ESAT Biology Practice Questions by Topic

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 (📌):
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: D. $\dfrac{1}{3}$

Fastest Approach (🚀):
Only the heterozygous branch can produce white, so the whole question collapses to $P(Bb)$ halved. Write down $\frac{2}{3}$, halve it, and stop.

Step-by-Step Breakdown:

1. Work out the first cross

$Bb \times Bb$ gives $1\,BB : 2\,Bb : 1\,bb$. As $B$ is completely dominant, $BB$ and $Bb$ are black and $bb$ is white.

2. Condition on the chosen rabbit being black

The rabbit was drawn from the black offspring only, so the $bb$ quarter is excluded. Of the three equally likely genotypes that remain, one is $BB$ and two are $Bb$, so $P(BB) = \frac{1}{3}$ and $P(Bb) = \frac{2}{3}$.

3. Send each branch through the second cross

The white partner is $bb$ and can donate only $b$. $BB \times bb$ gives all $Bb$, so $P(\text{white}) = 0$. $Bb \times bb$ gives $1\,Bb : 1\,bb$, so $P(\text{white}) = \frac{1}{2}$.

4. Combine

$P(\text{white}) = \frac{1}{3} \times 0 + \frac{2}{3} \times \frac{1}{2} = \frac{1}{3}$.

Sanity check: the answer must be below $\frac{1}{2}$, because the chosen parent might be $BB$ and give no white offspring at all, and above $\frac{1}{4}$, because it is more likely heterozygous than not.

Matches Option D.

Why the Other Options Are Wrong (❌):

• A. $\dfrac{1}{8}$ · Wrong second parent
  Takes $P(Bb) = \frac{1}{2}$ and then crosses the black rabbit with another heterozygote instead of with the white rabbit, so the second step contributes $\frac{1}{4}$ rather than $\frac{1}{2}$: $\frac{1}{2} \times \frac{1}{4} = \frac{1}{8}$.
• B. $\dfrac{1}{6}$ · Inverted conditional
  Inverts the conditional probabilities and uses $\frac{1}{3}$ for the heterozygote, which is actually $P(BB)$: $\frac{1}{3} \times \frac{1}{2} = \frac{1}{6}$.
• C. $\dfrac{1}{4}$ · Failure to condition
  Reads $P(Bb)$ straight off the unconditioned $1:2:1$ ratio as $\frac{1}{2}$, ignoring that the $bb$ quarter cannot be the black rabbit: $\frac{1}{2} \times \frac{1}{2} = \frac{1}{4}$.
• E. $\dfrac{1}{2}$ · Assumed genotype
  Assumes any black offspring of two heterozygotes must itself be heterozygous, so the cross is simply $Bb \times bb$ and the answer is the raw $\frac{1}{2}$.

Common Mistake (⚠️):
Using $P(Bb) = \frac{1}{2}$ for the chosen parent, because $Bb$ is half of the whole $1:2:1$ ratio, and forgetting that the $bb$ quarter was already ruled out by the statement that the rabbit is black. That gives $\frac{1}{2} \times \frac{1}{2} = \frac{1}{4}$ instead of $\frac{1}{3}$.

Takeaway (📌):
An offspring of two heterozygotes known only to show the dominant phenotype is two thirds heterozygous and one third homozygous dominant, not one half. Condition on the phenotype first, then weight the second cross by those two fractions.

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, $448 \times 3 = 1344$, gives the total.

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 (📌):
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.

Correct Answer: D. In P one nucleotide has been replaced by a different one; in Q one nucleotide has been lost.

Fastest Approach (🚀):
Read the two lengths before either sequence. With a single change in each strain, $1\,200$ against $1\,200$ has to be a replacement and $1\,199$ against $1\,200$ has to be a loss, and that fixes both halves of the answer at once.

