An original Thinka practice paper modelled on the structure and difficulty of the Nov 2025 (V2) Cambridge IGCSE Agriculture (0600) paper. Not affiliated with or reproduced from Cambridge.
Section A
Answer all questions in the spaces provided.
10 Question · 80 marks
Question 1 · structured
8 marks
(a) Describe how soil colour affects the soil temperature.
(b) Explain why waterlogged soils take longer to warm up in spring compared to well-drained soils.
(c) Suggest three methods a farmer could use to increase or maintain soil temperature in cold climates.
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Worked solution
(a) Darker soils absorb more solar radiation, which increases their temperature, whereas lighter-coloured soils reflect more sunlight and remain cooler.
(b) Water has a high specific heat capacity compared to dry soil particles. This means it requires a large amount of heat energy to raise the temperature of water. Therefore, waterlogged soils require more energy to warm up, and evaporation of excess water also cools the soil.
(c) 1. Applying organic mulches (like straw or leaf litter) or plastic sheeting over the soil. 2. Improving soil drainage (e.g., installing tile drains or digging ditches) to reduce water content. 3. Planting on ridges or raised beds to expose more soil surface to sunlight.
Marking scheme
(a) Max 2 marks: - Dark soils absorb more heat / light (1) - Light soils reflect more heat / light (1)
(b) Max 3 marks: - Water has a high specific heat capacity / requires more energy to heat up than air/soil (1) - Well-drained soils contain more air which heats up faster (1) - Evaporation of water from waterlogged soils has a cooling effect (1)
(c) Max 3 marks (1 mark for each valid method): - Use of plastic mulches / plastic covers (1) - Improving drainage / draining excess water (1) - Raised beds / ridging (1) - Planting windbreaks to reduce cold wind cooling (1)
Question 2 · structured
8 marks
(a) Define the term *transpiration pull*.
(b) Describe how environmental humidity affects the rate of transpiration in plants.
(c) Explain the difference between source and sink in the translocation of carbohydrates.
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Worked solution
(a) Transpiration pull is the suction force created inside the xylem vessels by the evaporation of water vapor from the leaves, which pulls water up from the roots against gravity.
(b) High humidity decreases the rate of transpiration because the water vapor concentration gradient between the inside of the leaf and the surrounding atmosphere is reduced, slowing down diffusion. Low humidity increases the rate.
(c) A source is an area of the plant where carbohydrates are produced (usually mature leaves through photosynthesis) or stored. A sink is an area where carbohydrates are actively consumed or stored (such as growing shoots, roots, fruits, or tubers).
Marking scheme
(a) Max 2 marks: - Evaporation of water from leaf creates a suction force / tension (1) - Pulls water column upward through the xylem (1)
(b) Max 3 marks: - High humidity reduces transpiration rate (1) - Lowers the water vapor concentration gradient between leaf interior and atmosphere (1) - Low humidity increases transpiration rate by steepening this gradient (1)
(c) Max 3 marks: - Source is where sucrose/sugar is produced / loaded into phloem (1) - Sink is where sucrose/sugar is unloaded / used for growth / stored (1) - Example of source (e.g., green leaves) and sink (e.g., roots/fruits) (1)
Question 3 · structured
8 marks
(a) State two reasons why ventilation is essential in a stored grain silo.
(b) A farmer has a concrete silo and a wooden store. Compare these two structures in terms of durability, resistance to pests, and ease of cleaning.
(c) Suggest three features of a modern grain store design that prevent water entry.
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Worked solution
(a) 1. To reduce heat build-up from grain respiration. 2. To lower moisture levels and prevent mold/fungal growth.
(b) Durability: Concrete is highly durable and long-lasting, whereas wood decays over time. Resistance to pests: Concrete is impenetrable to rodents and insects, while wood can be chewed through and harbors insects in cracks. Ease of cleaning: Concrete has smooth, non-porous surfaces that are easy to sweep and wash, while wood is porous and difficult to sterilize.
(c) 1. Overhanging eaves on the roof to shed rain away from walls. 2. A concrete foundation raised above ground level (with a damp-proof membrane). 3. Sealed joints and weatherproof doors/hatches.
