Biology Paper 2 Questions and Answers - MECS Cluster Joint Mock Exam

QUESTIONS

SECTION A (40 marks)
Answer all the questions in this section in the spaces provided.

1. The graph below shows the rate of transpiration of the same plants on two consecutive mornings, day 1 and 2.
   
   
   1.      
      1. Give two environmental factors that could account for the difference between day 1 and day 2 (2mks)
      2. Explain how the environmental factors named (a) (ii) above could have caused the difference between day 1 and day 2. (2mks)
   2. Name two forces involved in movement of water up the xylem. (2mks)
   3. Name the strengthening material found in the following tissue in a stem:
      1. Sclerenchyma (1mark)
      2. Collenchyma (1 mark)
2.      
   1. A young mother delivered a baby at Muranga Hospital, which was taken to the nursery shortly after delivery. When she was brought the baby later, she felt that it was the wrong baby. When she got home, she decided to contest the issue in a court of law. The blood tests showed that she was blood group AB and her husband was group O and the baby was blood group O. Use a genetic cross to find out if her claims were true. (5mks)
   2. The figure below is a diagram of a pair of homologous chromosomes during meiosis.
      
      
      1. Name the process shown above (1mk)
      2. Explain the effect of the process named in (b) (i) above on linked genes. (2mks)
3. The diagram below shows a vertical section through the part of a leaf of a mesophyte.
   
   
   1. Label cell A and organelle B (2marks)
   2. State two functions of the part labelled C. (2mks)
   3. Give two differences between the structure shown above and that of a floating hydrophyte. (2marks)
   4. Give two observable features that adapts the structure above to photosynthesis. (2 marks)
4. Below is a diagram of a mammalian nephron. Use it to answer the questions that follow.
   
   
   1. Name part 4. (1mark)
   2. Explain what happens to the concentration of sodium ions between 1 and 2. (2 marks)
   3.      
      1. Name the hormone that controls the amount of urine produced in the kidneys. (1 mark)
      2. How will the concentration of urine be affected at region 3 in the absence of the hormone mentioned in (c) (i) above. (2mks)
   4. What will happen at point 4 if there was partial constriction at point 5? (2marks)
5. The diagram below shows a food web. Study it and answer the questions that follow.
   
   
   1. Write two food chains with foxes as the quaternary consumer. (2mk)
   2. Name the organism with
      1. The highest biomass (1 mark)
      2. The highest number of predators (1 mark)
   3. State two possible effects on the ecosystem if kestrels migrated. (2 marks)
   4. Explain why primary productivity reduces with increase in depth in an aquatic ecosystem. (2marks)

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

6. During germination and growth of a cereal, the dry weight of endosperm, the embryo and total dry weight were determined at two-day intervals. The results are shown in the table below.
   

   Time after planting	Dry weight of endosperm	Dry weight of embryo (mg)	Total dry weight (mg)
   0	43	2	45
   2	40	2	42
   4	33	7	40
   6	20	17	37
   8	10	25	35
   10	6	33	39

   1. Using the same axes, draw graphs of dry weight of endosperm, embryo and the total dry weight against time. (8 marks)
   2. What is the total dry weight on day 5? (1mark)
   3. Account for:
      1. Decrease in dry weight for endosperm from day 0 to 10. (2 marks)
      2. Increase in dry weight of embryo from day 0 to 10. (2 marks)
      3. Decrease in total dry weight from day 0 to day 8. (2 marks)
      4. Increase in total dry weight after day 8 (1 mark)
   4. State two factors that causes seed dormancy in each of the following:
      1. Within a seed (2 mark)
      2. Outside a seed. (2 mark)
7.      
   1. A student sitting under a shade of a tree, on a sunny day, shifted the eyes from looking at an aero plane in the sky to reading a page on her book. Describe the changes that occurred in her eye. (15 marks)
   2. Explain how a neuron is adapted to its function. (5marks)
8.      
   1. Describe digestion of milk in the stomach (10 marks)
   2. Describe assimilation of the end products of digestion in mammals. (10 marks)

MARKING SCHEME

1.        
   1.      
      1. temperature
         humidity 
         wind 
         light intensity
         mark 1st two
      2. temperature- low temperature reduces latent heat of vaporization 
         humidity- high humidity decreases diffusion gradient
         wind- on a less windy day, water accumulates around the leaf reducing the diffusion gradient.
         Light intensity- low light intensity the stomata are partially open.
         (answer given in (ii) should match the answer given in (ii). part (i) and (ii) is tied)
   2. transpiration pull
      cohesion and adhesion force
      root pressure
      capillarity
      mark 1st two
   3.    
      1. lignin
      2. cellulose
2.    
   1. 
      
      Her claims were true. It was the wrong baby as they could only have a child with blood group A or B;
   2.      
      1. crossing over
      2. linked genes separate; and are transmitted on different chromosomes;
3.      
   1. A- epidermal cell rej epidermis
      B- chloroplast rej chloroplasts
   2. allows transpiration to take place
      allows gaseous exchange to occur.
   3. 
      

