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Question 1 Report
During a school rowing trial, a sports scientist records the pulse rate of a pupil who becomes breathless sooner than the rest of the team. The scientist uses Fig. 1 to discuss how the heart supplies the body cells with oxygen and glucose in the blood. The arrows show the direction of blood flow. The blood returning from the body has a lower oxygen concentration than blood leaving the lungs.
Fig. 1
(a) What is the name of the chamber labelled X? [1]
(b) What is the function of the structure labelled Y? [2]
(c) Describe the difference in oxygen concentration between the blood in vessel P and the blood in vessel Q. [2]
(d) Explain why a narrowed coronary artery can cause a person to stop exercising. [3]
(e) State two features of red blood cells that help them transport oxygen to body cells. [2]
(a) Chamber X is the left ventricle. [1]
(b) Structure Y is a valve. It prevents backflow of blood and ensures that blood flows in one direction through the heart. [2]
(c) Blood in vessel P has a lower oxygen concentration. Blood in vessel Q has a higher oxygen concentration because it has gained oxygen in the lungs. [2]
(d) A narrowed coronary artery means less blood reaches heart muscle. Less oxygen and glucose are supplied for aerobic respiration, so less energy is released for contraction. The person may become tired or breathless and have to stop exercising. [3]
(e) Red blood cells contain haemoglobin and have a biconcave shape that gives a large surface area. Other acceptable features are no nucleus, giving more space for haemoglobin, and flexibility to pass through narrow capillaries. Any two gain credit. [2]
(a) Chamber X is the left ventricle. [1]
(b) Structure Y is a valve. It prevents backflow of blood and ensures that blood flows in one direction through the heart. [2]
(c) Blood in vessel P has a lower oxygen concentration. Blood in vessel Q has a higher oxygen concentration because it has gained oxygen in the lungs. [2]
(d) A narrowed coronary artery means less blood reaches heart muscle. Less oxygen and glucose are supplied for aerobic respiration, so less energy is released for contraction. The person may become tired or breathless and have to stop exercising. [3]
(e) Red blood cells contain haemoglobin and have a biconcave shape that gives a large surface area. Other acceptable features are no nucleus, giving more space for haemoglobin, and flexibility to pass through narrow capillaries. Any two gain credit. [2]
Question 2 Report
The table below shows measurements from a microscope calibration activity. A student used a stage micrometer, with divisions of 0.01 mm, to calibrate an eyepiece graticule at two objective lenses. Fig. 1 shows a microscopic image of a strand of pondweed. The student wants to measure cells accurately before comparing how cell size changes along the strand. The black scale bar on the image represents an actual distance of 100 µm.
| Objective lens | Eyepiece divisions equal to 0.10 mm |
|---|---|
| ×10 | 20 |
| ×40 | 80 |
(a) What is the value of one eyepiece division at objective lens ×10, in mm? [1]
(b) Complete the calculation to give the value of one division at ×40 in µm. [2]
(c) Give one reason why calibration should be repeated when the objective lens is changed. [1]
Table 1 shows the width of cells observed in three prepared slides at a school laboratory. Each student used the same eyepiece graticule and calibration slide. Fig. 1 shows the outline of a bacterium from the yoghurt sample. Bacteria are single-celled organisms. They have a cell membrane, cytoplasm and genetic material, but their genetic material is not enclosed in a nucleus. The student also observed plant cells from spinach and animal cells from a fish gill.