Marking scheme
(a) Max 2 marks (1 mark for each point): - Prevents moisture accumulation / condensation (1) - Cools the grain / removes heat of respiration (1) - Prevents mold/fungal growth (1)
(b) Max 3 marks (1 mark for each comparison aspect): - Durability: Concrete lasts much longer / does not rot compared to wood (1) - Pest resistance: Concrete prevents rodent entry / has no cracks for insects, unlike wood (1) - Cleaning: Concrete is easier to wash/sanitize than porous wood (1)
(c) Max 3 marks (1 mark for each feature): - Sloped/pitched metal roof (1) - Overhanging roof eaves (1) - Raised concrete floor slab / damp-proof course (1) - Waterproof sealants on joints (1)
Question 4 · structured
8 marks
(a) State the main function of the rumen in a sheep.
(b) Explain why a non-ruminant animal, such as a pig, cannot digest large quantities of cellulose.
(c) Describe the role of the abomasum in ruminant digestion.
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Worked solution
(a) The rumen acts as a fermentation vat where anaerobic microbes (bacteria, protozoa, and fungi) break down cellulose and fibrous materials into volatile fatty acids.
(b) Non-ruminants like pigs lack a multi-chambered stomach and do not possess the necessary symbiotic microbial population in their stomach/small intestine to produce cellulase enzymes to break down cellulose cell walls.
(c) The abomasum is the 'true' glandular stomach of the ruminant. It secretes hydrochloric acid and digestive enzymes (like pepsin) to digest microbial proteins and any remaining feed bypass proteins chemically.
Marking scheme
(a) Max 2 marks: - Microbial fermentation of cellulose / fiber (1) - Production of volatile fatty acids / microbial protein (1)
(b) Max 3 marks: - Pigs have a simple/single-chambered stomach (1) - Lack the symbiotic microflora (bacteria/protozoa) in the upper digestive tract (1) - Cannot produce the cellulase enzyme required to break down cellulose (1)
(c) Max 3 marks: - Acts as the glandular/true stomach (1) - Secretes gastric juices / hydrochloric acid / pepsin (1) - Chemically digests proteins / microbes coming from previous chambers (1)
Question 5 · structured
8 marks
(a) Distinguish between a maintenance ration and a production ration.
(b) A farmer wants to formulate a feed mix containing 16% crude protein using maize (9% crude protein) and soyabean meal (44% crude protein). Calculate the ratio in which maize and soyabean meal must be mixed using a Pearson square.
(c) State two minerals essential for laying hens and outline their function.
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Worked solution
(a) A maintenance ration provides the minimum nutrients required to keep an animal alive, healthy, and at a constant body weight without gain or loss. A production ration is the extra feed given above maintenance to support productive functions like growth, lactation, pregnancy, or egg laying.
(b) Using the Pearson Square: Center: 16% Top Left (Maize): 9% Bottom Left (Soyabean): 44% Top Right (Soyabean parts): \(|9 - 16| = 7\) parts Bottom Right (Maize parts): \(|44 - 16| = 28\) parts Ratio of Maize to Soyabean is 28 : 7, which simplifies to 4 : 1.
(c) Calcium and Phosphorus. Calcium is essential for eggshell formation and bone strength. Phosphorus works with calcium for bone mineralization.
Marking scheme
(a) Max 2 marks: - Maintenance ration keeps animal at constant weight / supports basic life functions (1) - Production ration provides extra nutrients for milk, meat, eggs, or work (1)
(b) Max 4 marks: - Correct setup of Pearson square values (1) - Calculated parts for maize: 28 parts (1) - Calculated parts for soyabean: 7 parts (1) - Simplified ratio of 4:1 (or 28:7) (1)
(c) Max 2 marks: - Calcium (0.5) for eggshell formation / bone strength (0.5) - Phosphorus (0.5) for bone development / energy metabolism (0.5)
Question 6 · structured
8 marks
(a) Define the terms *heterozygous* and *recessive*.
(b) Use a genetic diagram to show the genotype and phenotype ratios of the offspring from a cross between a heterozygous polled (hornless) ram (Hh) and a horned ewe (hh). Assume the polled allele (H) is dominant over the horned allele (h).
(c) State one economic benefit to a farmer of breeding polled (hornless) sheep instead of horned sheep.
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Worked solution
(a) Heterozygous: Having two different alleles for a particular gene (e.g., Hh). Recessive: An allele that is only expressed in the phenotype when homozygous (e.g., hh).
(c) Polled sheep are safer to handle, cause less physical injury to other livestock in crowded conditions, and do not get their horns caught in fences.
