BIOLOGY FORM 2 MARKING SCHEME
NAME: ……………………………………………….ADM NO…………………….
SCHOOL: ……………………………………………...SIGNATURE …………………
DATE: …………………………………………………
231
BIOLOGY
THEORY
FORM 2
END OF TERM 3 2023
Time: 2 1/2 Hours
Answer ALL the questions in this section in the spaces provided on the question paper.
1. The pyramid below shows the arrangement of the taxonomic units starting with the largest.
Indicate the name of the missing taxonomic unit. (3mks)
Kingdom
Kingdom
Phylum/Division
Phylum/division
Class
Class
Order
Order
Family
Family
Genus
species
2. State the kingdom to which the following organisms belong. (2mks)
(ii) Plasmodium, the microorganisms that cause malaria.
Protoctista;
(iii) Bacteria found in mammalian alimentary canal.
Monera;
3. Plasmodium vivax and Plasmodium ovale are transmitted by a mosquito. State with a reason whether the
two organisms can interbreed. (2mks)
No; Because they belong to same genus but different species therefore their offspring is infertile;
4. Apart from insulin, name another hormone involved in regulating glucose level in blood.
(1mk)
Glucagon;
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5. (a) Name the compound that stores energy released during oxidation of glucose. (1mk)
Adenosine triphosphate/ATP;
6. (a) What is homeostasis. (1mk)
Ability of organisms to maintain a stable/constant internal tissue fluid environment despite
fluctuations in external environment;
(b)Name two processes in the human body in which homeostasis is involved. (2mks)
Thermoregulation;- Osmoregulation; Regulation of blood sugar; regulation of pH of tissue fluid; (1 st 2)
7. Name two waste materials excreted by both the skin and the kidneys. (2mks)
Water; Salts; Urea; (1st 2)
8. Explain why plants do not require specialized excretory organs. (2mks)
Some wastes eg gases easily diffuse out of the plant tissues; Some waste products are mainly made
from carbohydrates and hence are not as harmful as proteinous materials; Some waste products are
formed slowly hence little accumulation of wastes; Plants are less active; Some waste products e.g
CO2 and O2 are re-used in the plant cells; Some waste products are stored in non-toxic form in leaves,
flowers, fruits and bark, then drop off;
(any 2)
9. Adult elephants flap their ears twice as much as their calves in order to cool their bodies when it is hot.
Explain. (2mks)
The surface area to volume ratio is higher in calves than in adults(acc. converse); hence adults retain
more heat than calves hence need to have other ways to lose heat;
10. (a)What is respiratory quotient. (1mk)
Ratio showing the relationship between the amount of carbon (IV) oxide produced against the amount
of oxygen used in respiration;
(b) State two activities in the digestive system that depend on respiration. (2mks)
Peristalsis due to contraction of smooth muscles; Secretion of digestive enzymes; Churning of food
due to contraction of muscles to mix food; Absorption of digested food by active transport;
11. State one guideline that should be followed when typing scientific names. (1mk)
The name should be underlined as two separate words/name should be italicized; Generic name
should begin with a capital letter while specific name should start with a small letter;
12. Name the field of science that specializes in the study of chemical substances in an organism and the
reactions in which they take part. (1mk)
Biochemistry;
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13. Name two plant processes in which diffusion plays an important role (2mks)
Absorption of mineral salts by root hairs from the soil;
Translocation of food from leaves to other parts of the plant;
movement of salts from one cell to the next;
14. Explain how wilting assist plants to reduce excessive water loss to the atmosphere. (2mks)
Wilting results to drooping of leaves, hence reducing surface area exposed for transpiration to
occur;
15. State two characteristics that make alveolus and buccal cavity suitable for gaseous exchange. (2mks)
- Supplied with a rich network of blood capillaries to transport the diffused gases and hence
maintain a steep concentration gradient;
- Thin epithelium to shorten diffusion distance;
- Moist surfaces to dissolve the gases before they diffuse;
16. The following are photomicrographs of two common organelles
J K
a) State two ways in which reaction in organelle K is beneficial to reactions in J. (2 marks)
Supplies Oxygen (which is a raw material for reactions in J);
Supplies glucose which; (which is a raw material for reactions in J);
b) Name the structure that increases surface area for reactions in organelle J. (1 mark)
Crista;
17. Amina was given orange juice, reagent K, boiling tube, measuring cylinder, test tube holder and source of
heat. She was instructed to test for reducing sugar in the juice.
