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2020 HSC Biology Exam Paper Solutions

In this post, our Science Team gives you the answer to the 2020 HSC Biology paper.

Our Biology Team has been hard at work putting together the solutions to the 2020 HSC Biology Exam. Read the solutions below and see how you went!


2020 HSC Biology Exam Paper Solutions


Section 1: Multiple Choice

1B Curling up is an action that the animal has taken.
2D Sexual reproduction involves the fusion of gametes.
3C The egg is released (ovulation), then fertilised by sperm. The developing blastocyst implants on the uterus wall and the outer layer eventually becomes the placenta.
4C Plasmodium is the pathogen that causes the disease, the mosquito is the vector that transmits the plasmodium into the host.
5B Prokaryotes have circular DNA, eukaryotes have linear DNA.
6C Other options are too costly or ineffective.
7A The number of microbes counted will be dependent on which food is sampled. An untouched agar plate is used to check for contamination.
8C Many non-infectious diseases are genetic.
9C The aim is to lower the incidence of skin cancer, so the incidence of skin cancer should be recorded.
10A The farmer is introducing a new allele into the population.
11A Light will bend as it passes from air through the dense cornea.
12B Cloning is the most effective way to produce many individuals with the same favourable traits.
13BGene cloning is the replication of a gene.
14DThe heterozygous phenotype results in some receptors, but not as many as a normal cell. This represents an intermediate phenotype.
15C Drug C inhibits 70% of viral entry at this dose, with minimal toxicity.
16B Adenine always matches with thymine and guanine always matches with cyanine to form a double helix. Thus, adenine and guanine will make up 50% of the bases.
17B Mutations that occur on non-coding DNA are not expressed as proteins and will not affect phenotype.
18D Based on the pedigrees, D is the only individual that shares DNA with both I and II.
19A A illustrates metaphase II of meiosis in a cell with 6 chromosomes. The cell starts with 6 chromosomes (3 homologous pairs) the daughter cells will end up with 3 only.
20A Trait A is similar in adoptive siblings that do not share much genetics. Trait B is very similar in identical twins with identical genetics. Trait C has little similarity between close relations or adoptive siblings.


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Section 2: Short response

Question 21

As Cholera can be spread by contaminated water, boiling water before drinking it or using it to wash food will kill the bacteria and prevent them from infecting the body. To prevent contamination of food with the bacteria, hands should be washed thoroughly with soap and water before preparing food or after going to the bathroom. Visitors and residents of cholera prone areas should be vaccinated as this will eliminate the disease in the body before illness can occur, which further prevents the spread.


Question 22

The benefit of using pharmaceuticals such as antibiotics and antivirals is that recovery from the disease occurs faster and reduces the chance the disease will spread to others.

A major limitation of using these drugs is that the pathogen (e.g. bacteria or virus) can become resistant to the drug and it will no longer work. The drugs act as a selection pressure and any individual bacteria or viruses that have mutations giving them resistance to the drug will survive and reproduce. This results in a population that is resistant and can spread to other people.


Question 23a

This is a point mutation, specifically a base substitution.


Question 23b

Chromosome mutations change the number or structure of chromosomes. For example, an organism with trisomy has one extra chromosome.


Question 24a

2020 Biology HSC Solutions 24a graph

Question 24b

See dotted line above.


Question 24c

The role of the kidney is to remove urea from the blood, maintain salt and water balance, remove toxins and regulate pH. This is achieved through active and passive transport. When the kidney stops functioning, haemodialysis can compensate. Blood is pumped out of the body into a dialyser which has a semipermeable membrane. Waste such as urea and excess salt diffuses across the membrane into the dialysate solution which flows in the opposite direction and is removed. Haemodialysis is a lifesaving treatment but relies on diffusion (passive transport) and so is not as effective as a kidney which can actively transport molecules against a concentration gradient and reabsorb useful substances.


Question 25a

The mean for both groups is very similar (43 and 40) which suggests that there is no difference between the two modes of fertilisation, but no conclusion can be made from the data as there is too much variation (i.e. the standard deviation is too large).


Question 25b

A larger number of species need to be included in each mode of fertilisation group as currently 6 species included for each group is an insufficient sample size and the data is not representative. For example, if invertebrate species such as coral were included in the external fertilisation group it would dramatically increase the average.


Question 25c

Most species that use external fertilisation demonstrate less parental care of the offspring, and so they produce many offspring but do not invest time, resources and energy in them. With a high number of offspring produced there is a high chance that some will survive regardless.


