GAMSAT Pink Part I

Gold Standard GAMSAT’s Worked Solutions to ACER’s GAMSAT Practice Test 3

Pink eBook, Part I

For detailed commentary, see the related Gold Standard GAMSAT video covering that particular question (YouTube or GAMSAT-prep.com). There are many different ways to solve problems in science. The ones presented in this document do not always match the ones presented in the videos.

We do not have Section 1 worked solutions and we do not have Section 3 written solutions for other exams. If you have another way to solve a problem, please feel free to contribute at www.gamsat-prep.com.

We believe that all students should own ACER’s 5 eBooks which should be purchased directly from ACER. This is an adjunct to your studies and in no way can be used to replace the original documents. Occasionally, this document will include minor direct representations of ACER’s exam based on fair use copyright legislation.

Following each of the MCQs, you can find cross-referencing to the current edition of the Gold Standard GAMSAT books to explore further information about that topic. We hope that you will find this document helpful for your GAMSAT preparation.

1.

This question requires you to have a basic understanding of the basic terminology and physiology of the heart.

The atrioventricular valves (mitral and tricuspid), like their name suggests, allows blood to flow from the atrium to the ventricles. The semi-lunar valves (aortic and pulmonary) allow blood to flow from the ventricles to the great vessels (aorta and pulmonary artery, respectively).

Ventricular systole is the duration during which the ventricles are contracting, squeezing blood from the ventricles into the great vessels. Conversely, ventricular diastole is when the ventricles relax, allowing blood to flow from the atria into the ventricles.

Therefore, during ventricular systole, the aortic and pulmonary valves open allowing blood to flow in to the great arteries.

The answer is B.

Background information: GS BIO 7.2, 7.3

2.

Similar to the above question, ACER is testing basic knowledge of the heart.

Oxygenated blood flows from the lungs into the left atrium where it passes to the left ventricle and is eventually pumped into the aorta. Conversely, deoxygenated blood flows into the right atrium where it passes to the right ventricle and is pumped into the pulmonary artery.

You should recognise that since S drains in to the right side of the heart that this carries deoxygenated blood. Deoxygenated blood from the right side is then pumped into P. Conversely, oxygenated blood from the lungs, R, drains in to the left side and is then pumped in to Q.

The answer is Q and R, which is option D.


3.

ACER here is not looking to test your knowledge of heart murmurs; here they are wanting you to build on your knowledge from the previous questions.

Here we are asked about aortic regurgitation which we are told is caused by the aortic valve not closing properly. In question 1 we stated that the aortic valve opens during systole (the ventricles pump blood in to the aorta). This means the aortic valve is usually closed during diastole. Aortic regurgitation is caused by improper closing of the aortic valve – we can infer that the murmur will occur during diastole. Good examination strategy will allow you to reduce the options now to II or IV.

The strategy must now be how to separate the two answers. ACER gives some information in the paragraph preceding the image. The grey box refers to the sound caused by the pulmonary valve closing. The striped box refers to the aortic valve closing. We should refer back to the definition of aortic regurgitation – failure of the aortic valve to close properly. This means that there would not be any sound elicited from the aortic valve, leaving only the shutting of the pulmonary valve to be heard.

We can therefore eliminate IV, and the answer is II – option B.


4.

Again, ACER is wanting you to extrapolate the answer from the definition given in the paragraph and previous questions.

We are given a definition of the thin vertical lines being – less pronounced sounds of lower pitch, of which the length of the lines indicates the intensity. This combined with the definitions of the various murmurs and diagrams should point to the lines representing the sound of turbulent flow of blood.

From the preceding questions we determined that during ventricular diastole blood usually flows from the atria to the ventricles. Erroneous blood flow during diastole would therefore occur in blood travelling back over the semi-lunar valves (aortic and pulmonary) leading to blood flowing back from the arteries into the ventricles. Option D.


5.

This is a basic ACER question requiring you to plug in to the given equation.

The period, T, of the “short vocal track length” looks similar to the period of the “long vocal track length”. This gives a period of 8 ms.

Since the answers are separated by orders of magnitude – we can simplify 8ms to 10ms to make calculation easier. A trick I liked to use for scientific notation is to remember that if you have a “10x10-3” on one side of a fraction, you can change it to other side if you also switch the ‘sign’ (+/-) of the exponent. In the instance below 10x10-3 in the denominator can be turned into 1/10 x 103 in the numerator.

𝑓 =1

10𝑥10&'= 1𝑥10'

10= 100010

≈ 100𝐻𝑧

Therefore, the answer is closest to 100 Hz – we can infer the answer will be 125 Hz, which is option D.

