HSC Biology 2019

[3 marks]. The three missing steps, in order, are:

1. The rise in body temperature is detected by the hypothalamus (the body's thermostat in the brain).
2. The hypothalamus sends nerve impulses to the blood vessels in the skin, causing vasodilation (widening), increasing blood flow to the surface so more heat is lost.
3. The hypothalamus sends nerve impulses to the sweat glands, which produce sweat; evaporation of sweat removes heat and cools the skin.

The combined effect (increased blood flow plus evaporative cooling) returns the body temperature to normal.

Marker's note. Use precise terminology (vasodilation, hypothalamus, evaporative cooling) rather than general descriptions. Show the hypothalamus detecting the internal change, and explain that it is the evaporation of sweat, not the sweat itself, that cools the skin.

[3 marks].

Marker's note. Keep the table concise and answer each row for both columns. The key contrast is heritability: somatic mutations cannot reach offspring, germ-line mutations can.

[5 marks]. Educational programs reduce the incidence of non-infectious diseases by raising awareness of the causes and risk factors and prompting people to change their behaviour.

• Skin cancer (melanoma). The "Slip, Slop, Slap, Seek, Slide" campaign teaches that UV radiation damages skin-cell DNA. As more people cover up, apply sunscreen and seek shade, exposure to UV falls, so fewer people accumulate the mutations that cause melanoma and the incidence drops over time.
• Lung cancer. Anti-smoking programs such as "Quit", with graphic warnings and advertising, explain that tobacco smoke contains carcinogens. As smoking rates fall, fewer people are exposed and the incidence of lung cancer declines.

In each case the program changes a behaviour that is the cause of the disease, which lowers exposure and therefore lowers the number of new cases (the incidence).

Marker's note. Use at least two specific, named educational programs (not screening, genetic testing or legislation, which were a common error) and make the link explicit: program to awareness to behaviour change to reduced exposure to lower incidence.

[5 marks]. Biodiversity is the variety of species in an ecosystem and the genetic diversity within those species. Two agricultural biotechnologies can reduce it:

• Genetic engineering (Bt corn). A gene from Bacillus thuringiensis is inserted so the corn makes a toxin that kills caterpillars that eat it. This reduces the caterpillar population, removing food for the species that feed on them and disrupting food chains, so species diversity in the ecosystem can fall. It also lowers genetic diversity within the pest population, since only resistant individuals survive to breed.
• Artificial insemination (dairy cattle). Semen from one elite bull is used to inseminate very many cows. The resulting offspring are all half-siblings, so genetic diversity within the herd is greatly reduced, leaving it more vulnerable to disease and less able to adapt.

Both technologies narrow variety at the species level or the genetic level, which is a loss of biodiversity.

Marker's note. Define biodiversity at both the species and genetic level, name two specific biotechnologies, and link each one directly to a loss of biodiversity rather than making general comments.

(a) [1 mark]. There is only a single sex chromosome (one X with no matching X or Y), so a whole sex chromosome is missing.

(b) [4 marks]. This results from non-disjunction during meiosis followed by normal fertilisation:

1. During meiosis (homologous chromosomes in meiosis I, or sister chromatids in meiosis II), the sex chromosomes fail to separate correctly (non-disjunction).
2. This produces one gamete with no sex chromosome and another with an extra one.
3. At fertilisation, the gamete lacking a sex chromosome fuses with a normal gamete carrying one X.
4. The resulting individual therefore has only one sex chromosome, matching the karyotype shown.

Marker's note. Name non-disjunction and place it in meiosis (not mitosis, which does not make gametes). Give specific detail for both meiosis and fertilisation, and link the gamete lacking a chromosome to the single sex chromosome in the offspring.

[5 marks]. The map shows lactose tolerance varies between adult populations, being high in parts of northern Europe but low in groups such as Indigenous Australians.

This variation can be explained by evolution through natural selection, with the presence of milk in the diet as the selective pressure. A mutation in the DNA controlling the lactase gene keeps the gene switched on past early childhood, so the enzyme lactase is still produced in adults. In populations that herded animals and drank milk, adults with this mutation gained extra nutrition, so they were more likely to survive and reproduce. They passed the allele to their offspring, and over generations its frequency rose, making lactose tolerance common in those populations.

In populations without a reliable milk supply, the mutation gave no survival advantage, so it was not selected for and stayed rare, leaving most adults lactose intolerant. The map therefore reflects different selective histories acting on the same lactase mutation.

Marker's note. Refer to the map (identify the variation), then give the cause in terms of both DNA (a mutation keeping the lactase gene active) and evolution by natural selection (milk as the selective pressure). Avoid Lamarckian wording such as the trait developing because it was needed.

(a) [1 mark]. Asexual reproduction.

(b) [4 marks].

