Human Body Systems

QUESTION 1 2 marks Criterion A

Easy

The diagram shows the human digestive system. Two accessory organs, which release secretions into the small intestine through ducts (dashed lines), are labelled P and Q.

a. Name organ P.

[1]

b. Name organ Q.

[1]

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(a)

P is the liver. It produces bile, which is stored in the gall bladder before being released into the small intestine.

(b)

Q is the pancreas. It produces digestive enzymes (amylase, protease and lipase) which are released into the small intestine.

QUESTION 2 2 marks Criterion A

Easy

Saliva contains the enzyme amylase.

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Amylase is a carbohydrase enzyme that begins the chemical digestion of starch, breaking it down into simpler sugars (maltose). It is released into the mouth from the salivary glands and mixed with food as it is chewed, starting chemical digestion before the food is even swallowed.

QUESTION 3 2 marks Criterion A

Easy

The stomach produces hydrochloric acid.

a. State the approximate pH of the stomach.

[1]

b. State one function of this acid.

[1]

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(a)

The stomach is strongly acidic, with a pH of about 2.

(b)

It kills harmful bacteria that have been swallowed with food, and it provides the low pH needed to activate the protein-digesting enzyme pepsin.

QUESTION 4 4 marks Criterion A

Medium

The table shows information about three digestive enzymes. The row for amylase has been completed as an example.

Enzyme	Produced by	Substrate	Product(s)
Amylase	salivary glands / pancreas	starch	sugar (maltose)
Protease	stomach / pancreas	?	?
Lipase	pancreas	?	?

a. State the substrate and product(s) for protease.

[2]

b. State the substrate and product(s) for lipase.

[2]

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(a)

Substrate: protein. Product(s): amino acids.

(b)

Substrate: fat (lipid). Product(s): fatty acids and glycerol.

QUESTION 5 5 marks Criterion A

Medium

The diagram shows a cross-section through one villus in the wall of the small intestine.

a. Name the structures labelled A, B and C.

[3]

b. Explain how structure D helps the small intestine absorb digested food efficiently.

[2]

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(a)

A = microvilli (tiny folds on the surface of each villus cell).

B = outer wall / cell membrane, only one cell thick.

C = lacteal, a small vessel that absorbs digested fats.

(b)

D is the capillary network surrounding the villus. It carries digested glucose and amino acids away into the bloodstream as fast as they are absorbed, keeping the concentration of these nutrients low inside the villus. This maintains a steep concentration gradient between the gut contents and the blood, so diffusion (and active transport) of nutrients continues quickly.

QUESTION 6 4 marks Criterion A

Medium

One slice of bread contains approximately $25\,\text{g}$ of carbohydrate, $5\,\text{g}$ of protein and $2\,\text{g}$ of fat. Use the energy values: carbohydrate $= 4\,\text{kcal/g}$, protein $= 4\,\text{kcal/g}$, fat $= 9\,\text{kcal/g}$.

a. Calculate the energy contributed by each of the three nutrients.

[2]

b. Calculate the total energy content of one slice of bread.

[2]

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(a)

Carbohydrate: $25 \times 4 = 100\,\text{kcal}$

Protein: $5 \times 4 = 20\,\text{kcal}$

Fat: $2 \times 9 = 18\,\text{kcal}$

(b)

$$ 100 + 20 + 18 = 138 $$

Answer: $138\,\text{kcal}$

QUESTION 7 5 marks Criterion A

Hard

Without villi and microvilli, the internal surface of the small intestine would have an absorbing surface area of only about $0.4\,\text{m}^2$. Villi and microvilli increase this surface area by a factor of approximately $30$.

a. Calculate the approximate absorbing surface area of the small intestine, including villi and microvilli.

[2]

b. Explain why such a large surface area, together with a good blood supply, allows the small intestine to absorb digested food efficiently.

[3]

Show complete worked solution

(a)

$$ 0.4 \times 30 = 12 $$

Answer: approximately $12\,\text{m}^2$.

(b)

A larger surface area means many more villi cells are in contact with digested food at once, so many more molecules of glucose, amino acids, fatty acids and glycerol can diffuse (or be actively transported) into the blood at the same time, increasing the overall rate of absorption. Combined with a good, dense blood supply constantly carrying absorbed nutrients away, a steep concentration gradient is maintained between the gut contents and the blood, which keeps diffusion happening as quickly as possible.

QUESTION 8 7 marks Criterion B

Medium

You want to investigate how temperature affects the rate at which amylase digests starch. You will use iodine solution (which turns blue-black in the presence of starch, and stays orange-brown once all the starch has been digested) to test samples over time.

a. State the independent and dependent variables.

[2]

b. State two variables you would need to control, and explain why for one of them.

[2]

c. Describe a method, including the equipment you would use, to collect the data needed.

[3]

Show complete worked solution

(a)

Independent variable: temperature of the starch–amylase mixture (e.g. $10^\circ\text{C}, 20^\circ\text{C}, 30^\circ\text{C}, 40^\circ\text{C}, 50^\circ\text{C}$, controlled with a water bath). Dependent variable: time taken for the iodine test to show a negative result (all starch digested).

(b)

Control the concentration and volume of both the starch solution and the amylase solution used each time, and the pH of the mixture. Why control concentration/volume: a more concentrated starch solution, or more amylase, would change the reaction rate for reasons unrelated to temperature, making the comparison between temperatures unfair.

(c)

Place equal volumes of starch solution and amylase solution in separate test tubes in a water bath set to the first test temperature, and leave for a few minutes so both reach that temperature.
Mix the two solutions in a spotting tile well and immediately start a stopwatch.
Every $30\,\text{s}$, use a dropper to place one drop of the mixture into a fresh well of iodine solution on the spotting tile.
Record the time at which the iodine solution first stays orange-brown (no more blue-black colour) — this means all the starch has been digested.
Repeat at each of the other temperatures, keeping all other variables the same, and repeat each temperature at least twice to calculate a mean time.

QUESTION 9 6 marks Criterion B

Medium

A student's hypothesis is: "Protease digests protein fastest at pH 8, similar to conditions in the small intestine." The student has a cloudy suspension of egg white (albumin) protein, which turns clear as the protein is digested.

a. Identify the independent, dependent, and one controlled variable for an investigation to test this.

[3]

b. Explain how the data collected (clearing time at each pH) could be used to test the hypothesis.

[3]

Show complete worked solution

(a)

Independent: pH of the protease mixture (e.g. pH $2, 4, 6, 8, 10$, set using buffer solutions). Dependent: time taken for the cloudy egg-white suspension to turn clear. Controlled: temperature (kept constant in a water bath) — a different temperature would change the reaction rate regardless of pH, making the test unfair.

(b)

For each pH tested, the time for the suspension to turn clear would be recorded (or repeated and averaged). A shorter clearing time means a faster reaction rate. If the hypothesis is correct, the shortest clearing time (fastest rate) should occur at pH $8$, with longer times at the more acidic and more alkaline pH values tested either side of it — plotting clearing time (or $1/\text{time}$, the rate) against pH would show a clear peak at pH $8$ if the hypothesis is supported.

