Biology 9201 | Paper 2 Mock 01 | Written Paper 2
Question 1 Report
A marine reserve shows a coral reef that has recently been invaded by lionfish. Fig. 1 is a line-art view of a diver recording fish above the coral. Lionfish eat many small reef fish, including young parrotfish. Parrotfish feed on algae growing on coral surfaces.
Table 1 compares counts made at six reef sites inside the reserve. Each count covered 100 m2 of reef and was completed at the same water temperature.
| Site type | Mean lionfish count | Mean parrotfish count | Algae covering coral / % |
|---|---|---|---|
| Sites with lionfish removed | 1 | 34 | 12 |
| Sites with many lionfish | 12 | 9 | 47 |
(a) Describe the difference in parrotfish count between the two site types. [2]
(b) Use the information to suggest why algae cover is greater at sites with many lionfish. [2]
(c) Suggest one reason why the researchers counted fish at several sites. [2]
(d) Name one way divers could help control the invasive lionfish population. [2]
Labelled answer diagram:
(a) There were fewer parrotfish at sites with many lionfish: 9 compared with 34 at sites where lionfish were removed. This is a difference of \(34-9=25\) fish. [2]
(b) Lionfish eat parrotfish. With fewer parrotfish grazing on algae, less algae is eaten and algae cover increases. [2]
(c) Counting at several sites reduces the effect of chance variation between individual sites. It allows a mean to be calculated and improves reliability. [2]
(d) Divers could catch lionfish using nets or spears and remove them from the reef. Preventing release of pet lionfish would also help control the invasive population. [2]
Answer Details
Labelled answer diagram:
(a) There were fewer parrotfish at sites with many lionfish: 9 compared with 34 at sites where lionfish were removed. This is a difference of \(34-9=25\) fish. [2]
(b) Lionfish eat parrotfish. With fewer parrotfish grazing on algae, less algae is eaten and algae cover increases. [2]
(c) Counting at several sites reduces the effect of chance variation between individual sites. It allows a mean to be calculated and improves reliability. [2]
(d) Divers could catch lionfish using nets or spears and remove them from the reef. Preventing release of pet lionfish would also help control the invasive population. [2]
Question 2 Report
A brain scan shows the result of a small stroke in a 68-year-old patient. The blockage reduced blood flow to area P. Doctors asked the patient to squeeze each hand, speak a short sentence and identify objects placed in the left hand. Fig. 1 is a simplified side view of the brain. The cerebrum receives sensory information and controls many voluntary actions. The brain stem controls automatic functions such as breathing. The patient’s blood glucose and oxygen concentration were normal when the tests were carried out.
(a) Which major region of the brain contains area P? [2]
(b) Suggest why a blockage in a blood vessel can damage brain cells. [2]
(c) Describe one test the doctors could use to check voluntary control of the patient’s left hand. [2]
(d) What is the advantage of using a brain scan as well as observing the patient’s symptoms? [2]
(a) Area P is in the cerebrum [1]. Cerebral cortex is also acceptable [1].
(b) A blockage reduces the blood supply, so brain cells receive less oxygen or glucose [1]. Aerobic respiration releases less energy, so cells cannot function normally and may die [1].
(c) Ask the patient to squeeze a hand, lift fingers or pick up an object [1]. Compare movement, strength or accuracy with the right hand [1]. This gives a direct comparison of voluntary control on the two sides.
(d) A brain scan locates the damaged area [1]. This allows symptoms to be linked to a particular brain region, helping diagnosis and treatment [1].
Answer Details
(a) Area P is in the cerebrum [1]. Cerebral cortex is also acceptable [1].
(b) A blockage reduces the blood supply, so brain cells receive less oxygen or glucose [1]. Aerobic respiration releases less energy, so cells cannot function normally and may die [1].
(c) Ask the patient to squeeze a hand, lift fingers or pick up an object [1]. Compare movement, strength or accuracy with the right hand [1]. This gives a direct comparison of voluntary control on the two sides.
(d) A brain scan locates the damaged area [1]. This allows symptoms to be linked to a particular brain region, helping diagnosis and treatment [1].
