These are original Marlbridge practice questions, not past exam questions. The mark points are a suggested Marlbridge rubric to help you check your work; they are not an official scoring guideline. Formulas: cube SA = 6s², V = s³; sphere SA = 4πr², V = (4/3)πr³; rectangular solid SA = 2lh + 2lw + 2wh, V = lwh; cylinder SA = 2πrh + 2πr², V = πr²h. A calculator is assumed. All data sets and organisms are fictional.
Question 1 (multiple choice · foundation)
A spherical cell grows until its radius has doubled. Which row correctly describes the changes?
- (A) Surface area × 2, volume × 4, SA/V halves
- (B) Surface area × 4, volume × 8, SA/V halves
- (C) Surface area × 4, volume × 8, SA/V doubles
- (D) Surface area × 2, volume × 2, SA/V stays the same
Answer and explanation
Answer: (B). Area depends on r², so it rises by 2² = 4. Volume depends on r³, so it rises by 2³ = 8. SA/V changes by 4 ÷ 8 = ½, so it halves (check: SA/V = 3/r, and r doubled).
- (A) uses the wrong powers: it treats area as depending on r and volume on r².
- (C) has the right factors but divides the wrong way: 4 ÷ 8 is a half, not 2.
- (D) treats area and volume as scaling in the same way, which is the very idea this topic disproves.
Question 2 (multiple choice · core)
A fictional rod-shaped bacterium is modelled as a cylinder of radius 0.5 µm and length (height) 3.0 µm. What is its SA/V?
- (A) 2.7 µm⁻¹
- (B) 4.0 µm⁻¹
- (C) 4.3 µm⁻¹
- (D) 4.7 µm⁻¹
Answer and explanation
Answer: (D). SA = 2πrh + 2πr² = 2π(0.5)(3.0) + 2π(0.5)² = 9.425 + 1.571 = 11.00 µm². V = πr²h = π(0.25)(3.0) = 2.356 µm³. SA/V = 11.00 ÷ 2.356 = 4.67, so 4.7 µm⁻¹.
- (A) uses the diameter (1.0 µm) as the radius: SA/V = 25.13 ÷ 9.425 = 2.67.
- (B) leaves out both ends: 9.425 ÷ 2.356 = 4.0.
- (C) includes only one end (πr² instead of 2πr²): 10.21 ÷ 2.356 = 4.33.
Question 3 (multiple choice · core)
A shrew and a deer are both mammals that keep a steady body temperature. Which statement best explains why the shrew has the higher metabolic rate per gram of body mass?
- (A) The shrew has a higher SA/V, so it loses heat faster for its mass and must release more energy per gram to stay warm.
- (B) The shrew has a lower SA/V, so it holds more heat and needs more energy to cool down.
- (C) The shrew has more surface area in total, so it needs more energy in total than the deer.
- (D) The shrew’s cells are much larger than the deer’s, so each cell needs more energy.
Answer and explanation
Answer: (A). A small body has a large surface for its mass, so it exchanges heat with the surroundings proportionally faster. To replace the heat it loses, each gram of the shrew must release energy faster than each gram of the deer.
- (B) reverses the SA/V comparison; smaller bodies have the higher ratio.
- (C) is false (the deer’s total surface is far larger) and talks about total energy, not energy per gram.
- (D) is wrong because body size differences come mostly from cell number, not cell size; cells in both are small.
Question 4 (data analysis · core)
A student made cubes of agar coloured pink by an indicator that turns colourless in acid, with sides of 1 cm, 2 cm and 3 cm, and placed them in dilute acid. After 10 minutes she cut each cube in half. In every cube the acid had moved 0.40 cm in from each face, leaving a pink cube-shaped core in the centre.
(a) Calculate SA, V and SA/V for each cube. (b) Calculate the volume of the pink core in each cube, and the percentage of each cube’s volume reached by the acid. (c) Use your results to explain why cells are small. (d) State one way this model differs from a real cell.
