ESAT Practice Set 1A Biology Worked Solutions

Correct Answer: E. 72

Fastest Approach (🚀):
Do not split the glucose at all. The aerobic equation uses six oxygen and makes six carbon dioxide, so mole for mole the gas out equals the oxygen in, and the whole question is $24\times3=72\,\text{mmol}$. Dividing by $6$ and multiplying by $6$ again is the step to skip.

Step-by-Step Breakdown:

1. Let the oxygen decide how much glucose takes the aerobic route

The aerobic equation consumes $6$ moles of oxygen for each mole of glucose, so $24\,\text{mmol}$ of oxygen a minute can serve $24\div6=4\,\text{mmol}$ of glucose a minute. The muscles are using $28\,\text{mmol}$ a minute in total, so the remaining $28-4=24\,\text{mmol}$ a minute has to be respired anaerobically.

2. Take the carbon dioxide from the aerobic route

The same equation releases $6$ moles of carbon dioxide per mole of glucose, so the aerobic route gives $6\times4=24\,\text{mmol}$ a minute. Over $3$ minutes that is $24\times3=72\,\text{mmol}$.

3. Check what the anaerobic route adds

Read the second equation: glucose becomes lactate, and there is no carbon dioxide anywhere on its right-hand side. The $24\,\text{mmol}$ of glucose fermented each minute therefore adds nothing at all to the gas collected, and the total stands at $72\,\text{mmol}$.

Sanity check on the carbon: the $4\,\text{mmol}$ of glucose oxidised each minute carries $24\,\text{mmol}$ of carbon atoms, and all of it leaves as $24\,\text{mmol}$ of carbon dioxide. The $24\,\text{mmol}$ fermented carries $144\,\text{mmol}$ of carbon, and every atom of it stays behind in the $48\,\text{mmol}$ of lactate made each minute, which is why the muscle accumulates lactate rather than gas.

The key is $72\,\text{mmol}$.

Why the Other Options Are Wrong (❌):

• A. 24 · Rate reported as a total
  This is the carbon dioxide released in one minute, $6\times4=24\,\text{mmol}$, reported without multiplying by the $3$ minutes the question asks about. It is also the oxygen figure printed in the stem, which makes it doubly tempting to write down and stop.
• B. 84 · Coefficient of six dropped
  This takes one mole of carbon dioxide per mole of glucose and applies it to all the fuel: $28\times3=84\,\text{mmol}$. The aerobic equation releases six per glucose, not one, and the fermented share releases none, so both halves of the reasoning are wrong at once.
• C. 216 · Alcoholic fermentation used for muscle
  This adds two moles of carbon dioxide for every mole of glucose fermented, $72+2\times24\times3=216\,\text{mmol}$, which is the yield of alcoholic fermentation in yeast. The equation printed for these muscles makes lactate and nothing else, so its carbon never leaves as gas.
• D. 504 · Oxygen limit ignored
  This sends the whole $28\,\text{mmol}$ a minute down the aerobic route: $28\times6\times3=504\,\text{mmol}$. Only $24\,\text{mmol}$ of oxygen arrives each minute, which is enough for $4\,\text{mmol}$ of glucose, so $24\,\text{mmol}$ a minute cannot take that route however much glucose the muscle holds.

Common Mistake (⚠️):
Assuming that respiration of any kind releases carbon dioxide, so the $24\,\text{mmol}$ of glucose fermented each minute swells the total. In muscle the anaerobic equation stops at lactate and the carbon stays locked in it; only the glucose that meets oxygen sends its carbon out as gas, which is why the answer is set by the oxygen supply of $24\,\text{mmol}$ a minute and not by the $28\,\text{mmol}$ of fuel used.

Takeaway (📌):
The Cheat Code: When two routes share one fuel, let the scarce reagent size the first route, then read the second route's equation for what it does and does not make. Oxygen delivery fixes the aerobic share, and lactate fermentation contributes no carbon dioxide at all.