Question 3 Report
A technician investigated a mineral deficiency in bean plants. Fig. 1 shows two leaves, one from each treatment. Table 1 gives mean leaf area and the colour observation after 20 days.
| treatment | mean leaf area / cm2 | leaf colour |
|---|---|---|
| complete minerals | 34 | green |
| no magnesium | 18 | yellow |
(a) Which treatment produced yellow leaves? [1]
(b) Use Table 1 to calculate the difference in mean leaf area. [1]
(c) Suggest why a lack of magnesium reduces photosynthesis. [1]
(d) Describe how reduced photosynthesis can reduce plant growth. [1]
Fig. 1 shows an aquatic snail in a sealed jar with a water plant. A student used an indicator that changes colour when carbon dioxide is present. Table 1 shows the colour after 20 minutes.
| jar condition | indicator colour |
|---|---|
| lamp on | purple |
| lamp off | yellow |
(a) Name the process in the snail that releases carbon dioxide. [1]
(b) Use Table 1 to state which jar had the lower carbon dioxide concentration. [1]
(c) Describe why the colour in the lamp-on jar changed. [2]
Mineral deficiency
- (a) The treatment with no magnesium produced yellow leaves. [1]
- (b) \[34\text{ cm}^2-18\text{ cm}^2=16\text{ cm}^2\] The difference is 16 cm2. [1]
- (c) Magnesium is needed to make chlorophyll. A lack of magnesium reduces chlorophyll production and so reduces light absorption for photosynthesis. [1]
- (d) Reduced photosynthesis produces less glucose or carbohydrate, leaving less material or energy available for growth. [1]
Plant and snail jar
- (a) The snail releases carbon dioxide by respiration. [1]
- (b) The lamp-on, purple jar had the lower carbon dioxide concentration. [1]
- (c) With the lamp on, plant cells photosynthesised and removed carbon dioxide from the water or air in the jar. [2]
Answer Details
Mineral deficiency
- (a) The treatment with no magnesium produced yellow leaves. [1]
- (b) \[34\text{ cm}^2-18\text{ cm}^2=16\text{ cm}^2\] The difference is 16 cm2. [1]
- (c) Magnesium is needed to make chlorophyll. A lack of magnesium reduces chlorophyll production and so reduces light absorption for photosynthesis. [1]
- (d) Reduced photosynthesis produces less glucose or carbohydrate, leaving less material or energy available for growth. [1]
Plant and snail jar
- (a) The snail releases carbon dioxide by respiration. [1]
- (b) The lamp-on, purple jar had the lower carbon dioxide concentration. [1]
- (c) With the lamp on, plant cells photosynthesised and removed carbon dioxide from the water or air in the jar. [2]
Question 4 Report
A zoo keeps a group of parrots in a large aviary. Fig. 1 shows two enrichment devices added for 20 days: a puzzle box containing seeds and branches that can be stripped of bark. Keepers record the time spent feather-plucking before and after enrichment.
(a) Describe how the puzzle box may change the parrots' daily behaviour. [2]
(b) Suggest why reduced feather-plucking may indicate improved welfare. [2]
(c) Use the information to suggest one measurement the keepers should make before adding the devices. [1]
A trainer uses a clicker while teaching a rescue dog to touch a target with its nose. Fig. 1 shows the sequence used in each training session. Food is given immediately after the click.
(a) Name the type of learning shown in Fig. 1. [1]
(b) Describe how the food reward changes the likelihood of the dog touching the target again. [2]
(c) Suggest why the food should be given immediately after the click. [2]
Parrots: (a) The puzzle box encourages more time searching for and manipulating food [1], leaving less time for inactive or repetitive behaviour [1].
(b) Feather-plucking can be a sign of stress or boredom [1]. A lower rate suggests the birds are less stressed or more stimulated, indicating improved welfare [1].
(c) Measure time spent feather-plucking before enrichment, providing a baseline rate for comparison [1].
Dog: (a) This is operant conditioning [1].
(b) Food is a reward, or positive reinforcement [1], so the dog is more likely to repeat target-touching [1].
(c) Immediate food helps the dog associate the click and target-touching behaviour with reward [1]. With a delay, it may associate food with a different behaviour [1].
Answer Details
Parrots: (a) The puzzle box encourages more time searching for and manipulating food [1], leaving less time for inactive or repetitive behaviour [1].
(b) Feather-plucking can be a sign of stress or boredom [1]. A lower rate suggests the birds are less stressed or more stimulated, indicating improved welfare [1].
(c) Measure time spent feather-plucking before enrichment, providing a baseline rate for comparison [1].
