GAMSAT Section III Practice Test 2022 Sample Questions Answers Section III: Reasoning in Biological and Physical Science. The GAMSAT Reasoning in Biological and Physical Sciences Sample Questions contains examples of the kind of materials and questions you can expect to find in the Graduate Australian Medical School Admissions Test (GAMSAT).
GAMSAT Section III Practice Test
|Test Name||GAMSAT Practice Test 2022|
|Subject||Reasoning in Biological and Physical Science|
|Question Type||Sample Question Answers|
Section III: Reasoning in Biological and Physical Sciences
Questions 1 – 6 Peak arterial blood pressure is generated when the ventricles of the heart contract, forcing blood into the arterial system. This peak blood pressure is called systolic pressure. The base arterial blood pressure (i.e. the minimum between the peaks) is known as diastolic pressure.
Figure 1 shows the typical (mean) effect on the systolic and diastolic blood pressure of changing from a ‘normal’ diet (containing moderate levels of fruits and vegetables, and dairy products that are not fat-reduced) for:
- I people with high normal blood pressure, who are changed to a diet containing a relatively high proportion of fruits and vegetables;
- II people with high normal blood pressure, who are changed to a diet containing a relatively high proportion of fruits and vegetables, and fat-reduced dairy products; and
- III people with moderately raised blood pressure (i.e. mild hypertensives), who are changed to a diet containing a relatively high proportion of fruits and vegetables, and fat-reduced dairy products.
Suppose that before the dietary change the mean blood pressures (systolic/diastolic) of the three groups were:
Q1. Which one of the following is likely to be the best estimate of the systolic pressure of a high normal person on a combination diet?
- A. 125 mm Hg
- B. 130 mm Hg
- C. 135 mm Hg
- D. 140 mm Hg
Q2. Of the following, systolic blood pressure is likely to increase most for
- A. mild hypertensive people on a normal diet changed to a combination diet.
- B. mild hypertensive people on a combination diet changed to a normal diet.
- C. people with high normal blood pressure on a normal diet changed to a combination diet.
- D. people with high normal blood pressure on a combination diet changed to a normal diet.
Q3. Which one of the following is likely to be the best estimate of the diastolic pressure of a mild hypertensive person on a fruit and vegetable diet?
- A 86 mm Hg
- B 92 mm Hg
- C 96 mm Hg
- D 100 mm Hg
Q4. Of the following, diastolic blood pressure is likely to increase most for a typical
- A. mild hypertensive person on a combination diet changed to a fruit and vegetable diet.
- B. mild hypertensive person on a fruit and vegetable diet changed to a combination diet.
- C. person with high normal blood pressure on a combination diet changed to a fruit and vegetable diet.
- D. person with high normal blood pressure on a fruit and vegetable diet changed to a combination diet.
Questions 5 and 6 refer to the following additional information:
Blood pressure is commonly measured by placing an inflatable cuff around the arm above the elbow, inflating the cuff to a pressure that occludes arterial blood flow (i.e. closes the arteries) and then gradually decreasing the pressure. As the pressure applied by the cuff declines, a stethoscope applied to the arm is used to monitor arterial sounds. Initially no sound is heard. As the pressure applied declines, however, a pulse becomes audible. The pressure at which the pulse is first audible is the systolic pressure. With further decline in cuff pressure, the pulse sound becomes less pronounced and eventually disappears. The pressure when the pulse is first inaudible again is the diastolic pressure.
Q5. The pulse sound disappears mainly because the
- A. force of ventricular contraction declines as cuff pressure declines.
- B. diastolic pressure declines below a certain level.
- C. systolic pressure equals diastolic pressure.
- D. arterial occlusion ceases altogether.
Q6. Suppose the cuff pressure was 140 mm Hg for a typical mild hypertensive person on a normal diet. The pulse sound would
- A. be heard throughout most of the cardiac cycle.
- B. be heard once, briefly, during each cardiac cycle.
- C. not be heard because the cuff pressure is too high.
- D. not be heard because the cuff pressure is too low
Questions 7 – 9
Figure 1 shows a side view (not to scale) of a quarter-circle ‘slide-slot’ that is held in a fixed position on a horizontal table. The radius of curvature of the curved surface (RST) is 8 cm.
A solid ball of radius 1 cm and mass 50 g is held at rest, touching the slide slot at point R. The ball is released and allowed to slide down the surface RST under the influence of gravity, and without rolling. Point S is halfway along the surface between points R and T.
Assume the ball slides in a vertical plane, there are no frictional forces present, and the gravitational acceleration (g) of a free-falling object is 10 m s−2
Q7. As the ball slides down the curved surface from point R to point T, the total mechanical energy (sum of kinetic and gravitational potential energies) of the ball
- A. decreases only.
