2022 Mock Examination Chemistry Paper 3 with Answers-Kapsabet High School
SECTION A (40 marks)
Answer all the questions in this section the spaces provided
- (a) The skin, respiratory surfaces, and alimentary canal are possible sites through which micro- organisms may gain entry to the human body.
For each of these sites, describe the mechanisms that prevent the entry of micro-organisms.
(a) The skin. (3 marks)
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(b) The respiratory system. (3 marks)
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- In an experiment to investigate the functioning of the mammalian kidney, samples were taken by micropipette from different regions.
The diagram below shows the sample sites, labeled 1 to 6.
Each sample was analyzed to determine the concentration of glucose, protein, urea and sodium ions.
The flow rate was also measured at each of the sample sites. The results are shown in the table below.
Samples sites within the kidney Concentration (g dm-3)
Flow rate
(cm3 min-1)
Protein Glucose Sodium
ions Urea
- Plasma in afferent arteriole 80 1.5 34 0.3 600.0
- Filtrate in Bowman’s capsule 0 1.2 34 0.3 125.0
- End of proximal convoluted tubule 0 0 34 1.6 25.0
- Bottom of loop of Henle 0 0 70 1.8 1.5
- Beginning of distal convoluted
tubule 0 0 30 1.8 1.5 - Beginning of collecting duct 0 0 2.2 2.2 1.3
Use the information in this table and your own understanding to answer the following questions.
(a) Explain the changes in the composition of proteins and glucose between the plasma in the afferent arteriole (sample site 1) and the end of the proximal convoluted tubule (sample site 3). (2 marks)
● Protein
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● Glucose
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(b) Comment on the changes in sodium ion concentration in the different sample regions. (2 marks)
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(c) Explain the changes in urea concentration as it moves along the nephron. (2 marks)
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(d) Suggest an explanation for the fall in the flow rate as fluid moves from the plasma into and then along the nephron. (1mark)
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(e) The experiment was carried out at 37 °C. When the experiment was repeated at 30 °C, the glucose concentration at the end of the proximal convoluted tubule was 0.15 g dm–3. Suggest an explanation for this result. (1mark) ………………………………………………………………………………………………………
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- (a) The bar charts show the percentages of a human population with each type of blood group and the percentages of a cattle population with and without horns.
Which type of variation is shown in each population? (1mark)
Human:
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Cattle:
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(b) Albinism (lack of skin pigmentation) in humans is caused by two recessive alleles.
A phenotypically normal (non-albino) couple have three children; the first two are non albino, the third is an albino.
In your answer, use “A” for the dominant allele and “a” for the recessive allele.
(i) What are the genotypes of the parents? (1mark)
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(ii) Is there a possibility that their next child will be an albino?
Explain your answer (2 marks)
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(iii) The albino child eventually marries a non-albino whose father was an albino.
What is the probability that their first child will be an albino? Show all working. (4marks)
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- A potometer is a device for investigating the rate of transpiration.
Prior to setting up, the potometer and the stem of a leafy shoot are immersed in water.
Under water, the bottom centimeter of the stem is cut off and the cut end inserted into the plastic tubing.
The apparatus is removed from the water, a bubble of air allowed to enter the open end of the capillary tube and that end then inserted into a beaker of water.
The completed set-up for a simple potometer is shown below.
(a) What assumption is made when this apparatus is used to investigate the rate of transpiration? (1mark) ………………………………………………………………………………………………………
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(b) Explain each of the following. (i) Why it is necessary to cut the leafy shoot and fit it into the photometer under water (1mark)
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(ii) How the bubble of air is introduced into the capillary tube. (1mark)
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(iii) Why a syringe is attached. (1mark)
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(iv) Why the set-up is left for 15 minutes before taking readings. (1mark)
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(c) The table below shows some results recorded using the apparatus.
Time
(minutes) Distance travelled by bubble (mm)
“Normal” room
conditions Covered with clear
plastic bag Covered with black
plastic bag
0 0 0 0
2 18 10 4
4 36 19 8
6 55 29 11
8 74 38 15
10 90 48 18(i) Account for the results shown in the table. (2 marks)
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(ii) In ‘normal’ room conditions, the distance moved by the bubble was 90 mm during 10 minutes.
The capillary tube has a cross sectional area of 0.8mm 2.
Calculate the rate of movement in mm3 minute–1. (Show your working in the space below.) (1mark) ………………………………………………………………………………………………………
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- (a) Describe how each of the following structures adapt a bony fish to locomotion in water.
(i) Scales. (2 marks)
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(ii) Myotomes (2 marks)
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(b) State two adaptations of the synovial joints in man. (4 marks)
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SECTION B (40 marks)
Answer question 6 (compulsory) and either question 7 or 8 in the spaces provided after question 8.
