Biology 9201 | Paper 2 Mock 01 | Written Paper 2
Question 1 Report
Fig. 1 shows a choice chamber used by a student to observe woodlice. One side contains damp leaf litter and the other side contains dry leaf litter. The student placed 20 woodlice in the centre and counted the animals after 10 minutes.
Table 1 shows the results from four trials.
| trial | number on damp side after 10 min |
|---|---|
| 1 | 16 |
| 2 | 15 |
| 3 | 17 |
| 4 | 16 |
(a) Describe the response shown by the woodlice. [2]
(b) Suggest why a damp habitat increases the chance of survival of a woodlouse. [2]
Fig. 1 shows a young chimpanzee watching an adult use a stone to crack a hard nut. The young chimp later picks up a stone and attempts the same action. Both chimpanzees are in a forest group.
(a) Name the type of learning shown by the young chimpanzee. [1]
(b) Describe how this learning differs from an inherited behaviour. [2]
(c) Suggest one advantage to a young chimp of learning this behaviour from an adult. [2]
Woodlice: (a) More woodlice moved to or remained on the damp side [1]. This shows a preference for moist conditions, a directional response to humidity [1].
(b) Damp conditions reduce water loss from the body [1]. Avoiding dehydration allows cells to function and improves survival [1].
Chimpanzees: (a) The young chimp shows imitation, or observational learning [1].
(b) This behaviour is acquired by experience or observing another animal [1]. It is not controlled only by genes and is not necessarily present at birth [1].
(c) Copying an adult lets the young chimp learn a successful technique more quickly [1], allowing it to obtain food from hard nuts and gain energy [1].
Answer Details
Woodlice: (a) More woodlice moved to or remained on the damp side [1]. This shows a preference for moist conditions, a directional response to humidity [1].
(b) Damp conditions reduce water loss from the body [1]. Avoiding dehydration allows cells to function and improves survival [1].
Chimpanzees: (a) The young chimp shows imitation, or observational learning [1].
(b) This behaviour is acquired by experience or observing another animal [1]. It is not controlled only by genes and is not necessarily present at birth [1].
(c) Copying an adult lets the young chimp learn a successful technique more quickly [1], allowing it to obtain food from hard nuts and gain energy [1].
Question 2 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 3 Report
This figure shows a vertical farm shelf with red and blue LED lamps above basil plants. A student measured oxygen concentration in the air around the plants. Table 1 gives measurements after 30 minutes.
| lamp setting | oxygen change / % |
|---|---|
| red only | +0.4 |
| red and blue | +0.9 |
(a) Use Table 1 to state the lamp setting with the larger oxygen increase. [1]
(b) Suggest why oxygen concentration rose around the plants. [1]
(c) Describe one use of glucose made by basil cells. [1]
(d) Suggest why the farm controls lamp colour as well as lamp brightness. [1]
The diagram shows a school laboratory test for oxygen using a glowing splint. The student collected gas from a water plant under a funnel. Table 1 gives repeated results from three identical plants.
