ESAT Paper 3 Biology Sample Questions

Correct Answer: E. 520

Step-by-Step Breakdown:

1. Find the number of doublings

$40000 \div 2500 = 16$, and $16 = 2^{4}$, so the population has doubled 4 times in $40$ hours.

2. Convert doublings into one cycle length

Every cell divides once per doubling, so one cell cycle takes $40 \div 4 = 10$ hours, which is $600$ minutes.

3. Find the fraction of cells in interphase

The mitotic cells number $30 + 12 + 8 + 10 = 60$, so the interphase cells number $450 - 60 = 390$, a fraction $\frac{390}{450}$ of the sample.

4. Convert the fraction into a time

Interphase lasts $\frac{390}{450} \times 600 = 520$ minutes.

Sanity check: mitosis then occupies the remaining $80$ minutes, and $520 + 80 = 600$ minutes, the full cycle. Interphase taking the great majority of the cycle is exactly what is expected.

The key is $520$.

Why the Other Options Are Wrong (❌):

• A. 40 · Wrong stage timed
  Times prophase alone instead of the whole of interphase: $30/450$ of the sample is in prophase, and that fraction of the $600$ minute cycle is $40$ minutes. Prophase is one stage inside mitosis, not the long interval between divisions that the question asks about.
• B. 80 · Complement of the fraction
  Adds the four mitotic counts, $30 + 12 + 8 + 10 = 60$, and uses them as the fraction: $\frac{60}{450} \times 600 = 80$ minutes. That is the time spent in mitosis, the complement of what was asked, and the two together make the $600$ minute cycle.
• C. 416 · Miscounted doublings
  Reads the $16$-fold rise as 5 rounds of division rather than 4, giving a cycle of $60 \times 40 \div 5 = 480$ minutes, then $\frac{390}{450} \times 480 = 416$ minutes. The fold increase is $2^{4}$, not $2^{5}$.
• D. 2080 · Cycle length not derived
  Never converts the observation period into one cycle and applies the interphase fraction to all $40$ hours: $\frac{390}{450} \times 2400 = 2080$ minutes. That is longer than the cycle itself, which lasts only $600$ minutes.
• F. 560 · Prophase left inside interphase
  Removes only the metaphase, anaphase and telophase cells from the sample and leaves the $30$ prophase cells counted as interphase, which adds $\frac{30}{450} \times 600 = 40$ minutes to the total. Prophase is the first stage of mitosis, not the tail of interphase, so all four counts come out of the $450$: the interphase share is $\frac{390}{450}$ and the time is $520$ minutes.
• G. 600 · Cycle length reported, stage share never applied
  Stops once the cycle length is out. The $16$-fold rise is $2^{4}$, so $40$ hours holds 4 cycles and one cycle is $60 \times 40 \div 4 = 600$ minutes. That is the whole cycle, mitosis included. Only $\frac{390}{450}$ of it is interphase, which is $520$ minutes, the other $80$ minutes being mitosis.
• H. 8320 · Doublings multiplied in instead of divided out
  Turns the $40$ hour period into a cycle by multiplying by the 4 doublings rather than dividing by them, taking one cycle as $60 \times 40 \times 4$ minutes and then $\frac{390}{450}$ of it. A population that doubles 4 times fits 4 cycles into those $40$ hours, so one cycle is shorter than the period and not longer: $60 \times 40 \div 4 = 600$ minutes, of which interphase takes $520$.

Common Mistake (⚠️):
Using the $40$ hour observation period as the cell cycle length. Those $40$ hours cover 4 successive divisions, so they are 4 cell cycles rather than one, and treating them as one inflates every stage duration by a factor of 4: interphase comes out as $2080$ minutes instead of $520$.

Takeaway (📌):
A mitotic index converts counts into times only once you know the cycle length, and in a population where every cell divides and none dies, 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: C. 120

Step-by-Step Breakdown:

1. Find the ATP the aerobic route supplied

Oxidising $200\ \mu\text{mol}$ of glucose at $30$ mol of ATP per mole gives

$200 \times 30 = 6000 \ \mu\text{mol of ATP}$

2. Take that share off the demand

Glucose is the only substrate, so whatever the aerobic route did not supply came from fermentation:

$6240 - 6000 = 240 \ \mu\text{mol of ATP}$

3. Convert the fermentation ATP into glucose

Fermentation returns $2$ mol of ATP per mole of glucose, so the glucose fermented is

$\dfrac{240}{2} = 120 \ \mu\text{mol}$

Sanity check: the two routes account for $6000 + 240 = 6240\ \mu\text{mol}$ of ATP, which is the demand given, and $120$ is smaller than $3120$, half the demand, as it has to be once part of the ATP has an aerobic source.

