Biology Paper 2 Questions and Answers - Kapsabet Mock Exams 2021/2022

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INSTRUCTIONS TO CANDIDATES

  1. Write your name and index number in the spaces provided above.
  2. Sign and write the date of examination in the spaces provided above.
  3. This paper consists of two sections A and B
  4. Answer all the questions in section A in the spaces provided
  5. In section B answer question 6 (compulsory) and either question 7 or 8 in the spaces provided after question 8.

SECTION A 40 MARKS
Answer all the questions in this section in the space provided

  1. The diagram below shows a section through the mammalian skin
    1
    1. Name the parts labelled W and X (2mks)
      W………………………………………………….
      X …………………………………………………
    2. State the function of the parts labelled Y and Z (2mks)
    3. Explain the changes that occur in the skin when it is cold (4mks)
  2.      
    1. Eye colour in fruits flies is sex-linked. Red eye colour R is dominant to white eye colour r
      A heterozygous red –eyed female fly was crossed with a white eyed male
      1. Show the parental genotypes (1mk)
      2. By means of a genetic cross, determine the genotypic ratio of the offsprings (4mks)
      3. Explain why the actual phenotype ratio obtained from this cross could differ from the
        Expected (1mk)
    2. Name two disorders due to non-disjuction (2mks)
  3. The diagram below represents a feeding relationship in an ecosystem.
    2
    1. Name the type of ecosystem represented by the above food web (1mk)
    2. Name the organism in the food web that
      1. Is a producer (1mk)
      2. Occupies the highest tropic level. (1mk)
    3.      
      1. Write a food chain that ends with the hawk as a quarternary consumer. (1mk)
      2. State two short terms effects on the above ecosystem if all the small fish were killed (2mks)
    4.      
      1. How does oil spills lead to death of fish? (1mk)
      2. Name one other cause of water pollution apart from oil spills. (1mk)
  4. The experiment below was set – up to investigate some physiological processes. The glucose solution was first boiled then cooled. The set up was left for 24 hours
    3
    1. Suggest two aims of the experiment. (2mks)
    2.    
      1. State the expected observations after 24 hours (2mks)
      2. Explain your observations in a (i) above (2mks)
      3. Why was glucose solution boiled then cooled? (1mk)
    3. Suggest a control for the above experiment. (1mk)
  5. A group of students set up an experiment to investigate a certain physiological process. The set up was as shown in the diagram below.
    7
    After some time, the students observed that the level of sugar solution had risen
    1. What physiological process was being investigated. (1mk)
    2. Account for the rise in the level of sugar solution in this experiment. (4mks)
    3.    
      1. State the results that the students would obtain if they repeated the experiment using a piece of boiled pawpaw. (1mk)
        Time after Planting(days) Dry weight of endosperm(mg) Dry weight of embryo(mg) Total dry weight(mg)
      2. Give a reason for your answer (2mks)

        SECTION B (40 MARKS)
        Answer questions 6 (compulsory) and either questions 7 or 8 in the spaces provided questions
  6. During germination and growth of a cereal, the dry weight of endosperm, the embryo and the total dry weight were determined at two day intervals. The results are shown in the table below:
    0 43 2 45
    2 40 2 42
    4 33 7 40
    6 50 17 37
    8 10 25 35
    10 6 33 39
    1. Using the same axes, draw graphs of dry weight of endosperm, embryo and the total dry weight against time. (7mks)
      4
    2. What was the total dry weight on day 5 (1mark)
    3. Account for
      1. Decrease in dry weight of endosperm from 0 to 10 (2mks)
      2. Increase in dry weight of embryo from day 0 to day 10 (2mks)
      3. Decrease in total dry weight from day 0 to day 8 (1mk)
      4. Increase in total dry weight after day 8 (1mk)
    4. State two factors within the seed and two outside the seed that cause dormancy
      1. Within the seed. (2mks)
      2. Outside the seed (2marks)
    5. Give two characteristics of meristematic cells (2mks)
  7.              
    1. Describe the process of fertilization in flowering plants (15mks)
    2. State the changes that take place in a flower after fertilization (5mks)
  8.        
    1. Describe the mechanism of inhalation in man. (10mks)
    2. Using photosynthesis theory explain the mechanics of opening of stomata. (10mks)