| Slide | Mean cell width / µm |
|---|---|
| Yoghurt bacterium | 2 |
| Fish gill cell | 18 |
| Spinach leaf cell | 55 |
(a) What is the region labelled A? [1]
(b) Give one structure found in a spinach leaf cell but not in the bacterium. [1]
(c) Complete this sentence: the fish gill cell is ______ times wider than the bacterium. [1]
(d) Describe one difference between the genetic material in the bacterium and in a fish gill cell. [1]
Calibration investigation, (a) \[0.10\text{ mm}\div20=0.005\text{ mm}\] One eyepiece division at ×10 is 0.005 mm. [1]
Calibration investigation, (b) \[0.10\text{ mm}\div80=0.00125\text{ mm}\] Since \(1\text{ mm}=1000\text{ µm}\): \[0.00125\text{ mm}=1.25\text{ µm}\] One division at ×40 is 1.25 µm. [2]
Calibration investigation, (c) Calibration must be repeated because each objective lens gives a different magnification and therefore each eyepiece division represents a different actual size. [1]
Bacterium investigation, (a) Region A is genetic material, or DNA. [1]
Bacterium investigation, (b) A spinach cell, unlike the bacterium, may contain a nucleus, chloroplast, or large permanent vacuole. [1]
Bacterium investigation, (c) \[18\text{ µm}\div2\text{ µm}=9\] The fish gill cell is 9 times wider. [1]
Bacterium investigation, (d) Bacterial DNA is free in the cytoplasm, whereas DNA in a fish gill cell is enclosed inside a nucleus. [1]
Calibration investigation, (a) \[0.10\text{ mm}\div20=0.005\text{ mm}\] One eyepiece division at ×10 is 0.005 mm. [1]
Calibration investigation, (b) \[0.10\text{ mm}\div80=0.00125\text{ mm}\] Since \(1\text{ mm}=1000\text{ µm}\): \[0.00125\text{ mm}=1.25\text{ µm}\] One division at ×40 is 1.25 µm. [2]
Calibration investigation, (c) Calibration must be repeated because each objective lens gives a different magnification and therefore each eyepiece division represents a different actual size. [1]
Bacterium investigation, (a) Region A is genetic material, or DNA. [1]
Bacterium investigation, (b) A spinach cell, unlike the bacterium, may contain a nucleus, chloroplast, or large permanent vacuole. [1]
Bacterium investigation, (c) \[18\text{ µm}\div2\text{ µm}=9\] The fish gill cell is 9 times wider. [1]
Bacterium investigation, (d) Bacterial DNA is free in the cytoplasm, whereas DNA in a fish gill cell is enclosed inside a nucleus. [1]
Question 3 Report
Fig. 1 shows the front teeth and cheek teeth of a grazing rabbit skull found by a wildlife ranger. The ranger compared the teeth with the food eaten by the rabbit. Rabbits cut plant stems with incisors and then grind leaves before swallowing. Most cellulose in a leaf cannot be digested by human enzymes, but microorganisms in a rabbit's gut help process some plant material.
(a) Describe how the cheek teeth shown are suited to the rabbit's food. [2]
(b) What is the main function of the incisors? [1]
(c) Give the name of the carbohydrate that forms much of a plant cell wall. [1]
(a) The cheek teeth have broad, flat surfaces. These surfaces are suited to crushing and grinding leaves and stems before swallowing. [2]
(b) Incisors are used for cutting or biting off plant material. [1]
(c) The carbohydrate forming much of a plant cell wall is cellulose. [1]
(a) The cheek teeth have broad, flat surfaces. These surfaces are suited to crushing and grinding leaves and stems before swallowing. [2]
(b) Incisors are used for cutting or biting off plant material. [1]
(c) The carbohydrate forming much of a plant cell wall is cellulose. [1]
Question 4 Report
Fig. 1 shows a rooftop garden with flowering plants, aphids, ladybirds and robins. The gardener notices that several leaves have been damaged by aphids. Table 1 gives estimated energy at each trophic level in one part of the garden.
| Organisms | Energy / kJ m-2 |
|---|---|
| flowering plants | 15 500 |
| aphids | 1400 |
| ladybirds | 155 |
| robins | 18 |
(a) Draw a food chain using all four groups of organisms. [2]
(b) What is the percentage energy transfer from aphids to ladybirds? [1]
(c) Describe the overall pattern shown by the energy values. [1]
(d) Give one reason why ladybirds do not receive all of the energy stored in aphids. [1]
Fig. 1 shows a cattle farm using anaerobic digesters. Cow manure enters a sealed tank. Microorganisms break down the material and methane is collected to heat farm buildings. Digestate is spread on fields.
