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Question 1 Report
3 Fig. 3.1 shows a maize cob with grains of two different colours. In maize, purple grain colour (P) is dominant over yellow grain colour (p). A farmer crosses a homozygous purple-grained plant with a yellow-grained plant to produce an F1 generation.
Fig. 3.1
(a) State the genotype of the yellow-grained parent plant. [1]
(b) State the phenotype of the F1 offspring. [1]
(c) Give the genotype of all F1 plants. [1]
(d) The F1 plants are allowed to cross-pollinate with each other to produce an F2 generation.
(i) State the gametes produced by an F1 plant. [1]
(ii) Draw a Punnett square to show the expected genotypes of the F2 offspring. [2]
(iii) State the expected phenotypic ratio in the F2 generation. [1]
(e) Explain why a farmer might carry out a test cross on a purple-grained maize plant. [2]
(f) Describe how a test cross is carried out and state how the results would show whether the purple-grained plant is homozygous or heterozygous. [3]
Labelled answer diagram:
(a) The genotype of the yellow-grained parent plant is pp. [1] Yellow grain colour is the recessive trait, so the plant must carry two copies of the recessive allele to express it. There is no dominant P allele present to mask the yellow phenotype.
(b) The phenotype of all F1 offspring is purple-grained. [1] Since one parent is homozygous dominant (PP) and the other is homozygous recessive (pp), every F1 individual inherits one P allele and one p allele, making them all Pp. The dominant P allele produces the purple colour.
(c) The genotype of all F1 plants is Pp (heterozygous). [1]
(d)(i) The gametes produced by an F1 plant (Pp) are P and p. [1] During meiosis, the two alleles separate into different gametes, so each gamete carries only one allele.
(d)(ii) Punnett square for the F1 cross (Pp x Pp):
| P | p | |
|---|---|---|
| P | PP | Pp |
| p | Pp | pp |
The gametes P and p are shown across the top (from one parent) and down the side (from the other parent). [1] The four offspring genotypes are correctly placed: PP, Pp, Pp, and pp. [1]
(d)(iii) The expected phenotypic ratio in the F2 generation is 3 purple : 1 yellow (3:1). [1] Three of the four genotypes (PP, Pp, Pp) express the dominant purple phenotype; only pp expresses the recessive yellow phenotype.
(e) A farmer might carry out a test cross on a purple-grained plant because:
(f) How a test cross is carried out and interpreted:
Labelled answer diagram:
(a) The genotype of the yellow-grained parent plant is pp. [1] Yellow grain colour is the recessive trait, so the plant must carry two copies of the recessive allele to express it. There is no dominant P allele present to mask the yellow phenotype.
(b) The phenotype of all F1 offspring is purple-grained. [1] Since one parent is homozygous dominant (PP) and the other is homozygous recessive (pp), every F1 individual inherits one P allele and one p allele, making them all Pp. The dominant P allele produces the purple colour.
(c) The genotype of all F1 plants is Pp (heterozygous). [1]
(d)(i) The gametes produced by an F1 plant (Pp) are P and p. [1] During meiosis, the two alleles separate into different gametes, so each gamete carries only one allele.
(d)(ii) Punnett square for the F1 cross (Pp x Pp):
| P | p | |
|---|---|---|
| P | PP | Pp |
| p | Pp | pp |
The gametes P and p are shown across the top (from one parent) and down the side (from the other parent). [1] The four offspring genotypes are correctly placed: PP, Pp, Pp, and pp. [1]
(d)(iii) The expected phenotypic ratio in the F2 generation is 3 purple : 1 yellow (3:1). [1] Three of the four genotypes (PP, Pp, Pp) express the dominant purple phenotype; only pp expresses the recessive yellow phenotype.
(e) A farmer might carry out a test cross on a purple-grained plant because:
(f) How a test cross is carried out and interpreted:
Question 2 Report
| Method | Time to prepare 1 hectare (hours) | Cost per hectare ($) | Maize yield (tonnes per hectare) |
|---|---|---|---|
| Hand hoe (cutlass and hoe) | 80 | 120 | 2.8 |
| Animal-drawn plough | 20 | 200 | 3.5 |
| Tractor-drawn plough | 4 | 450 | 4.2 |
2 Table 2.1 shows the results of a trial on a farm. Three different methods of land preparation were compared. The farmer recorded the time taken, cost and maize yield for each method.
