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. Data for every question: ATP hydrolysis releases 30 kJ per mole (written −30 kJ mol⁻¹); a + sign means energy is required and a − sign means energy is released. All organisms, pathways and data sets are fictional.
Question 1 (multiple choice · foundation)
A seedling grows from a seed into a plant with thousands of ordered cells. Which statement about this is consistent with the laws of thermodynamics?
- (A) The seedling creates new energy as it grows, so its order can increase.
- (B) The seedling becomes more ordered, while the heat and simple molecules it releases increase the disorder of its surroundings by a larger amount.
- (C) Living things are exempt from the second law of thermodynamics while they are alive.
- (D) The seedling converts all the light energy it absorbs into ordered structures, with no energy lost.
Answer and explanation
Answer: (B). An organism is an open system. It can increase its own order only by releasing heat and simple molecules such as CO₂ and H₂O, which raise the entropy of the surroundings. The total entropy still increases.
- (A) breaks the first law: energy cannot be created, only converted.
- (C) is false: every process in living things obeys the second law.
- (D) is impossible: every energy conversion loses some energy as heat.
Question 2 (multiple choice · core)
A reaction that builds a fictional cell-wall component requires +68 kJ per mole. What is the smallest number of ATP molecules whose hydrolysis can drive one mole of this reaction?
- (A) 2
- (B) 2.27
- (C) 3
- (D) 4
Answer and explanation
Answer: (C). Two ATP release 2 × 30 = 60 kJ mol⁻¹, which is less than 68, so the overall change would still be +8 kJ mol⁻¹. Three ATP release 90 kJ mol⁻¹, and the overall change is +68 − 90 = −22 kJ mol⁻¹, so the coupled process releases energy and can proceed.
- (A) rounds 68 ÷ 30 = 2.27 down. Two ATP do not supply enough energy.
- (B) is 68 ÷ 30. A cell cannot hydrolyse a fraction of an ATP molecule.
- (D) would work, but it is not the smallest number.
Question 3 (multiple choice · core)
Cells break down glucose through a long sequence of enzyme-catalysed steps rather than in one reaction. Which statement best explains the advantage?
- (A) The sequence of steps releases more total energy than a single reaction would.
- (B) Each step releases a small amount of energy, so much of it can be captured in ATP, and each step can be regulated through its own enzyme.
- (C) Splitting the reaction into steps means no energy is lost as heat.
- (D) Each step requires energy, so the cell can control how much ATP it uses.
Answer and explanation
Answer: (B). Small, separate releases can be captured instead of all being lost at once, and having one enzyme per step lets the cell control the rate.
- (A) breaks the first law: the start and end points are the same, so the total energy released is the same.
- (C) breaks the second law: every step releases some heat.
- (D) is wrong: the breakdown of glucose releases energy overall; that is why it can make ATP.
Question 4 (data analysis · core)
In a fictional bacterium, a pathway converts W into Z in three steps, each catalysed by its own enzyme:
W → (enzyme 1) → X → (enzyme 2) → Y → (enzyme 3) → Z
A researcher measures each compound before and 30 minutes after adding compound Q to the culture.
| Compound | W | X | Y | Z |
|---|---|---|---|---|
| Before Q / µmol g⁻¹ | 2.0 | 1.5 | 1.2 | 3.0 |
| After Q / µmol g⁻¹ | 2.1 | 6.8 | 0.2 | 0.4 |
(a) Calculate how many times higher the concentration of X is after adding Q, and the percentage decrease in Z. (b) Identify the enzyme most likely inhibited by Q. Justify your answer with the data. (c) Explain why W hardly changes.
Worked solution and suggested Marlbridge rubric
(a) X: 6.8 ÷ 1.5 = 4.5 times higher (4.53). Z: (3.0 − 0.4) ÷ 3.0 × 100 = 86.7% decrease (about 87%).
(b) Enzyme 2 (X → Y). X, the substrate of enzyme 2, builds up, while Y, its product, falls (1.2 → 0.2). Because each product is the reactant of the next step, less Y means less Z is made, so Z falls too.
(c) Enzyme 1 still works, so W is still converted to X. W is upstream of the block. (X keeps accumulating because it is made but no longer used.) The small rise in W, 2.0 to 2.1, is within what you would expect from normal variation.
| Point | What earns it |
|---|---|
| 1 | Both calculations correct (about 4.5 times; about 87% decrease) |
| 1 | Enzyme 2 identified |
| 1 | Justified with X rising and Y falling |
| 1 | Explains Z falls because Y is the reactant for the next step, and W unchanged because enzyme 1 still works |
Do not award the second point for enzyme 3: if enzyme 3 were blocked, Y would rise, not fall.
Question 5 (calculation · core)
A fictional insect larva is kept at 20 °C. Each day it eats food containing 240 J of chemical energy. 60 J leaves in its droppings and 150 J is released as heat by respiration. The rest is stored as new tissue.
(a) Calculate the energy stored each day, as a value and as a percentage of the intake. (b) At 30 °C, its heat loss from respiration rises to 195 J per day, while intake and droppings stay the same. Calculate the new daily balance and describe what happens to the larva’s body. (c) Assume the droppings always carry 25% of the intake. Calculate the minimum daily intake at 30 °C for the larva not to lose stored energy.
