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Question 1 Report
(a) Name the causative agent of East Coast fever in cattle. [1]
(b) Describe how East Coast fever is transmitted to cattle. [3]
(c) A dairy farmer notices that several cows have swollen lymph nodes, high fever and are losing condition rapidly.
(i) Suggest two other signs the farmer might observe in cattle affected by East Coast fever. [2]
(ii) Explain two preventive measures the farmer should adopt to protect the remaining herd. [4]
(a) The causative agent of East Coast fever is Theileria parva, a protozoan parasite [1].
(b) Transmission of East Coast fever (3 marks):
(c)(i) Two other signs of East Coast fever besides swollen lymph nodes and fever (any two for [2]):
Also acceptable: diarrhoea; loss of appetite; rough coat; anaemia (pale mucous membranes).
(c)(ii) Two preventive measures explained (each for [2], total [4]):
Also acceptable: rotational grazing to reduce tick burdens on pasture; vaccination where available (infection-and-treatment immunisation method).
(a) The causative agent of East Coast fever is Theileria parva, a protozoan parasite [1].
(b) Transmission of East Coast fever (3 marks):
(c)(i) Two other signs of East Coast fever besides swollen lymph nodes and fever (any two for [2]):
Also acceptable: diarrhoea; loss of appetite; rough coat; anaemia (pale mucous membranes).
(c)(ii) Two preventive measures explained (each for [2], total [4]):
Also acceptable: rotational grazing to reduce tick burdens on pasture; vaccination where available (infection-and-treatment immunisation method).
Question 2 Report
A poultry farmer keeps 500 layers in a free-range system. Over a two-week period the farmer records the number of birds showing respiratory symptoms each day.
(a) The diagram shows the respiratory system of a chicken with structures labelled P, Q and R.
(i) Name the structures P, Q and R. [3]
(ii) Explain how Newcastle disease affects the respiratory system of poultry. [3]
(b) Describe how Newcastle disease spreads between birds on a farm. [3]
(c) The farmer wants to protect the remaining healthy birds.
(i) Suggest a vaccination programme the farmer should follow to prevent future outbreaks of Newcastle disease. [3]
(ii) Explain three biosecurity measures, apart from vaccination, that would reduce the risk of disease entering the poultry unit. [3]
Labelled answer diagram:
(a)(i) The structures labelled in the diagram of the chicken respiratory system:
(a)(ii) How Newcastle disease affects the respiratory system (3 marks):
(b) How Newcastle disease spreads between birds (3 marks):
(c)(i) A suitable vaccination programme for Newcastle disease (3 marks):
(c)(ii) Three biosecurity measures apart from vaccination (any three for [3]):
Also acceptable: netting open sides to exclude wild birds; cleaning and disinfecting feeders, drinkers and equipment regularly.
Labelled answer diagram:
(a)(i) The structures labelled in the diagram of the chicken respiratory system:
(a)(ii) How Newcastle disease affects the respiratory system (3 marks):
(b) How Newcastle disease spreads between birds (3 marks):
(c)(i) A suitable vaccination programme for Newcastle disease (3 marks):
(c)(ii) Three biosecurity measures apart from vaccination (any three for [3]):
Also acceptable: netting open sides to exclude wild birds; cleaning and disinfecting feeders, drinkers and equipment regularly.
Question 3 Report
(a) A dairy farmer plans to build a milking parlour for 40 cows.
(i) Name two types of milking parlour layout the farmer could use. [2]
(ii) State three hygiene practices that should be followed during milking. [3]
(b) Explain why the floor of the milking parlour should be made of concrete with a smooth, sloped surface. [3]
(c) Describe how the waste water and manure from the parlour should be managed to prevent pollution. [3]
(d) Suggest two reasons why a milking parlour is more efficient than hand milking in the field. [2]
[Total: 13]
(a)(i) Two types of milking parlour layout [2]:
Other acceptable answers: tandem parlour (cows stand in a line, one behind another) or abreast parlour (cows stand side by side facing the same direction).
(a)(ii) Three hygiene practices during milking [3]:
Also acceptable: milker wears clean gloves/has clean hands; foremilk is discarded or tested with a strip cup to check for mastitis.
