The S6 Biology II 2024 National Examination, sat on 25 July 2024 under NESA, covers ecology, physiology, cell biology, genetics, and microbiology for the BCG, MCB, PCB, and ANP combinations. Below are fully worked solutions to every question, with reasoning explained step by step.
You may also find our S6 Chemistry II 2024 worked solutions, S6 Physics II 2024 worked solutions, S6 Mathematics II 2024 worked solutions, and S6 Geography II worked solutions useful, all organized alongside every other past-paper article on our NESA past papers worked solutions hub.
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Paper Overview: S6 Biology II 012
The paper has two sections. Section A is compulsory and worth 70 marks, covering parasitism and symbiosis, hormones, respiration, excretion, microbiology, population ecology, biomolecules, and cell division. Section B requires candidates to attempt only 3 of 5 longer questions worth 30 marks, covering enzymes, metabolism, muscle contraction, asexual reproduction, and genetics. You have 3 hours to complete the paper.
Section A: Attempt All Questions (70 marks)
Question 1: Facultative vs Obligatory Parasites
An obligatory parasite can only survive by living on or in a host, so it is entirely dependent on one type of relationship. A facultative parasite, by contrast, can survive both as a parasite and by other means, such as living freely in the environment or feeding on dead organic matter, which gives it far more flexibility. This flexibility, specifically the ability to switch between different feeding strategies, is exactly what makes facultative parasites harder to control: even if you eliminate their host, they can persist by feeding another way.
Answer: c) Can change the mode of feeding.
Question 2: Termite and Unicellular Organism Relationship
Consider what each organism gains. The unicellular organism gets a safe, protected home inside the termite's gut. The termite gains the ability to digest wood, since it cannot break down cellulose on its own, and the unicellular organism does this for it. Both species benefit, and neither is harmed.
Answer: c) Beneficial to both species. This describes mutualism, a relationship where both partners gain a benefit.
Question 3: Identifying Ecological Relationships
A frog eating small insects is a straightforward feeding relationship where one organism kills and eats another. Blood-sucking lice living among a bird's feathers feed on the bird while harming it, without killing it outright. Tuis feeding on nectar benefit from the flower's resource while the flower benefits from pollination. Ants feeding on an already-dead cricket are consuming dead organic matter rather than hunting living prey. A small crab living in a mussel's shell and stealing its food, without harming the mussel, benefits from the relationship while the mussel is unaffected.
Answer:
Frog feeding on small insects → Predation
Blood-sucking lice living among the feathers of a bird → Parasitism
Tuis feeding on nectar from kowhai flowers → Mutualism
Ants feeding on a dead cricket → Scavenging (decomposition/detritivory)
A small crab living in a mussel's shell, stealing its food but causing no damage → Commensalism
Question 4: Chemical Control in Animals (Fill in the Blanks)
Chemical control in the body works through chemical messengers released by specific glands, traveling through the body to reach a distant target.
Answer: Chemical control in animals is brought about by hormones. These chemicals are usually either protein or steroid/amine. They are produced by endocrine glands which release them into the blood stream where they travel around the body until they reach the target organ. This is where they bring about a response.
Question 5: Respiratory Quotient
The reaction given is C₁₈H₃₆O₂ + 26O₂ → 18CO₂ + 18H₂O.
The respiratory quotient (RQ) is defined as the volume of carbon dioxide produced divided by the volume of oxygen consumed:
$$RQ = \frac{CO_2 \text{ produced}}{O_2 \text{ consumed}} = \frac{18}{26} \approx 0.69$$Answer: a) 0.7. This value is consistent with a fat (lipid) being the respiratory substrate, since fats typically give an RQ close to 0.7, compared to carbohydrates, which give an RQ close to 1.0. The molecular formula given, C₁₈H₃₆O₂, is in fact stearic acid, a fatty acid, which fits this pattern exactly.