Dog: (a) This is operant conditioning [1].
(b) Food is a reward, or positive reinforcement [1], so the dog is more likely to repeat target-touching [1].
(c) Immediate food helps the dog associate the click and target-touching behaviour with reward [1]. With a delay, it may associate food with a different behaviour [1].
Question 5 Report
Fig. 1 shows a woodland food web close to a public footpath. Oak trees produce acorns. Wood mice eat acorns, tawny owls eat mice, and foxes can eat both mice and owls. In autumn, volunteers counted acorns in sample squares and recorded wood mouse droppings as an estimate of mouse activity. They want to know whether a year with poor acorn production could affect predators. The woodland also contains decomposers that return mineral ions to the soil for plants.
| Year | Mean acorns per m2 | Mean mouse droppings per m2 |
|---|---|---|
| 1 | 74 | 31 |
| 2 | 28 | 12 |
| 3 | 65 | 27 |
(a) Use Fig. 1 to state one organism that eats wood mice. [1]
(b) Describe the relationship between acorn number and mouse-dropping number in Table 1. [2]
(c) Suggest why a low acorn year could reduce tawny owl breeding success. [2]
Fig. 1 shows a seal pup on a rocky shore. Its mother feeds at sea and transfers energy to the pup in milk. The pup has a thick layer of fat under its skin. Researchers measured the pup's mass every five days without separating it from its mother for long. They also recorded air temperature. During one period, storms prevented the mother from feeding for two days. The pup remained on land and used stored food molecules in respiration.
The pup's mass rose from 18.0 kg to 21.5 kg while its mother fed regularly. After the storms, its mass fell by 0.8 kg. The researchers did not conclude that all mass change was fat because water content can also change.
(a) What substance in milk can provide energy for respiration? [1]
(b) Describe the mass change while the mother fed regularly. [2]
(c) Suggest why the pup's mass fell during the storm period. [2]
Woodland food web
(a) A tawny owl eats wood mice [1]. A fox is also acceptable.
(b) More acorns are associated with more mouse droppings [1]. Year 2 has the lowest values for both measures, with 28 acorns per m2 and 12 droppings per m2 [1].
(c) A low acorn year may lead to fewer mice [1]. Tawny owls then have less food and energy for producing eggs and rearing young, reducing breeding success [1].
Seal pup
(a) Fat can provide energy for respiration [1]. Sugar or protein in milk are also acceptable.
(b) While the mother fed regularly, the pup's mass increased from 18.0 kg to 21.5 kg [1]. This is an increase of \[21.5-18.0=3.5\text{ kg}\] [1].
(c) Storms prevented the mother feeding, so the pup received less or no milk [1]. It used stored fat or other food molecules in respiration for energy, causing its mass to fall [1].
Answer Details
Woodland food web
(a) A tawny owl eats wood mice [1]. A fox is also acceptable.
(b) More acorns are associated with more mouse droppings [1]. Year 2 has the lowest values for both measures, with 28 acorns per m2 and 12 droppings per m2 [1].
(c) A low acorn year may lead to fewer mice [1]. Tawny owls then have less food and energy for producing eggs and rearing young, reducing breeding success [1].
Seal pup
(a) Fat can provide energy for respiration [1]. Sugar or protein in milk are also acceptable.
(b) While the mother fed regularly, the pup's mass increased from 18.0 kg to 21.5 kg [1]. This is an increase of \[21.5-18.0=3.5\text{ kg}\] [1].
(c) Storms prevented the mother feeding, so the pup received less or no milk [1]. It used stored fat or other food molecules in respiration for energy, causing its mass to fall [1].
Question 6 Report
A coastal restoration team shows a young mangrove plant growing where sea water enters a river estuary. The plant absorbs water through roots in mud, but the salty water makes uptake difficult. Some mangroves lose salt through special leaf glands. Fig. 1 shows a leaf with a salt gland on its upper surface and a stoma on its lower surface. The team measured the mass of salt crystals collected from leaves after 24 hours. They also recorded the mean number of open stomata at 10:00 and at 14:00. These measurements help the team judge whether the plants can conserve water while obtaining carbon dioxide.
