S6 Biology II 2025 Past Paper: Full Worked Solutions

These are the complete worked solutions for the S6 Biology II  NESA National Examination, 2024–2025, sat on 16 July 2025. This paper applies to combinations BCG, MCB, PCB, and ANP.

Every question is rewritten in full, followed immediately by the answer and a clear explanation. Read each solution carefully  understanding the reasoning is what prepares you for the next paper, not just memorising the letter.

If any topic here needs deeper work, book a Biology tutor on Mathrone Academy. You can also revise alongside the S6 Chemistry II 2025 worked solutions and the S6 Physics 2025 worked solutions to cover your full combination.

TO GO ALONG WITH ANSWER, DOWNLOAD THE QUESTIONPAPER HERE

Section A: Attempt ALL Questions (70 marks)

Question 1 (1 mark)

Which of the following is the correct order of taxonomic ranks from highest to lowest?

• a) Kingdom > Phylum > Family > Order > Species
• b) Domain > Kingdom > Class > Species > Order
• c) Domain > Kingdom > Phylum > Class > Order
• d) Phylum > Kingdom > Class > Family > Species

 Answer: c) Domain > Kingdom > Phylum > Class > Order

The full correct hierarchy from highest (broadest) to lowest (most specific) is: Domain → Kingdom → Phylum → Class → Order → Family → Genus → Species. Option c is the only one that lists ranks in the correct descending order without skipping or misplacing any. Option a skips Domain and places Order after Family. Option b places Species above Order, which is wrong. Option d places Phylum above Kingdom.

Question 2 (1 mark)

What term is used to describe the number of individuals of a species per unit area or volume?

• a) Population size
• b) Habitat
• c) Community structure
• d) Population density

Answer: d) Population density

Population density is defined as the number of individuals of a given species per unit area (for terrestrial organisms) or unit volume (for aquatic organisms). Population size is simply the total count of individuals with no reference to space. Habitat refers to the physical environment where a species lives. Community structure describes the composition and organisation of multiple species in an area.

Question 3 (1 mark)

How might conventional breeding lead to the formation of giraffes with long necks from those with short necks?

• a) By introducing random genetic mutations
• b) By selecting and reproducing the animals with favourable traits
• c) By altering the gene sequence in the body cells
• d) By introducing new species into the population

Answer: b) By selecting and reproducing the animals with favourable traits

Conventional (selective) breeding works by identifying individuals with a desired trait, in this case, longer necks — and using only those individuals for reproduction. Over successive generations, the frequency of the long-neck alleles increases in the population. This mirrors natural selection but is directed by human choice. Introducing mutations (a) describes mutagenesis, not conventional breeding. Altering gene sequences (c) describes genetic engineering. Introducing new species (d) is irrelevant to selective breeding.

Question 4 (1 mark)

If an image of a bacterium appears 50 times larger than its actual size, what is the magnification of the image?

• a) 5×
• b) 5000×
• c) 50×
• d) 500×

Answer: c) 50×

Magnification is calculated as:

$$\text{Magnification}=\frac{\text{Image size}}{\text{Actual size}}$$

If the image appears 50 times larger than the actual object, the magnification is simply 50×. No conversion is needed here because the question directly states the image is 50 times larger, the magnification is 50×.

Question 5: Enzyme digestion table (6 marks)

Complete the table by choosing appropriate words from the list: Amino acids, amylase, cellulose, fatty acids, hydrochloric acid, lipase, protein, starch, water.

Explanations:

• Fat → Lipase → Glycerol + Fatty acids: Lipase is produced by the pancreas and small intestine. It hydrolyses (breaks) the ester bonds in triglycerides (fats), producing one glycerol molecule and three fatty acid molecules.
• Protein → Protease → Amino acids: Proteases (e.g. pepsin in the stomach, trypsin in the small intestine) cleave peptide bonds between amino acids, releasing individual amino acids that are absorbed into the bloodstream.
• Starch → Amylase → Maltose: Amylase (produced in the salivary glands and pancreas) hydrolyses the glycosidic bonds in starch, producing the disaccharide maltose. Maltose is then further broken down to glucose by maltase.

Note: Cellulose, hydrochloric acid, and water were distractors. Cellulose is a structural carbohydrate not digested by humans (no cellulase enzyme). HCl acidifies the stomach to activate pepsinogen but is not an enzyme. Water participates in hydrolysis reactions but is not the enzyme.