Question 6: Excretory Organs and Nitrogenous Waste
Different animal groups excrete nitrogenous waste using different organs and in different chemical forms, largely depending on how much water they can afford to lose. Aquatic and simple organisms tend to excrete ammonia directly, since it is highly toxic but easily diluted in surrounding water. Terrestrial animals that need to conserve water typically convert ammonia into a less toxic form, such as urea or uric acid, before excreting it.
Answer:
Reptile — Excretory organ: Kidney | Excretory substrate: Uric acid
Paramecium — Excretory organ: Contractile vacuole | Excretory substrate: Ammonia
Earthworm — Excretory organ: Nephridia | Excretory substrate: Urea
Cockroach — Excretory organ: Malpighian tubules | Excretory substrate: Uric acid
Question 7: Substances Not Transported by Arthropod Blood
Arthropods have an open circulatory system, where blood, called haemolymph, bathes the tissues directly rather than always flowing through dedicated vessels. Arthropods breathe using a separate system entirely, a network of tracheae that deliver oxygen directly to tissues without needing the blood to carry respiratory gases at all. Nutrients, hormones, and nitrogenous wastes are all still transported by the haemolymph.
Answer: c) Respiratory gas.
Question 8: Matching Staining Techniques
Step by step: Gram staining distinguishes bacteria based on their cell wall structure. Gram-positive bacteria retain the crystal violet stain and appear purple/blue, while Gram-negative bacteria lose this stain during washing and instead take up the pink/red counterstain. Spore staining uses specific dyes, such as malachite green, to detect resistant bacterial endospores. Differential staining broadly refers to any technique using multiple staining reactions to distinguish between different structures or cell types.
Answer:
1. Gram-positive bacteria → ii. Stain blue or purple
2. Gram-negative bacteria → i. Stain red or pink
3. Spore staining → iii. Malachite Green and Safranin
4. Differential staining → iv. Multiple staining reactions are used
Question 9: Population Ecology Vocabulary
Answer:
a) The carrying capacity is the maximum population size that can be supported in an area without harming the environment.
b) Populations gain individuals through births and immigration.
c) Under ideal conditions, populations can grow at higher maximum rates.
Question 10: Comparing Monosaccharides and Amino Acids
Monosaccharides are simple sugars, made only of carbon, hydrogen, and oxygen; they contain no nitrogen. Amino acids, by contrast, always contain nitrogen as part of their amine group. Both can be polymerised into larger macromolecules (monosaccharides into polysaccharides like starch or cellulose, amino acids into proteins). Nucleic acid hydrolysis releases nucleotides, not monosaccharides or amino acids directly, so this is marked as incorrect for both. Both monosaccharides and amino acids are generally soluble in water, not insoluble. Monosaccharides are indeed linked by glycosidic bonds to form polysaccharides; amino acids are linked by peptide bonds instead, not glycosidic bonds. Cellulose is a polysaccharide made of glucose monosaccharide units, so its complete hydrolysis does release monosaccharides; it does not release amino acids at all. Monosaccharides always contain carbon, hydrogen, and oxygen; amino acids contain those three elements too, but also nitrogen (and sometimes sulfur), so this statement holds cleanly for monosaccharides but is incomplete for amino acids.
Answer:
Always contain nitrogen: Monosaccharides ✗ | Amino acids ✓
May be polymerised into macromolecules: Monosaccharides ✓ | Amino acids ✓
Released by complete hydrolysis of nucleic acids: Monosaccharides ✗ | Amino acids ✗
Insoluble in water: Monosaccharides ✗ | Amino acids ✗
May be linked by glycosidic bonds: Monosaccharides ✓ | Amino acids ✗
Released by complete hydrolysis of cellulose: Monosaccharides ✓ | Amino acids ✗
Always contain carbon, hydrogen and oxygen: Monosaccharides ✓ | Amino acids ✓
Question 11: Roles of Human Hormones
a) Testosterone: The primary male sex hormone, responsible for the development of male secondary sexual characteristics (such as facial hair and a deeper voice), sperm production, and muscle growth.
b) Prolactin: Stimulates and maintains milk production (lactation) in the mammary glands following childbirth.
c) Oestrogen: The primary female sex hormone, responsible for the development of female secondary sexual characteristics, regulation of the menstrual cycle, and maintaining the uterine lining.
d) Oxytocin: Stimulates uterine contractions during childbirth and triggers the "milk let-down" reflex during breastfeeding; it is also associated with bonding and social attachment.