(b) Why the floor should be concrete with a smooth, sloped surface [3]:
(c) Waste water and manure management to prevent pollution [3]:
(d) Two reasons a milking parlour is more efficient than hand milking in the field [2]:
Also acceptable: milk can be cooled quickly using in-parlour refrigeration; cows can be fed concentrates during milking, improving nutrition management.
(a)(i) Two types of milking parlour layout [2]:
Other acceptable answers: tandem parlour (cows stand in a line, one behind another) or abreast parlour (cows stand side by side facing the same direction).
(a)(ii) Three hygiene practices during milking [3]:
Also acceptable: milker wears clean gloves/has clean hands; foremilk is discarded or tested with a strip cup to check for mastitis.
(b) Why the floor should be concrete with a smooth, sloped surface [3]:
(c) Waste water and manure management to prevent pollution [3]:
(d) Two reasons a milking parlour is more efficient than hand milking in the field [2]:
Also acceptable: milk can be cooled quickly using in-parlour refrigeration; cows can be fed concentrates during milking, improving nutrition management.
Question 4 Report
(a) State what is meant by the term line breeding. [2]
(b) A pig breeder uses line breeding to improve the litter size of a herd of sows.
(i) Describe how the breeder would carry out line breeding over three generations to increase litter size. [4]
(ii) State one risk associated with line breeding if it is continued for too many generations. [1]
(c) The breeder also wants to improve the feed conversion ratio of the pigs.
The chart shows the feed conversion ratio of five boars tested over a 90-day feeding trial.
Which boar has the best feed conversion ratio? Explain your answer. [2]
(d) Suggest why a farmer might choose to buy hybrid breeding stock from a commercial breeder rather than carrying out a selective breeding programme on the farm. [2]
(e) State two characteristics, other than litter size and feed conversion ratio, that a pig farmer might select for. [2]
(f) Explain one way in which selective breeding of farm animals differs from natural selection. [2]
(a) Line breeding is a form of selective breeding in which animals that are related, but not as closely as parent-offspring or full siblings, are mated together (e.g. half-siblings, cousins, uncle-niece). [1] The aim is to concentrate the genes of a particular outstanding ancestor in the offspring, increasing the likelihood that the ancestor's desirable traits are inherited homozygously. [1]
(b)(i) How the breeder would carry out line breeding over three generations to increase litter size:
(b)(ii) One risk of continuing line breeding for too many generations is inbreeding depression. As relatedness accumulates, harmful recessive alleles become homozygous, leading to reduced fertility, smaller litters, lower disease resistance, and poorer overall performance. [1]
(c) Boar D has the best feed conversion ratio at 2.8. [1] A lower feed conversion ratio means the animal converts feed into body mass more efficiently - it needs fewer kilograms of feed to gain one kilogram of body weight. Therefore Boar D is the most feed-efficient, reducing feed costs per kilogram of pork produced. [1]
(d) A farmer might choose to buy hybrid breeding stock from a commercial breeder rather than running an on-farm programme because:
(e) Two other characteristics a pig farmer might select for:
(f) One way selective breeding differs from natural selection:
In selective breeding, humans deliberately choose which animals are allowed to reproduce, based on traits the farmer considers economically desirable (such as litter size or growth rate). [1] In natural selection, the environment determines which animals survive and reproduce. Traits that aid survival and reproduction in that environment are favoured, regardless of whether they are useful to humans. Selective breeding is therefore directed towards human goals, while natural selection is driven by environmental fitness. [1]
(a) Line breeding is a form of selective breeding in which animals that are related, but not as closely as parent-offspring or full siblings, are mated together (e.g. half-siblings, cousins, uncle-niece). [1] The aim is to concentrate the genes of a particular outstanding ancestor in the offspring, increasing the likelihood that the ancestor's desirable traits are inherited homozygously. [1]
(b)(i) How the breeder would carry out line breeding over three generations to increase litter size:
(b)(ii) One risk of continuing line breeding for too many generations is inbreeding depression. As relatedness accumulates, harmful recessive alleles become homozygous, leading to reduced fertility, smaller litters, lower disease resistance, and poorer overall performance. [1]
(c) Boar D has the best feed conversion ratio at 2.8. [1] A lower feed conversion ratio means the animal converts feed into body mass more efficiently - it needs fewer kilograms of feed to gain one kilogram of body weight. Therefore Boar D is the most feed-efficient, reducing feed costs per kilogram of pork produced. [1]
(d) A farmer might choose to buy hybrid breeding stock from a commercial breeder rather than running an on-farm programme because:
(e) Two other characteristics a pig farmer might select for:
(f) One way selective breeding differs from natural selection:
In selective breeding, humans deliberately choose which animals are allowed to reproduce, based on traits the farmer considers economically desirable (such as litter size or growth rate). [1] In natural selection, the environment determines which animals survive and reproduce. Traits that aid survival and reproduction in that environment are favoured, regardless of whether they are useful to humans. Selective breeding is therefore directed towards human goals, while natural selection is driven by environmental fitness. [1]
Question 5 Report
| Sow | Number of litters per year | Total piglets born alive | Total piglets weaned |
|---|---|---|---|
| A | 2 | 24 | 22 |
| B | 2 | 20 | 18 |
| C | 2 | 22 | 14 |
| D | 1 | 11 | 10 |
(a) A pig farmer wants to improve the reproductive performance of the breeding sows on the farm.