Question 6:  Growth and Development: True or False (5 marks)

State whether each of the following statements is True or False.

Question 7 (1 mark)

Which of these organisms uses cilia for locomotion?

• a) Earthworm
• b) Paramecium
• c) Grasshopper
• d) Snail

Answer: b) Paramecium

Paramecium is a single-celled protist covered in hundreds of short hair-like structures called cilia. These beat in coordinated waves to propel the organism through water and also create feeding currents. Earthworms use circular and longitudinal muscles with setae (bristles). Grasshoppers use jointed legs and wings. Snails use muscular foot contractions and mucus secretion.

Question 8: Chromosome Numbers: True or False (5 marks)

State whether each of the following statements is True or False.

Question 9: Cell Division Scenario (5 marks)

A living cell in the body of a female grasshopper undergoing a certain type of division aimed at producing haploid daughter cells was treated with a metabolic poison as the cell was traversing through a certain stage of the cell cycle. It was later found that the cell could not go through the remaining stages of cell division.

a) What type of cell division was the cell undergoing?

• (i) Mitosis
• (ii) Meiosis
• (iii) Binary fission
• (iv) Budding

Answer: (ii) Meiosis

The question states the division was "aimed at producing haploid daughter cells." Only meiosis produces haploid (n) cells from diploid (2n) parent cells. Mitosis produces diploid daughter cells identical to the parent. Binary fission and budding are forms of asexual reproduction in prokaryotes and some eukaryotes respectively, and neither specifically produces haploid cells.

b) From which part of the grasshopper's body was the cell obtained?

• (i) Muscle
• (ii) Brain
• (iii) Ovary
• (iv) Heart

Answer: (iii) Ovary

Meiosis in female animals occurs in the ovaries, producing eggs (ova) — haploid female gametes. Muscle, brain, and heart cells undergo mitosis (for growth and repair), not meiosis. Since the grasshopper is female, only the ovary would contain cells undergoing meiosis to produce haploid gametes.

c) Which stage of cell division was the cell undergoing when treated with the metabolic poison?

• (i) Metaphase
• (ii) Anaphase
• (iii) Prophase I
• (iv) Telophase

Answer: (iii) Prophase I

The key clue is that the cell "could not go through the remaining stages." Prophase I is the most complex and longest stage of meiosis. It is during Prophase I that homologous chromosomes pair up (synapsis) and crossing over occurs. A metabolic poison disrupting Prophase I would prevent the cell from progressing further. If the poison acted at metaphase or later, most of the critical preparatory work would already be complete.

d) Why was the cell unable to go through the remaining stages after treatment?

• (i) The poison interfered with chromosome replication
• (ii) The poison prevented chromosome alignment during metaphase
• (iii) The poison disrupted the pairing of homologous chromosomes in Prophase I
• (iv) The poison prevented the formation of the nuclear envelope

Answer: (iii) The poison disrupted the pairing of homologous chromosomes in Prophase I

During Prophase I, homologous chromosomes must pair (synapsis) to form bivalents. This pairing is essential for the cell to proceed to Metaphase I, where bivalents align at the equator. If the poison disrupts synapsis, the cell cannot complete Prophase I and division halts. This is consistent with the cell being stopped in Prophase I as identified in part (c).

e) What is the main biological importance of meiosis in a female grasshopper?

• (i) To produce identical cells for growth
• (ii) To ensure genetic variation in offspring
• (iii) To repair damaged cells
• (iv) To produce diploid gametes

Answer: (ii) To ensure genetic variation in offspring

Meiosis has two key biological functions: producing haploid gametes (enabling sexual reproduction) and generating genetic variation through crossing over (Prophase I) and independent assortment (Metaphase I). Option (iv) is wrong — meiosis produces haploid gametes, not diploid ones. Option (i) describes mitosis. Option (iii) also describes mitosis.

Question 10: Transport in Plants: True or False (5 marks)

Question 11: Respiratory Surfaces: True or False (5 marks)

Question 12 (1 mark)

A grayling butterfly will normally fly toward the sun. This is an example of:

• a) Phototropism
• b) Kinesis
• c) Migration
• d) Taxis

Answer: d) Taxis

Taxis is a directed movement of a whole organism toward or away from a stimulus. Flying toward the sun (a light source) is specifically positive phototaxis. Phototropism (a) is directional growth of a plant toward light,  not movement of an animal. Kinesis (b) is a non-directional change in movement rate or turning frequency in response to a stimulus (not directed). Migration (c) is a large-scale seasonal movement between habitats, not a stimulus-response directional movement.