Question 12: Negative Impacts of Monoculture and Intensive Livestock Farming
Both practices intensify land use in ways that can degrade the surrounding ecosystem over time.
Answer: (1) Loss of biodiversity: growing a single crop species over large areas (monoculture) removes habitat diversity, reducing the range of insects, plants, and wildlife an area can support, while intensive livestock farming often requires clearing natural habitat for grazing or feed production. (2) Soil degradation and pollution: monoculture depletes specific soil nutrients repeatedly without natural replenishment, often requiring heavy fertilizer use that can run off into waterways, while intensive livestock farming produces large volumes of concentrated animal waste that can contaminate soil and water sources if not properly managed.
Question 13: Mark-Recapture Population Estimate
John marked 90 butterflies. A week later, he caught 80 butterflies, of which 16 were already marked.
This is a classic mark-recapture (Lincoln Index) problem. The logic is that the proportion of marked butterflies in the second catch should reflect the proportion of marked butterflies in the whole population:
$$\text{Population size} = \frac{(\text{number marked first catch}) \times (\text{number in second catch})}{\text{number marked in second catch}}$$ $$\text{Population size} = \frac{90 \times 80}{16} = \frac{7200}{16} = 450$$Answer: The estimated population size is 450 butterflies.
Question 14: Negative Feedback in Blood Glucose Control
a) Stimuli, receptors, and effectors:
Stimulus: A change in blood glucose concentration, either a rise (after eating) or a fall (during fasting or exercise).
Receptor: Cells in the pancreas, specifically the islets of Langerhans, which detect changes in blood glucose concentration directly.
Effector: The liver (and muscle cells) respond to hormonal signals by converting glucose to glycogen (when glucose is high) or converting glycogen back to glucose (when glucose is low), directly changing blood glucose levels.
b) How negative feedback is involved: When blood glucose rises above the normal set point, the pancreas detects this and releases insulin, which stimulates liver and muscle cells to absorb glucose from the blood and store it as glycogen, bringing glucose levels back down toward normal. When blood glucose falls below the set point, the pancreas instead releases glucagon, which stimulates the liver to break down stored glycogen back into glucose, releasing it into the blood and raising levels back toward normal. In both cases, the response acts to counteract the original change, which is the defining feature of negative feedback: the output of the system works against the initial stimulus to restore the original set point.
Question 15: Stages of Mitotic Cell Division
The figure shows two stages: (a) chromosomes aligned at the equator, and (b) chromosomes appearing to be pulled apart toward opposite poles.
a) Naming the stages: (a) Metaphase, where chromosomes align individually along the equatorial plane (metaphase plate) of the cell, attached to spindle fibers from both poles. (b) Anaphase, where sister chromatids are pulled apart and move toward opposite poles of the cell.
b) Significance of mitosis in living things: Mitosis produces two genetically identical daughter cells from a single parent cell, which allows organisms to grow by increasing cell number, to repair damaged tissues by replacing lost or injured cells, and, in single-celled organisms, to reproduce asexually while maintaining a constant chromosome number across generations.
Question 16: Disadvantages and Risks of Gene Technology
Answer: (1) Unintended ecological consequences: genetically modified organisms released into the environment could potentially crossbreed with wild relatives, disrupt local ecosystems, or outcompete native species in unpredictable ways. (2) Health and safety uncertainties: the long-term effects of consuming genetically modified foods or of gene therapies on human health are not always fully understood, raising ongoing safety concerns. (3) Ethical and social concerns: gene technology raises difficult ethical questions, such as the potential for genetic discrimination, unequal access to genetic treatments between wealthy and poor populations, and concerns about modifying human embryos.