Step-by-Step Breakdown:

1. Compare each length with the wild type's $1\,200$ nucleotides.

A mutation is a change to the sequence of nucleotides, so the length of that sequence is the first thing to compare. Strain P's gene is $1\,200$ nucleotides, exactly the wild-type length, so no nucleotide has been gained or lost. Strain Q's gene is $1\,199$ nucleotides, one fewer, so exactly one nucleotide has been lost.

2. Read the disagreement in strain P.

P matches the wild type at every position but $640$, where a different base sits. With the length unchanged, that single disagreement is one nucleotide replaced by a different one, the rest of the sequence being untouched.

3. Read the shift in strain Q.

Q agrees with the wild type as far as position $640$. Beyond that, Q's nucleotide at any position is the wild-type nucleotide one place further along, so the wild type's nucleotide at position $641$ has no counterpart in Q and each later nucleotide has moved one place forward: $1\,199 - 640 = 559$ of them, filling positions $641$ to $1\,199$ of Q. That is one nucleotide lost at position $641$. It cannot be a run of replacements, because replacing nucleotides leaves the number of them unchanged.

Sanity check: taking one nucleotide out of $1\,200$ leaves $1\,199$, which is the length measured in Q.

Matches Option D.

Why the Other Options Are Wrong (❌):

• A. In P one nucleotide has been added; in Q one nucleotide has been lost. · A replaced base read as an added one
  Q is read correctly, but an added nucleotide would leave P's gene $1\,200 + 1 = 1\,201$ nucleotides long. P is $1\,200$, the wild-type length, so nothing has been added to it and the base at position $640$ has been swapped for another.
• B. In P one nucleotide has been lost; in Q one nucleotide has been added. · Both length comparisons reversed
  This has P losing a nucleotide although its length is unchanged at $1\,200$, and Q gaining one although its length has fallen from $1\,200$ to $1\,199$. A gain in Q would have made it $1\,201$ nucleotides long.
• C. In P one nucleotide has been replaced by a different one; in Q one nucleotide has been added. · Direction of Q's length change reversed
  P is read correctly. Q's $1\,199$ nucleotides are one fewer than $1\,200$, not one more, so a nucleotide has gone rather than arrived. The phrase "one place further along" means that each of the wild type's later nucleotides now sits one place earlier, closing the gap left behind.
• E. In P one nucleotide has been replaced by a different one; in Q every nucleotide beyond position $640$ has been replaced. · A shift mistaken for wholesale replacement
  It is true that Q's bases from position $641$ on mostly fail to match the wild-type bases at the same positions, which is what tempts this answer, but replacements, however many of them, leave the number of nucleotides untouched, so Q would still be $1\,200$ nucleotides long. Only a loss accounts for the measured $1\,199$.
• F. In P the sequence is unchanged; in Q one nucleotide has been lost. · A mutation assumed to require a change of length
  Q is read correctly. A different base at position $640$ is itself a change to the sequence of nucleotides, which is what a mutation is: the sequence is defined by which nucleotide sits at each position, not only by how many there are.
• G. In P one nucleotide has been added; in Q every nucleotide beyond position $640$ has been replaced. · Both lengths ignored
  This combines an addition in P, which would give $1\,201$ nucleotides, with a run of replacements in Q, which would leave Q at $1\,200$. The lengths given in the question, $1\,200$ and $1\,199$, rule out both halves.
• H. In P one nucleotide has been replaced by a different one; in Q the sequence is unchanged. · Comparison stopped at the first agreement
  P is read correctly. Q's first $640$ nucleotides do match the wild type's, but the comparison has to run to the end of the gene: Q is one nucleotide shorter and its nucleotides from $641$ on are displaced, so its sequence has changed.

Common Mistake (⚠️):
Comparing the two sequences position by position and counting every position that fails to match. From position $641$ on, none of Q's nucleotides sits opposite the wild-type nucleotide it matches, which looks like hundreds of separate changes. It is one nucleotide missing and the rest of the sequence closing up behind it, which is why the gene is one nucleotide shorter and not the same length.