- It was suspected that a pollution incident involving slurry had occurred in a local river.
Oxygen content of the water in the river was measured, both upstream and downstream from the suspected slurry (raw sewage) leak.
Samples were taken at seven points along the river and the results are shown in the graph below.
Distance along the stream (m) 0 20 40 60 80 100 120
Oxygen concentration (arbitrary
units) 7.0 7.0 1.6 2.0 3.4 5.0 7.0
(a) Plot a graph of this data. (7 marks)
(b) From the graph determine: (i) the distance along the stream where the slurry leak occurred. (1mark) ………………………………………………………………………………………………………
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(ii) the least oxygen concentration and the distance when it occurred. (2 marks)
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(c) Account for the shape of the graph between:
(i) 20m – 40m along the stream. (3 marks)
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(ii) 60m – 120m along the stream. (3 marks)
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(d) Waterways can also be polluted by fertilizer run- off.
The effects of fertilizer run-off and pollution by slurry are different in some ways.
State and explain two of these differences. (3 marks)
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- (a) Describe the adaptations of the essential parts of entomophilous flowers to pollination. (6 marks)
(b) Using a named example, describe the events from pollination to double fertilization. (14 marks)
- (a) Describe how the mammalian eye is adapted for accommodation. (6 marks)
(b) Describe the mechanism of hearing in man. (14 marks)
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Marking Scheme
SECTION A (40 marks)
Answer all the questions in this section the spaces provided
- (a) The skin, respiratory surfaces, and alimentary canal are possible sites through which micro- organisms may gain entry to the human body.
For each of these sites, describe the mechanisms that prevent the entry of micro-organisms.
(a) The skin. (3 marks)
Secretion of sebum which is antiseptic;
Secretion of sweat which is antiseptic;
Cornified layer is made of dead cells which prevents microbial entry;
(b) The respiratory system. (3 marks)
Mucus on the lining of the trachea traps dust particles and microorganisms;
Hairs in the nostrils traps dust particles and microorganisms;
Sensory cells in the olfactory epithelium ensure only clean air is inhaled;
Macrophage cells in the alveoli engulf and digest pathogens;
(c) The alimentary canal. (2 marks)
Hydrochloric acid in the stomach kills any microorganisms swallowed with food;
Parts of the alimentary canal like the stomach have goblet cells that secrete a mucoid lining;
- In an experiment to investigate the functioning of the mammalian kidney, samples were taken by micropipette from different regions.
The diagram below shows the sample sites, labeled 1 to 6.
Each sample was analyzed to determine the concentration of glucose, protein, urea and sodium ions.
The flow rate was also measured at each of the sample sites. The results are shown in the table below.
Samples sites within the kidney Concentration (g dm-3)
Flow rate
(cm3 min-1)
Protein Glucose Sodium
ions Urea
- Plasma in afferent arteriole 80 1.5 34 0.3 600.0
- Filtrate in Bowman’s capsule 0 1.2 34 0.3 125.0
- End of proximal convoluted tubule 0 0 34 1.6 25.0
- Bottom of loop of Henle 0 0 70 1.8 1.5
- Beginning of distal convoluted
tubule 0 0 30 1.8 1.5 - Beginning of collecting duct 0 0 2.2 2.2 1.3
Use the information in this table and your own understanding to answer the following questions.
(a) Explain the changes in the composition of proteins and glucose between the plasma in the afferent arteriole (sample site 1) and the end of the proximal convoluted tubule (sample site 3). (2 marks)
● Protein
Basement membrane is the filter; most proteins are too big to pass through so through the tiny pores in the glomerular capillaries;
● Glucose
Glucose being a smaller molecule all passes through the membrane; but all the glucose is reabsorbed by active transport back into the bloodstream
(b) Comment on the changes in sodium ion concentration in the different sample regions. (2 marks)
Sodium ions is a similar proportion in both the filtrate and the plasma because all the Na+ in the plasma is filtered off.
Sodium ions concentration at the end of proximal convoluted tubule is same to that in plasma and filtrate in Bowman’s capsule because sodium ions and water are reabsorbed in proportional amounts through the proximal tubule.