| time | mean open stomata per mm2 | salt crystal mass per leaf / mg |
|---|---|---|
| 10:00 | 64 | 3.1 |
| 14:00 | 29 | 5.8 |
(a) Describe the change in mean number of open stomata between 10:00 and 14:00. [2]
(b) Suggest why fewer stomata are open at 14:00. [2]
(c) Use Fig. 1 and Table 1 to suggest how the leaf helps the mangrove survive in salty conditions. [3]
(d) Name the process by which water enters root hair cells when the root cell has a lower water potential than the mud water. [1]
(e) Which dissolved substance is transported in phloem from a leaf to a root? [1]
- (a) The mean number of open stomata decreases from \(64\) to \(29\) per mm2, a decrease of \(35\) per mm2. [2]
- (b) Conditions at 14:00 are likely to be hotter or drier. Closing stomata reduces water loss by transpiration. [2]
- (c) Salt glands remove salt from the leaf. The rise in salt crystal mass from \(3.1\) mg to \(5.8\) mg supports this. Stomata can close to limit water loss, helping the mangrove conserve water in salty conditions. [3]
- (d) Water enters root hair cells by osmosis. [1]
- (e) The dissolved substance transported in phloem from leaf to root is sucrose. [1]
Answer Details
- (a) The mean number of open stomata decreases from \(64\) to \(29\) per mm2, a decrease of \(35\) per mm2. [2]
- (b) Conditions at 14:00 are likely to be hotter or drier. Closing stomata reduces water loss by transpiration. [2]
- (c) Salt glands remove salt from the leaf. The rise in salt crystal mass from \(3.1\) mg to \(5.8\) mg supports this. Stomata can close to limit water loss, helping the mangrove conserve water in salty conditions. [3]
- (d) Water enters root hair cells by osmosis. [1]
- (e) The dissolved substance transported in phloem from leaf to root is sucrose. [1]
Question 7 Report
This diagram is used in a clinic to show inheritance of cystic fibrosis. The condition is caused by recessive allele f. Both prospective parents have had a DNA test and are known to be carriers. Fig. 1 shows the genetic cross prepared by the nurse.
(a) Use Fig. 1 to state the probability that one child will have cystic fibrosis. [1]
(b) Which genotype gives a child who is a carrier but does not have cystic fibrosis? [1]
(c) Describe why the result for one pregnancy does not change the probability for the next pregnancy. [3]
A sports medicine team investigates an inherited condition that affects collagen in tendons. Fig. 1 shows normal collagen fibres and fibres from a person with a mutation. The person had repeated tendon injuries during exercise, despite following the same training plan as teammates.
(a) Describe one visible difference between the two sets of fibres in Fig. 1. [1]
(b) Suggest how a mutation in a collagen gene can increase the chance of tendon injury. [2]
(c) Use the term protein to describe why the gene change affects the tendon. [2]
Cystic fibrosis inheritance
- (a) One of the four outcomes is \(ff\), so the probability of cystic fibrosis is \(\frac{1}{4}\), or 25%. [1]
- (b) \(Ff\) gives a carrier who does not have cystic fibrosis. [1]
- (c) Each pregnancy is independent because gametes are made independently each time. Each parent can again pass on either \(F\) or \(f\), and fertilisation is random. Previous offspring do not change the parents' genotypes or the probability for a new pregnancy. [3]
Collagen mutation
- (a) The mutated fibres are less regularly aligned and include an altered pattern compared with the regular normal fibres. [1]
- (b) Altered collagen fibres have reduced strength or elasticity. The tendon is therefore more likely to tear when a force acts on it. [2]
- (c) A gene contains instructions for making a protein. A change in the DNA can produce collagen protein with a changed structure, affecting the tendon. [2]
Answer Details
Cystic fibrosis inheritance
- (a) One of the four outcomes is \(ff\), so the probability of cystic fibrosis is \(\frac{1}{4}\), or 25%. [1]
- (b) \(Ff\) gives a carrier who does not have cystic fibrosis. [1]
- (c) Each pregnancy is independent because gametes are made independently each time. Each parent can again pass on either \(F\) or \(f\), and fertilisation is random. Previous offspring do not change the parents' genotypes or the probability for a new pregnancy. [3]
Collagen mutation
- (a) The mutated fibres are less regularly aligned and include an altered pattern compared with the regular normal fibres. [1]
- (b) Altered collagen fibres have reduced strength or elasticity. The tendon is therefore more likely to tear when a force acts on it. [2]
- (c) A gene contains instructions for making a protein. A change in the DNA can produce collagen protein with a changed structure, affecting the tendon. [2]
Question 8 Report
Table 1 shows results from an investigation into the effect of standing still on blood flow in the lower leg. A Doppler ultrasound probe was used to estimate blood velocity in a vein. Each volunteer was tested while sitting, after standing still for 20 minutes and after walking for five minutes. The investigation was carried out in a room at 20 degrees C. The researchers were interested in how leg muscle contractions and vein valves help return blood to the heart. Long periods of immobility can increase the risk of a blood clot forming in a leg vein.