- B. increases only.
- C. remains constant.
- D. first increases, reaching a maximum value at point S, and then decreases.
Q8. The decrease in gravitational potential energy of the ball, as it slides from point R to point T, is closest to
- A. 5.0 × 10−2 J.
- B. 4.0 × 10−2 J.
- C. 3.5 × 10−2 J.
- D. 3.0 × 10−2 J.
Q9. The ball is allowed once more to slide down the slide-slot, starting at rest from point R. However, this time the slide-slot is unconstrained horizontally so that it is free to slide on the table. While sliding, the ball is always in contact with the slide-slot. The mass of the slide-slot is 500 g
Once again, assume the ball slides in a vertical plane, there are no frictional forces present, and that the gravitational acceleration (g) of a free-falling object is 10 m s−2.
As the ball slides down the curved surface, from point R to point T, the total mechanical energy (sum of kinetic and gravitational potential energies) of the ball
- A. decreases only.
- B. increases only.
- C. remains constant.
- D. first increases, reaching a maximum value at point S, and then decreases.
Questions 10 and 11
If a nonvolatile and unreactive solute is added to a solvent, the vapour pressure above the resulting solution is less than the vapour pressure above the pure solvent at the same temperature. This is called Raoult’s Law and can be expressed as
Q10. A consequence of Raoult’s Law would be that the
- A. solute is less soluble than would otherwise be expected.
- B. solute is more soluble than would otherwise be expected.
- C. boiling point of the solution is lower than that of the pure solvent.
- D. boiling point of the solution is higher than that of the pure solvent.
Q11. A negative deviation from Raoult’s Law is said to occur if the vapour pressure above a solution is less than the predicted value.
A negative deviation would be expected to occur if
- A. the solvent vapour is not an ideal gas.
- B. a solvent molecule has a greater mass than a solute molecule.
- C. a solvent molecule has a smaller mass than a solute molecule.
- D. the solute molecules are strongly attracted to the solvent molecules.
Questions 12 – 16
Young sockeye salmon of equal size were made to swim against a current so that their speed through the water increased from 0 cm/s to a top sustainable speed (tss) of 78 cm/s. At the tss, fish swim as fast as possible without going into oxygen debt (i.e., without using significant anaerobic respiration).
To swim faster than the tss, or if for some other reason there is insufficient oxygen to enable aerobic metabolism to supply energy needs, the salmon need to use anaerobic respiration. Anaerobic processes can only be used significantly for brief periods.
Figure 1(a) shows how the (average) oxygen uptake of the fish varied with speed at a water temperature of 15 °C. At point M the fish were swimming at tss.
Figure 1(b), which is associated with Figure 1(a), indicates the rate of oxygen uptake at various water temperatures for the fish at rest (line 1) and swimming at various speeds (lines 2, 3 and 4) up to the tss (line 5).
In relation to Figure 1(b), note that:
(1) the concentration of oxygen molecules in water decreases as water temperature rises;
(2) oxygen uptake increases with temperature because, as the temperature rises, a fish’s general metabolic rate increases and, in addition, the fish puts more effort into flushing water past its gills to get diminishing oxygen from the water; and
(3) for each point on one of the five lines the speed of the fish is the same (e.g. 0 cm/s for line 1, 78 cm/s for line 5).
Q12. For Figure 1(a), oxygen uptake increased with speed because
- A. water temperature changed.
- B. the fish used anaerobic respiration.
- C. the metabolic rate of the fish changed.
- D. gas exchange occurred at organs other than the gills.
Q13. According to Figure 1(a), which one of the following is the best estimate of how many milligrams of oxygen is taken up in five minutes by a 100 g salmon swimming at 60 cm/s?
- A. 1
- B. 4
- C. 100
- D. 400
Q14. Which one of the following is the best estimate of the speed corresponding to line 3 in Figure 1(b)?
- A. 30 cm/s
- B. 40 cm/s
- C. 50 cm/s
- D. It is not possible tell which of A, B or C is best.
Q15. Which one of the following best helps to explain why, for line 3, oxygen uptake at 20 °C is greater than at 10 °C?
Compared with the case at 10 °C, at 20 °C
- A. The fish swims faster.
- B. Anaerobic respiration is greater.
- C. More energy is needed to achieve the tss.
- D. More energy is used for activities other than swimming
For line 5 of Figure 1(b), Q10 is greatest when X is the temperature corresponding to
- A. L
- B. M
- C. N
- D. P
Organic acids that have a hydroxy group and a carboxylic acid group attached to two adjacent carbon atoms are often called by their colloquial name ß-hydroxyacids. This is because the hydroxy group is on the second carbon atom from the acid group. An example is mevalonic acid, HOCH2 -CH2 -COH(CH2 )-CH2 -COOH, which is a precursor to the production of cholesterol in the body.