Takeaway (📌):
Count the nucleotides before you read them. The same length with one position different is a replacement; one nucleotide shorter, with everything past the break shifted a place forward, is a loss.

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 (📌):
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: 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 (📌):
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. 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 (📌):
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. The gland releases a chemical into the blood, which the blood carries to the organ, where it acts.

Fastest Approach (🚀):
Read the two controls before the biology. Cut nerves removes every option that needs a nerve, and an organ that is silent until the gland is stimulated removes every option that does not begin at the gland.

Step-by-Step Breakdown:

1. Rule out a nervous route.

Every nerve to the gland and to the organ has been cut, so no impulse can pass between them along a nerve. The organ is also inactive while the gland is left alone, even though blood is flowing through the tube the whole time, so whatever produces the response appears only when the gland is stimulated.

2. Show that the blood is the route.

Clamping the tube removes the one remaining connection between the two chambers, and the same stimulation then produces nothing at all. The signal must therefore be something the gland puts into the blood, and it must reach the organ by being carried there in the blood. The delay of about twenty seconds fits transport by flowing blood rather than conduction along a fibre.

3. Name what the substance is.

A substance made in a gland, released into the blood, and carried in the blood to another organ where it produces an effect is a hormone, and a gland that works this way is an endocrine gland. The two results together support that description of the gland and its secretion.

Matches Option B.

Why the Other Options Are Wrong (❌):

• A. Impulses travel from the gland to the organ along nerve fibres running inside the connecting tube. · ignores a stated control
  Assumes that a signal passing from one structure to another must be a nerve impulse, although a delay of about twenty seconds is far longer than conduction along a fibre would take. Every nerve to both structures has been cut, and the tube carries blood, so there is no fibre along which an impulse could pass from one chamber to the other.
• C. The organ releases a chemical that passes back down the tube and switches the gland on. · reads the pathway backwards
  Reverses the direction of the effect. The tube carries blood from the gland to the organ, and it is the gland that is stimulated, so a substance travelling from organ to gland cannot be what produces the organ's response.
• D. The organ responds only to a fresh supply of oxygen, and the gland plays no part in the response. · attributes the result to the blood supply
  Explains the clamp result by loss of oxygen and drops the gland from the account. The organ is already receiving flowing blood while the gland is unstimulated, and does nothing, so oxygen supply on its own does not produce the response.
• E. The gland secretes into a duct that empties its contents onto the surface of the organ. · confuses endocrine with exocrine secretion
  Describes a gland that pours its secretion down a duct onto a surface. The only connection between the chambers here carries blood, and it is closing that blood route which abolishes the response.
• F. The stimulus travels along the tube as a pressure wave in the blood, and the organ responds to it. · treats the signal as mechanical rather than chemical
  Takes the clamp result as evidence that any disturbance carried by the tube would serve. A pressure change in a fluid-filled tube reaches the far end almost at once, whereas the organ responds about twenty seconds after the gland is stimulated, which is the time a substance takes to be carried there in flowing blood.

Common Mistake (⚠️):
Concluding from the clamp alone that the blood is responsible. The clamp shows only that the connection matters; it is the silent organ before stimulation that shows the gland has to be made active first, and both results are needed before the secretion can be pinned on the gland.

Takeaway (📌):
An endocrine gland signals by putting a chemical into the blood, so the blood supply between gland and target is the whole pathway. Cut the blood link and the message stops, however healthy both structures are.

Correct Answer: G. heart rate rises, glucose rises, muscle flow rises, gut flow falls

Fastest Approach (🚀):
Adrenaline raises heart rate and raises blood glucose, so strike every row whose first two entries are not both a rise. Of the rows left, take the only one that moves blood towards the muscles and away from the gut.

Step-by-Step Breakdown:

1. Identify the hormone and what it is preparing for.

An unexpected alarm is a sudden threat, and the sharp rise in blood adrenaline is the adrenal glands responding to it. Adrenaline prepares the body for vigorous action, which means getting more oxygen and more fuel to the muscles that would carry that action out, and getting them there before the action begins.

