HSC Biology 2022

(a) [2 marks]. Influenza virus particles released when an infected person coughs or sneezes travel through the air in droplets and are inhaled by another person, who is then infected. This is direct (droplet) transmission from one host to another.

(b) [3 marks]. Suitable named pathogens for each branch:

Marker's note. Give an actual outline of a route of transmission (name the mode, such as direct, indirect or vector, and how it works), not just the word "airborne". In the key, classify by the distinguishing feature shown (presence or absence of a nucleus, cellular or non-cellular) and use precise terms such as fungi or protozoa rather than "eukaryotes".

[3 marks]. Because crossing pure dark purple with pure white gives an intermediate (violet) colour, this is incomplete dominance: the heterozygote is its own phenotype.

Let dark purple be PP, white be WW, and violet (heterozygous) be PW. Cross PW x PW:

Genotypes 1 PP : 2 PW : 1 WW, so the phenotypic ratio is 1 dark purple : 2 violet : 1 white.

Marker's note. Recognise incomplete dominance (the parents were homozygous, so the violet plants are heterozygous), draw and label the Punnett square with a key, and read the genotype ratio straight off as a phenotype ratio (1:2:1, not 3:1).

(a) [2 marks]. Artificial pollination is a manual, human-controlled process. Pollen is collected from the anther (the male part) of a chosen donor plant and deliberately transferred, usually with a brush, onto the stigma (the female part) of a chosen recipient plant, so the breeder controls which two plants cross.

(b) [2 marks]. If pollen from one plant is used to pollinate many plants, and this is repeated over generations, the offspring share more of the same alleles. This reduces the genetic diversity of the population, which makes it more vulnerable to disease or environmental change because fewer individuals carry useful variant alleles.

Marker's note. In (a) make clear it is a deliberate human-controlled transfer and name the correct reproductive parts (anther to stigma). In (b) give a logical outcome with reasoning, most commonly the reduction in genetic diversity and its consequence.

(a) [3 marks]. Plot maternal age (years) on the x-axis and prevalence (per 1000 births) on the y-axis, with even scales. Plot (20, 1.5), (30, 3), (35, 8), (40, 22), (45, 38) and draw a smooth curved line of best fit through the points. Do not join them dot-to-dot, and do not extrapolate the line back to zero.

(b) [2 marks]. As maternal age increases, the prevalence of chromosomal abnormalities increases. The rise is not steady: it is slow up to about age 30, then increases more and more steeply (an increasing rate of increase), rising from about 3 per 1000 at age 30 to 38 per 1000 at age 45.

(c) [3 marks]. One type of chromosomal mutation is a numerical abnormality (aneuploidy), where a cell has one more or one fewer chromosome than the normal diploid number. It is caused by non-disjunction during meiosis: a homologous pair (or sister chromatids) fails to separate, so one gamete receives an extra chromosome and another receives one too few. If such a gamete is fertilised, the offspring has an abnormal chromosome number, for example trisomy 21.

Marker's note. Label both axes with the headings and units given, choose a sensible scale, plot accurately and use a smooth curved line of best fit (not dot-to-dot, and do not extend to zero). In (b) describe how the rate of increase changes and quote data. In (c) name a genuine chromosomal mutation and explain its cause (non-disjunction); a point mutation is not a chromosomal mutation.

[6 marks]

Similarity

In both, new variation can arise from mutation - a random error during DNA replication changes the base sequence and so can produce a new allele in the offspring of either type of reproduction.

Asexual reproduction

Offspring are produced by mitosis from one parent and are genetically identical clones, so apart from mutation there is no source of variation. The only variation among the offspring comes from copying errors (mutation) during mitosis.

Sexual reproduction

This generates much more variation because gametes are made by meiosis and then combine at fertilisation. Three extra sources of variation act:

• Crossing over swaps segments between homologous chromosomes, making new allele combinations.
• Independent assortment randomly sorts maternal and paternal chromosomes into gametes.
• Random fertilisation combines two genetically unique gametes from two parents.

So asexual reproduction varies only by mutation, while sexual reproduction adds crossing over, independent assortment and random fertilisation, giving far greater variation.

Marker's note. Identify mutation as the shared source, then distinguish the mechanisms: mitosis (clones) for asexual versus meiosis plus fertilisation for sexual. Focus on the sources of variation (crossing over, independent assortment, random fertilisation), not just on describing the two modes of reproduction.

[5 marks]. Aim: test whether Jelly Bush honey inhibits bacterial growth. Independent variable: presence of honey on the disc. Dependent variable: the zone of inhibition (no growth) around the disc. Controlled variables: bacterial species and amount, agar type, temperature and incubation time.