Marking scheme
(a) Max 2 marks: - Heterozygous: having two different alleles of a gene (1) - Recessive: allele only expressed when dominant allele is absent / homozygous state (1)
(b) Max 4 marks: - Correct parental genotypes and gametes shown (1) - Correct Punnett square layout / offspring genotypes (Hh and hh) (1) - Correct genotype ratio 1:1 / 50% Hh and 50% hh (1) - Correct phenotype ratio 1:1 / 50% polled and 50% horned (1)
(c) Max 2 marks (any one valid benefit): - Reduces risk of injury to handlers / other sheep (1) - Reduces carcass bruising / damage to meat quality (1) - Less chance of animals getting trapped in wire fences (1)
Question 7 · structured
8 marks
(a) Explain how weeds reduce crop yields through competition.
(b) Contrast contact herbicides with systemic herbicides in terms of how they work inside the plant.
(c) State two safety precautions a farm worker must take when applying chemical herbicides.
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Worked solution
(a) Weeds compete directly with crops for limited resources. They absorb water and essential soil nutrients, block solar radiation needed for crop photosynthesis, and take up physical space, restricting root and shoot development.
(b) Contact herbicides only kill the parts of the weed that they touch directly, causing localized tissue damage, and do not move within the plant. Systemic herbicides are absorbed by the leaves or roots and are translocated throughout the plant’s vascular system (phloem/xylem) to kill the entire weed, including the roots.
(c) 1. Wear Personal Protective Equipment (PPE) such as a face mask, gloves, and protective overalls. 2. Avoid spraying on windy days to prevent drift onto non-target areas or self-inhalation.
Marking scheme
(a) Max 3 marks: - Compete for soil nutrients / fertilizers (1) - Compete for water in the soil (1) - Compete for sunlight / shade the crop (1) - Compete for space (1)
(b) Max 3 marks: - Contact herbicides kill only tissues touched / do not translocate (1) - Systemic herbicides are absorbed and transported through vascular bundle / phloem / xylem (1) - Systemic herbicides kill the entire plant including roots (1)
(c) Max 2 marks (1 mark for each valid safety precaution): - Wear gloves / face shield / protective boots / overalls (1) - Do not spray against the wind direction / avoid windy days (1) - Wash skin and clothes thoroughly after application (1) - Triple-rinse containers and dispose of safely (1)
Question 8 · structured
8 marks
(a) Explain the purpose of a corner straining post assembly in a wire fence.
(b) Describe how treating wooden fence posts with preservatives prevents decay.
(c) Explain three factors a farmer should consider when choosing the type of fence to construct for livestock.
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Worked solution
(a) A corner straining post anchors the fence and absorbs the high tension exerted by the stretched fence wires. The diagonal strut transfers this lateral tension downward into the ground, preventing the post from leaning or pulling out.
(b) Preservatives (such as creosote or copper-based compounds) penetrate the wood fibers. They act as toxins to wood-boring insects, termites, and fungal microorganisms (rot), thereby preventing biological degradation of the wood.
(c) 1. Type and size of livestock (large animals need taller, stronger posts/wires, while sheep require mesh to prevent escape). 2. Cost and availability of materials (barbed wire is cheaper but might injure valuable animals, while post-and-rail is expensive). 3. Longevity and environmental conditions (resistance to termites, rot, or rusting in humid conditions).
Marking scheme
(a) Max 2 marks: - Anchors and tensions the fence lines (1) - Strut transfers lateral tension to the ground to keep post upright (1)
(c) Max 4 marks (1 mark for identifying a factor, 1 mark for explaining it, max 3 factors): - Class/species of animal: e.g. bulls need heavy posts/strong wire while poultry need small mesh (1) - Cost of materials/labor: e.g. barbed wire is cheaper but electric fences have high initial setup costs (1) - Topography/soil type: e.g. rocky soils make post-digging difficult, suggesting driven posts or electric wiring (1) - Durability/maintenance: e.g. treated wood or metal posts resist rot longer than untreated wood (1)
Question 9 · structured
8 marks
The ruminant digestive system is specially adapted to digest complex plant materials.
(a) State two functions of the rumen in a ruminant animal. [2]
(b) Explain how symbiotic microorganisms within the rumen allow the animal to utilize cellulose from grass. [3]
(c) (i) Name the stomach chamber in a ruminant that corresponds to the simple stomach of a non-ruminant. [1]
(ii) Explain the digestive process that occurs in the chamber named in (c)(i). [2]
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Worked solution
(a) Functions of the rumen include: 1. Acting as a large fermentation chamber where microbes break down complex carbohydrates. 2. Serving as a temporary storage compartment for swallowed vegetation to be regurgitated later.