(i) State the average gestation period of a sow. [1]
(ii) A sow produced 10 piglets in her first litter. Calculate how many piglets the farmer could expect from this sow over three years, assuming she farrows twice a year and litter size remains constant. Show your working. [2]
(b) Describe four management practices that help to ensure a high survival rate of piglets in the first week after farrowing. [4]
(c) Explain the process of pregnancy diagnosis in a sow, including two methods that can be used. [4]
(d) Table 3.1 shows the reproductive performance of four sows on the farm over one year.
Table 3.1
(i) Which sow had the best overall reproductive performance? Give a reason for your answer. [2]
(ii) Suggest why Sow C had a lower number of piglets weaned compared to the number born alive. [2]
(a)(i) The average gestation period of a sow is approximately 114 days, often remembered as "3 months, 3 weeks, and 3 days". [1]
(a)(ii) Calculating expected piglets over three years:
Farrowings per year = 2
Over 3 years: \( 2 \times 3 = 6 \) farrowings [1]
Total piglets = \( 6 \times 10 = \mathbf{60} \) piglets [1]
(b) Four management practices to ensure high piglet survival in the first week:
(c) Pregnancy diagnosis determines whether a sow has conceived after mating. Two methods:
(d)(i) Sow A had the best overall reproductive performance. [1] She produced the highest number of piglets weaned (22 from 24 born alive across two litters), meaning 91.7% of her piglets survived to weaning. This combines both high prolificacy (24 born alive) and excellent mothering ability or management conditions (only 2 pre-weaning deaths). [1]
(d)(ii) Sow C had only 14 piglets weaned from 22 born alive (a loss of 8 piglets, or 36%), which is significantly worse than the other sows. Possible reasons include:
(a)(i) The average gestation period of a sow is approximately 114 days, often remembered as "3 months, 3 weeks, and 3 days". [1]
(a)(ii) Calculating expected piglets over three years:
Farrowings per year = 2
Over 3 years: \( 2 \times 3 = 6 \) farrowings [1]
Total piglets = \( 6 \times 10 = \mathbf{60} \) piglets [1]
(b) Four management practices to ensure high piglet survival in the first week:
(c) Pregnancy diagnosis determines whether a sow has conceived after mating. Two methods:
(d)(i) Sow A had the best overall reproductive performance. [1] She produced the highest number of piglets weaned (22 from 24 born alive across two litters), meaning 91.7% of her piglets survived to weaning. This combines both high prolificacy (24 born alive) and excellent mothering ability or management conditions (only 2 pre-weaning deaths). [1]
(d)(ii) Sow C had only 14 piglets weaned from 22 born alive (a loss of 8 piglets, or 36%), which is significantly worse than the other sows. Possible reasons include:
Question 6 Report
Fig. 2.1 shows a cross-section of a hand hoe blade as seen from the side.
(a) On Fig. 2.1, label the following parts of the hoe:
(i) the cutting edge [1]
(ii) the tang [1]
(iii) the socket where the handle fits [1]
(b) The handle of the hoe is made from wood.
