Biology Paper 3 Questions and Answers With Confidential- Kenya High Post Mock 2023 Exams

INSTRUCTIONS

• Answer all questions
• You are required to spend the first 15 minutes of the 1 ¾ hours allowed for this paper reading the whole paper carefully before commencing your work.

QUESTIONS

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. (1mark)
   2. Pour 2ml of liquid X into a test-tube. Add equal amounts of Benedict’s solution boil the mixture. Record your observation (1mark)
   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.
      • 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.
      • To set —up C, add the third portion of substance Q. Add 8 drops of 2M
        hydrochloric acid and shake.
        Place the three set-ups in a warm water bath maintained at 37°C for 30minutes.
        Cool the set-ups by dipping the boiling tubes in cold water
        Place 2m1 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.
        Record your observations (3marks)
        Set-up A
        Set-up B
        Set-up C
   4. Account for your observations in the set-up (3marks)
      Set-up A
      Set-up B
      Set-up C
   5. Give the most likely identity of substance Q (1mark)
   6. Why was the water bath maintained at 37°C (1mark)
   7. What is the fate of the product of set up A in an organism? (2marks)
2. Below are photographs of a variety of invertebrates. Examine them and answer the questions that follow.
   
   
   1. Complete the dichotomous key given below. (5mks)
      1.  
         1. Animal with three pairs of legs…………………………………………….………..go to 2
         2. b......................................................................................................................................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. (8marks)
      

      Specimen	Steps followed	Identity
      K1
      K2
      K3
      K4
      K5
      K6
      K7
      K8

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

CONFIDENTIAL

Question 1:

•  Two dry cells and a cell holder
•  An ammeter
• A voltmeter
• Metre rule mounted with a nichrome wire( 28 gauge)
• A switch
• 8 connecting wires

Question 2

• Metre rule
• A knife edge
• 50g mass and a 100g mass
• Two pieces of threads each 30 cm long
• 100 ml of Water in a beaker
• 100 ml of glycerine(Liquid L) in a beaker
• Tissue paper
• 4 optical Pins
• semi-circular glass block
• two thumb pins
• soft board
• plain papers
• Vernier calipers

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.  2. a
           b.Animal with more than three pairs of legs ……….....................................................go to 5
      3. a   
          b. Animal with two pairs of wings ………………….......................................................go to 4
      5 a.Animal with four pairs of walking legs…………..................go to 6
         b.
      6 a. Animal with an tennae……………...............Crustacea
        b.
      7.a.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.  
      1.  
         1. Self pollination
         2. Cross pollination
         3. Double fertilization
      2.  
         • K – Stigma
         • P – Antipodal cells
         • R – Synergids. Rej: Synergid
         • W – Diploid zygote. Rej: Zygote, Embryo.
      3. Meiosis
      4. Pollen grains
   2.  
      1. Petals – Brightly colored; to attract insects; (for cross pollination)
      2. Filaments – Long; to expose the anthers to the agents of cross pollination.
   3. Gametophyte