Biology Paper 2 Questions and Answers - Murang'a County Mocks 2020/2021

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INSTRUCTIONS TO CANDIDATES

  • This paper consists of two sections A and B.
  • Answer ALL questions in section A
  • Answer question 6 (compulsory) and either question 7 or 8 in section B.

SECTION A (40 marks)

Answer all questions in this section in the spaces provided

  1. An experiment was set up to investigate a factor in autotrophism in green plants.
    autotrophismbiop2q1 JMkxV
    Vaseline was applied at joint between the cork and the mouth of glass bottle and set up was left under sunlight for 6 hours.
    1. Explain why it was necessary to apply Vaseline. (1 mark)
    2. Explain why it was necessary to cover the pot with polythene paper. (1 mark)
    3. What was the purpose of including the small animal? (2marks)
      1. What would happen to the small animal if the set up was left overnight in darkness? (1mark)
      2. Account for the answer above (1 mark)
    4. Explain why organisms in phylum Arthropoda die when Vaseline is applied on its thorax. (2marks)
  2.  
    1. In a field study to estimate the population of grasshoppers in the school field of 0.4 km2, 60 grasshoppers were caught using sweep nets, marked with red paint and released back to the field. The following day students went back with their sweep nets and caught 100 grasshoppers, of which 20 were found to be already marked.
      1. Calculate the population size of grasshoppers in the field. (2 marks)
      2. Calculate the population density of the grasshoppers in the field. (2 marks)
      3. What two factors would maintain the population of grasshoppers at the carrying capacity? (2 marks)
    2. Giving an example, state what is meant by the term symbiosis. (2 marks)
  3. The figure below shows the embryo sac before fertilization.
    embryosacbiop2q3 mur W0krV
    1. Identify the structures labeled A and B  (2mks)
      A
      B
    2. identify the structures labeled in the diagram that will develop into the following after fertilization (2mks)
      1. Embryo
      2. Endosperm
    3. State the ploidy of each of the following nuclei after fertilisation  (2mks)
      1. C
      2. D
    4. Briefly outline the process of double fertilisation in flowering plant (2mks)
  4. In an experiment, a black mouse was mated with a brown mouse; all the off-springs were  black. The off-springs grew and were allowed to mate with one another. The total number of (F2) generation off-springs was 96.
    1. Using the letter symbols capital letter B for the gene of black colour and small b for brown colour, Work out the genotype of the F1 generation. (3mrks)
    2. From the information above, work out the following for the F2 generation.
      1. Genotypic ratio. (2mrks)
      2. Phenotypic ratio.(1mrk)
      3. The total number of brown mice (2mrks)
  1. The diagram below shows part of gaseous exchange system in an insect. Study it and answer the questions that follows.
    gaseousexchangeinins hzmEe
    1. What is the structural adaptations of the parts labeled A and B to their functions   (2mks)
      A
      B

      1. Name the parts of the following animals that carry out the same functions as part B above (2mks)
      2. Tilapia fish
    2. Name the structures used for gaseous exchange in plant growing in waterlogged soils (1mk)
    3.  
      1. Give two reasons why accumulation of lactic acid during vigorous exercise leads to an increase of heart beat (2mks)
      2. In what form is oxygen transported from lungs to the tissues (1mk)

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 the quantities of urine and sweat vary with external temperature

    External temperature(oC )

    Urine (cm3/hr)

    Sweat (cm3/hr )

    0

    100

    5

    5

    90

    6

    10

    80

    10

    15

    70

    20

    20

    60

    30

    25

    50

    60

    30

    40

    120

    35

    30

    200

    1. On the grid provided, plot the quantities of urine and sweat produced against external temperature  (7 marks)
    2. At what temperature is the amount of sweat and urine produced equal? (1 mark)
    3. What happens to the amount of sweat produced as the temperature rises? Explain your observation (3 marks)
    4. Explain the observation made on the amount of urine produced. (3 marks)
    5. How are the following parts of the mammalian skin adapted for temperature regulation during cold weather? (6 marks)
      Hair:
      Sweat glands
      Blood vessels
  2.  
    1. Describe the opening and closing of the stomata using the photosynthetic theory. (10marks)
    2. Describe blood sugar regulation in mammals.  (10marks)
  3.  
    1. Describe how urea is formed in the liver cells from excess amino acid (5mks)
    2. Describe the roles of hormones in the growth and development in plants (15mks)

MARKING SCHEME

  1.  
    1. To prevent entry of gases   1mk
    2. Ensure soil microbes do not interfere with gas volume in glass bottles  1mk
    3. Consume oxygen released when the plant photosynthesis. Release  from respiration to be used for photosynthesis by the plant   2mks
      1. Small animal would die  1mk
      2. Lack of oxygen for respiration  1mk
    4. Vaseline blocks spiracle, thus no inhalation  2mks
  2.  
    1.  
      1. Population = first marked x second capture      100x60 ;    =    300 grasshoppers;
                                   Marked captured                     20         
      2. Population density =  total population =   300 ;   =     750; grasshoppers /km2
                                                Area               0.4
        • Competition;
        • death of those not suitably adapted;
        • Predation;
    2. Rhizobium in root nodules of leguminous plants; acc other relevant examples
      Is an association between two organisms of different species where both benefit;
  3.  
    1.  
      • A polen tube
      • B antipodal cells
    2.  
      1. C
      2. D
    3.  
      1. Diploid
      2. Triploid
    4.  one male nucleus fuses with the egg cell nucleus to form a diploid zygote while the other male nucleus fuses with the polar nuclei to form a triploid endosperm nucleus.
  4.  

