HSC Biology 2021

(a) [2 marks]. Label any TWO prokaryotic features visible on the cell, for example the cell wall (rigid outer boundary), the plasma membrane, and the free DNA / nucleoid (no membrane-bound nucleus). The absence of membrane-bound organelles is also acceptable.

(b) [2 marks]. The two missing flowchart boxes are:

1. Box 1 (after isolating the bacterium from Pig 1): culture and grow (multiply) the isolated bacterium in the laboratory.
2. Box 2 (after feeding the bacteria to healthy Pig 2): the healthy pig develops the same disease (diarrhoea).

This completes Koch's postulates: isolate from a sick animal, culture it, give it to a healthy animal that then gets the disease, and re-isolate the same bacterium from Pig 2 to compare with Pig 1.

(c) [3 marks].

• Farm 1 (antibiotics): benefit - it clears the infection quickly (recovery within two weeks). Limitation - repeated use selects for antibiotic resistance, so it loses effectiveness over time and treats rather than prevents.
• Farm 2 (hygiene / pest removal): benefit - removing the rats and mice removes a disease reservoir, a long-term, preventive solution with no resistance problem. Limitation - it acts slowly (three months) and does not cure animals that are already sick.

Marker's note. Label real bacterial features with clear leader lines. In (b) apply Koch's postulates to the flowchart, not just generic steps. In (c) develop the table with your own knowledge rather than copying it, and keep prevention (Farm 2) distinct from treatment (Farm 1), giving a benefit and a limitation for each.

[3 marks]. Let B = black (dominant), b = white (recessive). The black rabbit's mother is white (bb), so she could only pass a b allele; the black rabbit must therefore be heterozygous Bb. The white partner is bb.

Cross Bb x bb:

Offspring are 2 Bb (black) : 2 bb (white). Phenotypic ratio black : white = 1 : 1.

Marker's note. Deduce the black rabbit is heterozygous from its white mother, draw a labelled Punnett square with a key, and quote the phenotypic ratio (1 black : 1 white), not just the genotypic ratio.

[3 marks]. Strip 2. Its control line is present, so the test has worked correctly and the result is valid. It shows no test line, meaning LH is not high, so there has been no LH surge and ovulation is not occurring. Strips 3 and 4 lack a control line, so those tests are invalid and cannot be trusted regardless of the test line.

Marker's note. Engage with the whole stimulus, not just the picture. Distinguish the control line (shows the test worked, gives validity) from the test line (shows the LH result), and pick the strip that is both valid and negative.

(a) [3 marks]. It is likely a somatic mutation. The trait is dominant, yet it appears in the affected male while his parents and his offspring are unaffected. A germ-line mutation arises in the cells that form gametes, so a dominant germ-line change would usually be inherited and seen in his offspring. Because it is not passed on, the mutation most likely occurred in body (somatic) cells of that individual only.

(b) [4 marks].

• Mutation B occurs in Twin 1 before cell differentiation, so it ends up in both Twin 1's somatic cells and germ (gamete-forming) cells. Twin 1 shows the effect in their body, and because it is in the germ line it can be passed to Twin 1's offspring. Twin 2 is unaffected.
• Mutation C occurs in Twin 2 after differentiation, in cells leading only to somatic cells. So it affects Twin 2's body cells but is absent from their germ cells, meaning Twin 2 cannot pass it to offspring. Twin 1 is unaffected by Mutation C.

Marker's note. Use the correct terms (somatic, germ-line, autosomal, dominant). In (a) justify from the pedigree (present in the affected man but not in parents or offspring). In (b) tie the timing of each mutation relative to cell differentiation to whether it reaches the germ line and so whether it can be inherited.

