The Ultimate Guide to GAMSAT Section 3

GAMSAT

17/1/2024

Please note that GAMSAT ‘Section III: Reasoning in Biological and Physical Sciences’ has been renamed to ‘Reasoning in Biological and Physical Sciences Section’. To ensure this article is easy to follow, we’ll refer to this section as ‘Section 3’ throughout. Read about the latest changes to the GAMSAT.

In our GAMSAT Section 3 guide, we explore what types of questions you can expect to find in Section 3, common topics covered, and how best to approach preparation. We also provide free example questions with explanations from GAMSAT experts, so you can understand exactly how to tackle this section.

As you begin your GAMSAT preparation journey, we recommend going through as many GAMSAT resources as you can, including our GAMSAT FAQ, top GAMSAT preparation advice and tips, and GAMSAT preparation mistakes to avoid

If you haven’t explored the other articles in this series, make sure you check them out: 

Once you’ve read through our ultimate guides, and are further into your preparation journey, learn how to overcome GAMSAT time pressure.

Table of contents

GAMSAT Section 3 structure
What does GAMSAT Section 3 test?
GAMSAT Section 3 topics
How should I approach preparation for GAMSAT Section 3?
Free example questions for GAMSAT Section 3

GAMSAT Section 3 structure

GAMSAT section

Number of GAMSAT questions

Section time

Time per question

Reasoning in Biological and Physical Sciences

75 multiple choice questions

150 minutes

2 minutes

What does GAMSAT Section 3 test?

Graphic of student reading a science book

GAMSAT Section 3 tests your problem-solving skills and reasoning within a scientific context. 

Section 3 questions are split as follows:

  • Biology: 40% (30 questions)
  • Chemistry: 40% (30 questions)
  • Physics: 20% (15 questions)

Questions in each subject are put into sets which consist of a shared stimulus followed by 1–7 questions. On average, there are around 4 questions per set, with most sets containing 2–5 questions. Sets will be randomly mixed throughout this section so it's important to train switching your brain between different subjects as you progress through the test.

According to the official ACER GAMSAT Information Booklet, you’ll be expected to have the equivalent basic knowledge of a first year undergraduate in biology and chemistry, and an A-level student in physics. Despite the fact that you need prerequisite knowledge for this particular section, Section 3 is mainly testing how you use reasoning and problem-solving skills to analyse and interpret data provided, often in an unfamiliar context. You’ll need to show dental or medical schools that you possess these skills as they’re essential for a career in medicine or dentistry.

Like Section 2, this section can be worrisome for many, even for those with a pure science background. This is because the GAMSAT science section is worth 50% of your overall GAMSAT score as it’s weighted twice for most universities. This highlights the importance of preparing effectively for Section 3, so make sure you factor this into your GAMSAT study schedule.

GAMSAT Section 3 topics

Since there’s no official GAMSAT syllabus, it can be difficult for students to know which topics to revise and prioritise. Below we’ve collated some common biology, physics and chemistry topics that you can use to help prepare for Section 3, along with our free cheat sheets. Remember, your goal is not to become an expert in all fields. Your goal is to gain a fundamental understanding across a broad range of topics which you can apply to unfamiliar situations in the exam. 

Infographic covering common topics for GAMSAT Section 3

How should I approach preparation for GAMSAT Section 3?

Different approaches will be necessary to tackle this section of the exam. For instance:

  • In biology, you will need to interpret and analyse the information provided in the stimulus and apply logical reasoning and problem-solving skills to arrive at an answer. 
  • In chemistry and physics, you’ll often be required to apply scientific knowledge or principles, most of which will be provided in the stimulus.

It will be challenging to answer questions efficiently if you haven’t thoroughly revised the science subject matter at the level required. However, some students make the mistake of confusing GAMSAT Section 3 with a recall test. While there could be questions that directly test your recall of science knowledge, these types of questions have become increasingly rare in recent years.

For example, your understanding of key concepts about acids and bases will be essential to solving some chemistry stimuli and questions, but you won’t need to recall pKa values of common acids. Therefore, the focus should be on applying reasoning and problem-solving skills to GAMSAT Section 3 questions. 

Instead of spending a large amount of time compiling GAMSAT chemistry/physics/biology revision notes, it would be more advantageous to go through Section 3 GAMSAT practice questions and revise topics as they come up. By using a trusted preparation resource for Section 3, over time you should become very familiar with the format and style of this section, and how best to approach answering questions. This is when you can introduce timed conditions in mock exams to ensure you’re simulating the GAMSAT in every practice test. 