(a) State which method of land preparation in Table 2.1 took the longest time. [1]
(b) Name one hand tool, other than a cutlass and hoe, that could be used for land preparation. [1]
(c) Describe two advantages of using a tractor-drawn plough compared to a hand hoe for preparing land. [2]
(d) A small-scale farmer has a 2-hectare plot and limited money.
(i) Suggest which method in Table 2.1 would be most suitable for this farmer. Give a reason for your answer. [2]
(ii) State two ways this farmer could reduce the cost of using animal-drawn equipment. [2]
(e) Explain why the plot prepared with a tractor-drawn plough gave the highest maize yield. [2]
(a) The hand hoe method (cutlass and hoe) took the longest time at 80 hours per hectare [1].
(b) One hand tool other than a cutlass and hoe that could be used for land preparation: machete (or rake, spade, shovel, pickaxe, mattock) [1].
(c) Two advantages of using a tractor-drawn plough compared to a hand hoe:
Another acceptable advantage: requires less human physical effort and labour.
(d)(i) For a small-scale farmer with a 2-hectare plot and limited money, the animal-drawn plough would be most suitable [1]. It offers a good balance between cost and speed: it is much faster than the hand hoe (20 hours vs 80 hours per hectare) while costing significantly less than the tractor method ($200 vs $450 per hectare) [1].
(d)(ii) Two ways the farmer could reduce the cost of using animal-drawn equipment:
Other acceptable methods: maintain the implements regularly to avoid costly repairs, and join a cooperative to share costs.
(e) The plot prepared with a tractor-drawn plough gave the highest maize yield because the tractor ploughs deeper and more thoroughly than the other methods [1]. This deeper tillage improves soil aeration, water infiltration, and root penetration, allowing the maize plants to access more nutrients and water from a greater volume of soil [1].
(a) The hand hoe method (cutlass and hoe) took the longest time at 80 hours per hectare [1].
(b) One hand tool other than a cutlass and hoe that could be used for land preparation: machete (or rake, spade, shovel, pickaxe, mattock) [1].
(c) Two advantages of using a tractor-drawn plough compared to a hand hoe:
Another acceptable advantage: requires less human physical effort and labour.
(d)(i) For a small-scale farmer with a 2-hectare plot and limited money, the animal-drawn plough would be most suitable [1]. It offers a good balance between cost and speed: it is much faster than the hand hoe (20 hours vs 80 hours per hectare) while costing significantly less than the tractor method ($200 vs $450 per hectare) [1].
(d)(ii) Two ways the farmer could reduce the cost of using animal-drawn equipment:
Other acceptable methods: maintain the implements regularly to avoid costly repairs, and join a cooperative to share costs.
(e) The plot prepared with a tractor-drawn plough gave the highest maize yield because the tractor ploughs deeper and more thoroughly than the other methods [1]. This deeper tillage improves soil aeration, water infiltration, and root penetration, allowing the maize plants to access more nutrients and water from a greater volume of soil [1].
Question 3 Report
| Crop plant | Petal colour | Scent | Pollen type | Stigma |
|---|---|---|---|---|
| Plant A (Cowpea) | Purple / white | Faint sweet | Sticky, large grains | Small, sticky |
| Plant B (Tomato) | Yellow | Mild | Sticky, moderate | Short, sticky |
| Plant C (Maize) | Green / no petals | None | Light, smooth, abundant | Long silk, feathery |
| Plant D (Mango) | Cream / yellow | Strong sweet | Sticky | Small, sticky |
2 Table 2.1 shows information about the flowers of four different crop plants grown on a mixed farm.
Table 2.1
(a) State which crop plant in Table 2.1 is most likely to be wind-pollinated. Give a reason for your answer. [2]
(b) Name the type of pollination that occurs when pollen from Plant B lands on the stigma of a different flower on the same Plant B. [1]
(c) A farmer wishes to produce hybrid maize seed. Explain how the farmer could carry out controlled cross-pollination between two selected parent plants. [4]
(d) State two advantages of using hybrid seed on the farm. [2]
(a) The crop most likely to be wind-pollinated is Plant C (maize). [1]
Reason: maize has green flowers with no petals, produces no scent, has light smooth abundant pollen grains, and a long feathery stigma (silk). [1] All of these are classic adaptations for wind pollination: no need to attract insects, light pollen carried by air currents, and a large feathery stigma to catch airborne pollen.
