HSC Biology 2020

[3 marks]. One strategy per mark, each linked to its effect on the bacterium:

1. Treat and purify drinking water (e.g. chlorination or boiling) so the Vibrio cholerae bacteria in it are killed before anyone drinks it.
2. Dispose of sewage safely and away from water supplies, so faeces carrying the bacteria do not contaminate the food and water people consume.
3. Wash hands with soap after going to the toilet and before handling food, removing bacteria from the hands so they are not transferred to the mouth.

Marker's note. Give strategies specific to cholera, and link each to its purpose (boil the water to kill the bacterium, not just boil the water). Prevention, not treatment: vaccinating to build herd immunity prevents spread, whereas treating cases with antibiotics does not.

[3 marks]. Benefit: antibiotics treat bacterial infections by inhibiting bacterial growth or killing the bacteria (for example, stopping cell-wall synthesis), so the infection is cleared and the patient recovers.

Limitation: antibiotic resistance in bacteria is becoming increasingly common, so the bacteria survive treatment and the antibiotic loses its effectiveness against that disease. Antibiotics also do not work on viral infections.

Marker's note. Link the pharmaceutical to the correct pathogen (antibiotics treat bacteria, not viruses) and to the treatment of the disease. It is the bacteria that develop resistance, not the patient. Do not confuse antibiotics with antibodies.

(a) [1 mark]. A point mutation (base substitution), where one base is replaced by another.

(b) [2 marks]. A chromosomal mutation affects part of or a whole chromosome rather than a single base. For example, a non-disjunction event changes the number of chromosomes in the genome, so the cell gains or loses a whole chromosome.

Marker's note. In (a) name the type (point or substitution mutation), do not just describe the diagram. In (b) outline a genuinely different type and link it to an effect; chromosomal mutations alter part or all of a chromosome, not a specific base sequence. Do not list germline or somatic, or name mutagens such as UV.

(a) [2 marks]. Plot year on the x-axis and GFR on the y-axis, label both axes (GFR in mL per minute), use an even scale that fills the grid, and plot the nine points accurately.

(b) [2 marks]. Draw a line of best fit through the points and extrapolate it forward to where GFR reaches 15 mL per minute. The recent decline (roughly 10 to 15 mL per minute each year after 2016) projects the line to cross 15 mL per minute at about 2021. Mark that crossing point and read the year off the x-axis.

(c) [3 marks]. A key function of the kidneys is to remove nitrogenous waste such as urea from the blood; losing kidney function lets urea build up to toxic levels. In dialysis, the patient's blood flows on one side of a selectively permeable membrane and dialysate fluid flows on the other. Because urea is at a high concentration in the blood and a low concentration in the dialysate, urea diffuses down its concentration gradient across the membrane out of the blood. This removes the waste the failing kidneys can no longer remove, compensating for the lost function.

Marker's note. In (a) and (b) use the full grid with an even scale, draw a line of best fit (not just dot-to-dot), extrapolate, and mark the predicted year on the graph itself. In (c) use precise terms (diffusion not osmosis, urea not ammonia) and relate the process to a real kidney function, rather than describing the social setting of dialysis.

(a) [3 marks]

The data do not support the hypothesis; the conclusion is that there is no clear difference. The mean number of young is almost the same for internal (43) and external (40) fertilisation, a difference far smaller than the variation within each group. The standard deviations are very large (plus or minus 55 and plus or minus 32), so the ranges overlap heavily and the data are inconclusive. There is no evidence that external fertilisation produces more young.

(b) [2 marks]

Include many more species in each group. The students used only a handful of species, so a few high or low values (outliers) strongly skew each mean. Testing a much larger, randomly chosen sample of species reduces the effect of outliers and makes any real difference between the two modes easier to detect, improving the validity of the test of the hypothesis.

(c) [2 marks]

External fertilisation lets the animal release very large numbers of gametes at once, which can be fertilised together in the surrounding water. This produces many offspring in one event, so even though many are lost, enough survive to maintain the population. The parent also avoids the energy cost of internal gestation.

Marker's note. In (a) draw a clear conclusion and justify it with the means and the standard deviations, recognising the data are inconclusive. In (b) link the improvement to the hypothesis. In (c) give a feature of external fertilisation that is advantageous and link cause to effect on the species; do not confuse external fertilisation with asexual reproduction.

(a) [2 marks]. Both parents are orange yet they produce yellow offspring. The yellow trait must have been present but hidden in both parents and only shows when an offspring inherits two yellow alleles, so yellow is recessive and orange is dominant.

(b) [4 marks]. Cross the orange parent I with the yellow parent II and look at the colours of male versus female offspring.

If colour is autosomal, parent II is yellow (genotype aa) and parent I is orange (AA or Aa). If I is Aa, then about half the offspring are yellow and half orange, with yellow and orange spread equally across both sexes.

If colour is sex-linked (on the X), parent I is the orange male X^A Y and parent II the yellow female X^a X^a. Then every son receives X^a from the mother and is yellow, while every daughter receives X^A from the father and is orange.

