Biology Paper 2 Questions and Answers - Form 3 Mid Term 1 Exams 2021

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SECTION A (60MKS)

Answer all the questions in this section

  1. The graph below shows the effect of injecting one unit of insulin into a person. The concentration of glucose in the blood is measured at regular intervals.
             graph showing the effect of injecting insulin
    1. Give the lowest value of blood glucose observed and the time it was recorded. (1mk)
    2. Explain the fall in blood glucose level.(2mks)
    3. Name the mechanism that led to the increase in blood glucose level when it had been falling. (1mk)
    4. Name the hormone responsible for the conversion of glycogen to glucose. (½mk)
    5. State the effect of each of the following in human beings.
      1. Too much glucose in the blood (1mk)
      2. Very little glucose in the blood. (1mk)
  2. The diagram below shows a smear of blood on a microscope slide.
             smear of blood on a microscope slide
    1. Identify the structures labeled A, B and C. (1½mks)
    2. State the importance of the large number of structures A in the blood smear. (1mk)
    3. Name the process by which structure D would engulf C and state its importance.(1½mks)
    4. State one adaptation of the structure labeled A to its function.(1mk)
  3. The flow chart below shows a food web in a terrestrial ecosystem.
             flow chart of a food web
    1. From the food web, construct a food chain with five organisms.(1mk)
    2. Name the trophic level occupied by:
      1. Hawks(½mk)
      2. Cane toads(½mk)
    3. What would happen if leopards were introduced into the ecosystem? (2mks)
  4. Describe the processes that occur in the chest cavity during inspiration. (6mks)
  5. The graph below represents the effect of temperature on the rate of photosynthesis.
            graph on effects of temperature on the rate of photosynthesis
    1. On the diagram, label the axes. (1mk)
    2. Comment on the general trend of the graph. (2mks)
    3. List two other factors that may affect the shape of the graph. (2mks)
  6. The graph below shows the relationship between number of herbivores and carnivores in a park.
    graph showing herbivores and carnivores relationship
    1. Identify the curve representing the herbivores. Give a reason for your answer. (1½mks)
    2. Suggest a reason for the slope of graph x between points A and B. (2mks)
    3.  
      1. Name the relationship between the two types of organisms as portrayed by the graph.(1mk)
      2. State the significance of the relationship you have stated in (c)(i) above. (1mk)
    4. Describe the long term effect on the parks ecosystem if the species of the carnivores were to become extinct. (2mks)
  7. Use the diagram of the leaves below to construct dichotomous keys. Identify the steps you followed to identify leaves O, P, Q, R and S  (12mks)
    diagram of leaves. for dichotomous key
  8. In an experiment, germinating pea seeds were put in a retort flask which was placed in a beaker containing potassium hydroxide solution as shown in diagram A below. At the end of the experiment, the results were as shown in duiagram B.
    diagram on germinating pea seeds
    1. Suggest the aim of the experiment. (1mk)
    2. State the observable chnages that occurred as shown in the diagram. (2mks)
    3. Account for the changes noted in (b) above. (3mks)
    4. Name the chemical process taking place in the peas. (1mk)
    5. How would a control experiment be set. (1mk)
  9. A group of students set up an experiment to investigate a certain physiological process as shown in the figure below. After some time, the students observed that the level of the sugar solution had risen.
    experiment on physiological processes
    1. What physiological process was being investigated?  (1mk)
    2. Account for the rise in the sugar solution in the experiment. (2mks)
    3. Suggest with a reason the results that the students would obtain if they repeated the experiment using a piece of boiled potato. (1mk)
    4. Explain why the cells of the potato above, would not burst when immersed in distilled water and left for some time. (2mks)

SECTION B (20MKS)- Compulsory.

  1. Leaves were collected from the plant of a certain species growing in a shaded site and a plant from the same species growing in an open site. The surface area of each leaf was worked out. The results obtained are shown in the table below.
                 Surface area of leaves (cm3)  
     Shaded site  Open site 
     21
     14
     16
     18
     19
     21
     19
     22
     18
     16
     13
     22
     21
     23
     19
     18
     15
     15
     17
     18
     17
     17
     19
     13
     14
     21
     13
     16
     13
     16
     12
     14
     22
     20
     Mean surface area = x1   Mean surface area = x2
     
    1. Calculate the mean score x1 and x2 (2mks)
    2. Suggest one reason for the differences in the mean surface areas between the leaves from the two sites. Explain your answer. (2mks)
    3. Briefly state the adaptations of plant leaves to a desert habitat. (6mks)
    4. The leaves of a plant exposed directly to sunlight are often thicker than leaves found in the shade. Suggest two reasons for this observation. (2mks)
    5. How does the observation in (d) improve the efficiency of leaves exposed to direct sunlight? (2mks)
    6. Apart from photosynthesis, state two other functions of  a leaf. (2mks)
    7. State how a leaf is adapted for the functions you have stated in (f) above (3mks)
    8. Some plants have rolled leaves. Explain the importance of such leaves to the plant. (1mk)

SECTION C (2OMKS)
Select and answer only one questions in this section.