Free example questions for GAMSAT Section 3

Below you will find five sample questions for GAMSAT Section 3, with explanations for each question further down in the article.

Questions 1 – 5

The bacterial ribosome is targeted by over 60% of naturally derived antibiotics. Compromising protein synthesis makes metabolism, signalling, motility and many other crucial cellular functions impossible. Targets and mechanisms of action for selected ribosome-targeting antibiotics are described in Table 1 and illustrated in Figure 1.

GAMSAT Section 3 example question – figure 1

Table 1

Antibiotic class

Target subunit

Mechanism of action

Aminoglycosides

30S

Stabilise bulged-out conformation of 16S rRNA

Tetracyclines

30S

Occupy anticodon binding space near the A site

Chloramphenicols

50S

Occupy aminoacyl binding space near the A site; inhibit ribosomal protein translation; hinder 50S subunit formation

Macrolides

50S

Block nascent protein elongation at the E site, induce premature termination

Sparsomycin

50S

Increase aminoacyl-tRNA affinity to P site, causing premature translocation

Type I PrAMPs

50S

Prevent tRNA transfer from A to P site

Type II PrAMPs

50S

Prevent dissociation of release factors from the ribosome, stalling translating ribosomes and depleting the pool of free release factors for other ribosomes

The 30S subunit of the ribosome contains a 16S ribosomal RNA strand, which has a conserved loop of three unpaired adenine residues. When codon–anticodon pairing is correct, this loop bulges out and allows translation to proceed.

Release factors are necessary to recognise STOP codons in mRNA and initiate dissociation of the synthesised protein from the ribosome.

PrAMP stands for proline-rich antimicrobial peptide.

The envelope of Gram-positive bacteria consists of a thick cell wall and one inner plasma membrane. In Gram-negative bacteria, this envelope consists of a thin cell wall and two – inner and outer – plasma membranes.

1. The graph below shows the rate of new protein synthesis in an untreated Escherichia coli culture (dashed line) and E. coli treated with three different classes of antibiotics (A, B, C). Which types of antibiotics could have been used to treat the cultures?

GAMSAT Section 3 example question – graph

A A – type II PrAMP, B – sparsomycin, C – tetracycline
B
A – sparsomycin, B – aminoglycoside, C – tetracycline
C
A – macrolide, B – type II PrAMP, C - chloramphenicol
D
A – sparsomycin, B – chloramphenicol, C – type I PrAMP

2. Which of the following adaptations would improve survival of bacteria in presence of type II PrAMPs?

A synthesis of a PrAMP importer
B
overexpression of release factors
C
loss of the release factor–ribosome binding site
D
increase in release factor–ribosome binding affinity

3. Aminoglycoside uptake into bacteria is an energy-demanding process. As a result, aminoglycosides are less effective against

A motile bacteria.
B
anaerobic bacteria.
C
Gram-positive bacteria.
D
colony-forming bacteria.

4. Consider the following statements:

I. Prokaryotes use certain codons more or less frequently than eukaryotes.
II. Prokaryotic and eukaryotic ribosomes contain different rRNA and ribosomal proteins.
III. Prokaryotes and eukaryotes express different membrane transport proteins.

Which statement(s) explain why tetracycline is effective against bacteria but harmless to humans?

A Statements I, II and III explain tetracycline specificity.
B
Statements I and III, but not statement II, explain tetracycline specificity.
C
Statements II and III, but not statement I, explain tetracycline specificity.
D
Statement II, but not statements I and III, explain tetracycline specificity.

5. A bacterium gains several mutations that cause a complete structural rearrangement of the A site within the 50S ribosomal subunit while still maintaining functionality. This bacterium is likely to be

A non-viable.
B
susceptible to type I PrAMPs.
C
resistant to chloramphenicol.
D
resistant to tetracyclines and aminoglycosides.

Explanations for questions 1 – 5

1. B

Analyse the graph to determine the likely effects of each antibiotic:

Antibiotic A – there is an increased rate of protein synthesis compared to an untreated bacterial culture. This can be caused by increased rate of protein translation, which can only be induced by sparsomycin due to its ability to induce premature translation of the ribosome. An error here would be to consider type II PrAMPs as they cause translation of extended proteins; but longer peptide sequences do not equate to more proteins. Thus, sparsomycin is the only possible option, and answers A and C can be eliminated.

Antibiotic B – the rate of protein synthesis is nearly equal to the rate in an untreated bacterial culture. This occurs if the antibiotic is not designed to hinder protein synthesis, but rather to suppress bacterial growth by producing nonfunctional proteins. Such proteins can harbour amino acid mutations, which would be induced by aminoglycosides; they can be too short, which would be induced by macrolides; or too long, which would be induced by type II PrAMPs. This only leaves answer B.