(a) [3 marks]. Plot frequency (Hz) on the x-axis and minimum detectable volume (dB) on the y-axis. Plot the right-ear points (250, 5), (500, 9), (1000, 9), (2000, 5), (4000, 9), (8000, 20) and the left-ear points (250, 55), (500, 60), (1000, 75), (2000, 75), (4000, 80), (8000, 100). Use small, precise markers and a key with two different symbols (one per ear); do not join the points or add a line of best fit.

(b) [2 marks]. The right ear has normal hearing - it detects every frequency at low volumes (about 5 to 20 dB, in the normal range). The left ear has a hearing deficit - it needs much louder sound (55 to 100 dB) to detect the same frequencies, and the volume needed rises as frequency increases, so the higher frequencies are hardest for the left ear.

(c) [3 marks]. A bone conduction implant (or bone conduction hearing aid) is suitable. The outer ear is fully blocked, so a normal aid that sends sound down the ear canal cannot help, but the cochlea still works. A bone conduction device picks up sound with a microphone, converts it to vibrations in a processor, and passes those vibrations through the skull bone directly to the working cochlea, bypassing the blocked outer ear. This restores useful hearing, which is why it is the right choice here.

Marker's note. In (a) plot precisely with small markers and a clear key, reading the dB axis carefully; do not draw a line of best fit. In (b) give one conclusion per ear without repeating yourself, and avoid claiming the left ear is deaf. In (c) match the technology to the stimulus (blocked outer ear, working cochlea), and explain how it works as the justification.

[4 marks]. As habitat is destroyed, the zebra populations become smaller and more isolated, reducing gene flow between groups. Small, isolated populations breed among close relatives (inbreeding), which shrinks the gene pool. Many mutations are recessive and hidden in carriers; inbreeding raises the chance that an offspring inherits two copies of the same recessive allele, so the allele is expressed and the foal shows a phenotype (such as the spotted coat) different from its parents. Over generations, the reduced gene pool and accumulation of these recessive alleles increase the frequency of such offspring.

Marker's note. Link the reduced gene pool and isolation to inbreeding, then to the expression of two recessive alleles, then to the changed phenotype. Use the stimulus rather than inventing natural selection or cross-species hybridisation; a smaller gene pool amplifies existing mutations, it does not cause them.

[3 marks]. Read the bands as alleles. The mother, father and Child 2 each show two bands, so they are heterozygous carriers (Aa) - unaffected. Child 1 shows a single band matching the normal allele, so is homozygous normal (AA). Child 3 shows a single band, in the position of the recessive (sickle) allele that the carriers also carry, so Child 3 is homozygous recessive (aa). Therefore Child 3 will have sickle cell anaemia, having inherited a recessive allele from each carrier parent.

Marker's note. Make a definitive statement (Child 3 will have sickle cell anaemia), not a probabilistic one. Read a single band as homozygous and two bands as heterozygous, and justify directly from the profile - this is determined by the bands, not by a Punnett square.

(a) [3 marks]. mRNA carries a complementary copy of a gene's coding sequence out of the nucleus to the ribosomes. During transcription, RNA polymerase builds the mRNA against the DNA template; the mRNA then acts as the template for translation. Each codon (three nucleotides) on the mRNA specifies one amino acid, so the order of codons sets the order of amino acids added to the growing polypeptide chain.

(b) [5 marks]. The vaccine's mRNA enters body cells and is translated at the ribosomes, so the cells make the viral spike protein. Because the spike protein is foreign (not coded for by the human genome), it acts as an antigen. This triggers a specific adaptive response: helper T cells activate B cells, which become plasma cells producing antibodies against the spike protein, and cytotoxic T cells respond. After the response, memory B and memory T cells specific to the spike protein remain. If the person later meets the real virus, its surface spike proteins are recognised by these memory cells, which mount a rapid, large secondary response that clears the virus before illness develops. This self-made, lasting protection is active immunity.

Marker's note. In (a) use correct terms (codon, complementary, template, ribosome) and keep mRNA's role distinct from DNA replication. In (b) be clear the mRNA is not itself the antigen - it leads the body to make the spike protein, which is the antigen - and link this to memory B and T cells and a faster future response (active, not passive, immunity).

