Name...Index No /2 BIOLOGY PAPER 2 (THEORY) March/April 2 HOURS THE LAINAKU JOINT ASSESSMENT 2011
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1 Name.....Index No.... Candidate s signature. Dates /2 BIOLOGY PAPER 2 (THEORY) March/April 2 HOURS THE LAINAKU JOINT ASSESSMENT 2011 INSTRUCTIONS TO CANDITATES: a) Write your name and index number in the spaces provide above. b) Sign and write the dates of examination in the spaces provided above. c) This paper consist of TWO sections; A and B. d) Answer all questions in section A in the spaces provided. e) In section B answer question 6(compulsory) and either question 7 or 8 in the spaces provided after the question. f) This paper consist of 11 printed pages g) Candidates should check the question paper to ascertain that all the pages are printed as indicated and that no questions are missing. FOR EXAMINERS USE ONLY. Section Question Maximum score Candidates score A
2 B Total score Study the flow chart below of a process that takes place in both plants and animals. Glucose Pyruvic acid Enzyme controlled chemical Reaction X in Product Z Enzyme controlled reaction Organelle Y of cell in absence of Oxygen in both plants and animals. (a) What process is represented by the flow chart? (b) (i) In the above process, name the chemical reaction represented by X (ii) Name the part of the cell where the enzyme controlled reactions in (b) (i) above take place (c) Name the product Z in (i) Plants (ii) Animals (d) What would be the fate of Pyruvic acid if Oxygen supply is availed in the mitochondrion of an animal cell? (2marks) 2
3 (e) What is meant by the term Oxygen debt? 2. The diagram below shows circulatory system in insects. Use it to answer the questions that follow. (a) Name the type of circulatory system shown above. (b) Identify parts M, S, T M - S - T - (3marks) (c) Using lines and arrows draw on the diagram above, the direction of blood flow between M, T and S. (d) State two main differences between blood in insects and mammals. Insects Mammals (e) What is the disadvantage of the circulatory system illustrated above to the insect? 3
4 3. The diagram below represents energy flow within an ecosystem. Use it to answer the questions that follow. Heat lost to environment 2,000KJ 1,500KJ 320KJ Producers A Primary 500KJ Secondary B Consumers Consumers Tertiary Consumer s 1,500KJ 1,000KJ 100KJ C (a) Define the term ecosystem X (b) Name the process through which: (i) Producers convert sun s energy into chemical energy. (ii) Living organisms convert chemical energy into heat energy lost to the environment. (c) Identify organisms x. (d) Determine the amount of energy represented by: (i) A (show your working) (ii) B (show your working). 4
5 (e) What is the total amount of energy present within the producers? Show your working. (f) If 75% of the energy in the tertiary consumers is lost as heat, calculate the amount of energy represented by C. 4. The data below represents concentrations of Ions in the cellsap of a plant and of the water in the pond in which it grew. Na+ K+ Mg 2+ Ca 2+ Cl - SO 4 2- Cell sap of cell Pond of water (a) Name the process involved in the uptake of Ions from the pondwater to cellsap. (b) What effect would the following changes have on this process? (i) Reduced Oxygen supply (ii) Increased concentration of glucose. (c) Other than uptake of mineral ions, state Two roles played by this process in living organisms. (d) How is the rate of diffusion affected by:- (i) Temperature 5
6 (ii) Concentration (iii) Surface Area to volume ratio 5. Other than determining a person s blood group as A, B, AB, or O, the rhesus status of the blood has to be established too. Presence of rhesus antigen in blood makes a person s blood to be rhesus positive while its absence makes a person s blood rhesus negative. Rhesus factor is inherited genetically, whereby rhesus positive (Rh + ) is dorminant while (Rh - ) is recessive. The pedigree below shows inheritance of Rhesus factor in the family of certain former U.S. Senator. Use it to answer the questions that follow Key Rh + lady Rh - lady Rh + man Rh - man (a) Given that 1, 2 is the senator and his wife, and 3,4,5, are their children and 8,9,10,11, their grandchildren, using punnet square work out the genetic cross between persons 6 and 3. (Use Rh + as symbol for dorminant allele and Rh - for recessive allele) (4mks) 6
7 (b) Offspring 10 and offspring 11 are the first and second born respectively of marriage between 5 and 7. Offspring 11 however died immediately after birth. Explain why? (c) Explain how test cross is done. SECTION B Compulsory Question 6. An experiment was carried out to investigate the effects of hormones on growth of lateral buds of three plants. The shoots were treated as follows: Shoot A - Apical bud removed Shoot B - Apical bud removed and gibberellic acid placed on the cut shoot. Shoot C - Apical bud left intact. The lengths of branches developing from lateral buds were determined at regular intervals and results obtained were as shown in the table below. Time in days Length of branches in millimeters Shoot A Shoot B Shoot C
8 (a) Using the same axis draw graph to show length of branches against time. (8mks) (b) What was the length of the branch in shoot B on the 7 th day? (c) Account for the results obtained in the experiment for each of the three shoots. Shoot A: (6mks) 8
9 Shoot B: Shoot C: (d) Why was shoot C included in the experiment? (e) State two ways in which gibberellic acid is important in Agriculture. (f) State two physiological functions or processes that are brought about by the application of gibberellic acid on plants. Answer either Qn 7 or 8 Not both 7. How is the male reproductive system adapted to perform its functions? (20mks) 8. Discuss the various evidences of organic evolution. (20mks) 9
KIRINYAGA CENTRAL DISTRICT JOINT EXAMINATION
NAME INDEX NO... SCHOOL... CANDIDATE S SIGNATURE... DATE.. 231/1 BIOLOGY PAPER 1 (THEORY) JULY/AUGUST, 2013 TIME: 2 HOURS KIRINYAGA CENTRAL DISTRICT JOINT EXAMINATION - 2013 Kenya Certificate of Secondary
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NAME INDEX NO... SCHOOL... CANDIDATE S SIGNATURE... DATE.. 231/1 BIOLOGY PAPER 1 (THEORY) JULY/AUGUST, 2013 TIME: 2 HOURS KIRINYAGA CENTRAL DISTRICT JOINT EXAMINATION - 2013 Kenya Certificate of Secondary
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NAME:.. DATE:... ADM NO:......... CANDIDATE S SIGNATURE:.. 231/2 BIOLOGY PAPER 2 (THEORY) TIME: 2 HOURS TERM TWO FORM THREE INSTRUCTIONS TO CANDIDATES 1. Write your name and Admission number in the spaces
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