Question 13: Red Blood Cells: True or False (5 marks)

Question 14 (1 mark)

How many ATP molecules are produced from one pyruvate molecule during glycolysis?

• a) 0 ATP
• b) 1 ATP
• c) 2 ATP
• d) 3 ATP

Answer: a) 0 ATP

This question asks specifically about ATP produced per pyruvate during glycolysis. Glycolysis converts one glucose molecule into two pyruvate molecules, producing a net gain of 2 ATP total per glucose. Dividing by 2 pyruvate molecules gives 1 ATP per pyruvate... however, the question asks about pyruvate production specifically. During glycolysis, the ATP is produced from glucose → 2 pyruvate as a whole-molecule event, not attributed per pyruvate individually.

⚠️ Important correction: The standard answer for "net ATP per glucose in glycolysis" is 2 ATP. Per pyruvate produced, that is 1 ATP. However, option b) 1 ATP is the most defensible answer here. If the examiner intended the full-glucose yield divided by two pyruvates = 1 ATP per pyruvate, then b) 1 ATP is correct. If the answer key states a) 0 ATP, this would only be correct if referring to ATP per pyruvate in the sense that ATP is produced per glucose (not per pyruvate as a unit). The most biologically accurate answer is b) 1 ATP per pyruvate.

Question 15 (1 mark)

Using the equation for triolein respiration: 2C₅₇H₁₀₄O₆ + 160O₂ → 104H₂O + 114CO₂, what is the respiratory quotient (RQ) for the respiration of triolein?

• a) 0.6
• b) 0.7
• c) 0.8
• d) 1.0

Answer: b) 0.7

The Respiratory Quotient is calculated as:

$$RQ=\frac{{\text{Volume of CO}}_2\text{ produced}}{{\text{Volume of O}}_2\text{ consumed}}$$

From the equation: CO₂ produced = 114 molecules, O₂ consumed = 160 molecules.

$$RQ=\frac{114}{160}=0.7125\approx \mathbf{0.7}$$

An RQ of approximately 0.7 is characteristic of fat (lipid) respiration. Carbohydrates give RQ = 1.0, proteins give approximately 0.8–0.9. Fats contain more hydrogen and less oxygen than carbohydrates per carbon, requiring more oxygen to oxidise — hence RQ < 1.

Question 16 (1 mark)

The kangaroo rat can tolerate much higher concentrations of sodium ions in its bloodstream than humans. How does the blood entering the nephron of the kangaroo rat undergo filtration to remove excess sodium ions?

• a) The blood is filtered in the glomerulus where sodium ions are actively transported out into the urine
• b) Sodium ions are reabsorbed in the proximal convoluted tubule and water is retained
• c) The nephron only filters water, and sodium ions remain in the bloodstream
• d) Excess sodium is filtered into the filtrate at the distal convoluted tubule

Answer: d) Excess sodium is filtered into the filtrate at the distal convoluted tubule

In the normal mammalian nephron, sodium is freely filtered at the glomerulus and then largely reabsorbed. In the kangaroo rat, which needs to excrete excess sodium, the distal convoluted tubule (DCT) is where fine regulation of sodium occurs  less reabsorption takes place there, allowing excess sodium to remain in the filtrate and be excreted in concentrated urine. Option (a) is incorrect because glomerular filtration is a passive pressure-driven process, not active transport. Option (b) describes sodium reabsorption, the opposite of what is needed for excess excretion. Option (c) is anatomically incorrect , the nephron filters much more than just water.

Question 17 (1 mark)

The area of the eye that contains photoreceptors and neurons is the:

• a) Cornea
• b) Retina
• c) Lens
• d) Iris

Answer: b) Retina

The retina is the innermost light-sensitive layer at the back of the eye. It contains two types of photoreceptors: rods (sensitive to low light intensity, responsible for black-and-white vision) and cones (sensitive to colour, requiring bright light). Bipolar neurons and ganglion cells in the retina process and transmit signals via the optic nerve to the brain. The cornea refracts light. The lens focuses it. The iris controls pupil size.

Question 18 (1 mark)

Which of the following would occur at the onset of an action potential?

