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100+ Free SACE Stage 2 Agricultural Production Practice Questions
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2026 Statistics
Key Facts: SACE Stage 2 Agricultural Production Exam
70%
Weighting of School-based Assessment in SACE Stage 2 Agricultural Production
SACE Board Subject Outline
30%
Weighting of External Assessment (Agricultural Enterprise Investigation)
SACE Board Subject Outline
5 Core Areas
Plant Systems, Animal Systems, Soil Science, Pest Management, and Agribusiness
SACE Stage 2 Outline
C- Grade
Minimum satisfactory achievement threshold for SACE accreditation
SACE Board Grading Scale
SACE Stage 2 Agricultural Production evaluates student knowledge of plant and animal production systems, soil science, pest management, and agribusiness economics under the SACE Board of South Australia. This 100-question prep module offers targeted multiple-choice practice covering key subject outline topics and quantitative calculations.
Sample SACE Stage 2 Agricultural Production Practice Questions
Try these sample questions to test your SACE Stage 2 Agricultural Production exam readiness. Each question includes a detailed explanation. Start the interactive quiz above for the full 100+ question experience with AI tutoring.
1Which plant enzyme directly fixes atmospheric carbon dioxide during the Calvin cycle of C3 crops such as wheat and barley?
A.Phosphoenolpyruvate carboxylase (PEP carboxylase)
B.Ribulose-1,5-bisphosphate carboxylase-oxygenase (RuBisCO)
C.Nitrate reductase
D.ATP synthase
Explanation: RuBisCO is the primary enzyme responsible for catalyzing the initial step of carbon fixation in C3 plants by combining CO2 with ribulose-1,5-bisphosphate. PEP carboxylase is used by C4 and CAM plants for initial CO2 fixation, nitrate reductase converts nitrate to nitrite, and ATP synthase synthesizes ATP during photophosphorylation.
2During periods of high daytime temperatures and drought stress, why do C3 crops experience a significant decline in photosynthetic efficiency compared to C4 crops?
A.RuBisCO oxygenase activity increases relative to carboxylase activity, leading to photorespiration
B.Stomata open wider to increase evaporative cooling, depleting cellular ATP stores
C.Chlorophyll b degrades rapidly, preventing light absorption in photosystem II
D.PEP carboxylase becomes irreversibly denatured at temperatures above 25°C
Explanation: In C3 plants, stomatal closure under drought stress reduces internal CO2 concentrations. At elevated temperatures, RuBisCO binds oxygen instead of CO2, initiating photorespiration which consumes energy and releases fixed carbon without producing ATP.
3A grower needs to apply 92 kg of elemental nitrogen (N) per hectare to a canola crop prior to stem elongation. If using Urea (46% N), how many kilograms of Urea fertilizer must be applied per hectare?
A.150 kg/ha
B.200 kg/ha
C.250 kg/ha
D.423 kg/ha
Explanation: To calculate the fertilizer application rate: Required Fertilizer = Target Element Rate / (% Element / 100). Here, 92 kg N / 0.46 = 200 kg of Urea per hectare.
4Which plant macronutrient plays a central role in stomatal regulation, osmoregulation, and enzyme activation, and is frequently deficient in sandy South Australian soils?
A.Phosphorus (P)
B.Potassium (K)
C.Calcium (Ca)
D.Molybdenum (Mo)
Explanation: Potassium (K+) is essential for maintaining turgor pressure in guard cells to control stomatal opening and closing, regulating cellular water balance, and activating metabolic enzymes. Sandy soils with low cation exchange capacity are prone to potassium leaching.
5Monoammonium Phosphate (MAP) fertilizer has an N:P:K analysis of 10:22:0. If an agronomic recommendation specifies 22 kg of elemental phosphorus (P) per hectare at sowing, how much nitrogen (N) is applied simultaneously when MAP is applied at the required rate?
A.10 kg N/ha
B.22 kg N/ha
C.44 kg N/ha
D.100 kg N/ha
Explanation: MAP contains 10% N and 22% P. To apply 22 kg P/ha, the MAP application rate is 22 / 0.22 = 100 kg MAP/ha. At 100 kg MAP/ha, the amount of nitrogen delivered is 100 * 0.10 = 10 kg N/ha.
6What phenomenon causes F1 hybrid grain crops (such as hybrid canola or maize) to exhibit superior yield, vigor, and stress tolerance compared to their pure-breeding parent lines?
A.Inbreeding depression
B.Heterosis (hybrid vigor)
C.Genetic drift
D.Somaclonal variation
Explanation: Heterosis, or hybrid vigor, describes the performance superiority of F1 heterozygous offspring over the average or best of their homozygous inbred parents, resulting from maskings of deleterious recessive alleles and beneficial gene interactions.
7Why must commercial growers purchase fresh certified F1 hybrid seed each season rather than keeping saved seed from the harvested F1 crop?
A.F2 grain is genetically sterile and cannot germinate under field conditions
B.Genetic segregation in the F2 generation results in high phenotypic variability and loss of hybrid vigor
C.F2 seed automatically reverts to the maternal parent phenotype without expressing paternal traits
D.Saved seed accumulates viral toxins that permanently suppress seedling germination
Explanation: According to Mendelian genetics, saving seed from an F1 hybrid produces an F2 generation that segregate for heterozygous and homozygous combinations. This leads to unpredictable plant height, uneven maturity, variable yield, and loss of uniform hybrid vigor.
8A grain sample has a germination test result of 95% and a seed purity rating of 98%. What is the Pure Live Seed (PLS) percentage of this batch?
A.93.1%
B.95.0%
C.96.5%
D.98.0%
Explanation: Pure Live Seed percentage is calculated as: PLS % = (% Germination * % Purity) / 100. Substituting the values: (95 * 98) / 100 = 93.1%. This ensures growers adjust seeding rates for non-viable seeds and inert matter.
9If a pasture seed batch has a Pure Live Seed (PLS) rating of 80% and the target planting rate of pure live seed is 10 kg/ha, what gross seeding rate of the bulk seed batch must be sown per hectare?
A.8.0 kg/ha
B.10.0 kg/ha
C.12.5 kg/ha
D.18.0 kg/ha
Explanation: Gross Seeding Rate = Target PLS Rate / (PLS % / 100) = 10 kg / 0.80 = 12.5 kg of bulk seed per hectare to achieve the desired establishment density.
10Which plant hormone is responsible for promoting apical dominance, cell elongation, and adventitious root initiation in horticultural crops?
A.Abscisic acid (ABA)
B.Auxin (Indole-3-acetic acid / IAA)
C.Ethylene
D.Gibberellin
Explanation: Auxins (such as IAA and synthetic IBA/NAA) drive cell elongation, maintain apical dominance by inhibiting lateral bud growth, and stimulate root formation on stem cuttings.
About the SACE Stage 2 Agricultural Production Practice Questions
Verified exam format metadata for SACE Stage 2 Agricultural Production External Assessment is pending. The practice questions above remain available while official exam length, timing, passing score, fee, and administrator details are reviewed.