Biology Paper 3 Questions and Answers - Cekenas Mock Exams 2022

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QUESTIONS

  1. You are provided with a specimen T that has been obtained from a plant.
    1.      
      1. Identify the part of plant that is T. (1mk)
      2. Give one reason for your answer in a (i) above. (1mk)
    2. Cut specimen T into two halves to obtain its cross sections. Squeeze the juice from specimen T into the beaker provided. Using the reagents provided, test the food substances present in specimen T. record your results in the table below. (9mks)
      Food Procedure Observation Conclusion
             
    3. Specimen T is dispersed in nature by animals. Give two importance of fruits and seed dispersal. (2mks)
  2. You are provided with specimen J. examine it and answer the questions below.
    1.      
      1. Identify the subdivision of the plant from which J was obtained. (1mk)
      2. Give a reason. (1mk)
    2. State the class of J. (1mk)
      Reason. (1mk)…………
    3.      
      1. What is the mode of pollination of J. (1mk)
      2. Give a reason for your answer in (i) above. (1mk)
    4. Describe the following parts: (2mks)
      1. Corolla
      2. Androecium
    5. Carefully open T and remove the petals. Remove one stamen and make a well-labelled diagram. Indicate the magnification. (4mks)
  3. Study the photograph below and answer the questions that follow:
    1
    1. Name the three bones labelled A, B and C. (3mks)
    2. Name the cavity Z and state its role. (2mks)
    3. Name the structure labelled T and state the role it plays in female mammals. (2mks)
    4. In the photograph show with letter K where ball and socket joint is formed and name the bone that forms the joint. (2mks)
    5. Give one name of the structure formed by bones C, G and T. (1mk)
    6.        
      1. Where is it found in the mammalian body. (1mk)
      2. The actual length of the specimen from points X and Y is 20.8cm. Calculate the magnification of the photograph. (2mks)


MARKING SCHEME

  1.      
    1.      
      1. Fruit (1mk)
      2. Has two scars (1mk)

    2. Food Procedure Observation Conclusion
      Vitamin C (Ascorbic acid) reject. Vit.C or wrong spelling Put DCPIP in a test tube, add food substance (the juice) dropwise.

      DCPIP colour is decolourised
      DCPIP colour disappears

      Vitamin C (Ascorbic acid) is present 
      Non reducing sugar Put juice/ food substance in a Test tube.
      Add a few drops of hydrochloric acid and heat, add few drops of sodium hydrogen carbonate until fizzing stops, and add some drops of Benedict’s solution and heat.

      Colour turns orange or colour changes from blue, green, yellow and orange;
      NB: Colour sequence must be correct
      Non reducing sugar present;
      Protein (s) To the food substance add few drops of sodium hydroxide and (shake), add copper sulphate (drop by drop) No colour change;/ blue colour persists Absence of proteins
      (9mks)
      NB: No mark for food substance
    3. Importance of dispersal
      1. Colonise new place/ area
      2. Reduce overcrowding/ reducing competition
      3. Reduce the spreading of diseases.
  2.      
    1. Angiospermaphyta (1mk)
      Reason – it’s a flower (1mk)
    2. Dicotyledonae (1mk)
      Reasons 5 stigma, 5 petals, / 5 floral parts. (1mk)
    3. Insect
      Brightly coloured petals/ anthers. (1mk)
    4. 5 petals; fused from the base but free at tips.
      • Petals are star shaped;
      • Net veined; purple in colour; (2mks)
        Androecium – 5 anthers; free surrounding style;
      • Each anther has short style;
      • Anthers below the stigma;
      • Anthers yellow in colour. (2mks)

    5. 2
      L = 3mks
      D = 1mk
  3.      
    1. A – Lumbar vertebra reject. Lumbar bone, lumbar vertebrae
      B- Sacrum
      C- Pubic bone/pubis
    2. Z- Obturator foramen (1mk)
      Passage of blood vessels/ nerves/ muscles (1mk)
    3. T- Pubis Symphysis
      Role – During birth, it opens for easy passage of new born. (1mk)
    4. Femur
    5. Pelvic girdle
    6.      
      1. Pelvic region/ Hip region
      2. Drawing length = 8 ± cm = X0.4
        Actual length      20.8cm
        (X0.399 - X0.41)

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