  1.  
    1. Explain how gills of  a fish are adapted to the process of gaseous exchange. (5mks)
    2. Describe the mechanism of gaseous exchange in the gills of a bony fish. (15mks)
  2. Explain how the mammalian skin is adapted to perform its functions. (20mks)

Marking Scheme

  1.  
    1.  
      • 60mg/100cm3;(½mk)
      • time when observed –after 30 minutes(½mk)
    2. Insulin stimulated liver cells to convert some glucose to glucogen ;while some of the glucose was oxidized to provide energy; (2mks)
    3. Negative feedback mechanism;(1mk)
    4. Glucagon;(½mk)
    5.  
      1. overproduction of energy by tissues which can burn them;(1mk)
      2. reduction in production of energy in the tissues which may kill the tissues;(1mk)
  2.  
    1. A- red blood cell; Rej.red cells.
      B –White blood cell;Rej.white cells.
      C- an antigen /bacterium /virus/fungus;(@½ mk =1½mks)
    2. To increase the total surface area of oxygen transport;(1mk)
    3. Phagocytosis;(½mk)
      • Helps to destroy disease causing micro organisms;(1mk)
    4.  
      • Presence of hemoglobin with a high affinity for oxygen;
      • Absence of a nucleus /lacks most of the organelles to pack more hemoglobin;
      • Presence of enzyme carbonic anhydase to enhance loading of carbon (iv) oxide gas;
      • They are small and able to squeeze through narrow capillaries; (max 1=1mk)
  3.  
    1. Green plants → grasshoppers → lizards → snakes
                                                                       ↓
                                                                    Hawks
      Green plants → Mice → Cane toads → snakes
                                                                 ↓
                                                              Hawks
    2.  
      1. Quartenary consumer;(½mk)
      2. Secondary consumer ;(½mk)
    3.  
      • The leopards would compete with lions for antelope
      • the antelopes would reduce in number;
      • more grass would be available for mice and grasshoppers
      • Leopards and lions would later decrease in number due to migration or death due starvation;(any 2=2mks)
  4. ribs move upwards and outwards;
    1. External intercostals muscles contract; and the internal intercostal muscles relax.
    2. The diaphragm flattens;
    3. Volume of the chest cavity falls as pressure rises; hence air rushes into the lungs; each 1mk=6mks
  5.  
    1.  
      • x-temperature(½mk)
      • y-rate of photosynthesis(½mk)
    2. rate of photosynthesis increases as temperature increases ;upto the optimum temperature;and starts to fall; due to denaturation of enzymes by the high temperature @½=2mks
    3.  
      • Light intensity;
      • conc.of carbon(iv) oxide;
      • amount of chlorophyll in the leaves;(max 2 = 2mks)
  6.  
    1. Y ;(½mk)
      • it starts with a higher population than x,the predator;(1mk)
    2. the number /pop. of predators dropped due to increased competition for food ;due to decrease in number of prey;(max 2=2mks)
    3.  
      1. Prey –predator/predation;(1mk)
      2. helps to maintain the population of both prey and predators at the carrying capacity of habitat; weak organisms are removed from both populations ;(by natural selection) (Accept one=1mk)
    4.  
      • Pop. of herbivores would increase due to reduced predation;
      • Most of the vegetarian would be eaten by the herbivores resulting in soil erosion;
      • Eventually the pop. of herbivores would fall due to death or migration due to competition for food.(any 2=2mks)
  7.  
    1.  
      1. Leaf simple……. Go to 2;(1mk)
      2. Leaf compound……. Go to 3(1mk)
    2.  
      1. Leaf with a serrated margin …………………. S;(1mk)
      2. Leaf with a smooth margin….. go to 4(1mk)
    3.  
      1. Leaf with three leaflets…………………P(1mk)
      2. Leaf with more than three leaflets……………Q ;(1mk)
    4.  
      1. Leaf with a pointed apex………………………… R(1mk)
      2. Leaf with a rounded apex……………………..R(1mk)
         Leaf    Steps 
         O
         P
         Q
         R
         1a,2b,4a;(1mk) 
         1b,3a;(1mk)
         1b,3b;(1mk)
         1a,2b,4b;(1mk)
  8.  
    1. To show that germination seeds use oxygen and give out carbon(iv) oxide;(1mk)
    2. The level of potassium hydroxide rose in the retort flask; and dropped in the beaker;(2mks)