QUESTION 10 8 marks Criterion B

Hard

A student is testing how quickly amylase digests starch by removing a drop of the reaction mixture every $30\,\text{s}$ and judging, by eye, whether it still turns blue-black with iodine. Repeating the same experiment gave very different "digestion complete" times each time.

a. Identify the main source of error in this method that would explain the differences between repeats.

[2]

b. Suggest an improvement to the method that would reduce this error, and explain how it works.

[3]

c. Explain how the improved method would produce more reliable results.

[3]

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(a)

The main source of error is judging the colour change by eye — deciding exactly when the mixture has "stopped turning blue-black" is subjective, and the $30\,\text{s}$ sampling interval means the true end-point could fall anywhere within that $30\,\text{s}$ window.

(b)

Use a colorimeter connected to a data logger, sampling the mixture's colour/light absorbance automatically at short, regular intervals (e.g. every $2$–$3\,\text{s}$). The colorimeter measures the amount of light transmitted through the sample objectively, removing human judgement of colour entirely, and the shorter sampling interval pinpoints the end-point far more precisely than sampling every $30\,\text{s}$ by eye.

(c)

Because the colorimeter removes subjective human judgement and records data far more frequently, repeated trials at the same temperature should give end-point times that are much closer together, rather than being scattered by differences in how each reading was judged. This means the reaction rate calculated from the data would be a far more accurate and repeatable estimate of amylase's true activity at that temperature.

QUESTION 11 3 marks Criterion C

Easy

A food sample was tested for the four main nutrients. The results are shown in the table.

Test	Result
Iodine test	stays orange-brown
Benedict's test (heated)	turns brick-red
Biuret test	stays blue
Ethanol emulsion test	cloudy white layer forms

a. State which nutrients are present in the sample, using the results in the table.

[2]

b. State which nutrients are absent, and explain how you know.

[1]

Show complete worked solution

(a)

Reducing sugar is present (Benedict's test turned brick-red) and fat is present (ethanol emulsion test formed a cloudy white layer).

(b)

Starch is absent (iodine stayed orange-brown, not blue-black) and protein is absent (Biuret test stayed blue instead of turning purple/lilac).

QUESTION 12 4 marks Criterion C

Medium

A student measured the mass of glucose (mg) produced as amylase digested starch, at $2$-minute intervals.

Time (min)	0	2	4	6	8	10
Glucose produced (mg)	0	8	16	32	32	40

a. Identify the anomalous reading in this table.

[1]

b. Explain how you identified it.

[2]

c. State what you should do with this reading before analysing the trend.

[1]

Show complete worked solution

(a)

The reading at $t = 6\,\text{min}$ ($32\,\text{mg}$) is anomalous.

(b)

Every other reading increases by exactly $8\,\text{mg}$ each interval ($0, 8, 16, \ldots, 32, 40$). If the pattern had continued, the reading at $t=6\,\text{min}$ should have been $24\,\text{mg}$, not $32\,\text{mg}$ — it breaks the otherwise-consistent trend (and is identical to the following reading, which should be higher, not equal).

(c)

Exclude (discount) it from the analysis, and, if possible, repeat that measurement to check whether $32\,\text{mg}$ was a genuine result or an error, before drawing a line of best fit through the remaining consistent data.

QUESTION 13 5 marks Criterion C

Medium

The graph shows how the rate of starch digestion by amylase changes with pH.

a. Describe the trend shown by the graph.

[2]

b. State the optimum pH for amylase shown by this graph.

[1]

c. Explain, in terms of enzyme structure, why the rate falls at pH values far from the optimum.

[2]

Show complete worked solution

(a)

The rate of reaction increases as pH rises from $2$ up to a peak at pH $7$, then decreases as pH continues to rise from $7$ to $10$. The rate is low at both very acidic and very alkaline pH values, and highest at pH $7$.

(b)

The optimum pH is pH 7 (neutral), matching the roughly neutral conditions of saliva in the mouth.

(c)

At pH values far from the optimum, the shape of the enzyme's active site is changed (the enzyme becomes denatured) because the pH affects the bonds holding the enzyme's structure together. The starch substrate molecule no longer fits into the distorted active site, so far fewer successful enzyme–substrate collisions occur, and the reaction rate drops.

QUESTION 14 4 marks Criterion C

Medium

A student repeated the amylase digestion experiment at $35^\circ\text{C}$ four times, recording the time for the iodine test to become negative.

Trial	1	2	3	4
Time (s)	45	48	44	47

a. Calculate the mean digestion time.

[2]

b. Calculate the rate of reaction, in $\text{s}^{-1}$, using the mean time (rate $= 1/\text{time}$).

[2]

Show complete worked solution

(a)

$$ \text{mean} = \frac{45+48+44+47}{4} = \frac{184}{4} = 46\,\text{s} $$

(b)

$$ \text{rate} = \frac{1}{46} = 0.0217\,\text{s}^{-1} \ (\text{3 s.f.}) $$

QUESTION 15 6 marks Criterion C

Medium

A study measured average gut transit time (the time food takes to pass through the digestive system) for people eating different amounts of dietary fibre per day.

Fibre intake (g/day)	5	15	25	35	45
Mean transit time (hours)	72	54	38	30	28

a. Describe the relationship shown by the data.

[2]

b. Using the pattern in the data, estimate the transit time for someone eating $20\,\text{g}$ of fibre per day.

[2]

c. Using the data, explain why doctors often recommend a high-fibre diet.

[2]

Show complete worked solution

(a)

As fibre intake increases, mean transit time decreases — a negative correlation. The decrease is steep at first (between $5$ and $25\,\text{g/day}$) but levels off at higher fibre intakes (only a small further decrease between $35$ and $45\,\text{g/day}$).

(b)

$20\,\text{g/day}$ lies halfway between $15\,\text{g/day}$ ($54\,\text{h}$) and $25\,\text{g/day}$ ($38\,\text{h}$), so a reasonable estimate is approximately $\frac{54+38}{2} \approx 46\,\text{hours}$.

(c)

The data show that higher fibre intake is clearly associated with food moving through the gut faster (shorter transit time). Fibre is not digested, so it adds bulk and helps push other gut contents along; a shorter transit time means waste (and any harmful substances within it) spends less time in contact with the gut wall, which is linked to a lower risk of constipation and bowel disease.

QUESTION 16 7 marks Criterion C

Hard

A student's hypothesis is: "Amylase digests starch faster as temperature increases up to $37^\circ\text{C}$." The student measured digestion time (time for the iodine test to become negative) in triplicate at three temperatures.

Temperature (°C)	Trial 1 (s)	Trial 2 (s)	Trial 3 (s)
20	140	146	138
37	42	45	40
50	210	205	215

a. Calculate the mean digestion time at each temperature.

[3]

b. Evaluate whether these results support the student's hypothesis. Explain what happens beyond $37^\circ\text{C}$.

[4]

Show complete worked solution

(a)

$20^\circ\text{C}$: $\dfrac{140+146+138}{3} = \dfrac{424}{3} = 141\,\text{s}$ (3 s.f.)

$37^\circ\text{C}$: $\dfrac{42+45+40}{3} = \dfrac{127}{3} = 42.3\,\text{s}$ (3 s.f.)

