’)
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Describe (with examples) how the solubility of molecules affects their mode of transport within the blood
• Ionic compounds (e.g. salt) dissociate in water and are transported within blood plasma in a dissolved state
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• Glucose and other monosaccharides are water soluble and hence are transported freely within blood plasma
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• Amino acids are zwitterions and can be freely transported within blood plasma in an ionised state
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• Oxygen is soluble in water, but only in low amounts (most oxygen is complexed to haemoglobin in red blood cells)
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• Lipids (fats and cholesterol) are non-polar and insoluble in water (they are transported in blood as lipoproteins)
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 2.3 Carbohydrates and Lipids

Carbohydrates

Outline the structural organisation of carbohydrates (monomer → polymer)

Monosaccharides are joined together by condensation reactions to form polysaccharides (water produced)
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The covalent bond connecting the monosaccharides together is called a glycosidic linkage
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Polysaccharides can be digested into smaller oligosaccharides via hydrolysis reactions (water is required)
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List three examples of monosaccharides, disaccharides and polysaccharides

Glucose, Galactose, Fructose (Hint: Gives Good Flavour)
Monosaccharides: ………………………………………………………………………………………………………………………...
Lactose, Sucrose, Maltose (Hint: Length Supports Movement)
Disaccharides: ………………………………………………………………………………………………………………………...
Cellulose, Glycogen, Starch (Hint: Can Get Stored)
Polysaccharides: ………………………………………………………………………………………………………………………...

List three functions for carbohydrates within cells

Short-term energy source (e.g. glycogen and starch)
1. …………………………………………………………………………………………………………………………………………………...
Structural component (e.g. cellulose in plant cell walls)
2. …………………………………………………………………………………………………………………………………………………...
Cell recognition and signalling (e.g. glycoproteins in plasma membranes)
3. …………………………………………………………………………………………………………………………………………………...

Complete the following table comparing different polysaccharides of glucose

Monomer Bonding Shape Function

ß-glucose 1-4 linkages linear (sheets) structural component of

Cellulose
plant cell walls

short-term energy storage

Amylose α-glucose 1-4 linkages linear (helical)
in plants (type of starch)

1-4 linkages
less branched short-term energy storage
Amylopectin α-glucose AND
(~per 20 units) in plants (type of starch)
1-6 linkages

1-4 linkages
more branched short-term energy storage
Glycogen α-glucose AND
(~per 10 units) in animal cells
1-6 linkages
Lipids

Identify the main classes of lipids by completing the flow chart below

Lipids

Simple lipids Compound lipids Derived lipids

Esters of fatty acids Esters of fatty acids and other group Composed of hydrocarbon ring

triglycerides, waxes phospholipids, sphingolipids, etc. steroids, cholesterol

List five different functions of lipids

Storage of energy (e.g. triglycerides)
S ……………………………………………………………………………………………………………
Hormonal role (e.g. steroids)
H ……………………………………………………………………………………………………………
Insulation (e.g. sphingolipids)
I ……………………………………………………………………………………………………………
Protection of organs (e.g. triglycerides, waxes)
P ……………………………………………………………………………………………………………
Structural component (e.g. phospholipids)
S ……………………………………………………………………………………………………………

Compare the use of carbohydrates and lipids (triglycerides) as energy sources

triglycerides = long-term energy storage ; carbohydrates = short-term energy storage
Storage: …………………………………………………………………………………………………………………………………………
triglycerides have less effect on osmotic pressure (they are hydrophobic / not a solute)
Osmotic Effect: ………………………………………………………………………………………………………………………………
carbohydrates are easier to digest (more readily consumed and can be digested anaerobically)
Digestion: ………………………………………………………………………………………………………………………………………
lipid digestion has a higher ATP yield (~twice as much energy per gram)
ATP Yield: ………………………………………………………………………………………………………………………………………
triglycerides are insoluble in water and hence harder to transport
Solubility: ………………………………………………………………………………………………………………………………………

Explain why carbohydrates and lipids are used preferentially to proteins as an energy source
Protein digestion produces nitrogenous wastes (due to the presence of an amine group)
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Nitrogenous wastes are toxic to cells (must be removed via excretion by the kidneys)
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Differentiate between saturated and unsaturated fatty acids

Saturated fatty acids do NOT have double bonds (possess maximal amount of H atoms)
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Unsaturated fatty acids possess double bonds (may be monounsaturated or polyunsaturated)
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Contrast cis and trans isomers