(b) When pollen from Plant B (tomato) lands on the stigma of a different flower on the same plant, this is self-pollination. [1] Self-pollination is the transfer of pollen from an anther to a stigma of the same plant, whether within the same flower or between different flowers on the same individual.
(c) How a farmer carries out controlled cross-pollination to produce hybrid maize seed (4 marks):
This ensures the resulting seed carries genetic material from both chosen parents, producing F1 hybrid seed with predictable traits.
(d) Two advantages of using hybrid seed:
Other acceptable answers: improved disease resistance; better quality produce.
(a) The crop most likely to be wind-pollinated is Plant C (maize). [1]
Reason: maize has green flowers with no petals, produces no scent, has light smooth abundant pollen grains, and a long feathery stigma (silk). [1] All of these are classic adaptations for wind pollination: no need to attract insects, light pollen carried by air currents, and a large feathery stigma to catch airborne pollen.
(b) When pollen from Plant B (tomato) lands on the stigma of a different flower on the same plant, this is self-pollination. [1] Self-pollination is the transfer of pollen from an anther to a stigma of the same plant, whether within the same flower or between different flowers on the same individual.
(c) How a farmer carries out controlled cross-pollination to produce hybrid maize seed (4 marks):
This ensures the resulting seed carries genetic material from both chosen parents, producing F1 hybrid seed with predictable traits.
(d) Two advantages of using hybrid seed:
Other acceptable answers: improved disease resistance; better quality produce.
Question 4 Report
4 Fig. 4.1 shows the digestive tract of a non-ruminant animal such as a pig.
(a) Name the parts labelled P, Q and R in Fig. 4.1. [3]
(b) State the main site where protein digestion occurs in the pig. [1]
(c) Give two examples of concentrate feeds that are commonly used in pig rations. [2]
(d) Explain why pigs require a higher proportion of concentrate feed in their diet compared to cattle. [2]
(e) State two signs that a pig is suffering from a protein deficiency in its diet. [2]
(f) Suggest why it is important to provide clean water at all times for pigs in a fattening unit. [1]
Labelled answer diagram:
(a) Parts of the pig digestive tract [3]
(b) Main site of protein digestion in the pig [1]
The stomach (where pepsin begins breaking polypeptide chains) and especially the small intestine (where pancreatic trypsin and chymotrypsin complete digestion to amino acids). [1]
(c) Two concentrate feeds used in pig rations [2]
Any two of:
(d) Why pigs need more concentrate feed than cattle [2]
(e) Two signs of protein deficiency in pigs [2]
Any two of:
(f) Importance of clean water for pigs in a fattening unit [1]
Water is essential for digestion, metabolic processes, and temperature regulation. [1] A lack of clean water reduces feed intake and slows growth, directly undermining the fattening objective.
Labelled answer diagram:
(a) Parts of the pig digestive tract [3]
(b) Main site of protein digestion in the pig [1]
The stomach (where pepsin begins breaking polypeptide chains) and especially the small intestine (where pancreatic trypsin and chymotrypsin complete digestion to amino acids). [1]
(c) Two concentrate feeds used in pig rations [2]
Any two of:
(d) Why pigs need more concentrate feed than cattle [2]
(e) Two signs of protein deficiency in pigs [2]
Any two of:
(f) Importance of clean water for pigs in a fattening unit [1]
Water is essential for digestion, metabolic processes, and temperature regulation. [1] A lack of clean water reduces feed intake and slows growth, directly undermining the fattening objective.
Question 5 Report
5 A farmer notices that one of her fields has very shallow soil with fragments of rock visible just below the surface. The field is located on a hillside where the underlying rock is sandstone.
(a) State two mineral particles that are commonly found in sandstone. [2]
(b) Explain why the soil on this hillside is shallow compared to soil on the flat land at the bottom of the hill. [3]
(c) Name the type of weathering that occurs when minerals in the rock react with acidic rainwater. [1]
(d) Give two ways in which the farmer could improve the depth and fertility of this shallow soil over time. [2]
(a) Two mineral particles commonly found in sandstone:
Mica is also acceptable. Sandstone is a sedimentary rock composed primarily of sand-sized mineral grains cemented together.
(b) The soil on this hillside is shallow compared to soil on the flat land at the bottom of the hill because:
(c) The type of weathering that occurs when minerals in the rock react with acidic rainwater is chemical weathering [1]. Carbonic acid (formed when carbon dioxide dissolves in rainwater) and other acids dissolve susceptible minerals, gradually breaking down the rock structure.