• a) Potassium ions enter
• b) Sodium ions leave
• c) Potassium ions leave
• d) Sodium ions enter

Answer: d) Sodium ions enter

At the onset (depolarisation phase) of an action potential, voltage-gated sodium (Na⁺) channels open. Na⁺ ions rush into the neurone down their electrochemical gradient, causing the membrane potential to rise from approximately −70 mV (resting) to about +40 mV. Potassium ions (K⁺) leave the cell later, during repolarisation, when voltage-gated K⁺ channels open to restore the negative resting potential.

Question 19:  Parts of Angiosperm and Mammal (10 marks)

Name the parts of an angiosperm and of a mammal which perform the following roles:

Question 20:Parts of a Flower After Fertilisation (5 marks)

Which part of a flower becomes:

Question 21 (1 mark)

Which of the following is true about culturing bacteria compared to culturing viruses?

• a) Bacteria can grow on artificial media, while viruses need living cells
• b) Bacteria do not require any medium to grow
• c) Bacteria only need a host cell to grow
• d) Viruses can replicate in any environment without the need for a host

Answer: a) Bacteria can grow on artificial media, while viruses need living cells

Bacteria are living cells with their own metabolic machinery. They can be cultured on nutrient agar or broth in a laboratory without a host. Viruses are obligate intracellular parasites — they lack ribosomes, cannot generate ATP, and have no independent metabolic processes. They can only replicate inside living host cells (cell cultures, embryonated eggs, or living organisms). Options b, c, and d contradict these facts.

Question 22 (1 mark)

Which of the following is the result of yeast fermentation in bread making?

• a) Decreased dough volume
• b) Formation of a dense texture
• c) Expansion of the dough due to gas production
• d) Reduction in dough flavour

Answer: c) Expansion of the dough due to gas production

Yeast (Saccharomyces cerevisiae) carries out anaerobic fermentation of sugars in the dough:

$${\text{C}}_6{\text{H}}_{12}{\text{O}}_6\to 2{\text{C}}_2{\text{H}}_5\text{OH}+2{\text{CO}}_2$$

The carbon dioxide (CO₂) gas produced becomes trapped in the gluten network of the dough, forming bubbles that cause the dough to rise and expand. The ethanol produced also contributes to flavour (it evaporates during baking). Fermentation increases volume (not decreases it), creates a light texture (not dense), and actually enhances flavour.

Question 23 (1 mark)

In veterinary science, how can genetic engineering help improve animal health?

• a) By cloning animals to ensure they are disease-free
• b) By genetically modifying animals to resist specific diseases
• c) By altering animal behaviour through gene editing
• d) By replacing natural animal vaccines with artificial compounds

Answer: b) By genetically modifying animals to resist specific diseases

Genetic engineering can introduce or modify specific genes in animals to confer disease resistance, for example, editing out receptor genes that pathogens use to enter cells, or enhancing immune response genes. This has been applied in livestock research to produce animals resistant to diseases like foot-and-mouth disease or African swine fever. Cloning (a) reproduces an existing genotype but does not guarantee disease resistance. Behaviour alteration (c) is not a mainstream application of genetic engineering for health. Option (d) misrepresents how vaccines work.

Question 24 (1 mark)

Which one of the following least favours the emergence of new species?

• a) Selective breeding
• b) Stable environmental conditions
• c) Artificial selection
• d) Gene flow in a population

Answer: b) Stable environmental conditions

Speciation (formation of new species) is driven by genetic divergence, usually accelerated by environmental change, isolation, or selection pressure. Stable environmental conditions exert little selective pressure and allow populations to remain in equilibrium  there is no driver for divergence. Selective breeding (a) and artificial selection (c) can rapidly change gene frequencies and traits. Gene flow (d) actually reduces divergence between populations (making speciation less likely), but the question asks which least favours species emergence, and stable conditions most strongly inhibit it by removing the selection pressure that drives change.

Question 25: Evolutionary Terms Matching (4 marks)

Match each evolutionary term in Column A with its correct description in Column B.

Section B: Worked Solutions (Attempt any THREE)

Question 26: Deforestation: Impacts and Mitigation (10 marks)

How does deforestation impact the environment and society, and what measures can be taken to mitigate its negative effects?