Background information: GS GM 2.2; PHY 7.1.2, 7.2.2, 8.3; BIO 9.2, 12.1, 12.3

6.

This is a graphical interpretation question.

We first need to compare what two graphs in question look like - using the passage for assistance.

The images shown on the right are annotated to show the fundamental sound (circled) and the resonances (length shown by the horizontal line). Now, assessing the options given in the question:

• A – the period of the sound looks similar, in fact this is what we based our calculation in question 5.

• B – the wording of the question is important; we want to know what properties the short vocal tract has compared to the long vocal tract. This option states the short vocal tract decays more slowly. This is alluding to how quickly the fundamental sound returns to zero. We can see that in the image the short vocal tract length returns more quickly to zero, and option B is incorrect.

• C – This makes option C correct. Option C is stating that the short vocal tract returns to zero faster than the long vocal tract. This is shown in the diagram where the resonant frequencies have a larger amplitude than the resonant frequencies of the short vocal tract.


7.

The answer can be inferred via comparison of the difference between long and short vocal tract length and also short and fast glottal pulse rate.

We know from our above answer that the longer vocal tract length results in a longer lasting resonance. Similarly, the increases glottal pulse rate results in a decreased period.

Comparison Length of vocal tract Glottal Pulse rate

Change observed Resonance lasts longer Period decreases

The combined effect of these two would then be to: decrease the period and cause resonance to last longer. This is option A.


8.

This question requires use of the Henderson–Hasselbalch equation.

The question states the values of pH, pKa, and the concentration of citric acid. So, we should rearrange to find the concentration of its base (A-).

𝐻,𝐴& = 𝐻'𝐴 ∙ 10(01&023)

We can now substitute the given parameters (rounding to the closest whole number will make calculations easy for us) in to the equation and solve for the H2A-. In this instance we have simplified the pKa of equation 1 from 2.93 to 3.0

𝐻,𝐴& = 0.1𝑀 ∙ 10(7&')

𝐻,𝐴& = 0.1𝑀 ∙ 10(,) = 100𝑀

In this manner we now know the concentration of both H3A and H2A- which correspond to answers A and B. We will continue to find the answers HA2- and A3- (which correspond to answers C and D). Since we know the value of H2A- we can use this to find HA2-. The pKa for equation 2 is 4.3 which has been simplified to 4.

𝐻𝐴,& = 𝐻,𝐴& ∙ 10(01&023)

𝐻𝐴,& = 100𝑀 ∙ 10(7&8) = 100𝑀 ∙ 10 = 1000𝑀

Now, substituting HA2- to find A3-. The pKa for this equation is 5.21. Since this is greater than the pH we can infer that this will cause the exponent of 10 to be negative. This will mean that the value of A3- will be smaller than that of HA2-. If you arrive at this conclusion we can skips this equation and save some time (since we know it will be smaller than the previous answer). The working is shown below to demonstrate what is stated above.

𝐴'& = 𝐻𝐴,& ∙ 10(01&023)

𝐴'& = 1000𝑀 ∙ 10(&9.,) ≈ 𝑙𝑒𝑠𝑠𝑡ℎ𝑎𝑛1000𝑀

Now that we have the concentrations of all of the species we can answer the question. The species with the largest concentration is 𝐻𝐴,& = 1000. This is the citric acid species which has 2 H atoms removed. This is shown in option C.

Different solution shown in our YouTube video.

Background information: GM 3.7, 3.8; CHM 6.1-6.8

9.

This is another question which ACER tests your ability to both utilise information from a previous questions and also use equations correctly.

From above we know that are H2A- and HA2- respectively. We need to find what pH values these two are equal.

From the equation:

𝑝𝐻 = 𝑝𝐾𝑎 +𝑙𝑜𝑔F9 G𝐻𝐴,&

𝐻,𝐴&H

When the two species are equal, the value in the brackets will = 1. This means that the equation will simplify to:

𝑝𝐻 = 𝑝𝐾𝑎 +𝑙𝑜𝑔F9(1)

From GM 3.7, you will remember that log(1) = 0. So, the pH that the two species are equal is the pKa for the reaction.

In this instance the pKa is 4.28, and therefore the answer is D.


10.

This question states that if temperature increases, deprotonated species decreases. Again, we need to examine with this will mean in the context of the Henderson–Hasselbalch equation. Deprotonated species lie in the numerator of the log(A-/HA).

We can answer this question by considering how this will affect every reaction. If every numerator is larger than the denominator – the log part of the equation will increase. To keep the pH constant, the pKa must

therefore decrease. This scenario applies to all the reactions since the H-H equation always has the deprotonated species in the numerator.