The key is $120$.

Why the Other Options Are Wrong (❌):

• A. 3120 · Aerobic share never removed
  Divides the whole demand by the fermentation yield: $6240 \div 2 = 3120$. That treats every mole of ATP as though fermentation had made it, but $200\ \mu\text{mol}$ of glucose was oxidised aerobically and supplied $6000\ \mu\text{mol}$ of the ATP, so that share comes off the demand before anything is divided.
• B. 3000 · Wrong ATP share divided
  Divides the aerobic ATP instead of the fermentation ATP: $6000 \div 2 = 3000$. Those $6000\ \mu\text{mol}$ are already paid for by the $200\ \mu\text{mol}$ of glucose named in the stem; the ATP still waiting to be accounted for is $6240 - 6000 = 240\ \mu\text{mol}$.
• D. 60 · Gross glycolytic yield used
  Divides the fermentation ATP by the gross yield rather than the net one: $240 \div 4 = 60$. Glycolysis does make $4$ ATP per glucose, but two of them are spent activating the sugar, so a fermenting cell keeps only $2$ mol of ATP per mole of glucose.
• E. 200 · Aerobic glucose quoted
  Quotes the glucose figure the stem already supplies. The $200\ \mu\text{mol}$ is the glucose oxidised aerobically, which is what produced $6000$ of the $6240\ \mu\text{mol}$ of ATP; the fermented glucose is a separate quantity and has to be recovered from the $240\ \mu\text{mol}$ that the aerobic route did not supply.

Common Mistake (⚠️):
Dividing the whole demand by the fermentation yield. That gives $6240 \div 2 = 3120\ \mu\text{mol}$ and treats every mole of ATP as fermentation ATP, when $6000\ \mu\text{mol}$ of it came from the $200\ \mu\text{mol}$ of glucose oxidised aerobically. The aerobic share is removed first, and only the remaining $240\ \mu\text{mol}$ is divided.

Takeaway (📌):
Glucose is never read off an ATP total directly. Give each share of the ATP to the route that made it, then divide each share by that route's own yield per mole of glucose, because the same mole of ATP costs very different amounts of sugar depending on which route paid for it.

Correct Answer: E. 6

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 $16$ 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 (❌):

• A. 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.
• B. 9.6 · Impulses counted as breaths
  Takes the $16$ impulses in a burst as $16$ inspirations a minute: $0.6 \times 16 = 9.6$. 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.
• C. 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.
• D. 96 · Each impulse counted as a contraction
  Gets the burst rate right but gives every impulse its own contraction: $16 \times 10 = 160$ contractions a minute, then $0.6 \times 160 = 96$. One whole burst, all $16$ impulses of it, produces a single contraction.
• F. 3.6 · Burst interval read as the breathing rate
  Reads a burst every $6\ \text{s}$ as $6$ inspirations a minute and multiplies by it: $0.6 \times 6$. The $6$ counts seconds per inspiration, so it belongs underneath rather than on top: a minute holds $60 \div 6 = 10$ inspirations, and $0.6 \times 10 = 6$.
• G. 36 · Sixty applied without dividing by the interval
  Turns a per-inspiration volume into a per-minute volume by multiplying by $60$ and never uses the interval at all: $0.6 \times 60$. That is the volume a minute would hold if a contraction arrived once a second. One arrives every $6\ \text{s}$, so the minute holds $60 \div 6 = 10$ of them, giving $0.6 \times 10 = 6$.
• H. 0.006 · Converted as far as cubic metres
  Divides by $1000$ once too often, as though the tidal volume had to reach cubic metres: $1\ \mathrm{m}^3$ is $10^6\ \text{cm}^3$, but the answer is wanted in $\mathrm{dm}^3$ and $1\ \mathrm{dm}^3$ is only $1000\ \text{cm}^3$. So $600 \div 1000 = 0.6\ \mathrm{dm}^3$ an inspiration, and $0.6 \times 10 = 6$, a thousand times this option.