2. Take the heart and the fuel.

Heart rate rises, because a faster heart moves more blood each minute and so delivers more oxygen and glucose to the tissues. Blood glucose concentration rises, because adrenaline makes the liver break stored glycogen down into glucose and release it. Neither change can be put down to movement, since she stays seated, nor to another hormone, since none of the others changed.

3. Take the two blood flows.

Preparing the muscles means sharing the blood out differently, not raising every flow alike. Adrenaline widens the vessels supplying skeletal muscle and narrows those supplying the gut, so flow through the leg muscles rises while flow through the gut wall falls. The response is made in advance of the action, so the muscle vessels widen even though she has not yet moved.

The four directions are therefore rise, rise, rise, fall.

Matches Option G.

Why the Other Options Are Wrong (❌):

• A. heart rate rises, glucose rises, muscle flow rises, gut flow rises · misses the redistribution
  Gets the heart, the glucose and the muscle right, then treats adrenaline as raising every blood flow at once. The extra flow to the muscles is obtained by actively narrowing the vessels supplying the gut, so gut flow falls rather than rising.
• B. heart rate rises, glucose rises, muscle flow falls, gut flow rises · reverses the redistribution
  Sends more blood to the gut and less to the muscles, which is the pattern of rest and digestion. Adrenaline does the opposite: digestion is postponed so that the muscles can be supplied.
• C. heart rate rises, glucose rises, muscle flow falls, gut flow falls · treats adrenaline as narrowing every vessel
  Takes adrenaline to be a narrowing agent everywhere, so both flows fall. It does narrow the vessels of the gut and the skin, but it widens those supplying skeletal muscle, and a whole body narrowing would starve the very muscles the response is preparing.
• D. heart rate rises, glucose rises, muscle flow unchanged, gut flow falls · waits for the movement before the blood shifts
  Holds the muscle flow steady on the grounds that she never stands up, so nothing is demanded of her legs. The point of the response is that it comes first: the vessels widen on the alarm so that the muscles are already supplied if the action follows.
• E. heart rate rises, glucose falls, muscle flow rises, gut flow falls · confuses adrenaline with the hormone that clears glucose
  Takes blood glucose to fall, which is the action of the hormone that moves glucose out of the blood into store. Adrenaline works the other way, releasing glucose from liver glycogen, so glucose rises.
• F. heart rate rises, glucose unchanged, muscle flow rises, gut flow falls · takes adrenaline to act on the circulation only
  Treats adrenaline as a hormone of the heart and vessels alone, leaving the fuel supply untouched while she sits still. It also acts on the liver, converting stored glycogen to glucose, so the concentration in the blood rises before any movement calls on it.
• H. heart rate falls, glucose rises, muscle flow rises, gut flow falls · takes adrenaline to slow the heart
  Attributes a falling heart rate to adrenaline, which is the resting state rather than the response to a fright. The other three directions are right, so only the first entry is wrong.

Common Mistake (⚠️):
Reasoning that because she never moves, the blood flow to her leg muscles has no reason to change. The adrenaline response is anticipatory: it prepares the muscles for action that may never happen, which is why the flows shift on the sound of the alarm rather than on the first step taken.

Takeaway (📌):
Adrenaline both raises supply and redirects it. Heart rate and blood glucose go up together, and blood is diverted towards the muscles and away from organs like the gut, so muscle flow rises while gut flow falls.

Correct Answer: E. $68\,000$ plants

Fastest Approach (🚀):
Get the two densities in your head first, $100 \div 20 = 5$ and $20 \div 10 = 2$, and split $16\,000\ \text{m}^{2}$ into $12\,000$ and $4\,000$; the whole question is then $5 \times 12 + 2 \times 4$ in thousands.