| activity | mean blood velocity in leg vein / cm per second |
|---|---|
| sitting | 8.2 |
| standing still for 20 minutes | 4.1 |
| walking for 5 minutes | 13.6 |
(a) Which activity produced the lowest mean blood velocity? [1]
(b) Describe the effect of walking on blood velocity compared with standing still. [1]
(c) Suggest why walking increases blood flow in leg veins. [2]
A nurse recorded the pulse rates of a patient receiving treatment for dehydration after a stomach illness. The patient had lost water and salts, reducing the volume of plasma in the blood. Table 1 shows measurements taken before and after the patient was given an oral rehydration drink. Fig. 1 shows a simplified pressure cuff used to measure blood pressure. The cuff is wrapped around the upper arm and inflated. The nurse also checked that the patient was producing urine, because this indicates that the kidneys are receiving enough blood.
| measurement | before drink | one hour after drink |
|---|---|---|
| pulse rate / beats per minute | 112 | 86 |
| systolic pressure / mmHg | 96 | 112 |
(a) What happened to pulse rate after the drink? [1]
(b) Suggest why dehydration can lower blood pressure. [1]
(c) Describe why the rehydration drink can improve circulation. [2]
(a) The lowest mean blood velocity occurred after standing still for 20 minutes. [1]
(b) Walking increased mean blood velocity from 4.1 to 13.6 cm per second. [1]
(c) When walking, leg muscles contract and squeeze the veins. Vein valves prevent backflow, so blood is moved towards the heart. [2]
(a) Pulse rate decreased from 112 to 86 beats per minute after the drink. [1]
(b) Dehydration reduces the amount of water in plasma, lowering total blood volume and therefore blood pressure. [1]
(c) Water from the rehydration drink is absorbed into the blood. This increases blood volume and pressure, allowing blood to reach body cells more effectively. [2]
Answer Details
(a) The lowest mean blood velocity occurred after standing still for 20 minutes. [1]
(b) Walking increased mean blood velocity from 4.1 to 13.6 cm per second. [1]
(c) When walking, leg muscles contract and squeeze the veins. Vein valves prevent backflow, so blood is moved towards the heart. [2]
(a) Pulse rate decreased from 112 to 86 beats per minute after the drink. [1]
(b) Dehydration reduces the amount of water in plasma, lowering total blood volume and therefore blood pressure. [1]
(c) Water from the rehydration drink is absorbed into the blood. This increases blood volume and pressure, allowing blood to reach body cells more effectively. [2]
Question 9 Report
Fig. 1 shows a forensic scientist examining fly eggs and maggots on a dead rabbit found in grassland. The scientist recorded the air temperature and the stage of the flies to estimate when the rabbit died.
| day | mean air temperature / degrees C | observation |
|---|---|---|
| 1 | 20 | eggs present |
| 3 | 21 | small maggots present |
| 6 | 19 | large maggots present |
(a) Name the feeding relationship between the maggots and the rabbit tissue. [1]
(b) Describe how the fly observations changed from day 1 to day 6. [2]
(c) Suggest why temperature is recorded when estimating the time of death. [1]
The diagram shows a peat bog with moss plants at the surface and waterlogged peat below. A conservation team measured the depth of new peat formed in one year. The bog water is acidic and contains little oxygen.