a) Name reagent K (1 mark)
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Benedict’s Solution; reject Benedicts/benedict’s
b)State the steps she was to follow in this experiment. (3 marks)
Place (2ml of) juice in a boiling tube; Add (2ml of) Benedict’s solution/Reagent K; Boil/Heat;
c) State the observation she made for a positive result. (1 mark)
(K changed from Blue to) Green to Yellow to Orange to Brown;
reject if wrong sequence given
18. (a) Name two raw materials for the dark stage process of photosynthesis. (2 marks)
Carbon (IV) oxide, hydrogen atoms
(b) The set up shows an experiment to investigate photosynthesis.
Gas collected
Water containing sodium
hydrogen carbonate
At the start Aquatic plants After the
experiment
(i) What gas was collected in the test tube? (1 mark)
Oxygen gas
(ii) What was the role of sodium hydrogen carbonate in the experiment? (1 marks)
dissociate in water to provide carbon (IV) oxide which is necessary for photosynthesis;
19. The following are dental formulae of a dog and rabbit, state two differences between them. (2 marks)
Dog: i 3 c 1 pm 4 m 2
3 1 4 3
Rabbit: i 2 c 0 pm 3 m 3
1 0 2 3
Dog Rabbit
- Presence of canine - Absence of canines/presence of
diastema
- Has more teeth (42) - Has fewer teeth. (28)
Has fewer molars (2) on the upper jaw Has more molars (3) on the upper jaw
Has more premolars (4) on the lower Has fewer premolars (3) on the upper
jaw jaw
Has more incisors (3) on the upper jaw Has fewer incisors (2) on the upper
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jaw
Has more incisors (3) on the lower jaw Has fewer incisors (1) on the lower
jaw
Any two correct
20. A form one students set up the experiment below and left it to stand for 20minutes.
(a) State the observations made by the student at the end of the experiment. (2mks)
The starch solution turned blue black/black;
Iodine solution remained brown;
(b) Account for the observations stated in (a) above. (3mks)
Iodine molecules moved into starch solution across the tubing; by diffusion; turning it to blue-
black/black since they are too small; starch molecules were too large such that they could not
move across the tubing into iodine solution;
(c) Suggest the nature of the visking tubing. (1mk)
Semipermeable;
(d) i. Which physiological process was being investigated in this experiment. (1mk)
Diffusion;
ii. State one factor which affects the process being investigated. (1mk)
- Size of molecules;
- Thickness of membrane;
- Temperature
- Diffusion gradient
- Surface area to volume ratio
21. The diagram below represents the structure of a light microscope. Study it and answer the
questions that follow.
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a) Name the parts of the microscope labelled A,D and E (3 mks)
A: eyepiece lens;
D: objective lens;
E; condenser
b) State the functions of each of the parts marked C and G. (2 mks)
C: Move the body tube over a short distance for sharp focus of the image
G: Directs/ reflects light to the stage/ object
22. The diagram below illustrates the structure of the kidney nephron.
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(a) Name the part labeled E. (1mk)
Glomerulus; Rj. sp.mistake
(b) How is the part labeled F adapted to its functions. (2mks)
It is long; to increase the surface area for reabsorption of water; It is lined with a network of blood capillaries;
to enhance reabsorption of water; it is U-Shaped; to bring about counter-current multiplier effect to
concentrate salts in the medulla to bring about reabsorption of water; (1st two) Rj. If adaptation is wrong.