Question 26a

Two parents with the orange phenotype have had some offspring with yellow and some with orange. Since two orange parents have produced some yellow offspring, yellow must be recessive.


Question 26b

If the inheritance pattern is autosomal recessive, then Individual I may be homozygous dominant (OO) or heterozygous dominant (Oo). Individual II must be homozygous recessive (oo). In this case, a cross will result in all orange offspring or 50% orange and 50% yellow with no difference between male and female offspring.




If the inheritance pattern is sex-linked then individual I only has one copy of the dominant allele as males only carry one copy of the X chromosome. A cross between individuals I and II will result in all orange female offspring and all yellow male offspring. This will confirm that the inheritance pattern is sex-linked.


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Question 27a

Age, sex, education and socioeconomic status were taken into account and the study was conducted over a long time period (11 years).


Question 27b

The data in the graphs does support the hypothesis as the survival % was lower at mid-range and high-range concentrations of arsenic exposure compared to low range exposure over the 11-year study period, in both males and females. At lower concentrations of exposure <90 ug males had a lower survival rate than females which indicates there may be some other factor causing mortality in males (e.g. higher risk of accident). The data also indicates that females are more sensitive to the arsenic in the range 90-223 ug as they had much lower survival rates in this range of exposure. While the data does support the hypothesis, other factors should be investigated to establish why females are more sensitive to mid-range exposure and why males have a lower survival rate at low exposure. Survival % data for a control group that was not exposed to any arsenic would improve the validity of the study.


Question 28a

In the model shown above, before crossing over occurs, each homologous chromosome is composed of one paternal and one maternal chromatid. In reality, during interphase, the DNA is replicated and then condenses in prophase to produce one chromosome composed of two identical maternal chromatids and one chromosome composed of two identical paternal chromatids.


Question 28b

The process of meiosis maintains genetic variation as it produces offspring that are genetically unique, i.e. they have a different combination of alleles compared to each other or to their parents. This is achieved through crossing over, where segments of maternal and paternal chromosomes are swapped over creating new combinations of alleles. It is also enabled by random segregation where the four chromatids for each chromosome pair are randomly allocated to the four daughter cells.


 Question 29

Mutation can affect the gene pool of populations by introducing new alleles. While mutations that occur in non-coding regions are unlikely to have an effect on the organism, mutations that occur on a coding gene can produce a new allele of the gene which results in a new phenotype. If this new phenotype is beneficial to the organism, they are more likely to survive and reproduce. Their descendants also have a higher chance of surviving and thus the allele will become more common in the population over time (evolution).

Genetic drift can also lead to evolution. In small populations, allele frequency can change quickly due to random external factors such as natural disasters. The event does not act as a selection pressure, but by chance may remove individuals with a specific allele. This allele is then lost from the population.


Question 30

Genetic technologies have had a significant impact on the management of both infectious and non-infectious diseases. The capability to sequence DNA in itself has revolutionised management of disease as technologies such as sanger sequencing have allowed us to identify gene sequences that cause non-infectious disease such as phenylketonuria, which results in early diagnosis and treatment. This technology also allows us to sequence the genomes of viruses. Diagnostic tools such as real-time PCR can then be used to detect the presence of viral RNA/DNA in mucus samples (e.g. COVID-19).

Recombinant DNA technology has enabled scientists to introduce foreign genes (from a different species) into bacteria, plants and animals. This has allowed for the creation of agricultural plants that are resistant to infectious disease, due to an added gene. Introducing human genes into bacteria in this way causes them to produce human protein products for medicine, such as the hormone insulin which is used to treat diabetes. A beta-carotene gene from corn has been introduced into a rice plant to produce transgenic golden rice. Consuming the rice will prevent the vitamin A deficiency that causes blindness.

CRISPR-Cas9 is a recently discovered guide RNA-enzyme complex that is capable of finding a specific location on DNA and cutting it. This technology makes it easier to cut DNA to switch off a gene (knock-out) or cut can be used to insert a transgene. Switching off genes in mice allows us to study the function of the gene in an animal that has similar systems to humans. This is used in medicine to understand how gene disfunction can cause disease. CRISPR-Cas9 has been adapted to produce gene drives which not only modify a gene in an individual, but also modify all copies of the gene in their future offspring. In this way, it may be possible to modify mosquito populations, so they eventually become extinct or are not capable of carrying the Malaria parasite. This has the potential to save millions of lives.