(a) Name the plant process that removes carbon dioxide from the atmosphere. [1]
(b) Suggest why dead moss decays slowly below the water level. [2]
(c) What is the effect on atmospheric carbon dioxide when peat is drained and burned? [1]
(a) The maggots show detritivory, meaning feeding on dead organic material. [1]
(b) Eggs were present first, then small maggots were observed, followed by large maggots by day 6. [2]
(c) Temperature is recorded because it affects the rate of insect development and decay, so it affects estimates of time since death. [1]
(a) Moss plants remove carbon dioxide from the atmosphere by photosynthesis. [1]
(b) Waterlogged peat has little oxygen. Aerobic decomposers cannot respire effectively in these conditions, so decay is slow. [2]
(c) If peat is drained and burned, atmospheric carbon dioxide increases. [1]
Answer Details
(a) The maggots show detritivory, meaning feeding on dead organic material. [1]
(b) Eggs were present first, then small maggots were observed, followed by large maggots by day 6. [2]
(c) Temperature is recorded because it affects the rate of insect development and decay, so it affects estimates of time since death. [1]
(a) Moss plants remove carbon dioxide from the atmosphere by photosynthesis. [1]
(b) Waterlogged peat has little oxygen. Aerobic decomposers cannot respire effectively in these conditions, so decay is slow. [2]
(c) If peat is drained and burned, atmospheric carbon dioxide increases. [1]
Question 10 Report
The figure shows a chromatogram made from spinach leaf extract. A student used a solvent to separate pigments before investigating which colours of light gave the greatest oxygen output.
(a) Describe what the separated bands show about spinach leaf extract. [2]
(b) Suggest why a plant has pigments other than chlorophyll. [1]
(c) Use your knowledge of photosynthesis to suggest why green light often gives a lower oxygen output than red light. [2]
(d) Name the gas used by chloroplasts to make glucose. [1]
(e) Suggest one safety precaution when using an organic solvent. [1]
- (a) The separate bands show that spinach leaf extract contains more than one pigment. The pigments moved different distances because they have different solubilities in the solvent or different attractions to the paper. [2]
- (b) Other pigments allow the plant to absorb additional wavelengths of light. [1]
- (c) Green light is mainly reflected or transmitted rather than absorbed by chlorophyll. Less light energy is therefore available for photosynthesis, so less oxygen is produced. [2]
- (d) The gas used by chloroplasts to make glucose is carbon dioxide. [1]
- (e) Keep the organic solvent away from flames. Goggles or good ventilation are also acceptable precautions. [1]
Answer Details
- (a) The separate bands show that spinach leaf extract contains more than one pigment. The pigments moved different distances because they have different solubilities in the solvent or different attractions to the paper. [2]
- (b) Other pigments allow the plant to absorb additional wavelengths of light. [1]
- (c) Green light is mainly reflected or transmitted rather than absorbed by chlorophyll. Less light energy is therefore available for photosynthesis, so less oxygen is produced. [2]
- (d) The gas used by chloroplasts to make glucose is carbon dioxide. [1]
- (e) Keep the organic solvent away from flames. Goggles or good ventilation are also acceptable precautions. [1]
Question 11 Report
Fig. 1 shows a young patient receiving a leaflet about cystic fibrosis screening. The condition is caused by a recessive allele. The patient's parents are both known carriers, but neither has the condition. The leaflet includes a Punnett square showing possible fertilisation events. Cells from the developing fetus could be tested for the allele, but the family is also told that genetic counselling is available before any decision is made. The condition affects mucus in the lungs, so oxygen uptake during exercise can be reduced in some people.
(a) Name the genotype of a carrier parent. Use F and f. [1]
(b) Use the Punnett square to calculate the probability that one child will have cystic fibrosis. [2]
(c) Suggest why a genetic counsellor should not tell the parents what decision to make. [1]
Labelled answer diagram:
(a) A carrier has genotype Ff [1]. The dominant allele means the person does not have cystic fibrosis, but the recessive allele can be passed on.
(b) Two carrier parents produce the possible genotypes FF, Ff, Ff and ff. One of the four possible genotypes is affected, ff [1]. Therefore the probability that one child will have cystic fibrosis is \(\frac14\), or 25% [1]. Each pregnancy is an independent event.
(c) The decision is personal and ethical, so a genetic counsellor should provide unbiased information rather than tell the parents what decision to make [1].
Answer Details
Labelled answer diagram:
(a) A carrier has genotype Ff [1]. The dominant allele means the person does not have cystic fibrosis, but the recessive allele can be passed on.
(b) Two carrier parents produce the possible genotypes FF, Ff, Ff and ff. One of the four possible genotypes is affected, ff [1]. Therefore the probability that one child will have cystic fibrosis is \(\frac14\), or 25% [1]. Each pregnancy is an independent event.
(c) The decision is personal and ethical, so a genetic counsellor should provide unbiased information rather than tell the parents what decision to make [1].