Correct Answer: A. 16

Fastest Approach (🚀):
Work from the all-fermented case. Fermenting the whole $20\ \text{mol}$ would give only $2 \times 20 = 40\ \text{mol}$ of carbon dioxide, a shortfall of $48\ \text{mol}$ against the $88\ \text{mol}$ collected, and each mole switched to the aerobic route closes $4\ \text{mol}$ of that gap. So $12\ \text{mol}$ went aerobic, $8\ \text{mol}$ fermented, and doubling the fermented figure gives $16\ \text{mol}$ of ethanol.

Step-by-Step Breakdown:

1. Split the glucose between the two routes

Let $a$ be the moles of glucose respired aerobically and $b$ the moles fermented. Every mole of glucose took one route or the other, so

$a + b = 20$

2. Balance the carbon dioxide

The aerobic equation releases $6$ moles of carbon dioxide per mole of glucose and the fermentation equation releases $2$, so

$6a + 2b = 88$

3. Solve the pair

Substituting $b = 20 - a$ gives $6a + 2(20 - a) = 88$, so $4a = 88 - 40 = 48$, giving $a = 12$ and $b = 20 - 12 = 8$.

4. Turn the fermented glucose into ethanol

Fermentation gives two moles of ethanol per mole of glucose:

$2 \times 8 = 16\ \text{mol}$

Sanity check: the aerobic route released $6 \times 12 = 72\ \text{mol}$ of carbon dioxide and the fermentation route $2 \times 8 = 16\ \text{mol}$, which together give the $88\ \text{mol}$ stated. The oxygen taken up was $6 \times 12 = 72\ \text{mol}$.

The key is $16\ \text{mol}$.

Why the Other Options Are Wrong (❌):

• B. 8 · Coefficient dropped
  Solves the split correctly to $8\ \text{mol}$ of glucose fermented, then reports that figure as the ethanol and drops the $2$ in $\text{C}_6\text{H}_{12}\text{O}_6 \rightarrow 2\text{C}_2\text{H}_5\text{OH} + 2\text{CO}_2$. Each mole of glucose fermented gives two moles of ethanol, so the ethanol is $2 \times 8 = 16\ \text{mol}$.
• C. 12 · Wrong route reported
  Quotes the glucose that went down the aerobic route, $12\ \text{mol}$. That route makes carbon dioxide and water and no ethanol at all, so it cannot be the answer: the ethanol comes from the $8\ \text{mol}$ that fermented, at two moles of ethanol for each.
• D. 72 · Oxygen quoted instead of ethanol
  Splits the glucose correctly and then answers with the wrong quantity. The aerobic equation takes six moles of oxygen per mole of glucose, so the oxygen used is $6 \times 12 = 72\ \text{mol}$. The question asks for ethanol, which comes only from the $8\ \text{mol}$ that fermented.
• E. 20 · Even split assumed
  Assumes the glucose divided evenly between the two routes, $10\ \text{mol}$ each, giving $2 \times 10 = 20\ \text{mol}$ of ethanol. An even split would have released $6 \times 10 + 2 \times 10 = 80\ \text{mol}$ of carbon dioxide, not the $88\ \text{mol}$ collected, so the split has to be solved for rather than guessed.

Common Mistake (⚠️):
Stopping one step early. The pair of balances gives the glucose fermented, $8\ \text{mol}$, and that figure looks like an answer, but the question asks for ethanol and the fermentation equation puts two moles of it into the vessel for every mole of glucose used: $2 \times 8 = 16\ \text{mol}$.

Takeaway (📌):
The Cheat Code: When two pathways draw on the same substrate and make the same product in different ratios, write one balance for the substrate and one for the product, then solve the pair. The quantity actually asked for is usually one stoichiometric step beyond the split, so do not stop at the moles of glucose you have just found.

Correct Answer: D. 324

Fastest Approach (🚀):
Never count interphase cells. Add the four mitotic counts, $90 + 36 + 24 + 30 = 180$, subtract from $1800$ once, and pair that with the cycle length $360$ minutes that $48000/12000 = 4 = 2^{2}$ hands you in a single step.

Step-by-Step Breakdown:

1. Find the number of doublings

$48000 \div 12000 = 4$, and $4 = 2^{2}$, so the population has doubled 2 times in $12$ hours.

2. Convert doublings into one cycle length

Every cell divides once per doubling, so one cell cycle takes $12 \div 2 = 6$ hours, which is $360$ minutes.

