Biology 9201 | Paper 1 Mock 01 | Written Paper 1
Question 1 Report
Fig. 1 shows a red mangrove growing at the edge of a tropical estuary. Twice each day, seawater covers the muddy soil around the plant. The mud has little oxygen because water fills most of the air spaces. Some roots grow upwards above the mud. Salt water may enter root cells, so the plant must maintain the water concentration of its cells. A ranger is comparing this plant with a tree growing on dry land.
(a) Describe one feature visible in Fig. 1 that helps the mangrove obtain oxygen. [2]
(b) Explain why taking in salt water can cause a problem for mangrove cells. [2]
Table 1 shows growth of young bean plants grown in sterilised soil. In one set of pots, the roots were inoculated with mycorrhizal fungi. Fig. 1 shows fungal hyphae around and inside a root. The plants all received the same amount of water, light and carbon dioxide. Fungal hyphae increase the area from which mineral ions can be absorbed.
| Soil treatment | Mean dry mass after 28 days / g |
|---|---|
| No fungi added | 2.8 |
| Mycorrhizal fungi added | 4.6 |
(a) Describe the result shown in Table 1. [1]
(b) Explain how fungal hyphae can increase the growth of bean plants. [2]
(c) Give one substance supplied by the plant to the fungus. [1]
Labelled answer diagram:
Mangrove
- (a) The visible upward-growing roots project out of the mud. They are exposed to air, so oxygen can diffuse into them for respiration. [2]
- (b) Salt water is more concentrated, so has a lower water potential, than the cell contents. Water therefore leaves mangrove cells by osmosis; the cells lose water and may become plasmolysed. [2]
Mycorrhizal fungi and bean plants
- (a) Plants with mycorrhizal fungi had the greater mean dry mass: 4.6 g compared with 2.8 g without fungi. [1]
- (b) Fungal hyphae provide a large surface area, increasing absorption of mineral ions from soil. The plant uses these ions, for example to make proteins and new cells, so growth increases. [2]
- (c) The plant supplies glucose, or other sugars, to the fungus. [1]
Exam focus: In this mutualistic relationship, the fungus improves mineral-ion uptake and receives sugars made by photosynthesis.
Answer Details
Labelled answer diagram:
Mangrove
- (a) The visible upward-growing roots project out of the mud. They are exposed to air, so oxygen can diffuse into them for respiration. [2]
- (b) Salt water is more concentrated, so has a lower water potential, than the cell contents. Water therefore leaves mangrove cells by osmosis; the cells lose water and may become plasmolysed. [2]
Mycorrhizal fungi and bean plants
- (a) Plants with mycorrhizal fungi had the greater mean dry mass: 4.6 g compared with 2.8 g without fungi. [1]
- (b) Fungal hyphae provide a large surface area, increasing absorption of mineral ions from soil. The plant uses these ions, for example to make proteins and new cells, so growth increases. [2]
- (c) The plant supplies glucose, or other sugars, to the fungus. [1]
Exam focus: In this mutualistic relationship, the fungus improves mineral-ion uptake and receives sugars made by photosynthesis.
Question 2 Report
Fig. 1 shows a pedigree prepared by a clinic for a family affected by cystic fibrosis. Cystic fibrosis is caused by a recessive allele. Filled symbols show people with the disorder. A square represents a male and a circle represents a female. Individual IV-2 is expecting a baby with a partner who is known not to have cystic fibrosis.
(a) What is the genotype of individual II-2? Use C for the normal allele and c for the cystic fibrosis allele. [1]
(b) Describe the evidence in Fig. 1 that shows cystic fibrosis is recessive. [2]
(c) Complete a genetic diagram to show the possible genotypes of children produced by two carriers of cystic fibrosis. [3]
(d) Give the percentage chance that a child of two carriers will have cystic fibrosis. [1]
(e) When a couple are considering a genetic test, give one reason why a trained counsellor should discuss the result with them. [1]
(a) Individual II-2 is affected and cystic fibrosis is recessive, so their genotype is \(cc\). [1]
(b) The parents in generation I are unaffected but have an affected child. This shows that they can carry the cystic fibrosis allele without having the disorder, so the allele is recessive. [2]
(c) Two carriers are \(Cc\) and \(Cc\). Each produces gametes carrying \(C\) or \(c\).