(d) Two ways the farmer could improve the depth and fertility of this shallow soil over time:
Other acceptable methods: plant cover crops or grass to reduce erosion and add organic matter through root decay, and apply mulch to protect the surface and encourage earthworm activity.
(a) Two mineral particles commonly found in sandstone:
Mica is also acceptable. Sandstone is a sedimentary rock composed primarily of sand-sized mineral grains cemented together.
(b) The soil on this hillside is shallow compared to soil on the flat land at the bottom of the hill because:
(c) The type of weathering that occurs when minerals in the rock react with acidic rainwater is chemical weathering [1]. Carbonic acid (formed when carbon dioxide dissolves in rainwater) and other acids dissolve susceptible minerals, gradually breaking down the rock structure.
(d) Two ways the farmer could improve the depth and fertility of this shallow soil over time:
Other acceptable methods: plant cover crops or grass to reduce erosion and add organic matter through root decay, and apply mulch to protect the surface and encourage earthworm activity.
Question 6 Report
1 Fig. 1.1 shows the results of a ribbon test carried out on two soil samples, P and Q, collected from different parts of a farm.
(a) State what is meant by soil texture. [1]
(b) The ribbon formed from sample P was long and smooth. The ribbon from sample Q broke apart quickly and felt gritty.
(i) Name the soil type of sample P. [1]
(ii) Name the soil type of sample Q. [1]
(c) Explain why sample Q drains water faster than sample P. [2]
(d) The farmer wants to grow rice on one part of the farm. Suggest which sample, P or Q, would be more suitable for rice cultivation. Give a reason for your answer. [2]
(e) State one way the farmer could improve the water-holding ability of the soil where sample Q was taken. [1]
Labelled answer diagram:
(a) Meaning of soil texture [1]
Soil texture refers to the relative proportion of different-sized mineral particles - sand, silt, and clay - in a soil sample. [1] It determines many of the soil's physical properties including drainage, water-holding capacity, and workability.
(b)(i) Soil type of sample P [1]
Clay (clay soil). [1] The ribbon test produces a long, smooth ribbon from clay soil because the tiny, flat particles stick together when moist and can be moulded without breaking.
(b)(ii) Soil type of sample Q [1]
Sandy (sandy soil). [1] Sandy soil produces a ribbon that breaks apart quickly and feels gritty because the large, rounded particles do not bind together well.
(c) Why sample Q drains faster than sample P [2]
(d) Which sample is more suitable for rice, and why [2]
Sample P (clay soil) is more suitable. [1] Rice (paddy rice) is grown in flooded fields and requires the soil to retain standing water. Clay soil holds water well because its small particles and narrow pores resist drainage, maintaining the waterlogged conditions that rice needs for healthy growth. [1]
(e) One way to improve water-holding ability of sample Q [1]
Add organic matter (compost, manure, or humus) to the sandy soil. [1] Organic matter acts like a sponge, absorbing and retaining moisture in the large pore spaces between sand grains. Adding clay to the sandy soil is also acceptable.
Labelled answer diagram:
(a) Meaning of soil texture [1]
Soil texture refers to the relative proportion of different-sized mineral particles - sand, silt, and clay - in a soil sample. [1] It determines many of the soil's physical properties including drainage, water-holding capacity, and workability.
(b)(i) Soil type of sample P [1]
Clay (clay soil). [1] The ribbon test produces a long, smooth ribbon from clay soil because the tiny, flat particles stick together when moist and can be moulded without breaking.
(b)(ii) Soil type of sample Q [1]
Sandy (sandy soil). [1] Sandy soil produces a ribbon that breaks apart quickly and feels gritty because the large, rounded particles do not bind together well.
(c) Why sample Q drains faster than sample P [2]
(d) Which sample is more suitable for rice, and why [2]
Sample P (clay soil) is more suitable. [1] Rice (paddy rice) is grown in flooded fields and requires the soil to retain standing water. Clay soil holds water well because its small particles and narrow pores resist drainage, maintaining the waterlogged conditions that rice needs for healthy growth. [1]
(e) One way to improve water-holding ability of sample Q [1]
Add organic matter (compost, manure, or humus) to the sandy soil. [1] Organic matter acts like a sponge, absorbing and retaining moisture in the large pore spaces between sand grains. Adding clay to the sandy soil is also acceptable.
