Q1. In an ecosystem energy flows from the sun and is transferred in a series of organisms. The diagram represents different levels of energy levels of energy transfer.

Level D

Level C

Level B

Level A

(a) Insert the sun in the diagram and using arrows show the direction of energy transfer. (1mk)

(b) Name the trophic levels represented by D and B (2mks)

(c) Identify the general trend in the amount of energy along the path illustrated in (a) give reasons for the trend. (2mks)

(d)Explain short term effect of decreasing the number of individuals in the level C. (2mks)

Q2. In maize, yellow colour W is dominant over white colour. Describe how one would establish whether a given sample of yellow maize is pure or hybrid. Show your working. (4mks)

Q3. The sketch graph shows how the metabolic rate in man and lizard is affected by environmental temperature.

Suggest why the metabolic rate was high in man between 5^oc and 20^o (2mks)

Account for increase in metabolic rate in lizard as environmental temperature was increasing. (2mks)

(i) state two physiological processes that occur in man when environmental temperature rises above 35^o (2mks)

(ii) How would the lizard respond to similar temperature changes in (c) (1mk)

Q4. The apparatus shown was set up by a group of students.

What was the aim of the experiment? (1mk)

Explain the results obtained in the set up at the end of the experiment. (2mks)

State the expected results if the cellophane was replaced with a thin section of

(i) Raw arrowroot (1mk)

(ii) Boiled arrowroot (1mk)

(d) (i) Account for the results obtained in ( c) (ii). (1mk)

(ii) What is the equivalent of sucrose solution in plant tissue? (3mks)

Q5. The diagram represents a plant cell

Name a carbohydrate which forms the structure label S. (1mk)

(i) state the function of the part labeled R (2mks)

(ii) Suggest what would happen in the plant if the number of organelle labeled T is

reduced.

.(1mk)

Q6. The set of apparatus was assembled by a group of students to investigate some physiological process.

(i) give two aims of the experiment. (2mks)

(ii) Explain the observation expected after 24 hours. (2mks)

Before experiment, the glucose solution was boiled then cooled.

(i) Why was it necessary to boil the solution? (1mk)

(ii) What was the importance of oil layer in the experiment?(1mk)

Describe a control experiment for the set up. (1mk)

Suggest two industrial application of the process being investigated. (1mk)

SECTION B (40mks)

Answer question 7 (compulsory) and EITHER question 8 or 9 in the spaces provided after question 9

Q7 The table below represents body weight, metabolic rate and eaten per day by different mammals. (Compulsory)

Animal	Body weight	Metabolic rate (cm³) Oxygen/ghr	Food eaten per day (kg)
Rat	0.10	800	0.098
Hare	2.00	480	1.2
Dog	8.00	300	3.0
Man	60.00	150	4.0
Horse	200.0	100	15.0
Elephant	800.0	86	40.0

(i) Draw a graph of metabolic rate against body weight of the animals (6mks)

(ii) From the graph estimate the metabolic rate of an animal whose body weight is 35 kg (2mks)

Express the food eaten per day of the following mammals as a percentage of their body weight.

(i) Rat (2mks)

(ii) Elephant (2mks)

Account for the difference in percentage obtained in (b) (4mks)

Which of the six mammals would least suffer dehydration during a sunny day? Give reasons for your answers. (2mks)

A reptile would require less food than a mammal of the same weight. Explain. (2mks)

Q8. (a)Define the terms

(i)Transpiration (2mks)

(ii)Translocation (2mks)

(b)Explain five factors that affect the rate of transpiration in plants.(16mks)

Q9. (a) Define

(i) Chemical evolution (2mks)

(ii) Organic evolution (2mks)

(b) Describe the evidence of organic evolution (16mks)

BIOLOGY FORM 4 PAPER 2 MARKING SCHEME

Q1.

level D Level C Level B Level

A (1mk)

SUN

(All must be correct)

(b) B-secondary consumers; D-producers (2mks)

(c) Amount of energy decrease along the path (from D to A); some energy is lost through respiration unconverted materials and uneaten individuals. (2mks)

(d) Number of individuals in level B decrease leading to decrease at level A due to starvation/less energy available; number of individuals in level D would increase due to decrease of herbivores/grazers. (2mks)

Q2. (a)Back cross with white maize (ww)

Parent Pure breed X White

Genotype WW ww

Gametes W W w w

F1 genotype Ww Ww Ww Ww

NB all yellow NB:Gametes should be circled individually. (2mks)

Hybrid White

genotype Ww ww

Gametes W w w w

F1 genotype Ww Ww ww ww

2 yellow 2 white

Mixture of yellow and white (2mks)

(b) Male sex chromosome XY female XX.