$50^\circ\text{C}$: $\dfrac{210+205+215}{3} = \dfrac{630}{3} = 210\,\text{s}$

(b)

Between $20^\circ\text{C}$ and $37^\circ\text{C}$, the mean digestion time falls sharply (from $141\,\text{s}$ to $42.3\,\text{s}$), meaning the rate of reaction increases — this supports the hypothesis for this range. However, the hypothesis only claims faster digestion "up to $37^\circ\text{C}$", and the data at $50^\circ\text{C}$ (mean $210\,\text{s}$, much slower than at either lower temperature) is consistent with this, not a contradiction: above the optimum temperature the amylase enzyme begins to denature — its active site changes shape so it can no longer bind starch efficiently — so the rate falls sharply again rather than continuing to rise.

QUESTION 17 6 marks Criterion C

Hard

A student judged, by eye, when an iodine test first stopped turning blue-black, repeating the timing five times under the same conditions. The stopwatch used could be read to $\pm0.1\,\text{s}$.

Trial	1	2	3	4	5
Time (s)	58	63	55	67	60

a. Calculate the mean time.

[2]

b. Calculate the range of the results.

[1]

c. The stopwatch itself is precise to $\pm0.1\,\text{s}$, but the readings vary far more than this. Explain what this tells you about the main source of uncertainty in this experiment.

[3]

Show complete worked solution

(a)

$$ \text{mean} = \frac{58+63+55+67+60}{5} = \frac{303}{5} = 60.6\,\text{s} $$

(b)

$$ \text{range} = 67 - 55 = 12\,\text{s} $$

(c)

Since the readings vary by up to $12\,\text{s}$ — far more than the stopwatch's own $\pm0.1\,\text{s}$ precision — the main source of uncertainty is not the stopwatch, but the subjective judgement of exactly when the colour change happened (a human/reaction-time error in deciding the end-point and starting/stopping the watch). This is why repeating trials and averaging is essential when an end-point must be judged by eye.

QUESTION 18 6 marks Criterion D

Medium

A health survey compared two groups following different diets over one year.

	Low-fibre (mostly processed food)	High-fibre (wholegrains, fruit, vegetables)
Reporting regular constipation	35%	8%
Relative weekly food cost	100 (index)	130 (index)

Discuss one benefit and one drawback of switching to a high-fibre diet, using the data to support your discussion.

Show complete worked solution

Benefit: The data show a large difference in reported constipation — only $8\%$ of the high-fibre group compared with $35\%$ of the low-fibre group, more than a four-fold reduction. Fibre adds bulk to the gut contents and helps them move through the large intestine more easily, so a high-fibre diet clearly supports healthier, more regular digestion and is linked in general to a lower long-term risk of bowel disease.

Drawback: The weekly food cost index for the high-fibre diet is $30\%$ higher than for the low-fibre diet. Wholegrains, fresh fruit and vegetables can be more expensive and require more time to prepare and cook than cheap, processed alternatives, which may make a high-fibre diet harder to afford or maintain for some families, especially those on a limited budget or with little time to cook.

QUESTION 19 5 marks Criterion D

Medium

People with lactose intolerance cannot produce enough of the enzyme lactase to digest lactose (milk sugar), causing bloating and discomfort after drinking milk. Food scientists now produce "lactose-free" milk by treating ordinary milk with lactase enzyme before it is sold, breaking the lactose down in advance.

Discuss one benefit and one drawback of this use of enzyme technology in the food industry.

Show complete worked solution

Benefit: Lactose-free milk allows people with lactose intolerance to enjoy milk and dairy products, and the calcium, protein and vitamins they contain, without uncomfortable digestive symptoms — improving their quality of life and diet without needing to avoid dairy completely.

Drawback: Treating milk with lactase is an extra industrial process, which increases production costs, so lactose-free milk is usually more expensive than ordinary milk. It also does not treat any underlying digestive condition — it only avoids the symptoms of that one specific problem, and people may come to rely on processed alternatives rather than understanding or managing their condition more broadly.

QUESTION 20 6 marks Criterion D

Hard

Some patients cannot chew or swallow safely (for example, after a stroke or serious surgery). A feeding (nasogastric) tube can be passed through the nose directly into the stomach or small intestine, delivering a liquid nutrient mixture that bypasses the mouth and, if needed, part of the digestive tract.

Evaluate the impact of this technology, discussing both a benefit and a concern it raises.

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Benefit: Feeding tubes allow patients who cannot eat normally to still receive the nutrients, energy and water their body needs, preventing malnutrition and dehydration while they recover or manage a long-term condition. This can be life-saving, especially for patients who would otherwise be unable to take in any food at all.

Concern: Inserting and maintaining a tube carries a risk of infection at the entry site or further into the digestive tract, and the tube can cause discomfort or irritation. Beyond the physical risks, eating is often a social and enjoyable part of daily life, and long-term tube feeding removes this experience, which can affect a patient's mental wellbeing; it also requires trained staff and ongoing medical supervision, adding cost to healthcare systems.

QUESTION 1 2 marks Criterion A

Easy

The diagram shows the human breathing (respiratory) system.

a. Name the structure labelled X.

[1]

b. Name the structure labelled Y.

[1]

Show complete worked solution

(a)

X is an alveolus (air sac) — the site of gas exchange between the lungs and the blood.

(b)

Y is the diaphragm — the sheet of muscle beneath the lungs that contracts and flattens during inhalation.

QUESTION 2 2 marks Criterion A

Easy

The trachea and bronchi are lined with mucus and tiny hair-like cilia.

a. State the function of the mucus.

[1]

b. State the function of the cilia.

[1]

Show complete worked solution

(a)

Mucus traps dust, bacteria and other particles breathed in with the air, stopping them from reaching the lungs.

(b)

Cilia beat/sweep the mucus (with trapped particles) upward, away from the lungs and toward the throat, where it can be swallowed or coughed out.

QUESTION 3 2 marks Criterion A

Easy

Breathing in (inhalation) involves the diaphragm and the rib cage.

a. State what the diaphragm does during inhalation.

[1]

b. State what the rib cage does during inhalation.

[1]

Show complete worked solution

(a)

The diaphragm contracts and flattens (moves downward).

(b)

The intercostal muscles contract, moving the rib cage upward and outward.

QUESTION 4 5 marks Criterion A

Medium

The diagram shows a cluster of alveoli (air sacs) surrounded by a capillary network, where gas exchange takes place.

a. State three features of alveoli that adapt them for efficient gas exchange.

[3]

b. Explain why a short diffusion distance increases the rate of gas exchange.

[2]

Show complete worked solution

(a)

Millions of alveoli give the lungs a very large total surface area.
Alveoli walls are moist and only one cell thick, giving a short diffusion distance.
Alveoli have a dense capillary network, giving a good blood supply that constantly maintains a steep concentration gradient.

(b)

Diffusion is faster over shorter distances, since gas particles have less far to travel between the air in the alveolus and the blood in the capillary. With walls only one cell thick, oxygen and carbon dioxide molecules cross into and out of the blood much more quickly than they would through a thicker barrier, increasing the overall rate of gas exchange.

