The Graduate Medical School Admissions Test (GAMSAT) stands as one of the most challenging examinations for aspiring medical professionals. It scrutinizes candidates' abilities across a spectrum of subjects, with GAMSAT Biology being one of its core areas. As we approach 2024, understanding the structure and expectations of the March 2024 GAMSAT is crucial for effective preparation.
Deciphering GAMSAT Biology
GAMSAT Biology is an intricate blend of various skills. It's not merely about memorizing biological facts but demands a profound understanding of contextual scenarios, logical inferences, data interpretation, graphical analysis, and surprisingly, mathematical reasoning. This section, in essence, mirrors the analytical depth of GAMSAT Section 1. For more on the basics of GAMSAT Biology, see our detailed article.
What to Expect from GAMSAT Biology in 2024
In recent years, the exam has evolved. Instead of focusing solely on knowledge-based questions, the test now requires candidates to analyze textual data, relate it to graphical representations, and derive meaningful conclusions.
Take, for instance, a question that begins with the basics of blood groups but soon transitions into the intricate realm of genetics, prompting candidates to calculate the frequency of specific genotypes. While it may seem like a straightforward knowledge test, the real challenge lies in amalgamating pre-existing knowledge with a thorough comprehension of the internal mechanisms or rules inferred from the stem.
What level of biology is needed for the GAMSAT?
However, if you’re wondering what the pre-requisite knowledge for GAMSAT Biology is, it’s 1st year university level content. However, I would implore you to begin with questions and figure out your theory weaknesses from that 🙂
Essential Skills for GAMSAT Biology Mastery
To excel in this rigorous section of the GAMSAT, aspirants must possess:
- An Open Mind: It's paramount to process the given information without veering beyond the stem's confines.
- Digital Annotation Expertise: The ability to highlight and track pivotal statements and rules within the stem significantly boosts comprehension.
- Analytical Patience: Before plunging into the questions, it's essential to decode the core message.
- Graphical Analysis Prowess: Candidates must be adept at dissecting graphs and seamlessly shifting between diverse information formats.
How to Prepare for GAMSAT Biology in 2024
- Mock Examination: Kickstart your preparation with a GAMSAT mock biology exam to understand your baseline proficiency.
- Introspective Analysis: Merely marking answers won't suffice. Delve deep and categorize errors based on specific skills, such as recurrent misinterpretation of graphical data.
- Formulate a Strategy: Design a concrete game plan. If graphical analysis is a challenge, make it a habit to scrutinize the axes, title, legend/key, captions, context, and units rigorously.
- Implement and Adapt: Armed with your strategy, confront more GAMSAT practice questions, adapting and refining as you proceed.
- Re-evaluate: Assess the effectiveness of your approach. If there's a marked improvement, optimize your timing. If not, revisit your weaknesses and tweak your strategy.
Key Biology Topics with Sample Questions
In the realm of GAMSAT Biology, these are the fundamental topics you should cover:
- Cell Structure and Cellular Processes
- Enzymes and Reaction Rates
- Cellular Genetics and Protein Synthesis
- Population Genetics
- Evolutionary Biology
- Major Organ System Physiology
- Basic Principles of Anatomy
- Immunology
- Microbiology
- Principles of Experimental Design and Interpretation
For a comprehensive overview of the GAMSAT Syllabus for 2024, including detailed syllabi for Sections 1, 2, and 3, refer to our GAMSAT Syllabus Guide.
Sample GAMSAT Question On Immunology
Adaptive immunity is specific immunity that targets a particular threat: a particular virus, bacteria or parasite. It works in tandem with innate immunity, which is the first line of defence against pathogens. One of the components of adaptive immunity is the CD4+ T cell, also known as the T helper cell. There are various types of T helper cells, two of which are involved in the TH1/TH2 paradigm responsible for determining the outcome of intracellular bacterial & protozoan infection and helminth infection, respectively. The ability of the host to generate a particular effector T cell subset determines the outcome of infection. Inducing differentiation in favour of one response type typically suppresses the other response type.
The cytokine interferon gamma (IFN-γ) combats intracellular infection by activating macrophages. This eradicates microbes that go inside the cell, such as parasites or bacteria. IFN-γ also activates a process called opsonisation, whereby B cells coat parasites and bacteria with antibody rendering them recognisable to macrophages. IL-4 has several effector functions: it promotes intestinal mucus secretion and peristalsis, activates an alternative function of macrophages (enhanced tissue repair) and promotes production of IgE (a type of antibody) by B cells. When mast cells detect IgE, they degranulate, releasing large amounts of histamine to surrounding tissues. IL-5 activates eosinophils, which attack helminth parasites (large multicellular organisms) by releasing histamine through the process of degranulation. IL-13 carries out similar functions to both IL-4 and IL-5.
