INSTRUCTIONS TO CANDIDATES
- Write your name and Index Number in the spaces provided above.
- This paper consists of two sections. Section A and section B.
- Answer ALL 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
- This paper consists of 8 Printed pages. Candidates should check the question paper to ensure that all the papers are printed as indicated and no questions are missing.
SECTION A
- In a certain plant species which is normally green, a recessive gene for colour (n) causes the plant to be white when present in a homozygous state. Such plants die at early age. In heterozygous state, the plants are pale green in colour but grow to maturity.
- Suggest a reason for the early death of plants with homozygous recessive gene. (2 marks)
…………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………….. - If a normal green plant was crossed with a pale green plant, what would be the genotype of the F1 generation? (Show your working) (3 marks)
…………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………. - If seeds from the heterozygous plants were planted and the resulting plants allowed to self pollinate. Workout the phenotypic ratio of the plants that would grow to maturity. (2 marks)
……………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………… - Give an explanation for occurrence of the pale green colour in heterozygous plants. (1 mark)
……………………………………………………………………………………………………………………………………………………………………………………………………………………
- Suggest a reason for the early death of plants with homozygous recessive gene. (2 marks)
- Study the diagram below and answer the questions that follow.
- Name the tissue where the cells drawn above are found. (1 mark)
………………………………………………………………………………………………… - Identify cells A and B. (2 marks)
A………………………………………………………………………………………………………..
B……………………………………………………………………………………………………….. - Give two structural differences between cell A and cell B. (2 marks)
………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………… - Describe how structure C opens as explained by the photosynthetic theory. (3 marks)
………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………
- Name the tissue where the cells drawn above are found. (1 mark)
- Catalase is an enzyme present in all living tissues in both plants and animals. It breaks down toxic hydrogen peroxide produced during cellular metabolism into less toxic water and oxygen is evidenced by effervescence.
In an experiment 10 ml of hydrogen peroxide was put in different boiling tubes into which different specimens were put. The table below summarizes part of the results. Carefully analyze the table and answer the questions that follow.
The specimen Observation A Fresh liver A lot of bubbling almost violent B Boiled liver No bubbling C Fresh muscle tissue Vigorous bubbling less than tube A D Dry bean seed Very slow bubbling E Soaked bean seed Vigorous bubbling done intensity of tube C F 1 cm3 potato cube Moderate bubbling G 1 cm3 mashed potato Vigorous bubbling since intensity as in tube E - Compare & account for the rate of bubbling between
- Tube A and tube B. (2 marks)
……………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………… - Tube A and C (2 marks) ……………………………………………………………………………………………………………………………………………………………………………………………………………………………….……………………………………………………………………………………………………………….
- Tube D and tube E (2 marks)
……………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………… - Tube F and G (1 mark)
………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………
- Tube A and tube B. (2 marks)
- Write the equation for the reaction that produces the bubbling. (1 mark)
…………………………………………………………………………………………………………
- Compare & account for the rate of bubbling between
- The diagram below shows an instrument used in the laboratory.
- Name the apparatus shown above
…………………………………………………………………………… (1 mark) - Name the parts labeled Q , K and R (3 marks)
Q……………………………………………………
K……………………………………………………
R ………………………………………………….. - What are the functions of parts P, N and S. (3 marks)
P……………………………………………………………………………………………………
N…………………………………………………………………………………………………..
S……………………………………………………………………………………………………. - What is the formula of calculating linear magnification (1 mark)
………………………………………………………………………………………………………………………………………………………………………………………………………………
- Name the apparatus shown above
- Diagram below represents a germinating seedling.
- What is germination? (1 mark)
………………………………………………………………………………………………………………………………………………………………………………………………………………….. - Name the part labelled P, Q and R. (3 marks)
P………………………………………………………….
Q………………………………………………………….
R…………………………………………………………. - Identify the type of germination shown in the diagram. (1 mark)
………………………………………………………………………………………………………... - What is the role of the following in germination of the above seedling?
- Oxygen (1 mark)
………………………………………………………………………………………………………… - Enzymes (1 mark)
………………………………………………………………………………………………………… - Water (1 mark)
…………………………………………………………………………………………………………………………………………………………………………………………………………………..
- Oxygen (1 mark)
- What is germination? (1 mark)
SECTION B
Answer question 6 and either 7 or 8
- Some students used a model to demonstrate the effect of sweating on human body temperature. Two boiling tubes A and B were filled with hot water. The surface of tube A was continually wiped with a piece of cotton wool soaked in methylated spirit. The temperature of water in the tubes was taken at the start of the experiment and then at 5 minutes interval. The results obtained are as shown in the table below.
Time(in minutes) Temperature (ºC) in tubes A B 0 80 80 5 54 67 10 40 59 15 29 52 20 21 47 25 18 46 - On the same axis plot graphs of temperature of water in the tubes against time. (7 marks)
- At what rate was the water cooling in tube A? (2 marks)
…………………………………………………………………………………………………………………………………………………………………………………………………………………… - Why was tube B included in the set up? (1 mark)
…………………………………………………………………………………………………………………………………………………………………………………………………………………… - Account for the rate of cooling in tube A (3 marks)
…………………………………………………………………………………………………………………………………………………………………………………………………………………… - State two processes of heat loss in tube B. (2 marks)
…………………………………………………………………………………………………………………………………………………………………………………………………………………… - What would be the expected results if tube B was insulated? (1 mark)
……………………………………………………………………………………………………….. - What would the insulation be compare to in
- Birds ? (1 mark)
………………………………………………………………………………………………………… - Mammals? (1 mark)
………………………………………………………………………………………………………..
- Birds ? (1 mark)
- Name the structures in the human body that detect
- External temperature changes (1 mark)
…………………………………………………………………………………………………….. - Internal temperature changes (1 mark)
………………………………………………………………………………………………………..
