Biology 4BI1 | Paper 2 Mock 01 | Written Paper 2
Question 1 Report
Fig. 1 shows a prickly pear cactus growing in a dry botanical garden. The plant has a thick green stem, spines instead of broad leaves and a waxy outer surface. At night, pores in the stem can open to allow carbon dioxide to enter. During the day the pores are mostly closed, but the plant can still use light energy to make food. This reduces water loss in the hot, dry conditions.
(a) Name the process that uses carbon dioxide and light energy to make food in this plant. [1]
(b) Describe two features in Fig. 1 that reduce water loss. [2]
(c) Explain why carbon dioxide is needed for the cactus to make glucose. [1]
(d) Suggest two ways in which a cactus can use the glucose made by photosynthesis. [2]
(a) The process is photosynthesis. [1 mark]
(b) Two water-loss-reducing features are:
- Spines have a much smaller surface area than broad leaves, reducing evaporation. [1 mark]
- A waxy outer surface reduces evaporation of water. [1 mark]
A thick stem that stores water, and pores opening mainly at night, are also acceptable adaptations. At night temperatures are lower, so opening pores then reduces water loss.
(c) Carbon dioxide supplies the carbon atoms from which glucose is made in photosynthesis. [1 mark]
(d) Any two uses of glucose:
- used in respiration to release energy; [1 mark]
- converted to starch for storage; [1 mark]
- used to make cellulose for cell walls; [1 mark]
- converted into lipids or proteins when suitable additional substances are available. [1 mark]
Answer Details
(a) The process is photosynthesis. [1 mark]
(b) Two water-loss-reducing features are:
- Spines have a much smaller surface area than broad leaves, reducing evaporation. [1 mark]
- A waxy outer surface reduces evaporation of water. [1 mark]
A thick stem that stores water, and pores opening mainly at night, are also acceptable adaptations. At night temperatures are lower, so opening pores then reduces water loss.
(c) Carbon dioxide supplies the carbon atoms from which glucose is made in photosynthesis. [1 mark]
(d) Any two uses of glucose:
- used in respiration to release energy; [1 mark]
- converted to starch for storage; [1 mark]
- used to make cellulose for cell walls; [1 mark]
- converted into lipids or proteins when suitable additional substances are available. [1 mark]
Question 2 Report
A conservation group is growing young mangrove plants for a coastal restoration project. Fig. 1 shows one plant in waterlogged mud after a high tide. The mud contains much more water than air, so oxygen diffuses slowly to root cells. Some mangrove species have roots that project above the mud. A student plans to compare these plants with young mangroves whose root systems remain completely below the surface. The student will measure root growth and test the water around the roots for ethanol after seven days.
(a) Use Fig. 1 to describe how the exposed root structures could increase the oxygen supply to cells in the submerged roots. [3]
(b) What type of respiration is likely to occur in root cells if oxygen becomes unavailable? [2]
(c) Explain why prolonged anaerobic respiration can reduce the uptake of mineral ions by root hair cells. [3]
(d) Use the information in the scenario to describe a method the student could use to make a valid comparison between the two groups of plants. [5]
(e) State three variables that should be kept the same for both groups. [3]
(a) The exposed root structures contact the air, so oxygen can diffuse into them. Oxygen then moves through air spaces or tissues to the submerged root cells. This improves their oxygen supply in waterlogged mud. [3]
(b) If oxygen is unavailable, root cells carry out anaerobic respiration. In plant cells this produces ethanol and carbon dioxide, and releases less energy. [2]
(c) Anaerobic respiration releases less energy, so less ATP is available for active transport. Consequently, fewer mineral ions can be transported into root hair cells against a concentration gradient. [3]
(d) A valid comparison could be made as follows:
- Use similar young mangrove plants of the same species.
- Place one group with exposed roots and a second group with roots fully covered by waterlogged mud.
- Use several plants in each group.
- Keep both groups for seven days and measure root growth using the same method.
- Test water or root samples for ethanol using the same test, then compare mean results for the groups.