Cis Isomer Trans Isomer

H atoms on the same side of the double bond H atoms on opposite sides of the double bond
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Double bond creates kink in fatty acid chain Double bond does NOT create kink in chain
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Are loosely packed and usually liquid Are tightly packed and usually solid
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Occurs commonly in nature Occurs in processed foods
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Generally considered good for health Generally considered bad for health
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Identify the role of high density and low density lipoproteins

Low density lipoproteins (LDLs) transport cholesterol from the liver to the rest of the body
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High density lipoproteins (HDLs) scavenge excess cholesterol and return it to the liver for disposal
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Outline the effect of different types of fatty acids on lipoprotein levels

Cis fats raise levels of HDL (lowers blood cholesterol)
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Saturated fats raise levels of LDL (raises blood cholesterol)
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Trans fats raise levels of LDL and lower levels of HDL (significantly raise blood cholesterol)
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Cis fats are generally considered good for health, saturated and trans fats are considered bad for health
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Describe the health implications associated with high levels of blood cholesterol
High cholesterol levels lead to hardening and narrowing of the arteries (atherosclerosis)
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The cholesterol forms fat deposits in the arterial lining, leading to the development of plaques
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If a plaque ruptures, blood clotting will cause the vessel to become blocked
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If coronary arteries become blocked, coronary heart disease will result (including heart attacks, etc.)
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Discuss the evidence for (and against) health risks associated with diets rich in lipids
Evidence for:
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A positive correlation exists between intake of saturated fats and incidence of CHD
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Intervention studies demonstrate that lowering intake of saturated fats reduces onset of CHD
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In patients who have died from CHD, fatty deposits with high levels of trans fats were found in arteries
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Evidence against:
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Certain populations have high fat intakes but low rates of CHD (e.g. Maasai tribe in Africa)
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Genetic factors play a significant role (blood cholesterol levels only show a weak association to diet)
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Results from intervention studies are influenced by cohort size and composition, as well as study duration
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Use the information included below to assess the weight of individuals according to BMI categories

Underweight Normal Weight Overweight Obese (Class I) Obese (Class II) Obese (Class III)

<18.5 18.5 – 24.9 25.0 – 29.9 30.0 – 34.9 35.0 – 39.9 >40.0

Individual 1: Height = 175 cm Weight = 65 kg

65 ÷ (1.75 x 1.75) = 21.22 Normal weight

BMI: ……………………………………………………… Category: ……………………………………………………………..

Individual 2: Height = 190 cm Weight = 130 kg

130 ÷ (1.9 x 1.9) = 36.01 Obese (Class II)

BMI: ……………………………………………………… Category: ……………………………………………………………..

Individual 3: Height = 165 cm Weight = 50 kg

50 ÷ (1.65 x 1.65) = 18.37 Under weight

BMI: ……………………………………………………… Category: ……………………………………………………………..
 2.4 Proteins

Protein Structure

Draw the basic structure of a generalised amino acid

Outline the structural organisation of a polypeptide chain (monomer → polymer)

Amino acids are joined via condensation reactions to form dipeptides (a water molecule is produced)
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The covalent bond that connects amino acids together in a polypeptide chain is called a peptide bond
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Polypeptide synthesis occurs at the ribosome via the process of translation
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Draw the basic structure of a generalised dipeptide

Identify the number of different amino acids present in nature and how they are distinctive
There are 20 different amino acids present within nature
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They differ according to the chemical composition of their variable side chain (the ‘R’ group)
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Explain, with the aid of the diagram, the levels of protein structure and indicate their significance

Primary Structure:
The order / sequence of amino acids within a polypeptide chain
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Formed via peptide bonds between the amine and carboxyl groups of adjacent amino acids
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Primary structure determines all subsequent levels of protein structure
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Secondary Structure:
The folding of a polypeptide chain into repeating arrangements (alpha helices or beta-pleated sheets)
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Formed via hydrogen bonds between the amine and carboxyl groups of non-adjacent amino acids
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Tertiary Structure:
The overall three dimensional shape of the polypeptide chain
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Formed via a variety of bonds / interactions between the variable side chains (R groups)
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These interactions may include hydrogen bonds, ionic bonds, disulphide bridges or hydrophobic interactions
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Quaternary Structure:
The presence of multiple polypeptides or prosthetic groups to form a biologically active protein
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Not all proteins will have a quaternary structure
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Protein Function