Worked solution and suggested Marlbridge rubric
(a)
| Side / cm | SA / cm² | V / cm³ | SA/V / cm⁻¹ |
|---|---|---|---|
| 1 | 6 | 1 | 6.0 |
| 2 | 24 | 8 | 3.0 |
| 3 | 54 | 27 | 2.0 |
(b) The core’s side is s − 2 × 0.40 = s − 0.80 cm, because acid enters from both opposite faces.
| Side / cm | Core side / cm | Core V / cm³ | Reached by acid |
|---|---|---|---|
| 1 | 0.20 | 0.008 | (1 − 0.008) ÷ 1 = 99.2% |
| 2 | 1.20 | 1.728 | (8 − 1.728) ÷ 8 = 78.4% |
| 3 | 2.20 | 10.648 | (27 − 10.648) ÷ 27 = 60.6% |
(c) The higher the SA/V, the larger the fraction of the volume reached in the same time. A cell depends on substances entering through its surface; a small cell (high SA/V) has almost all its volume supplied quickly, while in a large cell much of the volume would be poorly supplied and wastes would build up.
(d) Any one: real cells use energy and transport proteins, not only diffusion; cells have membranes that let only some substances through; cells use the substances as they enter, so demand also matters; real cells are not cubes and may have folded surfaces; cells are micrometres across, not centimetres.
| Point | What earns it |
|---|---|
| 1 | All three SA/V values correct, with units |
| 1 | Core side s − 0.80 used (not s − 0.40) |
| 1 | All three percentages correct |
| 1 | Explanation links higher SA/V to a larger fraction supplied and a valid model limitation |
A common error is subtracting 0.40 only once; that gives cores of 0.60, 1.60 and 2.60 cm.
Question 5 (constructed response · core)
A fictional mutant of a small plant has normal roots but no root hairs. A researcher grows wild-type and mutant seedlings in the same nutrient solution containing phosphate ions.
(a) State a hypothesis about phosphate uptake. (b) Identify the independent variable, the dependent variable, and two variables that must be kept constant. (c) Explain why the dependent variable should be expressed per gram of root (or per cm of root), not per plant. (d) Predict the result and justify it using surface area.
Model answer and suggested Marlbridge rubric
(a) “Seedlings without root hairs take up phosphate more slowly, per gram of root, than wild-type seedlings.”
(b) Independent variable: presence or absence of root hairs (wild type versus mutant). Dependent variable: rate of phosphate uptake, for example the fall in phosphate concentration of the solution per hour, per gram of root. Controlled variables (any two): age of seedlings; temperature; light; volume and starting phosphate concentration of the solution; oxygen supply to the roots; time of measurement.
(c) If the two types have different amounts of root, uptake per plant would reflect root size as well as root hairs. Dividing by root mass (or length) compares the same amount of root, so any difference comes from the root hairs.
(d) Uptake should be lower in the mutant. Root hairs are long, thin extensions that greatly increase the surface area of the root in contact with the solution while adding little volume. With less membrane area, fewer phosphate ions can be absorbed per unit time.
| Point | What earns it |
|---|---|
| 1 | Testable hypothesis naming root hairs and phosphate uptake |
| 1 | Independent and dependent variables correct, with two valid controlled variables |
| 1 | Reason for standardising per gram or length of root |
| 1 | Prediction of lower uptake linked to reduced surface area for absorption |
Question 6 (constructed response · stretch)
Fictional data for five mammal species at rest:
| Species | V | W | X | Y | Z |
|---|---|---|---|---|---|
| Body mass / kg | 0.010 | 0.10 | 1.0 | 10 | 100 |
| Whole-body metabolic rate / W | 0.11 | 0.60 | 3.4 | 19 | 108 |
(a) Calculate the metabolic rate per kilogram for each species. (b) Describe the trend, using two calculated values. (c) A student says: “Species Z uses the most energy, so it has the highest metabolic rate.” Evaluate this statement. (d) Explain the trend in (b) using surface area-to-volume ratio and heat exchange.
