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2021 HSC Biology Exam Solutions

Want to see how you went in this Year's Biology HSC exam? Read on!

Our Biology team has been hard at work putting together the solutions for the 2021 HSC Biology Exam Paper. Read on to see how you fared!


2021 HSC Biology Exam Solutions

Keep in mind that these answers were not written under exam conditions and may be longer than required.


Section I: Multiple Choice

Question Answer Solution
1 C As the iodised salt successfully treated the condition, the patient most likely has a deficiency of iodine.
2 D Frog spawning involves a female laying eggs and a male fertilising them with sperm.
3 B Pollen is manually transferred from the anther of one plant to the stigma of another.
4 B Pharmaceuticals are an example of a treatment that can be used for non-infectious disease, the others are examples of prevention for infectious diseases.
5 C This meal will increase blood glucose levels. Insulin levels rise to trigger the storage of glucose to bring blood glucose levels back down.
6 A A single chromosome can contain thousands of genes, so a large deleted segment of a chromosome will affect many genes.
7 C In DNA replication A pairs with T and C pairs with G.
8 D A conclusion must relate to the research problem. While A may be true it is a statement for the discussion, not a conclusion of a study.
9 D It is the role of ribosomes to form polypeptide chains as part of protein synthesis.
10 A A is dominant and a single copy will produce a normal phenotype. a1 and a2 produce cystic fibrosis, but a1 has the highest frequency so a1/a1 is the most likely genotype of a cystic fibrosis patient.
11 B The transgenic crops have most likely pollinated their nearby relatives.
12 C HCG is released once the embryo has implanted. Fertilisation cannot occur until around 2 weeks after the end of the menstrual cycle when ovulation occurs.
13 A When there is higher availability of water the stomata can open more and release more water through transpiration. When there is less water available the stomata will partially close to reduce water loss.
14 A CJD can be caused by a mutation or by exposure to an infectious agent.
15 B Bonds have formed between adjacent thymine bases on a single strand of DNA. This is known as a thymine dimer.
16 C As the UV dose is increased the number of bacterial colonies decrease. There is no control plate.
17 B Sister chromatids separate in anaphase of mitosis.
18 D This pattern is consistent with a sex-linked recessive disorder if the female is a carrier (heterozygous).
19 C The graph indicates that the two alleles can produce three different phenotypes – fast, medium and slow alcohol metaboliser.
20 B The enzyme activity has decreased which has resulted in a build-up of acetaldehyde in mice with the ALDH2 allele.

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Section II: Extended Response

Question 21

(a) Label the cell wall and nucleoid (non-membrane bound DNA).

(b) Box 1: Bacteria are grown in pure culture. Box 2: Pig 2 develops the same symptoms as pig 1.

(c) The benefit of Farm 1 treating with antibiotics is that the pigs recover quickly, and the disease will not spread to other pigs. The limitation of this strategy is the likelihood that the bacteria will develop antibiotic resistance and the antibiotics will eventually no longer work on this disease.
The benefit of the strategy used by farm 2 is that it eliminates possible vectors/reservoirs of the disease from the farm so they cannot spread it. A limitation of this strategy is it does nothing to help the pigs that are already sick and spreading the disease. Ideally these two strategies would be used at the same time.

Question 22

Let B = black, b = white. If white is recessive then the white rabbit has the genotype bb and if the black rabbit has a mother with white fur then it must be heterozygous (Bb).

B b
b Bb bb
b Bb bb

The offspring will be 50% heterozygous black (Bb) and 50% homozygous white (bb), thus a ratio of black to white of 1:1.


Question 23

Test number two shows that ovulation is not occurring. In order to be valid the test must have a control line, so tests one and two are the only ones that are valid. Test number one has a control line and a test line, so it shows a positive result for LH and ovulation. Test two has a control line to confirm it is working, but there is no second test line. Therefore test 2 is a negative result for LH and ovulation is not occurring.


Question 24

(a) This trait is likely to be the result of a somatic mutation as it has not been inherited or passed on. The male with the trait has not inherited the trait from his parents as it is not possible for two homozygous recessive (unaffected) parents to have offspring with a dominant trait. If the affected male was homozygous dominant, they would pass the trait to all of their offspring and if they were heterozygous dominant then around half of their offspring would have the trait. However, the pedigree indicates that none of the offspring have the trait.