(i) State two qualities of wood that make it suitable for tool handles. [2]
(ii) Describe how a farmer should fit a new wooden handle to the hoe head. [2]
(c) Suggest why metal handles are not commonly used for hand hoes. [2]
[Total: 9]
Labelled answer diagram:
(a) The three parts to label on the cross-section of the hoe blade (Fig. 2.1):
(b)(i) Two qualities of wood that make it suitable for tool handles:
(b)(ii) Fitting a new wooden handle to the hoe head:
(c) Metal handles are uncommon for hand hoes because:
Labelled answer diagram:
(a) The three parts to label on the cross-section of the hoe blade (Fig. 2.1):
(b)(i) Two qualities of wood that make it suitable for tool handles:
(b)(ii) Fitting a new wooden handle to the hoe head:
(c) Metal handles are uncommon for hand hoes because:
Question 7 Report
(a) A poultry farmer feeds a flock of broiler chickens on a commercial ration containing maize, soybean meal and grit.
(i) Name the part of the chicken's digestive system where food is temporarily stored before entering the stomach. [1]
(ii) State the function of grit in the diet of poultry. [1]
(iii) Name the muscular organ in which grit is used. [1]
(b) Explain why poultry do not need to chew their food before swallowing. [2]
(c) Describe two ways in which the digestive system of a chicken differs from that of a pig. [4]
(a)(i) The part where food is temporarily stored before entering the stomach is the crop. [1] The crop is a thin-walled expandable pouch in the oesophagus that holds swallowed food and softens it with moisture before it moves into the glandular stomach.
(a)(ii) Grit serves as a grinding agent inside the gizzard: it mechanically crushes and breaks down food, compensating for the fact that poultry have no teeth. [1]
(a)(iii) The muscular organ in which grit is used is the gizzard (also called the ventriculus). [1]
(b) Poultry do not need to chew because they lack teeth entirely. [1] Instead, mechanical breakdown of food is accomplished in the gizzard, where thick muscular walls contract rhythmically to grind food against ingested grit particles. [1] This internal grinding replaces the function that teeth and mastication serve in mammals.
(c) Two differences between the chicken and pig digestive systems (paired comparisons): [4]
Other valid pairs: the chicken has a proventriculus (glandular stomach) that secretes acid and enzymes before the gizzard, while the pig has a single stomach combining these functions; the chicken has two caeca at the small-large intestine junction, while the pig has one caecum.
(a)(i) The part where food is temporarily stored before entering the stomach is the crop. [1] The crop is a thin-walled expandable pouch in the oesophagus that holds swallowed food and softens it with moisture before it moves into the glandular stomach.
(a)(ii) Grit serves as a grinding agent inside the gizzard: it mechanically crushes and breaks down food, compensating for the fact that poultry have no teeth. [1]
(a)(iii) The muscular organ in which grit is used is the gizzard (also called the ventriculus). [1]
(b) Poultry do not need to chew because they lack teeth entirely. [1] Instead, mechanical breakdown of food is accomplished in the gizzard, where thick muscular walls contract rhythmically to grind food against ingested grit particles. [1] This internal grinding replaces the function that teeth and mastication serve in mammals.
(c) Two differences between the chicken and pig digestive systems (paired comparisons): [4]
Other valid pairs: the chicken has a proventriculus (glandular stomach) that secretes acid and enzymes before the gizzard, while the pig has a single stomach combining these functions; the chicken has two caeca at the small-large intestine junction, while the pig has one caecum.
Question 8 Report
4 Wind erosion is a serious problem in semi-arid farming regions.
(a) State three conditions that make wind erosion more likely to occur. [3]
(b) Describe how a windbreak reduces wind erosion on a crop field. [3]
(c) A farmer plants a windbreak of Casuarina trees along the windward side of a 200 m wide field. The windbreak is 8 m tall. Research shows that a windbreak protects a distance of approximately 10 times its height on the leeward side.
(i) Calculate the distance protected by this windbreak. [1]
(ii) Suggest whether this single windbreak adequately protects the entire field. Explain your answer. [2]
[Total: 9]
(a) Three conditions that make wind erosion more likely [3]:
Also acceptable: strong or persistent winds; soil that has been finely cultivated (fine tilth is more vulnerable than cloddy soil).