As a final check, antibiotic C should be considered. In this case, protein synthesis rate is below that of an untreated bacterial culture. This would be caused by an antibiotic which directly blocks protein synthesis, which can be either tetracycline or type I PrAMP.

2. B

Answer B is correct: overexpression of the release factor would replenish the pool and compensate for the effects of type II PrAMPs.

Answer A is incorrect: a PrAMP importer would result in active uptake of the antibiotic, increasing bacterial susceptibility.

Answer C is incorrect: loss of the release factor–ribosome binding site would prevent release factor binding. As a result, protein translation would never terminate, resulting in synthesis of extended proteins. This would decrease survival of bacteria even in absence of type II PrAMPs.

Answer D is incorrect: an increase in the binding affinity between release factor and ribosome is effectively the same as adding a type II PrAMP, which stabilises the interaction between release factor and ribosome. Thus, survival of the bacteria would also be negatively affected.

3. B

Recall that aerobic respiration is much more energetically efficient than anaerobic respiration. Therefore, anaerobic bacteria will have lower energy sources that can be harnessed for aminoglycoside uptake, and these antibiotics will be less effective for anaerobic bacteria. Answer B is correct.

Bacterial motility is also an energetically demanding process. On one hand, motile bacteria are likely to also have enough energy to subvert for aminoglycoside uptake; on the other hand, this may not leave enough energy for aminoglycoside uptake. In light of both possibilities, bacterial motility would not be a definitive reason aminoglycoside uptake is more efficient in these types of bacteria. Answer A is incorrect.

Gram-positive bacteria are distinguished from Gram-negative bacteria by the structure of their cell envelope, as stated in the stimulus. Gram-positive bacteria may require more energy for aminoglycoside uptake due to the thicker cell wall; but they may also save energy as they only have one plasma membrane instead of two. Again, this is not a definitive reason and answer C is incorrect.

Colony formation is a result of bacterial division and secretion of substances that keep them together. It is also an energy-consuming process, but it improves bacterial survival. A bacterial colony may become more resistant to antibiotics, but this is due to its structural organisation (bacteria on the inside of the biofilm are shielded from the environment) rather than energy processing, thus this is also not a definitive explanation and answer D is incorrect.

4. C

All statements are factually true. However:

Statement I does not explain tetracycline specificity. Even if bacteria use certain codons more frequently than others, tetracycline is not a codon-specific inhibitor: it inhibits any tRNA binding to the 30S subunit.

Statement II is an explanation of tetracycline specificity. Due to differences in ribosome composition between bacteria and humans, even if tetracycline was present in the cell, it would be unable to bind to the eukaryotic ribosome.

Statement III is also a valid explanation for tetracycline specificity. Bacteria possess membrane transporters that take up tetracycline, while human cells do not. Thus, tetracycline is not present inside human cells.

5. C

Answer A is incorrect: the question stem states that the structurally rearranged 50S subunit is still functional, thus the bacterium should be perfectly viable.

Answer B is incorrect: type I PrAMPs prevent A to P transfer of tRNA. It is not clear to which part of the ribosome they bind (if any). If they bind near the A site, then the mutation should reduce binding ability and the bacterium would become resistant to type I PrAMPs; if they bind elsewhere, then the mutation should have no effect.

Answer C is correct: a structural rearrangement would prevent chloramphenicol binding to the A site of the 50S subunit, therefore this bacterium is likely resistant.

Answer D is incorrect: tetracycline and aminoglycosides act on the A site on the 30S subunit, so the rearrangement of the 50S subunit should have no effect.

Want more example questions? Check out our free GAMSAT Section 3 sample questions with explanations for every GAMSAT practice question

Succeed in the GAMSAT with Medify

Applying to dental or medical schools as a graduate can be a long and difficult process. To increase your chances of securing your place, you should implement an effective preparation strategy for every step of the admissions journey.

If you need a helping hand with GAMSAT preparation, look no further than our GAMSAT Online Course. It provides:

  • A robust Question Bank with 2,000+ original GAMSAT style questions
  • 7 realistic and unique mock GAMSAT exams designed to simulate the real test
  • 50+ in-depth tutorials with guidance from GAMSAT experts
  • Medify’s science GAMSAT syllabus with key topics for biology, chemistry and physics
Don’t forget to check out our ultimate guide to GAMSAT Section 1 and GAMSAT Section 2 if you haven’t already!

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