The answer is B.


11.

Here ACER is testing how you can infer from the text/image given. It is most important to consider the wording of the question – “what is the most variable” stage of the cell cycle.

There are some hints that point to the answer. Firstly, the passage states that “most cells are in G0, after they have exited G1” and that “many cells can subsequently return to G1”. Secondly, in the diagram there is an arrow, G0, coming out of G1. This arrow could be occuring at any point in G1 according to the passage. These hints combined, point to the answer being G1 – answer A.

Background information: BIO 1.3

12.

This is a standard ACER graphical interpretation question that relies on some knowledge within the passage.

It is important to note the key sentences in the passage – the level of fluorescence is proportional to amount of DNA. The best way to approach this question is process of elimination.

A – I would be the highest peak if this was true. Most cells are in G1 phase, and since the amplitude of the peaks is proportional to number of cells (y-axis), we can see that this is definitely the correct answer.

B – There is no indication from either the graph or the information given that this answer is explained by the peak I.

C – This statement is likely to be true (since cells have synthesised twice the DNA). However, this statement is not explained by the height of peak I. It is incorrect

D - Cells in G1 will fluoresce the least as the amount of DNA present is less than in the other phases. This is incorrect.


13.

This question can be inferred from the graph provided for the questions.

It can be seen that the largest peak, which we have identified as G1 (question 12), has the lowest fluorescence; a relative fluorescence of around 300 per cell. The question calls for the rel. fluorescence of a population of cells not actively dividing. That is most likely to be cells in G1. We are therefore looking for the picture in the answers that shows a peak at around the 300 mark. This is shown in option C.


14.

Before we begin answering the questions in this unit is important to first review how triangular grid images can be interpreted. Please refer to pages GM-86 to GM-88 in section 3 of the Maths chapter of the GS textbook.

This question requires use of a ruler or straight edge. It is important to remember to work parallel to the dotted lines in the diagram to find the correct solution

In this diagram we can now make a fair estimate as to the composition of the formulation.

• TPGS looks to be almost halfway between 15% and 30%. A fair estimate would be 22% • DOC-Na looks to be close to 15%. A fair estimate would be anything over 10% • Tyloxapol looks to be almost 30%. A fair estimate would be anything between 25% to 30%.

From these estimates the closest matching option is answer A.

Background Information: GM 3.9; Heaps Exams

15.

From the passage it is important to note that the region X corresponds to opaque macroemulsions, Y is nanosized emulsions, and Z is transparent microemulsions.

Formulation P contains 15% tyloxapol. To answer the question, you should draw a line (shown in red) from the 15% across the triangle (ensuring you follow the dotted lines). This will lead the line to cross both Y and Z domains. This means if a formulation has 15% tyloxapol it can be either nanosized emulsions (Y), or transparent microemulsions (Z).

Formulation Q contains 30% tyloxapol. The line in green is drawn from 30% tyloxapol to the appropriate edge. Following our previous example, the line traverses both the Y and Z domains.

Formulations P and Q can either be nanosized emulsions (Y) or transparent microemulsions (Z). They cannot be macroemulsions (X). Therefore the only appropriate answer is option C.


16.

The line above is drawn roughly at 40% DOC-Na. This line traverses through only X – opaque macroemulsions. The answer is option B.


17.

The Zone ‘Z’ is the zone at which microemulsions are made. The maximum percentage of DOC-Na which would permit such a formulation is given by the line shown in red. This is 30% - option B.


18.

The passage states that once a Th cell is activated, it secretes growth factors, IL-2, which stimulate the division of more Th cells which can then be activated in a similar manner. This sounds similar to an exponential expansion of Th cells.

The graph that most resembles this is option C.

Background Information: GM 3.7, 3.8; BIO 6.3.6, 6.3.7, 7.5, 8.2

19.

It is important to examine the wording of the question. The question asks for the best explanation for the value of the process above. We are told that this process is an immune response to a foreign antigen. In this case we should infer that the overall ‘goal’ of the immune response is to produce as many Th cells as possible to fight the invader. We can now examine the options.

A – There is a large increase in Th2 molecules but this does not match the ‘goal’.

B – It would be easy to put B as the answer since we are increasing the population of Th cells; however, the goal is not just produce Th cells – it is to produce Th cells specific to the foreign invader. Th cells are only useful if they can attach the foreign antigen.

C – Similar to A, C is incorrect.

D – Is the correct answer.


20.

For this question we should examine how each option will affect the population of Th cells. The passage contains all the information that we will need to infer the answer.