Questions
Question 1: Which of the following is a plausible statement regarding the role of TH1 or TH2?
- A. TH1 plays the primary role in resistance to helminthic infection as IFN-γ activated opsonisation is a process applicable to both bacteria and helminths.
- B. The involvement of the TH2 CD4+ T cell subset is critical in determining the outcome of helminthic infection.
- C. Resistance to bacterial and protozoan infection is impossible without the TH1 subset of CD4+ T cells.
- D. IL-4 both promotes the development of, and is a cytokine secreted by TH1 CD4+ T cells.
Question 2: If there was a helminthic infection, which of the following responses would not promote resistance against the pathogen?
- A. Secretion of IL-13 by Th2 CD4+ T cells.
- B. Increased peristalsis and intestinal mucous secretion.
- C. Degranulation of mast cells in response to IgE secreted by B cells
- D. Classical macrophage activation to combat intracellular parasites.
Solutions
Solution 1: B is the correct answer. A is incorrect as Th2 cells play a much larger role in resistance to helminth infection. C is incorrect as it is mentioned that some resistance to pathogens is offered by the innate immune system. D is incorrect as IL-4 is not secreted by Th1 CD4+ T cells.
Solution 2: D is correct – helminths are large multicellular parasites which cannot exist inside human cells. Hence classical macrophage activation would not aid defence against helminths. A-C describe different components of the Th2 CD4+ T cell response, which is primarily directed against large parasites.
Sample GAMSAT Question On Major Organ System Physiology
Metabolism is the body’s ability to consume energy in its normal daily functioning. It is affected by many aspects of human physiology including genetics, calorie consumption, hormones and comorbidities. Measuring metabolic rate is done by measuring energy expenditure over time per unit of mass. Measuring basal metabolic rate or BMR requires a strict set of conditions to determine the body’s metabolism in a completely energy neutral state. Increasing metabolism causes an increase in energy consumption which increases oxygen intake, heart rate (tachycardia) and muscle twitching. Decreasing metabolic rate has the opposite effect. Metabolic rate is increased by thyroid hormone, a hormone produced by the thyroid gland. It regulates energy expenditure on a cellular level, allowing the body to adapt to different scenarios in which different levels of energy saving are required.
Questions
Question 1: An increase in metabolic rate is most likely associated with which of the following?
- A. Increased pupil diameter
- B. Increased lung size
- C. Decreased tolerance to environmental heat
- D .Decreased tolerance to environmental cold
Question 2: Which of the following is the best explanation for the conditions necessary to measure BMR?
- A. Science must always be done in very strict conditions
- B. BMR considers only the least amount of energy spent by a person
- C. To minimise any extraneous energy expenditure of the body above baseline
- D. The equipment required is very sensitive to various variables
Solutions
Solution to 1: Increasing metabolic rate causes increase in heat production as a by-product, hence C. A is a function of the sympathetic nervous system. B is not true as lungs do not increase in size, oxygen consumption increases by increasing respiratory rate. D is the opposite of C, so untrue.