[4 marks]. The koalas use behavioural adaptations, changing posture with the ambient temperature shown in the graph. In mild conditions they were more often observed curled up, which minimises the body surface area exposed to the air and so reduces heat loss - keeping warm. In hot conditions they were more often observed leaning back to expose the ventral (belly) surface and hugging tree trunks. Because the stimulus shows trunks can be up to 9 degrees C cooler than the air, pressing the belly against the cool trunk uses it as a heat sink, conducting body heat away and increasing heat loss to stay cool. By switching posture (and surface area exposed) with the temperature, the koala balances heat gain and loss and keeps its temperature stable. Directing blood flow to the belly against the trunk is an acceptable physiological extension.

Marker's note. Link each posture to the ambient temperature in the graph and to a change in exposed surface area, then to heat loss or retention. Use the stimulus - especially the cool-trunk data - rather than describing adaptations in the abstract.

[7 marks]

The data both support and challenge the statement.

Supporting it (vaccine is specific)

the graphs show measles cases in the vaccinated group were consistently very low, sometimes zero, while the unvaccinated group had large, recurring outbreaks. The table agrees: only 2 vaccinated children died of measles versus 40 unvaccinated. As the two groups were matched for age, sex, dwelling, siblings and maternal education, this difference is most likely due to the vaccine protecting specifically against measles. Fever deaths were almost equal (22 vaccinated vs 25 unvaccinated), a difference of only 3, suggesting no protection against that unrelated cause - consistent with the statement.

Challenging it (wider protection)

the vaccinated group also had fewer deaths from diseases other than measles. Diarrhoea and dysentery deaths were 85 vaccinated vs 156 unvaccinated (about half), and oedema deaths 6 vs 21 (under one third). Total deaths were 115 vaccinated vs 242 unvaccinated, again about half. This suggests the vaccine had broader, non-specific benefits beyond measles, which contradicts the statement.

Judgement

overall the data show the measles vaccine clearly protected against measles, but the lower mortality from several other causes means the claim that it protects only against one specific disease needs qualification. The small oedema numbers mean that part needs further study.

Marker's note. Use every part of the stimulus and manipulate the data, not just restate it (for example "85 vs 156, about half" rather than "more in the unvaccinated"). Give arguments both for and against the statement, draw on the matched design, and reach a clear, qualified judgement.

[6 marks]

Interpreting the results: symptoms persisted even in the no-tablet period (score 8.0), and the placebo score (15.4) was almost as high as the statin score (16.3) and roughly double the no-tablet score. Because statin and placebo gave very similar scores, the side-effects appear to be largely a placebo (or nocebo) effect rather than caused by the statin's active ingredient.

Evaluating the design

the study has serious limitations. The sample of only 60 patients is far too small to generalise to the millions who take statins, so the results are not reliable. The patients self-scored their own symptoms subjectively from 0 to 100, which is not an accurate, objective measurement. There was also no separate control group for comparison. These flaws mean the study is not valid and firm conclusions cannot be drawn.

Judgement

the study is not valid and its results are inconclusive, but it is still useful because it suggests a much larger randomised controlled trial would be worthwhile to test whether statins genuinely cause these side-effects.

Marker's note. Address both the study design and the results. Use the terms reliability, accuracy and validity correctly, note the purpose of the placebo (to test whether the statin itself causes the symptoms), and link the design flaws (tiny sample, self-reported scores, no control) to the inconclusive outcome.

[5 marks]. Normally the hypothalamus releases releasing hormone, which makes the anterior pituitary secrete LH and FSH; these stimulate the testes (Leydig cells to make testosterone, Sertoli cells to support sperm production), and the testosterone and inhibin produced feed back negatively on the hypothalamus and pituitary to keep levels balanced.