List seven functions of protein, providing an example for each

Structure (e.g. spider silk, collagen)
S ………………………………………………………………………………………………
Hormones (e.g. insulin, glucagon)
H ………………………………………………………………………………………………
Immunity (e.g. antibodies)
I ………………………………………………………………………………………………
Transport (e.g. haemoglobin)
T ………………………………………………………………………………………………
Sensitivity (e.g. rhodopsin)
S ………………………………………………………………………………………………
Movement (e.g. actin, myosin)
M ………………………………………………………………………………………………
Enzymes (e.g. catalase, Rubisco)
E ………………………………………………………………………………………………

Define denaturation
A loss of protein structure resulting in a loss of biological activity
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Identify two factors that can cause protein denaturation and explain how this occurs
Temperature (Heat)
1. …………………………………………………………………………………………………………………………………………………...
Thermal energy breaks hydrogen bonds responsible for tertiary structure
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A loss of tertiary structure results in a loss of function
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pH
2. …………………………………………………………………………………………………………………………………………………...
Amino acids are zwitterions – possessing both positive and negative regions (amine and carboxyl groups)
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Altering the pH will change the charge and solubility of the protein (changing the structure)
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Outline the relationship between genes and polypeptides

Genes are sequences of DNA that encode polypeptides
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Typically, one gene equals one polypeptide – however exceptions exist (e.g. tRNA genes)
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Define proteome
The totality of proteins expressed within a cell, tissue or organism at a given time
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 2.5 Enzymes

Enzyme Action

Define the following terms

A globular protein that speeds up the rate of a chemical reaction by lowering the activation
Enzyme: …………………………………………………………………………………………………………………………………...
energy (i.e. a biological catalyst)
…………………………………………………………………………………………………………………………………………………..........
A compound that binds to an enzyme’s active site and is converted into a product
Substrate: …………………………………………………………………………………………………………………………………...
(the substrate is complementary in shape and chemical properties to the active site)
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Outline the steps in enzymatic binding and catalytic activity with the aid of the following diagram

Enzyme and substrate collide in an appropriate orientation

1. …………………………………………………………………………………………………………………………………………………...
(substrate binds to the enzyme’s active site)
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Enzyme and substrate form a complex, leading to catalysis
2. …………………………………………………………………………………………………………………………………………………...
(enzyme-substrate interaction shows specificity)
…………………………………………………………………………………………………………………………………………………...
Substrate is converted into a product
3. …………………………………………………………………………………………………………………………………………………...
(enzyme may stress the bonds within the substrate to catalyse this process)
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Enzyme and product dissociate
4. …………………………………………………………………………………………………………………………………………………...
(enzyme is not consumed by the reaction and may be re-used)
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Identify factors that can affect an enzyme’s molecular motion and collision with a substrate
Temperature - increases the kinetic energy of particles, leading to more frequent collisions
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Substrate concentration - increases frequency of collisions with enzyme
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Enzyme concentration - increases frequency of collision with substrate
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 Distinguish between the ‘lock and key’ and ‘induced fit’ models of enzyme specificity

Lock and Key Model:

Substrate and active site are complementary in both shape and chemical properties
…………………………………………………………………………………………………………………………………………………..........
Means that enzymes are specific for the reaction they catalyse
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Induced Fit Model:

Active site is not a perfect fit, but undergoes a conformational change in shape to accommodate substrate
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This stresses bonds in the substrate (promoting catalysis) and gives rise to broad specificity
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Explain the effect of different factors on enzyme activity

Enzyme Activity

Enzyme Activity

Enzyme Activity

Temperature pH Substrate Concentration

At low temperatures there is insufficient activation energy for reaction to proceed

Temperature: ………………………………………………………………………………………………………………………………….

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As temperatures increase, the rate of reaction increases (higher kinetic energy = more collisions)

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At a certain optimum temperature, the rate of reaction will plateau (i.e. peak efficiency)

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Above this temperature, the enzyme will begin to denature and rate of reaction will decrease

Enzymes have an optimal pH at which the rate of reaction is highest

pH: ………………………………………………………………………………………………………………………………………………….
At higher and lower pH levels, activity will decrease (leading to a bell-shaped curve)
…………………………………………………………………………………………………………………………………………………..........
This is because pH affects the charge and solubility of the enzyme
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The change in the enzyme’s chemical properties causes it to denature (lose tertiary structure)
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Enzyme activity increases as substrate concentration increases
Substrate Concentration: …………………………………………………………………………………………………………………
This is because there are more substrate particles, leading to a higher rate of successful collisions
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At a certain point, the rate of enzyme activity will plateau
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This is because all enzyme active sites are occupied (the solution is saturated)
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Enzyme Application