| plant | gas volume in 15 min / cm3 |
|---|---|
| 1 | 8.1 |
| 2 | 7.9 |
| 3 | 8.0 |
(a) Name the gas that relights a glowing splint. [1]
(b) Use Table 1 to calculate the mean gas volume. [1]
(c) Describe why repeating the measurement improves the investigation. [1]
(d) Suggest why the water temperature should be monitored. [1]
LED lamps and basil
- (a) The combined red and blue lamp setting produced the larger oxygen increase, \(+0.9\%\). [1]
- (b) Oxygen concentration rose because photosynthesis released oxygen. [1]
- (c) Glucose can be used in respiration, to make cellulose, or to make other substances for growth. [1]
- (d) Lamp colour is controlled because chlorophyll absorbs different wavelengths of light differently. [1]
Testing collected gas
- (a) The gas that relights a glowing splint is oxygen. [1]
- (b) \[\frac{8.1+7.9+8.0}{3}=8.0\text{ cm}^3\] The mean gas volume is \(8.0\text{ cm}^3\). [1]
- (c) Repeating measurements allows a mean to be calculated and helps identify anomalous results. [1]
- (d) Water temperature should be monitored because temperature affects enzyme-controlled photosynthesis. [1]
Answer Details
LED lamps and basil
- (a) The combined red and blue lamp setting produced the larger oxygen increase, \(+0.9\%\). [1]
- (b) Oxygen concentration rose because photosynthesis released oxygen. [1]
- (c) Glucose can be used in respiration, to make cellulose, or to make other substances for growth. [1]
- (d) Lamp colour is controlled because chlorophyll absorbs different wavelengths of light differently. [1]
Testing collected gas
- (a) The gas that relights a glowing splint is oxygen. [1]
- (b) \[\frac{8.1+7.9+8.0}{3}=8.0\text{ cm}^3\] The mean gas volume is \(8.0\text{ cm}^3\). [1]
- (c) Repeating measurements allows a mean to be calculated and helps identify anomalous results. [1]
- (d) Water temperature should be monitored because temperature affects enzyme-controlled photosynthesis. [1]
Question 4 Report
A grower noticed that some Colorado potato beetles remained alive after a potato crop had been sprayed. Fig. 1 shows an adult beetle feeding on a potato leaf. The grower had used the same insecticide for six years. Beetles were collected from three fields and kept in separate boxes with fresh potato leaves. Each box received a measured spray of insecticide. After 24 hours, the technician counted living beetles.
Table 1 shows the results. Field C had not received this insecticide in the previous six years because the farmer had used a different method to control pests. The three fields had similar temperature, water supply and potato plants. The insects were not tested on the same day.
| Field | Number of beetles at start | Number alive after 24 hours |
|---|---|---|
| A, sprayed for 6 years | 80 | 49 |
| B, sprayed for 6 years | 80 | 44 |
| C, not sprayed for 6 years | 80 | 7 |
The resistance allele can be passed from adult beetles to their eggs. A resistant beetle may have cells that make an enzyme which breaks down the insecticide. The grower is considering alternating insecticides and leaving some unsprayed strips of plants.
(a) Use Table 1 to calculate the percentage of beetles from field B that were killed by the insecticide. Show your working. [2]
(b) Describe the difference between the results from fields A and C. [2]
(c) Suggest why the result from field C provides evidence that some beetles from fields A and B are resistant. [2]
(d) Describe how repeated spraying changes a beetle population so that resistance becomes more common. [5]
(e) Suggest two reasons why alternating insecticides could slow the increase of resistant beetles. [3]
(a) First find the number killed:
\[80-44=36\text{ beetles}\]
Then calculate the percentage killed:
\[\frac{36}{80}\times100=45\%\]
Therefore, 45% of beetles from field B were killed. [2]
(b) More beetles from field A survived than from field C: 49 compared with 7. Therefore, fewer were killed in field A, 31 compared with 73 in field C. [2]
(c) The insecticide killed most beetles from field C, but many beetles from fields A and B survived exposure. This is evidence consistent with resistance in some A and B beetles, rather than all beetles being equally susceptible. [2]
(d) The beetle population has inherited variation, so some beetles carry a resistance allele. Spraying kills susceptible beetles, while resistant beetles survive. The survivors reproduce and lay eggs. Resistant parents pass the resistance allele to offspring, so the frequency of the resistance allele increases in later generations. [5]
(e) Alternating insecticides can slow resistance because a beetle resistant to one insecticide may be killed by a different insecticide. Different insecticides therefore impose different selection pressures, reducing the number of resistant survivors that reproduce after every treatment. Leaving unsprayed strips can also maintain susceptible beetles, which can breed with resistant beetles. [3]
Answer Details
(a) First find the number killed:
\[80-44=36\text{ beetles}\]
Then calculate the percentage killed:
\[\frac{36}{80}\times100=45\%\]
Therefore, 45% of beetles from field B were killed. [2]
(b) More beetles from field A survived than from field C: 49 compared with 7. Therefore, fewer were killed in field A, 31 compared with 73 in field C. [2]
(c) The insecticide killed most beetles from field C, but many beetles from fields A and B survived exposure. This is evidence consistent with resistance in some A and B beetles, rather than all beetles being equally susceptible. [2]
(d) The beetle population has inherited variation, so some beetles carry a resistance allele. Spraying kills susceptible beetles, while resistant beetles survive. The survivors reproduce and lay eggs. Resistant parents pass the resistance allele to offspring, so the frequency of the resistance allele increases in later generations. [5]
(e) Alternating insecticides can slow resistance because a beetle resistant to one insecticide may be killed by a different insecticide. Different insecticides therefore impose different selection pressures, reducing the number of resistant survivors that reproduce after every treatment. Leaving unsprayed strips can also maintain susceptible beetles, which can breed with resistant beetles. [3]
Question 5 Report
A dentist examined the diet record of a 10-year-old child with tooth decay. Fig. 1 shows a molar tooth with a cavity in its enamel.