Question 7 Report
3 A farmer harvests 500 kg of maize grain and stores it in a traditional granary made from wooden poles and a grass roof. After three months, the farmer opens some bags and finds small round holes in many of the grains, as well as fine dust at the bottom of the bags. Fig. 3.1 shows a damaged grain with an adult weevil beside it.
(a) State two signs that indicate stored grain has been attacked by insect pests. [2]
(b) Name two insects that are common pests of stored grain. [2]
(c) Explain why grain must be dried to a low moisture content before it is placed in storage. [2]
(d) Suggest two physical methods the farmer could use to protect the stored grain from insect attack. [2]
(e) State one advantage and one disadvantage of using chemical fumigants to treat stored grain. [2]
(f) Give two reasons why rodents are a serious problem in grain stores. [2]
Labelled answer diagram:
(a) Two signs of insect attack in stored grain:
Other acceptable signs: presence of live or dead insects; webbing on the grain surface (from grain moths); unpleasant musty smell; grain feels warm to the touch due to insect metabolic activity.
(b) Two common insects that attack stored grain:
Other acceptable answers: rice weevil (Sitophilus oryzae), flour beetle (Tribolium spp.), grain moth (Sitotroga cerealella).
(c) Why grain must be dried to low moisture before storage:
(d) Two physical methods to protect stored grain from insect attack:
Other acceptable methods: sieving and winnowing to physically remove insects; mixing grain with inert dust (ash, sand) to block air spaces and abrade insect cuticles; raising storage containers off the ground.
(e)
Advantage of fumigants: they can kill all life stages of insects (eggs, larvae, pupae and adults), including those hidden deep inside the grain kernels, and can treat large quantities of grain quickly. [1]
Disadvantage of fumigants: they are highly toxic to humans and animals and require careful handling with specialised equipment and training; improper use can cause serious poisoning. [1]
Other disadvantages: they leave no residual protection so grain can be re-infested after treatment; they may leave chemical residues on the grain.
(f) Two reasons rodents are a serious problem in grain stores:
Other acceptable answers: rodents carry and spread diseases (e.g. leptospirosis, salmonellosis); they reproduce rapidly so a small infestation can grow quickly into a large one.
Labelled answer diagram:
(a) Two signs of insect attack in stored grain:
Other acceptable signs: presence of live or dead insects; webbing on the grain surface (from grain moths); unpleasant musty smell; grain feels warm to the touch due to insect metabolic activity.
(b) Two common insects that attack stored grain:
Other acceptable answers: rice weevil (Sitophilus oryzae), flour beetle (Tribolium spp.), grain moth (Sitotroga cerealella).
(c) Why grain must be dried to low moisture before storage:
(d) Two physical methods to protect stored grain from insect attack:
Other acceptable methods: sieving and winnowing to physically remove insects; mixing grain with inert dust (ash, sand) to block air spaces and abrade insect cuticles; raising storage containers off the ground.
(e)
Advantage of fumigants: they can kill all life stages of insects (eggs, larvae, pupae and adults), including those hidden deep inside the grain kernels, and can treat large quantities of grain quickly. [1]
Disadvantage of fumigants: they are highly toxic to humans and animals and require careful handling with specialised equipment and training; improper use can cause serious poisoning. [1]
Other disadvantages: they leave no residual protection so grain can be re-infested after treatment; they may leave chemical residues on the grain.
(f) Two reasons rodents are a serious problem in grain stores:
Other acceptable answers: rodents carry and spread diseases (e.g. leptospirosis, salmonellosis); they reproduce rapidly so a small infestation can grow quickly into a large one.
Question 8 Report
2 A farmer noticed that the leaves of cassava plants on one side of the field were smaller and showed signs of yellowing, while those on the opposite side were healthy and green.
(a) Name the process by which mineral salts enter the root hair cells from the soil solution. [1]
(b) State two mineral elements that the yellowing plants are most likely lacking. [2]
(c) Explain how water moves from the soil into the root hair cell of the cassava plant. [3]
(d) Give one reason why the plants on one side of the field may have received fewer minerals than those on the other side. [1]
(e) Suggest what the farmer should do to correct the yellowing in the affected plants. [2]
(a) The process by which mineral salts enter root hair cells from the soil solution is active transport. [1] This requires energy from cellular respiration because minerals are moved against their concentration gradient (from a lower concentration in the soil solution to a higher concentration inside the cell).