Explain the benefits of immobilized enzymes in industry, with examples

Immobilised enzymes have been fixed to a static surface in order to increase enzyme efficiency
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• Enzyme concentrations are conserved, as the enzyme is not dissolved - hence it can be re-used
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• Product is more easily separated from enzyme, as the enzyme is fixed in position
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Examples of immobilised enzymes in industry include in food production, medicine and biofuel production
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Outline the production of lactose-free milk and its advantages

Production of Lactose-Free Milk:

Lactase is fixed to agarose beads (lactase purified from yeast or bacteria)
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Milk is passed over the immobilised enzyme, becoming lactose-free
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Advantages of Lactose-Free Milk:

Provides a source of milk and other dairy products for lactose-intolerant individuals
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Increases sweetness in the absence of artificial sweeteners (monosaccharides are sweeter in taste)
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Reduces the crystallisation of ice-creams (monosaccharides are more soluble, less likely to crystallise)
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Suggest ways of measuring the following enzyme-catalysed reactions

1. The conversion of hydrogen peroxide to oxygen gas (by catalase)

• Measure time taken for an enzyme-soaked disc to rise in solution
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• Measure gas formation via pressure change (with data logger) or displacement of a syringe cap
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2. The breakdown of starch into maltose (by diastase)

• Measure colour change with a colorimeter (starch is stained purple by iodine)
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• Measure volume change via dialysis tubing (starch is impermeable to tubing, maltose is not)
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3. The digestion of pectin in cell walls (by pectinase)

• Measure volume of liquid produced as a consequence of the breakdown in cell wall
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• Measure the digestion of pectin via percentage weight change
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 2.6 Structure of DNA and RNA

Nucleic Acid Structure

Draw the basic structure of a nucleotide, labelling its parts

Differentiate between the different types of nitrogenous bases

The four bases in DNA are: Guanine (G), Cytosine (C), Adenine (A) and Thymine (T)
…………………………………………………………………………………………………………………………………………………..........
RNA has Uracil (U) instead of thymine
…………………………………………………………………………………………………………………………………………………..........
G and A are double-ringed purines, while C, T and U are single-ringed pyrimidines
…………………………………………………………………………………………………………………………………………………..........
G and C pair via three hydrogen bonds; A and T/U pair via two hydrogen bonds
…………………………………………………………………………………………………………………………………………………..........

Compare the structure of DNA and RNA

DNA RNA

Sugar deoxyribose ribose

Bases Has thymine (G, C, A, T) Has uracil (G, C, A, U)

Strands Double stranded Usually single stranded

List three types of RNA

Messenger RNA (mRNA) - An RNA copy of a DNA gene
1. …………………………………………………………………………………………………………………………………………………...
Transfer RNA (tRNA) - Transfers amino acids to the ribosome
2. …………………………………………………………………………………………………………………………………………………...
Ribosomal RNA (rRNA) - The catalytic component of ribosomes
3. …………………………………………………………………………………………………………………………………………………...
Outline, with the aid of the diagram, how nucleotides are linked together to form DNA strands

• DNA nucleotides are linked by covalent phosphodiester bonds

…………………………………………………………………………………………...

(between the 5’-phosphate and 3’-hydroxyl of two nucleotides)

…………………………………………………………………………………………...

to form a long polynucleotide strand

…………………………………………………………………………………………...
• Two strands join together via complementary base pairing
…………………………………………………………………………………………...
(G pairs with C via 3 H bonds, A pairs with T via 2 H bonds)
…………………………………………………………………………………………...
• The two strands run antiparallel to each other in order for
…………………………………………………………………………………………...
the bases to face each other and form hydrogen bonds
…………………………………………………………………………………………...
• The double-stranded DNA then twists into a double helix
…………………………………………………………………………………………...
• Each twist occurs every ~10 nucleotides
…………………………………………………………………………………………...

…………………………………………………………………………………………...

…………………………………………………………………………………………...

List the contributions of the following scientists to the elucidation of DNA structure

Linus Pauling: Identified molecular distances and bond angles for basic molecular structures
………………………………………………………………………………………………………………………...