So if the yellow and orange offspring are split by sex (all yellow sons, all orange daughters), the inheritance is sex-linked; if the colours appear in both sexes equally, it is autosomal. Punnett squares for each case make the expected ratios clear.

Marker's note. In (a) use the pedigree to argue orange parents with yellow offspring means yellow is recessive, with correct terminology. In (b) explain the outcomes for both autosomal and sex-linked cases and relate the difference (a sex bias in colour) to the type of inheritance; Punnett squares and clear formats help.

(a) [2 marks]. Any two of: the very large sample size (58 406 people), which reduces the effect of chance; the long study period (11 years), which captures effects of chronic exposure; and controlling for confounding factors (age, sex, education and socioeconomic status) so differences in survival can be attributed to arsenic.

(b) [4 marks]. The data partly support the hypothesis. In both males and females, survival is highest in the group exposed to less than 90 micrograms per litre, and survival falls as average arsenic exposure rises, which suggests arsenic increases mortality. The dose-response is clearest in males, where each higher exposure band lowers survival; in females the two higher exposure bands give similar survival, hinting that other factors (such as nutrition or genetics) also act.

Against the hypothesis: although the sample is huge and survival declines progressively over the 11 years, the actual fall in survival is tiny, only about 0.1% or less, exaggerated by the narrow y-axis scale (99.90 to 100%). So the trend supports a link, but the effect size is very small.

Marker's note. In (a) take the features from the stimulus rather than speculating about controls not stated. In (b) discuss points for and against, using specific data trends and the misleading limited y-axis scale, and link them back to the hypothesis rather than just describing the graphs.

(a) [3 marks]. The homologous pair is drawn incorrectly. In a homologous pair one chromosome is maternal and the other paternal. Before crossing over, each chromosome has already replicated, so it consists of two identical sister chromatids; both chromatids of the maternal chromosome should be maternal, and both of the paternal chromosome paternal. The model wrongly shows a single chromosome made of one maternal and one paternal chromatid, confusing chromosomes with chromatids.

(b) [3 marks]. Meiosis increases genetic variation in several ways. In crossing over (prophase I), homologous chromosomes exchange segments, producing new combinations of alleles on each chromatid. In independent assortment, homologous pairs line up randomly at metaphase I and separate independently, so gametes receive many different combinations of maternal and paternal chromosomes. These processes mean the gametes, and therefore the offspring, are genetically varied rather than identical to the parent.

Marker's note. In (a) explain that paired homologues are one maternal and one paternal, and that after replication a chromosome's two chromatids must be identical (both maternal or both paternal); distinguish chromosomes from chromatids. In (b) explain at least one process (crossing over, independent assortment) and link it to increased variation; do not blur independent assortment, random alignment and random segregation.

[5 marks]. A gene pool is the total set of alleles in a population. Evolution is a change in the gene pool (allele frequencies) over generations. Two processes that change it:

• Gene flow is the movement of alleles into or out of a population, for example when a migrant breeds with residents and adds new alleles, or when individuals leave. This alters allele frequencies and can introduce variation that natural selection then acts on.
• Genetic drift is a random change in allele frequencies, especially marked in a small population. By chance, some alleles are passed on more than others (for example after a population crash), so allele frequencies shift and some alleles may be lost, independent of any advantage.

Both processes change which alleles are present and how common they are. Acted on over many generations, often together with mutation and natural selection, these changes accumulate so the population evolves.

Marker's note. Describe the main features of two genuine mechanisms (mutation, gene flow, genetic drift) and relate the change in the gene pool to evolution. Note that meiosis and fertilisation shuffle existing alleles but do not add or remove them, and that evolution occurs over many generations, not in one.

[7 marks]

Two genetic technologies, applied to both disease types:

Recombinant DNA technology (non-infectious disease, e.g. diabetes)

Type 1 diabetes is a non-infectious disease in which the pancreas no longer makes insulin. A human insulin gene is inserted into bacteria, which then transcribe and translate it to mass-produce human insulin. Diabetic patients inject this insulin to control their blood glucose, so the disease is managed and life-threatening complications are avoided.

Recombinant DNA technology (infectious disease, e.g. recombinant vaccines)

For an infectious disease such as hepatitis B, the gene for a viral surface protein (antigen) is inserted into yeast, which produce that antigen. The harmless antigen is used as a vaccine: it triggers active immunity and memory cells, so vaccinated people are protected and the spread of the pathogen through the population falls. (Bt crops engineered to resist insect pests, reducing crop disease, are another acceptable infectious-disease example.)

Overall impact

At the genetic level these technologies move a chosen gene into a host cell that then expresses the desired protein. This has transformed disease management, providing reliable supplies of treatments (insulin) and safe, specific vaccines, improving both the treatment of non-infectious disease and the prevention of infectious disease.

Marker's note. Be specific: outline two genuine genetic technologies and what happens at the genetic level, then link each to its impact on managing infectious and non-infectious disease. Distinguish a genetic technology from a generic biotechnology, and avoid grouping diseases or technologies with vague statements.

