BIOLOGY PAPER 2 - KCSE 2019 MOCK EXAMINATION - KAKAMEGA

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  1. The diagram below is a cross section through a part of human ileum.
    biology2q1
    1.  
      1. Identify the structure drawn above                                                                               1 mark
      2. Sate the significance of the structure shown above.                                                      1 mark
    2. Name the parts labelled A, B and C                                                                               3 marks
    3. Give the functions of the part labelled B and C                                                              2 marks
    4. Describe the function of goblet cells.                                                                              1 mark
  1. An experiment was carried out to find out the concentration of ions in the cell sap of an aquatic plant and that of the pond water in which they were found.
     

    Concentration in

    Ions

    Cell sap

    Pond water

    Na+

    50

    1.2

    K+

    49

    0.5

    Mg2+

    11

    3.0

    Ca2+

    13

    1.3

    Cl-

    101

    1.3

    SO42-

    13

    0.67


    1.  
      1. Name the process by which the aquatic plant absorbs ions from pond water.             1 mark
      2. State the four roles of the process you have named in (a)(i) above in a mammalian body.     4 marks
    2. Name the organelle that allows passage of ions in and out of the cell.                                        1 mark
    3. How can the rate of uptake of ions by the aquatic plant be increased.                                          2 marks
  2.  
    1. Differentiate between the mode of fertilization in higher plants and in mammals. 2 marks
    2. Give the functions of the placenta during pregnancy.                                                                 2 marks
    3. Describe events that occur during anaphase II of meiosis.                                                         1 mark
    4. Outline the role of acrosome in human spermatozoan.                                                                1 mark
  1. In a certain plant species, which is normally green, a recessive gene for colour (n) causes the plant to be white when present in the homozygous stat. Such plants die young at an early age. In the heterozygous state the plants are pale green and grow to maturity.
    1. Suggest a reason for the early death of the plants with the homozygous recessive genes. 2 marks
    2. If a normal green plant was crossed with the pale green plant, what will be the phenotypic ratio of F1 generation?            4 marks
    3. Give an explanation for the occurrence of pale green colour in the heterozygous state.              2 marks
  2. The diagram below represents one of the joints in the mammalian skeleton.
    biology2q5
    1. Name the type of joint shown in the diagram. 1 mark
    2. Name the parts labelled U and Z 2 marks
    3. Name two parts of the body where this type of joint is found.                            2 marks
    4. State the functions of the fluid found in W                                                                     2 marks
    5. Name the structure which attaches muscles to the bones at a joint.                                   1 mark

SECTION B; 40 MARKS
Answer question 6 (compulsory) and either question 7 or 8 in the spaces provided after question 8.

  1. The table below shows how quantities of sweat and urine vary with external temperature.

    External temperature oC

    0

    5

    10

    15

    20

    25

    30

    35

    Urine , cm3/hr

    100

    90

    80

    70

    60

    50

    40

    30

    Sweat , cm3/hr

    5

    6

    10

    15

    30

    60

    120

    125


    1. Using the same axes, draw quantities of urine and sweat produced against the external temperature. 8 mark
    2. Account for the amount of sweat produced as the temperature rises.                              3 marks
    3. Account for the amount of urine produced as the temperature rises.                               3 marks
    4. Explain how the following help in temperature regulation when it is cold.
      1. Hair                                                                                                                                     3 marks
      2. Blood vessels                                                                                                                     3 marks
  1.  
    1.  
      1. Give four modes of expressing food relationship in an ecosystem. 4 marks
      2. Explain how food as a factor regulate the population of animals in an ecosystem. 8 marks
    2. How are desert plants adapted to conserving water?                                                     8 marks
  2. Describe the;
    1. Process of inhalation in mammals                                                                                        10 marks
    2. Mechanism of opening and closing of stomata.                                                                 10 marks