Sodium ions concentration at bottom of loop of henle is very high because sodium ions are concentrated by the bottom of the loop of henle to provide an osmotic gradient through the medulla that ensures reabsorption of water by osmosis from the glomerular filtrate back to the body cells;
Concentration in the distal convoluted tubule and collecting duct decreases // is low because in the distal tubule and collecting tubule the sodium content of the plasma is adjusted (actively absorbed) (under hormonal control;
(c) Explain the changes in urea concentration as it moves along the nephron. (2 marks)
Absorption of water in the proximal tubule increases the urea concentration (fivefold/by 500%); later reabsorption of water further increases the urea concentration/ some urea is absorbed into the nephron;
(d) Suggest an explanation for the fall in the flow rate as fluid moves from the plasma into and then along the nephron. (1mark)
Only some of the fluid (about 20%) leaves the plasma thus reducing the volume and therefore the pressure is Bowman’s capsule is lower
Friction with the walls reduces the flow/hydrostatic pressure is reduced (as fluid moves through the nephron)
Flow rate falls as water is removed by osmosis;
(e) The experiment was carried out at 37 °C. When the experiment was repeated at 30 °C, the glucose concentration at the end of the proximal convoluted tubule was 0.15 g dm–3. Suggest an explanation for this result. (1mark)
Low temperatures reduce the rate of respiration/metabolic rate / reduces production of ATP and this causes a reduction in the rate of active transport/operation of carrier proteins – Rate of reabsorption is not quick enough for glucose to be fully removed from the filtrate;
- (a) The bar charts show the percentages of a human population with each type of blood group and the percentages of a cattle population with and without horns.
Which type of variation is shown in each population? (1mark)
Human:
Discontinuous;
Cattle:
Discontinuous;
(b) Albinism (lack of skin pigmentation) in humans is caused by two recessive alleles.
A phenotypically normal (non-albino) couple have three children; the first two are non albino, the third is an albino.
In your answer, use “A” for the dominant allele and “a” for the recessive allele.
(i) What are the genotypes of the parents? (1mark)
Aa
(ii) Is there a possibility that their next child will be an albino?
Explain your answer (2 marks)
Yes, the probability of having an albino at each birth is 25% / ¼ / 0.25;
(iii) The albino child eventually marries a non-albino whose father was an albino.
What is the probability that their first child will be an albino? Show all working. (4marks)
Crossing
- A potometer is a device for investigating the rate of transpiration.
Prior to setting up, the potometer and the stem of a leafy shoot are immersed in water.
Under water, the bottom centimeter of the stem is cut off and the cut end inserted into the plastic tubing.
The apparatus is removed from the water, a bubble of air allowed to enter the open end of the capillary tube and that end then inserted into a beaker of water.
The completed set-up for a simple potometer is shown below.
(a) What assumption is made when this apparatus is used to investigate the rate of transpiration? (1mark)
That the rate of transpiration is equal to the rate of water uptake (actually measured using the apparatus);
(b) Explain each of the following.
(i) Why it is necessary to cut the leafy shoot and fit it into the photometer under water (1mark)
To prevent air collecting in the xylem vessels/air locks preventing water uptake;
(ii) How the bubble of air is introduced into the capillary tube. (1mark)
The open end of the capillary tube is exposed to the air which is drawn up as the shoot takes up water
(iii) Why a syringe is attached. (1mark)
To enable the air bubble to be moved back to the origin;
(iv) Why the set-up is left for 15 minutes before taking readings. (1mark)
(iv) To allow the rate of transpiration to acclimatize to the surrounding conditions;
(c) The table below shows some results recorded using the apparatus.
Time
(minutes) Distance travelled by bubble (mm)
“Normal” room
conditions Covered with clear
plastic bag Covered with black
plastic bag
0 0 0 0
2 18 10 4
4 36 19 8
6 55 29 11
8 74 38 15
10 90 48 18(i) Account for the results shown in the table. (2 marks)
Transpiration is reduced when the plant is covered with a (clear) plastic bag since the air becomes more humid; and there are no air currents; humid air reduces the diffusion gradient of moisture out of the plant / no air movement allows diffusion shells to build up
Transpiration is further reduced when the plant is covered with a black plastic bag since the stomata close in the dark; thus the main route of water loss from the leaf is closed/only cuticular transpiration occurs;
(ii) In ‘normal’ room conditions, the distance moved by the bubble was 90 mm during 10 minutes.
The capillary tube has a cross sectional area of 0.8mm 2.
Calculate the rate of movement in mm3 minute–1. (Show your working in the space below.) (1mark)
90 × 0.8 = 72 mm3;
72 ÷ 10 = 7.2 mm3 min–1;
- (a) Describe how each of the following structures adapt a bony fish to locomotion in water.
(i) Scales. (2 marks)
Are overlapping; with free ends facing backwards; to reduce resistance during swimming in water; and prevent water from coming into contact with the skin;
(ii) Myotomes (2 marks)
) Contractile; to contact and relax; to allow for forward propulsion;
(b) State two adaptations of the synovial joints in man. (4 marks)
Have ligaments which are inelastic to hold adjacent bones at a joint;
Have articular cartilage which is hard yet soft; to absorb shock between adjacent bones;
Have synovial membrane which is secretory; to secrete synovial fluid; which lubricates the joint; and absorbs shock at the joint;
Have joint capsule, a casing that holds the joint together;
SECTION B (40 marks)
Answer question 6 (compulsory) and either question 7 or 8 in the spaces provided after question 8.