      Mesophyte leaf	Hydrophyte leaf
      Most stomata on the lower surface	Stomata on the upper surface
      Less number of stomata	More number of stomata
      No aerenchyma tissue	Has aerenchyma tissue

   4. presence of stomata to allow diffusion of carbon (IV) oxide into the leaf;
      palisade cells/spongy mesophyll have numerous chloroplasts that contain chlorophyll to trap light for photosynthesis;
      numerous palisade cells to increase the surface area for photosynthesis;
      spongy mesophyll cell irregularly arranged to create large airspaces to allow for circulation of carbon (IV) oxide;
      palisade cells closely packed for maximum trapping of sunlight;
      mark 1st two
4.        
   1. Glomerulus
   2. concentration reduces; because of active reabsorption of sodium ions at the ascending loop of henle;
   3.        
      1. Antidiuretic hormone; rej ADH acc Vasopressin
      2. the renal tubules will be less permeable; to water leading to less reabsorption of water; hence urine will become less concentrated/dilute;
         Max-2 mks
   4. more pressure (at the glomerulus); resulting in increased ultrafiltration;
5.      
   1. grass → herbivorous insects → carnivorous insects → toads and lizards → foxes
      grass → herbivorous insects → spiders → toads and lizards → foxes
      Mark any one
   2.      
      1. Grass
      2. carnivorous insects/herbivorous insects
   3. moles will increase in number
      stoats will increase in number
   4. light intensity decreases with depth; less light limits/reduces photosynthesis;
      temperature decreases with depth; lowering the rate of photosynthesis.
6.      
   1. scale -2 mks
      Plotting- 2 mks
      Axis- 1 mk
      Labelling 1 ½ mks
      Curve- 1 ½ mks
      
   2. 38.5 mg ± 0.5 (should be shown on the graph to score)
   3.         
      1. There is hydrolysis of starch into simple sugars which are transported to the embryo; hydrolysed food are oxidized to release energy for respiration; hence the decrease in dry weight.
      2. New materials are synthesized from protein; bringing about growth of embryo;
      3. The rate of respiration is faster; than that of synthesis of new materials for growth;
      4. First leaf has been formed and they carry out photosynthesis leading to growth;
   4.         
      1. Embryo not fully developed/ immature embryo
         Presence of abscisic acid/ germination inhibitors
         Low concentration of hormones that stimulate germination
         Low concentration of enzymes 
         Impermeable seed coat
         any one
      2. unfavourable temperature
         Absence of light/ particular wavelength
         Lack of water
         Lack of oxygen
         Any one
7.      
   1. when looking at the aeroplane in the sky, the ciliary muscles relax; this causes the suspensory ligaments to become taut; the lens becomes thinner; light rays from the object are less refracted; and focused on the retina; the radial muscles of the iris relax; circular muscles contract; the pupil becomes smaller; less light enters the eye;
      When going back to read a page of her book, the ciliary muscles contract; suspensory ligaments become loose; the lens becomes thicker; light rays from the object are greatly refracted; and focused on the retina; the radial muscles contract; circular muscles relax; the pupil becomes bigger; more light enters the eye;
      Total 17 mks max 15 mks
   2. Has a cell body that has a nucleus; to control the activities of the neurone;
      Has a myelin sheath that is made up of fatty materials; to insulate the axon;
      Axon is long/extended; to transmit nerve impulse;
      Has dendrites that are extensions on the cell body; to transmit nerve impulse to adjacent neurone;
      Has schwann cells that are secretory cells; that secrete myelin sheath;
      Has node of ranvier; that speed up transmission of nerve impulse/amplify the impulse;
      max 5 mks
8.      
   1. Presence of milk in the stomach stimulates secretion of hormone gastrin; Gastrin stimulates production of gastric juice; by gastric glands; the stomach wall muscles contract and relax; mixing the milk with the gastric juice; Gastric juice contains hydrochloric acid; pepsinogen; and prorennin; hydrochloric acid activates prorennin to rennin; and pepsinogen to pepsin; rennin coagulates milk protein/converts milk protein caseinogen to casein; enzyme pepsin breaks down casein to peptides; gastric juice also contains mucus; to lubricate the food;
      Student should describe enzyme activation before digestion,
      Conversion of milk protein before breakdown
      Total 14 mks    max 10mks
   2. Glucose; oxidized in cells to release energy; excess is converted to glycogen; for storage in the liver/ muscle cells; excess glucose is also converted to fats for storage;
      Amino acids; used in synthesis of proteins; for growth; repair of worn out tissue; some of the proteins form enzymes/ hormones; it is broken down and oxidized during periods of starvation to release energy; in the absence of glucose and lipids.
      Fatty acids and glycerol; oxidized to release energy in the absence of glucose; combined to form neutral fats; which are stored underneath the skin to insulate the body against cold;
      Total  15 mks max 10 mks