QUESTION 5 4 marks Criterion A

Medium

Muscle cells use aerobic respiration to release energy from glucose.

a. Write the word equation for aerobic respiration.

[2]

b. A runner's muscle cells produce $18\,\text{mol}$ of carbon dioxide during a race. Using the ratio of reactants and products in the equation, state how many mol of oxygen were used.

[2]

Show complete worked solution

(a)

$$ \text{glucose} + \text{oxygen} \rightarrow \text{carbon dioxide} + \text{water} \ (+ \text{energy}) $$

(b)

The balanced equation ($\text{C}_6\text{H}_{12}\text{O}_6 + 6\text{O}_2 \rightarrow 6\text{CO}_2 + 6\text{H}_2\text{O}$) shows oxygen and carbon dioxide are used/produced in a $1:1$ ratio. So $18\,\text{mol}$ of $\text{CO}_2$ produced means $18\,\text{mol}$ of $\text{O}_2$ were used.

QUESTION 6 5 marks Criterion A

Medium

At rest, a student breathes $16$ times per minute, with a tidal volume (volume of air taken in per breath) of $500\,\text{cm}^3$.

a. Calculate the student's minute ventilation (total volume of air breathed in one minute) at rest, in $\text{cm}^3$.

[2]

b. During exercise, the student's breathing rate rises to $24$ breaths/min and tidal volume rises to $800\,\text{cm}^3$. Calculate the new minute ventilation in litres, and the factor by which it has increased compared with rest.

[3]

Show complete worked solution

(a)

$$ 16 \times 500 = 8000\,\text{cm}^3 \ (= 8\,\text{L}) $$

(b)

$$ 24 \times 800 = 19\,200\,\text{cm}^3 = 19.2\,\text{L} $$$$ \text{factor} = \frac{19.2}{8} = 2.4 $$

Answer: minute ventilation rises to $19.2\,\text{L}$, a $2.4\times$ increase.

QUESTION 7 7 marks Criterion A

Hard

Human lungs contain approximately $3.5 \times 10^8$ (350 million) alveoli, each with an average surface area of about $2 \times 10^{-7}\,\text{m}^2$.

a. Explain four features of the alveoli and lungs that together adapt them for efficient gas exchange.

[4]

b. Calculate the total gas-exchange surface area of the lungs, showing your working in standard form.

[3]

Show complete worked solution

(a)

Enormous total surface area from hundreds of millions of alveoli.
Moist walls, which allow oxygen and carbon dioxide to dissolve before diffusing.
Walls only one cell thick, giving the shortest possible diffusion distance.
Dense capillary network around every alveolus, giving a good blood supply that keeps carrying gases away/to maintain a steep concentration gradient.

(b)

$$ 3.5 \times 10^8 \times 2 \times 10^{-7} = (3.5 \times 2) \times 10^{(8-7)} = 7 \times 10^1 $$

Answer: approximately $70\,\text{m}^2$ — roughly the area of half a tennis court.

QUESTION 8 7 marks Criterion B

Medium

You want to investigate how exercise intensity affects breathing rate.

a. State the independent and dependent variables.

[2]

b. State two variables you would need to control, and explain why for one of them.

[2]

c. Describe a method, including the equipment you would use, to collect the data needed.

[3]

Show complete worked solution

(a)

Independent variable: exercise intensity (e.g. walking, jogging, fast running, or step-ups at different rates). Dependent variable: breathing rate (breaths per minute).

(b)

Control the duration of exercise at each intensity and the length of rest given before each test (to return to resting breathing rate). Why control exercise duration: exercising for longer at any intensity would raise breathing rate further for reasons unrelated to intensity itself, making the comparison between intensities unfair.

(c)

Rest the participant for $5$ minutes, then count the number of breaths taken in $1$ minute using a stopwatch, to find the resting breathing rate.
Have the participant exercise at the first intensity (e.g. slow step-ups) for $3$ minutes.
Immediately after stopping, count breaths for $1$ minute using the stopwatch.
Allow a full rest period (e.g. $10$ minutes) for breathing rate to return to resting level, then repeat at the next, higher intensity.
Repeat each intensity at least twice and calculate a mean breathing rate.

QUESTION 9 6 marks Criterion B

Medium

A student's hypothesis is: "The higher the intensity of exercise, the longer it takes for breathing rate to return to its resting value (recovery time)."

a. Identify the independent, dependent, and one controlled variable for an investigation to test this.

[3]

b. Explain how recovery-time data collected at several intensities could be used to test the hypothesis.

[3]

Show complete worked solution

(a)

Independent: exercise intensity. Dependent: recovery time (time for breathing rate to return to the resting value after exercise stops). Controlled: duration of exercise at each intensity — exercising for a different length of time would itself change recovery time, regardless of intensity.

(b)

For each exercise intensity tested, the time taken for breathing rate (measured at regular intervals after exercise, e.g. every minute) to return to the pre-exercise resting rate would be recorded. If the hypothesis is correct, recovery time should be longer at higher intensities and shorter at lower intensities — plotting recovery time against exercise intensity should show a clear increasing trend if the data support the hypothesis.

QUESTION 10 8 marks Criterion B

Hard

A student counts a partner's breaths by watching their chest rise and fall while they exercise on a treadmill, but finds it very difficult to count accurately because breathing is fast, and having someone watch closely seems to make the partner breathe differently than normal.

a. Identify two sources of error in this method.

[2]

b. Suggest an improvement to the method that would reduce these errors, and explain how it works.

[3]

c. Explain how the improved method would produce more reliable data.

[3]

Show complete worked solution

(a)

(1) Human counting error — fast breathing is easy to miscount by watching the chest. (2) The observer effect — being watched closely may cause the partner to breathe consciously/differently than they normally would, changing the very thing being measured.

(b)

Use a chest-strap breathing-rate sensor connected to a data logger (or record on video for later, less-obtrusive counting). The sensor detects each breath automatically and electronically, removing human counting error, and because it is worn discreetly under clothing rather than being watched directly, it reduces the chance that the participant changes their breathing simply because they know they are being observed.

(c)

With counting error removed and the observer effect minimised, the breathing rate recorded is much more likely to reflect the participant's true, natural breathing rate during exercise. Repeated trials at the same intensity should then give very similar readings, rather than being scattered by inconsistent human counting or by the participant unconsciously altering their breathing because they are being watched.

QUESTION 11 3 marks Criterion C

Easy

Five students had their resting breathing rate measured.

Student	1	2	3	4	5
Breathing rate (breaths/min)	14	16	15	15	18

a. Calculate the mean breathing rate.

[2]

b. Comment on whether Student 5's reading of $18$ seems anomalous.

[1]

Show complete worked solution

(a)

$$ \text{mean} = \frac{14+16+15+15+18}{5} = \frac{78}{5} = 15.6\,\text{breaths/min} $$

(b)

It is a little higher than the others, but not extremely far from the group (only $2$–$4$ breaths/min above the rest), so it is not clearly anomalous — normal resting breathing rate varies between individuals, and $18$ is still within a typical resting range.

QUESTION 12 4 marks Criterion C

Medium

A student's breathing rate was recorded every two minutes during a period of steadily increasing exercise intensity.