(a) [6 marks]. Blood glucose is kept within narrow limits by negative feedback involving the pancreas and liver.

In the first hour (at rest) glucose, insulin and glucagon sit at steady resting levels. After the carbohydrate meal, glucose is absorbed from the gut and plasma glucose rises (Figure 1). The rising glucose stimulates the beta cells of the pancreas to release insulin, which is why insulin rises just after glucose (Figure 2). Insulin makes body cells take up glucose for metabolism and makes the liver store glucose as glycogen. As glucose is taken out of the blood, plasma glucose falls back toward resting level; the falling glucose removes the stimulus, so insulin also falls. This is negative feedback.

When glucose drops, the alpha cells of the pancreas release glucagon (Figure 3), which makes the liver break glycogen down to glucose and release it into the blood, raising glucose again. So glucagon falls while glucose is high and rises as glucose falls. The opposing actions of insulin and glucagon hold blood glucose around its set point.

(b) [3 marks]. Temperature change is detected by the hypothalamus in the brain, whereas glucose change is detected directly by the pancreas. The temperature response is coordinated largely through the nervous system (for example triggering shivering or sweating), while glucose is regulated through hormones (insulin and glucagon) in the bloodstream. So the two differ in both the control centre and the type of message used to respond.

Marker's note. In (a) refer to features of all three graphs and show detailed understanding of the roles of insulin, glucagon, glycogen, the alpha and beta cells, the pancreas and the liver in negative feedback. In (b) give clear differences, linking each component of glucose control to its equivalent in temperature control (control centre, body system, type of message).

(a) [2 marks]

Two features from the flow chart: the virus travels up the nerves to reach the salivary glands of the host, so infectious virus collects in the saliva; and the host then bites another animal, injecting that saliva through the skin, giving direct contact transmission to a new host.

(b)(i) [3 marks]

A cellular pathogen (such as a bacterium) is a complete cell with its own cytoplasm, ribosomes and machinery, and usually a large DNA genome, so it can carry out its own life processes. Diagram 1 shows the rabies virus is only a small particle: a short single-stranded RNA genome with a few proteins and a lipid bilayer, and no cytoplasm, ribosomes or cell membrane. Because it lacks the cellular structures needed to carry out its own metabolism and reproduction, and must use a host cell to reproduce, it cannot be classified as a cellular pathogen.

(b)(ii) [5 marks]

The viral RNA polymerase is made of the L and P proteins shown in Diagram 1, and it has two roles in reproduction (Diagram 2).

First, transcription: the RNA polymerase uses the viral RNA genome as a template to make complementary mRNA strands. Host-cell ribosomes then translate these mRNAs into the five viral proteins (G, M, N, P and L), including more L and P proteins.

Second, replication of the genome: the same RNA polymerase copies the viral RNA into a full-length complementary strand, which is then used as a template to make new copies of the original viral RNA genome for the offspring viruses.

The link between the two is that the newly translated L and P proteins are themselves more RNA polymerase, so the proteins made in transcription and translation go on to catalyse the genome replication. Thus RNA polymerase produces both the proteins and the RNA genomes needed to assemble new virus particles.

(c) [8 marks]. Without PEP, the rabies virus replicates at the bite site, then travels along the nerves to the central nervous system (the graphs show it reaching the CNS around day 7), causing fatal encephalitis. The patient has no memory cells for rabies, so their own antibody response (which only begins around day 7) is too slow to stop the virus before it reaches the CNS.

PEP works in two complementary ways shown in the data.

• Passive immunity (HRIG): the injected human rabies antibodies act immediately. They bind to and inactivate (neutralise) the virus at the entry site and flag it for destruction by phagocytes, lowering the virus concentration in the first days (the with-PEP graph shows virus falling around days 6 to 8). This buys time but is short-lived, lasting only about three weeks.
• Active immunity (vaccine): the vaccine contains inactivated, harmless rabies virus. Macrophages present its antigen to helper T cells, which activate specific B cells; these divide into plasma cells that secrete rabies-specific antibodies and into memory B cells. The graph shows the patient's own antibodies rising sharply from about day 7 and reaching a maximum near day 21.

As the vaccine-induced antibodies take over from the fading HRIG, virus at the entry site is neutralised before it can reach the CNS, so by about day 11 no virus remains and rabies (and death) is prevented. The two parts of PEP overlap in time: the HRIG controls the virus early while the vaccine builds lasting protection.

Marker's note. In (a) draw the features from the stimulus (bite essential for transmission; virus reaching the salivary glands). In (b)(i) use structural features (single-stranded RNA, no cell machinery) rather than processes. In (b)(ii) use the diagrams to show RNA polymerase drives both transcription/translation and RNA replication, and that the new L and P proteins are themselves polymerase. In (c) use both graphs and your own knowledge to explain passive (HRIG) and active (vaccine) immunity and the role of antibodies, rather than just describing the graphs.