| Item consumed per day | Number of times |
|---|---|
| Sweetened drink | 5 |
| Fruit | 2 |
| Water | 1 |
(a) Describe the evidence in Table 1 that the child often consumes sugar. [1]
(b) Suggest how bacteria in the mouth can cause the cavity shown in Fig. 1. [2]
(c) Name the type of digestion that chewing starts. [1]
(d) What is the benefit of breaking food into smaller pieces before enzymes act? [1]
A student made a model stomach from a sealed plastic bag. Fig. 1 shows food cubes, acid and protease being mixed by squeezing the bag.
After 20 minutes at 37 degrees C, the student tested the liquid for protein.
(a) Describe how squeezing the bag models the action of stomach muscles. [1]
(b) Suggest why the food cubes became smaller. [1]
(c) Use the model to describe two conditions needed by protease in the stomach. [2]
(d) What is the purpose of mixing food with digestive juices? [1]
Tooth decay
- (a) The child consumed a sweetened drink five times each day. This is evidence of frequent sugar consumption. [1]
- (b) Mouth bacteria feed on sugar and produce acid. The acid damages or dissolves tooth enamel, producing a cavity. [2]
- (c) Chewing starts mechanical digestion. [1]
- (d) Smaller food pieces have a larger surface area for enzyme action. [1]
Model stomach
- (a) Squeezing the bag churns and mixes the food, like contractions of stomach muscles. [1]
- (b) Protease digested protein in the food cubes, so the cubes became smaller. [1]
- (c) The model shows that protease needs a warm temperature, about 37 degrees C, and acidic conditions. [2]
- (d) Mixing food with digestive juices increases contact between food and enzymes. [1]
Key distinction: chewing and churning are mechanical digestion; protease breaking down protein is chemical digestion.
Answer Details
Tooth decay
- (a) The child consumed a sweetened drink five times each day. This is evidence of frequent sugar consumption. [1]
- (b) Mouth bacteria feed on sugar and produce acid. The acid damages or dissolves tooth enamel, producing a cavity. [2]
- (c) Chewing starts mechanical digestion. [1]
- (d) Smaller food pieces have a larger surface area for enzyme action. [1]
Model stomach
- (a) Squeezing the bag churns and mixes the food, like contractions of stomach muscles. [1]
- (b) Protease digested protein in the food cubes, so the cubes became smaller. [1]
- (c) The model shows that protease needs a warm temperature, about 37 degrees C, and acidic conditions. [2]
- (d) Mixing food with digestive juices increases contact between food and enzymes. [1]
Key distinction: chewing and churning are mechanical digestion; protease breaking down protein is chemical digestion.
Question 6 Report
Fig. 1 shows a simplified section through a human alveolus and a nearby blood capillary. The person has been exercising at a high altitude training camp. Air in the alveolus has a higher oxygen concentration than the blood arriving at the capillary. Red blood cells pass through the capillary one at a time.
The capillary walls and alveolar walls are each one cell thick. The athlete's doctor also measured the number of red blood cells in a blood sample. It was 5.8 million cells per mm3, compared with 5.1 million cells per mm3 before the training period.
(a) Describe the movement of oxygen shown by the arrow in Fig. 1. [2]
(b) Suggest two features of an alveolus that allow rapid gas exchange. [3]
(c) Use the blood data to calculate the increase in red blood cells per mm3. [2]
(d) Suggest how the increase in red blood cells could help the athlete during exercise. [2]
(a) Oxygen diffuses from the alveolar air into the blood. It moves down its concentration gradient, from higher oxygen concentration in the alveolus to lower oxygen concentration in the arriving blood. [2]
(b) Three features allowing rapid gas exchange are:
- a large surface area;
- walls one cell thick, giving a short diffusion distance;
- a good blood supply, which maintains a concentration gradient.