- External temperature changes (1 mark)
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- Differentiate between nervous system and endocrine system. (5 marks)
- Describe how hormones regulate the menstrual cycle in human being. (15 marks)
- How is the mammalian intestine adapted to its functions? (20 marks)
MARKING SCHEME
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- Homozygous recessive plants do not have chlorophyll/cannot photosynthesize;
- Parental phenotype Normal green Pale green
Plants plants
Parental genotypes NN x Nn ;
Punnet’s Square
Parental genotype NN x Nn;
♀ ♂ N N; N NN NN; n Nn Nn - Parental genotype Nn x Nn
3/4 x 100 = 75% ; Grow to maturity - Due to incomplete dominance of the gene for normal colour;
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- Epidermal tissue
- Cell A – epidermal cell
Cell B – Guard cell
Cell A Cell B Lack chloroplasts Contain chloroplasts Rectangular shaped Bean shaped No thickening on either walls Thickened on inner walls - During the day, guard cells (cell B) carry out photosynthesis forming glucose. Glucose formed increases osmotic pressure of guard cells making them to draw water from adjacent cells of epidermis by osmosis
Guard cells swell and expand leading to opening of stoma (c)
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- Tube A and B – liver cells have very high metabolic rates hence higher concentration of enzyme catalase.
- The resultant gas is oxygen.✓1mk
- However, boiling denatures the enzyme catalase in the liver cells hence no action on hydrogen peroxide.✓1mk
- Tube A & C –✓1mk liver cells have higher metabolic rate than muscle cells hence higher concentration of enxyme catalase.✓1mk
- Tube D & E - cells of bean seeds have lower metabolic rates than soaked beans hence a lower concentration of enzymes catalase ✓1mk
Dry conditions inactivate enzymes in the cells of dry bean seeds ✓1mk - F & G – crushing increases S.A of contact between enzymes catalase in a potato cells and hydrogen peroxide hence faster reaction than in the potato cubes.✓1mk
- Tube A and B – liver cells have very high metabolic rates hence higher concentration of enzyme catalase.
- Hydrogen —Catatlase→ peroxide Water + Oxygen✓1mk
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-
- Light microscope
- Q – objective lenses, Rj. Lense
K – Eye piece lens;
R – stage; - P- condenser – concentrate light from the source and directs it to the specimen
N – fine adjustment knob;
-Raises or lowers the body tube through small distance to bring the image into sharp focus;
S – Base
For support
Mg = (length of the drawing)
length of the actual object)(1mark)
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- Development of seed into a seedling
- P – Coleoptile
Q – Plumule
R – Root hair Rej. Root hairs - Hypogeal germination
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- Used during respiration to provide energy required for cell division and growth.
- Converts insoluble food forms into soluble form
Plays role in breakdown and oxidation of food substances - Activates enzymes
Softens the seed coat
Hydrolyses and dissolves food materials
Provides medium for enzymes to act
Medium of transport of soluble food to growing parts
- 80-18=62=2.48℃/minute✓2mks
25 25 - Control experiment✓1mk
- The rate was faster in tube A because it’s surface was being wiped with cotton wool having methylated spirit.✓1mk The methylated spirit absorbed latent heat of vaporization from tube releasing it to the air, ✓1mk therefore encourages the cooling of the tube content.✓1mk
- Conduction✓1mk
Radiation✓1mk - would have low rate of heat loss✓1mk
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- Feathers✓1mk
- Fur✓1mk
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- Receptors in the skin✓1mk
- Hypothalamus✓1mk
- 80-18=62=2.48℃/minute✓2mks
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Nervous system Hormonal system Nerve impulse evoke response Chemical substance/ hormone evoke a response; Impulse transmitted through nerves Chemical substance /hormone transmitted through blood; Response immediate Response slow; Effects rapid and short-lived Effects long-lasting; Location of the response confined to effectors Location of the response wide-spread; - Immediately after menstruation; the pituitary gland is stimulated to secrete follicle stimulating hormone (FSH); which stimulates the ovary leading to start of development of a new Graafian follicle; its high concentration stimulates follicle cells of the ovary; to secrete oestogen hormone; Oestrogen hormone brings about healing and repair of uterine wall; and its high concentration stimulates pituitary gland; to secrete Luteinising hormone (LH); LH brings about ovulation; causes the reorganization of remnants of Graafian follicle into corpus luteum; and stimulates this corpus luteum to secrete progesterone hormone; high concentration of LH inhibits secretion of oestrogen; Progesterone hormone leads to proliferation of uterine in readiness for implantation; When conception occurs, the progesterone hormone maintains pregnancy up to birth; and inhibits production of FSH; If no conception, progesterone level falls drastically leading to menstruation; [max 15 marks]
- Mammal intestines are long to allow food more time for digestion and absorption.
- They are coiled to increase surface area for digestion and absorption and have villi and microvilli to increase the surface area for absorption.
- The walls have glands that secrete enzymes for digestion.
- They have goblet cells that secrete mucus which protect the intestinal wall from being digested and reduce friction as food moves down.
- The alimentary canal has a circular and longitudinal muscles contraction and relaxation leads to peristalsis. This facilitates faster digestion.
- The intestine is well supplied with blood vessels to supply oxygen and remove digested food material hence maintain high diffusion gradient.
- It has a narrow lumen to bring food into close contact with the walls to enhance absorption.
- It has a thin epithelium to facilitate faster diffusion of digestion food material.
- Mammalian intestine has lacteals in villi for absorption and transportation of lipids.
- At the stomach, there are pyloric sphincter and cardiac sphincter muscles which prevent food from moving out of the stomach during churning. (1*2=20MKS)
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