Environmental Impacts

1. Loss of biodiversity. Forests contain more than half of the world's terrestrial species. When trees are cleared, habitats are destroyed and species lose food sources, shelter, and breeding grounds. Specialist species with narrow habitat requirements face local extinction. Fragmentation of remaining forest also isolates populations, reducing genetic diversity.

2. Climate change and increased CO₂. Trees absorb CO₂ through photosynthesis and store carbon in their biomass. When forests are cleared — especially by burning — stored carbon is released as CO₂. Deforestation accounts for approximately 10–15% of global CO₂ emissions. Fewer trees means less carbon absorption, accelerating global warming.

3. Disruption of the water cycle. Trees release water vapour through transpiration, contributing to cloud formation and rainfall. Deforestation reduces transpiration, lowering local humidity and rainfall patterns. This can turn previously fertile areas into drier landscapes, reducing water availability for rivers, agriculture, and communities downstream.

4. Soil erosion and degradation. Tree roots bind soil together. Without them, rainwater runs off rapidly, washing topsoil into rivers (increasing turbidity and sediment load) and leaving behind nutrient-poor, compacted soil. This makes the land less productive for future agriculture and increases flood risk downstream.

5. Loss of ecosystem services. Forests provide clean air and water filtration, regulate temperature, prevent floods, and maintain soil fertility. These services have enormous economic value but are lost when forests are cleared.

Social and Economic Impacts

6. Displacement of indigenous and forest-dependent communities. Many communities — including in Rwanda and across Africa — depend on forests for food, medicine, building materials, and cultural practices. Deforestation can destroy livelihoods and force displacement.

7. Reduced agricultural productivity. Paradoxically, while forests are often cleared for agriculture, the resulting soil degradation reduces long-term agricultural yields. Reduced rainfall from disrupted water cycles further threatens food security.

8. Increased flooding and landslides. Without forest cover to absorb rainwater, flash floods become more frequent and severe, damaging infrastructure and communities.

Mitigation Measures

1. Reforestation and afforestation. Planting new trees in deforested areas (reforestation) and establishing forests in areas that previously had none (afforestation) restores carbon sinks, stabilises soils, and rebuilds habitat. Rwanda's national reforestation programme is a regional example of this approach.

2. Sustainable forest management. Implementing selective logging (removing only mature trees and leaving the forest structure intact), certification schemes (e.g. FSC certification), and enforcing logging bans in protected areas reduces the rate of forest loss.

3. Agroforestry. Integrating trees into agricultural systems provides farmers with timber and income while maintaining some forest cover, soil protection, and carbon storage.

4. Government policy and law enforcement. Strengthening land-use laws, prosecuting illegal logging, establishing protected areas and national parks, and providing economic incentives for forest conservation (e.g. REDD+ carbon credit programmes) reduce deforestation rates at a systemic level.

5. Education and community engagement. Raising awareness of the ecological and economic value of forests among communities, schools, and businesses supports long-term conservation behaviour changes.

Question 27: Biuret Test Experiment for Protein Digestion (10 marks)

How could a student design an experiment using the Biuret test to monitor the completion of protein digestion into amino acids, and what role does the test play in this process?

Background: The Biuret Test

The Biuret test detects the presence of peptide bonds. When Biuret reagent (dilute copper sulphate in alkaline sodium hydroxide solution) is added to a solution containing proteins or polypeptides, the Cu²⁺ ions form a complex with peptide bonds, producing a purple/violet colour. If no peptide bonds are present (i.e. only amino acids are present), the solution remains blue. This makes it ideal for monitoring protein digestion  as digestion proceeds and peptide bonds are broken, the purple colour should fade.

Experimental Design

Aim: To monitor the progress of protein digestion by protease enzyme using the Biuret test.

Materials:

• Protein solution (e.g. egg albumin or casein dissolved in buffer)
• Protease enzyme solution (e.g. pepsin at pH 2, or trypsin at pH 8)
• Biuret reagent (1% CuSO₄ in 1% NaOH)
• Water bath at 37°C
• Test tubes, pipettes, timer, colorimeter or colour chart

Procedure:

1. Prepare the reaction mixture: Mix equal volumes of protein solution and protease enzyme solution in a test tube. Place in a water bath at 37°C (optimal enzyme temperature).
2. Take samples at intervals: At time 0, 5, 10, 15, 20, and 30 minutes, withdraw a 1 cm³ sample from the reaction mixture using a pipette.
3. Perform the Biuret test on each sample: Add 2 cm³ of Biuret reagent to each sample. Mix and observe the colour after 5 minutes.
4. Record results: Note the colour at each time point. A purple colour indicates proteins/polypeptides are still present. Blue indicates digestion is complete (only amino acids remain).
5. Control: Run a parallel test tube with protein solution but no enzyme (or boiled/denatured enzyme) to confirm that any colour change is due to enzymatic activity and not spontaneous breakdown.