MARKING SCHEME

  1.    
    1. W – Sebaceous gland;
      X – Erector pili muscles; (2mks)
    2. Y – Produces melanin which protects the body against U.V light/determines the skin colour;
      Z – Secrets sweat which evaporates to bring about cooling or
      Sweat also removes excretory products/excess salts/water (2mks)
    3. Vasoconstriction; hence less blood flows to the skin surface; reducing heat loss; no sweating; heat produced thought metabolisms/shivering; is retained in the body;
      6 marks max 4 mks
  2.    
    1.      
      1. XR Xr and Xr Y; (1mk)
        Both must be present
      2. Phenotype; Red eyed female White eyed male
        Genotype XR Xr    x     Xr y;
      3. \6
      4. Crossing over;
        Mutations; (1mk)
        Any one – 1mk
      5. Down’s syndrome; klinefelters syndrome; turners syndrome; (2 mks) first 2 – 2mks
  3.      
    1. Aquatic; (1mk)
    2.      
      1. Phytoplankton’s; (1mk)
      2. Hawks; (1mk)
    3.      
      1. Phytoplankton’s → zooplanktons → frogs → snakes → hawks
        Reject if arrow is not indicated
      2. Snakes would decrease (due to less food)
        Zooplanktons would increases (due to less predator) 3mks
    4. Oil clogs fish gills;
      Oil cuts off dissolved oxygen in water leading to suffocation
      Any one 1 mark
    5. Domestic effluents;
      Sewage;
      Silting;
      Industrial effluents;
      Agrochemicals;
      Any one 1 mark
  4.    
    1. To find out whether energy/heat is released in anaerobic respiration/fermentation; 1mk
      To investigate the gas produced during fermentation/anaerobic respiration; (1mk)
    2.      
      1. (Significant) rise in temperature; colour of bicarbonate indicator turns yellow; 2mks
      2. Yeast will respire aerobically releasing energy/and carbon (iv) oxide gas that turn indicator yellow; 1mk
      3. Expel/drive out oxygen; 1mk
  5.      
    1. Osmosis; (1mk)
    2.      
      • Sugar solution is hypertonic to the cell sap of pawpaw;
      • These cells lose water to sugar solution by osmosis;
      • These cells thus become more concentrated/hypertonic to the water in the beaker;
      • The cells then gain water by osmosis from the beaker;
      • Causing a rise in level of the sugar solution; (max 4mks)
    3.      
      1. The sugar solution level will not rise/remain the same/no change;(1mk)
      2. Boiling kills cells; making them osmotically inactive; (2mks)
    4. Use glucose solution without yeast cells/killed yeast cells; (1mk)
  6.    
    1. Graph
      5
    2. Total dry weight 38.5mg; acc ±0.5
    3.        
      1. Hydrolysis of starch into simple sugars/glucose which are translocated to the embryo;
        Oxidation/respiration of( simple sugars) to the embryo;
        CO2/energy/heat; acc water vapor
      2. New cells/tissues materials are synthesized (from proteins);bring about growth of embryo
      3. The rate of respiration is faster than that of synthesis of materials for growth;
      4. First leaf carried out photosynthesis; (leading to growth
    4.    
      1.      
        • Presence of absissic acid; (ABA)
        • Presence of germination inhibitors;
        • Embryo not fully developed/immature embryo;
        • Absence of hormones/enzymes that stimulate germination;
          Acc inactivity of hormones/enzymes inhibitors;
        • Impermeable seed coat;
          Acc for germination hormones such as cytokines, gibberellins;
      2. Unsuitable temperatures/lack of suitable/unfavorable temperatures; absence of light; lack of O2 Rej lack of air
        Lack of water
    5.      
      • Dense cytoplasm; thin cell walls
      • Absence of vacuoles (cell sap);
  7.      
    1. Fertilization is the fusion of the male and female nuclei in the embryo sac; this is preceded by the process of pollination which involves transfer of pollen grains from the anther to the stigma;
      Stigma secrets sticky substance; which causes adherence of pollen grains; and stimulates germination of pollen tube; pollen tube grows down the style deriving nutrients from the style tissues; the tube nucleus follows behind; generative nucleus divides mitotically to form two male nuclei; in the ovule the pollen tube penetrates the embryo sac and the tube nucleus disintegrates; one of the male nuclei gets in and fuses with the egg cell nucleus; to form a diploid zygote; the other male nuclei fuses with polar nucleus; to form a triploid primary endosperm nucleus; hence double fertilization in flowering plants;
    2. Corolla/stamens/style wither/dry and fall off;
      Calyx persists;
      Ovule develops into a seed;
      Zygote forms an embryo;
      Primary endosperm tissues develops into an endosperm;
      Ovary forms a fruit;
  8.      
    1. External intercostals muscles contract; internal intercostals muscle relax, this movement pulls the Rib cage move outwards; and upwards; Diaphragm muscles contract,which causes the Diaphragm to flatten; volume in thoraci cavity increases; pressure reduces.
      Atmospheric air enters the lungs; inflate (correct sequence to be followed)
    2. Guard cells have chloroplast which photosynthesis in the presence of light, to form sugar, the osmotic pressure of guard cell increases; water move from neighbouring cells into guard cells being thicker than outer walls. Causes the outer wall to stretch more resulting guard cells budging outwards.

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