This isolates root exposure as the variable being investigated and uses repeats to improve reliability. [5]
(e) Any three variables to control are starting size or age, species, volume or type of mud, water depth, light intensity, temperature, duration, and mineral-ion concentration or pH of the water. [3]
Answer Details
(a) The exposed root structures contact the air, so oxygen can diffuse into them. Oxygen then moves through air spaces or tissues to the submerged root cells. This improves their oxygen supply in waterlogged mud. [3]
(b) If oxygen is unavailable, root cells carry out anaerobic respiration. In plant cells this produces ethanol and carbon dioxide, and releases less energy. [2]
(c) Anaerobic respiration releases less energy, so less ATP is available for active transport. Consequently, fewer mineral ions can be transported into root hair cells against a concentration gradient. [3]
(d) A valid comparison could be made as follows:
- Use similar young mangrove plants of the same species.
- Place one group with exposed roots and a second group with roots fully covered by waterlogged mud.
- Use several plants in each group.
- Keep both groups for seven days and measure root growth using the same method.
- Test water or root samples for ethanol using the same test, then compare mean results for the groups.
This isolates root exposure as the variable being investigated and uses repeats to improve reliability. [5]
(e) Any three variables to control are starting size or age, species, volume or type of mud, water depth, light intensity, temperature, duration, and mineral-ion concentration or pH of the water. [3]
Question 3 Report
Table 2 shows the number of species recorded by scientists in two habitats on the same farm. Each habitat was searched on five dry mornings in June. The old hedge had not been cut for eight years, whereas the grass verge was mown every three weeks. Organisms were identified using field guides and placed into groups. The results show the variety of animals, plants and fungi in each place.
| group of organisms | old hedge | mown grass verge |
|---|---|---|
| flowering plants | 18 | 7 |
| ferns | 4 | 0 |
| fungi | 11 | 3 |
| insects | 26 | 14 |
| arachnids | 9 | 4 |
| birds | 7 | 5 |
(a) Calculate the total number of species recorded in the old hedge. [2]
(b) Use Table 2 to describe three differences between the variety of organisms in the two habitats. [3]
(c) Explain why the scientists used the same number of searches in each habitat. [3]
(d) Suggest two changes to the management of the grass verge that could increase the variety of organisms there. [4]
(e) Describe two limitations of using these results alone to decide which habitat has the greater biodiversity. [4]
(a)
\[18+4+11+26+9+7=75\]
The total number of species recorded in the old hedge is 75 species. [2 marks]
(b) Three valid comparisons are:
- The hedge has more flowering plant species, 18 compared with 7. [1 mark]
- Ferns were found in the hedge, 4 species, but none were recorded on the verge. [1 mark]
- The hedge has more fungi, 11 compared with 3. [1 mark]
The hedge also has more insects, 26 compared with 14, and more arachnids, 9 compared with 4; bird species are only slightly higher, 7 compared with 5.
(c) Using five searches in each habitat gives equal sampling effort and makes the comparison fair. [1 mark] Differences are then less likely to be caused by one habitat being searched more often. [1 mark] Equal numbers of searches allow totals or mean results to be compared more reliably. [1 mark]
(d) Four suitable changes are: mow less frequently [1 mark]; leave some areas uncut [1 mark]; sow native flowering plants [1 mark]; avoid herbicides and pesticides [1 mark]. Leaving dead wood or leaf litter, creating a pond or damp area, and planting shrubs or a hedge are also acceptable.
(e) The surveys were carried out only in June. [1 mark] Some organisms are seasonal and may not have been present then. [1 mark] Five searches may not represent the entire habitat. [1 mark] Organisms may have been missed or misidentified, and species number does not show the number of individuals of each species. [1 mark] Weather and differences in habitat area are further limitations.
Answer Details
(a)
\[18+4+11+26+9+7=75\]
The total number of species recorded in the old hedge is 75 species. [2 marks]
(b) Three valid comparisons are:
- The hedge has more flowering plant species, 18 compared with 7. [1 mark]
- Ferns were found in the hedge, 4 species, but none were recorded on the verge. [1 mark]
- The hedge has more fungi, 11 compared with 3. [1 mark]
The hedge also has more insects, 26 compared with 14, and more arachnids, 9 compared with 4; bird species are only slightly higher, 7 compared with 5.
(c) Using five searches in each habitat gives equal sampling effort and makes the comparison fair. [1 mark] Differences are then less likely to be caused by one habitat being searched more often. [1 mark] Equal numbers of searches allow totals or mean results to be compared more reliably. [1 mark]
(d) Four suitable changes are: mow less frequently [1 mark]; leave some areas uncut [1 mark]; sow native flowering plants [1 mark]; avoid herbicides and pesticides [1 mark]. Leaving dead wood or leaf litter, creating a pond or damp area, and planting shrubs or a hedge are also acceptable.