Phoebus Levene: Identified structure of a nucleotide (sugar, phosphate, base)

………………………………………………………………………………………………………………………...

Erwin Chargaff: Determined that there is an equal number of purines and pyrimidines
………………………………………………………………………………………………………………………...
Identified the organisation of DNA into a helical structure
Rosalind Franklin: ………………………………………………………………………………………………………………………...

Describe how Watson and Crick proposed the structure of DNA via model making
Watson and Crick used the above knowledge and constructed DNA models through trial and error
…………………………………………………………………………………………………………………………………………………..........
They demonstrated that DNA strands were antiparallel and formed a double helix
…………………………………………………………………………………………………………………………………………………..........
They also showed complementary base pairing
…………………………………………………………………………………………………………………………………………………..........

…………………………………………………………………………………………………………………………………………………..........
 2.7 DNA Replication, Transcription and Translation

DNA Replication

Explain how DNA replication is semi-conservative

DNA replication is semi-conservative because when a new double-stranded DNA molecule is formed:
…………………………………………………………………………………………………………………………………………………..........
• One strand is from the original template molecule (i.e. conserved)
…………………………………………………………………………………………………………………………………………………..........
• One strand is newly synthesised (i.e. not conserved)
…………………………………………………………………………………………………………………………………………………..........

Outline how the results of the Meselson-Stahl experiment supported semi-conservative DNA replication
Meselson and Stahl treated DNA with a heavier nitrogen isotope (15N) and then replicated in the presence
…………………………………………………………………………………………………………………………………………………..........
of a lighter nitrogen isotope (14N) - so template DNA and newly synthesised DNA could be differentiated.
…………………………………………………………………………………………………………………………………………………..........
The results supported a semi-conservative model of DNA replication:
…………………………………………………………………………………………………………………………………………………..........
• After one division, all molecules contained both 15N and 14N
…………………………………………………………………………………………………………………………………………………..........
• After two divisions, some molecules contained both 15N and 14N, while other molecules only contained 14N
…………………………………………………………………………………………………………………………………………………..........

Describe the role of the following enzymes in DNA replication

Helicase unwinds and separates double-stranded DNA molecules
Helicase: …………………………………………………………………………………………………………………………………………
(by breaking the hydrogen bonds between the complementary base pairs)
…………………………………………………………………………………………………………………………………………………..........

…………………………………………………………………………………………………………………………………………………..........
DNA polymerase synthesises a new strand (complementary to the template strand)
DNA Polymerase: …………………………………………………………………………………………………………………………….
Nucleotides align opposite their partner, and DNA Pol III covalently joins them together
…………………………………………………………………………………………………………………………………………………..........
DNA Pol III synthesises a new strand in a 5’ - 3’ direction
…………………………………………………………………………………………………………………………………………………..........

Explain the significance of complementary base pairing in the conservation of base sequence
Free nucleotides can only align opposite their complementary base partner (A=T, G=C)
…………………………………………………………………………………………………………………………………………………..........
This means a newly synthesised strand will be identical to the complementary partner of a template strand
…………………………………………………………………………………………………………………………………………………..........
Hence, base sequence is conserved
…………………………………………………………………………………………………………………………………………………..........

State when and where DNA replication happens in a typical eukaryotic cell
DNA replication occurs within the nucleus, during the S phase of interphase
…………………………………………………………………………………………………………………………………………………..........
Describe the purpose and process of the polymerase chain reaction (PCR)
PCR is used to rapidly amplify minute quantities of DNA
………………………………………………………………………………………………………
It involves a thermal cycler and three repeating steps:
………………………………………………………………………………………………………
• Denaturation - DNA is heated to separate strands
………………………………………………………………………………………………………
• Annealing – Primers are introduced to designate copying points
………………………………………………………………………………………………………
• Elongation - Taq polymerase* synthesises new strand
………………………………………………………………………………………………………
These three steps double the amount of DNA, so a typical reaction of
………………………………………………………………………………………………………
30 cycles will produce over 1 billion copies of desired DNA sequence
………………………………………………………………………………………………………
* Taq polymerase is heat resistant and so doesn’t denature during PCR
………………………………………………………………………………………………………

Transcription

Define transcription
Transcription is the process by which an RNA sequence is produced from a DNA template (gene)
…………………………………………………………………………………………………………………………………………………..........

…………………………………………………………………………………………………………………………………………………..........