(b) Somatic cells are body cells and germ cells become sex cells. Mutation B will only affect twin 1 as it occurred after the twins had formed. This mutation will be found in the germ cells and somatic cells of twin one. As a result, Twin 1 may pass this mutation on to future offspring. Mutation C will only affect twin 2 as it occurred after the twins had formed. This mutation will only affect some somatic cells as it occurred after germ cells and somatic cells had formed. Twin 2 will not pass this mutation on to its offspring.


Question 25


2021 HSC Biology Exam Solutions Question 25 graph


(b) The right ear is functioning within the normal hearing range, while the left ear is significantly impaired. There is no profound hearing loss, but a hearing aid or other device may be required.

(b) The right ear is functioning within the normal hearing range, while the left ear is significantly impaired. There is no profound hearing loss, but a hearing aid or other device may be required.

(c) A blockage of the outer ear means that sound cannot enter the ear canal so a regular hearing aid would be inappropriate. A bone-anchored hearing aid (BAHA) can bypass the blockage (and the middle ear) by receiving sound outside the head and transmitting it to a device that transmits vibrations into the skull. The vibrations conduct through bone into the (fully functional) cochlea where they are converted into electrochemical signals which are sent to the brain.


Question 26

The fact that the offspring have different patterns from their parents suggests that these novel coat patterns are initially a result of random mutation. As the zebra population is becoming increasingly isolated and small, they are more vulnerable to genetic drift: in small populations, allele frequency is more likely to change due to random factors, some characteristics may be seen more often and others may be lost. For example, in a small population with low genetic diversity, high levels of inbreeding can result in more individuals being homozygous for rare recessive traits and so those traits appear more frequently. There is no evidence provided to suggest that these phenotypes are caused by an increase of mutagens in the environment.


Question 27

Child 3 will have sickle cell anaemia. Child 1 is homozygous and the two allele copies are the same length so they appear on top of each other to form the lower band in the DNA profile. Child 2 is heterozygous and has two different bands to represent the two alleles. Since child 1 is unaffected it must be the upper band that represents the allele that causes sickle cell anaemia. Child 3 has two copies (overlapping) of the top allele and will be affected by the disease. If child 2 is heterozygous and unaffected then the condition must be recessive.


Question 28

(a) mRNA has an important role in protein synthesis within human cells. During transcription, RNA polymerase unwinds the gene (DNA) and makes a copy by adding complementary RNA bases to the template strand. This gene copy is known as messenger RNA (mRNA) and it detaches from the DNA, leaving the nucleus to find a ribosome in the cytoplasm. The ribosome translates the codons on the RNA into a string of amino acids (polypeptide) that will eventually become a protein.

(b) mRNA vaccines are an example of artificially acquired active immunity, active immunity is when the body produces its own antibodies to defend itself against a pathogen. The mRNA contains the instructions for producing the spike protein of the virus and is introduced into the body through an injection. The mRNA enters the cells of the body and the codons are translated into a string of amino acids (polypeptide) by a ribosome. The polypeptide is folded and the end product is the same spike protein that is found on the virus.

The spike protein is presented on the surface of the cell or released into the body. The spike protein is recognised as foreign by the immune system (e.g. by a T-helper cell) and antibodies are produced against the spike protein. Memory cells that recognise the spike protein are also produced and they will launch a stronger secondary immune response if the actual virus is encountered in the future. Thus, someone who is vaccinated with the mRNA vaccine will have active immunity to the virus as they can produce their own antibodies against it.


Question 29

The koalas are exhibiting behavioural adaptations in order to maintain a stable body temperature. The Koalas change position depending on the temperature of the air. According to the graph, Koalas were only observed to curl up into a ball when it was mild. This posture would decrease the surface area of its body that is exposed to the cold air in order to retain heat. When the temperature is mild a koala is more likely to be found in a relaxed posture compared to when the temperature is hot. When it is hot the Koalas are more likely to be observed to be stretched out to increase the surface area of its body that is exposed to the air so that heat can be lost. When it is hot it is more likely that Koalas will be observed to ‘hug’ the tree, this is likely explained by the fact that the tree is cooler than the air and this allows heat to transfer from the body of the koala to the tree.


Question 30

There is an obvious difference in the number of cases of measles in the vaccinated and unvaccinated groups. The bar graph shows that the unvaccinated group had seasonal outbreaks of measles with a few hundred cases. The month of March in 1984 and 1985 had over 100 cases for example, while the vaccinated group never had more than around 10 cases in a month. The table shows that the unvaccinated group also had a higher number of deaths from measles (40) compared to the vaccinated group (2). The data supports the idea that the measles vaccine protects the population from measles.