    3. The germinating seeds respire using oxygen and give out carbon (iv) oxide;which is absorbed by potassium hydroxide; The level of potassium hydroxide rise to fill the space formally occupied by oxygen carbon(iv) oxide;(max 3=3mks)
    4. Germination
    5. Use boiled and disinfected peas/dry seeds ;(1mk)
  9.  
    1. Osmosis;(1mk)
    2. Water is hypotonic compared to sugar solution (OWTTE) hence water is drained from the beaker by osmosis into the sugar solution making it rise;(2mks)
    3. No obserrable change /no rise in sugar solution as boiling kills the cells /cells as osmotically dead/protoplasm is killed by boiling ;(1mk)
    4. They have rigid cellulose cell walls; that resist outward pressure even after filling with water;(2mks)
  10.  
    1. X1- sum of all surface areas = 18.53cm2;1/2 mk
                             17
      X2Sum of all surface areas =15.71cm2;1/2mk
                             17
    2. the amount of light the leaves receive;1mk
      • leave in shaded areas have larger surface areas to trap light of low intensity;1/2mk
      • Leaves in open sites have smaller surface areas as light is of high intensity ;(max2=2mks)
    3.  
      • Some have their stomata hidden in grooves;
      • some have leaves modified to thorns;
      • some have hairy leaves;
      • All have reduced no. of stomata;
      • Some have reversed stomatal mechanism;
      • Some have succulent leaves to store water;
      • Some have thick cuticles  (max 6=6mks)
        NB/All these adaptations help the plant to reduce water loss by transpiration.
    4. Leaves exposed to light have more palisade cells hence more chloroplasts and are thicker than those growing in shade;(2mks)
    5. Plants exposed to light have higher rates of photosynthesis;-Due to the greater concentration of chloroplasts; (max 2mks)
    6.  
      • gaseous exchange;
      • Transpiration; (@1mk =2mks)
    7.  
      • Presence of stomata;
      • Thinness;
      • Air spaces;
      • Large surface areas (except for xerophytes) any 2=2mks)
    8. Rolling of leaves reduces the surface area of transpiration ;(1mk)
  11.  
    1.  
      • Rich blood supply to transport gases;
      • Thin membrane for faster diffusion of gases;
      • Presence of a gill arches to support more gill filaments;
      • Presence of gill rakers to trap solid particles in water.
      • Very many gill filaments to increase the surface area for gaseous exchange ;(@ 1mk =5mks)
    2. The mouth opens and the floor of the mouth is lowered ;hence volume of the buccal cavity increases and pressure falls; this causes water to rush into the mouth;
      opercula bulge outwards raising the volume; and a fall in the pressure of the buccal cavity.
      Water is forced to flow over the gills at high pressure; This is enabled by the higher external pressure; which presses both opercula initially before opening the mouth and the opercula.;
      The water entering the mouth has more oxygen; and low in carbon (iv) oxide; which ceates a steep diffusion gradient;
      Oxygen diffuses from the water to the blood in the capillaries and combines with hemoglobin;
      Carbon (iv) oxide diffuses from the blood capillaries into the water;(max 15=15mks)
  12. The granular layer contains living cells that give rise to new epidermis /whose cell replace the cornfield layer.
    The Malpighian layer secrets melanin; which protects the body against harmful ultra violet rays;
    Sebaceous glands produce sebum; which keeps the skin soft/is water proof /is mildly antiseptic;
    It has blood vessels /arterioles and capillaries; which dilate /vasodilation;to help blood lose heat at hot times; or constrict/vasoconstriction;to prevent heat loss when its cold.
    It has erector –pili muscles which contracts when it is cold making hair to stand upright; trapped air helps to insulate the body against heat loss;
    When it is hot, the erector –pili muscles relaxes and hairs lie flat; hence no air is trapped and excess heat is lost;
    Sweat glands produce sweat; the sweat on the skin surface evaporate hence cooling the body;
    The subcutaneous fat/adipose tissue; serves as an insulator to prevent heat loss when its cold;
    Nerve endings and sensory receptors;
    Perceive any changes around the body; hence detects any dangers ;(max 20mks)

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