Solution to 2: The basal metabolic rate is described as requiring a strict set of conditions to “determine the body’s metabolism in a completely energy neutral state”. This implies that the BMR is measuring a baseline level of energy, and as with all baselines, requires strict and controlled conditions for the sake of accuracy. A is a sweeping statement and nonsensical. The sensitivity of equipment does not provide a sufficient or logical explanation for the strict experimental set up, hence D is incorrect. B is incorrect as BMR is not the least amount of energy, it’s the baseline level of energy
Sample GAMSAT Question On Cellular Genetics and Protein Synthesis
Huntington’s disease (HD) is a hereditary neurodegenerative disorder. It is clinically characterised by progressive decline in motor and cognitive functions, as well as psychiatric disturbances. HD is caused by an expansion in the trinucleotide (CAG) repeats within exon 1 of the Huntington gene, which leads to an expanded polyglutamine stretch (>35) in the N-terminus of the huntingtin protein (htt). The disease is inherited in an autosomal dominant manner with age-dependent penetrance, whereby disease onset occurs earlier in patients with longer CAG repeats. HD is considered to arise predominantly from central nervous system degeneration that is due to both loss of normal htt function and toxic gain-of-function from an abnormal conformation of mutant htt. A schematic of the two modes of inheritance is shown below:
Questions
Question 1: The mutation of the Huntington gene will affect:
- A. The sequence of the RNA only
- B. The sequence of the RNA and protein
- C. The length and sequence of the RNA and protein
- D. The length and sequence of the RNA and the length, sequence and shape of the protein
Question 2: 4 In a patient who has been diagnosed with HD, the addition of 10 CAG repeats will:
- A. Move the reading frame of the gene, altering the amino acid sequence of the region encoded by exon 2
- B. Move the reading frame of the gene, but not cause any changes to the amino acid sequence of encoded by exon 2
- C. Not move the reading frame and not change the amino acid sequence encoded by exon 1
- D. Not move the reading frame
Solutions
Solution 1: The answer is D. An expansion in the CAG repeats of the huntingtin gene will increase the number of CAGs therefore increasing the RNA and protein length. The additional glutamine residues will alter the conformation (shape) of the protein. 4
Solution 2: The answer is D. In mammals, the genetic code is always read in a non-overlapping reading frame. Adding triplets to the DNA sequence will not change how the DNA is read. The answer is not A or B for this reason. The answer is not C because the amino acid sequence does change as per the Figure.
Sample GAMSAT Question On Evolutionary Biology
All life on Earth shares a common ancestor known as the last universal ancestor, which lived approximately 3.5–3.8 billion years ago. Repeated formation of new species (speciation), change within species (anagenesis), and loss of species (extinction) throughout the evolutionary history of life on Earth are demonstrated by shared sets of morphological and biochemical traits, including shared DNA sequences. These shared traits are more similar among species that share a more recent common ancestor, and can be used to reconstruct a biological "tree of life" based on evolutionary relationships (phylogenetics), using both existing species and fossils. In the mid-19th century, Charles Darwin formulated the scientific theory of evolution by natural selection, which described reproductive success as being achieved by the “fittest” organisms in a population. Natural selection within a population for a trait that can vary across a range of values, such as height, can be categorised into three different types. The first is directional selection, which is a shift in the average value of a trait over time—for example, organisms slowly getting taller. Secondly, disruptive selection is selection for extreme trait values and often results in two different values becoming most common, with selection against the average value. This would be when either short or tall organisms had an advantage, but not those of medium height. Finally, in stabilising selection there is selection against extreme trait values on both ends, which causes a decrease in variance around the average value and less diversity. This would, for example, cause organisms to slowly become all the same height. On each graph in Figure 3, the x-axis variable is the type of phenotypic trait and the y-axis variable is the number of organisms. Group A is the original population and Group B is the population after selection
Questions
Question 1: Disruptive selection is illustrated by
- A. Graph 1 in Figure 3
- B. Graph 2 in Figure 3
- C. Graph 3 in Figure 3
- D. Figure 4
Question 2: Evolution would not occur by natural selection if
- A. the population is homogenous with mutation being the only mechanism of individual genetic change
- B. the population is homogenous with asexual reproduction as the only mechanism of individual genetic exchange
- C. the population is heterogeneous with sexual recombination as the only mechanism of genetic exchange
- D. the population is heterogeneous with gene flow as the only mechanism of genetic exchange
Solutions
Solution to 1: C. Graph 1 in Figure 3 illustrates directional; Graph 2 illustrates stabilising; Graph 3 illustrates disruptive selection. Figure 4 illustrates mutations and natural selection.
Solution to 1: B. For natural selection to occur, there must be a driving force for variation as suggested in the passage Mutation can cause variation which can lead to natural selection with selection pressure. Sexual recombination and gene flow can cause variation, whereas asexual reproduction does not.
Sample GAMSAT Question On Microbiology
Drug resistance is an increasing issue amongst pathogenic bacteria, especially in a hospital setting. These microbes acquire resistance through various methods of gene transfer. Although the exact methods of resistance vary between drug to drug, there are similarities between the pathways involved. Five of these are listed below:
- Varying the flow of drugs in and out of the cell
- Altering the structure of the target of the drug
- Acquiring genes that encode enzymes which degrade the drug
- Blocking entry of the drug into the cell
- Creating a biofilm which reduces the concentration of drug around the microbes
Questions
Question 1: The drug penicillin is a drug which acts to block cell wall synthesis. Of the following, which option best describes a microbe with high resistance and a microbe with low resistance?
