Biology Paper 2 Questions and Answers - Kenya High Post Mock 2023 Exams

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  • This paper consists of TWO sections A and B.
  • Answer ALL questions in section A in the spaces provided
  • In section B answer question 6(compulsory) and either question 7 or 8 in the spaces provided after 



  1. A potted plant was placed in each of the following conditions for a period of one hour in the order given and transpiration in each hour was measured. The air temperature was 180C throughout the experiment.




    A)     Still air in sunlight and shade



    B)     Moving air only



    C)     Still air in bright sunlight



    D)     Still air in dark chamber



    1. Account for the rate of transpiration in
      1. Condition C (3mks)
      2. Condition D (3mks)
    2. Name the apparatus used to measure the rate of transpiration. (1mark)
    3. Give one modification in the stomata of xerophytes that reduce the rate of transpiration. (1 marks)
  2. Sickle cell anaemia is a disease in which people produce abnormal haemoglobin in their red blood cells. Letter H represents the gene for normal haemoglobin while letter S represent the gene for abnormal haemoglobin. Heterozygous individuals are said to have sickle cell trait.
    1. If both parents have sickle cell trait, work out the proportion of their offspring that have sickle cell anemia. (5marks)
    2. Explain why sickle cell trait is more prevalent in tropical countries than temperate countries (2marks)
    3. Name any other disease caused by gene mutation. (1mk)
  3. Equal volumes of three different sugar solutions were placed in visking tubings X, Y and Z. The tubings were placed in a beaker containing 5%sugar solution. The set up was left for two hours. The results were as shown in the diagram below.
    3 7
    1. Name the process being investigated (1mark)
    2. What is the significance of set up X in the experiment? (1mark)
    3. Account for the observations in set up Y and Z. (4mks)
    4. Name a structure in cells that can be compared to the visking tubing. (1mark)
    5. Explain how high temperature above 400C will affect the process being investigated in the cells of organism. (1marks)
  4. The diagrams below show changes in the life cycle of a flowering plant.
    4 1
    1. Name the parts labelled A, B and G. (3marks)
    2. State the function of the part labelled D. (1mark)
      1. What will part A develop into after fertilization? (1mark)
      2. Define the term parthenocarpy. (1mark)
    4. Name two features in flowering plants that prevent self-fertilization (2marks)
  5. The diagram below represents one of the joints in the mammalian body.
    5 3
    1. Name the type of joint shown in the diagram. (1mark)
    2. Name each of the parts labeled Z and U. (2marks)
    3. State two functions of the fluid found in W. (2marks)
    4. Identify the type of muscle found attached to bone Z. (1mark)
    5. State two differences between the muscle identified in (d) above and those found in the gut. (2marks)

Answer question 6 (compulsory) and either question 7 or 8 in the spaces provided after question 8.

  1. In an experiment to investigate a certain process in a given plant species, the rate of carbon (IV) oxide consumption and the rate of carbon (IV) oxide released were measured over a period of time of 

    Time of day











    CO2 consumption (mm3/min)











    CO2 release (mm3/min)











    1. On the same axis, plot graphs of volume carbon (IV) oxide against time (7mkS)
      SCTN B
    2. Name the biochemical process represented by:
      1. Carbon (IV) oxide consumption (1 mark)
      2. Carbon (IV) oxide released (1 mark)
    3. Account for the shape of the curve for:
      1. Carbon (IV) oxide consumption (3 marks)
      2. Carbon (IV) oxide released (3marks)
      1. What is meant by the term compensation point? (1marks)
      2. From the graph state the time of the day when the plant attains compensation point (2 mark)
    5. Explain how temperature affects the rate of CO2 consumption in a plant. (2 marks)
    1. Explain how the villi in the small intestines are adapted to their functions. (10 marks)
    2. Describe the photosynthetic theory as a mechanism of opening and closing of the stomata. (10marks)
    1. Explain how ultrafiltration occurs within the kidneys. (5marks)
    2. Describe water pollution under the following headings:
      1. Causes (5marks)
      2. Effects (5marks)
      3. Control measures. (5marks)



      1. There is highest rate of transpiration; because of high light intensity hence the stomata are fully opened; bright light increases the internal leaf temperature hence high rate of evaporation of water;
      2. There is lowest rate of transpiration; because there was high relative humidity hence a low saturation deficit; there was no light hence stomata are closed;
    2. Potometer;
    3. small stomata
      fewer number of stomata
      sunken stomata
      reversed stomatal rhythm any one 1mk
    2. in tropical countries, malaria is more prevalent; people with sickle cell trait are more resistant to malaria than normal individuals because plasmodium survive very poorly in sickled red blood cells;
    3. Haemophilia
      colour blindness
      albinism any one 1mk
    1. Osmosis
    2. a control experiment
    3. Y- solution Y is hypotonic to the 5% sugar solution in the beaker; it loses water by osmosis to the solution in the beaker; decreasing in volume.
      Z- Solution Z is hypertonic to the 5% sugar solution; it draws in water by osmosis from the beaker; hence increasing in volume.
    4. Cell membrane;
    5. the process stops as high temperature destroys the structure of the cell membrane;
    1. A- Polar nuclei;
      B-Egg cell;
      G- Tube nucleus;
    2. Transport male gametes to the embryo sac for double fertilization to occur;
      1. Endosperm;
      2. development of a fruit without fertilization.
    4. heterostyly
      self sterility any one – 1 mark
    1. Ball and socket joint;
    2. Z- Femur;
      U- Articular cartilage;
    3. Absorbs shock;
      Lubricates the joints, reducing friction between bone;
    4. Skeletal muscle;