A – IL-2 is only expressed once Th cells are activated. Th cells expressing IL-2 receptors prior to activation will not divide as there is no IL-2. There will be no change in the population.

B – Th cells are activated by IL-2 binding to IL-2 receptors. If there is an injection of IL-2, there will be increased division of Th cells. The population will increase.

C – IL-2 will bind to IL-2 receptors. If there is excess IL-2 receptors, there is increased competition for the IL-2. This means that fewer IL-2 molecules will bind to Th cells. Following this fewer Th cells will be activated – Th cell activation is being inhibited.

D – Increasing the number of receptors on Th cells will increase the likelihood of their activation when IL-2 is present. This will not lead to inhibition.


21.

This question tests how well you have memorised the all-important physics equation – Ohm’s Law.

The equation to use is 𝑉 = 𝐼𝑅 which can be rearranged into:

𝐼 =𝑉𝑅

We now have to find the values of I110V and I240V. Using the values given in the tables:

𝐼FF9L =110𝑉3000Ω

≈ 130

𝐼,89L =240𝑉2200Ω

≈ 111

We must now find the ratio of the two currents. The 240 V source has a larger current than the 110 V.

𝑅𝑎𝑡𝑖𝑜 = 111130

=3011

≈ 3: 1

The answer is C.

Background Information: GM 1.4; PHY 10.1, 10.3

22.

The Ein – Eout (simplified to ΔE) for H, Cl, and CN are provided below:

H Cl CN

ΔE 0 10 -2

The passage states that the highest magnitude sum of ΔE will prefer to rotate in the ‘in’ direction.

Because Cl has the only positive ΔE, the Cl group will always have the highest magnitude and therefore always rotate in. The answer is A.

23.

We need to again find the sum of magnitudes for each C connected to R groups to answer this question.

H Cl F3C

ΔE 0 10 -1

If we first consider the carbon that is circled on the left:

• Sum of ΔE = (F3C + Cl) = 10 – 1 = 9

Then consider the right-hand carbon

• Sum of ΔE = (Cl + H) = 10 + 0 = 10

This means that the right-hand side has a higher magnitude and will turn ‘in’. Note that because this reaction was photochemically promoted this means that the opposite end would ideally prefer to rotate ‘out’. However, outward rotation is favoured by a negative ΔE. In this case the ΔE is still positive (+9) and so the left hand groups will not rotate.

We need to look for the option that has:

• F3C and Cl in the same positions on the left-hand carbon • H and Cl switched on the right-hand carbon

This is shown in C and is the answer.

Background Information: ORG 4.1, 4.2

24.

The answer to this question involves similar logic to that which we applied above. First note that the H groups are not present. It so happens that in this question we can ignore them altogether but this is a common ‘trick’ in questions. Note again that this reaction is photochemically promoted.

OCH3 Cl CH3 CF3 NO2 CHO

ΔE 14 10 7 -1 -4 -5

Remember from the answer to question 23, only positive ΔE will rotate ‘in’. In the table above only three groups have a ΔE that is positive. Therefore, the answer is C.

Background Information: ORG 4.1, 4.2

25-27

This unit of questions can be tricky without understanding genetics and inheritance. If you struggled on these questions we urge you to read chapter 15 on genetics in the GS GAMSAT textbook before reviewing these questions.

There are a few key points from the passage which are important to consider for the entire unit:

• Due to random deactivation of X-chromosomes local patches of tissue can have either maternal or paternal features

• AED is inherited via the recessive allele on the X-chromosome = X-linked recessive • AED involves sweat gland distribution. Some individuals have patchy distribution, others have no

sweat glands.

First and foremost, we should consider which individuals have a patchy distribution of sweat glands and who have none.

1. Men receive only one X-chromosome – from the mother (receiving Y from father). Remember than AED is X-linked this means that if the allele is recessive on the receiving X-chromosome then this man will inherit AED, affecting all areas of tissue. It affects all areas of tissue because there are no working copies of the allele present in any cells in the body. Therefore, in the context of AED, men have NO sweat glands.

2. Women receive two X-chromosomes; one from the mother and one from the father. There are three scenarios to consider:

o No recessive alleles received (SS) – no AED o One recessive allele received from either mother or father (Ss) – patchy distribution due

to some areas of tissue sharing maternal X-chromosome and others with paternal X-chromosome

o Two recessive alleles received (ss) – no sweat glands at all.

Now we can begin to answer questions.

25.

As we have detailed above, option A, B and C are false as a patchy sweat gland distribution can only be found in women. Men have no sweat glands as the disease is X-linked recessive. Option C is the correct answer as explained above – only women can receive both a defective copy and working copy.