(a) What gas is used as the fuel collected from the digester? [1]
(b) Describe why the digester must be sealed to maintain anaerobic conditions. [1]
(c) Give one energy transfer that occurs when methane is burned. [1]
(d) What happens to much of the chemical energy in manure that is not stored in methane? [1]
Labelled answer diagram:
Rooftop food chain
(a) A correct food chain is flowering plants → aphids → ladybirds → robins. [2]
(b) \[\left(\frac{155}{1400}\right)\times100=11.1\%\] Therefore the percentage transfer is 11.1%, or 11%. [1]
(c) Energy decreases at each successive trophic level. [1]
(d) Ladybirds do not receive all aphid energy because aphids use energy in respiration and movement, and may lose energy in waste. [1]
Anaerobic digester
(a) The fuel gas is methane. [1]
(b) The digester must be sealed to prevent oxygen entering, maintaining anaerobic conditions. [1]
(c) When methane burns, chemical energy is transferred to thermal energy, or heat. [1]
(d) Much chemical energy not stored in methane is released as heat during respiration by microorganisms. [1]
Labelled answer diagram:
Rooftop food chain
(a) A correct food chain is flowering plants → aphids → ladybirds → robins. [2]
(b) \[\left(\frac{155}{1400}\right)\times100=11.1\%\] Therefore the percentage transfer is 11.1%, or 11%. [1]
(c) Energy decreases at each successive trophic level. [1]
(d) Ladybirds do not receive all aphid energy because aphids use energy in respiration and movement, and may lose energy in waste. [1]
Anaerobic digester
(a) The fuel gas is methane. [1]
(b) The digester must be sealed to prevent oxygen entering, maintaining anaerobic conditions. [1]
(c) When methane burns, chemical energy is transferred to thermal energy, or heat. [1]
(d) Much chemical energy not stored in methane is released as heat during respiration by microorganisms. [1]
Question 5 Report
The table below shows gas measurements from a school anaerobic digester. Food waste and animal manure were placed in the digester without oxygen. Fig. 1 shows the sealed tank and a pipe leading to a gas bag. The gas can be burned to heat water for washing equipment.
| day | methane collected / dm3 |
|---|---|
| 1 | 0.4 |
| 3 | 1.8 |
| 5 | 3.1 |
| 7 | 2.0 |
(a) When was the greatest volume of methane collected? [1]
(b) What is the main source of energy in the methane? [1]
(c) Describe one environmental advantage of using this methane rather than coal. [2]
The table below shows carbon stores measured in a restored peatland. Fig. 1 shows moss plants growing above waterlogged peat. The site manager blocked drainage channels in 2022 so that the peat remained wet. Carbon was measured in samples from the top 20 cm of soil.
| year | carbon in topsoil / kg m-2 |
|---|---|
| 2021 | 8.6 |
| 2022 | 8.8 |
| 2023 | 9.4 |
| 2024 | 10.1 |
(a) Describe the overall change in carbon stored from 2021 to 2024. [1]
(b) What gas is removed from the air by moss cells during photosynthesis? [1]
(c) Explain why keeping peat waterlogged can reduce the rate of decay. [2]
(d) Give one reason why repeating soil samples improves the reliability of the data. [1]
Labelled answer diagram:
Anaerobic digester
Peatland carbon store
Labelled answer diagram:
Anaerobic digester
Peatland carbon store
Question 6 Report
Fig. 1 shows cells from a sample of an unknown organism collected from a pond. The cells have a cell membrane, cytoplasm, a nucleus and a cell wall. They do not have chloroplasts. The sample was later identified as yeast, a fungus used by bakers. When yeast has glucose and warm water, it releases carbon dioxide. The gas causes bread dough to rise. A baker asks why yeast is not classed as a plant even though its cells have a wall.
(a) What is structure B? [1]
(b) Give one structure shown in Fig. 1 that is also found in an animal cell. [1]
(c) Describe why yeast cells cannot make their own food by photosynthesis. [1]
(d) Complete the word equation for anaerobic respiration in yeast: glucose → ethanol + ______. [1]
Fig. 1 shows cells from the leaf of a Venus flytrap plant. When an insect touches sensitive hairs, the two halves of the leaf close. Digestive enzymes are then released and the plant absorbs mineral ions from the insect. The cells shown are from a green part of the trap and contain chloroplasts. The plant still needs photosynthesis to make glucose, although it gains nitrogen-containing compounds from captured food.