Time (min)	0	2	4	6	8
Breathing rate (breaths/min)	14	18	22	35	30

a. Identify the anomalous reading in this table.

[1]

b. Explain how you identified it.

[2]

c. State what you should do with this reading before describing the overall trend.

[1]

Show complete worked solution

(a)

The reading at $t=6\,\text{min}$ ($35$ breaths/min) is anomalous.

(b)

Every other reading increases by exactly $4$ breaths/min each interval ($14, 18, 22, \ldots, 30$). If the pattern had continued, the reading at $t=6\,\text{min}$ should have been $26$, not $35$ — it breaks the otherwise-consistent, steady increase.

(c)

Exclude (discount) it from the analysis, and, if possible, repeat that measurement to check whether $35$ was a genuine result or a counting error, before drawing a line of best fit through the remaining consistent data.

QUESTION 13 6 marks Criterion C

Medium

The graph shows a student's lung volume over time. Exercise begins at $t = 12\,\text{s}$.

a. Using the graph, estimate the student's resting breathing rate (before $t=12\,\text{s}$), in breaths per minute.

[2]

b. Describe how the tidal volume (depth of each breath) changes once exercise begins.

[2]

c. Explain, in terms of gas exchange, why both breathing rate and depth increase during exercise.

[2]

Show complete worked solution

(a)

At rest, one full breath cycle (peak to peak) takes about $4\,\text{s}$, giving $\dfrac{60}{4} = 15$ breaths per minute.

(b)

The tidal volume (the height of each wave, i.e. the difference between the highest and lowest lung volume in a breath cycle) clearly increases once exercise begins — each breath becomes noticeably deeper, and the breathing cycles also become closer together (faster).

(c)

During exercise, muscle cells respire faster to release more energy, so they need more oxygen and produce more carbon dioxide than at rest. Breathing faster and more deeply increases the volume of air moved in and out of the lungs each minute (minute ventilation), supplying oxygen to the blood more quickly and removing carbon dioxide more quickly, to keep up with the muscles' increased demand.

QUESTION 14 4 marks Criterion C

Medium

Vital capacity (the maximum volume of air that can be breathed out after a full breath in) was measured for two groups.

Group	Mean vital capacity (L)
Non-smokers	4.8
Smokers (10 years)	3.6

a. Calculate the percentage reduction in vital capacity for the smoker group compared with the non-smoker group.

[2]

b. Suggest, in terms of the alveoli, why long-term smoking reduces vital capacity.

[2]

Show complete worked solution

(a)

$$ \frac{4.8-3.6}{4.8} \times 100 = \frac{1.2}{4.8} \times 100 = 25\% $$

(b)

Chemicals in cigarette smoke (such as tar) damage and can break down the thin walls between alveoli, destroying the delicate structure of the lungs. This reduces the total surface area available for air to be held and gas exchange to occur, so the lungs can hold and exchange less air overall, lowering vital capacity.

QUESTION 15 5 marks Criterion C

Medium

The table compares the approximate composition of inhaled and exhaled air.

Gas	Inhaled air (%)	Exhaled air (%)
Oxygen	21	16
Carbon dioxide	0.04	4
Nitrogen	78	78

a. Calculate the percentage decrease in oxygen between inhaled and exhaled air.

[2]

b. Explain, using the data, why the percentage of nitrogen stays about the same.

[1]

c. Explain what the data show is happening in the alveoli.

[2]

Show complete worked solution

(a)

$$ \frac{21-16}{21} \times 100 = \frac{5}{21} \times 100 = 23.8\% \ (\text{3 s.f.}) $$

(b)

Nitrogen is not used in respiration and is not produced by the body, so it simply passes in and out of the lungs unchanged — its percentage in the air is unaffected by gas exchange.

(c)

The fall in oxygen and the large rise in carbon dioxide show that oxygen is diffusing out of the alveoli into the blood (to be used in respiration), while carbon dioxide, a waste product of respiration, is diffusing from the blood into the alveoli to be breathed out. Both gases move down their own concentration gradient, in opposite directions.

QUESTION 16 7 marks Criterion C

Hard

A student's hypothesis is: "Fitter people recover to their resting breathing rate faster after exercise." Recovery times (minutes) were measured for individuals in two groups performing the same exercise.

Group	Recovery times (min)
Regularly active	3.0, 2.5, 3.5
Not regularly active	5.0, 4.5, 6.5

a. Calculate the mean recovery time for each group.

[2]

b. Evaluate whether these results support the hypothesis, discussing the spread of results within each group.

[5]

Show complete worked solution

(a)

Regularly active: $\dfrac{3.0+2.5+3.5}{3} = \dfrac{9.0}{3} = 3.0\,\text{min}$. Not regularly active: $\dfrac{5.0+4.5+6.5}{3} = \dfrac{16.0}{3} = 5.3\,\text{min}$ (3 s.f.).

(b)

The mean recovery time for the regularly active group ($3.0\,\text{min}$) is clearly shorter than for the not regularly active group ($5.3\,\text{min}$), which supports the hypothesis. However, the sample size is very small (only three people per group), and there is some spread within each group (regularly active: $2.5$–$3.5\,\text{min}$; not regularly active: $4.5$–$6.5\,\text{min}$) — the highest regularly-active time ($3.5$) and lowest not-regularly-active time ($4.5$) do not overlap here, which strengthens the conclusion, but a larger sample of participants would be needed before concluding confidently that fitness level, rather than individual variation, is the real cause of the difference.

QUESTION 17 6 marks Criterion C

Hard

A student measured how long they could hold their breath, repeating the test five times, with the stopwatch readable to $\pm0.1\,\text{s}$.

Trial	1	2	3	4	5
Breath-hold time (s)	48	52	45	58	50

a. Calculate the mean breath-hold time.

[2]

b. Calculate the range of the results.

[1]

c. The stopwatch is precise to $\pm0.1\,\text{s}$, but the results vary by far more than this. Explain what this tells you about the main source of uncertainty in this experiment.

[3]

Show complete worked solution

(a)

$$ \text{mean} = \frac{48+52+45+58+50}{5} = \frac{253}{5} = 50.6\,\text{s} $$

(b)

$$ \text{range} = 58-45 = 13\,\text{s} $$

(c)

Since the results vary by up to $13\,\text{s}$ — far more than the stopwatch's own $\pm0.1\,\text{s}$ precision — the main source of uncertainty is not the stopwatch, but genuine biological variation between attempts (e.g. differences in how deep a breath was taken before holding, motivation, or slight tiredness from repeated trials). This is why repeating trials and calculating a mean gives a more reliable estimate of true breath-hold ability than any single reading.

QUESTION 18 6 marks Criterion D

Medium

Many countries have introduced smoking bans in enclosed public places (such as restaurants and workplaces) to reduce exposure to secondhand smoke.

	Before smoking ban	5 years after ban
Adults reporting daily secondhand smoke exposure	31%	9%
Adult smoking rate	28%	19%

Discuss one benefit and one drawback of introducing smoking bans in public places, using the data to support your discussion.

