Biology Paper 2 - 2020 MOKASA JOINT MOCKS EXAMINATION (QUESTIONS AND ANSWERS)

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SECTION A: (40 MARKS)

Answer all questions in the spaces provided.

  1. The diagrams below show samples of blood obtained from two different persons A and B.

    1biop2

    1. What genetic disorder is person B suffering from? (1 mark)
    2. State one advantage and one disadvantage of the disorder exhibited in person A (2 mark)
    3. Work out the genotypes and phenotypes of the resulting offsprings of a marriage between person A and person B (5 marks)
  2. Study the diagram below and answer the questions that follow.
    2biop2
    1. Name the division to which the organism belongs giving two reasons for your answer (3 marks)
    2. Name the function of the parts labelled
      A (1 mark)
      B (1 mark)
    3. State three differences between the process of fertilization in the above named division and in a flowering plant. (3 marks)
  3. A biologist carried out a study to investigate the growth of a certain species of herbivorous fish and the factors influencing plant and animal life in four lakes A, B, C and D. The lakes were located in the same geographical area.

    Two of the lakes A and B were found to contain hard water due to the presence of high content of calcium salts. The mean body length of 2 year old fish, amount of plant life and invertebrates biomass in each lake were determined. The data was as shown in the table below.

    Lakes  Means of body length (cm)  Type of water  Amount of plant life  Invertebrate biomas g/cm3
    Insects Snails Crabs Worms
    A 31.2 Hard 1050 11 300 10 180
    B 28.6 Hard 950 72 100 9 90
    C 18.4 Soft 1.2 79 0 2 20
    D 16.3 Soft 0.5 99 0 1 10

    1. Describe the procedure that may have been used to determine the mean body length of the fish. (4 marks)
    2. What are the likely reasons for the difference in mean body length of the fish living in lakes A and D? (2 marks)
    3. Explain why primary producers have a higher biomass (2 marks)
  4. The diagram below is a section from the mammalian body. Study and use it to answer the questions that follow.
    4biop2
    1. Name the parts labelled;
      1. ___ (1 mark)
      2. ___ (1 mark)
      3. ___ (1 mark)
    2. Describe the process of milk letdown (5 marks)
  5.  
    1. In an experiment, food sample A was respired by an organism and the gaseous product was directed into a test tube containing calcium hydroxide solution through a glass capillary tube.
      The same experiment was repeated using the same amount of food sample B. It was noted that it takes 15 minutes for the gaseous product of food sample A to turn calcium hydroxide solution white and 50 minutes by gaseous product of food sample B to do the same.
      1. Suggest with a reason, the possible identity of food sample A and B (4 marks)
      2. Suggest the possible identity of the gaseous product of food samples A and B. (1 mark)
    2. Explain how anaerobic respiration has been applied in making of beer and wines. (3 marks)

SECTION B: (40 MARKS)

Answer question 6 (COMPULSORY) in the spaces provided and either question 7 or 8

  1. The table below shows results of an experiment in which small pieces of tradescantia stems were placed in different salt concentrations. After 6 hours they were removed from the solutions, wiped to dry and weighed. The results are as shown below. Study the table and answer the questions that follow.

    Salt concentration (mg) Percentage change in weight
    2.5 +11
    5.0 +8
    7.5 +5
    10.0 +3
    12.5 +2
    15.0 +2
    17.5 -2
    20.0 -8
    22.5 -9.5
    25.0 -11

    1.  
      1. Draw a graph of the percentage change in weight against salt concentration. (6 marks)
      2. From the graph determine the salt concentration that is equal to the concentration of the tradescantia cell sap. (1 mark)
    2. Account for the following changes in the weight.
      1. Percentage positive change (4marks)
      2. Percentage negative change (3 marks)
    3. Briefly describe how the above physiological process brings about upright posture in seedlings (3 marks)
    4.  
      1. Define the physiological process in (c) above (1 mark)
      2. State any two differences between the physiological process above and the physiological process that root hairs use to absorb mineral salts from a soil solution that is hypertonic to their cell saps (2 marks)
  2.  
    1. Explain the biological importance of abiotic factors in seed germination. (12 marks)
    2. Explain the following evidences of organic evolution.
      1. Comparative anatomy (5 marks)
      2. Geographical distribution (3 marks)
  3. In terms of homeostatic balance in the body, describe the function of the following body systems in regulation of blood sugar level. (20 marks)
    1. Digestive system
    2. Circulatory system
    3. Respiratory system
    4. Nervous system
    5. Hormonal system


Marking Scheme

  1.  
    1. Sickle cell anaemia;
    2. Advantage – Individuals having this trait hardly/rarely suffer from malaria;
      Disadvantages – Suffocation due to insufficient supply of oxygen during straneous activity;
    3.  
      1cbiomp2
  2.  
    1. Division: Bryophyta;
      Reasons: Have developed rhizoids for anchoring & absorbing water;
      They are thalloid;
      Show alternation of generation;
    2. A - Produces spores;
      B – Anchorage and absorption of water and mineral salts;