(a) What is structure A? [1]
(b) Give the name of the process that uses chloroplasts to make glucose. [1]
(c) Describe why the Venus flytrap gains an advantage from digesting insects even though its cells can make food. [2]
(d) Give one other structure present in these plant cells but not in animal cells. [1]
Yeast cells, (a) Structure B is the nucleus. [1]
Yeast cells, (b) A structure shown that is also found in animal cells is the cell membrane, cytoplasm, or nucleus. [1]
Yeast cells, (c) Yeast cannot photosynthesise because it has no chloroplasts. [1]
Yeast cells, (d) Anaerobic respiration in yeast is:
glucose → ethanol + carbon dioxide. [1]
Venus flytrap cells, (a) Structure A is a sensitive hair, also called a trigger hair. [1]
Venus flytrap cells, (b) The process using chloroplasts to make glucose is photosynthesis. [1]
Venus flytrap cells, (c) Digesting insects provides mineral ions or nitrogen-containing compounds. These are needed to make proteins and support healthy growth, even though photosynthesis supplies glucose. [2]
Venus flytrap cells, (d) A structure in these plant cells but not animal cells is a cell wall, chloroplast, or large permanent vacuole. [1]
Yeast cells, (a) Structure B is the nucleus. [1]
Yeast cells, (b) A structure shown that is also found in animal cells is the cell membrane, cytoplasm, or nucleus. [1]
Yeast cells, (c) Yeast cannot photosynthesise because it has no chloroplasts. [1]
Yeast cells, (d) Anaerobic respiration in yeast is:
glucose → ethanol + carbon dioxide. [1]
Venus flytrap cells, (a) Structure A is a sensitive hair, also called a trigger hair. [1]
Venus flytrap cells, (b) The process using chloroplasts to make glucose is photosynthesis. [1]
Venus flytrap cells, (c) Digesting insects provides mineral ions or nitrogen-containing compounds. These are needed to make proteins and support healthy growth, even though photosynthesis supplies glucose. [2]
Venus flytrap cells, (d) A structure in these plant cells but not animal cells is a cell wall, chloroplast, or large permanent vacuole. [1]
Question 7 Report
Fig. 1 shows a pedigree drawn during an investigation into Duchenne muscular dystrophy, DMD. DMD is an X-linked recessive disorder that causes progressive weakening of skeletal muscles. Filled squares show males with DMD. The laboratory has offered DNA testing to females in this family because some may carry the altered allele without having the disorder.
(a) Describe why DMD is more common in males than in females. [3]
(b) What is the most likely genotype of the mother of the affected male in generation II? Use XD and Xd. [2]
(c) Give the genotype of an affected male. [1]
(d) Complete a genetic diagram for a carrier female and an unaffected male. [3]
(e) Give one advantage of DNA testing for the unaffected females in this family. [1]
(a) DMD is more common in males because males have one \(X\) chromosome and one \(Y\) chromosome. A male with \(X^dY\) has only one copy of the X-linked gene, so \(X^d\) is expressed because there is no normal allele on a second X chromosome. [3]
(b) The mother of the affected male is most likely \(X^D X^d\). She is an unaffected carrier but can pass \(X^d\) to a son. [2]
(c) An affected male has genotype \(X^dY\). [1]
(d) A carrier female is \(X^D X^d\) and an unaffected male is \(X^D Y\).
| \(X^D\) | \(Y\) | |
|---|---|---|
| \(X^D\) | \(X^D X^D\) | \(X^D Y\) |
| \(X^d\) | \(X^D X^d\) | \(X^dY\) |
The gametes are \(X^D\), \(X^d\), \(X^D\), and \(Y\), giving offspring \(X^D X^D\), \(X^D X^d\), \(X^D Y\), and \(X^dY\). [3]
(e) DNA testing can identify unaffected females who are carriers and allow informed reproductive decisions. [1]
(a) DMD is more common in males because males have one \(X\) chromosome and one \(Y\) chromosome. A male with \(X^dY\) has only one copy of the X-linked gene, so \(X^d\) is expressed because there is no normal allele on a second X chromosome. [3]
(b) The mother of the affected male is most likely \(X^D X^d\). She is an unaffected carrier but can pass \(X^d\) to a son. [2]
(c) An affected male has genotype \(X^dY\). [1]
(d) A carrier female is \(X^D X^d\) and an unaffected male is \(X^D Y\).