    1. genotypebiop2ms q4 m LoAEY

      Rej. If it does not start from phenotype
             If X is not there at the genotype
             If X is placed at the gamates.
             If gametes are not circled
             If the circle has a tail or a gap- must be as smooth as possible.
             If the fusion lines do not touch the line of the circle or if it penetrates into the circle.
             If fusion lines join at the gamete.
             If fusion lines are not at the same level at the F1 generation.

    2.  
      1. F2 offspring’s are BB, Bb, Bb, bb:
        Genotypic   ratio = 1BB:2Bb:1bb;
        Rej. BB:Bb:bb
        1  : 2  :1
      2. Phenotypic ratio   -1brown: 3black ;
      3. ¼ x 96 = 24 ;
  5.  
    1.  
      1. A- presence of rings of chitin which keeps them always open
      2. B- Moist epithelium for dissolution of respiratory gases
    2.  
      1. Alveoli
      2. Gill filaments
    3. Pheumatophores
    4.  
      1. To increase supply of oxygen to breakdown lactic acid into carbon (IV) oxide, water and ATP; to remove carbon (IV) oxide ‘produced;
      2. Oxyhaemoglobin ACC HbO2/ HbO8 (2mks)
  6.  

    1. graphbiop2ms q6 mura tniDy
    2. 24oC +/- 0.5;
    3. Sweat production increases with increase in temperature high temperature increases the evaporation rate and hence more sweat converted to vapour; This uses latent heat of evaporation from body hence cooling.
    4. An increase in temperature deceases the amount of urine produced; this is due to sweating that raises the osmotic pressure of blood; A lot of water is then reabsorbed into the blood from the kidney tubules, resulting in production of little concentrated urine;
      • Hair - When cold erector pili muscles contract; causes hair to stand and trap air which acts as an insulator; less heat loss
      • Sweat glands - When cold sweat glands releases less sweat; hence less evaporation hence less heat loss.
      • Blood vessels - When cold, blood vessels constrict; less blood flows near skin surface reducing heat loss by radiation;
  7.  
    1. During the day in the presences of sunlight guard gill synthesize glucose/sugar from the photosynthesis process, the synthesized sugar accumulates in the guard cells, increasing their osmotic pressure(makes them hypertonic to the adjacent cells of the epidermis. By osmosis, guard cells, draw in water and bulge outwards opening, the stoma. During the night in the absence of sunlight; guards cell are unable to carry out photosynthesis hence sugar is converted into starch; starch lowers the osmotic pressure of the neighbouring cel, hence they low water by osmosis to the neighbouring epidermal cells become fluid hence closing the stomata  max 10mks
      • When blood sugar rises above normal the hypothalamus stimulates the pancreatic cells to secrete insulin hormone which travels through the blood stream to the liver when it stimulates the liver cells to
        1. convert excess glucose/sugar into glycogen
        2. Increase oxidation of sugar/glucose into energy, carbon(Iv)oxide and water
        3. Convert excess glucose/sugar into fats for storage in adipose tissues
        4. Inhibits conversion of glycogen into sugar
      • When blood sugar/glucose drops below normal; the hypothalamus stimulates the pancreatic cells to secrete hormone glucagon which travels through the blood stream to the liver; when it stimulates the liver cells to:
        1. Convert stored glycogen into sugar/glucose
        2. Decrease the oxidation of sugar/glucose
        3. Convert stored fat into sugar/glucose
      • Blood glucose is then restored back to normal levels  max 10mks                                                 
  8.  
    1. Amino acid are deaminated The amino group; combines with hydrogen ions forming ammonia; Ammonia enters ornithine cycle where it combines with carbon (IV) oxide forming urea;
      Remaining part of amino acid (carboxyl) is either converted to carbohydrates for storage in the liver; urea is transported to kidney by blood.
    2.  
      • AUXINS
        • Indole acetic acid /AA/Auxins
        • promotes cell division and cell elongation
        • promotes trophic responses
        • promotes formation of abscission layer/bring about leaf fall.
        • Promotes cell differentiation (of vascular tissue)
        • Promotes growth of adventitious roots (on stem)
        • Cause  epical dominance/ inhibits growth and development of lateral buds.
        • AA + Cytokinin induce formation of callus tissue (during healing of wounds)
      • GIBBERELLINS
        • promotecells division/cell elongation in dwarf varieties.
        • Parthenocapy after fertilization.
        • Promotes formation of side branches (of stem) and dormancy (in buds)
        • Inhibits growth of adventitious roots
        • Activates (hydrolytic) enzyme during germination/ promotes germination of seeds hence breaks seed dormancy.
        • Affect leaf expansion and shape/retard abscission.
      • CYTOKININS
        • breaks dormancy ( in same species)
        • promotes flowering in same species.
        • Promotes cell division (in presence of 1AA
        • Stabilizesprotein and chlorophyll.
        • Promotes roots formation
        • Low concentration encourage leaf senescence high concentrated protein increased cell enlargement
        • Promote flowering (in same species)
      • ETHYLENE/ETHENE
        • stimulates lateral bud development.
        • Ripening of banana/fruit.
        • Induces thickening of stems/inhibits stem elongation.
        • Causes abscission of leaves/fruits fall
        • High concentration of ABA causes stomata closure (by interfering with uptake of potassium ions.
      • ABSCISIC ACID ABA
        • Inhibits germination/growth of embryo/cause seed dormancy
        • Causes abscission of leaves/fruits/leaf fall.
        • Inhibits elongation growth inhibits sprouting of bud/ induces dormancy in bud.
      • FLORIGENS
        • promotes flowering.
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