Distinguish between sense and antisense strands

The antisense strand is the DNA strand that IS transcribed (complementary to eventual RNA sequence)
…………………………………………………………………………………………………………………………………………………..........
The sense strand is the strand that is NOT transcribed (identical to RNA sequence - except T in place of U)
…………………………………………………………………………………………………………………………………………………..........

Outline the role of RNA polymerase in the process of transcription

RNA polymerase unwinds and separates the double stranded DNA and then synthesises a new RNA strand
…………………………………………………………………………………………………………………………………………………..........
based on the antisense template – the RNA strand is then released and DNA double helix reforms
…………………………………………………………………………………………………………………………………………………..........
(when RNA polymerase separates the DNA strands, free nucleotides align opposite their complementary
…………………………………………………………………………………………………………………………………………………..........
base partners and RNA polymerase covalently joins them together)
…………………………………………………………………………………………………………………………………………………..........

Convert the following DNA sequence into an RNA transcript

T A C A A A T T C G T A C T G C A C T C C G G A A C A A C T
A U G U U U A A G C A U G A C G U G A G G C C U U G U U G A
…………………………………………………………………………………………………………………………………………………..........
Translation

Define translation
Translation is the process of protein synthesis, whereby genetic information encoded by mRNA is
…………………………………………………………………………………………………………………………………………………..........
translated into an amino acid sequence (i.e. polypeptide) at the ribosome
…………………………………………………………………………………………………………………………………………………..........

Describe the function of the genetic code (including degeneracy)

The genetic code is the set of rules by which information encoded by mRNA is translated into polypeptides
…………………………………………………………………………………………………………………………………………………..........
• It identifies the specific amino acid encoded by each triplet of mRNA bases (codon)
…………………………………………………………………………………………………………………………………………………..........
• There are 64 possible codon combinations (4 x 4 x 4), but only 20 possible amino acids
…………………………………………………………………………………………………………………………………………………..........
• This means some codons code for the same amino acid (degeneracy)
…………………………………………………………………………………………………………………………………………………..........

Identifying the key components of the process of translation

Messenger RNA (mRNA) - Contains the genetic instructions

M ………………………………………………………………………………………………

Ribosome – Site of translation

R ………………………………………………………………………………………………
Codon - Triplet of bases denoting a specific amino acid
C ………………………………………………………………………………………………
Anticodon - Complementary sequence on tRNA molecules
A ………………………………………………………………………………………………
Transfer RNA (tRNA) - Transfers amino acids to ribosome
T ………………………………………………………………………………………………
Amino acid - Monomeric component of a polypeptide chain
A ………………………………………………………………………………………………
Peptide bond - The covalent bond formed between amino acids
P ………………………………………………………………………………………………
Polypeptide - The end product of translation
P ……………………………………………………………………………………………… Hint: Mr Cat App

Summarise the process of translation

• Ribosome binds to mRNA and moves along it in a 5’ - 3’ direction, reading the sequence in codons
…………………………………………………………………………………………………………………………………………………..........

• Each codon encodes a specific amino acid, which is brought to the ribosome by tRNA molecules
…………………………………………………………………………………………………………………………………………………..........

• Each tRNA is specific for a particular codon due to the presence of a complementary anticodon
…………………………………………………………………………………………………………………………………………………..........
• The tRNA molecules bring the amino acids to the ribosome in an order determined by the codon sequence
…………………………………………………………………………………………………………………………………………………..........
• The ribosome moves along the mRNA, joining the amino acids together via peptide bonds
…………………………………………………………………………………………………………………………………………………..........
• Translation of a polypeptide begins at a START codon (AUG) and is terminated at a STOP codon
…………………………………………………………………………………………………………………………………………………..........

…………………………………………………………………………………………………………………………………………………..........
Use the genetic code to convert the following DNA
sequence into a polypeptide sequence

TAC AAA TTC GTA CTG CAC TCC GGA ACA ACT

RNA: AUG UUU AAG CAU GAC GUG AGG CCU UGU UGA
………………………………………………………………….
Met-Phe-Lys-His-Asp-Val-Arg-Pro-Cys-STOP
Protein: ……………………………………………………………..

Outline the effect of the following mutations:

His to Gln (missense mutation)
GTA → GTT: ………………………………………………………
Arg to Arg (silent mutation)
TCC → TCT: ………………………………………………………

Describe the different types of point mutations

Silent mutations do not change the polypeptide sequence (possible due to degeneracy of the genetic code)
…………………………………………………………………………………………………………………………………………………..........
Missense mutations involve a change in a single amino acid within the polypeptide sequence
…………………………………………………………………………………………………………………………………………………..........
Nonsense mutations create a STOP codon (thus prematurely terminating the polypeptide chain)
…………………………………………………………………………………………………………………………………………………..........