Expected Results:

• Time 0: Purple  protein is intact with many peptide bonds
• Time 5–15 min: Lighter purple partial digestion, fewer peptide bonds
• Time 20–30 min: Blue  digestion complete, only amino acids remain
• Control (no enzyme): Purple throughout  confirming enzyme is necessary

Role of the Biuret Test

The Biuret test acts as a qualitative (and semi-quantitative) indicator of protein digestion completion. By tracking the colour change from purple to blue over time, the student can determine when all peptide bonds have been hydrolysed. If a colorimeter is used to measure absorbance at 540 nm, the results become quantitative, allowing a rate of digestion to be calculated. This makes the Biuret test a practical and reliable tool for monitoring enzyme-catalysed protein hydrolysis in a laboratory setting.

Question 28: Freshwater Amoeba Osmoregulation (10 marks)

Why do freshwater amoebae need to constantly expel large volumes of water from their cells, and what mechanisms might they use to accomplish this task?

Why Water Constantly Enters

Freshwater amoebae (such as Amoeba proteus) live in an environment where the surrounding water has a very low solute concentration (hypotonic environment). The cytoplasm of the amoeba contains dissolved salts, proteins, sugars, and other solutes at a higher concentration than the surrounding fresh water. This creates a water potential gradient  the water potential of the cytoplasm is lower (more negative) than that of the surrounding water.

By the process of osmosis, water molecules move continuously down this water potential gradient  from the high water potential environment (surrounding water) into the lower water potential cytoplasm through the partially permeable plasma membrane. If unchecked, this inflow would cause the cell to swell and eventually burst (lysis).

The amoeba must therefore continuously expel the excess water entering by osmosis to maintain cell volume and internal concentration — a process called osmoregulation.

Mechanism: The Contractile Vacuole

The primary mechanism used by freshwater amoebae to expel excess water is the contractile vacuole:

1. Collection: The contractile vacuole is a specialised membrane-bound organelle that collects excess water from the cytoplasm. Small feeder vesicles or tubules surrounding the vacuole collect water from the cytosol and deliver it to the central vacuole.
2. Filling: The vacuole gradually fills with water, expanding as it does. This process is driven partly by osmosis (water moves from the cytoplasm into the vacuole) and partly by active transport of ions into the vacuole (creating an osmotic gradient that draws water in).
3. Expulsion: When the vacuole reaches a critical size, it moves to the plasma membrane, fuses with it, and expels the water to the outside of the cell — a process called systole. This requires energy (ATP) and involves contraction of associated proteins.
4. Reformation: The vacuole then reforms (diastole) and begins filling again. In freshwater amoebae, this cycle can occur every 5–10 minutes, depending on the osmotic conditions.

Energy Cost

The operation of the contractile vacuole is energetically expensive because it involves active transport of ions against concentration gradients and the mechanical work of membrane fusion and expulsion. This is why freshwater protists must maintain active metabolism and  they are constantly spending ATP simply to maintain cell volume against the osmotic influx of water.

Comparison with Marine Relatives

Marine amoebae or amoebae living in isotonic environments face no such challenge because the solute concentration of their surroundings matches their cytoplasm so, there is no net osmotic water inflow. These organisms either lack contractile vacuoles or use them far less frequently. This comparison reinforces that the freshwater environment specifically drives the need for constant water expulsion.

Question 29: The Calvin Cycle: Three Phases (10 marks)

Summarise the three phases of the Calvin cycle and explain how ATP and NADPH contribute to each phase.