Worked solution and suggested Marlbridge rubric
(a) Stored = 240 − 60 − 150 = 30 J per day. Percentage = 30 ÷ 240 × 100 = 12.5%.
(b) Balance = 240 − 60 − 195 = −15 J per day. Losses now exceed input. The larva stops growing and must break down stored molecules (such as fat) to supply the missing energy. If this continues long enough, it cannot maintain its cells and dies.
(c) Energy absorbed = 0.75 × intake. For no loss of stores, 0.75 × intake = 195 J, so intake = 195 ÷ 0.75 = 260 J per day.
| Point | What earns it |
|---|---|
| 1 | 30 J and 12.5% |
| 1 | −15 J per day with correct sign or clear statement of a deficit |
| 1 | Links deficit to using stores and to loss of order or death if prolonged |
| 1 | 260 J per day |
A common error in (c) is 195 + 60 = 255 J, which keeps the droppings at 60 J instead of 25% of the larger intake.
Question 6 (argumentation · core)
Researchers compare three organisms: a bacterium, an archaeon and a eukaryotic alga.
| Feature | Bacterium | Archaeon | Alga |
|---|---|---|---|
| ATP is the main energy carrier | yes | yes | yes |
| Glycolysis-type pathway | yes | yes (a few enzymes differ) | yes |
| ATP synthase driven by a proton gradient | yes, plasma membrane | yes, plasma membrane | yes, in mitochondria and chloroplasts |
(a) Make a claim about the evolutionary relationship of these organisms. (b) Support your claim with evidence from the table and reasoning. (c) A student argues: “These features are the same only because every cell needs energy.” Explain why this does not account for the evidence.
Model answer and suggested Marlbridge rubric
(a) The three organisms, from all three domains, share a common ancestor.
(b) All three use ATP, a glycolysis-type pathway and an ATP synthase powered by a proton gradient. These are complex, many-step systems. Inheriting them from one ancestor that already had them is far more likely than three lineages inventing the same machinery separately. Because the pathways are essential, most changes to them were harmful, so they were conserved over very long times. The few differing archaeal enzymes show change within a shared pathway, which fits descent with modification.
(c) Needing energy explains why each organism must have some way of getting it, but not why they use the same carrier, the same sequence of steps and the same kind of enzyme. Many other chemical routes could release energy. Sharing the same detailed route points to shared ancestry.
| Point | What earns it |
|---|---|
| 1 | Claim of common ancestry |
| 1 | Evidence: at least two shared features from the table, across all three domains |
| 1 | Reasoning: independent origin of the same complex pathway is unlikely, and/or essential pathways are conserved |
| 1 | Rebuttal: need for energy does not explain why the specific mechanism is the same |
Question 7 (calculation and reasoning · stretch)
A fictional four-step pathway converts compound J into compound N. The energy change for each step, in kJ per mole, is:
| Step | J → K | K → L | L → M | M → N |
|---|---|---|---|---|
| Energy change / kJ mol⁻¹ | +14 | −22 | +9 | −31 |
(a) Calculate the overall energy change from J to N. (b) Identify the steps that cannot run on their own. Calculate the energy change of step J → K when it is coupled to the hydrolysis of one ATP. (c) Use the two laws of thermodynamics to explain where the energy released by the whole pathway goes. (d) Explain one advantage of this four-step pathway over a single reaction from J to N.
Worked solution and suggested Marlbridge rubric
(a) Overall = 14 − 22 + 9 − 31 = −30 kJ mol⁻¹ (energy released).
(b) Steps J → K (+14) and L → M (+9) require energy. Coupled to one ATP, J → K becomes +14 − 30 = −16 kJ mol⁻¹, so it can proceed. (L → M coupled to one ATP would be +9 − 30 = −21 kJ mol⁻¹.)
(c) First law: the 30 kJ mol⁻¹ released is not destroyed; it is converted. Part of it may be captured, for example in ATP or electron carriers in the energy-releasing steps. Second law: no transfer is perfect, so some of the released energy becomes heat that spreads into the surroundings and increases their disorder.
(d) Any one: energy is released in smaller amounts that can be captured rather than lost all at once as heat; each step has its own enzyme, so the rate can be regulated at one step; the intermediates K, L and M can be used by other pathways.
| Point | What earns it |
|---|---|
| 1 | −30 kJ mol⁻¹ with correct sign |
| 1 | Both energy-requiring steps named and −16 kJ mol⁻¹ for coupled J → K |
| 1 | First law (conserved, converted) and second law (some lost as heat, entropy rises) |
| 1 | One valid advantage, explained |
A common error in (a) is adding the sizes without signs (14 + 22 + 9 + 31 = 76 kJ mol⁻¹).
How did you do?
- Q1 or Q7(c) wrong: re-read “Order, energy and the laws of thermodynamics” in the study guide.
- Q2 or Q7(b) wrong: rework Worked example 1 and check your signs.
- Q3 or Q4 wrong: re-read “Energy pathways run in steps” and look again at Figure 2.
- Q5 wrong: rework Worked example 2; keep every loss in the budget.
- Q6 incomplete: re-read Worked example 3, especially the reasoning and its limits.
Then tick off the topic checklist.