Background Information: BIO Chap 14, 15

26.

Be careful of wording from ACER – this question wants you to find what does not match the information given.

As detailed in passage ‘two’ above, there are three options: women can receive either SS, Ss, or ss. So option A is correct. In passage ‘one’ it is detailed that men can only receive one copy of the allele; if they receive ‘S’ there is no AED, if they receive ‘s’ they have AED. Therefore, in men with AED they cannot carry S – option B is incorrect. This argument helps to find option C is the correct answer – the information supports men receiving only ONE allele. Consistent with argument at the beginning of this paragraph, women can indeed have the Ss allele and have AED.


27.

To answer this question we should again visit the passages ‘1’ and ‘2’. Only two options result in offspring having AED with a total absence of sweat glands. This is either: a male receiving a recessive allele from the mother OR a female receiving TWO recessive alleles from the mother and the father.

This means that the answer is B.


28.

This question can be reasoned through by noting that each ‘row’ of numbers must equal each other on either side of the equation. Secondly, as a positron is positively charged the resultant element must have an atomic number of one less than fluoride-18 (so that either side of the arrow the charges = each other).

In the case of option A – the fluoride is changed into oxygen, which has one fewer protons in its nucleus. In addition, the mass number remains consistent such that mass is conserved (both sides add to 18).

Since all aspects of the equation balance – we know for certain that option A is the answer.

To provide an example of what is not correct, please see option B below.

Here we can see that although the mass numbers balance, the charges do not balance. There is more positive charge on the right (10) compared to 9 on the left.

Background Information: PHY 12.1-12.4

29.

All the information is provided for you, you need to use some clever manipulation and ensure that you keep track of the scientific notation and the exponents.

Firstly, find the number of annihilations per 20 minutes.

𝑁𝑢𝑚𝑏𝑒𝑟𝑜𝑓𝐴𝑛𝑛ℎ𝑖𝑙𝑎𝑡𝑖𝑜𝑛𝑠 = 10X𝑝𝑒𝑟𝑠𝑒𝑐𝑜𝑛𝑑 ∙ 20𝑚𝑖𝑛𝑢𝑡𝑒𝑠 ∙ 60𝑠𝑒𝑐𝑜𝑛𝑑𝑠

= 1.2 × 10F,

Next, we need to find the total energy of the annihilations in this time period.

𝐸^_`3a = 1.2 ∙ 10F, ∙ 10&F'𝐽 = 0.12𝐽

Now we calculate the dose received by a 60 kg human. Since 1 Gy = J/kg we must divide our answer above by 60 kg.

0.12𝐽60𝑘𝑔

= 0.002𝐺𝑦

Therefore the answer is D.

Background Information: GM 2.2; PHY 12.1-12.4

30.

This approach to this answer is very similar to question 28.

Firstly, try to balance the mass number on either side of the equation. We quickly find that we can rule out both options A and B as they do not both equal the same number on either side.

This leaves C and D. We can discern that D is the correct answer by looking at the balancing of the proton number (or charge).

Background Information: PHY 12.1-12.4

31.

**There is a typo in the question: it should say Which of I – V are shunts…”

For this question you must consider the normal human anatomy (which is shown in the diagram). We know that deoxygenated blood from the vena cavae is received into the right atrium and then right ventricle. It is then pumped into the pulmonary artery. The pulmonary veins bring the oxygenated blood into the left atrium and ventricle which then pumps blood into the aorta.

Consider each of the options I-V:

• I – Blood flows from RA to the LA • II – Blood flows from the RV to the Lungs • III – Blood flows from the lungs to the LA • IV – Blood flows from the LV to the tissues • V – Blood flows from the RV to the aorta.

I and V are the two shunts that do not match the diagram in the passage (or previous knowledge). The answer is B.

Background Information: BIO 7.2, 7.3, 14.7, Chapter 0 online

32.

First begin by drawing a sketch of the four heart chambers + inflow and outflow arrows. Beginning with the RA we know that it gives 46 to the LA. The LA then receives 12 from the lungs and the total of the two is pumped out by the LV – 58. Similarly 42 is pumped from the RV; 12 is pumped into the lungs (as 12 is pumped into the LA). However, 30 is shunted into the aorta. Theoretically it should join the 58 pumped out by the LV but only 73 is in the aorta – the difference of (30 + 58 – 73) V + the output of LV minus the flow in the aorta is the flow to the brain (superior tissues) which is 15. What every goes to the head must return via the SVC and so we know that the SVC volume is 15 too. Lastly, we use the same principle to note that the aortic volume (73) will be the same as the IVC – also 73.