One method of producing ß-hydroxyacids in the laboratory is to react an aldehyde or ketone with an α−bromo ester (the bromine is attached to the same carbon atom as the carboxylate group) and metallic zinc in ether. Subsequent hydrolysis produces the ß-hydroxyacid. Figure 1 shows the steps in this reaction.
• the bromo-compound must be an α-bromo alkyl or aryl ester, not an α-bromocarboxylic acid.
• R´, R˝ and R´˝ can each be any of H, an alkyl or an aryl group
Q17. Blood contains low levels of ß-hydroxybutyric acid (its more formal IUPAC name is 3-hydroxybutanoic acid).
Which of the following alkyl groups could be part of their respective reactants in the preparation of this
ß‑hydroxyacid by the reaction sequence described in Figure 1?
- A. R = R´ = R˝ = R´˝ = H
- B. R = CH3 and R´ = R˝ = R´˝ = H
- C. R = R˝ = CH3 and R´ = R´˝ = H
- D. R´ = R´˝ = CH3 and R = R˝ = H
Q18. The reaction sequence described in Figure 1 would produce 3-hydroxypropanoic acid if
- A. R = R´ = R˝ = R´˝ = H.
- B. R = CH3 and R´ = R˝ = R´˝ = H.
- C. R = R˝ = CH3 and R´ = R´˝ = H.
- D. R´ = R´˝ = CH3 and R = R˝ = H.
Q19. Which one of the following could be produced by the reaction sequence described in Figure 1?
- A. 4-hydroxybutanoic acid
- B. 3-hydroxy-3-methylpentanoic acid
- C. 4-hydroxy-3-methylpentanoic acid
- D. 4-hydroxy-4-methylhexanoic acid
Q20. In order to prepare the compound 3-hydroxy-4-ethylhexanoic acid by the reaction sequence described in Figure 1, the two reactants needed are
- A. methyl bromoacetate and 2-ethyl butantal.
- B. methyl bromoacetate and 2-hexanone.
- C. ethyl bromoacetate and 3-methyl butanal.
- D. ethyl bromoacetate and 3-hexanone.
Q21. Under the conditions of the reaction sequence described in Figure 1, the reaction between acetone and propyl bromoacetate would produce
- A. 2-ethyl-3-hydroxypropanoic acid.
- B. 2,2-dimethyl-3-hydroxybutanoic acid.
- C. 2,3-dimethyl-3-hydroxypropanoic acid.
- D. 3-methyl-3-hydroxybutanoic acid.
Question 22 Figure 1 depicts an observer viewing the images of four coloured beads that are formed (focused) on two vertical screens by a thin converging lens. The beads are held at two different heights — one height above the principal axis of the lens, the other height below it — by vertical needles that are pinned into the corners of a rectangular board. The rectangular board is aligned so that its centre is directly below the principal axis of the lens and its longest sides are parallel to this axis. The lens is held vertically between the board and the two screens. The screens rest on the table at different distances from the lens.
Q22. The images of the green, yellow, blue and red beads, in this order, are
- A. 4 1 2 3 .
- B. 1 4 3 2 .
- C. 3 2 1 4 .
- D. 2 3 4 1 .
Questions 23 – 26
As flatfish (Pleuronectiformes) develop, they lose their bilateral symmetry. One of their eyes migrates across the top of the head, so that both eyes end up adjacent on the same side of the head. As a result, the adult fish is able to lie flat on its side on the sea floor with both eyes facing up. In some species of flatfish the right eye migrates to the left side of the head (they are left-eyed); in other species the left eye migrates to the right side of the head (they are right-eyed).
In one species of flatfish, the starry flounder (Platichthys stellatus), different populations have different proportions of left-eyed and right-eyed individuals. Off the west coast of the USA the population is evenly divided between left-eyed and right-eyed individuals. However, in the north Pacific, midway between the USA and Japan, 70% of the population is left-eyed and in Japanese waters all the starry flounder are left-eyed
Consider the following hypotheses (I – III) about the starry flounder. The hypotheses assume that a single gene locus determines the final location of the eyes on the head.
I The Japanese fish are homozygous for a dominant left-eyed allele (LL), whereas west coast fish are homozygous for a neutral allele (ll) and have an equal chance of developing into either left- or right-eyed individuals.
II The Japanese fish are homozygous for the left-eyed allele (LL), whereas in the west coast fish both lefteyed (L) and right-eyed (R) alleles occur in equal proportions. Heterozygous individuals (LR) have an equal chance of developing into either the left- or right-eyed form.