(e) The surveys were carried out only in June. [1 mark] Some organisms are seasonal and may not have been present then. [1 mark] Five searches may not represent the entire habitat. [1 mark] Organisms may have been missed or misidentified, and species number does not show the number of individuals of each species. [1 mark] Weather and differences in habitat area are further limitations.
Question 4 Report
A horticulture student crossed two pea plants. One plant had black seed coats and genotype Bb. The other plant had white seed coats and genotype bb. Black is caused by the dominant allele B. Fig. 1 is a partially completed genetic cross used by the student before planting the offspring. The student will count the colours of seeds produced by many plants, because a small number of offspring may not show the expected proportion.
(a) Complete the four boxes in Fig. 1 with the possible genotypes of the offspring. [2]
(b) Use your completed cross to state the probability that one offspring plant has white seed coats. [1]
(c) Describe the expected ratio of black-seeded plants to white-seeded plants in a large sample of offspring. [1]
(d) Explain why a seed with genotype Bb has black seed coats. [2]
(a) The genetic cross is:
| Gamete B | Gamete b | |
|---|---|---|
| Gamete b | Bb | bb |
| Gamete b | Bb | bb |
The two boxes where \(B\) meets \(b\) contain \(Bb\). [1 mark] The two boxes where \(b\) meets \(b\) contain \(bb\). [1 mark]
(b) Two of the four equally likely genotype combinations are \(bb\), which gives white seed coats. Therefore the probability is \(\frac{2}{4}=\frac{1}{2}\), or 50%. [1 mark]
(c) In a large sample, the expected ratio is 1 black : 1 white. [1 mark]
(d) \(B\) is the dominant allele. [1 mark] One copy of \(B\) is sufficient to produce black seed coats, so in \(Bb\) it masks the effect of the recessive \(b\) allele. [1 mark]
Answer Details
(a) The genetic cross is:
| Gamete B | Gamete b | |
|---|---|---|
| Gamete b | Bb | bb |
| Gamete b | Bb | bb |
The two boxes where \(B\) meets \(b\) contain \(Bb\). [1 mark] The two boxes where \(b\) meets \(b\) contain \(bb\). [1 mark]
(b) Two of the four equally likely genotype combinations are \(bb\), which gives white seed coats. Therefore the probability is \(\frac{2}{4}=\frac{1}{2}\), or 50%. [1 mark]
(c) In a large sample, the expected ratio is 1 black : 1 white. [1 mark]
(d) \(B\) is the dominant allele. [1 mark] One copy of \(B\) is sufficient to produce black seed coats, so in \(Bb\) it masks the effect of the recessive \(b\) allele. [1 mark]
Question 5 Report
Fig. 1 shows a variegated leaf used by a teacher to investigate whether light and chlorophyll are needed before plants make starch. The plant was kept in darkness for 24 hours before the test. Opaque paper was then fixed over part of the leaf while the plant was left in bright light. The diagram shows the positions of the paper and the green and white areas before the leaf was removed.
(a) Name the carbohydrate for which the teacher will test the leaf. [1]
(b) Use Fig. 1 to describe how the leaf should be tested for this carbohydrate after it has been in light. [5]
(c) Describe the expected iodine test results for the exposed green region, the covered green region and the exposed white region. [3]
(d) Explain what these results show about the conditions needed for starch production in leaf cells. [3]
(a) The carbohydrate tested for is starch. [1]
(b) After exposure to light, remove the leaf from the plant [1]. Boil it in water [1], then heat it in ethanol in a water bath to remove chlorophyll [1]. Rinse or soften the leaf in water [1], then add iodine solution [1]. Ethanol is heated in a water bath rather than directly because it is flammable. [5]
(c) The exposed green region turns blue-black [1], showing starch. The covered green region remains brown or orange-brown [1], and the exposed white region also remains brown or orange-brown [1]. [3]
(d) Only the region that had both chlorophyll and light contains starch [1]. Green cells contain chlorophyll [1]. Therefore both light and chlorophyll are needed for starch to be made in leaf cells [1]. The initial 24 hours in darkness removes stored starch, so the result tests starch made during the experiment. [3]
Answer Details
(a) The carbohydrate tested for is starch. [1]
(b) After exposure to light, remove the leaf from the plant [1]. Boil it in water [1], then heat it in ethanol in a water bath to remove chlorophyll [1]. Rinse or soften the leaf in water [1], then add iodine solution [1]. Ethanol is heated in a water bath rather than directly because it is flammable. [5]
(c) The exposed green region turns blue-black [1], showing starch. The covered green region remains brown or orange-brown [1], and the exposed white region also remains brown or orange-brown [1]. [3]
(d) Only the region that had both chlorophyll and light contains starch [1]. Green cells contain chlorophyll [1]. Therefore both light and chlorophyll are needed for starch to be made in leaf cells [1]. The initial 24 hours in darkness removes stored starch, so the result tests starch made during the experiment. [3]
Question 6 Report
A dairy cooperative records information from cattle on four farms. Fig. 1 shows a cow being milked by an automatic unit. The table shows the mean results for daughters of four different bulls. The cooperative plans to use semen from selected bulls to improve milk production in its cattle. All cows were given the same type of feed, including silage made from plants, during the study.