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 (📌):
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. 16

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

$a + b = 14$

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

$6a + 2b = 52$

3. Solve the pair

Substituting $b = 14 - a$ gives $6a + 2(14 - a) = 52$, so $4a = 52 - 28 = 24$, giving $a = 6$ and $b = 14 - 6 = 8$.

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 $6 \times 6 = 36\ \text{mol}$ of carbon dioxide and the fermentation route $2 \times 8 = 16\ \text{mol}$, which together give the $52\ \text{mol}$ stated. The oxygen taken up was $6 \times 6 = 36\ \text{mol}$.

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

Why the Other Options Are Wrong (❌):

• A. 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. 6 · Wrong route reported
  Quotes the glucose that went down the aerobic route, $6\ \text{mol}$. 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. 36 · 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 $6 \times 6 = 36\ \text{mol}$. The question asks for ethanol, which comes only from the $8\ \text{mol}$ that fermented.
• E. 14 · Even split assumed
  Assumes the glucose divided evenly between the two routes, $7\ \text{mol}$ each, giving $2 \times 7 = 14\ \text{mol}$ of ethanol. An even split would have released $6 \times 7 + 2 \times 7 = 56\ \text{mol}$ of carbon dioxide, not the $52\ \text{mol}$ 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 (📌):
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: F. 108

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 $18\,\text{mmol}$ of oxygen a minute can serve $18\div6=3\,\text{mmol}$ of glucose a minute. The muscles are using $20\,\text{mmol}$ a minute in total, so the remaining $20-3=17\,\text{mmol}$ 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 $6\times3=18\,\text{mmol}$ a minute. Over $6$ minutes that is $18\times6=108\,\text{mmol}$.

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 $17\,\text{mmol}$ of glucose fermented each minute therefore adds nothing at all to the carbon dioxide carried away in the blood, and the total stands at $108\,\text{mmol}$.

Sanity check on the carbon: the $3\,\text{mmol}$ of glucose oxidised each minute carries $18\,\text{mmol}$ of carbon atoms, and all of it leaves as $18\,\text{mmol}$ of carbon dioxide. The $17\,\text{mmol}$ fermented carries $102\,\text{mmol}$ of carbon, and every atom of it stays behind in the $34\,\text{mmol}$ of lactate made each minute, which is why the muscle accumulates lactate rather than gas.

The key is $108\,\text{mmol}$.

Why the Other Options Are Wrong (❌):

• A. 18 · Rate reported as a total
  This is the carbon dioxide released in one minute, $6\times3=18\,\text{mmol}$, reported without multiplying by the $6$ 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. 120 · Coefficient of six dropped
  This takes one mole of carbon dioxide per mole of glucose and applies it to all the fuel: $20\times6=120\,\text{mmol}$. 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. 312 · Alcoholic fermentation used for muscle
  This adds two moles of carbon dioxide for every mole of glucose fermented, $108+2\times17\times6=312\,\text{mmol}$, 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. 720 · Oxygen limit ignored
  This sends the whole $20\,\text{mmol}$ a minute down the aerobic route: $20\times6\times6=720\,\text{mmol}$. Only $18\,\text{mmol}$ of oxygen arrives each minute, which is enough for $3\,\text{mmol}$ of glucose, so $17\,\text{mmol}$ a minute cannot take that route however much glucose the muscle holds.
• E. 648 · Oxygen rate read as the glucose rate
  This treats the $18\,\text{mmol}$ of oxygen a minute as the glucose going down the aerobic route and then applies the six from the equation to it: $6\times18\times6$, six times the true total of $108\,\text{mmol}$. The six in $6\text{O}_2$ and the six in $6\text{CO}_2$ belong to the same one mole of glucose, so $18\,\text{mmol}$ of oxygen covers $18\div6=3\,\text{mmol}$ of glucose and returns $18\,\text{mmol}$ of gas a minute, not six times that.

Common Mistake (⚠️):
Working from the fuel rather than from the oxygen. The stem prints $20\,\text{mmol}$ of glucose a minute and says the store never runs out, so it is tempting to send all of it down the aerobic equation; the oxygen arriving is only enough for $3\,\text{mmol}$ of it, and the $17\,\text{mmol}$ left over goes to lactate, which puts no carbon into the gas at all.

Takeaway (📌):
Carbon dioxide in muscle is an aerobic product and nothing else. Find the aerobic share from the oxygen, not from the fuel, and ignore the anaerobic share entirely: its carbon leaves the equation as lactate and never as carbon dioxide.