3. Find the fraction of cells in interphase

The mitotic cells number $90 + 36 + 24 + 30 = 180$, so the interphase cells number $1800 - 180 = 1620$, a fraction $\frac{1620}{1800}$ of the sample.

4. Convert the fraction into a time

Interphase lasts $\frac{1620}{1800} \times 360 = 324$ minutes.

Sanity check: mitosis then occupies the remaining $36$ minutes, and $324 + 36 = 360$ minutes, the full cycle. Interphase taking the great majority of the cycle is exactly what is expected.

The key is $324$.

Why the Other Options Are Wrong (❌):

• A. 18 · Wrong stage timed
  Times prophase alone instead of the whole of interphase: $90/1800$ of the sample is in prophase, and that fraction of the $360$ minute cycle is $18$ minutes. Prophase is one stage inside mitosis, not the long interval between divisions that the question asks about.
• B. 36 · Complement of the fraction
  Adds the four mitotic counts, $90 + 36 + 24 + 30 = 180$, and uses them as the fraction: $\frac{180}{1800} \times 360 = 36$ minutes. That is the time spent in mitosis, the complement of what was asked, and the two together make the $360$ minute cycle.
• C. 216 · Miscounted doublings
  Reads the $4$-fold rise as 3 rounds of division rather than 2, giving a cycle of $60 \times 12 \div 3 = 240$ minutes, then $\frac{1620}{1800} \times 240 = 216$ minutes. The fold increase is $2^{2}$, not $2^{3}$.
• E. 648 · Cycle length not derived
  Never converts the observation period into one cycle and applies the interphase fraction to all $12$ hours: $\frac{1620}{1800} \times 720 = 648$ minutes. That is longer than the cycle itself, which lasts only $360$ minutes.

Common Mistake (⚠️):
Using the $12$ hour observation period as the cell cycle length. Those $12$ hours cover 2 successive divisions, so they are 2 cell cycles rather than one, and treating them as one inflates every stage duration by a factor of 2: interphase comes out as $648$ minutes instead of $324$.

Takeaway (📌):
The Cheat Code: A mitotic index converts counts into times only once you know the cycle length, and in a dividing population the cycle length is the growth time divided by the number of doublings, which you read off the fold increase as a power of two.

Correct Answer: A. 6

Fastest Approach (🚀):
Do the conversion first, since $600 \div 1000 = 0.6$ is the easier of the two steps, then multiply by the $10$ inspirations that $60 \div 6$ gives. The impulse count $15$ plays no part in the answer at all.

Step-by-Step Breakdown:

1. Turn the burst interval into a breathing rate

One whole burst produces one contraction of the diaphragm, so one burst is one inspiration however many impulses the burst contains. A burst every $6\ \text{s}$ gives

$\dfrac{60}{6} = 10$ inspirations per minute.

The $15$ impulses describe the traffic along the motor neurone that drives a single contraction; they are not a count of breaths.

2. Put the volume into the units of the answer

$600\ \text{cm}^3 = \dfrac{600}{1000} = 0.6\ \mathrm{dm}^3$ drawn in per inspiration.

3. Combine volume per breath with breaths per minute

$0.6 \times 10 = 6$, so $6\ \mathrm{dm}^3$ of air is drawn in each minute.

Sanity check: a whole minute holds $10$ inspirations, so the answer must be larger than the $0.6\ \mathrm{dm}^3$ that one inspiration draws in, and it is. Leaving the volume in $\mathrm{cm}^3$ would give $6000$, a thousand times too large.

The key is $6$.