Outline the significance of a frameshift mutation

Frameshift mutations (insertions, deletions) change the reading frame (meaning all codons are changed)
…………………………………………………………………………………………………………………………………………………..........

…………………………………………………………………………………………………………………………………………………..........

Use the following image to explain how the universality of the genetic code allows for gene transfer

Insulin gene
Human Cell
Grow in Extract
Recombinant Transgenic culture insulin
Plasmid bacteria
Bacteria Plasmid

The genetic code is universal, meaning (almost) all organisms follow the same set of genetic instructions
…………………………………………………………………………………………………………………………………………………..........
This means that a DNA sequence from one organism can be successfully translated by another organism
…………………………………………………………………………………………………………………………………………………..........
The gene for insulin is extracted from human cells and inserted into bacterial cells (via recombinant plasmid)
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The bacteria can now produce human insulin (bacteria divide quickly, allowing for effective mass production)
…………………………………………………………………………………………………………………………………………………..........
This can be used to produce insulin treatments for type I diabetics
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 2.8 Cell Respiration

ATP Production

Define cell respiration

Cell respiration is the controlled release of energy from organic compounds to produce ATP
…………………………………………………………………………………………………………………………………………………..........

…………………………………………………………………………………………………………………………………………………..........

Describe how ATP acts as the energy currency for the cell
ATP is composed of a sugar and base (adenosine) connected to three phosphate groups
…………………………………………………………………………………………………………………………………………………..........
When a phosphate is cleaved (to form ADP + Pi), the energy stored in the bond is released
…………………………………………………………………………………………………………………………………………………..........

Compare anaerobic and aerobic respiration

Anaerobic Aerobic

Required Conditions Oxygen not required Oxygen required

Energy Yield Small yield (~2 ATP) High yield (~36 - 38 ATP)

Animals: Lactic acid

Products Carbon dioxide and water
Plants: Ethanol and carbon dioxide

Location Cytosol Cytosol and mitochondria

Outline the purpose (and products) of fermentation

The purpose of fermentation is to restore stocks of NAD+ needed for glycolysis (anaerobic respiration)
…………………………………………………………………………………………………………………………………………………..........
This allows ATP to continue to be produced in the absence of oxygen (otherwise respiration would stop)
…………………………………………………………………………………………………………………………………………………..........
Fermentation in animals reversibly converts pyruvate to lactic acid (and frees up NAD+)
…………………………………………………………………………………………………………………………………………………..........
Fermentation in plants reversibly converts pyruvate to ethanol and carbon dioxide (and frees up NAD+)
…………………………………………………………………………………………………………………………………………………..........

Describe the use of respirometry in measuring respiration rates

Respirometers measure respiration rate - either by amount of carbon dioxide produced or oxygen consumed
…………………………………………………………………………………………………………………………………………………..........
• A manometer can accomplish this by recording the pressure change (as a moving bubble in solution) that
…………………………………………………………………………………………………………………………………………………..........
occurs when oxygen is consumed (must include a carbon dioxide absorbant)
…………………………………………………………………………………………………………………………………………………..........
 2.9 Photosynthesis

Light Absorption

Define photosynthesis
Photosynthesis is the process by which cells use light energy to synthesise organic compounds from
…………………………………………………………………………………………………………………………………………………..........
inorganic molecules
…………………………………………………………………………………………………………………………………………………..........

Identify the range of wavelengths absorbed via photosynthesis

Photosynthesis uses the visible spectrum (white light), which ranges from 400 nm (violet) to 700 nm (red)
…………………………………………………………………………………………………………………………………………………..........

State the main photosynthetic pigment in plants and identify where they are found
Chlorophyll (a) is the main photosynthetic pigment and it is found in chloroplasts in plants
…………………………………………………………………………………………………………………………………………………..........

Outline the difference in absorption of red, green and blue light by plants
Plants predominantly absorb red and blue light, but reflect green light
…………………………………………………………………………………………………………………………………………………..........