(c) State two physiological mechanisms of controlling the human body temperature during a cold
day. (2mks)
- Shivering;
- Vasoconstriction/blood diverted to shunt system;
- hair raises/become erect;
- increased metabolic rate; (first 3)
- sweat glands stop producing sweat
23. Penina counted 10 onion epidermal cells along the diameter of field of view of a light microscope. The
field of view had a radius of 2.5mm.
a)Determine the diameter of one cell in micrometers. (2 marks)
10 cells measure 5mm
Thus 1 cell will measure (1cell X 5mm)/10cells
= 0.5mm;
1 mm = 1000micrometers
Thus 0.5mm = (0.5mm X 1000micrometers)/1mm
= 500micrometers;
b)If she used Eye piece lens magnification of X15 and Objective lens magnification of X40, determine
the following:
i) Total Magnification used. (2 marks)
Total Mag = Eye piece Lens Magnification X Objective Lens Magnification
= X15 x X40; reject if the symbol ‘X’ is Missing
=X600; reject x600/600X/600
ii) Actual diameter of one cell. (2 marks)
Total Mag = Object Length/Actual Length
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Actual Length = Object length/Total Magnification
= 500micrometers ÷ X600;
= 0.833micrometers;
reject Missing Units/Wrong symbol for ‘Micro’
a) Give two reasons why this method of cell size estimation is not accurate. (2 marks)
Cells not linearly arranged; Cells are of different sizes;
24. The diagram below shows a set up that was used to demonstrate fermentation.
Sucrose solution was boiled and oil added on top of it. The glucose was then allowed to cool before yeast
suspension was added.
a) Why was the Sucrose solution boiled before adding the yeast suspension? (1mk)
To expel all the air inside the solution;
b) What was the importance of cooling the Sucrose solution before adding the yeast suspension?
(1mk)
To prevent killing the yeast cells/denature yeast enzymes by heat;
c) What was the use of the oil in the experiment? (1mk)
To prevent re-entry of air into the solution;
d) What observation would be made in test tube B. (1mk)
White precipitate/suspension;
e) Suggest a control for the experiment. (1mk)
An experimental set up without yeast suspension/boiled yeast suspension;
25. Apart from presence or absence of oxygen, state two other differences between aerobic and anaerobic
respiration in plants. (2mks)
Aerobic respiration Anaerobic respiration
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High amount of energy is released /2882kJ/ 38 Low amount of energy released /210 kJ /2ATP
ATP molecules per molecule of glucose. molecules per molecule of glucose
Substrate completely broken down (to CO2, Incomplete breakdown of substrate to ethanol,
water and energy) CO2, and some energy.
Occurs both in the cytoplasm and mitochodria Occurs only in cytoplasm
26. The appearance of a plant cell after being placed in Solution Q is as shown below
a) Which physiological process is responsible for the appearance of the cell above? (1mk)
Osmosis;
b) Account for the appearance of the cell (3mks)
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Solution Q was hypertonic to cell sap of the cell; the cell hence lost water by osmosis; and shrunk
and became plasmolysed;
c) Explain how the cell shown can be made to resume its normal appearancE. (1mk)
Placing it in distilled water / hypotonic solution to its cell sap;
27. (a). Explain how the following factors increases the rate of photosynthesis (5 mks)
Carbon (IV) 0xide concentration
Carbon (IV) 0xide is a raw material for photosynthesis; an increase in Carbon (IV) 0xide concentration lead to
increase in the rate of photosynthesis; up to a given optimum;.
Light intensity.
Light provide energy needed for photosynthesis/photolysis/light stage; the rate of photosynthesis increases as
the light intensity increases; up to an optimum level;
Temperature
Increase in temperature increases the rate of photosynthesis; up to optimum temperature;
(b). Describe the process of digestion of food in the duodenum. (15mks)
Chyme from the stomach enters the duodenum through pyloric sphincter; Arrival of the food in the
duodenum stimulate the duodenal wall to secrete secretin; and cholecystokinin hormones; secretin
hormone stimulate the liver cells to secrete bile juice; and pancreas to secrete mineral salts;
cholecystokinin hormone stimulate the liver cells to secrete pancreatic enzymes; and gall bladder to
release bile juice into the duodenum; pancreatic juice contains sodium hydrogen carbonate; which
neutralizes acidic chyme; and provide alkaline medium ;suitable for activity of pancreatic enzymes.
Pancreatic amylase; breakdown remaining starch to maltose; inactive trypsinogen is activated by
enterokinase; to trypsin; trypsin digest protein to peptides; trypsin is secreted in inactive form to
prevent it from digesting the cells of the gland that secrete it; pancreatic lipase; digest lipids to fatty
acids and glycerol; bile contains bile salts sodium glycocholate; and sodium taurocholate; which
emulsifies lipids;
21mks max15
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