Question 17: Xylem and Phloem Structure
a) Structure of a xylem vessel: A xylem vessel is a long, hollow tube formed from dead cells joined end to end, with their end walls broken down to create a continuous channel. The cell walls are thickened and strengthened with lignin, which provides structural support and allows the vessel to withstand the tension created as water is pulled upward through the plant.
b) Why xylem vessel walls have pits: Pits are thin, unlignified regions in the otherwise thickened cell wall that allow water and dissolved minerals to move sideways between adjacent xylem vessels, providing an alternative route around any blockages (such as air bubbles) and allowing water to reach surrounding living cells.
c) Why sieve tube elements have only a thin layer of cytoplasm: Sieve tube elements are living cells that transport sugars (via phloem), and having only a thin peripheral layer of cytoplasm, rather than a dense cell interior, minimizes resistance to the flow of sugar solution moving through the central lumen of the cell, allowing for more efficient mass flow of nutrients along the phloem.
Question 18: Cell Membranes and Glucose Uptake
a) Why muscle cells need to take up glucose rapidly: Muscle cells rely heavily on glucose as fuel for cellular respiration to generate ATP for muscle contraction, especially during exercise, when energy demand rises sharply. Rapid glucose uptake ensures the muscle has an immediate, readily available energy supply to sustain repeated contractions without delay.
c) Why cell membranes are described as partially permeable rather than semi-permeable: The term "semi-permeable" implies that a membrane allows only certain sizes of molecule through, based purely on size, like a simple sieve. In reality, cell membranes are selective in a more complex way, using specific channel and carrier proteins to control which particular molecules pass through, regardless of their size alone, and can even actively transport specific molecules against their concentration gradient. "Partially permeable" better reflects this selective, protein-mediated control over what crosses the membrane, rather than a purely size-based filter.
Section B: Attempt Only Three Questions (30 marks)
Question 19: Enzyme Specificity and Mechanism
a) Why enzymes are so specific: Each enzyme has an active site with a precise three-dimensional shape and chemical properties that complement only one specific substrate (or a small group of closely related substrates). Since only a molecule with the correct shape and chemical groups can bind properly to this active site, each enzyme can only catalyse one particular reaction, or a narrow range of similar reactions.
b) Why all enzymes are protein molecules: Proteins can fold into highly specific, complex three-dimensional shapes, determined by their amino acid sequence, which allows them to form the precisely shaped active sites needed for substrate specificity. This complex folding capability, along with the chemical diversity of the twenty amino acid side chains available to form bonds and interactions with a substrate, is what makes proteins uniquely suited to the catalytic role enzymes require.
c) Lock-and-key vs induced-fit hypothesis: The lock-and-key hypothesis proposes that an enzyme's active site is a rigid, fixed shape that already perfectly matches its substrate, like a key fitting a lock exactly, with no change in shape required. The induced-fit hypothesis, by contrast, proposes that the active site is flexible, and changes shape slightly as the substrate binds, molding itself around the substrate to form a closer, more precise fit than existed before binding occurred.
d) How enzymes reduce activation energy: Enzymes provide an alternative reaction pathway with a lower activation energy by binding the substrate in a way that strains or destabilizes its chemical bonds, brings reacting molecules into close proximity and correct orientation, or provides a favorable chemical microenvironment, all of which make it easier for the reaction to proceed compared to the uncatalyzed pathway.
e) Why this reduction in activation energy is essential to living organisms: Without enzymes, most biochemical reactions needed for life would occur far too slowly at normal body temperature to sustain life, since the activation energy barrier would rarely be overcome. By lowering this barrier, enzymes allow essential reactions, such as those in respiration, digestion, and DNA replication, to proceed rapidly enough at normal physiological temperatures to support the organism's survival, without needing dangerously high body temperatures to speed reactions up instead.