Why the Other Options Are Wrong (❌):

• B. 6000 · Volume conversion omitted
  Multiplies the tidal volume by the breathing rate and labels the result in cubic decimetres: $600 \times 10 = 6000$. The tidal volume is measured in $\mathrm{cm}^3$ and $1000\ \mathrm{cm}^3 = 1\ \mathrm{dm}^3$, so it has to be divided by $1000$ before the multiplication, which makes this option exactly $1000$ times the true figure.
• C. 9 · Impulses counted as breaths
  Takes the $15$ impulses in a burst as $15$ inspirations a minute: $0.6 \times 15 = 9$. Every impulse in one burst serves the same single contraction, so a burst counts once however many impulses it carries, and the minute holds $10$ bursts.
• D. 0.6 · Volume per breath reported as volume per minute
  Converts the tidal volume and stops there: $600 \div 1000 = 0.6$. That is the volume drawn in by one contraction of the diaphragm, and it still has to be multiplied by the $10$ inspirations a minute contains.
• E. 90 · Each impulse counted as a contraction
  Gets the burst rate right but gives every impulse its own contraction: $15 \times 10 = 150$ contractions a minute, then $0.6 \times 150 = 90$. One whole burst, all $15$ impulses of it, produces a single contraction.

Common Mistake (⚠️):
Leaving the tidal volume in cubic centimetres. $600\ \text{cm}^3$ is $0.6\ \mathrm{dm}^3$, because $1\ \mathrm{dm}^3 = 1000\ \mathrm{cm}^3$, and multiplying $600$ by the $10$ inspirations a minute holds gives $6000$, a thousand times the true figure.

Takeaway (📌):
The Cheat Code: Turn an interval into a per-minute rate before anything else, and convert the volume into the units the answer asks for at the same time. Any per-minute total is then the amount moved by one event multiplied by the number of events in a minute, and a count of impulses inside one event is not a count of events.

Correct Answer: B. $1344$

Fastest Approach (🚀):
$246$ divides by three, so stay in triplets throughout. Gene $M$ is $265$ triplets, gene $N$ is $246 \div 3 = 82$ triplets fewer, so $183$, and the two come to $265 + 183 = 448$ triplets. One multiplication at the end, $448 \times 3 = 1344$, replaces three separate ones.

Step-by-Step Breakdown:

1. Turn the chain coded by gene $M$ into a length of DNA

Each amino acid in a chain is specified by one triplet, and a triplet is three nucleotides. The chain coded by gene $M$ holds $265$ amino acids, so the coding stretch of gene $M$ is $265 \times 3 = 795$ nucleotides.

2. Take the stated difference off, in nucleotides

The difference of $246$ is given as a count of nucleotides, so it comes off the $795$ rather than off the $265$:
$$795 - 246 = 549.$$
Gene $N$'s stretch is $549$ nucleotides, which is $549 \div 3 = 183$ triplets, so the chain it codes for is $183$ amino acids long.

3. Add the two coding stretches

$$795 + 549 = 1344.$$
Between them the two stretches contain $1344$ nucleotides.

Check: the two chains hold $265 + 183 = 448$ amino acids, and $448 \times 3 = 1344$, the same figure, as it has to be when every amino acid is specified by three nucleotides.

Matches Option B.

Why the Other Options Are Wrong (❌):

• A. $795$ · Stopped at one gene
  $265 \times 3 = 795$ is the coding stretch of gene $M$ on its own. Gene $N$'s stretch, $795 - 246 = 549$, still has to be added to answer what the two contain between them.
• C. $284$ · Triplet code never applied
  Treating the $265$ amino acids as $265$ nucleotides makes gene $N$'s stretch $265 - 246 = 19$, and $265 + 19 = 284$. Each amino acid needs three nucleotides, so gene $M$'s stretch is $795$, not $265$.
• D. $549$ · Stopped at the shorter gene
  $795 - 246 = 549$ is gene $N$'s coding stretch by itself. The question asks for the two stretches together, so gene $M$'s $795$ must be added to it.
• E. $448$ · Amino acids reported, not nucleotides
  $549 \div 3 = 183$ amino acids in the chain from gene $N$, and $265 + 183 = 448$ amino acids in the whole protein. That is a count of amino acids; the question asks for nucleotides, which is three times as many.

Common Mistake (⚠️):
Subtracting before converting. The $246$ counts nucleotides and the $265$ counts amino acids, so $265 - 246 = 19$ subtracts one kind of thing from another and means nothing. Converting first, to $795$ nucleotides, is what makes the subtraction legitimate.

Takeaway (📌):
The Cheat Code: Nucleotides and amino acids stand in a fixed three to one ratio along a coding stretch, so convert every quantity into the same one of the two units before you add or subtract anything.