Question 31a

Blood glucose levels must be maintained in the blood within a certain level to prevent illness. This homeostasis of blood glucose is maintained via a negative feedback loop. When blood glucose levels increase (stimulus), this is detected by the pancreas (receptor/effector) which then releases insulin (response). Insulin is a hormone that causes the glucose molecules in the blood to be converted into glycogen and stored in the cells of the body, including the liver. This is visible in the graph as the carbohydrate meal causes a spike in glucose levels soon after (Figure 1), with a corresponding spike in insulin (Figure 2). These then gradually reduce over time as the glucose in the blood is naturally used up.

When blood glucose levels are low (stimulus) the pancreas (receptor/effector) releases the hormone glucagon (response). This causes the stored glycogen to be broken down into glucose, in order to bring the blood glucose level back up. This can be seen in the figures as when the glucose levels are low (0-1h Figure 1), the glucagon levels are high (0-1h Figure 3).


Question 31b

While maintenance of temperature and maintenance of glucose levels are both achieved with a negative feedback loop, the maintenance of temperature involves the nervous system while glucose is maintained using hormones only. When body temperature is too high or low it is detected by receptor cells which then send a signal via sensory neurons to the central nervous system (CNS). The CNS will process the information and send a signal via motor neurons to an effector (muscle or gland) that will enact a response (e.g. shivering when cold, sweating when hot).


Question 32a

The virus enters salivary glands and is found in the saliva. The saliva is now a route of exit out of the body for the virus (6). The virus can then enter the tissue of the next host through bites which are a mode of entry that bypasses the skin barrier (1).


Question 32b

(i) The virus cannot be classified as a cellular pathogen as it is not composed of a living cell. While the virus has some structures found in cells (lipid bilayer, polymerase, RNA) there are no organelles such as ribosomes which are required for a cell to make the products it needs to grow, survive and replicate. The virus, therefore, will not be able to reproduce on its own, therefore it is not cellular and not considered to be living.

(ii) RNA polymerase is an enzyme that creates an RNA copy of a DNA or RNA sequence by attaching complementary RNA which then detaches from the original DNA/RNA. In the case of the viral RNA polymerase, it is delivered into the cell along with the viral RNA. It then transcribes the viral RNA to form viral mRNA. This mRNA contains viral genes which will then go through translation by a ribosome in the cell. The ribosomes in the cell will produce viral proteins (L,P,M,G and N) which will be assembled to form new viruses. The proteins L and P will assemble to form more RNA polymerase. The extra copies of RNA polymerase allow for faster transcription of the Viral RNA, and also for many copies of the complementary vRNA to be produced. These copies will become the RNA within the new viruses that are assembled. Thus, the viral RNA polymerase has two roles, to transcribe the viral RNA for protein synthesis and also to make copies that form the genome of the new viruses.


Question 32c

PEP prevents rabies from developing by taking advantage of the body’s adaptive immune response. Normally when the body is infected with a virus, the antigens on the surface of the virus are detected by the immune system and this triggers specific B-cells to begin cloning. Some of the B cells become plasma cells and begin producing antibodies. These antibodies have specific binding sites that match the antigen. The antibodies attach to the virus to prevent virus entry into cells and attract macrophages. Killer T-cells detect infected body cells and destroy them before the disease can spread.

Normally it would take around a week for the immune system to build up a sufficient number of specific immune cells. For rabies, this is not fast enough to prevent the virus from infecting the nervous system and eventually entering the salivary glands (first graph part (c)). PEP solves this problem by introducing human antibodies (HRIGs) against the virus directly into the blood stream where they will bind to the virus and help the immune system get rid of it (artificially acquired, passive immunity). By the time the HRIG treatments are stopped, the vaccine included in the PEP has taken effect and the immune system has started producing its own antibodies (artificially acquired, active immunity).

The vaccine will include dead/weakened rabies virus or antigens extracted from the rabies virus. An example of an antigen that might be used in the vaccine would be the G protein in Diagram 1 from part (b). After 7 days the body begins to make its own antibodies against the virus and this can be seen in the lower graph for part (c), the body maintains a high level of antibodies until the virus is destroyed and no longer found in the body. With PEP the viral infection never becomes full-blown rabies illness.


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Written by Matrix Science Team

The Matrix Science Team are teachers and tutors with a passion for Science and a dedication to seeing Matrix Students achieving their academic goals.


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