- It was suspected that a pollution incident involving slurry had occurred in a local river.
Oxygen content of the water in the river was measured, both upstream and downstream from the suspected slurry (raw sewage) leak.
Samples were taken at seven points along the river and the results are shown in the graph below.
Distance along the stream (m) 0 20 40 60 80 100 120
Oxygen concentration (arbitrary
units) 7.0 7.0 1.6 2.0 3.4 5.0 7.0
(a) Plot a graph of this data. (7 marks)
Marking:
Scale = 1mark
Labeling = 2 marks
Plotting = 2 marks
Curve = 2 marks)
(b) From the graph determine:
(i) the distance along the stream where the slurry leak occurred. (1mark)
50m + 2;
Reject without units
(ii) the least oxygen concentration and the distance when it occurred. (2 marks)
1.8 + 0.1units
Reject without units
(c) Account for the shape of the graph between:
(i) 20m – 40m along the stream. (3 marks)
Sharp decline in oxygen concentration; slurry leakage at 20m released raw sewage into the river; the slurry / raw sewage contained organic matter and microorganisms; the microbes like bacteria and protozoa broke down the organic matter utilizing dissolved oxygen in the water for respiration;
(ii) 60m – 120m along the stream. (3 marks)
Gradual rise in amount of dissolved oxygen until the original concentration before slurry leakage; amount of organic matter in the river decreases distance downstream; less oxygen is used in aerobic respiration; slurry breakdown releases more minerals ions like nitrates, phosphates etc in water; which lead to increased growth of water plants; the water plants carry out photosynthesis releasing oxygen in water;
(d) Waterways can also be polluted by fertilizer run- off.
The effects of fertilizer run-off and pollution by slurry are different in some ways.
State and explain two of these differences. (3 marks)
Run-off fertilizers increases the amount of mineral ions in water; leading to excessive of growth of water plants and algae //algal blooms; hence eutrophication (explanation of eutrophication);
Run –off fertilizers may alter pH of the water hence negatively affecting organisms in the water;
- (a) Describe the adaptations of the essential parts of entomophilous flowers to pollination.(6 marks)
They have small anthers, which are firmly attached to the filament; to resists breakage when insects make contact;
They have large, heavy, and rough pollen grains; to stick onto the body of the insects.
They have stigmas, which are small, sticky and occur inside the flower; so that pollen on insects’ bodies easily stick onto them;
Note: essential parts = parts used in reproduction
(b) Using a named example, describe the events from pollination to double fertilization. (14 marks)
Once the pollen grains land on the stigma, the stigma produces a sticky chemical substance; that provides a suitable medium for generation of the pollen tube from the pollen grains;
As the pollen tube grows down the style; it receives nutrients from the surrounding style tissues;
The tube nucleus takes its position just behind the tip of the pollen tube;
The generative nucleus divides mitotically; to form two male nuclei;
The pollen tube enters the ovule through the micropyle; and bursts open creating a clear way for the passage of the two male / generative nuclei;
Once the pollen tube bursts, the pollen tube nucleus degenerate;
One of the male (generative) nuclei fuses with the polar nuclei; to form a triploid primary endosperm nucleus;
The second male (generative) nucleus fuses with the egg cell nucleus; to form a diploid zygote;
- (a) Describe how the mammalian eye is adapted for accommodation. (6 marks)
The iris; is opaque and contractile; to control light intensity/ amount of light entering the eye (by controlling the size of the pupil);
Ciliary body; contains ciliary muscles; which are contractile; for controlling the curvature and hence focal length of the lens.
Suspensory ligaments; are fibrous; to hold the lens in position.
Have a transparent; biconvex lens; to allow light to go through; to refract the light; and to focus the light.
External eye muscles; is contractile; to move the eyeball (within the socket).
(b) Describe the mechanism of hearing in man. (14 marks)
When sound waves are trapped by the pinna, they are directed inwards into the auditory canal and then to the ear drum.
Wax in the auditory canal trap dust particles and microorganisms in air.
The sound waves cause the eardrum to vibrate with the same intensity as the sound waves; thereby converting the sound waves to vibrations.
The vibrating eardrum causes the ear oscicles (malleus, incus, and stapes) to vibrate inturn.
The oscicles amplify and transmits the vibrations to the oval window (fenestra ovalis).
The vibrations of the oval window are transmitted to the fluid (perilymph) in the cochlea i.e. within the vestibular canal of cochlea.
The perilymph transmits vibrations into the endolymph;
Displacement / movement of the endolymph causes the movement of the sensory hair cells within the organ of corti;
This movement of the hair generates impulses in the sensory hair cells.
The impulses are transmitted into the brain (cerebrum) via the auditory nerve.
The brain then translates the impulses to perception of sound.