1. Prepare a single yeast suspension and mix it so the cells are evenly distributed.
2. Measure 5 mL of suspension into each of nine test tubes (the same volume each time).
3. Set up three water baths at the chosen temperatures (for example 15 degrees C, 25 degrees C and 35 degrees C); the temperature is the independent variable.
4. Place three test tubes in each water bath (replicates) and incubate for 2 hours.
5. Take a drop from each tube onto a slide and count the yeast cells under a microscope using a mm grid; this cell count is the dependent variable.
6. Keep constant the volume and concentration of suspension, incubation time and equipment, then compare the mean count at each temperature.

Marker's note. State explicitly how the dependent variable is measured (count cells on a grid under the microscope, not just observe growth), give specific controlled variables with quantities, and include replication (for example three tubes per temperature) for reliability.

(a) [2 marks]. The mother has neither disease and is homozygous, so she must be homozygous unaffected: h (unaffected for Huntington's, recessive) and R (unaffected for Stargardt, dominant). The patient is Hh Rr, so the father supplied the other alleles: H and r.

(b) [4 marks]. Model the homologous pair with the loci in a fixed order, for example H/h at one locus and R/r at another on chromosome 4. The patient's pair carries H and r on one chromosome (from the father) and h and R on the other (from the mother).

• Before crossing over: one chromosome H r, the homologue h R, each shown as two sister chromatids.
• Crossing over swaps a segment between the homologues, producing recombinant chromatids H R and h r alongside the parental H r and h R.
• After separation, the four possible gametes are H r, h R, H R and h r.
• Include a key: H = Huntington's allele, h = unaffected, R = unaffected (Stargardt), r = Stargardt allele.

Marker's note. Distinguish allele from genotype, and use all the stem information to fix the parental alleles in (a). In (b) model meiosis simply and clearly: alleles at specific loci on the chromosomes, the genotype carried through each stage, and a clearly shown crossover with a key.

[3 marks]. Plants regulate water loss by controlling the opening of the stomata (pores on the leaf), which is driven by the guard cells. When the plant's internal water level falls, the guard cells lose water and become flaccid, which closes the stomata and reduces water loss by transpiration. This is under hormonal control: the stress hormone abscisic acid is produced when water is low and triggers the guard cells to close the stomata. When water is plentiful the guard cells become turgid, the stomata open, and transpiration can resume. By detecting the internal water level and adjusting stomatal opening, the plant keeps its internal water balance stable.

Marker's note. This is about plant homeostasis, not adaptations or photosynthesis. Show detection of a change in the internal environment (not the external temperature) and describe the response, including turgid and flaccid guard cells opening and closing the stomata and the link to transpiration.

[5 marks].

Graph A (seed shape), 3:1 ratio. A 3:1 F2 ratio is typical of simple dominant and recessive inheritance. The pure-breeding parents were RR (round) and rr (wrinkled), so all the F1 were Rr (round). Crossing two F1:

Key: R = round (dominant), r = wrinkled (recessive). This gives 3 round : 1 wrinkled, matching Graph A.

Graph B (feather colour), 1:2:1 ratio. Three phenotypes in a 1:2:1 ratio indicates codominance (or incomplete dominance): the heterozygote shows a distinct phenotype. The pure-breeding parents were F(B)F(B) (black) and F(W)F(W) (white), so the F1 were all F(B)F(W). Crossing two F1:

Key: F(B) = black, F(W) = white. This gives 1 black : 2 black-and-white : 1 white, matching Graph B.

Marker's note. Draw a correct Punnett square for each trait with a clear key, identify the modes of inheritance (dominant/recessive for A, codominance or incomplete dominance for B), and link the genotypes to the phenotype ratios shown in the graphs.

(a) [2 marks]. The bacterium Helicobacter pylori, which causes stomach ulcers, has a flagellum. The flagellum lets it swim and burrow through the thick mucus lining of the stomach wall, so it can reach and colonise the epithelium beneath, entering and establishing in the host despite the acidic stomach.

(b) [3 marks]. The mode of transmission affects how quickly and how widely a disease spreads.

• Pathogens spread by airborne droplets (for example the influenza virus in coughs and sneezes) pass directly between people, so infection rates are high and spread is fast, especially in crowded areas.
• Pathogens that need a vector or intermediate host (for example malaria carried by mosquitoes) can only spread where that vector is present, so the spread is slower and limited to particular regions.

So a disease with easy direct transmission spreads faster and further than one that depends on a vector.

Marker's note. In (a) name a specific pathogen (not just a disease) and an adaptation for entry, and distinguish entry from immune evasion. In (b) describe more than one mode of transmission and give specific disease examples, linking the mode to how the disease spreads through a population.

(a) [3 marks].

• Dengue fever: the distribution expanded markedly between 1950 and 2010, spreading into many more regions such as South America and Africa.
• Malaria: the number of countries with reported cases fell from 140 in 1900 to 88 by 2010, so its distribution shrank. The population at risk rose (from 0.9 to 3.4 billion) as world population grew, but as a percentage of the global population it fell from 75% to about 50%.

So dengue spread wider while malaria became more geographically contained, even though more people were at risk in absolute terms.