Question 2: Several subspecies of Escherichia coli were discovered to have acquired multiple methods of resistance to penicillin when sampled at various timepoints after being placed in a medium containing the antibiotic. Which of the following populations is least resistant to penicillin?
- A. Strain A, sampled three days after introduction to the medium, had acquired genes which increased cell wall production with an altered structure
- B. Strain B, sampled one week after introduction to the medium, had decreased numbers of inward active transporters on its outer membrane and had increased its level of cell wall production
- C. Strain C, sampled after 24 hours of being in the medium, had acquired a plasmid encoding a beta lactamase which inhibits penicillin action in the medium
- D. Strain D, sampled after one week of being in the medium, had acquired an altered structure to its cell wall and increased production of an efflux transporter on its outer plasma membrane
Solutions
Solution to 1: D. Gram negatives are generally more resistant because they have an extra (second) plasma membrane, and it makes it harder for a drug to penetrate. Having an altered cell wall structure will alter drug targets and decrease the potency of the drug. A is wrong because second plasma membrane generally increases resistance. B is wrong because less cell wall production means less resistance. C is wrong because the cell wall is not inside the cell so efflux does nothing.
Solution to 2: C. Strain C would be least resistant as it relies on plasmid acquisition which is a chance event. Also, beta-lactamases only decrease levels of active drug which is less effective than altering the target structure. The population of strain C has only been there for 24 hours so potentially not all of the population is resistant. A and D both have altered cell wall structure which means penicillin does not exert its effects. B seems viable but two components to this option make it likely to be more resistant than strain C: the fact that it increases cell wall production as well as the fact the strain itself survived and stayed in the medium for the longest period of time (7 days).
Sample GAMSAT Question On Cell Structure and Cellular Processes
During translation, mRNA is "read" according to the genetic code, which translates the nucleotide sequence in DNA to the amino acid sequence in proteins (Figure 2). The mRNA sequence is used as a template to assemble the chain of amino acids that form a protein, with each codon of three nucleotide bases on the mRNA corresponding to a particular amino acid. In eukaryotic cells, ribosomes are composed of two subunits, the large (60S) subunit and the small (40S) subunit (where S is the Svedberg unit – a measure of particle mass based on its rate of sedimentation in a centrifuge). The ribosomal subunits contain proteins and specialized RNA molecules such as tRNA and rRNA. The tRNA molecules are adaptor molecules—they have one end that can read the triplet code in the mRNA through complementary base-pairing and another end that attaches to a specific amino acid.
Although methionine (Met) is the first amino acid incorporated into any new protein, it is not always the first amino acid in mature proteins—in many proteins, methionine is removed after translation. In fact, if a large number of proteins are sequenced and compared with their known gene sequences, methionine (or formylmethionine) occurs at the N-terminus of all of their gene sequence. However, not all amino acids are equally likely to occur second in the chain, and the second amino acid can influence whether the initial methionine is enzymatically removed. For example, many proteins begin with methionine followed by alanine. In both prokaryotes and eukaryotes, these proteins tend to have the methionine removed, so that alanine becomes the N-terminal amino acid. However, if the second amino acid is lysine, which is also frequently the case, methionine is not removed.
Questions
Question 1: It can be deduced from the passage that:
- A. Proteins with alanine as the second amino acid generally have methionine as the first amino acid
- B. Many proteins begin with methionine followed by lysine
- C. The second amino acid will determine whether methionine is enzymatically removed or not
- D. All proteins have methionine followed by alanine as the second amino acid
Question 2: A tRNA attached to Histidine could have which of the following anticodons also attached to it from the 5’ to 3’ end?
- A. CAU
- B. UAC
- C. GUG
- D. ATG
Solutions
Solution to Question 1:
- A is wrong: in the stem, it says that if methionine is followed by alanine, then methionine is generally removed enzymatically.
- B is correct: from the stem, we can deduce that many proteins who begin with methionine (hence, methionine hasn’t been enzymatically removed), will often follow with lysine.
- C is incorrect: the stem says that the second amino acid “influences” whether the methionine is enzymatically removed, it is not an absolute factor.
- D is incorrect: The stem doesn’t suggest this. It talks about how the second amino acid can influence enzymatically removed of methionine, meaning that the second amino acid cannot exclusively be alanine.
Solution to Question 2: C. The RNA codon table shows the codon sequence on the mRNA. The sequence of the anti-codon of the tRNA must therefore be complementary. Histidine on the mRNA can either show as CAU or CAC. Therefore, the anti-
codon sequence can either be GUA or GUG, giving option C.