      Skeletal muscle                            Gut muscle/ smooth muscle

      Has striations

      Lacks striations ;



      Cylindrical shaped

      Spindle shaped;


      6A 1
      1. Photosynthesis;
      2. Respiration;
      1. Rate of carbon (IV) oxide consumption increases from 6 am to 12 pm; due to increase in light intensity; the rate remains constant and at maximum between 12pm and 2pm; because there is optimum light intensity and other factors are limiting;past 2 pm the rate decreases due to decrease in light intensity;
      2. Rate of carbon (IV) oxide release decreases from 6 am to 12 pm; because some of the carbon (IV) released is used in photosynthesis; the rate increases from 2 pm to 8pm; because rate of photosynthesis is decreasing as light intensity decreases; after which it remains constant since no photosynthesis is taking place to consume the carbon(IV) oxide;
      1. the point at which the rate of respiration and the rate of photosynthesis are equal.
      2. 7.12 am ± 12 minutes (7.00 am- 7.24 am)
        5.36 pm ± 12 minutes (5.24 pm- 5.48 pm)
    5. increase in temperature increases the rate of carbon (IV) oxide consumption upto optimum; temperature above optimum denature enzymes; reducing the rate of carbon (IV) oxide consumption; very low temperature inactivate enzymes; thus reducing the rate of carbon (IV) consumption; Total 5 mks max 2mks
    1. Have thin epithelium; to shorten diffusion distance of digested food;
      Numerous blood capillaries; to transport glucose and amino acids;
      Have lacteal; for absorption of fatty acids and glycerol;
      Have numerous microvilli; to increase the surface area for absorption of digested food;
      The epithelium has goblet cells; which secrete mucus for lubrication;
      They are numerous; to increase the surface area for absorption; mks
    2. during the day guard cell carry out photosynthesis and manufacture glucose; this increases the osmotic pressure of the sap vacuole; which becomes higher than that of the neighbouring epidermal cells; guard cells take in water by osmosis; and become turgid; the outer thin wall stretches easily pulling the thicker inner wall outward; thus the stomata opens; at night there is no light and no photosynthesis; plant cells respire using up more glucose; the osmotic pressure of sap vacuole of guard cells reduces; becoming lower than that of the neighbouring epidermal cells; they then become flaccid; pulling together the thick inner walls; and stomata closes; total 13 marks max 10 marks
    1. Ultrafiltration occurs at the glomerulus; the glomerulus is a fine network of blood capillaries enclosed by the Bowman’s capsule; the afferent arteriole which supplies blood to the glomerulus; is wider than the efferent arteriole; which takes blood away from the glomerulus; this creates resistance to blood flow; and back pressure at the glomerulus; this builds a high pressure within the glomerulus; that forces the liquid part of blood and dissolved substances of low molecular sizes (such as urea, glucose, salts and amino acids) out through the pores in the glomerulus capillaries into the cavity of the bowman’s capsule; The filtrate formed is called glomerular filtrate; large sized molecules in plasma(such as red blood cells and plasma proteins) are not filtered; because they are too large to pass through the capillaries; Total 12 mks, max 5mks
      1. Causes
        discharge of untreated sewage from urban centers into water bodies;
        industrial effluents from industries being discharged into rivers, dams and lake;
        Large scale discharge of oil into oceans by oil tanker accidents;
        Discharge of hot water from industries into water bodies;
        Agricultural chemicals get washed into water bodies;
        Soil erosion leading to siltation;
        Lead from use of lead pipes and tank in domestic water supply;
        Mercury from industries that manufacture chlorine, sodium hydroxide;
        Total 8mks max 5mks
      2. Effects
        Epidemics of waterborne disease e.g cholera due to untreated sewage;
        Breakdown of organic matter in raw sewage cause eutrophication;
        Industrial effluents kill aquatic organisms like fish due to poisoning and eutrophication;
        Oil clogs gills of fish causing their death by preventing gaseous exchange;
        Oil also clogs feathers of marine birds leading to difficulty in flight; oil coats phytoplanktons till they die; oil coats the water surface thus reducing oxygen concentration in water hence suffocation and death of aquatic plants and animals;
        Nitrates and phosphate fertilizer cause eutrophication;
        Hot water reduces concentration of dissolved oxygen and carbon (IV) oxide hence suffocates and kills aquatic organisms; the high temperatures may also kill the organism directly;
        Mercury and lead results into physiological poisoning; that may result to defects like skin cancer or even death (lethal); Total 11mks Max 5mks
      3. Control Measures
        Sewage treatment and proper disposal;
        Treatment of industrial wastes before discharge into water;
        Encourage the use of unleaded petrol;
        Enforcing relevant legislation and providing heavy offenders;
        Carry out environmental impact assessment before setting of industries;
        Encouraging organic farming/ controlled use of agrochemicals;
        Appropriate soil control measures should be put in place;
        Total 7mks max 5 maks
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