BIOLOGY PAPER 3 - 2019 KCSE KASSU JOINT MOCK EXAMS (QUESTIONS AND ANSWERS)

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  1. You are provided with liquid X and substance Q
    1. Place three drops of liquid X onto a white tile. Add four drops of iodine solution and record your observation.       (lmk)
    2. Pour 2ml of liquid X into a test-tube. Add equal amounts of Benedict’s solution boil the mixture. Record your observation   (lmk)
    3. Label three boiling tubes as set-ups A, B, and C. Place 3m1of liquid X into each of the set-ups. Divide substance Q into three equal portions.
      To set-up A, add one portion of substance Q and shake.
      1. Place the second portion of substance Q into a test tube. Add 1ml of water to it and boil for four minutes. Add it to set-up B and shake.
      2. To set —up C, add the third portion of substance Q. Add 8 drops of 2M hydrochloric acid and shake. 
      3. Place the three set-ups in a warm water bath maintained at 37°C for 30minutes.
      4. Cool the set-ups by dipping the boiling tubes in cold water
      5. Place 2ml of the contents of each set-up into three separate test tubes. Add equal amount of Benedict’s solution to each of the three test-tubes and boil.
    4. Record your observations             (3mks)
      Set-up A
      Set-up B
      Set-up C
    5. Account for your observations in the set-up        (3mks)
      Set-up A
      Set-up B
      Set-up C
    6. Give the most likely identity of substance Q          (1mk)
    7. Why was the water bath maintained at 37°C           (lmk)
    8. What is the fate of the product of set up A in an organism?                          (2mks)
  1. Below are photographs of a variety of invertebrates. Examine them and answer the questions that follow.
    Webp.net compress image 47
    1. Complete the dichotomous key given below. (5mks)
      1.  
        1. Animal with three pairs of legs…………………………………………….………..go to 2
        2. _______________________...................................................go to 5
      2.  
        1. Animal with wings…………………………………………………………..……….go to 3
        2. Animal without wings………………………………………………………...Hymenoptera
      3.  
        1. Animal with one pair of wings……………………………………………………….Diptera
        2. ___________________  ......................................................go to 4
      4.  
        1. Fore wings hard……………………………….…………………………………..Orthoptera
        2. Fore wings membranous…………………………………………………………….Odonata
      5.  
        1. __________________________....................................................go to 6
        2. Animal with more than four pairs of walking legs……………………………………go to 7
      6.  
        1. _____________________..................................................Crustacea
        2. Animal without antennae……………………………………………………….…..Arachnida
      7.  
        1. ______________________________________………….……………………..Chilopoda
        2. Animal with two pairs of legs in each body segment …………………………..Diplopoda
    2. Use the dichotomous key to identify the specimen in the photographs above. (8mks)

      Specimen

      Steps followed

      Identity

      K1

         

      K2

         

      K3

         

      K4

         

      K5

         

      K6

         

      K7

         

      K8

         
  1. All members of plant division Spermatophyta exhibit alternation of generation. The photographs below show stages in the growth and development of a spermatophyte.
    Webp.net compress image 48
    1. Processes I, II and III. (3mks)
      I
      II
      III
    2. Structures K, P, R and W. (4mks)
      K
      P
      R
      W
    3. The cell division process that occurs in structures J. (1mk)
    4. The products of the process named in (iii) above. (1mk)
      1. Explain the role of the following in promoting process II in the flowering plants.
        Petals (2mks)
      2. The photographs above represents one of the phases in alternation of generations in spermatophytes. Name the phase. (1mk)


MARKING SCHEME

  1.  
    1. Blue black/black dark blue colour is formed
    2. No colour change/colour of Benedict’s solution remains;
      Rej: No change /no reaction/ no observation /nothing happens
    3. Set-up A- colour changes from blue to green to yellow to orange/brown;
      Set-up B: No colour change/ colour of Benedicts’ solution remains;
      Rej- No change/no reaction/no observation/nothing happens
      Set-up C- No colour change/colour of Benedict’s solution remains;
      Rej- No change no reaction/ no observation/ nothing happens
    4. Set-up A — Enzyme amylase/diastase/invertase (in Q); digests /hydrolysis/breaks down/converts starch (in liquid X); to reducing sugar/maltose;
      Set-up B: boiling denatures/destroys enzymes amylase/diastase/invertase; hence starch is not converted to reducing sugar/maltose;
      Set up C:- Hydrochloric acid provides unfavorable pH for enzyme amylase diastase/invertase; hence starch is not converted to reducing sugar/maltose;
    5. Enzyme amylase/diastase/invertase;
    6. To provide optimum temperature for enzymatic reaction
    7. Broken down to glucose. Glucose is oxidesed to provide energy during cellular respiration In excess it is converted to glycogen in animals or starch for storage in plants.
  2.  
    1. Animal with more than three pairs of legs ……….....................................................go to 5
      Animal with two pairs of wings ………………….......................................................go to 4
      Animal with four pairs of walking legs …………........................................................go to 6
      Animal with antennae ……………….……………...................................................Crustacea
      Animal with one pair of legs in each body segment………….……………………..Chilopoda
      1 x 5 = 5mks
    2.  

      Specimen

      Steps followed

      Identity

      K1

      1a, 2a, 3b, 4a

      Orthoptera

      K2

      1b, 5a, 6b

      Arachnida

      K3

      1a, 2b

      Hymenoptera

      K4

      1a, 2a,3a

      Diptera

      K5

      1a, 2a, 2b, 4b

      Odonatan

      K6

      1b, 5b, 7b

      Diplopoda

      K7

      1b, 5a, 6a

      Crustacea

      K8

      1b, 5b, 7b

      Chilopoda


      1 x 8 = 8mks; Award a mark for identity only when the steps followed are correct.
  3.  
    1. I – Self pollination
      II – Cross pollination
      III – Double fertilization
    2. K – Stigma
      P – Antipodal cells
      R – Synergids. Rej: Synergid
      W – Diploid zygote. Rej: Zygote, Embryo.
    3. Meiosis
      Pollen grains
    4. Petals – Brightly colored; to attract insects; (for cross pollination)
      Filaments – Long; to expose the anthers to the agents of cross pollination.
    5. Gametophyte                                     

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