(b) How a windbreak reduces wind erosion [3]:
(c)(i) Distance protected by the windbreak [1]:
\[ 10 \times 8 = 80 \text{ m} \] [1]
(c)(ii) Whether the single windbreak adequately protects the entire field [2]:
No, the windbreak protects only 80 m, but the field is 200 m wide. [1] The far side of the field (200 - 80 = 120 m from the windbreak) remains exposed to wind erosion. A second windbreak approximately 80 m from the first, or additional protective measures such as cover crops or mulching, would be needed to protect the entire field. [1]
(a) Three conditions that make wind erosion more likely [3]:
Also acceptable: strong or persistent winds; soil that has been finely cultivated (fine tilth is more vulnerable than cloddy soil).
(b) How a windbreak reduces wind erosion [3]:
(c)(i) Distance protected by the windbreak [1]:
\[ 10 \times 8 = 80 \text{ m} \] [1]
(c)(ii) Whether the single windbreak adequately protects the entire field [2]:
No, the windbreak protects only 80 m, but the field is 200 m wide. [1] The far side of the field (200 - 80 = 120 m from the windbreak) remains exposed to wind erosion. A second windbreak approximately 80 m from the first, or additional protective measures such as cover crops or mulching, would be needed to protect the entire field. [1]
Question 9 Report
(a) Fig. 1.1 shows the nitrogen cycle in an agricultural system.
Fig. 1.1
(i) Name the type of bacteria found in the root nodules of legumes that convert atmospheric nitrogen into a form usable by plants. [1]
(ii) State the name of the process carried out by these bacteria. [1]
(iii) Name the process by which nitrates in the soil are converted back to atmospheric nitrogen. [1]
(b) Explain how growing a legume crop such as groundnuts and then ploughing it into the soil benefits the next crop planted in that field. [3]
(c) State two ways in which nitrogen can be lost from agricultural soils. [2]
[Total: 8]
Labelled answer diagram:
(a)(i) The bacteria found in the root nodules of legumes that convert atmospheric nitrogen into a plant-usable form are Rhizobium. [1]
(a)(ii) The process is called nitrogen fixation (biological nitrogen fixation). [1] Rhizobium bacteria use the enzyme nitrogenase to break the strong triple bond in atmospheric nitrogen gas (\(\text{N}_2\)) and convert it into ammonia (\(\text{NH}_3\)), which is then available to the host plant.
(a)(iii) The process by which nitrates in the soil are converted back to atmospheric nitrogen is denitrification. [1] This is carried out by anaerobic bacteria in waterlogged or poorly aerated soils, and it represents a loss of available nitrogen from the soil.
(b) How growing and ploughing in a legume crop benefits the next crop: [3]
(c) Two ways nitrogen can be lost from agricultural soils (any two): [2]
Other acceptable answers: denitrification by anaerobic bacteria in waterlogged soils; erosion of nitrogen-rich topsoil by wind or water; volatilisation of ammonia from surface-applied manure or urea fertiliser.
Labelled answer diagram:
(a)(i) The bacteria found in the root nodules of legumes that convert atmospheric nitrogen into a plant-usable form are Rhizobium. [1]
(a)(ii) The process is called nitrogen fixation (biological nitrogen fixation). [1] Rhizobium bacteria use the enzyme nitrogenase to break the strong triple bond in atmospheric nitrogen gas (\(\text{N}_2\)) and convert it into ammonia (\(\text{NH}_3\)), which is then available to the host plant.
(a)(iii) The process by which nitrates in the soil are converted back to atmospheric nitrogen is denitrification. [1] This is carried out by anaerobic bacteria in waterlogged or poorly aerated soils, and it represents a loss of available nitrogen from the soil.
(b) How growing and ploughing in a legume crop benefits the next crop: [3]
(c) Two ways nitrogen can be lost from agricultural soils (any two): [2]
Other acceptable answers: denitrification by anaerobic bacteria in waterlogged soils; erosion of nitrogen-rich topsoil by wind or water; volatilisation of ammonia from surface-applied manure or urea fertiliser.