Step-by-Step Breakdown:

1. Work out how much ground each set of quadrats searched, and the density it gives.

On the dry ground each quadrat is a square of side $2\ \text{m}$, so it covers $4\ \text{m}^{2}$ and the five together searched $20\ \text{m}^{2}$. They held $100$ plants, so the dry ground carries $100 \div 20 = 5\ \text{plants m}^{-2}$. On the bog each quadrat covers $1\ \text{m}^{2}$ and the ten together searched $10\ \text{m}^{2}$, holding $20$ plants, so the bog carries $20 \div 10 = 2\ \text{plants m}^{-2}$. A density has to come from the ground covered, not from the number of quadrats.

2. Put the two zones into square metres.

The heath covers $1.6 \times 10\,000 = 16\,000\ \text{m}^{2}$. Three quarters of that is dry sandy ground, $12\,000\ \text{m}^{2}$, and the remaining quarter is bog, $4\,000\ \text{m}^{2}$.

3. Scale each zone on its own density and add.

Dry ground: $12\,000 \times 5 = 60\,000$ plants. Bog: $4\,000 \times 2 = 8\,000$ plants. Heath: $60\,000 + 8\,000 = 68\,000$ plants. Across the whole $16\,000\ \text{m}^{2}$ that is $68\,000 \div 16\,000 = 4.25\ \text{plants m}^{-2}$.

Sanity check: $4.25\ \text{plants m}^{-2}$ for the heath as a whole falls between the $2\ \text{plants m}^{-2}$ measured on the bog and the $5\ \text{plants m}^{-2}$ measured on the dry ground, and nearer the dry figure, which covers three quarters of the ground.

Matches Option E.

Why the Other Options Are Wrong (❌):

• A. $44\,000$ plants · Densities attached to the wrong zones
  Puts the bog's $2\ \text{plants m}^{-2}$ on the dry ground and the dry ground's $5\ \text{plants m}^{-2}$ on the bog: $12\,000 \times 2 = 24\,000$ and $4\,000 \times 5 = 20\,000$, giving $44\,000$. The larger quadrats, and the count of $100$, were on the dry ground.
• B. $56\,000$ plants · Densities averaged rather than weighted by area
  Takes the mean of the two densities, $(5 + 2) \div 2 = 3.5\ \text{plants m}^{-2}$, and spreads it over the heath: $16\,000 \times 3.5 = 56\,000$. A plain mean would be right only if the two zones covered equal areas, and the dry ground covers three times the bog.
• C. $60\,000$ plants · Second zone left out
  Stops after the dry ground: $12\,000 \times 5 = 60\,000$. The bog is part of the heath and holds bell heather at $2\ \text{plants m}^{-2}$ over $4\,000\ \text{m}^{2}$, which is a further $8\,000$ plants.
• D. $64\,000$ plants · Samples pooled instead of each zone scaled on its own
  Adds the counts and the searched areas before dividing: $120 \div 30 = 4\ \text{plants m}^{-2}$, then $16\,000 \times 4 = 64\,000$. That weights the zones by the ground searched in each rather than by the ground each occupies on the heath.
• F. $248\,000$ plants · Divided by the number of quadrats rather than the ground they covered
  Treats every quadrat as one unit of area: $100 \div 5 = 20$ and $20 \div 10 = 2$, then $12\,000 \times 20 = 240\,000$ plus $4\,000 \times 2 = 8\,000$, giving $248\,000$. Each dry quadrat covered $4\ \text{m}^{2}$, so the five covered $20\ \text{m}^{2}$, not $5$.

Common Mistake (⚠️):
Pooling the two samples as though they were one: $120$ plants in $30\ \text{m}^{2}$ of searched ground gives $4\ \text{plants m}^{-2}$ and $64\,000$ for the heath. A pooled figure weights the two zones by how much ground was searched in each, which was two thirds dry, instead of by how much of the heath each covers, which is three quarters dry.

Takeaway (📌):
Density before area, and one zone at a time. Divide each count by the ground its own quadrats covered, multiply each density by the area of its own zone, then add.