Biology Paper 2 Questions and Answers - Lainaku II Joint Mock Examination 2023

Instructions to candidates

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

SECTION A (40 MARKS)

Answer all questions in this section in the spaces provided.

1. In certain breeds of rats, a pure breeding of black rat was crossed with a pure breeding white rat. The offspring had a coat with white and black strands that appeared grey in colour.
   1. Using letter B to represent the gene for black coat colour and W to represent the gene for white coat colour, work out the genotypic ratio of F2 generation (5 mks)
   2. Give one example in human beings where genes behave the same as described in (a) above (1 mk)
   3.  
      1. What is polyploidy (1 mk)
      2. Give one advantage of polyploidy in organisms (1 mks)
2. In a practical investigation a raw potato was peeled, mashed into a fine paste and was treated as shown in the set-up below
   
   1. Name the physiological process being investigated (1 mk)
   2. State the expected colour change of the solution inside and outside the visking tubing after 40 minutes
      inside………… (1 mk)
      outside…………(1 mk)
   3. Explain the observation made in (b) above (2 mks)
   4. Distinguish between the following terms
      1. Haemolysis and plasmolysis (2 mks)
      2. Give the role osmosis in plants ( 1 mk)
3. The diagram below represents the nitrogen cycle. Study it and answer the questions that follow.
   
   1. Name the process represented by B. (1 mk)
   2. Identify organisms responsible for;
      1. process A ………………………………………………………………….. (1 mk)
      2. process D ……………………………………………………………………. (1 mk)
      3. Name the compound represented by C. (1 mk)
   3. Name the method of population estimation in the following;
      1. When studying the population of fish in a dam. (1mk)
      2. When studying cabbages in one acre of land. (1mk)
   4. Name two waterborne diseases and causative agents that affect human beings from kingdom
      Monera. (2mks)
      

      Disease	Causative agent

4. Study the diagram below and answer the question that follow
   
   1.  
      1. Identify the mode of feeding of the organism. (1 mk)
      2. Name the parts labelled A and B . (2 mks)
   2. How is the tooth S in the photograph above adapted for its functions ( 1 mk)
   3. What are the functions of the part labelled A. (2 mks)
   4. Distinguish between the terms homodont and heterodont (2 mks)
5. The diagram below represents a transverse section through a plant organ
   
   1.  
      1. From which organ was the section obtained (1 mk)
      2. Give two reasons for your answer in (a) above (2 mks)
   2. How is part C adapted to its functions (2 mks)
   3. State two structural differences between an arteries and veins (2 mks)
   4. Give one defect that affect the arteries in human beings (1 mk)

SECTION B   (40 MARKS)

Answer question 6 (compulsory) and either question 7 or 8

1. Two sets of bean seeds were germinated, set A was placed in normal daylight condition in the laboratory while set B was placed in a dark cupboard. After a few days later the shoots were measured twice daily and their mean lengths recorded as shown in the table below.
   

   Time in hours	0	12	24	36	48	60	72	84
   Set A length (mm)	13	15	21	24	29	32	48	55
   Set B length (mm)

   
   
   1. Using a suitable scale draw the graphs of the mean lengths in set A and B against time (8 mks)
   2. From the graph state the mean shoot length of each set of seedlings at the 54th hour (1 mk)
   3. Account for the difference of curve B and A . (3mks)
   4. Explain what would happen to set-up B if it were allowed to continue to grow under condition of darkness. (4mks)
   5. State three external conditions which should be constant for both set-ups (3 mks)
   6. Give one survival value of phototropism in plants. (1 mk)
2.  
   1. Giving relevant examples, explain how seeds and fruits are adapted to animal
      Dispersal (6 mks)
   2. Describe the process of double fertilization in flowering plants (14 mks)
3.  
   1. Distinguish between excretion and secretion ( 2 mks)
   2. Explain the main homeostatic function of the mammalian liver (18 mks)

MARKING SCHEME.

1. (a)  

   (b)  - ABO blood group; (1mk)
         - Sickle cell trait; (first correct)

   (c) (i) A condition in cells of an organism where the cells have more than one pair of (homologous) chromosomes; (1 mks)
       (ii) - Increased yields;
            - Resistance to pest;
            - Resistance diseases
            - Resistance to drought; (first correct)

2. (a) Diffusion; (1mk)
         inside……colour turns blue-black; (1 mk)
        outside……iodine colour remains/the colour remains brown/yellow; 1 mk)
    (c) Iodine molecules were small and diffused across the visking tubing; these molecules moving from regions of high concentration to region of            low concentration; (they reacted with starch forming a substance which is blue-black) (2 marks)
    (d) i) haemolysis is the bursting of red blood cells when placed in hypotonic solution they gain water and due to lack of cell wall they eventually                 burst; while plasmolysis is shrinking of plant cell cytoplasm away from the wall as a result of it being placed in hypertonic solution there by               losing water by osmosis; (2mks)
         ii) - Closing and opening of stomata; (1mk)
             - Feeding in insectivorous plants;
             - Absorption of water by root hair cells; (any one correct)

3. (a) Feeding; (1mk)
    (b) (i) Nitrogen fixing bacteria (Symbiotic-Rhizobium)/Non-symbiotic (azotobacter and Clostridium); Algae (nostoc, anabaena, and chlorella)                        (2mks)
         (ii) Denitrifying bacteria (Thiobacillus denitrificans and Pseudomonas spp, and Fungi);
        (iii) Nitrates; (1mk)
    (c) (i) Capture recapture method; (1mk)
         (ii) Line transect/belt transect; (1mk)