While the data in the table suggests that the vaccine against measles was also protecting against other diseases that cause Diarrhoea and dysentery or Oedema (as the number of deaths was lower in the vaccinated children), this is more likely to be because these children are also vaccinated against other diseases or have other lifestyle factors such as hygiene that are protecting them.  Pathogens that cause disease have specific proteins or carbohydrates on their surface that the immune system uses to recognize the pathogen as a foreign invader. These molecules trigger an immune response (antibody production) and so are referred to as antigens (antibody generators). A vaccine contains the antigen/s from a specific disease. When the vaccine introduces the antigen into the body, the body will launch a primary immune response that results in the production of memory cells. These memory cells will launch a stronger secondary immune response when the live pathogen is encountered as the antigen on the surface of the pathogen will be encountered. However, only the specific antigen will be recognised by the memory cells. Other diseases will have different antigens that these memory cells will not recognize.


Question 31

Both tablet treatments had around twice the symptom score of the no tablet treatments. However, the study shows that there was very little difference in symptoms when taking the statin tablets and taking the placebo tablets, which suggests that the symptoms were psychological. This implies that some of the symptoms patients report in reaction to the statins are in fact imagined. It is possible that some patients could be convinced to stay on the statins or be persuaded that the benefits of the drug outweigh the side effects. It appears from the table that about 0.9% of patients were possibly having symptoms from the statin medication and these cases should be investigated further. It could also be investigated to see what underlying condition may be causing symptoms when no tablets are being taken. In terms of sample size, 60 patients is a reasonable amount, although the study could be repeated to improve reliability. Asking patients to rate their symptoms is subjective as different individuals might give the same symptoms a different rating. A score that records the number of days with the presence or absence of symptoms might be an alternative. The treatment of ‘no tablets’ acted as a control but including a separate control group of people that did not have the same history would be useful for comparison. Overall, this is an effective study with important findings.


Question 32

(a) This process is gene amplification using recombinant plasmids in bacteria. (1) First the target gene is identified and cut out using a restriction enzyme. Then the same restriction enzyme is used to cut a bacterial plasmid. (2) The target gene is inserted into the plasmid and annealed using DNA polymerase. (3) The plasmid is inserted back into the bacterium which is allowed to reproduce by binary fission. In this way, the target gene is copied and then (4) the gene copies are cut out of the plasmids.

(b) One potential benefit of using transgenic salmon in aquaculture is that they reach market size faster than standard salmon. Transgenic salmon first reach market size just before they reach two years of age, while standard salmon don’t reach market size until they are almost 2.5 years old. By the time standard salmon reach market size at around 3.25kg, transgenic salmon are already close to 5 kg. This would increase the yield for the farmer.

(c) Transgenic species have had a gene from a different species (transgene) inserted into their genome through the process of genetic engineering. This form of biotechnology is used to give the target species a specific protein and characteristic that they didn’t have before. In the case of the transgenic salmon, the transgene is dominant and will produce the transgenic characteristics in any individual carrying even a single copy of the gene.

While adding a new gene to a population initially increases genetic diversity of that population, there are concerns about how transgenic species can impact biodiversity if they are allowed to escape into the wild. As they have a growth hormone and antifreeze protein the transgenic salmon have a strong genetic advantage over wild fish. If the fish could escape into the wild populations, they would grow faster and larger, which means they could potentially outcompete the wild salmon and even other fish species that compete with them for food. This could even eliminate wild salmon or other species which would reduce species diversity and have unknown impacts on the food web. They could also interbreed with wild salmon, which would have a similar outcome by passing on the dominant transgenes to other fish. These captive animals could also introduce new diseases into wild populations.

In the production of these transgenic fish, a variety of methods have been employed to prevent the fish from escaping and to prevent them breeding if they do escape. To start with, using Inland tanks reduces the risk that fish will escape into waterways and eventually escape into the wild to breed with each other or interbreed with wild fish. The breeding females are kept in quarantine, which means they are completely isolated with no chance of escaping.

Producing infertile triploid fish (XXX) means that even if they do escape, these fish cannot interbreed with each other or with wild salmon. The sister chromatids are prevented from separating in meiosis II by pressure shock treatment and so the egg cells are diploid. When they are fertilised by a haploid (transgenic) sperm, the resulting offspring will be triploid. Having three sex chromosomes prevents the development of sex organs in the fish. Any individual fish that escapes into the wild will live out its normal life span but will never reproduce. These techniques should be effective at preventing any impacts on biodiversity by these transgenic fish.


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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