| \(X^D\) | \(Y\) | |
|---|---|---|
| \(X^D\) | \(X^D X^D\) | \(X^D Y\) |
| \(X^d\) | \(X^D X^d\) | \(X^dY\) |
The gametes are \(X^D\), \(X^d\), \(X^D\), and \(Y\), giving offspring \(X^D X^D\), \(X^D X^d\), \(X^D Y\), and \(X^dY\). [3]
(e) DNA testing can identify unaffected females who are carriers and allow informed reproductive decisions. [1]
Question 8 Report
This field study is about temperature regulation in desert mammals. Fig. 1 shows the outer surface of a fennec fox ear with many blood vessels close to the skin. The fox was observed at a conservation centre on days with different air temperatures. Blood flowing through vessels near the surface can transfer thermal energy to the surroundings. The body must keep the temperature of its cells within a narrow range so that enzymes work effectively. Table 1 shows measurements taken while the fox was resting in shade.
| Air temperature / degrees C | Ear surface temperature / degrees C | Blood flow in ear vessels / arbitrary units |
|---|---|---|
| 18 | 24 | 3 |
| 28 | 32 | 6 |
| 38 | 40 | 10 |
(a) What is the name of the process that keeps body conditions stable? [1]
(b) Describe the change in blood flow in ear vessels as air temperature rises. [1]
(c) Give the increase in ear surface temperature between 18 degrees C and 38 degrees C. [1]
(d) Explain how increased blood flow near the ear surface can cool the fox. [3]
(e) Draw arrows on Fig. 1 to show heat moving from blood vessels to the air. [1]
(f) When air temperature is low, what happens to the diameter of skin arterioles? [1]
(g) Give one response, other than changing blood flow, that a mammal can use when it is too hot. [1]
(h) Complete this sentence: an increase in skin blood flow is called vasodilation of the __________. [1]
Question 9 Report
This diagram is of an alveolus and a nearby blood capillary in a human lung. Air in the alveolus has a higher concentration of oxygen than the blood arriving at the capillary. The thin walls allow rapid exchange. Red blood cells carry oxygen away from the lung to body cells, including cells in a leaf-like network of muscles around the chest.
(a) What process moves oxygen from the alveolus into the blood? [1]
(b) Give one feature of an alveolus that makes gas exchange rapid. [1]
(c) Describe what happens to the oxygen concentration in the blood as it flows past the alveolus. [2]
Fig. 1 shows a microscope field from a culture of a single-celled organism living in pond water. The organism was placed in water containing glucose. A researcher added methylene blue, which is blue when oxygen is present and becomes colourless when oxygen is used. Table 1 shows the time taken for the dye to become colourless at different temperatures.
| Temperature / degrees C | Time for dye to become colourless / min |
|---|---|
| 12 | 18 |
| 22 | 8 |
| 32 | 5 |
(a) Describe the result as temperature increases from 12 to 32 degrees C. [1]
(b) What does a shorter time show about the rate of respiration? [1]
(c) Give two variables that should be kept the same in each sample. [2]
(d) Explain why the researcher should repeat each temperature test. [1]
Labelled answer diagram:
Alveolus and capillary
(a) Oxygen moves from an alveolus into the blood by diffusion. [1]
(b) An alveolus has a wall one cell thick, giving a short diffusion distance. A large surface area, moist lining or good blood supply are also acceptable. [1]
(c) Oxygen diffuses into the blood as it flows past the alveolus, so the oxygen concentration in the blood increases. [2]
Single-celled organism investigation
(a) As temperature increases from 12°C to 32°C, the time for the dye to become colourless decreases from 18 minutes to 5 minutes. [1]
(b) A shorter time means oxygen is used more quickly, so respiration is faster. [1]
(c) Two variables to keep the same are the number of organisms and the glucose concentration. Water volume, and the volume or concentration of methylene blue, are also acceptable. [2]
(d) Repeating each test helps identify anomalous results, allows a mean to be calculated and improves reliability. [1]
Labelled answer diagram:
Alveolus and capillary
(a) Oxygen moves from an alveolus into the blood by diffusion. [1]
(b) An alveolus has a wall one cell thick, giving a short diffusion distance. A large surface area, moist lining or good blood supply are also acceptable. [1]
(c) Oxygen diffuses into the blood as it flows past the alveolus, so the oxygen concentration in the blood increases. [2]
Single-celled organism investigation
(a) As temperature increases from 12°C to 32°C, the time for the dye to become colourless decreases from 18 minutes to 5 minutes. [1]
(b) A shorter time means oxygen is used more quickly, so respiration is faster. [1]
(c) Two variables to keep the same are the number of organisms and the glucose concentration. Water volume, and the volume or concentration of methylene blue, are also acceptable. [2]
(d) Repeating each test helps identify anomalous results, allows a mean to be calculated and improves reliability. [1]
Question 10 Report
Fig. 1 shows an investigation in which scientists release marked beetles into two grassland areas, then recapture beetles after a dry summer. The beetles vary in water loss because their body coverings are controlled by different alleles.