| \(C\) | \(c\) | |
|---|---|---|
| \(C\) | \(CC\) | \(Cc\) |
| \(c\) | \(Cc\) | \(cc\) |
Parents \(Cc\) and \(Cc\), correct gametes, and offspring \(CC\), \(Cc\), \(Cc\), \(cc\) each gain a marking point. [3]
(d) One of the four equally likely genotypes is \(cc\), so the probability of cystic fibrosis is \(\frac14=25\%\). [1]
(e) Genetic counselling is important because a result may cause anxiety, may affect relatives, and can help the couple understand their reproductive choices. Any one of these is creditworthy. [1]
Answer Details
(a) Individual II-2 is affected and cystic fibrosis is recessive, so their genotype is \(cc\). [1]
(b) The parents in generation I are unaffected but have an affected child. This shows that they can carry the cystic fibrosis allele without having the disorder, so the allele is recessive. [2]
(c) Two carriers are \(Cc\) and \(Cc\). Each produces gametes carrying \(C\) or \(c\).
| \(C\) | \(c\) | |
|---|---|---|
| \(C\) | \(CC\) | \(Cc\) |
| \(c\) | \(Cc\) | \(cc\) |
Parents \(Cc\) and \(Cc\), correct gametes, and offspring \(CC\), \(Cc\), \(Cc\), \(cc\) each gain a marking point. [3]
(d) One of the four equally likely genotypes is \(cc\), so the probability of cystic fibrosis is \(\frac14=25\%\). [1]
(e) Genetic counselling is important because a result may cause anxiety, may affect relatives, and can help the couple understand their reproductive choices. Any one of these is creditworthy. [1]
Question 3 Report
Fig. 1 shows part of a family tree used by a cancer genetics team. Some people have developed breast or ovarian cancer linked to a changed BRCA1 allele. The changed allele increases risk but does not guarantee that cancer will develop, because cancer also involves further mutations in body cells and environmental factors. Individual III-2 is an adult who has not had cancer and is considering a test for the family allele.
(a) What type of allele is most consistent with the pattern in Fig. 1? [1]
(b) Describe two features of Fig. 1 that support your answer. [2]
(c) Give the probability that III-2 inherited the changed BRCA1 allele if their affected parent is heterozygous. [1]
(d) Explain why a positive BRCA1 result cannot predict with certainty that III-2 will develop cancer. [3]
(e) Describe two ways in which regular screening could help a person with a changed BRCA1 allele. [2]
(f) When a result is shared, give one reason why it may also be important to inform close blood relatives. [1]
(g) Give one ethical issue that a genetics team should consider before testing an adult for this allele. [2]
(a) The pattern is most consistent with a dominant allele. [1]
(b) Affected people occur in successive generations, and an affected parent has an affected child. The pedigree also shows that both males and females can inherit the allele. Any two relevant observations gain credit. [2]
(c) If the affected parent is heterozygous, there is a \(50\%\) probability that III-2 inherited the changed allele. [1]
(d) A changed BRCA1 allele increases susceptibility to cancer; it does not guarantee cancer. Further mutations in body cells may be needed before cancer develops, and environmental or lifestyle factors also affect risk. [3]
(e) Regular screening can detect a tumour at an early stage. Earlier treatment may improve the outcome or allow treatment before the tumour spreads. [2]
(f) Close blood relatives may also have inherited the changed allele, so informing them allows them to seek testing or screening. [1]
(g) The adult must give informed consent. A genetics team should also consider that a result may cause anxiety or affect insurance and family relationships. [2]
Answer Details
(a) The pattern is most consistent with a dominant allele. [1]
(b) Affected people occur in successive generations, and an affected parent has an affected child. The pedigree also shows that both males and females can inherit the allele. Any two relevant observations gain credit. [2]
(c) If the affected parent is heterozygous, there is a \(50\%\) probability that III-2 inherited the changed allele. [1]
(d) A changed BRCA1 allele increases susceptibility to cancer; it does not guarantee cancer. Further mutations in body cells may be needed before cancer develops, and environmental or lifestyle factors also affect risk. [3]
(e) Regular screening can detect a tumour at an early stage. Earlier treatment may improve the outcome or allow treatment before the tumour spreads. [2]
(f) Close blood relatives may also have inherited the changed allele, so informing them allows them to seek testing or screening. [1]
(g) The adult must give informed consent. A genetics team should also consider that a result may cause anxiety or affect insurance and family relationships. [2]
Question 4 Report
This figure shows a farmer selecting the largest seeds from a crop of wheat plants for planting next year. Fig. 1 uses arrows to show that only selected plants provide seeds. The farmer wants plants that produce more food grain.
