ESAT Paper 3 Biology Sample Questions

Correct Answer: C. 120

Step-by-Step Breakdown:

1. Find the ATP the aerobic route supplied

Oxidising $200$ mol of glucose at $30$ mol of ATP per mole gives

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

2. Take that share off the demand

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

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

3. Convert the fermentation ATP into glucose

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

${\displaystyle \frac{240}{2}}=120\text{mol}$

Sanity check: the two routes account for $6000+240=6240$ 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: $6240÷2=3120$. That treats every mole of ATP as though fermentation had made it, but $200$ mol of glucose was oxidised aerobically and supplied $6000$ 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: $6000÷2=3000$. Those $6000$ mol are already paid for by the $200$ mol of glucose named in the stem; the ATP still waiting to be accounted for is $6240-6000=240$ mol.
• D. 60 · Gross glycolytic yield used
  Divides the fermentation ATP by the gross yield rather than the net one: $240÷4=60$. 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 $200$ mol is the glucose oxidised aerobically, which is what produced $6000$ of the $6240$ mol of ATP; the fermented glucose is a separate quantity and has to be recovered from the $240$ mol that the aerobic route did not supply.

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

${\displaystyle \frac{60}{6}}=10$ 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

$600{\text{cm}}^3={\displaystyle \frac{600}{1000}}=0.6{\mathrm{d}\mathrm{m}}^3$ drawn in per inspiration.

3. Combine volume per breath with breaths per minute

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

Sanity check: a whole minute holds $10$ inspirations, so the answer must be larger than the $0.6{\mathrm{d}\mathrm{m}}^3$ that one inspiration draws in, and it is. Leaving the volume in ${\mathrm{c}\mathrm{m}}^3$ 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 ${\mathrm{c}\mathrm{m}}^3$ and $1000{\mathrm{c}\mathrm{m}}^3=1{\mathrm{d}\mathrm{m}}^3$, 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: $600÷1000=0.6$. 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 $60÷6=10$ 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 $60÷6=10$ 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: $1m^3$ is ${10}^6{\text{cm}}^3$, but the answer is wanted in ${\mathrm{d}\mathrm{m}}^3$ and $1{\mathrm{d}\mathrm{m}}^3$ is only $1000{\text{cm}}^3$. So $600÷1000=0.6{\mathrm{d}\mathrm{m}}^3$ 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 $0.6{\mathrm{d}\mathrm{m}}^3$, because $1{\mathrm{d}\mathrm{m}}^3=1000{\mathrm{c}\mathrm{m}}^3$, 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: E. $0.215\mu m^{-1}$

Step-by-Step Breakdown:

1. Add up the membrane the cell carries

The cell has no wall, so every part of the cuboid's surface is membrane. The six faces come in three pairs. The base and the top measure $16\times 40=640\mu m^2$ each, two of the side faces measure $16\times 50=800\mu m^2$ each, and the other two measure $40\times 50=2000\mu m^2$ each. Together:

$2\times (640+800+2000)=6880\mu m^2$

Leaving any pair out makes this area too small, and the base and the top are as much membrane as the faces a drawing of a box shows.

2. Work out the volume that membrane encloses

The cytoplasm fills the box, so the volume is $16\times 40\times 50=32000\mu m^3$.

3. Divide the area by the volume

$6880÷32000$ gives $0.215\mu m^{-1}$.

The cancelled form gets there in one line: ${\displaystyle \frac{2(ab+ac+bc)}{abc}}$ is ${\displaystyle \frac{2}{a}}+{\displaystyle \frac{2}{b}}+{\displaystyle \frac{2}{c}}$, which here is ${\displaystyle \frac{2}{16}}+{\displaystyle \frac{2}{40}}+{\displaystyle \frac{2}{50}}$.

Sanity check: an area divided by a volume leaves one over a length, and the answer is quoted in $\mu m^{-1}$.

The key is $0.215\mu m^{-1}$.

Why the Other Options Are Wrong (❌):