When the athlete continuously takes anabolic steroids (which act like testosterone), the body senses persistently high testosterone. Through the negative feedback loop shown, this strongly inhibits the hypothalamus and anterior pituitary, so much less releasing hormone, LH and FSH are released. With little LH and FSH reaching the testes, the testes are no longer stimulated: the Leydig cells make far less of their own testosterone and the Sertoli cells reduce sperm production, so inhibin also falls. With prolonged use the testes receive almost no stimulation, shrink and lose normal function (reduced sperm and reduced natural testosterone production).

Marker's note. Interpret the flowchart correctly, including the negative (inhibitory) signs. Explain cause and effect: external steroid raises feedback, suppressing LH and FSH, which reduces testicular stimulation, sperm production and natural hormone output. Keep negative feedback distinct from positive feedback.

(a) [3 marks]

Steps 1 to 4 build and clone the transgene by recombinant DNA technology. The desired sequences (the Chinook growth-hormone coding sequence and the Ocean Pout antifreeze promoter) are cut out and joined, using restriction enzymes and ligase, to make the transgene. This transgene is inserted into a bacterial plasmid (a vector), and the recombinant plasmid is taken up by bacteria. As the bacteria reproduce, they copy the plasmid and the transgene, producing many identical copies (gene cloning / amplification) ready to be introduced into salmon eggs.

(b) [3 marks]

A benefit is faster growth to market size, lowering production cost. The graph shows the transgenic salmon grow faster, especially in the first two years, and reach market size in about their second year, whereas standard salmon take roughly five months longer. Reaching market size sooner means fewer months of feeding and tank maintenance per fish, so feed and running costs fall and farms can produce more fish in the same time.

(c) [9 marks]

Biodiversity is the genetic variety within a species and the variety of species in an ecosystem. Transgenic salmon threaten it: if they escaped, their growth-hormone advantage could let them outcompete and reduce wild salmon (lowering species diversity), and interbreeding would spread the transgene and shrink the wild gene pool (lowering genetic diversity). The techniques are designed to prevent this.

• Techniques 1 and 5 (quarantine and inland tanks) are physical isolation. Keeping breeding stock in quarantine and growing offspring in inland tanks stops transgenic fish escaping into wild waterways, so they cannot interbreed with or outcompete wild salmon. This directly protects wild populations and their biodiversity.
• Technique 2 (hormone-induced sex reversal) turns genetically female (XX) fish into sperm producers. Because these fish are XX, all their sperm carry an X chromosome, never a Y, so the next generation cannot include males - a reproductive control that limits self-sustaining transgenic populations.
• Technique 3 (fertilising wild-type eggs) crosses the transgenic sperm with eggs from wild-type salmon. Using wild-type eggs introduces fresh genetic variation, avoids inbreeding of the transgenic stock and gives hybrid vigour, which preserves variety within the farmed line.
• Technique 4 (pressure shock, triploid XXX) blocks meiosis II to make triploid offspring. Triploid fish are sterile, so even if one escaped it could not breed and pass the transgene into wild populations. All offspring are female and XXX fish cannot develop sex organs, adding a further reproductive barrier.

Overall: physical isolation (1, 5), the all-female sperm and triploid sterility (2, 4) and the use of wild-type eggs (3) work together so the transgene cannot establish in the wild, protecting wild salmon diversity, while crossing with wild-type stock preserves genetic variety within the transgenic line. The combination makes escape into and breeding with wild populations extremely unlikely, safeguarding biodiversity at both the genetic and species levels.

Marker's note. In (a) explain (cause and effect) the recombinant steps and name the role of the plasmid and bacteria in cloning the gene. In (b) show cause and effect and quote comparative data from the graph (transgenic reaches market size about five months sooner). In (c) analyse each technique in turn, in a logical order, saying how it raises or lowers biodiversity, refer to both wild-type and transgenic populations, and address all five techniques rather than just stating they protect biodiversity.