4. (a) (i) Herbivorous/Herbivory; (1mk)
         (ii) A -; Diastema;
              B - Horny pad; (2mks)
    (b) Has cusps/ridges to enable the tooth to grind food; (1mks)
    (c) - Enables turning of the tongue to roll food in the mouth during chewing;
         - Stores food temporarily during chewing;
         - Helps in removing of unwanted wastes from the mouth; (2mks)
    (d) Homodont is a type of dentition having teeth that are of same type, shape and size while heterodont is type of detention of teeth                               differentiated into incisors, canines, molars and premolars; (2mks)

5. (a) (i)Dicotyledonous root; (1mk)
         (ii) - Presence of star shaped xylem structure;
              - Presence of endodermis;
    (b) - Elongated/long to enable it reach water in the soil;
         - Having large sap vacuole with dissolved substance to enable it to absorb water by osmosis;
         - Has a thin membrane to reduce distance covered hence facilitating faster absorption; (2mks)
     (c) (2mks)

    (d ) Arteriosclerosis; (reject incorrect spelling)(1mk)

SECTION B
6 (a)    (8 mks)

Graph mark distribution.
Scales – (2 mks)
Plot (2 Marks)
Curves- (1 mks)
Identification – (1mk)
Axes – (2mks)

 (b) Set A – 31 ± 1 (2mks)
      Set B – 57 ± 1

 (c) Account for the difference of curve B and A (3 mks)

There is a faster growth rate in plant B than plant A; this is because to plant B was growing very fast to obtain light; light is used for photosynthesis to make food; plant A had the necessary light intensity so there was no need for faster growth;

 (d) Explain what would happen to set-up B if it were allowed to continue to grow under condition of darkness (4mks)

The plant in set up B becomes weak in darkness; and eventually die; the food reserve in the cotyledon would be exhausted; this is due to absence of light necessary for photosynthesis;

 (e) State three external conditions which should be constant for both set-ups (3 mks
- Amount of water;
- Temperature;
- Oxygen concentration;
 (f) - Exposes leaves for photosynthesis;
     - Enables roots to absorb water and mineral salts from soil;

7) (a) Giving relevant examples, explain how seeds and fruits are adapted to human dispersal.
          Adaptation of animal dispersed seeds and fruits
- Some fruits and seeds are hooked (sticky). These hooks stick on the fur/hair (or clothes in man) of passing animals; this enables them to be carried and dispersed away from the parent plant e.g Biden pilosa;

- Fruits that are succulent/juicy (fleshy pericarp) are eaten by animals and their seeds; discarded some distance from parent plants e.g mangoes, oranges or any other;

- Some fruits have brightly coloured epicarp which attracts animals; facilitating feeding and hence dispersed e.g mangoes/any other;

- Some fruits are scented/aromatic i.e sweet smell which attracts animals; (The animals feed on fruits and discard them away from parent plant) e.g banana/any other;

- Some fruits have seeds with hard/indigestible coats (testa) that are resistant to digestive enzymes; (The fruit together with the seeds are swallowed the seeds being indigestible are passed out with faeces while they are still capable of germination) e.g passion fruit/any other;
- Other seeds covered with sticky (mucoid)material are carried on the body parts of animals; and consequently discarded away from parent plant e.g passion seeds; (accept any correct example). (Three correct pints with adaptations and examples)

(b) the process of double fertilization in flowering plants

The pollen grains land and stick onto stigma; the stigma secretes/produces a sticky substance that causes pollen grains to adhere onto it; the pollen grains are also stimulated to develop/germinate a pollen tube; the pollen tube grows along the style tissues/cells; the pollen tube is nourished by style cells; as pollen tube grows downwards the tube nucleus occupies the position just behind the tip; it is followed by generative nucleus which divides mitotically; into two male nuclei which follow tube nucleus; on reaching the ovary the pollen tubes enters the embryo sac via the micropyle; the tube nucleus then disintegrates, leaving a clear passage for entry of two male nuclei; one of male nuclei fuses with both polar nuclei; in the embryo sac to form a triploid nucleus/primary endosperm nucleus; the other male nuclei fuses with egg cell/ovum; to form a diploid zygote which later develops embryo;

8) (a) Distinguish between excretion and secretion

          - Excretion is the removal of toxic waste products of metabolism from the body during chemical reactions in the cells;
          - Secretion is the production and release from cells or glands of certain substance which are useful to the body e.g enzymes;

    (b) Explain the main homeostatic function of mammalian liver
         - Regulation of temperature/Thermoregulation;
         - Heat is generated during chemical reactions in the liver cells/exothermic reactions generate heat in the liver cells; the heat is then distributed to the body raising the temperature to the normal whenever it falls below normal;

Regulation of blood sugar level;

The normal amount of blood sugar is (about 90mg/100ml of blood); increase in blood sugar level (above the normal) is detected by pancreatic cells; which secretes insulin; insulin stimulates the liver cells; to lower the blood sugar level by conversion of excess glucose to glycogen; conversion of excess glucose to fats; and increased oxidation/ respiration of glucose to yield energy; (until blood glucose level is brought back to normal.

Decrease in blood sugar level below the normal; is detected by the pancreas; which secretes the hormone glucagon; glucagon stimulates stimulate liver cells; to convert stored glycogen to glucose; convert amino acids to glucose; reduce oxidation of glucose to release energy (until blood glucose/sugar level is brought back to normal);

Deamination of excess amino acids;
Excess amino acids are deaminated where the amino group is removed; then converted to ammonia; the toxic ammonia combines with carbon (iv)oxide to form the less toxic urea; urea is then excreted in urine through the kidney/ excreted through sweat in skin;

Detoxification;
The liver converts toxic substances into less toxic substances; the toxic substances e.g.hydrogen peroxide produced in body cells is detoxified to water and oxygen gas;  (25 marks maximum 18)