FORM 3 BIOLOGY PAPER 2
END OF TERM 2
Answer all the questions in SECTION A in the spaces provided.
- In section B answer question 6 (Compulsory) in the spaces provided and either question 7 or 8 in the space provided after question 8.
SECTION A (40MKS)
- The two cells shown below are obtained from two different potato cylinders which were immersed in two different solutions P and Q.
- (i) Name the structures labelled A and C (2mks)
A –
B –
(ii) State the function of structure B. (1mk)
- Suggest the identity of the solution Q (1mk)
- Account for the change in Cell 1 above (2mks)
- State two importance of the physiological process being demonstrated above in living organisms. (2mks)
- Study the following food web and answer the questions that follow.
- (i) Name the organisms that occupy the second trophic level (2mks)
(ii) What is the other name for the second trophic level (1mk)
- Write down two food chains from the web that
- End with hawks as tertiary consumer (1mk)
- End with hawks as Quaternary consumer. (1mk)
- Giving reasons state;
- The organism with largest biomass (1mk)
- The organism with least biomass (1mk)
- Name the source of energy in the above ecosystem. (1mk)
- The diagram below represents a longitudinal section through the ileum wall.
- Identify the structure labeled A and B (2mks)
A –
B –
- State one function of A and B (2mks)
A –
B –
- State two functions of the ileum. (2mks)
- Explain the role of the liver in digestion. (2mks)
- The equation below represents a metabolic process that occurs in a certain organ in the mammalian body.
Ammonia enzymes Organic compound Q + water
Carbon (IV) Oxide
- Name the process represented in the equation. (1mk)
- Name the organ in which the process occurs. (1mk)
- Why is the process important to the mammal? (1mk)
- Identify the organic compound Q. (1mk)
- What happens to organic compound Q ? (1mk)
- A person was found to pass large volume of dilute urine frequently. Name the; (3mks)
- Disease the person was suffering from.
- Hormone that was deficient.
- The gland that secretes the above hormone.
- Some glucose was boiled and cooled in a boiling tube. Some yeast was added and a layer of oil put on top. The set up below was used.
- Why was the glucose solution boiled before the experiment. (1mk)
- What is the use of the oil film in the experiment ? (1mk)
- Name the process being investigated by the above experiment. (1mk)
- State what happens to the lime water as the experiment proceeds to the end. (1mk)
- Explain what would happen if the temperature of glucose and yeast was raised beyond 450 (2mks)
- State two industrial applications of the process being investigated above in the experiment. (2mks)
SECTION B (40MKS)
Question 6 (Compulsory) and either question 7 or 8.
- In an ecological study, a locust population and that of crows was estimated in a grassland area over a period of one year. The results were tabulated as shown below:-
| Months | J | F | M | A | M | J | J | A | S | O | N | D |
| Number of locusts | 90 | 20 | 11 | 25 | 200 | 450 | 652 | 15 | 10 | 35 | 192 | 456 |
| Number of crows(birds) | 4 | 2 | 0 | 1 | 8 | 16 | 22 | 2 | 1 | 1 | 5 | 15 |
| Amount of rainfall | 20 | 0 | 55 | 350 | 520 | 400 | 350 | 10 | 25 | 190 | 256 | 350 |
- Draw a graph of population of locusts and crows (birds) against time. (8mks)
- (i) State the relationship between rainfall and locust population. (1mk)
(ii) Account for the relationship you have stated in (b) (i) above (1mk)
- What happens on the populations of locusts and crows in the months of January to
March ? Give a reason. 2mks)
- State one method used to estimate the population of locust. (1mk)
- (i) State the trophic level of the; (2mks)
Locusts –
Crows –
(ii) Construct a simple complete food chain involving these organisms (2mks).
- If the locusts were removed from the food chain, what would be its effect ? (1mk)
- Define the following terms (2mks)
(i)Biomass
(ii)Ecosystem
- Describe how hydrophytes and xerophytes are adapted to deal with environmental problems in the regions where they grow. (20mks)
- Describe how the mammalian skin regulates body temperature.(20mks)
FORM 3 BIOLOGY PP 2 MARKING SCHEME
END OF TERM 2
1(a)(i) A – Nucleus
C – Cell wall
(ii) – Maintain the shape of the cell;
- Providing support to herbaceous plants; (any 1)
- Stores sugar and salts;
(b) Hypotonic solution / dilute solution / dilute sugar / salt solution.
(c) The potato cell sap was lowly concentrated than the surrounding solution; hence lost
water molecules by osmosis (through semi-permeable membrane) to become plasmolysed.