Question 10 Report
| Bull | Number of calves | Average daily weight gain / kg |
|---|---|---|
| A | 10 | 0.95 |
| B | 10 | 1.10 |
| C | 10 | 1.35 |
| D | 10 | 1.05 |
(a) The table shows the average daily weight gain of calves sired by four different bulls over a 200-day growth trial. Each bull was mated with a similar group of ten cows.
(i) State which bull should be selected for future breeding. Give a reason for your answer. [2]
(ii) Name the type of test being carried out in this trial. [1]
(iii) Suggest why each bull was mated with a similar group of cows rather than all with the same cows. [1]
(b) The farmer decides to use artificial insemination (AI) with semen from the selected bull.
State two advantages of using AI instead of keeping the bull on the farm for natural mating. [2]
(c) Explain why the farmer should introduce an unrelated bull or use semen from a different bloodline after several generations of selection. [3]
(a)(i) Bull C should be selected for future breeding because its calves achieved the highest average daily weight gain of 1.35 kg. [1] This demonstrates that Bull C transmits genes for superior growth rate to its offspring, which will translate into heavier animals at market age, higher carcass weights, and greater profitability. [1]
(a)(ii) The type of test being carried out is a progeny test. [1] It evaluates the breeding value of each bull based on the measured performance of its offspring rather than the bull's own characteristics.
(a)(iii) Each bull was mated with a similar but separate group of cows rather than the same cows to ensure the comparison is fair. [1] If all bulls mated the same cows sequentially, differences in the cows' age, condition, or milk production between matings could confound the results. Using similar groups ensures that any differences in calf performance are attributable to the bull's genetics, not the dam's.
(b) Two advantages of using AI instead of keeping the bull on the farm:
(c) After several generations of selecting from the same genetic line, the farmer should introduce unrelated genetics because:
(a)(i) Bull C should be selected for future breeding because its calves achieved the highest average daily weight gain of 1.35 kg. [1] This demonstrates that Bull C transmits genes for superior growth rate to its offspring, which will translate into heavier animals at market age, higher carcass weights, and greater profitability. [1]
(a)(ii) The type of test being carried out is a progeny test. [1] It evaluates the breeding value of each bull based on the measured performance of its offspring rather than the bull's own characteristics.
(a)(iii) Each bull was mated with a similar but separate group of cows rather than the same cows to ensure the comparison is fair. [1] If all bulls mated the same cows sequentially, differences in the cows' age, condition, or milk production between matings could confound the results. Using similar groups ensures that any differences in calf performance are attributable to the bull's genetics, not the dam's.
(b) Two advantages of using AI instead of keeping the bull on the farm:
(c) After several generations of selecting from the same genetic line, the farmer should introduce unrelated genetics because:
Question 11 Report
| Ear type | Number of F2 piglets |
|---|---|
| Erect ears | 34 |
| Floppy ears | 14 |
| Total | 48 |
(a) In a breed of pig, the allele for erect ears (E) is dominant over the allele for floppy ears (e).
A pig farmer crossed a homozygous erect-eared boar with a floppy-eared sow. All of the F1 piglets had erect ears.
(i) State the genotypes of the boar and the sow. [2]
(ii) State the genotype of the F1 piglets. [1]
(iii) Two F1 piglets were crossed. Using a Punnett square, determine the expected genotype and phenotype ratios of the F2 generation. [4]
(b) The farmer recorded the ear type of 48 F2 piglets. The results are shown in the table.
(i) Calculate the ratio of erect-eared to floppy-eared piglets. Show your working. [2]
(ii) Suggest why this ratio is not exactly 3 : 1. [1]
(c) Explain what is meant by the term heterozygous. [1]
(d) Describe how a farmer could use a test cross to find out whether an erect-eared pig is homozygous or heterozygous. [4]
(a)(i) The genotypes of the parents are: boar = EE (homozygous dominant) [1]; sow = ee (homozygous recessive). [1] The boar is stated to be homozygous erect-eared, and the sow has floppy ears which can only occur with the genotype ee.
(a)(ii) The genotype of the F1 piglets is Ee (heterozygous). [1] Each F1 piglet inherits one E allele from the EE father and one e allele from the ee mother. All F1 piglets have erect ears because the dominant E allele masks the recessive e.