(a) Describe one characteristic that could vary between individual beetles. [1]
(b) Explain why beetles that lose less water may become more common after several dry summers. [3]
(c) Give two features of the investigation that would make its conclusion more reliable. [2]
(a) One characteristic that could vary is body colour. Body size, rate of water loss and thickness of body covering are also acceptable. [1]
(b) Beetles that lose less water are more likely to survive drought. They then reproduce more successfully and pass the alleles associated with lower water loss to their offspring. Over generations, these alleles become more frequent, so such beetles become more common. [3]
(c) Reliability could be improved by using many beetles and repeating the investigation in several areas or years. Random sampling and comparison with a wetter control area are also valid. [2]
(a) One characteristic that could vary is body colour. Body size, rate of water loss and thickness of body covering are also acceptable. [1]
(b) Beetles that lose less water are more likely to survive drought. They then reproduce more successfully and pass the alleles associated with lower water loss to their offspring. Over generations, these alleles become more frequent, so such beetles become more common. [3]
(c) Reliability could be improved by using many beetles and repeating the investigation in several areas or years. Random sampling and comparison with a wetter control area are also valid. [2]
Question 11 Report
Fig. 1 shows a pedigree from an optometry practice. Some males in the family have red-green colour blindness. The condition is caused by a recessive allele carried on the X chromosome. The normal allele is written XN and the colour-blind allele is written Xn. Individual II-2 has normal colour vision but has a colour-blind son.
(a) What is the genotype of individual II-2? [2]
(b) Describe why the colour-blind child in Fig. 1 is male. [2]
(c) Give the allele inherited from the father by the colour-blind son. [1]
(d) Complete a genetic diagram for a carrier mother and a father with normal colour vision. [2]
(e) Give the probability that a son of this couple will be colour blind. [1]
(a) Individual II-2 has normal vision but has a colour-blind son. She must therefore be a carrier: \(X^N X^n\). [2]
(b) A son receives a \(Y\) chromosome from his father. He has only one \(X\) chromosome, inherited from his mother, so a single \(X^n\) allele is expressed because there is no normal allele on a second \(X\) chromosome. [2]
(c) The colour-blind son inherited the \(Y\) allele-bearing chromosome from his father. [1]
(d) A carrier mother is \(X^N X^n\) and a father with normal colour vision is \(X^N Y\).
| \(X^N\) | \(Y\) | |
|---|---|---|
| \(X^N\) | \(X^N X^N\) | \(X^N Y\) |
| \(X^n\) | \(X^N X^n\) | \(X^nY\) |
The possible offspring are \(X^N X^N\), \(X^N X^n\), \(X^N Y\), and \(X^nY\). [2]
(e) Of the possible sons, half inherit \(X^n\) from the mother. The probability that a son is colour blind is \(50\%\), or one half. [1]
(a) Individual II-2 has normal vision but has a colour-blind son. She must therefore be a carrier: \(X^N X^n\). [2]
(b) A son receives a \(Y\) chromosome from his father. He has only one \(X\) chromosome, inherited from his mother, so a single \(X^n\) allele is expressed because there is no normal allele on a second \(X\) chromosome. [2]
(c) The colour-blind son inherited the \(Y\) allele-bearing chromosome from his father. [1]
(d) A carrier mother is \(X^N X^n\) and a father with normal colour vision is \(X^N Y\).
| \(X^N\) | \(Y\) | |
|---|---|---|
| \(X^N\) | \(X^N X^N\) | \(X^N Y\) |
| \(X^n\) | \(X^N X^n\) | \(X^nY\) |
The possible offspring are \(X^N X^N\), \(X^N X^n\), \(X^N Y\), and \(X^nY\). [2]
(e) Of the possible sons, half inherit \(X^n\) from the mother. The probability that a son is colour blind is \(50\%\), or one half. [1]
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