Biology Paper 2 Questions and Answers - Asumbi Girls Mock Examinations 2022

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  1. The diagram below represents the direction of flow of blood in a gill capillary. The percentage of oxygen in solution at position A, B, P, Q and R is given in the table below.
    Position Oxygen concentration in solution (%) Haemoglobin saturation with oxygen (%)
    A 10  
    B 7  
    P 4 55
    Q 7 85
    1. Why is the oxygen percentage low at P?
    2. Using evidence from the data given, suggest what will happen to oxygen in the water at point B. (3mks)
    3. Name the organ into which blood coming from the capillary at Q flows.  (1mk)
    4. Suppose the flow of blood in the capillary illustrated above was in the opposite direction, explain the disadvantage it would have to the fish. (2mks)
    5. Name the principle where the blood flows in the opposite direction to another fluid. (1mk)
  2. Below is a diagram of a poorly planned town showing some building and facilities.
    1. Giving evidence from the diagram, state two likely sources of water pollution. (2mks)
    2. State three ways that the positioning of the refuse pit and sewage works pose danger to the residence of the town.  (3mks)
    3. Residents living close to the marsh are likely to suffer from malaria. Explain. (1mk)
    4. Suggest two control measures to overcome water pollution in the area.(2mks)
  3. The diagram below represents a maize seedling.
    1. Name the structure labeled A and C(2mks)
    2. State the functions of parts labeled A, B and C(3mks)
    3. Name the type of germination exhibited by maize(1mk)
    4. Name two conditions necessary for seed germination other than water and oxygen.(1mk)
    5. What is the role of oxygen in seed germination? (1mk)
  4. The figure below shows part of a human skeleton.
    1. Which part of the human skeleton is it? (1mk)    
    2. On the diagram label by name three types of joints. (3mks)
    3. Label the S, T and P.  (3mks)
    4. Which two bones on the diagram manufactures red blood cells? (1mk) 
  5. In maize the gene for purple colour is dominant to the gene for white colour. A pure breeding maize plant with purple grains was crossed with a heterozygous plant.
    1. Using letter G to represent the gene for purple colour, work out the genotypes of the offspring. (4mks)
    2. State the phenotype of the offspring.(1mk)
    3. What is genetic engineering? (1mk)
    4. Gene for smooth seed coat is dominant over gene for wrinkled seed coat. Two heterozygous pea plants with smooth seed coats were crossed and produced a total of 14640 seeds. How many seeds had wrinkled seed coat? Show your calculations. (2mks)
  6. The diagram below is obtained from measurements of growth in the leaf petiole of a certain plant. The relative growth rate is calculated and the data is obtained as shown below
    Time in days 0 1 2 3 4 5 6 7 8 9
    Relative growth rate(cm/day) 0 0.1 0.3 0.8 2.0 4.0 4.5 3.5 0.2 0
    1. Plot a graph of relative growth rate against time. (5mks)
    2. State two functions of a leaf petiole. (2mks)
    3. State two characteristics of cells found in the region of cell division. (2mks)
    4. Account for the shape of the curve between the following days  (3mks)
      1. 2 - 5
      2. 6 - 8 (3mks)
      3. 8 - 9
    5. Distinguish between primary growth and secondary growth in a flowering plant. (2mks)
  7. How are flowers adapted to wind and insect pollination? (20mks)
    1. Name factors that affects the enzyme controlled reactions. (6mks)
    2. Explain the factors that affect the rate of enzyme activity. (14mks)