Question 20: Metabolism of Proteins and Lipids
a) How proteins and lipids are metabolized for energy: Proteins are first broken down into individual amino acids; the amino group is removed through a process called deamination, producing ammonia (which is converted to urea for excretion) and a carbon-containing residue called a keto acid, which can then enter the respiratory pathway (often via the Krebs cycle) to be oxidized for energy. Lipids are broken down into glycerol and fatty acids; glycerol can be converted into a compound that enters glycolysis, while fatty acids undergo beta-oxidation, breaking them down into two-carbon acetyl groups that enter the Krebs cycle directly as acetyl-CoA, generating a large amount of energy due to the high proportion of energy-rich carbon-hydrogen bonds in fatty acid chains.
b) Why the body does not primarily use fats to produce energy, despite fats providing more energy than carbohydrates: Although fats release more energy per gram than carbohydrates, they cannot be broken down and metabolized as quickly as carbohydrates, since fat metabolism (beta-oxidation) is a slower, more complex process, and fats require oxygen to be metabolized efficiently, meaning they cannot supply energy quickly under anaerobic conditions the way glucose can via anaerobic glycolysis. Carbohydrates, particularly glucose, can be broken down rapidly and can also supply at least some energy without oxygen, making them the preferred, faster-access fuel source, especially for tissues like muscle during intense activity or the brain, which relies almost exclusively on glucose under normal conditions.
c) Major differences between cellular respiration and photosynthesis: Cellular respiration breaks down glucose to release energy, producing carbon dioxide and water as byproducts, while photosynthesis builds glucose using energy from light, consuming carbon dioxide and water as raw materials, essentially the reverse overall process. Respiration occurs in the mitochondria (and cytoplasm for glycolysis) of virtually all living cells, while photosynthesis occurs only in chloroplasts, found only in plant cells and some other photosynthetic organisms. Respiration releases energy (is exergonic/catabolic), while photosynthesis stores energy (is endergonic/anabolic). Respiration occurs continuously, at all times, while photosynthesis only occurs in the presence of light. Respiration consumes oxygen and releases carbon dioxide, while photosynthesis consumes carbon dioxide and releases oxygen.
Question 21: Sarcomere Structure and Muscle Contraction
The diagrams show a sarcomere at different stages of contraction, from A (most stretched/relaxed) through to D (most contracted).
a) Naming the parts labeled P, Q, and R: P refers to the thick filaments, made of the protein myosin. Q refers to the thin filaments, made of the protein actin. R refers to the region of overlap between the actin and myosin filaments, where actin-myosin cross-bridges can form during contraction.
b) Why there are no actin-myosin cross-bridges visible in diagram A: In diagram A, the sarcomere is shown in its most stretched or relaxed state, meaning the thin (actin) and thick (myosin) filaments barely overlap, or do not overlap at all. Since cross-bridges can only form where the myosin heads are physically close enough to bind to actin, no overlap means no cross-bridges can form at this stage.
c) Which diagram shows the greatest force, and why: Muscle force is directly related to the number of actin-myosin cross-bridges that can form, which depends on the degree of overlap between the thick and thin filaments. According to the sliding filament model's length-tension relationship, force is greatest at an intermediate degree of overlap, where the maximum possible number of cross-bridges can form, neither too stretched (too little overlap, few cross-bridges possible) nor too fully contracted (filaments overlapping so much that they begin to interfere with each other or the thick filaments hit the Z-lines, reducing effective cross-bridge formation). Diagram C, showing a substantial but not maximal degree of overlap, would typically represent the state of greatest force generation, since it allows for the greatest number of properly aligned, functional cross-bridges without the filaments overlapping so much that they obstruct each other.
d) i) Why a muscle cannot pull itself back to its original relaxed length: Muscle contraction works through actin-myosin cross-bridges actively pulling the thin filaments inward using ATP-powered myosin head movements, a mechanism that only works in one direction, shortening the sarcomere. There is no equivalent active mechanism within the muscle itself to push or pull the filaments back outward to lengthen the sarcomere again; muscles can only actively shorten, never actively lengthen themselves.
ii) How the muscle in diagram D could be returned to the state in diagram A: The muscle must be stretched back to its original length by an external, opposing force, most commonly the contraction of an antagonistic muscle (a muscle that pulls in the opposite direction across the same joint), which passively stretches the relaxed muscle back out to its longer resting length.