    3. Bryophytes Flowering plants
      Antheridia release mobile sperm which swim in water to fertilize the egg in the archegonia Pollen tubes carrying male gametes grow down the style to ovary, to bring about double fertilization.
      Male gametes are produced in antheridia while female egg in archegonia Pollen grains in anthers and eggs are produced in ovary
      Zygote formed develops into young sporophyte plant which grows upwards while still attached on the female gametophyte plant The formed zygote develops into a seed in a fruit.
  1.  
    1. Fish were caught, age determined, 2 years old retained; length measured and recorded. This was done repeatedly until a large number were measured.
      The total length divided by number of fish.  (4 marks)
    2. Lake A has hard water with more calcium necessary for bone formation. Fish in A grow faster and greater bone length than fish in Lake D.
      Lake A has more food which fish eat than Lake D.
    3. receives light energy directly from the sun, and energy is lost from one trophic level to the next.
  2.  
    1. I           Lobule/Alveoli;
      II          Lacteferous duct;
      III        Lacteferous sinus;
    2. When the environment is positive; (e.g. small of baby etc.) the hypothalamus relays; impulses to the pituitary gland; which in turn releases oxytocin hormone;
      Oxytocin hormone causes alveoli to contract forcing milk into ducts which conduct milk to reservoirs;
  3.  
    1. Food sample A – Carbohydrates/glucose;
      Reason:          Quickly broken down/easily oxidized to release energy, water and carbon (IV) oxide;
      Food Sample B: Lipid/protein;
      Reason:          Slowly broken down slowly oxidized to release energy, water and carbon (IV) oxide gas;
    2.  
      1. Carbon (IV) oxide;
      2. Sugar in fruits or barley seeds; is fermented/broken down anaerobically; to produce ethanol in wines;
  4.  
    1.  
      1.  
      2. 16.25% salt;
    2.  
      1. The salt solution is hypotonic; to the cell sap of the tradescantia, water molecules moved from the solution into the cells by osmosis; making the cells turgid hence increase in size;
      2. The salt solution is hypertonic; to the cell sap of the tradescantia, water molecules moved from the cells into the solution by osmosis; making the cells flaccid to reduce in size;
    3. The plants absorb water from the soil by osmosis; the water is transported to cells; the cells become turgid making the plant upright;
      1. Movement of water molecules from a region of low concentration to a region of high across a semi permeable membrane.
      2.  
        Osmosis Diffusion
        Involves movement of water

        Involves movement of other substances apart From water

        Should be across a semi permeable membrane Should not be across a semi permeable membrane.
  5.  
    1. Water; Activation of enzymes to hydrolyze food materials;
      Transport of hydrolyzed food;
      Dissolve the stored food;
      Softens the seed coat to allow for imbibition;

      Oxygen: For aerobic respiration to supply energy (for formation of new cells and tissues).

      Optimum/favorable temperature
      Germination enzymes require optimum temperature for maximum activity.
      Low temperature inactivate the enzymes.
      High temperatures beyond optimum denatures/destroy enzymes
      The embryo is also killed at high temperatures.
      pH ;change from optimum pH denatures enzymes responsible for germination. Optimum pH activates the enzymes and the rate of germination will be faster.

      Light: Light induces the formation of gibberellins which stimulate seed germination by breaking seed dormancy.

      It is also a requirement of photosynthesis by the young growing leaves of the developing seedlings.

    2.  
      1. Comparative anatomy

        The study of internal structures reveal that certain body structures or organs found in different species of organisms have similar internal structure or plan; eg pentadactyl limbs in vertebrates.

        Such similar structures with similar origin but have undergone structural modifications to perform different functions in different organisms are called homologous structures, and the basis of their origin is called divergent evolution.

        Some organs have different embryonic origin but perform similar functions hence such organs are called analogous structures, and basis of their origin is called convergent evolution.

        Some body parts or structures become vestigial in the course of development and reduce in size due to disuse.

      2. Geographical distribution

        initially organisms occupied one big land mass, drifting of centinents isolated organisms from common ancestry leading to speciation/formation of new species due to adaptive radiation.


  6.   Digestive system

    Ingested carbohydrates are digested to glucose in the gut; glucose is absorbed into the bloodstream; raising blood glucose levels; Pancreas secretes insulin; thus stimulates liver cells to convert excess sugar to glycogen, fats and increase rate of respiration; restoring the level to normal; (6marks)

    Transport system

    Capillaries in small intestines join the hepatic portal vein which transports absorbed glucose to the liver directly;

    Transport system also transport hormones insulin and glucagon from the pancreas to the liver; transports glycogen from liver to muscles in the body for storage; hepatic vein transport glucose from the liver to other parts of the body for respiration;   (4marks)

    Respiratory system

    When level of glucose in blood increases beyond normal, insulin stimulate liver cells; to increase breakdown of excess glucose to water carbon(IV)oxide and energy; hence lowering the concentration of sugar in blood to normal levels ;

    When the level of glucose in blood falls below normal, pancreas is stimulated to secrete glucagon hormone; which stimulate liver cells to reduce breakdown /metabolism of glucose into water , carbon(IV)oxide and energy; restoring the amount of glucose in blood to normal.(5marks)

    Nervous system

    The hypothalamus detects levels of blood sugar in blood; sends impulses to the pituitary gland; which generates impulses and sends to the pancreas; to secrete insulin/glucagon; to stimulate liver cells to lower or raise blood sugar; (5marks)

    Hormonal system

    The pancreas as an endocrine gland secretes ;( the following hormones involved in regulation of blood glucose levels)

    Insulin stimulates liver cells; (to convert excess glucose to fat or glycogen thus lowering its concentration)

    Glucagon stimulates liver cells ;( to carry out activities that release glucose to blood thus increasing its concentration)

    Adrenaline hormone; secreted by adrenal glands; increase hydrolysis of glycogen into glucose; and inhibits release of insulin resulting to increase in blood sugar level. (6marks)

Maximum 20 marks       

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