• A. $0.175\mu m^{-1}$ · Base and top left out
  Counts only the four faces around the side of the box, $2\times 16\times 50+2\times 40\times 50=5600\mu m^2$, and divides that by $32000\mu m^3$ to get $0.175\mu m^{-1}$. The base and the top are membrane as well, and they add $1280\mu m^2$ between them.
• B. $0.04\mu m^{-1}$ · Only the base and the top counted
  Takes the membrane as the base and the top alone, $2\times 16\times 40=1280\mu m^2$, which gives $0.04\mu m^{-1}$. The four faces around the side carry $5600\mu m^2$ of membrane and belong in the total too.
• C. $0.43\mu m^{-1}$ · Each face counted twice
  Treats the inner and the outer surface of the membrane as two separate areas, doubling the total to $13760\mu m^2$ and giving $0.43\mu m^{-1}$. One face carries one area of membrane, whichever side of it is looked at.
• D. $0.375\mu m^{-1}$ · Cube shortcut applied to the first edge
  Uses the cube result $6s^2/s^3=6/s$ on the edge quoted first: $6÷16$ gives $0.375\mu m^{-1}$. This cell is not a cube. Its edges are $16\mu m$, $40\mu m$ and $50\mu m$, and only when all three are equal does one edge fix the ratio on its own.
• F. $0.12\mu m^{-1}$ · Six faces the size of the base
  Works out the base, $16\times 40=640\mu m^2$, and multiplies it by six as though every face matched it: $6\times 640=3840\mu m^2$, and dividing by $32000\mu m^3$ gives $0.12\mu m^{-1}$. Only two of the six faces are the size of the base; the other four measure $16$ by $50$ and $40$ by $50$.
• G. $4.651\mu m^{-1}$ · Ratio inverted, volume per unit area
  Divides the volume by the membrane area, $32000÷6880$, taken to three decimal places where it runs on. That is volume per unit area of membrane: the reciprocal of what was asked, and a length in $\mu m$ rather than one over a length. Area per unit volume puts the area on top, $6880\mu m^2$ over $32000\mu m^3$, which gives $0.215\mu m^{-1}$.

Common Mistake (⚠️):
Counting only the four faces that a drawing of a box shows around its side and forgetting the base and the top. Those two faces carry $1280\mu m^2$ of membrane between them, so leaving them out gives $0.175\mu m^{-1}$ in place of $0.215\mu m^{-1}$, and it describes a cell whose cytoplasm would be open to the surroundings above and below.

Takeaway (📌):
For any cuboid, surface area over volume cancels to ${\displaystyle \frac{2}{a}}+{\displaystyle \frac{2}{b}}+{\displaystyle \frac{2}{c}}$: take the reciprocal of each edge, add them, double. Neither the area nor the volume ever has to be written down, and the cube result $6/s$ is only the special case where the three edges are equal.

Correct Answer: A. $31.25\%$

Step-by-Step Breakdown:

1. Total the DNA held in the plasmids

Each plasmid is a separate ring of $5000$ base pairs and the cell carries 125 of them. The copies are identical, but they are physically distinct rings, so between them they hold

$$125\times 5000$$$$=625000\text{base pairs}.$$

Using the length of one ring here is the commonest slip, because it throws away the copy number the stem takes the trouble to give.

2. Compare that total with the chromosome, then convert to a percentage

The chromosome holds $2\times {10}^6$ base pairs. Both figures are counted in base pairs, so they divide directly, and multiplying by $100$ turns the fraction into a percentage:

$${\displaystyle \frac{625000}{2\times {10}^6}}\times 100\%$$

which comes to $31.25\%$.

Sanity check: both quantities are given in the same unit, base pairs on each side of the division, so no conversion belongs anywhere in the working. Slipping one side alone into individual bases is what puts an answer out by a factor of two, in one direction or the other.

The key is $31.25\%$.

Why the Other Options Are Wrong (❌):

• B. $0.3125\%$ · Fraction never converted to a percentage
  Divides correctly and then stops one step early. $625000÷(2\times {10}^6)$ is the FRACTION of the chromosome that the plasmid DNA matches, and it is written down as $0.3125\%$ without being multiplied by $100$ first. A fraction and a percentage of the same quantity differ by exactly that factor of $100$.
• C. $0.25\%$ · Uses one plasmid and drops the copy number
  Takes the $5000$ base pairs of a single ring as the whole plasmid contribution: $5000÷(2\times {10}^6)$, which is $0.25\%$. That is the share one plasmid holds, and this cell carries 125 of them, each a separate ring of DNA.
• D. $15.625\%$ · Chromosome counted in bases, plasmids left in base pairs
  Turns the chromosome's base pairs into individual bases, doubling it to $4\times {10}^6$, while leaving the plasmid total in base pairs. Dividing $625000$ by that doubled figure gives $15.625\%$. Both sides of a ratio have to be counted in the same unit, and the stem already gives both in base pairs.
• E. $62.5\%$ · Plasmid total counted in bases, chromosome left in base pairs
  The same unit slip made on the other side of the division: the plasmid total is doubled to $1250000$ individual bases while the chromosome stays at $2\times {10}^6$ base pairs, which gives $62.5\%$. Both figures are already in the unit the stem states them in, so neither needs converting.
• F. $3.2\%$ · Divides the wrong way round
  Divides the chromosome by the plasmid total rather than the other way about, $(2\times {10}^6)÷625000$, and reads the quotient as a percentage, $3.2\%$. That quotient is really the number of times the chromosome exceeds the DNA held in the plasmids.

Common Mistake (⚠️):
Comparing one plasmid with the chromosome and ignoring the copy number. Copy number is the whole point of a plasmid as a piece of cell biology: 125 identical rings hold 125 times the DNA of one, and here that factor is the whole difference between $0.25\%$ and $31.25\%$.

Takeaway (📌):
Plasmid DNA per cell is the length of one ring multiplied by the number of rings, never the length of one ring on its own. Multiply first, divide second, and multiply by $100$ last.