Question 22: Asexual Reproduction
a) Binary fission in Amoeba: The Amoeba's single nucleus first duplicates its DNA and divides into two identical nuclei through mitosis. The cytoplasm then constricts and divides into two, with each half containing one of the two nuclei, producing two genetically identical daughter amoebae of roughly equal size, each capable of independent life immediately after division.
b) Budding in yeast: A small outgrowth, or bud, forms on the surface of the parent yeast cell. The parent cell's nucleus divides by mitosis, and one of the resulting nuclei migrates into the developing bud. The bud continues to grow while still attached to the parent, eventually pinching off to become a separate, genetically identical daughter cell, though in some cases the bud may remain attached, forming a small chain or cluster of cells.
c) Spore formation in Rhizopus nigricans: Rhizopus (a type of bread mould fungus) produces specialized upright hyphae called sporangiophores, each topped with a swollen structure called a sporangium. Inside the sporangium, the nucleus undergoes repeated mitotic divisions to produce many haploid spores. When mature, the sporangium wall ruptures, releasing large numbers of lightweight spores into the air, which can be dispersed by wind and germinate into new fungal hyphae if they land in a suitable environment.
Question 23: Genetics Cross in Hamsters
The cross: Black coat colour (B) is dominant over white (b). Rough coat (R) is dominant over smooth (r). Cross a hamster that is heterozygous black and homozygous rough (BbRR) with one that is heterozygous black and heterozygous rough (BbRr).
First, determine the gametes each parent can produce. The BbRR parent can only pass on R for the coat texture gene, since it is homozygous (RR), so its possible gametes are BR and bR only. The BbRr parent, being heterozygous at both genes, can produce four different gametes: BR, Br, bR, and br.
Setting up a Punnett square with the BbRR parent's two gamete types across the top and the BbRr parent's four gamete types down the side, and combining every pair, produces the following offspring genotypes and their frequencies out of 8 total combinations:
- BBRR: 1/8
- BBRr: 1/8
- BbRR: 2/8
- BbRr: 2/8
- bbRR: 1/8
- bbRr: 1/8
Converting genotypes to phenotypes: Since the BbRR parent contributes only R gametes, none of the offspring can inherit two recessive r alleles, meaning no offspring can be smooth-coated; every single offspring will show the rough coat phenotype. For coat colour, any offspring with at least one B allele will be black, and only the bbRR and bbRr genotypes (2 out of 8) will be white.
Answer: The offspring phenotype ratio is 6 Black, Rough : 2 White, Rough, which simplifies to a 3 Black Rough : 1 White Rough ratio. No smooth-coated offspring are possible from this particular cross, since one parent could not contribute a recessive "r" allele.
Study Tips for S6 Biology II
A recurring theme across this paper is that Biology exam answers reward precise, mechanistic reasoning over vague description. Notice how nearly every "explain why" question, from enzyme specificity to muscle contraction to negative feedback, is really asking you to trace a cause-and-effect chain step by step, rather than simply naming a concept. Practicing that habit of reasoning through the mechanism, not just recalling the term, will carry over well to unfamiliar Biology questions on exam day.
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⚠️ Important 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, REB-approved answer guide, or official Cambridge/Pearson mark scheme. 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 examination board publications for authoritative marking guidance. Mathrone Academy accepts no responsibility for any discrepancies between these solutions and official results.