A moist surface, allowing gases to dissolve, is also acceptable. [3]
(c) \[5.8\text{ million cells mm}^{-3}-5.1\text{ million cells mm}^{-3}=0.7\text{ million cells mm}^{-3}\] The increase was 0.7 million red blood cells per mm3. [2]
(d) More red blood cells contain more haemoglobin, so more oxygen can be carried in the blood. Muscles can then carry out more aerobic respiration and release more energy during exercise. [2]
Answer Details
(a) Oxygen diffuses from the alveolar air into the blood. It moves down its concentration gradient, from higher oxygen concentration in the alveolus to lower oxygen concentration in the arriving blood. [2]
(b) Three features allowing rapid gas exchange are:
- a large surface area;
- walls one cell thick, giving a short diffusion distance;
- a good blood supply, which maintains a concentration gradient.
A moist surface, allowing gases to dissolve, is also acceptable. [3]
(c) \[5.8\text{ million cells mm}^{-3}-5.1\text{ million cells mm}^{-3}=0.7\text{ million cells mm}^{-3}\] The increase was 0.7 million red blood cells per mm3. [2]
(d) More red blood cells contain more haemoglobin, so more oxygen can be carried in the blood. Muscles can then carry out more aerobic respiration and release more energy during exercise. [2]
Question 7 Report
A sports scientist used the apparatus in Fig. 1 to compare pulse rate before and after a 10 minute rowing test. A light sensor was clipped to the student's finger. Each pulse of blood through an artery changed the amount of light reaching the sensor. The computer displayed a pulse trace. Before exercise the student was sitting quietly. Immediately after rowing, the trace peaks were closer together. The student breathed more deeply after the test because active muscle cells needed more oxygen and produced more carbon dioxide.
(a) Name the measurement recorded by the sensor. [1]
(b) Describe the change in pulse rate after rowing. [1]
(c) Suggest why the heart rate increases during the rowing test. [2]
(a) The sensor records pulse rate, which reflects heart rate. [1]
(b) The pulse rate increases after rowing because the peaks are closer together in the trace. [1]
(c) Working muscle cells need more oxygen and glucose for respiration. Increasing heart rate makes circulation faster, so these materials are supplied more rapidly and carbon dioxide is removed more rapidly. [2]
Answer Details
(a) The sensor records pulse rate, which reflects heart rate. [1]
(b) The pulse rate increases after rowing because the peaks are closer together in the trace. [1]
(c) Working muscle cells need more oxygen and glucose for respiration. Increasing heart rate makes circulation faster, so these materials are supplied more rapidly and carbon dioxide is removed more rapidly. [2]
Question 8 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 9 Report
Fig. 1 shows a small island where medium ground finches feed mainly on seeds. In one dry year, most small soft seeds failed to form. A conservation team measured the beak depth of adult birds that survived until the following breeding season. The same team recorded the seeds available in two areas of the island.
| seed type | mean seed width / mm | number found before drought | number found after drought |
|---|---|---|---|
| small, soft | 2.1 | 840 | 90 |
| large, hard | 5.8 | 170 | 150 |
Beak depth varies between individual finches and is inherited. Finches with deeper beaks can crack hard seeds more easily. No finch changes its inherited beak depth because it needs food. The team plans to return in 20 years to compare the population.
(a) Use Table 1 to describe the change in the availability of the two seed types. [2]
(b) Suggest which finches were more likely to survive the drought. [1]
(c) Describe how the mean beak depth of the population may change over several generations. [3]
(d) What evidence would support the idea that beak depth is inherited? [2]
Labelled answer diagram:
(a) Small soft seeds decreased greatly from 840 to 90 [1]. Large hard seeds changed little, from 170 to 150 [1].
(b) Finches with deeper or larger beaks were more likely to survive [1].
(c) Deeper-beaked finches can obtain more hard seeds and are more likely to survive [1]. They reproduce more successfully and pass alleles for deeper beaks to offspring [1]. Over generations, mean beak depth increases [1].