(d) – Opening and closing of stomata.
– Absorption of water by root hairs.
– Absorption of water in intestines.
– Reabsorption of water in kidney nephron. (any 2)
– Feeding in insectivorous plants.
– Movement of water from cell to cell.
– Osmoregulation.
2(a) Cartepillars
- Aphids
Mice each (1/2 mk)
Slugs
- Primary consumers
(b)(i) Plants Cartepillars Insectivorous hawks ;
Birds
(ii) Plants Slugs Frogs Snakes Hawks (any 1)
Plants Aphids Beetles Insectivorous Hawks
Plants
(c)(i) Largest Biomass – plants
Directly obtain energy from the sun.
(ii) Least biomass – Hawks
- Loss of energy in form of heat, respiration, defaceation, excretion.
- Sun / solar energy.
3(a) A – Villus
B – Lacteal
(b) A – Increases surface area for maximum digestion and absorption.
B – Absorption of fatty acids and glycerol.
(c) – Final digestion of food.
– Absorption of soluble products of digestion.
(d) Produces bile juice which contain bile salts that emulsify fats; and neutralizes the acidic chyme from the stomach;
4(a) Deamination
(b) Liver
(c) – Removal of excess amino acids.
– Availing of energy in the body. (any 1)
– Formation of glycogen / fats for storage.
(award any one)
(d) Urea
(e) It is transported to the kidney where it is excreted.
(f) (i) Diabetes inspidus
(ii) Antidiuretic Hormone / Vasopressin
(iii) Pituitary gland.
5(a) To remove / expel dissolved air from the glucose solution.
(b) To prevent entry of oxygen into the yeast – glucose solution.
(c) Anaerobic respiration.
(d) Becomes white precipitate.
(e) High temperature kills the yeast cells; hence the reaction stops;
(f) – Making of beer / Brewing / Ethanol / Alcohol.
– Baking industry / Raising of the dough. (any 2)
6(a)
- Axes labelling (2mks)
Scale (x and y – axis) (2mks)
Curve plotting – each (1mk) total 2mks
Curve labeling – each (1mk) total 2mks
(b)(i)The population of locusts increase with increase in the amount of rainfall.
(ii)Increased amount of food;
Improved breeding conditions;
- The population of both decreases;
Less food availability to locusts and crows.
- Capture – Recapture method.
- (i) Locust – Primary consumers
Crows – Secondary consumers
(ii) Grass Locusts Crows
- Grass would increase
crows would reduce
- (i) Biomass – The total dry weight of organisms at a particular trophic level.
- Ecosystem – A natural unit composed of abiotic and biotic factors whose
interactions lead to a self-sustaining system.
- Xerophytes
- Have thick cuticle to prevent cuticular transpiration.
- Have reduced stomata on lower leaf surface to lower transpiration.
- Fold or curl leaves in dry weather to protect stomata from direct sunlight.
- Have reversed stomatal opening rhythm where they open at night to reduce water loss.
- Have succulent stems and leaves to store water for use in dry season.
- Some have superficial roots to absorb light showers of rain.
- Some are deep rooted to absorb water from water table.
- Shedding of leaves during dry season.
- Some have short life cycles and survive as underground perennating structures or seeds during drought.
- Hairy leaves.
- Sunken stomata 1 x 10 = 10mks.
Hydrophytes
- Submerged plants have dissected leaves to increase surface area for maximum light absorption.
- Emerged plants have broad leaves with stomata on upper surface to increase transpiration.
- Have aerenchyma tissue to increase buoyancy and for gaseous exchange.
- Floating water plants have raised flower for pollination.
- Poorly developed roots that lack root hairs to reduce absorption of water.
- Some submerged plants have sensitive chloroplasts that photosynthesise in low light intensities. 5 x 2 = 10mks
- High body temperature above normal:
- Sweat glands; Produce sweat; water in the sweat evaporates / sweat evaporates; absorbing latent heat of vaporization producing a cooling effect;
- Hairs lie flat; due to relaxation of erector pilli muscles; no / little air is trapped; hence increased heat loss from the body;
- Blood vessels / arterioles; vasodilate / dilate; more blood flow to the skin hence loss of heat from the body, by radiation and convection;
When body temperature is low below normal:
- Sweat glands produce less / no sweat; no latent heat is absorbed; more heat retained in the body;
- The hairs stand upright / erect; to trap air between them; that insulate the body against heat loss; more heat retained in the body.
- Blood vessels / arterioles vasoconstrict / constrict;
Less blood flow to the skin; reduces heat loss / more heat is retained in the body;
(20mks)
(;) means a marking point.