(a)(iii) Crossing two F1 piglets (Ee x Ee):
Parents: Ee x Ee
Gametes from each parent: E and e
| E | e | |
|---|---|---|
| E | EE | Ee |
| e | Ee | ee |
Offspring genotypes: EE, Ee, Ee, ee [1]
Expected genotype ratio: 1 EE : 2 Ee : 1 ee [1]
Expected phenotype ratio: 3 erect ears : 1 floppy ears [1] (EE and Ee both show the dominant erect-ear phenotype; only ee shows floppy ears.) [1]
(b)(i) Calculating the ratio of erect-eared to floppy-eared piglets from the data:
Erect ears = 34, Floppy ears = 14
\[ \text{Ratio} = \frac{34}{14} : \frac{14}{14} = 2.4 : 1 \] [1] for method (dividing both by the smaller number), [1] for arriving at approximately 2.4 : 1.
(b)(ii) The ratio is not exactly 3:1 because fertilisation is a random event, and the sample size (48 piglets) is relatively small. [1] Chance variation means the actual numbers will fluctuate around the expected ratio; a perfect 3:1 would require 36 erect : 12 floppy, but random sampling produced 34:14 instead.
(c) Heterozygous means having two different alleles for a particular gene - one dominant and one recessive (e.g. Ee). [1] The dominant allele determines the phenotype in a heterozygous individual.
(d) How to use a test cross to determine whether an erect-eared pig is homozygous or heterozygous:
(a)(i) The genotypes of the parents are: boar = EE (homozygous dominant) [1]; sow = ee (homozygous recessive). [1] The boar is stated to be homozygous erect-eared, and the sow has floppy ears which can only occur with the genotype ee.
(a)(ii) The genotype of the F1 piglets is Ee (heterozygous). [1] Each F1 piglet inherits one E allele from the EE father and one e allele from the ee mother. All F1 piglets have erect ears because the dominant E allele masks the recessive e.
(a)(iii) Crossing two F1 piglets (Ee x Ee):
Parents: Ee x Ee
Gametes from each parent: E and e
| E | e | |
|---|---|---|
| E | EE | Ee |
| e | Ee | ee |
Offspring genotypes: EE, Ee, Ee, ee [1]
Expected genotype ratio: 1 EE : 2 Ee : 1 ee [1]
Expected phenotype ratio: 3 erect ears : 1 floppy ears [1] (EE and Ee both show the dominant erect-ear phenotype; only ee shows floppy ears.) [1]
(b)(i) Calculating the ratio of erect-eared to floppy-eared piglets from the data:
Erect ears = 34, Floppy ears = 14
\[ \text{Ratio} = \frac{34}{14} : \frac{14}{14} = 2.4 : 1 \] [1] for method (dividing both by the smaller number), [1] for arriving at approximately 2.4 : 1.
(b)(ii) The ratio is not exactly 3:1 because fertilisation is a random event, and the sample size (48 piglets) is relatively small. [1] Chance variation means the actual numbers will fluctuate around the expected ratio; a perfect 3:1 would require 36 erect : 12 floppy, but random sampling produced 34:14 instead.
(c) Heterozygous means having two different alleles for a particular gene - one dominant and one recessive (e.g. Ee). [1] The dominant allele determines the phenotype in a heterozygous individual.
(d) How to use a test cross to determine whether an erect-eared pig is homozygous or heterozygous:
Question 12 Report
7 Fig. 7.1 shows a section of farmland where sheet erosion has removed the topsoil, leaving a lighter-coloured subsoil exposed.
(a) Describe what is meant by the term sheet erosion. [2]
(b) Explain why sheet erosion is often more difficult for a farmer to detect than gully erosion. [2]
(c) State three ways in which losing topsoil through sheet erosion reduces crop yields. [3]
[Total: 7]
Labelled answer diagram:
(a) Definition of sheet erosion [2]:
Sheet erosion is the uniform removal of a thin layer of topsoil over a wide area by shallow overland flow. [1] It is caused by rainwater that flows as a broad, shallow sheet rather than concentrating into defined channels or rills. [1]
(b) Why sheet erosion is more difficult to detect than gully erosion [2]:
(c) Three ways losing topsoil through sheet erosion reduces crop yields [3]:
Also acceptable: applied fertilisers are washed away before crops absorb them; seed and young seedlings may be displaced or buried by moving soil.