III There are three alleles of the controlling gene: the left-eyed allele (L) is dominant to both the right-eyed allele (l) and the neutral allele (l’ ). Further, the right-eyed allele is dominant to the neutral allele. Fish that are homozygous neutral (l’l’ ) have an equal chance of developing into right- or left-eyed individuals.
Q23. From hypothesis I it can be concluded that the offspring (F1) of a cross between a Japanese starry flounder and a right-eyed west coast starry flounder
- A. would include both left-eyed and right-eyed flounder.
- B. would be right-eyed flounder only.
- C. would be left-eyed flounder only.
- D. cannot be specifically predicted.
Q24. According to hypothesis III, many crosses between ll’ and Ll’ individuals would result in offspring (F1) in proportions closest to
- A 85% left-eyed : 15% right-eyed.
- B 75% left-eyed : 25% right-eyed.
- C 60% left-eyed : 40% right-eyed.
- D 50% left-eyed : 50% right-eyed.
Q25. Consider hypothesis II.
Assume equal frequencies of left- and right-eyed alleles in the west coast population and that there is no impediment to the occurrence of any possible genotype. From hypothesis II it can be concluded that the offspring (F1) of many crosses between Japanese starry flounder and randomly selected west coast flounder would be
- A. three left-eyed flounder for every one right-eyed flounder.
- B. seven left-eyed flounder for every one right-eyed flounder.
- C. equal proportions of left- and right-eyed flounder.
- D. none of the above.
Q26. When a left-eyed mid-Pacific female and a right-eyed west coast male were crossed and the offspring were all kept in standard laboratory conditions on the west coast of the USA, 50% of the thousands of offspring were right-eyed and 50% were left-eyed.
Of the following this result completely rules out
- A hypothesis I.
- B hypothesis II.
- C hypothesis III.
- D none of the hypotheses.
Questions 27 – 30
The male Australian brush turkey (Alectura lathami) assembles an incubation mound consisting of litter, which is typically one metre high by five metres in diameter. As the litter decomposes, mostly due to the action of fungi, the temperature in the mound rises, stabilising around 33˚C at a distance of 60 centimetres below the top of the mound. This is where the eggs are located. At mound temperatures below the stable incubation temperature, the rate of heat production in the mound is faster than the rate of heat loss from the surface of the mound, but the reverse is true above the incubating temperature.
In a typical mound, the fungi and other microorganisms utilise oxygen at the rate of 20 litres per hour, many times the rate of consumption by the eggs. As a consequence, the air in the spaces around the eggs contains 17% O2 and 4% CO2 , which presents a problem for effective gas exchange and the maintenance of appropriate CO2 and O2 levels in the embryo. (In normal air, the O2 level is 21% and the CO2 level is 0.03%.)
The brush turkey eggshell is, relative to eggs of comparable size, half as thick as would be expected. Within the egg, the embryo is immersed in fluid, and gases reach it by diffusion across the chorioallantoic membranes.
After construction, the female digs trenches in the mound, into which she can lay eggs. In a single season, a brush turkey can lay 50 eggs. Unlike most birds, once the eggs are laid, brush turkeys do not care for their young.
The brush turkey embryo, like that of other birds, has an egg tooth, which can be used to break the shell. However, it primarily uses its large feet to break free. Brush turkeys emerge from their eggs in a highly developed state and dig through the litter to the surface. Having primary feathers, the birds are able to fly in the first day or two posthatching.
Figure 1 indicates metabolic rate (joules per second) and oxygen consumption (mL per hour) of embryos within the eggs of two mound-building galliform birds (brush turkey and mallee fowl), which do not provide parental care, and of a typical galliform bird, which do care for their young. The curves cover the period until the bird emerges from the egg.
Q27. According to Figure 1, of the following, the greatest total amount of oxygen consumption
- A. occurs in the mallee fowl between day 50 and day 60.
- B. occurs in the brush turkey between day 15 and day 35.
- C. occurs in the galliform bird between day 1 and day 28.
- D. cannot be determined from the data provided.
Q28. According to Figure 1, of the following, the estimate that is closest to the total amount of energy produced by the mallee fowl during its 60 day incubation is
- A. 50 kilojoules.
- B. 200 kilojoules.
- C. 700 kilojoules.
- D. 2000 kilojoules.
Q29. Which one of the following features present in the brush turkey embryo or egg is least likely to specifically suit the bird to its unusual incubation environment or parenting approach?
- A. egg tooth
- B. thin eggshell
- C. large feet suitable for digging
- D. relatively large yolk supply compared to other birds
Q30. In which one of the following graphs could the Y axis most suitably be labelled heat produced in the mound minus heat lost from the mound and the X axis most suitably be labelled mound temperature at location of egg minus stable incubation temperature at location of egg?
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