The Calvin cycle (also called the light-independent reactions or the dark reactions) takes place in the stroma of the chloroplast. It uses the ATP and NADPH produced during the light-dependent reactions to fix atmospheric CO₂ into organic compounds. The cycle has three phases:

Phase 1: Carbon Fixation

CO₂ from the atmosphere enters the leaf through stomata and diffuses into the stroma. Here, each CO₂ molecule is combined with a 5-carbon acceptor molecule called ribulose-1,5-bisphosphate (RuBP) by the enzyme RuBisCO (ribulose-1,5-bisphosphate carboxylase/oxygenase). The unstable 6-carbon intermediate immediately splits into two molecules of 3-phosphoglycerate (3-PGA), a 3-carbon compound.

$${\text{CO}}_2+\text{RuBP}\stackrel{\text{RuBisCO}}{\to }2\times \text{3-PGA}$$

Role of ATP and NADPH in Phase 1: Neither ATP nor NADPH is directly consumed in carbon fixation itself. The energy input at this stage is solely the chemical energy stored in RuBP from the previous cycle turn. However, the fixation step is the entry point that makes subsequent energy input meaningful.

Phase 2: Reduction

The 3-PGA molecules produced in Phase 1 are converted into glyceraldehyde-3-phosphate (G3P), a 3-carbon sugar — the first stable organic product of the Calvin cycle. This involves two steps:

1. 3-PGA is phosphorylated (a phosphate group is added) using ATP, producing 1,3-bisphosphoglycerate (1,3-BPG).
2. 1,3-BPG is then reduced using NADPH (which donates hydrogen/electrons), producing G3P and releasing inorganic phosphate (Pᵢ) and NADP⁺.

$$3\text{-PGA}\stackrel{\text{ATP}}{\to }1,3\text{-BPG}\stackrel{\text{NADPH}}{\to }\text{G3P}$$

Role of ATP and NADPH in Phase 2: This is where both ATP and NADPH are consumed. ATP provides the phosphate group and energy needed for phosphorylation. NADPH provides the electrons (reducing power) needed to reduce the phosphorylated intermediate to G3P. This phase directly converts inorganic carbon into organic carbon — the core of carbon fixation.

Phase 3: Regeneration of RuBP

Most of the G3P produced (5 out of every 6 molecules) is used to regenerate RuBP, allowing the cycle to continue. This involves a series of enzymatic reactions that rearrange carbon skeletons, ultimately converting 5-carbon intermediates back into RuBP using energy from ATP.

$$5\times \text{G3P}\stackrel{\text{ATP}}{\to }3\times \text{RuBP}$$

The remaining 1 G3P molecule per cycle turn exits the cycle and is used to synthesise glucose, sucrose, starch, fatty acids, amino acids, and other organic compounds the plant needs.

Role of ATP in Phase 3: ATP is required to phosphorylate the 5-carbon intermediate ribulose-5-phosphate (Ru5P) to regenerate RuBP. Without this ATP input, the cycle would run out of CO₂ acceptor and stop. NADPH is not directly consumed in Phase 3.

Summary Table

For every 3 CO₂ molecules fixed, the Calvin cycle consumes 9 ATP and 6 NADPH and produces one net G3P molecule. Two G3P molecules are needed to make one glucose molecule, so producing one glucose requires 18 ATP and 12 NADPH overall.

Question 30: Genetics: Dihybrid Cross (10 marks)

In peas, C (coloured seed) is dominant over c (colourless). W (starchy endosperm) is dominant over w (waxy). Pure breeding coloured starchy plants were crossed with pure breeding colourless waxy plants.

a) Genotype and phenotype of F₁ individuals (2 marks)

Pure breeding coloured starchy parent: CCWW
Pure breeding colourless waxy parent: ccww

Cross: CCWW × ccww

All F₁ individuals receive one C and one W allele from the CCWW parent, and one c and one w allele from the ccww parent.

F₁ Genotype: CcWw

F₁ Phenotype: Coloured seeds with starchy endosperm (both dominant alleles are expressed)

b) F₁ (CcWw) × ccww testcross — expected phenotype ratio (6 marks)

F₁ plant CcWw produces four types of gametes (by independent assortment):

• CW
• Cw
• cW
• cw

The ccww parent produces only one type of gamete: cw

Cross: CcWw × ccww

Expected ratio (assuming independent assortment):

Coloured Starchy : Coloured Waxy : Colourless Starchy : Colourless Waxy = 1 : 1 : 1 : 1 (i.e. 25% each)

c) Why do the observed results differ from the expected 1:1:1:1 ratio? (2 marks)

Observed results:

• Coloured starchy: 37%
• Coloured waxy: 16%
• Colourless starchy: 14%
• Colourless waxy: 33%

The parental combinations (coloured starchy and colourless waxy) appear in higher frequencies (37% and 33%) than expected, while the recombinant types (coloured waxy and colourless starchy) appear less frequently (16% and 14%).