This means that the volume into the RA (88) is greater than the output of the LV (58). The answer is B.


33.

It is best to estimate this answer. The oxygen saturation in V is 52. From the LV it is likely to be closer to 62. The saturation in the aorta must be between these two values. Since there is 43(58 – 15) flowing from the LV it is more likely to be closer to 62 than to 52. A safe estimate would be just more than halfway between the two saturations – 58/59 would be a suitable assumption.

We can actually stop here – all the other options give saturations that are outside these boundaries. Option A (70%), option B (64%), and D (52%). Option D could be a distractor, but it is important to remember that the blood hasn’t reach the tissue until it is past the SA. This way we can discriminate that C is the correct option.


34.

Our notation from question 32 has paid off – luckily, we have calculate this answer already. The answer is B, 15%.

Background Information: BIO 7.2, 7.3, Chapter 0 online

35.

This question is almost a trick question! ATP in this diagram is actually just recycled from ATP and phosphate groups not produced from scratch. This means that there will be no change in carbon groups of the ADP/ATP molecules. Therefore the answer is D.

Background Information: BIO 4.4-4.10; ORG 12.5

36.

It is critical that you consider the entire diagram when determining the number of ATP produced by a mole of glucose. A common pitfall would be to only count the ATP produced in one reaction involving glucose (glucose to pyruvate).

½ a mole of glucose yields the following moles of ATP and NADH:

ATP NADH Which reaction it came from

2 1 Glucose to pyruvate

0 1 Pyruvate to acetate

1 3 Citrate to oxaloacetate

This yields a total of 3 ATP and 5 NADH. Note that 1 NADH yields 3 ATP. This means that ½ a mole of glucose produces a total of 18 ATP and therefore 1 mole would produce 36 ATP.

Now consider 1 mole of ethanol:

ATP NADH Which reaction it came from

0 1 Ethanol to ethanal

0 1 Ethanal to acetate

1 3 Citrate to oxaloacetate

This yields a total of 1 ATP and 5 NADH. Since 1 NADH yields 3 ATP one mole of ethanol produces 16 ATP. 36 ATP (produced by glucose) is more than twice, but less than four times 16 ATP (produced by 1 mole of ethanol). The answer is C.

Background Information: BIO 4.4-4.10; ORG 12.5

37.

You are given the equation below:

1𝑓= (𝑛 − 1) ∙ g

1𝑅F−1𝑅,h

There is one key piece of information from the passage that must be taken into consideration before you attempt to answer the question. Note the ‘sign convention for the radii of curvature’. We have a biconvex lens and so we need to account for R1 being positive and R2 being negative.

So we have: R1 = 60 cm, R2 = -60 cm, n = 1.75. Inputting these into the equation and then solving:

1𝑓= (1.75 − 1) ∙ g

160𝑐𝑚

−−1

60𝑐𝑚h

1𝑓= g

34h ∙ g

130𝑐𝑚

h

1𝑓=

140𝑐𝑚

Therefore, the answer is B, 40 cm.

Background Information: GM 1.4, PHY 11.5

38.

It is important to note that this question is asking what the effect would be on flipping the lens.

We can rule out both A and B; in the passage the sign convention (i.e the inverse of the focal length) only changes if the lens changes from converging to diverging or vice versa. In this instance the lens properties do not change; the only thing that changes is which side of the lens receives incoming light first.

We can safely state that the answer is C. This is because the properties of the lens have not changed. I we were to recalculate the focal length when the lens has been reversed we would find that the focal length would not change and that the light would be refracted in the same way. Thus, D is false and C is correct.


39.

To find the refractive index let us rearrange the equation given to find n:

1(𝑛 − 1)

= 𝑓 ∙ g1𝑅F−1𝑅,h

The focal length f, is the same magnitude as the radius of curvature for the symmetrical lens. Therefore, if we let this length be ‘x’ and substitute this in and simplify:

1(𝑛 − 1)

= 𝑥 ∙ g1𝑥−−1𝑥h

1(𝑛 − 1)

= 𝑥 ∙ g2𝑥h

The x terms cancel and you are left with

1(𝑛 − 1)

= 2

Finally, this can be rearranged to find n.

12= 𝑛 − 1

𝑛 = 1.5

The answer is C.


40.

Process of elimination is the best strategy here.

Option A is incorrect as the passage states that the volatile propanone in Fred’s breath can be converted to an alcohol, 2-propanol. In this case we cannot say for certain whether his car would start. Similarly, the half-life of the organic compound is irrelevant in the context of this question as we are unsure of the concentration of alcohol in his blood/being exhaled.