(b) The resident bacteria, protozoa, and fungi in the rumen produce the enzyme cellulase, which the host animal cannot produce itself. This enzyme breaks down tough plant cellulose into volatile fatty acids (VFAs). These VFAs are absorbed directly through the rumen wall and serve as the animal's primary energy source.
(c) (i) Abomasum (ii) In the abomasum, hydrochloric acid and digestive enzymes are secreted. This acidic environment kills the incoming microorganisms from the forestomachs, allowing them to be digested as high-quality microbial protein.
Marking scheme
(a) Award 1 mark for each valid function of the rumen up to [2]: - Storage of large volumes of feed / grass - Fermentation of cellulose / plant matter by microbes - Production of volatile fatty acids (VFAs) - Physical mixing of feed
(b) Award 1 mark for each point explained up to [3]: - Microbes produce / secrete the enzyme cellulase [1] - Cellulase breaks down plant cellulose into volatile fatty acids (VFAs) / simpler compounds [1] - VFAs are absorbed through the rumen wall to provide energy [1] - Microorganisms act as a microbial protein source when digested later [1]
(c) (i) Award 1 mark for: - Abomasum [1] (Reject: rumen, reticulum, omasum)
(ii) Award 1 mark for each point explained up to [2]: - Secretes gastric juice containing hydrochloric acid (HCl) / enzymes [1] - Chemical breakdown of proteins / digestion of microbes [1]
Question 10 · structured
8 marks
The ruminant digestive system is specially adapted to digest complex plant materials.
(a) State two functions of the rumen in a ruminant animal. [2]
(b) Explain how symbiotic microorganisms within the rumen allow the animal to utilize cellulose from grass. [3]
(c) (i) Name the stomach chamber in a ruminant that corresponds to the simple stomach of a non-ruminant. [1]
(ii) Explain the digestive process that occurs in the chamber named in (c)(i). [2]
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Worked solution
(a) Functions of the rumen include: 1. Acting as a large fermentation chamber where microbes break down complex carbohydrates. 2. Serving as a temporary storage compartment for swallowed vegetation to be regurgitated later.
(b) The resident bacteria, protozoa, and fungi in the rumen produce the enzyme cellulase, which the host animal cannot produce itself. This enzyme breaks down tough plant cellulose into volatile fatty acids (VFAs). These VFAs are absorbed directly through the rumen wall and serve as the animal's primary energy source.
(c) (i) Abomasum (ii) In the abomasum, hydrochloric acid and digestive enzymes are secreted. This acidic environment kills the incoming microorganisms from the forestomachs, allowing them to be digested as high-quality microbial protein.
Marking scheme
(a) Award 1 mark for each valid function of the rumen up to [2]: - Storage of large volumes of feed / grass - Fermentation of cellulose / plant matter by microbes - Production of volatile fatty acids (VFAs) - Physical mixing of feed
(b) Award 1 mark for each point explained up to [3]: - Microbes produce / secrete the enzyme cellulase [1] - Cellulase breaks down plant cellulose into volatile fatty acids (VFAs) / simpler compounds [1] - VFAs are absorbed through the rumen wall to provide energy [1] - Microorganisms act as a microbial protein source when digested later [1]
(c) (i) Award 1 mark for: - Abomasum [1] (Reject: rumen, reticulum, omasum)
(ii) Award 1 mark for each point explained up to [2]: - Secretes gastric juice containing hydrochloric acid (HCl) / enzymes [1] - Chemical breakdown of proteins / digestion of microbes [1]
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Answer any two questions. Each question is worth 15 marks.
2 Question · 30 marks
Question 1 · Essay
15 marks
10 (a) Describe the methods a farmer could use to harvest and store rainwater on a farm. [5]
(b) Compare the advantages and disadvantages of using a drip irrigation system rather than an overhead sprinkler system. [4]
(c) Discuss the consequences of poor soil drainage on crop growth and describe how a farmer can improve drainage. [6]
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Worked solution
(a) Rainwater harvesting and storage methods include using roof gutter systems on farm buildings to direct water into tanks, constructing surface reservoirs or dams, and lining ponds with clay or plastic to prevent water seepage.
(b) Drip irrigation directly applies water to the root zone, reducing evaporative loss, leaf disease, and weed growth compared to overhead sprinklers. However, drip systems have a high installation cost and can easily suffer from clogged emitters.
(c) Poor drainage leads to waterlogging, which fills soil pores with water, depriving roots of oxygen and causing root rot. Farmers can improve drainage by digging open ditches, laying underground perforated pipes, subsoiling to break hardpans, or adding organic matter to clay soils.