(a) What term describes this method of changing a population of plants? [1]
(b) Describe why the farmer should use seeds from several large plants, rather than one plant. [2]
(c) Give one disadvantage of producing wheat plants with very little genetic variation. [1]
Table 1 shows measurements from an investigation into variation in leaf length in a greenhouse population of bean plants. Fig. 1 shows how a student measured each leaf from the base of the blade to its tip.
| Leaf length / cm | Number of plants |
|---|---|
| 5.0-5.9 | 3 |
| 6.0-6.9 | 11 |
| 7.0-7.9 | 14 |
| 8.0-8.9 | 7 |
(a) Describe the type of variation in leaf length shown in Table 1. [2]
(b) Give one environmental factor that could affect leaf length in these plants. [1]
(c) Explain why leaf length is not determined only by genes. [1]
Selective breeding in wheat
(a) This method is selective breeding, also called artificial selection. [1]
(b) Seeds should be selected from several large plants because these plants may provide a wider range of alleles. This reduces the risk that all offspring inherit the same harmful characteristic. [2]
(c) With very little genetic variation, a disease or environmental change could affect all the plants similarly. [1]
Leaf length in bean plants
(a) Leaf length shows continuous variation. There is a range of values, including intermediate lengths, rather than separate categories. [2]
(b) One environmental factor is light intensity. Water supply, mineral ion availability and temperature are also acceptable. [1]
(c) Leaf length is not determined only by genes because environmental conditions can affect leaf growth. [1]
Answer Details
Selective breeding in wheat
(a) This method is selective breeding, also called artificial selection. [1]
(b) Seeds should be selected from several large plants because these plants may provide a wider range of alleles. This reduces the risk that all offspring inherit the same harmful characteristic. [2]
(c) With very little genetic variation, a disease or environmental change could affect all the plants similarly. [1]
Leaf length in bean plants
(a) Leaf length shows continuous variation. There is a range of values, including intermediate lengths, rather than separate categories. [2]
(b) One environmental factor is light intensity. Water supply, mineral ion availability and temperature are also acceptable. [1]
(c) Leaf length is not determined only by genes because environmental conditions can affect leaf growth. [1]
Question 5 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]
- (a) The process maintaining stable body conditions is homeostasis. [1]
- (b) Blood flow in the ear vessels increases as air temperature rises, from 3 to 10 arbitrary units. [1]
- (c) Increase in ear surface temperature \(=40-24=16\degree\text{C}\). [1]
- (d) Increased blood flow brings more warm blood close to the ear skin surface. Heat transfers from the blood to the cooler surrounding air, preventing body temperature from rising too much. [3]
- (e) The arrows show heat moving from blood vessels, through the skin, to the surrounding air. [1]
- (f) In low air temperature, skin arterioles constrict and become narrower. [1]
- (g) A mammal may sweat, pant, rest in shade, or reduce activity when too hot. [1]
- (h) Increased skin blood flow is vasodilation of the arterioles. [1]
Answer Details
- (a) The process maintaining stable body conditions is homeostasis. [1]
- (b) Blood flow in the ear vessels increases as air temperature rises, from 3 to 10 arbitrary units. [1]
- (c) Increase in ear surface temperature \(=40-24=16\degree\text{C}\). [1]
- (d) Increased blood flow brings more warm blood close to the ear skin surface. Heat transfers from the blood to the cooler surrounding air, preventing body temperature from rising too much. [3]
- (e) The arrows show heat moving from blood vessels, through the skin, to the surrounding air. [1]
- (f) In low air temperature, skin arterioles constrict and become narrower. [1]
- (g) A mammal may sweat, pant, rest in shade, or reduce activity when too hot. [1]
- (h) Increased skin blood flow is vasodilation of the arterioles. [1]
Question 6 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]
Answer Details
(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 7 Report
A plant breeder grew orchid tissue in sterile boxes. Fig. 1 shows one original cell producing a small group of cells. Table 1 gives the number of plantlets formed from one piece of tissue.
