Biology Paper 2 Questions and Answers - Maranda High School Mock Exams 2023

Instructions to Candidates

• This paper consists of two sections; A and B. Answer all the questions in section A in the spaces provided.
• In section B answer question 6 (compulsory) and either question 7 or 8 in the spaces provided after question 8

SECTION A (40 MARKS)

Answer all the questions in this section in the spaces provided

1.  
   1. A female with sickle cell trait marries normal man. The allele for sickle cell is Hbs and the normal allele is HbA. Determine the probability that their first born will have the sickle cell trait. Show your working (5 marks)
   2. State how inbreeding leads to reduced hybrid vigour (1 mark)
   3. Explain how the sex of a male child is determined in human beings (2 marks)
2. The diagram below represents a plant in a particular division. Use it to answer the question that follow.
   
   1.  
      1. Identify the class to which the plant belongs (1 mark)
      2. Give two observable characteristics that support your answer in 3(a) i) above. (2 marks)
   2. State three differences between members of Kingdom Monera and members of Kingdom Plantae (3 marks)
   3. Name the structures for locomotion in the following organisms (2 marks)
      1. Chlamydomonas
      2. Amoeba
3.  
   1. Explain why a bony fish dies shortly after being removed from water. (4 marks)
   2. How is the spiracle of an insect adapted to its function? (1 mark)
   3. State one limitation of photosynthetic theory in explaining the opening and closing of stomata.   (1 mark)
   4. List two adaptations of the aerenchyma tissue to gaseous exchange. (2 marks)
4. The diagram below represents a simple reflex arc in an animal
   
   1. Name neuron labelled A and B. (2 marks)
   2. Identify the gap labeled C and state its importance in the nervous system. (2 marks)
      Identity……………………………………………………………………………………………...
      Importance……………………………………………………………………………………………
   3. Name one example of a chemical substance that enables impulse to pass across gap C. (1 mark)
   4.  
      1. How is the aqueous humour adapted to its function? (2 marks)
      2. State one advantage of binocular vision. (1 mark)
5. In an investigation a student extracted three pieces of paw paw cylinders using a cork borer. The cylinders were cut back to 50mm length and placed in a beaker containing a solution. The results after 40 minutes were as shown in the table below.
   

   Feature	Result
   Average length of cylinders	44m
   Stiffness of cylinders	Flexible

   1. Account for the results in the table above (3 marks)
   2. What would be a suitable control set-up for the investigation (2 marks)
   3.  
      1. Define the term diffusion gradient (1 mark) 
      2. Differentiate between hypotonic solution and hypertonic solution (2 marks)

SECTION B: (40 MARKS)

Answer question 6(compulsory) and either question 7 or 8 in the space provided after question 8

6. A scientist carried out an investigation to find out the population growth of mice under laboratory conditions. Twenty young mice were placed in a cage. The results obtained from the investigation were as shown in the table below.
   

   Time in months	0	2	4	6	7	10	12	16	18
   Number of mice	20	20	65	115	310	455	450	145	160

   1. On the grid provided, draw a graph of the number of mice against time. (6marks)
   2. Account for the changes in mice population between:
      1. 0 to 2 months (2 marks)
      2. 2 to 6 months (2 marks)
      3. 6 to 10 months (2 marks)
      4. 10 to 12 months (2 marks)
   3.  
      1. Between which two months was the population change greatest ? (1mark)
      2. Calculate the population change over the period in (c) i) above. (2 marks)
   4. What change in population would be expected if the investigation was continued to the 19th month.? (1 mark)
   5. To obtain the observed results state two variables that were kept constant during the investigation. (2 marks)
7. Describe how various factors affect the rate of enzyme action (20 marks)
8.  
   1. How does excretion take place in plants? (4 marks)
   2. Describe the role of the human skin in Homeostasis (16 marks)

MARKING SCHEME

1.  
   1.  
      
      Probability - ½/ 0.5 / 50%
   2. Chances of recessive/ defective genes being combined increase hence weaker offspring;
   3. Male produces sperm cell with X or Y chromosomes; when a sperm containing Y chromosome fuses with an ovum a male child is formed;
2.  
   1.  
      1. Coniferales;
      2.  
         • Presence of cones;
         • Needlike leaves;
   2.  
      