(d) Evidence would be that offspring of deep-beaked parents tend also to have deep beaks [1]. This pattern should remain when offspring are raised under similar conditions, or beak depth should be associated with parental genotype [1].
Answer Details
Labelled answer diagram:
(a) Small soft seeds decreased greatly from 840 to 90 [1]. Large hard seeds changed little, from 170 to 150 [1].
(b) Finches with deeper or larger beaks were more likely to survive [1].
(c) Deeper-beaked finches can obtain more hard seeds and are more likely to survive [1]. They reproduce more successfully and pass alleles for deeper beaks to offspring [1]. Over generations, mean beak depth increases [1].
(d) Evidence would be that offspring of deep-beaked parents tend also to have deep beaks [1]. This pattern should remain when offspring are raised under similar conditions, or beak depth should be associated with parental genotype [1].
Question 10 Report
Fig. 1 shows a pedigree for a family attending a hospital genetics clinic. Some family members have a recessive inherited condition that affects the ability of blood cells to carry oxygen during exercise. A filled symbol represents a person with the condition. The clinic has confirmed that the allele causing the condition is recessive and is not sex-linked. A couple, labelled parents in generation III, ask about the possible genotypes of their future eggs and sperm. The nurse also explains that environmental factors such as temperature do not change the alleles a person inherited.
(a) Use Fig. 1 to state which sex is affected in generation II. [1]
(b) Suggest the genotype of an unaffected parent who has an affected child. Use A and a. [2]
(c) Describe why brothers and sisters can inherit different combinations of alleles from the same parents. [3]
Labelled answer diagram:
(a) The affected person in generation II is male. [1]
(b) An unaffected parent with an affected child is likely to have genotype \(Aa\). They are unaffected because \(A\) is dominant, but they must carry recessive allele \(a\) to pass it to the affected \(aa\) child. [2]
(c) During meiosis, chromosome pairs separate. Gametes receive different combinations of alleles through independent assortment. Fertilisation is random, so different sperm can fuse with different eggs, giving siblings different allele combinations. [3]
Answer Details
Labelled answer diagram:
(a) The affected person in generation II is male. [1]
(b) An unaffected parent with an affected child is likely to have genotype \(Aa\). They are unaffected because \(A\) is dominant, but they must carry recessive allele \(a\) to pass it to the affected \(aa\) child. [2]
(c) During meiosis, chromosome pairs separate. Gametes receive different combinations of alleles through independent assortment. Fertilisation is random, so different sperm can fuse with different eggs, giving siblings different allele combinations. [3]
Question 11 Report
Fig. 1 shows a simplified diagram of organs involved when the concentration of glucose in human blood changes. A student traced the path of hormones after eating a meal containing rice. The blood glucose concentration rose above the usual level for a short time. Organ X releases a hormone into the blood. The hormone acts on cells in organ Y. The student knows that body cells need glucose for respiration, but a very high concentration can damage blood vessels.
(a) Name the organs labelled X and Y. [2]
(b) Describe how the hormone released by X reduces blood glucose concentration after the meal. [3]
(c) Suggest why controlling blood glucose concentration is important for a human during exercise. [3]
(a) Organ X is the pancreas. Organ Y is the liver. [2]
(b) After the meal, the pancreas releases insulin into the blood. Insulin causes liver and muscle cells to take up glucose from the blood. The glucose is converted to glycogen for storage, reducing blood glucose concentration. [3]
(c) Blood glucose must be controlled during exercise so cells have glucose for respiration. Respiration releases energy for muscle contraction. Control also prevents blood glucose becoming dangerously high or low, either of which can disrupt normal cell function. [3]
Answer Details
(a) Organ X is the pancreas. Organ Y is the liver. [2]
(b) After the meal, the pancreas releases insulin into the blood. Insulin causes liver and muscle cells to take up glucose from the blood. The glucose is converted to glycogen for storage, reducing blood glucose concentration. [3]
(c) Blood glucose must be controlled during exercise so cells have glucose for respiration. Respiration releases energy for muscle contraction. Control also prevents blood glucose becoming dangerously high or low, either of which can disrupt normal cell function. [3]