Sample GAMSAT Question On Population Genetics
Hereditary Haemochromotosis (HH) is a genetic disease characterized by excessive intestinal absorption of dietary iron resulting in an increase in total body iron. This results in excessive iron accumulation in various tissues, mainly liver, adrenals, pancreas, heart, skin, and joints. Hereditary haemochromatosis is most commonly due to a mutation of the HFE gene on chromosome 6, which codes for a protein that participates in the regulation of iron absorption. There are two common mutations of the HFE gene, C282Y and H63D. The C282Y allele is a transition point mutation from guanine to adenine at nucleotide 845 that replaces the cysteine residue at position 282 with a tyrosine residue. HH is the most common genetic condition of Caucasian populations with one in 200 individuals at genetic risk for this Disease.
Hardy Weinberg equation: p2 + 2pq + q2 = 1
Questions
Question 1: Which is the mode of inheritance of Haemochromatosis?
- A Autosomal dominant
- B Autosomal recessive
- C X-linked dominant
- D X-linked recessive
Question 2: Assuming that haemochromatosis fits the Hardy-Weinberg equilibrium, what is the
frequency of the disease allele in the Caucasian population?
- A 1/6
- B 1/10
- C 1/14
- D 1/30
Solutions
Solution to 1: B. Hereditary Haemochromotosis (HH) is the most common genetic condition of Caucasian populations with one in 200 individuals at genetic risk for this Disease. This implies that the frequency of individuals with the disease is q^2=1/1200 = 0.005. Therefore, q = 0.07. HH is a genetic disease characterized by excessive intestinal absorption of dietary iron resulting in an increase in total body iron. This results in excessive iron accumulation in various tissues. These clinical features only appear in homozygous individuals, therefore the inheritance of HH is autosomal recessive.
Solution to 2: C. The frequency of the disease allele (q) is the square root of the frequency of affected individuals (q^2). Given that 1 in 200 individuals are affected,
q^2=1/1200. Therefore, q = 1/14.
Sample GAMSAT Question On Basic Principles of Anatomy
The heart of Heterodontus portusjacksoni differs from the mammalian heart in that there is only one ventricle, despite still having four chambers. Blood flows from the sinus venosus to the atrium to the ventricle and conus arteriosus before entering the arteries. A schematic of the shark heart and an electrocardiogram are shown in the figure below.
An electrocardiogram (ECG) records the electrical activity of the heart over a period of time, with various ‘waves’ signifying particular depolarisation or repolarisation events. Electrical activity in a particular region of the heart typically stimulates contraction in that region soon after. For instance, the P wave corresponds with atrial depolarisation, which is followed immediately by atrial systole (contraction). The QRS complex, on the other hand, represents ventricular depolarisation and hence ventricular systole (signalled by a noticeable rise in ventricular pressure).
Questions
Question 1: Which of the structures in the diagram of the heart above corresponds to the ventricle?
- A. 1
- B. 2
- C. 3
- D. 4
Question 2: The B wave on the ECG represents electrical activation of the conus arteriosus, passing up this structure rather slowly. What is the most likely function of this?
- A. To equalise the pressures of the conus arteriosus and that of the ventral aorta
- B. To enable final contraction of the ventricular myocardium before ventricular repolarisation (represented by the T wave)
- C. To close the AV valve and prevent backflow of blood
- D. To increase the pressure of the conus and prolong aortic flow during ventricular diastole
Solutions
Solution to 1: B. The ventricle in sharks corresponds to structure 2 as described in the accompanying text and diagram.
Solution to 2: A. The B wave corresponds to the activation of the conus arteriosus. Its primary function is to equalize the pressures of the conus arteriosus and that of the ventral aorta.
Final Words: Navigating the GAMSAT Labyrinth in 2024
GAMSAT's iterative nature demands continuous reflection and adaptation. With each cycle, refine your strategies, internalize feedback, and embark on the journey again. Beyond this cyclic process, it's crucial to understand that the GAMSAT stems often present incomplete or fragmented information. Herein lies the real test. It challenges you to understand the boundaries of your knowledge, interpolate data judiciously, and make informed decisions. Such skills aren't innate but are honed over time, demanding rigorous practice and a cultivation of a diverse skill set.
In conclusion, GAMSAT Biology in 2024, like its predecessors, will test not just what you know, but how you apply what you know.
Where to From Here?
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