In determining between options C or D we should consider how the BAIID measures the blood alcohol content. The BAIID oxidises the volatile alcohol and allows it to produce a current. Propanone is a ketone. 2-propanol, by comparison is an alcohol and thus can be readily oxidised. The answer is C.

Background Information: ORG Chapter 6, 7

41.

Following the explanation for question 40 and from the passage, we have deduced that 2-propanol can be oxidised. Its production is due to the fact that the propanone is exhaled from the body. There is nothing to suggest that Fred has drunk any alcoholic beverages (see paragraph 2 of the passage) and so the answer must be B.


42.

If Fred maintains his diet (which contains no ethanol), 2 – propanol will continue to be produced, and since BAIID can only detect an alcohol, it will only detect 2-propanol from Fred. The answer is C.


43.

The wording of the question is tricky. The air is inhaled at 25 degrees. First read the dotted line at 25 degrees, reading the ‘water in air’ from the y-axis for that temperature (roughly 5 mg/L). Next follow the red line from the dotted to the solid – this represents the transition to 100% humidity within the human body. Next follow the curve up to normal human temperature, 38 degrees Celsius. Lastly, read across to the y-axis and note the ‘water in air’ (roughly 45 mg/L). The answer is C, 40 mg.

Background Information: GM 2.2, Chapter 3; BIO Chapter 0 online

44.

We know from the graphs given that as the temperature of air increases humidity has more of an effect on adding water from inhaled air. Therefore, the more humid the air, the more water is inhaled. This eliminates option B.

The key statement from the passage is the description that cooling allows water vapour to condense on the surface of the nasal passage – it follows that more cooling of the air will result in more water being recovered (option C is incorrect, and option D is correct). This logic helps us to eliminate A. If there was less cooling (higher ambient air temperature) there would be a greater effect of water loss in exhaled air.


45.

This question requires similar logic to question 43. Firstly, read the ‘water in air’ at 30 degrees with 25% humidity (roughly 8). Then read the 100% humidity graph at a temperature of 27 degrees (roughly 28). Therefore, the answer is the difference between the two (20 mg). The answer is A.


46.

Make sure to note the not.

We need to find which of the following would NOT improve heat exchange.

- Option A would not improve heat exchange. There would be less time for the heat exchange and therefore there would be less heat exchange

- Option B would improve heat exchange. Narrow passages would mean the air would be closer to the heat source (nasal passages)

- Option C would improve heat exchange as there would be more surface area for heat to exchange. - Option D would improve heat exchange. There is a small amount of previous knowledge here.

Counter-current heating is a very efficient form of heating.

Background Information: GM 2.2, Chapter 3; BIO Chapter 0 online; BIO 10.3, 12.2

47.

This question does not require previous knowledge – it is more pattern recognition.

2-chloropropanoic acid has an H value of ‘45.2’. 2-butanoic acid has a value of ’45.7’. The most likely value for 2-chloropentanoic acid is likely to be only 0.4-0.5 more units above from 45.7. the answer is most likely 46.1, A.

Background Information: ORG Chapter 8

48.

This question does not require previous knowledge – it is more pattern recognition.

If we track the values in the table that are close to 60.7 we find that they are all contained by the 3-chloro(acid) series. The next acid in the series would be a pentanoic acid – therefore the answer is 3-chloropentanoic acid (A).


49.

Regard only the 3-chloro part of the names.

Therefore, I took 3-chloro-2-methylbutanoic acid to be 3-chlorobutanoic acid. This simplification was justified as ACER doesn’t really give you any other information you can use to solve the equation.

Using this information:

We can see that the H value for 2-chlorobutanoic acid is less than the value for 3-cholorobutanoic acid which is turn less than the value for 4-chlorobutanoic acid. This leads to the conclusion that the answer is B. Option C is not right following the above logic and nothing from the information given suggests that D is the answer.


50.

Almost a trick question! Candidates could be tempted to read off the graph where it says “80” on the y-axis and take “hours of light per 24-hour day” associated with the ‘80’ mark. Instead the information in the passage states the y -axis depicts diapause which is the stopping of growth. The question asks for when 80% of the population grows – which equates to 20% diapause. This between 14.5 hours and 15.5 hours and the answer is C.

Background Information: GM Chapter 3; BIO Chapter 0 online

51.

The largest percentage growing means the smallest percentage in diapause (not growing). Of the 4 answer choices, answer choice A has the smallest percentage as per the diagram: 55%, and thus it is the correct answer.

Background Information: GM Chapter 3; BIO Chapter 0 online

52.