1. Put on gloves, safety glasses and a lab coat, and work near a flame using sterile (aseptic) technique.
2. Inoculate several nutrient agar plates with the same amount of one pure bacterial culture, spread evenly over the surface.
3. On each test plate place a sterile paper disc soaked in Jelly Bush honey in the centre.
4. On a matching control plate place a sterile disc soaked in distilled water (no honey).
5. Seal all plates with tape and incubate at about 25 degrees C for 48 hours. Do not reopen the plates after incubation.
6. Without opening the plates, measure the clear zone of inhibition (no growth) around each disc.
7. Repeat the whole experiment at least three times and average the zones.

A larger zone of inhibition around the honey discs than the water discs shows the honey inhibits bacterial growth.

Marker's note. The plates must be inoculated with a pure bacterial culture (not just honey added to bare agar), there must be a valid control and repetition, and the procedure must stay safe - never reopen plates after incubating bacteria. State the variables explicitly.

[6 marks]

Screening (1991). After the screening program began, incidence roughly halved (from about 14 to about 7 per 100 000), so screening appears effective: it detects pre-cancerous cells early for treatment before cancer develops. However, incidence was already falling slightly before 1991, and other factors may have changed at the same time, so the screening program alone cannot be proven to be the sole cause.

HPV vaccination (2007 girls, 2013 boys)

Because HPV causes most cervical cancers, vaccinating girls should prevent infection, and vaccinating boys reduces circulating virus and protects girls indirectly (herd protection). However, the data do not yet show a clear extra fall in incidence after 2007, partly because the vaccine is given to school-aged students whereas cervical cancer takes years to develop in older women, so the effect would not appear in this time frame.

Judgement

The campaigns, especially screening, have very likely contributed to the large drop in incidence, but there is insufficient time and data to assess the vaccination programs separately. Overall the combined public health approach appears successful, but its full effect must be judged over more years.

Marker's note. Use the whole stimulus (the text describing each program and the data on the graph), link each campaign to the incidence around the year it started, and reach a clear judgement. Recognise the limitation that there is too little time since vaccination to see its effect, and that correlation here is not proof of cause.

(a) [1 mark]

Model 2 - the semi-conservative model, where each new DNA molecule has one original (parent) strand and one new strand.

(b) [3 marks]

An enzyme (helicase) unwinds and "unzips" the double helix at a replication fork, breaking the hydrogen bonds between the bases to expose two single template strands. Free nucleotides pair with the exposed bases by complementary base pairing (A with T, C with G), and an enzyme (DNA polymerase) joins them into a new strand along each template. The result is two identical double-stranded DNA molecules, each made of one original strand and one new strand (semi-conservative).

(c) [3 marks]

Prokaryotic and eukaryotic DNA differ in their packaging, not in the molecule itself:

• Prokaryotic DNA is a single circular chromosome; eukaryotic DNA is in several linear chromosomes.
• Prokaryotic DNA lies free in the cytoplasm; eukaryotic DNA is enclosed in a nucleus.
• Prokaryotic DNA is not wrapped around histone proteins and carries few introns; eukaryotic DNA is tightly coiled around histones and contains many introns.

Marker's note. In (b) describe the key steps (unzipping, complementary base pairing against each template, joining the new nucleotides) and make the point that two identical semi-conservative molecules result; naming specific enzymes is not required, but if you name them, name them correctly. In (c) compare DNA structure and packaging (circular versus linear, histones, introns), not just where the DNA is located.

(a) [2 marks]. As the percentage of Bt cotton grown rose after 2002, the cotton yield increased. Bt cotton makes its own insecticidal protein that kills cotton bollworm, so the crop suffered less bollworm damage, allowing more plants to survive and grow and raising the yield per hectare.

(b) [5 marks]. The data partly support Bt cotton for disease control. Early on it worked well: as Bt cotton spread, insecticide used against bollworm fell to almost zero while yield rose, showing Bt cotton controlled the bollworm pest effectively and reduced spraying.

However, the data also show a limitation. Once most of the crop was Bt cotton, insecticide use against hemiptera climbed sharply, eventually higher than total insecticide use before Bt cotton. Removing the bollworm reduced competition, so hemiptera (which Bt toxin does not kill) became the dominant pest and now had to be controlled with more spraying.

So the data support Bt cotton as a control for one specific pest in the short term, but not as a complete or lasting solution: a new pest replaced the old one, and spraying returned. Bt cotton needs to be combined with other pest management to keep yields up.

Marker's note. In (a) give a clear cause-and-effect link (more Bt cotton -> bollworm killed -> less damage -> higher yield). In (b) use all the graphs, recognise that bollworm spraying fell to zero while hemiptera spraying rose because the niche was opened up, weigh both the benefit and the limitation, and make a judgement on the extent of support.

[7 marks]

Because malaria needs the mosquito vector to spread, the disease occurs mainly where the vector lives, and the two maps overlap across much of Africa, South America and South Asia. But there are two kinds of difference to explain.