    1. Blood coming from the body has supplied the tissue cells with oxygen/ oxygen has diffused out of capillary into the tissue fluid;
    2. Oxygen concentration at B reduces/will be at lowest; because blood at P has a lower oxygen concertation creating a diffusing gradient; From A to B there is a diffusion gradient hence at B much oxygen has diffused into the from water into the blood;
    3. Heart
    4. Amount of exchange of respiratory gasses between blood and water would reduce; because of reduced diffusion gradient;
    5. Counter flow system.
    1. Industrial effluents because the factory is close to the river.
      Domestic effluents since some houses are next to the river and the marsh.
      1. Drinking water contamination since the water works is close to the sewage works;
      2. Spread of diseases to residents nearby because many pathogens and diseases vectors are found in the sewage works and refuse pit near the houses;
      3. Air pollution because the prevailing wind carries unpleasant smell over the town;
    3. The swamp is the breeding site for mosquito larva hence adult mosquitos may spread malaria to residents living nearby.
      • Treating industrial and domestic effluents before releases it into the water.
      • Cooling the water from the factories to avoid raising water temperature in the river and swamp.
      • Carry out environmental impact assessment before establishing factories/industries; etc
    1. A – Coleoptile
      B – Remaining of seed / seed.
    2. A - Protects the delicate plumule as it pushes through the soil during germination.
      B – Contains stored food which is hydrolyzed and used by the germinating seed / seedling before the seedling starts to photosynthesize.
      C – Anchors the seedling.
    3. Absorbs water and mineral ions.
      • Hypogeal germination.
      • Optimum temperature / warmth.
      • Growth hormones.
      • Viability of seed
      • Enzymes
    5. It oxidizes the stored food in the seed to give energy for growth/synthesis of new materials; (ANY TWO)
    1. Appendicular;
    3. S – Tibia
      T – Patella
      P – Lumbar verterae
    4. vertebrae
  5. G -  Gene for Purple
    g – Gene for white
    2. All purple colored grained maize plants.
    3. Deliberate modification of characteristics of an organism by manipulating the DNA / Gene, by transferring genes from one organism to another;
    4. Phenotypic ratio; 3 smooth seed coat: 1 wrinkled seed coat
      Seeds with wrinkled coats = ¼ x 14640; = 3660;
    2. Supports the leaf in position to trap light for photosynthesis;
      Contains xylem tissue to transport water and dissolved mineral salts to the leaf for photosynthesis, and phloem to transport synthesized food to the rest parts of a plant;
      • Dense cytoplasm;
      • Thin cell wall;
      • Numerous mitochondria;
      • Very small / No sap vacuole;
      1. 2-5 There is a fast growth rate; because the leaf is very young; and cells are actively dividing and elongating;
      2. 6-8 There is reducing rate or relative growth; of the petiole. This is because the cells if the petiole/leaf are no longer dividing and elongating; instead the cells are becoming differentiated;
      3. 8-9 Growth ceases; /very little growth because all cells are differentiated; and has formed permanent tissues;
    5. Primary growth takes place at shoot tip and root tip leading to increase in length due to activity of apical meristems;
      Secondary growth leads to increase in girth in stems due to activity of lateral meristems / vascular cambium;   
  7. The flowers have the following features:
    Insect pollination / Entomophilous flowers
    • Are scented to attract insects
    • Have small sticky, stigma that occur inside the flower for pollen grains to stick on it.
    • Have nectaries to secrate nectar; nectar acts as a bait to attract insects
    • Have nectar guides to guide the insects to the nectaries.
    • Have special shaped corolla to provide landing platform i.e. tubular or funnel sheped corolla to increase chance of contact by insects.
    • Large / heavy and rough / sticky / spiny / spiky pollen grains which stick on the body of insects on stigma.
    • Large; conspicuous flowers with brightly colored petal, bracts or inflorescence to attract insects.
    • Anthers are small and firmly attached to filament to ensure insect brush against the anthers as they crawl into the flower hence collect as many pollen grains.
    • Stigmas are small, sticky and occur inside the flower, so that pollen grain from insect body can stick onto it.
    • Another’s are situated inside the flower to ensure that they are into contact with the insects.
    • Mimicry to attract insects / flowers mimic female insects which attract male insects for mating e.g orchids. 
      N/B-   1 mark per point
      Max 10mks
      Wind pollinated / Anemophilous flowers
    • Anthers and stigma hang outside the flowers to increase chances of pollination; style / filament is long to expose stigma / anthers.
    • Stigma is hairy / feathery / branched / long to increase surface are over which pollen grains land / to trap pollen grains.
    • Pollen grains are smooth / dry / light / small to be easily carried by wind; large amount of pollen grains to increase chances of pollination.
    • Flowers are small with inconspicuous petals, bracts or inflorescence.
    • Flowers not scented and lack nectar.
    • Anthers are large and loosely attached to flexible filaments to enable them sway easily to release pollen grains. This ensures that pollen grains released readily when wind blows.
    • Pollen grains may have structures which contain air to increase buoyancy, flowers have long stalks holding them out in the wind.   
      Max 10mks
    1. Temperature; PH value; co-factors; enzyme and co-enzymes; enzyme concentration; substrate concertation; metabolic poison / inhibitors;
      Max 6mks
      • Temperature- increase in temperature increases rate of enzymatic activity; up to an optimum where enzymes work at best hence maximum / highest rate of reaction;
        low temperature makes the enzymes less active; high temperatures above 400c denatures enzymes; enzymatic activity reduces and eventually stops;
      • PH – Enzymes work best at optimum PH; extreme PH denatures enzymes;
        Some enzymes act best in acidic or basic medium while others work best at neutral pH; Optimum PH should be maintained.
      • Enzyme concentration – Increase in concentration increases enzymatic activity as there is more active sites; to combine with substrate hence an increase in reaction;
      • Co-enzymes – activates enzymes; increasing rate of activity;
      • Substrate concentration – increase in substrate concentration increases enzymatic activity; up to certain level where enzymes become a limiting factor;
      • Inhibitors –  They compete with substrate for active sites or combines permanently with active sites of enzymes; They slow down or stop the rate of reaction; 
        Max 14mks

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