This can be a pretty tricky question. At first it is not obvious how to figure the answer. We are directed to look at the sequences separately but it is important to look at them together and observe the similarities. If we place the sequences together we find that there are common bases:

A T T G A T A T A T T G

This leaves four bases (in the correct order) which are common to both sequences and four bases which are found only once in either of the sequences. This means there are eight total distinct bases (out of the possible twelve). This leads us to infer that the above fragments represent at least 8/12 of the total sequence. The answer is A.

Background Information: BIO 1.2.2, 1.3, Chapter 3

53.

First write down the given information:

Total compound to be dissolved 2.3 mol

Compound dissolved by II 1.2 mol

Total remaining compound 1.1 mol

We can infer that the total remaining compound (1.1 mol) will be dissolved by solvents I and III in a ratio of 15:40.

If we calculate the concentrations for compound dissolved in I and III. The concentration dissolved in solvent one is given by the ratio of ‘solvent I’ (15) over the dissolved compound in both solvents (15 + 40).

1.1𝑚𝑜𝑙 ∙15

15 + 40

We can make this ratio slightly easier to solve without a calculator multiplying 1.1 by 100 to get 110 mol. For concentration dissolved in solvent III this then becomes:

110𝑚𝑜𝑙 ∙1555

= 30𝑚𝑜𝑙

Following the same for solvent III

110𝑚𝑜𝑙 ∙4055

= 80𝑚𝑜𝑙

Remember that the concentration of compound in solvent II was 1.2 mol. If we also multiply this by 100 (like we did to 1.1 mol) then we get a concentration of 120 mol. Now placing the ratio’s side by side and simplifying we get the answer, A.

30: 120: 80 = 3: 12: 8

Background Information: GM 1.4; CHM 5.3

54.

Dihomozygous red-eyed males X cis-dihybrid females

Rb/Rb X RB/rb

Genes on the same chromosome that are in close proximity are often inherited together, meaning that while this question involved 2 genes, it is easier to solve than it appears. The confusing part in this question is the 50% recombination in the females, which means that 50% of the time the genes are not inherited together in the females and they in fact crossover. However, this is actually solvable with a simple Punnett square, as the male genes are both Rb. This means that regardless of recombination in the female the outcome in the offspring will be the same. This is shown in the extended Punnett square below which accounts for the recombination:

RB (25%) rb (25%) Rb (25%) rB (25%)

Rb (25%) Rb/RB

Maroon

Rb/rb

Red

Rb/Rb

Red

RbrB

Maroon

Rb (25%) Rb/RB

Maroon

Rb/ rb

Red

Rb/Rb

Red

RbrB

Maroon

Rb (25%) Rb/RB

Maroon

Rb/rb

Red

Rb/Rb

Red

RbrB

Maroon

Rb (25%) Rb/RB

Maroon

Rb/rb

Red

Rb/Rb

Red

RbrB

Maroon

Ratio: 8 Maroon: 8 Red

= 1:1 ratio

= Equal proportions of maroon-eyed and red-eyed

Now, let’s simplify the Punnett square:

RB rb

Rb Rb/RB

Maroon

Rbrb

Red

Rb Rb/RB

Maroon

Rbrb

Red

Ratio: 2 Maroon: 2 Red

= 1:1 ratio

= Equal proportions of maroon-eyed and red-eyed

The answer is D.

*Note that with different combinations of male genotypes or different percentages of recombination, an extended Punnett square is required.

Background Information: BIO Chapters 14, 15

55.

trans-dihybrid male X white-eyed female

Rb/rB X rb/rb

Once again, this question is easier than it appears. The female recombination can be ignored, as the genotype is rb/rb, meaning that recombination will yield the same genotype. Thus, it is solvable with a simple punnett square.

rb rb

Rb Rbrb

Red

Rbrb

Red

Rb rBrb

Brown

rBrb

Brown

Ratio: 2 Red: 2 Brown

= 1:1 ratio

= Equal proportions of red-eyed and brown-eyed

Thus, the only eye colours the progeny can have are red and brown. The answer is C.


56.

A. Brown eyed phenotypes must be homozygous for the rd allele. This is correct because Rd is co-dominant with Bw. This means that if any Rd alleles are present, it will be expressed phenotypically, either in maroon eyes (when in combination with a Bw allele) or as red eyes (in the absence of Bw allele).

B. This is incorrect, because at least one Bw allele must be present to express brown eye colour in the phenotype.

C. This is incorrect, because it cannot be homozygous for brown as brown must be expressed in the phenotype to have brown eyes.

D. This is incorrect, because it cannot be heterozygous for red, or red would be expressed in the phenotype.


GAMSAT Pink Part I

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