This pattern is the result of genetic linkage combined with crossing over.

The genes for seed colour (C/c) and endosperm type (W/w) are located on the same chromosome (they are linked). When genes are linked, they do not assort independently,  they tend to be inherited together, which is why parental combinations appear more frequently. The independent assortment assumed in part (b) does not apply.

The recombinant phenotypes (coloured waxy and colourless starchy) do appear — but at lower frequency — because crossing over (recombination) during Prophase I of meiosis occasionally separates the linked alleles onto different chromatids, creating recombinant gametes. The frequency of recombinant offspring (~30% total) can be used to estimate the map distance between the two genes in centimorgans (cM).

Study Resources

Biology II covers a wide range of topics. If any section needs more work before your next examination, book a Biology tutor on Mathrone Academy for targeted past paper practice. Check the 2026 national exams study guide for the full timetable and the study techniques guide for revision strategies that work.

Also available: S6 Chemistry II 2025, S6 Physics 2025, and S6 Computer Science 2025 worked solutions.

Frequently Asked Questions

Which combinations sit S6 Biology II (012) in Rwanda?

Biology II paper 012 is sat by students in BCG (Biology-Chemistry-Geography), MCB (Mathematics-Chemistry-Biology), PCB (Physics-Chemistry-Biology), and ANP (Associate Nursing Program).

What is the difference between taxis and kinesis?

Taxis is a directed movement of an organism toward or away from a stimulus the direction of movement is determined by the direction of the stimulus. Kinesis is a non-directional response, the organism changes its speed of movement or frequency of turning in response to a stimulus, but the direction of movement is random. A woodlouse moving faster in dry conditions (kinesis) versus a moth flying toward light (positive phototaxis) illustrates the difference.

What is double fertilisation in angiosperms?

Double fertilisation is unique to flowering plants. Two male gametes from the pollen tube both participate in fertilisation. The first male gamete fuses with the egg cell to form the diploid (2n) zygote, which becomes the embryo. The second male gamete fuses with the two polar nuclei in the embryo sac to form the triploid (3n) primary endosperm nucleus, which develops into the nutrient-rich endosperm that feeds the germinating seedling.

What is the respiratory quotient (RQ) and what does it tell us?

The RQ is the ratio of CO₂ produced to O₂ consumed during cellular respiration. It reveals the type of substrate being respired: RQ = 1.0 means carbohydrates are being used; RQ ≈ 0.7 means fats are being respired; RQ ≈ 0.8–0.9 means proteins are the substrate. Values above 1.0 indicate anaerobic respiration is also occurring (more CO₂ produced than O₂ consumed).

What is genetic linkage and how does crossing over affect it?

Genes are linked when they are located on the same chromosome. Linked genes tend to be inherited together and do not follow Mendel's Law of Independent Assortment. However, during Prophase I of meiosis, crossing over (recombination) between non-sister chromatids of homologous chromosomes can separate linked alleles, producing recombinant gametes. The closer together two genes are on a chromosome, the less likely crossing over is to occur between them, and the more tightly they are linked.

Why do mature mammalian red blood cells lack a nucleus?

During RBC maturation from reticulocytes, the nucleus is extruded. Losing the nucleus creates more internal space for haemoglobin (allowing each cell to carry more oxygen), makes the cell more flexible (easier to deform in narrow capillaries), and reduces the cell's own oxygen consumption (no nucleus = no transcription/translation = lower metabolic demand). The trade-off is that RBCs cannot repair themselves or divide, they survive approximately 120 days before being broken down in the spleen and liver.

Where can I find more S6 past papers with worked solutions on Mathrone Academy?

Full worked solutions are available for S6 Mathematics II 2025, S6 Chemistry II 2025, S6 Physics 2025, and S6 Computer Science 2025. You can also book a tutor for live past paper sessions tailored to your combination.

⚠️ Disclaimer

The solutions on this page are prepared by the Mathrone Academy team for revision and learning purposes only. This is not an official NESA marking scheme or REB-approved answer guide. While every effort has been made to ensure accuracy, answers and explanations may differ from the official examiners' marking guide. Always refer to your school teacher or the official NESA publications for authoritative marking guidance. Mathrone Academy accepts no responsibility for any discrepancies between these solutions and official results. 

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