| Bull | Number of daughters tested | Mean milk volume / dm3 per day | Daughters treated for mastitis / % |
|---|---|---|---|
| P | 42 | 23.8 | 8 |
| Q | 39 | 25.1 | 26 |
| R | 44 | 24.6 | 6 |
| S | 41 | 22.9 | 5 |
(a) Use Table 1 to name the bull whose daughters had the greatest mean milk volume. [1]
(b) Calculate the difference in mean milk volume between the daughters of bulls Q and S. [2]
(c) Explain why bull Q may not be the best bull to use for selective breeding. [3]
(d) Use Table 1 to suggest which bull the cooperative should select. Explain your answer. [4]
(e) Describe two ways the cooperative could make this comparison more reliable. [2]
(f) Explain why farmers may select cattle for a low chance of mastitis as well as for high milk yield. [2]
(a) Bull Q had daughters with the greatest mean milk volume: \(25.1\text{ dm}^3\) per day. [1]
(b)
\[25.1-22.9=2.2\text{ dm}^3\text{ per day}\]
The difference is \(2.2\text{ dm}^3\) per day. [2]
(c) Although Q's daughters have the highest milk volume, they also have the highest percentage treated for mastitis, \(26\%\). Mastitis is an infection or disease of the udder. It can reduce milk production, increase veterinary and treatment costs, and harm animal welfare, so Q may not be the best breeding choice. [3]
(d) The cooperative should select bull R. Its daughters have a high mean milk volume of \(24.6\text{ dm}^3\) per day and a low mastitis-treatment percentage of \(6\%\). Among the higher-yielding bulls, this is the lowest mastitis value, giving the best balance between high milk yield and health. [4]
(e) Reliability could be improved by testing more daughters and repeating the comparison in different years or on more farms. Keeping feed and living conditions the same, and calculating means from repeat measurements, would also improve the comparison. [2]
(f) Selecting cattle with a low chance of mastitis produces healthier cows and improves welfare. It also reduces treatment costs and avoids loss of milk production caused by disease. [2]
Answer Details
(a) Bull Q had daughters with the greatest mean milk volume: \(25.1\text{ dm}^3\) per day. [1]
(b)
\[25.1-22.9=2.2\text{ dm}^3\text{ per day}\]
The difference is \(2.2\text{ dm}^3\) per day. [2]
(c) Although Q's daughters have the highest milk volume, they also have the highest percentage treated for mastitis, \(26\%\). Mastitis is an infection or disease of the udder. It can reduce milk production, increase veterinary and treatment costs, and harm animal welfare, so Q may not be the best breeding choice. [3]
(d) The cooperative should select bull R. Its daughters have a high mean milk volume of \(24.6\text{ dm}^3\) per day and a low mastitis-treatment percentage of \(6\%\). Among the higher-yielding bulls, this is the lowest mastitis value, giving the best balance between high milk yield and health. [4]
(e) Reliability could be improved by testing more daughters and repeating the comparison in different years or on more farms. Keeping feed and living conditions the same, and calculating means from repeat measurements, would also improve the comparison. [2]
(f) Selecting cattle with a low chance of mastitis produces healthier cows and improves welfare. It also reduces treatment costs and avoids loss of milk production caused by disease. [2]