Labelled answer diagram:
(a) Definition of sheet erosion [2]:
Sheet erosion is the uniform removal of a thin layer of topsoil over a wide area by shallow overland flow. [1] It is caused by rainwater that flows as a broad, shallow sheet rather than concentrating into defined channels or rills. [1]
(b) Why sheet erosion is more difficult to detect than gully erosion [2]:
(c) Three ways losing topsoil through sheet erosion reduces crop yields [3]:
Also acceptable: applied fertilisers are washed away before crops absorb them; seed and young seedlings may be displaced or buried by moving soil.
Question 13 Report
| Farm animal | Gestation period (days) | Typical litter size |
|---|---|---|
| Cattle | 283 | 1 |
| Sheep | 150 | 1 - 3 |
| Goat | 150 | 1 - 3 |
| Pig | 114 | 8 - 14 |
(a) Table 1.1 shows the gestation periods and typical litter sizes for four farm animals.
Table 1.1
(i) Using Table 1.1, state which animal has the longest gestation period. [1]
(ii) Calculate the difference in gestation period between cattle and sheep. Show your working. [2]
(iii) Suggest why pigs produce larger litters than cattle. [2]
(b) Explain why a farmer should keep accurate breeding records for the herd. [3]
(a)(i) The animal with the longest gestation period is cattle, at 283 days. [1] This is read directly from the table; sheep and goats both have 150 days, and pigs have 114 days.
(a)(ii) The difference in gestation period between cattle and sheep:
\[ 283 - 150 = 133 \text{ days} \]
Cattle gestation = 283 days [1]; sheep gestation = 150 days; difference = 133 days. [1]
(a)(iii) Pigs produce larger litters than cattle because:
(b) A farmer should keep accurate breeding records for the following reasons:
(a)(i) The animal with the longest gestation period is cattle, at 283 days. [1] This is read directly from the table; sheep and goats both have 150 days, and pigs have 114 days.
(a)(ii) The difference in gestation period between cattle and sheep:
\[ 283 - 150 = 133 \text{ days} \]
Cattle gestation = 283 days [1]; sheep gestation = 150 days; difference = 133 days. [1]
(a)(iii) Pigs produce larger litters than cattle because:
(b) A farmer should keep accurate breeding records for the following reasons:
Question 14 Report
(a) Explain what is meant by green manuring. [2]
(b) A farmer in a tropical region grows cowpeas as a green manure crop before planting cassava.
(i) Describe two benefits that the cowpea crop provides to the soil before it is ploughed in. [2]
(ii) Explain why ploughing in the cowpea plants at the flowering stage is more beneficial than waiting until the seeds have matured. [2]
(c) Apart from green manuring, suggest two other methods a farmer can use to add organic matter to the soil. [2]
[Total: 8]
(a) Green manuring is the practice of growing a crop (usually a legume) specifically for the purpose of ploughing it back into the soil while the plants are still green and actively growing, [1] in order to add organic matter and nutrients to the soil and improve its fertility. [1]
(b)(i) Two benefits the cowpea crop provides to the soil before it is ploughed in (any two): [2]
(b)(ii) Why ploughing in at the flowering stage is more beneficial than waiting until seed maturity: [2]
(c) Two other methods of adding organic matter to the soil, apart from green manuring (any two): [2]
Other acceptable answers: mulching with crop residues; applying biosolids / sewage sludge.
(a) Green manuring is the practice of growing a crop (usually a legume) specifically for the purpose of ploughing it back into the soil while the plants are still green and actively growing, [1] in order to add organic matter and nutrients to the soil and improve its fertility. [1]
(b)(i) Two benefits the cowpea crop provides to the soil before it is ploughed in (any two): [2]
(b)(ii) Why ploughing in at the flowering stage is more beneficial than waiting until seed maturity: [2]
(c) Two other methods of adding organic matter to the soil, apart from green manuring (any two): [2]
Other acceptable answers: mulching with crop residues; applying biosolids / sewage sludge.
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