Draw the action spectrum and absorption spectrum and provide a comparison between the two

Action Spectrum Absorption Spectrum

Amount of light absorbed
Rate of Photosynthesis

400 nm 700 nm 400 nm 700 nm

The range of wavelengths used by a plant for photosynthesis

Action Spectrum: ……………………………………………………………………………………………………………………………..
(red and blue is used for a high photosynthetic rate, green is not used)
…………………………………………………………………………………………………………………………………………………..........
The range of wavelengths absorbed by a plant (each pigment has its own spectrum)
Absorption Spectrum: ………………………………………………………………………………………………………………………
(for chlorophyll a: red and blue are absorbed, green is reflected)
…………………………………………………………………………………………………………………………………………………..........
Photosynthetic Action

Complete the following diagram outlining the key components of photosynthesis

LIGHT DEPENDENT
REACTIONS Light energy

Water Chlorophyll Chemical energy

e–

Oxygen gas Hydrogen NADPH ATP

Carbon dioxide gas Carbon fixation

LIGHT INDEPENDENT Carbon compounds

REACTIONS

Summarise the key events of the two stages of photosynthesis

Light Dependent Reactions:

Light is absorbed by chlorophyll, which triggers the production of ATP (via photophosphorylation)
…………………………………………………………………………………………………………………………………………………..........
Light is also responsible for the photolysis of water, which produces hydrogen (carried by NADPH)
…………………………………………………………………………………………………………………………………………………..........
The photolysis of water also produces oxygen gas as a by-product
…………………………………………………………………………………………………………………………………………………..........

Light Independent Reactions:

The products of the light dependent reaction (NADPH and ATP) are used in the light independent reactions
…………………………………………………………………………………………………………………………………………………..........

These reactions fix carbon (from carbon dioxide) to form organic molecules (such as carbohydrates)
…………………………………………………………………………………………………………………………………………………..........

Outline how photosynthetic pigments can be separated via chromatography

Pigments are dissolved in a fluid (mobile phase)
……………………………………………………………………………………………………………………………………………
Fluid is then passed through a static material (stationary phase)
……………………………………………………………………………………………………………………………………………
Different pigments pass through the static material at different speeds, becoming separated
……………………………………………………………………………………………………………………………………………
A retardation factor (Rf) can be assigned to each pigment (to allow for identification)
……………………………………………………………………………………………………………………………………………
 Measuring Photosynthesis

Outline the law of limiting factors

A reaction will only proceed as fast as the component nearest its minimum value (rate-limiting factor)
…………………………………………………………………………………………………………………………………………………..........

List three conditions by which photosynthesis can be detected and suggest how they can be measured
Oxygen uptake - measure by pressure change (data logger) or displacement of syringe cap
1. …………………………………………………………………………………………………………………………………………………...

…………………………………………………………………………………………………………………………………………………...
Carbon dioxide production - measure by a change in pH if plant is in solution
2. …………………………………………………………………………………………………………………………………………………...
(carbon dioxide dissolves in water to form carbonic acid = decreases pH)
…………………………………………………………………………………………………………………………………………………...
Change in biomass - measure by a change in weight
3. …………………………………………………………………………………………………………………………………………………...
(plant sample must first be dessicated to account for change in mass due to water)
…………………………………………………………………………………………………………………………………………………...

Explain the effect of different factors on the rate of photosynthesis

Rate of Photosynthesis

Rate of Photosynthesis

Rate of Photosynthesis

Temperature Light Intensity CO2 Concentration

Temperature affects enzyme activity (photosynthesis controlled by enzymes)

Temperature: ………………………………………………………………………………………………………………………………….
Activity initially increases as a rise in kinetic energy results in more frequent collisions
…………………………………………………………………………………………………………………………………………………..........
Above a certain temperature, activity decreases as enzymes become denatured
…………………………………………………………………………………………………………………………………………………..........

Increasing light levels results in more photoactivation of chlorophyll (= increased activity)

Light Intensity: ………………………………………………………………………………………………………………………………..

At a certain intensity, all chlorophyll are photoactivated and so activity plateaus

…………………………………………………………………………………………………………………………………………………..........

(i.e. plant is saturated with light)

…………………………………………………………………………………………………………………………………………………..........

Carbon dioxide is fixed (via the enzyme Rubisco) to form organic compounds
CO2 Concentration: ……….…………………………………………………………………………………………………………………

Activity increases at higher concentrations as there are more frequent collisions

…………………………………………………………………………………………………………………………………………………..........
Above a certain level activity will plateau as all enzyme active sites are occupied
…………………………………………………………………………………………………………………………………………………..........