      Kingdom Monera	Kingdom plantae
      cell wall made of microprotein;	Cell wall made of cellulose,
      Prokaryotic	Eukaryotic
      Heterotrphic	Autotrophic
      Cell lack some organelles e.g mitochonria	Cells have most cell organelles e.g mitochondria

   3.  
      1. Flagella; acc Flagellum
      2. Pseudopodia; acc pseudopodium
3.  
   1. Fish uses dissolved oxygen for gaseous exchange; gill filament epithelium dries up; gill filaments clump together; surface area for gaseous exchange is reduced; oxygen lacks moist surface for dissolution causing suffocation; (hence death)
   2. (Muscular) valve to control it's opening and closing;// Hairs to reduce water loss by evaporation;
   3. Does not take into account plants that undergo reversed stomatol rythm;
   4.  
      • Cell loosely packed creating large air spaces that store air for gaseous exchange;
      • Cells thin walled to reduce the diffusion distance;
4.  
   1.  
      • A - relay;
      • B - Motor;
   2.  
      • Identity : Synapse;
      • Importance: Allows transmission of an impulse from one neuron to another; // Ensures an impulse is transmitted in one direction 
   3. Acetylcholinie; Noradrenaline -
      • Fluid that exerts hydrostatic press; hence maintaining spherical shape of eyeball;
   4.  
      1.  
         • Contains oxygen & nutrients; that nourish the cornea & lens;
         • Transparent; to allow light to pass through;
      2.  
         • Wider field of view;
         • Damage to one eye is compensated for by the other eye;
5.  
   1.  
      • The solution was hypertonic to cell sap of cells in the potato cylinder; acc converse.
      • Cellls lost water by osmosis; shrunk and became flaccid;
   2.  
      • Pawpaw cylinder of the same size/ length; placed in an isotonic solution;
      • Boiled pawpaw cylinders off same size/ length; placed in a similar/ hypertonic solution;
   3.  
      1. Difference in the concentration of molecules between a region of high concentration and a region of low concentration;/ between two regions.
      2. Hypotonic solution is a solution with more water/solvent molecules than the adjacent soln when separated by a semipermeable membrane while hypertonic is a solution with more solute molecules than the adjacent soln when the two are separated by a semi-permeable membrane.
6.  
   1.  
      
   2.  
      1. Constant/ no change in population; mice still maturing/ have not given birth;
      2. Gradual/ slow population growth; Few mice have reached sexual maturity;
      3. Rapid/faster rate of population growth/ exponential many rats sexually matures/ reproducing/ availability of food/ space/ no competition/ higher birth rate than death rate/ no diseases;
      4. Slight population declines/ decreases; competition is high/ food is limiting/ space is limiting/ accumulation of toxic wastes/ diseases/ higher death rate than birth rate;
   3.  
      1. 6 and 8;
      2. 370(Aug) − 115(June) = 255 = 127.5 ≅ mice/month;
                        2                        2
         NB 370 ± 5
   4. Population would increase;
   5. Nutrients; Food; Space/ cage size; water; 
7.  
   1. Temperature
      • Increase in  temperature upto optimum range increases the rate of enzyme action; above optimum range the rate decreases; since enzymes are denatured; below optimum range the rate decreases; since enzymes become inactive;.
   2. pH
      • Some enzymes work best in acidic conditions while others work best in basic conditions; Extreme acidity/akalinity denatures enzymes hence reducing the rate of enzyme action;
   3. Specificity;
      • A particular enzyme will only act on a particular substrate;
   4. Substrate concentration; Enzyme concentration;
      • Increase in substrate concentration increases the rate of enzyme action upto a certain level; due to presence of enough active sites; Further increase in substrate concentration results to constant/ no increase in the rate of enzyme action; since all active sites of the enzymes have been occupied; 
   5. Co factors and Coenzymes;
      
      • They activate enzymes hence increasing the rate of enzyme action;
   6. Enzyme Inhibitors;
      • Competitive enzyme inhibitors compete with substrates for active sites of enzymes; hence slowing down the rate of enzyme action; Non competitive inhibitors combine permanently with enzyme molecules hence blocking the active sites; this prevents substrates from interracting with enzymes hence slowing down the rate of reaction;
8.  
   1.  
      • Diffusion of Carbon(IV) Oxide and oxygen through the stomata & lenticels;
      • Some wastes are stored/ deposited in plant tissues in non toxic forms; some of these tissues/ organs drop off; from plants i.e leaves, flowers, fruits e.g caffeine, nicotine.
      • Some wastes are released by transpiration through stomata & lenticels e.g water vapour;
      • Others are released by guttation through hydathodes e.g excess water;
      • Others are released through exudation;
   2.  
      • When body temperature is lowered below normal; arterioles in the skin constrict; blood is diverted to a shunt system; less blood flows to the skin hence less heat is lost; when body temperature is raised above normal; arterioles in the skin dilate; more blood flows to the skin; hence more heat is lost by convection & radiation;
      • When body temperature is lowered below normal; erector pili muscles contract, hairs stand erect; more air is trapped, air ia a bad conductor and insulates body against heat loss;
      • When body temperature is raised above normal; erector pili muscles relax, hairs lie on skin; less air; is trapped, more heat is lost;
      • When boby temperature is lowered below normal; less fluids are absorbed by sweat glands; less sweating, less vaporisation of water;
      • When body temperature is raised above normal; sweat glands are more stimulated and more sweat is produced; water in sweat evaporates and takes up heat from the body; body temperature is lowered;
      • Through sweating, excess mineral salts and excess water are gotten rid off from the body;