| Week | Number of plantlets |
|---|---|
| 2 | 4 |
| 4 | 16 |
| 6 | 64 |
(a) What process produces the extra cells in Fig. 1? [1]
(b) Describe the trend in the number of plantlets. [1]
(c) Give two advantages of producing orchid plants in this way. [2]
The table below shows results from a study of radiation exposure and cell division in cultured human blood cells. Fig. 1 shows a cell with chromosomes arranged at the centre before they separate.
| Radiation dose / arbitrary units | Cells completing mitosis / % |
|---|---|
| 0 | 94 |
| 2 | 71 |
| 4 | 39 |
(a) Describe the relationship shown in Table 1. [1]
(b) What is the name of the stage in Fig. 1 when chromosomes line up at the cell centre? [1]
(c) Explain why radiation that damages DNA can reduce successful mitosis. [2]
(d) Give one reason why this experiment should use cells from the same tissue type. [1]
Orchid tissue culture
- (a) Mitosis produces the extra cells. [1]
- (b) The number of plantlets increases and doubles every two weeks: 4, 16, then 64. [1]
- (c) Two advantages are that many plants can be made quickly and the plants are genetically identical with desired features. Conserving rare plants is also acceptable. [2]
Radiation and mitosis
- (a) As radiation dose increases, the percentage of cells completing mitosis decreases, from 94% at zero dose to 39% at dose 4. [1]
- (b) Chromosomes lined up at the cell centre are in metaphase. [1]
- (c) Radiation can damage DNA, so it may not be copied correctly. This can stop cell division or produce abnormal daughter cells. [2]
- (d) Cells should come from the same tissue type because cells from different tissues naturally divide at different rates. [1]
Answer Details
Orchid tissue culture
- (a) Mitosis produces the extra cells. [1]
- (b) The number of plantlets increases and doubles every two weeks: 4, 16, then 64. [1]
- (c) Two advantages are that many plants can be made quickly and the plants are genetically identical with desired features. Conserving rare plants is also acceptable. [2]
Radiation and mitosis
- (a) As radiation dose increases, the percentage of cells completing mitosis decreases, from 94% at zero dose to 39% at dose 4. [1]
- (b) Chromosomes lined up at the cell centre are in metaphase. [1]
- (c) Radiation can damage DNA, so it may not be copied correctly. This can stop cell division or produce abnormal daughter cells. [2]
- (d) Cells should come from the same tissue type because cells from different tissues naturally divide at different rates. [1]
Question 8 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]
Answer Details
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 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]
Answer Details
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
This animal is a dromedary camel used by a wildlife team to carry equipment across a desert. Fig. 1 shows the camel during the hottest part of the day. Temperatures can exceed 40 degrees C and there is little surface water or plant food. The team records that the camel produces a small volume of concentrated urine and can travel for several days without drinking. Its long legs keep much of its body away from hot sand.
(a) Give one advantage of producing concentrated urine in this habitat. [1]
(b) Describe two visible adaptations in Fig. 1 that reduce the effect of the desert environment. [2]
(c) Explain how the hump can help the camel when food is scarce. [2]
Labelled answer diagram:
(a) Concentrated urine reduces water loss, conserving water in the camel's body. [1]
(b) Long legs keep much of the body away from hot sand, reducing heat gain. Broad feet prevent the camel sinking into sand. Eyelashes protecting eyes from sand and a pale coat reflecting heat are also acceptable visible adaptations. [2]
(c) The hump stores fat. When food is scarce, the fat can be respired to release energy. [2]
Answer Details
Labelled answer diagram:
(a) Concentrated urine reduces water loss, conserving water in the camel's body. [1]
(b) Long legs keep much of the body away from hot sand, reducing heat gain. Broad feet prevent the camel sinking into sand. Eyelashes protecting eyes from sand and a pale coat reflecting heat are also acceptable visible adaptations. [2]
(c) The hump stores fat. When food is scarce, the fat can be respired to release energy. [2]
Question 11 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]
Answer Details
(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]