Show complete worked solution

Benefit: The data show a large fall in secondhand smoke exposure, from $31\%$ to $9\%$ of adults, alongside a drop in the adult smoking rate itself from $28\%$ to $19\%$. Reducing exposure to secondhand smoke lowers the risk of respiratory conditions (such as bronchitis and reduced lung function) in non-smokers, particularly benefiting workers and children who previously had little choice but to breathe in smoke-filled air in public places.

Drawback: Smoking bans restrict what people are legally allowed to do with a legal product, which some people see as an unfair restriction on personal freedom, and businesses such as bars that previously allowed smoking may lose some customers or income as a result. Enforcing the ban also requires ongoing resources (inspections, fines) from local authorities.

QUESTION 19 5 marks Criterion D

Medium

In cities with high air pollution, many residents wear filtering face masks and use smartphone apps that report the daily air quality index (AQI).

Daily AQI category	Hospital respiratory admissions (per day, city average)
Good	12
Hazardous	47

Discuss one benefit and one drawback of relying on masks and air quality apps as a response to air pollution, using the data to support your discussion.

Show complete worked solution

Benefit: Hospital respiratory admissions are nearly four times higher on hazardous-AQI days ($47$) than on good-AQI days ($12$), showing polluted air clearly harms the respiratory system. Air quality apps let people check pollution levels and choose to wear a filtering mask, or avoid strenuous outdoor exercise, on the worst days, directly protecting their lungs (particularly for people with asthma or other respiratory conditions).

Drawback: Masks and apps only protect the individuals who use them correctly and consistently, and can be uncomfortable, expensive to replace regularly, or simply forgotten. Crucially, this approach treats only the symptoms of the problem for those who can access and afford it — it does nothing to reduce the pollution itself, so hospital admissions on hazardous days remain high for the wider population, including those without access to masks or apps.

QUESTION 20 6 marks Criterion D

Hard

Mechanical ventilators can breathe for a patient whose lungs cannot work well enough on their own, for example during major surgery or severe illness. Demand for ventilators rose sharply during the COVID-19 pandemic, when hospitals in many countries reported shortages.

Evaluate the impact of ventilator technology, discussing both a benefit and a concern it raises.

Show complete worked solution

Benefit: Ventilators keep patients alive when their own breathing cannot supply enough oxygen to the body, buying crucial time for the underlying illness or injury to be treated or for the lungs to heal. Without this technology, many patients with severe respiratory failure would not survive.

Concern: Ventilators are expensive, complex machines that require specially trained staff to operate safely, so hospitals — especially in poorer regions, or during a sudden surge in demand such as a pandemic — may not have enough available for every patient who needs one, forcing very difficult decisions about who receives treatment. Being on a ventilator for a long time also carries its own risks, such as ventilator-associated lung infections.

QUESTION 1 2 marks Criterion A

Easy

The diagram shows a simplified cross-section of the human heart, viewed as if facing you (so the patient's right side is on your left).

a. Name chamber A.

[1]

b. Name chamber D.

[1]

Show complete worked solution

(a)

A is the right atrium — it receives deoxygenated blood returning from the body via the vena cava.

(b)

D is the left ventricle — it pumps oxygenated blood out to the whole body via the aorta.

QUESTION 2 3 marks Criterion A

Easy

Blood is made up of plasma and several types of cells.

a. State the function of red blood cells.

[1]

b. State the function of white blood cells.

[1]

c. State the function of platelets.

[1]

Show complete worked solution

(a)

Red blood cells carry oxygen around the body, bound to the protein haemoglobin.

(b)

White blood cells are part of the immune system — they defend the body against pathogens (e.g. by engulfing bacteria or producing antibodies).

(c)

Platelets help the blood to clot at a wound, sealing the damaged blood vessel and preventing excessive blood loss and the entry of pathogens.

QUESTION 3 3 marks Criterion A

Easy

The circulatory system contains three main types of blood vessel.

a. Name the type of vessel that carries blood away from the heart.

[1]

b. Name the type of vessel that carries blood back to the heart.

[1]

c. Name the type of vessel with walls only one cell thick, allowing exchange of substances with body tissues.

[1]

Show complete worked solution

(a)

Arteries.

(b)

Veins.

(c)

Capillaries.

QUESTION 4 4 marks Criterion A

Medium

The diagram shows cross-sections of two blood vessels, A and B, drawn to different scales.

a. Identify which vessel, A or B, is an artery, and explain your reasoning using the diagram.

[2]

b. State the function of the valves shown in vessel B.

[2]

Show complete worked solution

(a)

Vessel A is the artery. It has a much thicker, more muscular wall (relative to the width of its central lumen) than vessel B — arteries need thick, muscular, elastic walls to withstand the high pressure of blood being pumped directly from the heart.

(b)

Valves in veins prevent blood from flowing backward, ensuring blood — which is under much lower pressure in veins — keeps moving in one direction, back toward the heart, especially against gravity in the limbs.

QUESTION 5 4 marks Criterion A

Medium

Humans have a double circulatory system, meaning blood passes through the heart twice on each full circuit of the body.

Show complete worked solution

Starting from the vena cava, blood follows this pathway: vena cava ? right atrium ? right ventricle ? pulmonary artery ? lungs (gas exchange: picks up oxygen, releases carbon dioxide) ? pulmonary vein ? left atrium ? left ventricle ? aorta ? body (delivers oxygen to tissues, collects carbon dioxide) ? vena cava, and the cycle repeats. The first loop (right side of the heart to the lungs and back) is the pulmonary circulation; the second loop (left side of the heart around the rest of the body and back) is the systemic circulation.

QUESTION 6 5 marks Criterion A

Medium

A person's resting heart rate is $72$ beats per minute, and each heartbeat pumps a stroke volume of $70\,\text{cm}^3$ of blood.

a. Calculate the person's resting cardiac output (volume of blood pumped per minute), in $\text{cm}^3/\text{min}$.

[2]

b. During exercise, heart rate rises to $150$ beats per minute and stroke volume rises to $100\,\text{cm}^3$. Calculate the new cardiac output, in $\text{L/min}$.

[3]

Show complete worked solution

(a)

$$ \text{cardiac output} = \text{heart rate} \times \text{stroke volume} = 72 \times 70 = 5040\,\text{cm}^3/\text{min} $$

(b)

$$ 150 \times 100 = 15\,000\,\text{cm}^3/\text{min} = 15\,\text{L/min} $$

QUESTION 7 6 marks Criterion A

Hard

At rest, a red blood cell takes about $20\,\text{s}$ to complete one full circuit of the body.

a. Calculate how many times a red blood cell circulates the body in one hour.

[2]

b. Using the idea of double circulation, explain why the left ventricle has a much thicker, more muscular wall than the right ventricle.

[4]

Show complete worked solution

(a)

$$ \frac{3600\,\text{s}}{20\,\text{s}} = 180 \ \text{circuits per hour} $$

(b)

In double circulation, the right ventricle only pumps blood a short distance to the nearby lungs (pulmonary circulation) against relatively low resistance, so it does not need to generate very high pressure. The left ventricle, however, must pump blood all the way around the rest of the body (systemic circulation) — a much longer journey, against much greater total resistance from many more blood vessels — so it needs to generate a far higher pressure to push blood that whole distance. A thicker, more muscular wall allows the left ventricle to contract with greater force, generating this higher pressure.