Vector present, disease absent (for example Europe and North America)

The mosquito occurs here yet malaria does not circulate. This is most likely because effective public health and wealth break transmission: bed nets, insecticide spraying, draining of breeding sites, and prompt diagnosis and antimalarial drugs stop the parasite reaching new hosts even though the vector is present. Imported cases from travel are treated quickly and do not spread.

Disease present, vector apparently absent (for example parts of northern Africa)

Here malaria cases occur where Map 1 shows little vector. This may be because the vector data are incomplete for those regions, or because infected people keep travelling in and out, maintaining the parasite in the population even where vector numbers are low.

So the distribution of a vector-borne disease depends not only on where the vector lives but on human factors - public health control measures, wealth, drug availability and movement of infected people - which can break or maintain transmission independently of the vector's range.

Marker's note. Refer to specific regions on both maps and explain both mismatches: vector present but no disease (control measures, wealth, treatment) and disease present but little vector shown (incomplete data or travel maintaining cases). Discuss, weighing several reasons, rather than just stating that the maps differ.

(a)(i) [4 marks]. The method has real strengths. The sample is large (over 100 000 women total) and women were grouped by their exposure to cigarette smoke, with a valid control group (non-smoker women with non-smoker husbands). The 14-year follow-up is long enough for a slow disease like lung cancer to appear, so the study can validly link exposure to mortality.

Its weaknesses are in the assumptions. Grouping by husband's smoking assumes every woman in a group had similar exposure (similar time with the smoker and similar amount smoked), which in reality varies widely. Other lung cancer causes (such as air pollution or occupation) were not controlled. The large cohort tends to average these effects out, so overall the method is reasonably valid but would be stronger with measured individual exposure.

(a)(ii) [3 marks]. Two justified conclusions:

• Exposure to cigarette smoke increases the risk of dying from lung cancer. Mortality rises from 8.7 (no exposure) to 15.5 for women with smoking husbands (passive exposure) and to 32.8 for women who smoke (active exposure) - a clear dose-like increase.
• Smoke is not the only cause of lung cancer. Even unexposed women had a mortality of 8.7 per 100 000, so other factors must also cause lung cancer.

(b)(i) [4 marks]. Flow chart for EGFR protein synthesis:

EGFR gene (DNA) -> Transcription in the nucleus: RNA polymerase reads the template strand and builds mRNA by complementary base pairing (A with U, C with G) -> mRNA leaves the nucleus to a ribosome -> Translation: the ribosome reads mRNA codons; each tRNA anticodon pairs with a codon and brings the matching amino acid -> amino acids join by peptide bonds into a polypeptide chain -> the polypeptide folds into the functional EGFR protein.

(b)(ii) [4 marks]. A mutation in the EGFR gene changes its DNA base sequence and so its codons. If the change is in a coding region, it alters the amino acid sequence of the polypeptide. A different amino acid sequence changes how the protein folds, altering its three-dimensional structure - including its enzyme (active site) and receptor regions. A faulty active site or receptor can leave the protein permanently switched on, so it no longer regulates cell division correctly. Because EGFR normally controls when a cell divides, this drives uncontrolled cell division, and uncontrolled division is cancer - here, lung cancer.

Marker's note. In (a)(i) give a judgement supported by features of the methodology (sample size, control group, follow-up time) and weigh the assumptions about exposure. In (a)(ii) state conclusions and back each with the specific mortality figures. In (b)(i) the flow chart must cover transcription (DNA to mRNA), translation (mRNA, ribosome, tRNA) and folding into the EGFR protein - keep it specific to EGFR, not generic. In (b)(ii) link DNA change -> altered amino acids -> altered structure and function -> loss of control of cell division -> uncontrolled division -> cancer.

[7 marks].

Mutation is the original source of the allele: a random change in DNA first produced it in one population. The very high frequency in South Asians (60.3%) compared with the others suggests the allele arose and built up mainly in that ancestral population rather than appearing equally everywhere.

Natural selection can change its frequency where the relevant virus exerts a selection pressure. If the virus caused severe, often fatal inflammation, carriers of the allele would be less likely to survive and reproduce, so selection would lower its frequency - consistent with the very low frequencies in African (2.4%) and East Asian (1.8%) populations. Where the virus was rare or the allele also conferred some benefit, selection against it would be weak, keeping frequency higher.

Genetic drift changes allele frequencies by chance, especially in small founding populations. Random differences in which individuals survived and bred could have raised the allele frequency in some ancestral groups and lowered it in others independently of any advantage.

Gene flow is the movement of alleles between populations through migration and interbreeding. Limited gene flow between these geographically separated ancestries meant the different frequencies were not evened out, so the large differences (60.3% to 1.8%) were maintained; more migration would have made the frequencies more similar.

Marker's note. Define and explain each of the four processes and link each to the data in the table (especially the very high South Asian frequency and very low African and East Asian frequencies). Use cause-and-effect statements and refer to the stimulus throughout; do not confuse gene flow (migration between populations) with genetic drift (chance change within a population).