(b) [7 marks]. Both diseases depend on mosquito vectors, so factors affecting the mosquitoes and their hosts drive the changes in distribution.

• Increased air travel and trade over the last century let infected people and mosquitoes move quickly around the world, helping dengue reach new continents (the marked spread shown on the 2010 map).
• Population growth and urbanisation (global population rose from 1.2 to 6.8 billion in the table) raised the density of human hosts and created new urban breeding habitats for Aedes mosquitoes, supporting dengue's expansion and the rising number at risk of malaria.
• Vector control (insecticide spraying of water bodies, drainage and quarantine) has been applied successfully against Anopheles, which helps explain why the number of malaria countries shrank from 140 to 88 even as population grew.
• Medical advances differ between the two diseases. Antimalarial drugs and control programs have contained malaria's distribution, but there is no widely used dengue vaccine and the virus can evolve quickly, so dengue kept spreading. This matches the data: malaria became more contained while dengue expanded.

Overall, travel, population growth and urbanisation favoured spread, while targeted vector control and medical intervention contained malaria more than dengue.

Marker's note. In (a) read both the table and the maps and give genuine trends for each disease, using terms such as incidence and distribution correctly. In (b) identify several factors, analyse how each affects distribution for both diseases, and tie the analysis directly to figures in the data rather than asserting in general.

(a) [3 marks]. Proteins are made by transcription then translation:

1. Transcription (in the nucleus): the DNA of the gene unwinds and a complementary mRNA strand is built against the template by RNA polymerase.
2. The mRNA leaves the nucleus and moves to a ribosome.
3. Translation: each mRNA codon is matched by a tRNA carrying a complementary anticodon and a specific amino acid; the amino acids are joined by peptide bonds in order to build the polypeptide (the amyloid beta protein).

(b)(i) [2 marks]. Alleles are different versions of the same gene. Multiple alleles means that three or more different versions of that gene exist in the population (such as e2, e3 and e4 of APOE), although any one individual still carries only two.

(b)(ii) [4 marks]. The data show the e2 and e4 alleles have opposite effects on risk.

• e2 lowers risk: e2/e2 and e2/e3 are both 40% less likely than average, so e2 is protective and appears to mask e3.
• e4 raises risk strongly, in a dose-dependent way: one e4 (e3/e4) is 3.2 times more likely, while two e4 (e4/e4) is 14.9 times more likely, so each extra e4 sharply increases risk.
• e3 is the baseline: e3/e3 gives average risk, and e3 partly tempers e4 (e3/e4 at 3.2 times is far below e4/e4 at 14.9 times).
• The alleles interact: e2/e4 is 2.6 times more likely, so the protective e2 reduces but cannot fully cancel the harmful e4.

So e2 reduces risk, e3 is neutral, and e4 greatly increases risk, with the genotype combination determining the overall effect.

(c) [3 marks]

Over the years of follow-up, the untreated HSV group has the highest and steadily rising cumulative risk of dementia, increasing sharply after about 10 years. The treated HSV group also rises, with a sharp increase after 10 years, but to less than half the risk of the untreated group. The group without HSV shows only a very small increase in risk across the whole period. So untreated HSV is associated with the greatest dementia risk, antiviral treatment lowers that risk, and no HSV gives the lowest risk.

(d) [8 marks]

Whether Alzheimer's disease (AD) is infectious or non-infectious depends on whether a transmissible pathogen causes it, and the data point both ways.

Evidence it could be infectious

Infectious diseases are caused by pathogens passed between people, and a cause is established using Koch's postulates. The HSV study (part c) shows untreated HSV infection is associated with a much higher cumulative risk of dementia, and that antiviral treatment lowers that risk. The study's strengths (a large sample of over 8000 HSV patients with about 25 000 matched controls, a long follow-up, and matched controls) add validity and suggest a viral pathogen may contribute.

Evidence it is non-infectious

Non-infectious diseases are not spread between people and arise from factors such as genes or environment. AD results from the build-up of amyloid beta, which is made by the cell's own protein-synthesis machinery (part a) under the control of genes. The APOE data (part b) show that inherited alleles strongly change the risk (e4/e4 is 14.9 times more likely, e2 is protective), which is clear evidence of a genetic, non-infectious basis.

Judgement

The HSV association is only correlation, not proof: Koch's postulates are not satisfied, the study measured dementia generally rather than confirmed AD, and AD has a strong inherited genetic component. On balance the evidence leans towards non-infectious because of the genetic basis, but because a viral pathogen may also raise the risk, AD cannot be classified with certainty from the information given; it appears to have both genetic and possibly infectious contributing factors.

Marker's note. In (a) outline both transcription and translation with correct terms. In (b)(i) define multiple alleles without confusing them with polygenic inheritance. In (b)(ii) analyse the actual numbers, including allele interactions, not just restating the table. In (c) describe all three lines and their relationship. In (d) define both disease categories precisely, use data from every part of the question to support the judgement, weigh the evidence (validity of the HSV study, Koch's postulates, the genetic data) and reach a justified evaluation.