QUESTION 8 7 marks Criterion B

Medium

You want to investigate how exercise intensity affects pulse rate.

a. State the independent and dependent variables.

[2]

b. State two variables you would need to control, and explain why for one of them.

[2]

c. Describe a method, including the equipment you would use, to collect the data needed.

[3]

Show complete worked solution

(a)

Independent variable: exercise intensity (e.g. walking, jogging, star jumps at increasing speeds). Dependent variable: pulse rate (beats per minute).

(b)

Control the duration of exercise at each intensity and the rest period allowed before the next test. Why control exercise duration: exercising for longer at any intensity would itself raise pulse rate further, regardless of intensity, making comparisons between intensities unfair.

(c)

Rest for $5$ minutes, then measure resting pulse by counting beats felt at the wrist for $15\,\text{s}$ (using a stopwatch) and multiplying by $4$.
Exercise at the first intensity for $3$ minutes, then immediately measure pulse the same way.
Rest fully (e.g. $10$ minutes, until pulse returns to resting rate) before repeating at a higher intensity.
Repeat each intensity at least twice and calculate a mean pulse rate.

QUESTION 9 6 marks Criterion B

Medium

A student's hypothesis is: "Drinking a caffeinated soft drink increases resting heart rate."

a. Identify the independent, dependent, and one controlled variable for an investigation to test this.

[3]

b. Explain how comparing data from a caffeine group and a control group would test the hypothesis.

[3]

Show complete worked solution

(a)

Independent: whether the drink contains caffeine or not (caffeinated drink vs. an identical caffeine-free drink). Dependent: resting heart rate, measured before and after drinking. Controlled: volume of drink given to each participant — a different volume could affect heart rate for reasons unrelated to caffeine.

(b)

One group drinks the caffeinated drink, while a control group drinks an identical-looking, identical-volume drink without caffeine; heart rate is measured for both groups before and $30$ minutes after drinking. If the hypothesis is correct, the caffeine group's heart rate should increase noticeably more than the control group's — comparing the change in heart rate between the two groups isolates the effect of caffeine itself from any other factor (such as simply having drunk something, or normal time-based changes).

QUESTION 10 8 marks Criterion B

Hard

A student measures pulse rate by counting beats at the wrist for $15\,\text{s}$ and multiplying by $4$, but presses down hard while searching for the pulse, and sometimes loses count partway through. Repeat measurements under the same conditions give very different results.

a. Identify two sources of error in this method.

[2]

b. Suggest an improvement to the method that would reduce these errors, and explain how it works.

[3]

c. Explain how the improved method would produce more reliable results.

[3]

Show complete worked solution

(a)

(1) Pressing too hard on the artery can partially block blood flow, distorting the pulse felt or making it harder to count accurately. (2) Human counting error — losing count of individual beats, especially at a fast pulse rate.

(b)

Use an electronic heart-rate monitor or pulse oximeter (e.g. a finger clip sensor or chest strap), which detects each heartbeat automatically and displays or logs the rate digitally. This removes the need to physically press on the artery to find a pulse, and removes human counting error entirely, since the beats are counted electronically.

(c)

Because the sensor detects every heartbeat consistently and automatically, without depending on how hard a finger presses or how well a person counts, repeated measurements taken under the same conditions should be very close together rather than scattered. This means the heart rate values recorded would be a far more accurate and repeatable reflection of the person's true heart rate.

QUESTION 11 3 marks Criterion C

Easy

Five students had their resting pulse rate measured.

Student	1	2	3	4	5
Pulse rate (bpm)	68	72	70	71	75

a. Calculate the mean pulse rate.

[2]

b. Calculate the range of the results.

[1]

Show complete worked solution

(a)

$$ \text{mean} = \frac{68+72+70+71+75}{5} = \frac{356}{5} = 71.2\,\text{bpm} $$

(b)

$$ \text{range} = 75-68 = 7\,\text{bpm} $$

QUESTION 12 4 marks Criterion C

Medium

A student's pulse rate was recorded every minute during recovery after exercise.

Time (min)	0	1	2	3	4
Pulse rate (bpm)	150	138	126	98	102

a. Identify the anomalous reading in this table.

[1]

b. Explain how you identified it.

[2]

c. State what you should do with this reading before describing the recovery trend.

[1]

Show complete worked solution

(a)

The reading at $t=3\,\text{min}$ ($98$ bpm) is anomalous.

(b)

Every other reading decreases by exactly $12$ bpm each minute ($150, 138, 126, \ldots, 102$). If the pattern had continued, the reading at $t=3\,\text{min}$ should have been $114$, not $98$ — it breaks the otherwise-consistent, steady decline (and is lower than the following reading, which should continue to fall, not rise back up).

(c)

Exclude (discount) it from the analysis, and, if possible, repeat that measurement to check whether $98$ was a genuine reading or a counting error, before drawing a line of best fit through the remaining consistent data.

QUESTION 13 5 marks Criterion C

Medium

The graph shows a student's pulse rate before, during and after a period of exercise (exercise takes place between $t=2\,\text{min}$ and $t=6\,\text{min}$).

a. Using the graph, state the student's resting pulse rate and their peak pulse rate.

[2]

b. Calculate the increase from resting to peak pulse rate.

[1]

c. Describe the shape of the recovery section of the graph (after $t=6\,\text{min}$), and explain what a faster recovery generally indicates.

[2]

Show complete worked solution

(a)

Resting pulse rate $\approx 70\,\text{bpm}$. Peak pulse rate $\approx 150\,\text{bpm}$ (reached during exercise, from $t=4$ to $t=6\,\text{min}$).

(b)

$$ 150-70 = 80\,\text{bpm} $$

(c)

After $t=6\,\text{min}$, pulse rate falls fairly steeply at first, then more gradually, curving back down toward (but not quite reaching) the original resting rate by $t=12\,\text{min}$. A faster return to resting pulse rate generally indicates a fitter cardiovascular system, which can supply the body's oxygen demand and clear the effects of exercise more efficiently.

QUESTION 14 5 marks Criterion C

Medium

Resting heart rate was measured for two groups.

Group	Mean resting heart rate (bpm)
Trained athletes	52
Non-athletes	74

a. Calculate the difference between the two group means.

[1]

b. Calculate the percentage by which the athletes' mean heart rate is lower than the non-athletes' mean.

[2]

c. Suggest, in terms of stroke volume, why trained athletes tend to have a lower resting heart rate.

[2]

Show complete worked solution

(a)

$$ 74-52 = 22\,\text{bpm} $$

(b)

$$ \frac{74-52}{74} \times 100 = \frac{22}{74} \times 100 = 29.7\% \ (\text{3 s.f.}) $$

(c)

Regular training strengthens the heart muscle, so the left ventricle can contract more powerfully and pump a larger stroke volume with each beat. Since cardiac output (heart rate $\times$ stroke volume) needs to supply the same amount of blood at rest either way, an athlete's heart does not need to beat as often to deliver that same volume of blood, giving a lower resting heart rate.