Model answer and suggested Marlbridge rubric
(a) Divide each rate by mass: V 0.11 ÷ 0.010 = 11 W kg⁻¹; W 0.60 ÷ 0.10 = 6.0 W kg⁻¹; X 3.4 W kg⁻¹; Y 19 ÷ 10 = 1.9 W kg⁻¹; Z 108 ÷ 100 = 1.08 W kg⁻¹ (about 1.1).
(b) As body mass increases, metabolic rate per kilogram decreases: from 11 W kg⁻¹ in the 0.010 kg species to 1.08 W kg⁻¹ in the 100 kg species, about ten times lower (11 ÷ 1.08 = 10.2).
(c) Partly true. Z has the highest whole-body rate (108 W, about 980 times species V’s). But per kilogram, Z has the lowest rate. The statement does not say which measure it means, and per unit mass is the fair comparison between bodies of very different sizes.
(d) Smaller animals have a higher SA/V. A small body therefore loses heat to its surroundings faster for each kilogram of mass. To keep a steady body temperature, each kilogram must release energy faster, so metabolic rate per unit mass is higher. Larger animals have a lower SA/V, lose heat proportionally more slowly and need less energy per kilogram.
| Point | What earns it |
|---|---|
| 1 | All five rates per kilogram correct |
| 1 | Trend described with two calculated values |
| 1 | Distinguishes whole-body rate from rate per unit mass |
| 1 | Higher SA/V linked to faster heat loss per unit mass and to a higher metabolic rate per unit mass |
Question 7 (calculation · stretch)
In a fictional model, each µm³ of cytoplasm needs 0.012 units of a nutrient per minute, and each µm² of membrane can take in at most 0.030 units per minute.
(a) Show that supply can just meet demand when SA/V = 0.4 µm⁻¹. (b) Calculate the largest side a cube-shaped cell can have. (c) A cube-shaped cell has a side of 20 µm. Show by calculation that it cannot meet its demand. (d) The same volume of cytoplasm is reshaped into a flat box 40 µm × 40 µm × 5 µm. Calculate its SA/V and state whether it can now meet its demand.
Worked solution
(a) Supply = SA × 0.030; demand = V × 0.012. Setting them equal: SA × 0.030 = V × 0.012, so SA/V = 0.012 ÷ 0.030 = 0.4 µm⁻¹.
(b) For a cube, SA/V = 6/s. 6/s = 0.4 gives s = 15 µm.
(c) SA = 6 × 20² = 2400 µm²; V = 20³ = 8000 µm³; SA/V = 2400 ÷ 8000 = 0.30 µm⁻¹, below 0.4. Supply = 2400 × 0.030 = 72 units min⁻¹; demand = 8000 × 0.012 = 96 units min⁻¹. Supply is only 75% of demand.
(d) V = 40 × 40 × 5 = 8000 µm³ (unchanged). SA = 2(40 × 40) + 2(40 × 5) + 2(40 × 5) = 3200 + 400 + 400 = 4000 µm². SA/V = 4000 ÷ 8000 = 0.50 µm⁻¹, above 0.4, so it can meet demand (supply 120 against demand 96 units min⁻¹).
Suggested mark points (4): 1 for the derivation in (a); 1 for s = 15 µm; 1 for the cube’s 0.30 µm⁻¹ (or supply 72 < demand 96); 1 for the flat cell’s 0.50 µm⁻¹ with the conclusion. Interpretation: changing shape, not size, solved the problem, which is why many exchange cells are flat or folded.
How did you do?
- Q1 or Q2 wrong: rework the cube table and Figure 1 in the study guide; for cylinders, include both ends and use the radius.
- Q4 or Q7 wrong: rework Worked example 1, especially the supply-and-demand step.
- Q5 incomplete: re-read “Ways to keep SA/V high” and Worked example 2 on how projections add area.
- Q3 or Q6 wrong: re-read “From cells to whole organisms”: smaller bodies have a higher SA/V and a higher metabolic rate per unit mass.
Then tick off the topic checklist.