Sperm

Ova

Girl

Boy

Girl

Boy

50% boys, 50% girls/equal chances (2mks)

Q3. (a) Environmental temperature lower than body temperature more heat was being generated to maintain body temperature. (2mks)

(b) The lizard is poikilothermic, enzymes were inactive at low temperature, their activity increased as temperature increased because enzymes were activated.(2mks)

(ii)Aestivation, migration to shade, burrowing, sand bathing, (any one correct)(1mk)

Q4. (a) To demonstrate osmosis in non -living tissue (1mk)

(b) Level of sucrose rose in thistle funnel, water drawn into funnel by osmosis,since sucrose solution was hypertonic to/more concentrated than distilled water. (2mks)

(ii) No observable change/sucrose level remained the same.(1mk)

(d) (i) Cell membrane in arrow root destroyed hence no osmosis (1mk)

(ii) Cell sap/dissolve sugars and salts in vacuole.(1mk)

Q5 (a) Cellulose (1mk)

(b) (i) Store sugar/salts/food; create osmotic gradient for osmosis; cause cell turgidity(2mks)

(ii) Rate of photosynthesis would reduce/inadequate food produced.(1mk)

( c ) Cell wall; chloroplast(2mks)

Q6.(a) (i) to show energy is released in anaerobic respiration; to show carbon( IV) ; oxide is produced in anaerobic respiration. (2mks)

(ii) Increase in temperature since energy is released colour of indicator; changed to yellow due to acidity/carbon. (IV)oxide released changed indicator to yellow. (1mk)

(b) (i) Expel dissolved oxygen; (1mk)

(ii) Prevent entry of air/oxygen into glucose solution.(1mk)

( c) use glucose solution without yeast/used killed yeast cells.(1mk)

(d) Ethanol production in breweries, breads production in bakeries.(1mk)

Q7.

(i) Labeling axis (2mks)

Scale (2mks)

Curve (1mk)

Plotting (1mk)

(ii) 205cm³ oxygen /ghr ^-1_–⁺_{5 evidence from gragh (2mkS)}

_{(b) (i) RAT food eaten X 100% 0.098 X100 =98% (2mks)}

Body weight 0.1

(ii)Elephant 40 X 100 =5% (2mks)

⁸⁰⁰

(C)Rat eat more food per unit body weight than elephant; Rat has larger surface area to volume ratio; therefore tend t0 lose more heat per unit body weight; oxidize more food/faster to maintain body temperature(4mks)

(d) Elephant: least surface area to volume ratio; lose water very slowly since some tissues are far from body surface; metabolic rate very slow; hence little water lost through sweating per unit body weight(2mks)

(e)Reptile poikilothermic/depend on environmental temperature; mammal homiothermic/oxidize food to maintain body temperature (2mks)

8(a) (i) Transpiration is the process by which plants lose water to the atmosphere in form of water vapour

(ii)Translocation is the process by which soluble products of photosynthesis/simple sugars are transported from leaves ;to other parts of the plants through phloem

(b)Temperature: high temperature increases water evaporation from mesophyll cell; increase capacity of atmosphere to hold more water due to faster movement of molecules/ low temperatures reduces the rate of water evaporation; air capacity to hold more water hence low transpiration rate

Humidity:high humidity lower the saturation deficit /reduce ability of atmosphere to hold more water;hence low transpiration rate/low humidity offering great saturat

ion deficit;hence high rate of transpiration

Light intensity:at low light intensity stomata close; reduce surface area over which water is lost/low transpiration rate/stomata open at high light intensity;surface area hence increases transpiration rate

Wind/air current:fast air movement /strong air current sweep away saturated air around plant;increasing transpiration rate ;still air/weak air currents make water accumulate around the plant lowering transpiration rate

Size of stomata/number of stomata/leaf area:large/many stomata/leaves/large area; increase transpiration rate/few leaves/small/few stomata reduce surface area; hence low transpiration rate

Atmospheric pressure: low atmospheric pressure increase water evaporation; hence high rate of transpiration/high atmospheric pressure reduce water evaporation; hence low rate of transpiration [3marks] award only once in each condition/low or high. {max 16marks}

Media Team @Educationnewshub.co.ke October 10, 2024

8 minutes read