MARKING SCHEME

  1.  
    1.  
      1. Villi;   rej. Villus
      2.  Increase surface area for absorption of digested food;
    2. A- Epithelium;
      B- Lacteal;
      C – Blood capillaries;
    3. B- responsible for absorption of fats;   rej. Fatty acids and glycerol
      C- Important for transporting digested food;
      Secrete mucus to lubricate food; Form a protective layer for the gut wall to prevent it            from being digested;
  1.  
    1.  
      1. Active transport;
      2. Reabsorption of glucose/ salts in kidney tubules; Absorption of digested food      from the alimentary canal into the blood stream;   Excretion of waste materials from cell of the body; Accumulation of substances in the body to counter osmotic imbalance in saline Environment;                                                         
        STRICTLY IN MAMAMALS
    2. Cell membrane/Plasma membrane;
    3. Increasing supply of oxygen/ oxygen concentration;
      Increasing supply of glucose/ glucose concentration;
  2.  
    1. Fertilization in higher plants involves one male nucleus fusing with a functional egg nucleus to form a zygote; while the other male nucleus fuses with the polar nuclei to form a triploid cell; In mammals fertilization involves the fusion of the male nucleus and female ovum nucleus to form a zygote;
    2. Provide site for exchange/ diffusion of nutrients and waste products between the maternal blood and the foetal blood system;
      Secretes/ produces Progesterone hormone;
      Attaches the foetus to the mother`s uterus;
    3. Sister chromatids separate; Sister chromatids move to opposite ends of the pole/cell;
    4. Contain lytic enzymes which dissolves egg membrane;
  3.  
    1. The gene for white colour are present in homozygous state; the plant will fail to manufacture chlorophyll hence no photosynthesis; they will thus die as soon as the food reserves are depleted;
    2.  
      BIOLOGY2Q14B
    3. Gene for white and green colour are both dominant; hence they all express themselves equally producing pale green plants/ Codominance;
  1.  
    1. Ball and socket (joint);
    2. U- Articular cartilage;       Z- Femur;    
    3. Shoulder/ pectoral girdle;                   Hip/ pelvic girdle;
    4. Reduction of friction/ lubrication;
      Absorption of shock/ distributes pressure/ shock absorber;
    5. Tendon;
  2.  
    1. Labeling axes, A 2 marks
      Scale -           S 2 marks – for correct vertical and horizontal scale
      Plotting         P 2 marks for all points correctly plotted (missing or wrongly plotted         penalize fully.
      Curve          C 1 mark smooth curve passing through all the plotted points
      Identity of the curve, I     1/2 mark each for correctly identified curve
      Total 8 marks
    2. Sweat produced increases with increase in temperature; increase in temperatures increases vaporization of sweat; in which latent heat of vaporization is absorbed from the body;
    3. Increase in temperature reduces the amount of urine produced; because increased sweat increases the osmotic pressure of blood; more water is reabsorbed into the blood stream from the kidney tubules (resulting to production of little and concentrated urine);
    4.  
      1. Erector pilli muscles contract; hair stands erect trapping air; which insulate the body against heat loss;
      2. Blood vessels constrict; blood is diverted to a shunt system; les blood flows to the skin surface; and less heat is lost;
  3.  
    1.  
      1. Food web; Food chain; Pyramid of biomass; Pyramid of numbers;
      2. A lot of food; causes population increase; leading to high rate of reproduction; and immigration;
        Little food ; leads to stiff competition (for food) ; leading to low rate of reproduction; high rates of deaths; and emigration; thus reducing the population.
    2. Leaves are modified to spines/ thorns; to reduce surface area over which transpiration can occur; Shed their leaves during the dry season; to reduce the surface area exposed to transpiration; Leaves have thick , waxy cuticles; to minimize rate of cuticular transpiration. Leaves for some plants can roll or fold; to reduce rate of transpiration by not exposing stomata to environmental factors. Have sunken stomata; which accumulates moisture in sub-stomatal air spaces hence low diffusion gradient thus reducing transpiration rate. Reduced number of stomata; hence low rate of transpiration. Some plants have reversed stomatal rhythm; to prevent excessive water loss by transpiration. Possession of very deep roots; to absorb water from deep in the soil surface; Possession of parenchyma cells in swollen stems and leaves; for storage of water; Many leaves are sclerophylous/ possess resin coatings; to increase reflection of solar radiation ; hence lower transpiration rate.
      NB Award transpiration once
  4.  
    1. Process of inhalation in mammals
      External intercostal muscles contract; while internal intercostal muscles relax; Pulls ribs upwards and outwards; the diaphragm muscles contract; and the diaphragm flattens; This increases the volume of thoracic cavity; and decreases its pressure; atmospheric pressure being higher than thoracic cavity pressure; forces the air to rush into the lungs; (through the nose and trachea ) the lungs are inflated;
      Maximum 10 marks
    2. Mechanism of opening and closing of stomata
      During the day chloroplast of guard cells accumulate sugar/ glucose produced through the process of photosynthesis; Accumulated sugar/ glucose in the guard cells increases osmotic pressure of the cell sap of the guard cells; Water is drawn from neighbouring epidermal cells by osmosis; guard cells become turgid and bulges outward; this opens the stomata; At night, sugar / glucose levels which had accumulated in guard cells is converted to starch; Osmotic pressure of guard cells falls; the cells lose water to the neighbouring epidermal cells by osmosis; become flaccid; the guard cells are drawn towards one another; the stomata closes;
      Maximum 10 marks
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