Marking scheme
(a) Rainwater harvesting [Max 5 marks]: - Roof catchment / guttering systems on farm structures [1 mark] - Storage tanks (plastic, concrete, or metal) [1 mark] - Construction of farm dams / reservoirs [1 mark] - Lining storage ponds with clay or plastic liners to prevent seepage [1 mark] - Covering tanks or reservoirs to prevent evaporation and contamination [1 mark] - Runoff diversion channels directing surface flow to storage ponds [1 mark]
(b) Drip vs. Sprinkler Comparison [Max 4 marks]: - Advantage: Water applied directly to root zone / minimizes water loss through evaporation [1 mark] - Advantage: Leaves remain dry, reducing foliar fungal diseases [1 mark] - Advantage: Keeps inter-row soil dry, reducing weed germination [1 mark] - Disadvantage: High initial capital / installation costs [1 mark] - Disadvantage: Emitters easily blocked by silt or mineral deposits [1 mark]
(c) Drainage consequences and solutions [Max 6 marks]: Consequences [Max 3 marks]: - Waterlogging excludes oxygen from soil pores [1 mark] - Lack of oxygen prevents root respiration / leads to root rot or plant death [1 mark] - Decreased nutrient uptake because active transport requires energy [1 mark] - Cold soils (high heat capacity of water) which delays germination/growth [1 mark] Solutions [Max 3 marks]: - Digging open surface drainage channels/ditches [1 mark] - Installing subterranean perforated plastic/clay drainage pipes [1 mark] - Adding bulky organic matter to improve soil structure and porosity [1 mark] - Subsoiling to break up impermeable hardpans [1 mark] - Planting crops on raised beds or ridges [1 mark]
Question 2 · Essay
15 marks
11 (a) Describe the differences between a maintenance ration and a production ration, and outline how the nutrient requirements of a high-yielding dairy cow differ from those of a non-lactating mature cow. [5]
(b) Explain the role of micro-organisms in the rumen of a ruminant animal. [4]
(c) Discuss how a farmer can manage pastures to ensure a high-quality forage supply for grazing livestock throughout the dry season. [6]
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Worked solution
(a) A maintenance ration keeps an animal at a constant healthy weight without production, whereas a production ration provides extra nutrients for milk, meat, or pregnancy. High-yielding dairy cows require significantly more water, energy, protein, and minerals (like calcium) than non-lactating cows.
(b) Rumen micro-organisms produce cellulase to ferment plant cell walls into volatile fatty acids for energy. They also synthesize high-quality microbial proteins, B-complex vitamins, and vitamin K.
(c) Farmers can sustain dry-season forage by implementing rotational grazing, conserving wet-season pasture as hay or silage, cultivating drought-tolerant species like alfalfa, irrigating, or utilizing zero-grazing to prevent pasture damage.
Marking scheme
(a) Maintenance vs. Production + Lactation requirements [Max 5 marks]: - Definition of maintenance ration (sustains vital body functions and stable weight) [1 mark] - Definition of production ration (nutrients needed in addition to maintenance for production, e.g., milk/growth/pregnancy) [1 mark] - Dairy cow requires higher protein for milk synthesis [1 mark] - Dairy cow requires higher carbohydrate/energy intake for metabolic demands of milk production [1 mark] - Dairy cow requires increased calcium and phosphorus levels to prevent deficiency/milk fever [1 mark] - Dairy cow requires a much larger daily water intake [1 mark]
(b) Rumen micro-organisms [Max 4 marks]: - Secretion of cellulase to digest cellulose and hemicellulose in plant material [1 mark] - Fermentation of carbohydrates into volatile fatty acids (VFAs) as the primary energy source [1 mark] - Synthesis of microbial proteins which are digested later in the abomasum/small intestine [1 mark] - Synthesis of B-complex vitamins and Vitamin K [1 mark] - Utilization of non-protein nitrogen (e.g. urea) to build amino acids [1 mark]
(c) Pasture management for dry season [Max 6 marks]: - Rotational grazing to allow pasture rest, recovery, and accumulation of biomass [1 mark] - Conserving wet-season surplus grass as silage or hay [1 mark] - Cultivating drought-resistant forage crops or deep-rooted grass/legumes [1 mark] - Irrigation of high-value pasture zones to maintain growth [1 mark] - Application of nitrogen fertilisers before the dry season to maximize late growth [1 mark] - Implementing zero-grazing (cut-and-carry) to minimize waste and trampling damage [1 mark] - Supplementary feeding with concentrates or urea-treated crop residues [1 mark]
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