QUESTION 15 5 marks Criterion C

Medium

A patient's blood pressure was measured at four clinic visits after starting a new medication and diet plan.

Visit	1	2	3	4
Blood pressure (systolic/diastolic, mmHg)	150/95	145/92	138/88	130/85

a. Describe the trend shown for systolic blood pressure across the four visits.

[2]

b. Calculate the total decrease in systolic blood pressure from visit 1 to visit 4.

[1]

c. Suggest why it is important to measure blood pressure on multiple occasions rather than relying on a single reading.

[2]

Show complete worked solution

(a)

Systolic blood pressure decreases steadily at every visit, from $150\,\text{mmHg}$ at visit $1$ down to $130\,\text{mmHg}$ at visit $4$.

(b)

$$ 150-130 = 20\,\text{mmHg} $$

(c)

Blood pressure naturally varies from moment to moment (e.g. due to stress, recent activity, or simply the anxiety of being at a clinic — sometimes called "white coat syndrome"), so a single reading might not reflect a patient's true, typical blood pressure. Multiple readings over time reveal a genuine trend and make it possible to judge whether treatment is actually working, rather than being misled by one unusually high or low reading.

QUESTION 16 7 marks Criterion C

Hard

A student's hypothesis is: "Resting heart rate is lower in people who exercise regularly." Individual resting heart rates were recorded for two groups.

Group	Individual resting heart rates (bpm)
Regular exercisers	58, 62, 55, 65
Non-exercisers	70, 68, 85, 73

a. Calculate the mean resting heart rate for each group.

[2]

b. Evaluate whether these results support the hypothesis, discussing the spread and any overlap between the two groups.

[5]

Show complete worked solution

(a)

Regular exercisers: $\dfrac{58+62+55+65}{4} = \dfrac{240}{4} = 60\,\text{bpm}$. Non-exercisers: $\dfrac{70+68+85+73}{4} = \dfrac{296}{4} = 74\,\text{bpm}$.

(b)

The mean resting heart rate is clearly lower for regular exercisers ($60\,\text{bpm}$) than for non-exercisers ($74\,\text{bpm}$), a difference of $14\,\text{bpm}$, which supports the hypothesis. However, there is some individual variation within each group (exercisers range $55$–$65$; non-exercisers range $68$–$85$), and although the ranges do not overlap here, the highest exerciser value ($65$) is fairly close to the lowest non-exerciser value ($68$) — showing that individual factors other than exercise habit also affect resting heart rate. With only four people per group, a larger sample would be needed to be more confident the difference is due to regular exercise rather than chance or other differences between the individuals studied.

QUESTION 17 6 marks Criterion C

Hard

A student counted their pulse for $15\,\text{s}$ at a time (then multiplied by $4$ to get bpm), repeating the count five times in a row while resting. The stopwatch used was precise to $\pm0.1\,\text{s}$.

Trial	1	2	3	4	5
Beats counted in 15 s	18	19	17	22	18

a. Calculate the mean number of beats counted in $15\,\text{s}$.

[2]

b. Convert this mean value to a pulse rate in beats per minute (bpm).

[1]

c. The stopwatch is precise to $\pm0.1\,\text{s}$, but the beat counts vary more than this alone would explain. Explain what this tells you about the main source of uncertainty in this experiment.

[3]

Show complete worked solution

(a)

$$ \text{mean} = \frac{18+19+17+22+18}{5} = \frac{94}{5} = 18.8 $$

(b)

$$ 18.8 \times 4 = 75.2 \approx 75\,\text{bpm} $$

(c)

Since the number of beats counted varies by up to $5$ (from $17$ to $22$) — far more than could be explained by the stopwatch's own $\pm0.1\,\text{s}$ precision — the main source of uncertainty is not the stopwatch, but human error in feeling and counting each beat (missing a faint beat, or miscounting during the $15\,\text{s}$ window). This is why repeating trials and averaging gives a far more reliable estimate of true pulse rate than any single $15\,\text{s}$ count.

QUESTION 18 6 marks Criterion D

Medium

Donated blood is used to treat patients who have lost large amounts of blood (for example, in surgery or after an accident) or who have certain blood conditions.

	Value
People who will need a blood transfusion at some point in their life	about 1 in 3
Blood donations screened and found unsuitable for use (infection risk, etc.)	roughly 1-2%

Discuss one benefit and one drawback of blood donation and transfusion technology, using the data to support your discussion.

Show complete worked solution

Benefit: With around $1$ in $3$ people expected to need a transfusion at some point in their life, blood donation directly saves lives — supplying blood for emergency surgery, childbirth complications, cancer treatment and chronic conditions like severe anaemia that would otherwise be fatal or highly dangerous.

Drawback: Even though only roughly $1$–$2\%$ of donations are found unsuitable during screening, transfusions still carry some risk, such as an allergic/immune reaction if blood group is mismatched, or (historically, before modern screening) transmission of infections through blood. Maintaining a safe, well-matched blood supply also depends entirely on enough healthy volunteers regularly donating, and shortages of specific blood types can occur.

QUESTION 19 5 marks Criterion D

Medium

An artificial pacemaker is a small electronic device implanted under the skin near the heart, which sends electrical signals to correct an irregular heartbeat (arrhythmia). Its battery typically needs replacing, requiring further minor surgery, after about $10$ years.

Discuss one benefit and one drawback of artificial pacemaker technology.

Show complete worked solution

Benefit: A pacemaker keeps the heart beating at a safe, regular rate for patients whose own heart's electrical signals are unreliable, preventing dangerously slow or irregular heartbeats and allowing patients to live an active, normal life that would otherwise be impossible or very restricted.

Drawback: Implanting a pacemaker requires surgery, which carries risks such as infection, and because the battery needs replacing roughly every $10$ years, patients face repeated minor surgeries over their lifetime. The device and its ongoing medical monitoring also add cost to the patient or healthcare system.

QUESTION 20 6 marks Criterion D

Hard

Statins are medicines that lower LDL ("bad") cholesterol, used to reduce the risk of heart attacks and strokes caused by fatty deposits building up in artery walls.

	Heart attacks per 1000 people over 5 years
Without statins	42
With statins	29

Evaluate the impact of statins, discussing both a benefit and a concern raised by their widespread use, using the data to support your discussion.

Show complete worked solution

Benefit: The data show a clear reduction in heart attacks, from $42$ to $29$ per $1000$ people over five years — roughly a $31\%$ reduction. By lowering LDL cholesterol, statins reduce the build-up of fatty deposits inside artery walls, keeping arteries wider and reducing the chance of a blocked artery causing a heart attack or stroke, potentially saving many lives when prescribed to large numbers of at-risk people.

Concern: Statins can cause side effects in some patients (such as muscle pain), and require patients to take medication daily, often for the rest of their life. Some doctors are also concerned about "medicalizing" a lifestyle-related problem — prescribing a pill instead of addressing root causes such as diet, exercise and smoking, which could reduce heart attack risk without any medication at all, and long-term widespread prescribing adds a substantial ongoing cost to healthcare systems.

The Digestive System 20 questions

The Respiratory System 20 questions

The Circulatory System 20 questions