(b) Two mineral elements that the yellowing plants are most likely lacking:
Other acceptable answers: iron or sulphur, both of which are involved in chlorophyll synthesis or function.
(c) How water moves from the soil into the root hair cell:
(d) One reason plants on one side of the field may have received fewer minerals: the soil on that side may be more sandy, causing nutrients to leach away more quickly with drainage water. [1] Other acceptable reasons: uneven distribution of fertiliser during application; the previous crop on that side removed more nutrients; or drainage patterns differ across the field.
(e) To correct the yellowing, the farmer should:
Foliar application is faster-acting because nutrients are absorbed directly through the leaf epidermis, bypassing the slower root uptake pathway.
(a) The process by which mineral salts enter root hair cells from the soil solution is active transport. [1] This requires energy from cellular respiration because minerals are moved against their concentration gradient (from a lower concentration in the soil solution to a higher concentration inside the cell).
(b) Two mineral elements that the yellowing plants are most likely lacking:
Other acceptable answers: iron or sulphur, both of which are involved in chlorophyll synthesis or function.
(c) How water moves from the soil into the root hair cell:
(d) One reason plants on one side of the field may have received fewer minerals: the soil on that side may be more sandy, causing nutrients to leach away more quickly with drainage water. [1] Other acceptable reasons: uneven distribution of fertiliser during application; the previous crop on that side removed more nutrients; or drainage patterns differ across the field.
(e) To correct the yellowing, the farmer should:
Foliar application is faster-acting because nutrients are absorbed directly through the leaf epidermis, bypassing the slower root uptake pathway.
Question 9 Report
| Layer (from bottom) | Particle type | Depth of layer (mm) |
|---|---|---|
| 1 | Sand | 36 |
| 2 | Silt | 24 |
| 3 | Clay (suspended) | 20 |
4 Table 4.1 shows the results of a sedimentation test carried out on a soil sample collected from a vegetable plot. The farmer shook the sample with water in a measuring cylinder and left it to settle for 24 hours.
(a) State why the soil sample was shaken with water before the test. [1]
(b) Name the soil particle that settled at the bottom of the cylinder first. Give a reason for your answer. [2]
(c) The soil sample contained 45% sand, 30% silt and 25% clay. Name the textural class of this soil. [1]
(d) Explain why this type of soil is considered good for growing vegetables. [3]
(e) State two ways in which the addition of organic matter would improve this soil further. [2]
(a) Why the soil sample was shaken with water [1]
Shaking separates the different-sized particles from each other by breaking up soil aggregates (clumps). [1] This allows each particle type to settle individually according to its size and weight when the cylinder is left undisturbed.
(b) Particle that settled first, and why [2]
Sand settled at the bottom of the cylinder first. [1] Sand particles are the largest and heaviest of the three mineral fractions, so they sink through the water most rapidly under gravity. Silt settles next (medium-sized), and clay particles, being the smallest and lightest, remain suspended longest and settle last at the top. [1]
(c) Textural class of the soil (45% sand, 30% silt, 25% clay) [1]
Loam (loamy soil). [1] A loam contains a relatively balanced proportion of sand, silt, and clay, without any single fraction dominating to the point where it controls all the soil's properties.
(d) Why loam is good for growing vegetables [3]
Any three of:
(e) Two ways organic matter improves this soil [2]
Any two of:
(a) Why the soil sample was shaken with water [1]
Shaking separates the different-sized particles from each other by breaking up soil aggregates (clumps). [1] This allows each particle type to settle individually according to its size and weight when the cylinder is left undisturbed.
(b) Particle that settled first, and why [2]
Sand settled at the bottom of the cylinder first. [1] Sand particles are the largest and heaviest of the three mineral fractions, so they sink through the water most rapidly under gravity. Silt settles next (medium-sized), and clay particles, being the smallest and lightest, remain suspended longest and settle last at the top. [1]
(c) Textural class of the soil (45% sand, 30% silt, 25% clay) [1]
Loam (loamy soil). [1] A loam contains a relatively balanced proportion of sand, silt, and clay, without any single fraction dominating to the point where it controls all the soil's properties.
(d) Why loam is good for growing vegetables [3]
Any three of:
(e) Two ways organic matter improves this soil [2]
Any two of:
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