IB Biology Paper 1B practice paper. Higher level, 2025 format.

• so that no light could drive the photosystems / electron transport (to pump protons or make ATP), so any ATP formed must be due to the pH difference alone, OWTTE;

• in tube 1 the proton concentration inside the thylakoids was (10 000 times) higher than outside / there was a proton gradient from the lumen to the outside;
• this is the same direction of gradient as electron transport sets up in illuminated thylakoids (protons concentrated in the lumen);
• ATP was formed (98 nmol per mg) only when this gradient was present; tubes 2 and 3, with the same pH on both sides, formed almost none (3 and 2);
• tubes 2 and 3 show that a low pH alone / the soaking and transfer alone do not cause ATP synthesis, OWTTE;
• there was no light and no electron transport, so no high-energy compound could have been made by electron transport;
• so a proton gradient alone is enough to drive ATP synthesis, as protons flow out through ATP synthase, as the chemiosmotic model predicts, OWTTE;

• tube 4: protons leak out across the membrane without passing through ATP synthase, so the gradient is lost without making ATP;
• tube 5: the gradient is in the opposite direction (higher proton concentration outside the thylakoids), so there is no flow of protons out of the lumen through ATP synthase, OWTTE;
• (tube 4 shows that) the membrane must be impermeable to protons for a gradient to drive ATP synthesis;

Award one mark for each tube; the third point may replace the tube 4 point.

• protons flowing out of the lumen through ATP synthase lower the proton concentration inside, so the gradient disappears;
• in darkness nothing pumps protons back into the lumen, so the gradient is not restored;
• in the light, excited electrons from the photosystems pass along the chain of electron carriers, and the energy released pumps protons from the stroma into the lumen;
• photolysis of water inside the lumen releases protons, adding to the gradient;
• the small volume of the lumen means that pumping relatively few protons quickly rebuilds a steep gradient, OWTTE;

Content pivot. Award [2 max] for answers that only explain why the synthesis stopped, or only how it continues in light.

• jar 3 at 15 °C (1.6 µg g⁻¹ h⁻¹);
• the jar was not airtight / the lid leaked, so carbon dioxide escaped OR the sensor was faulty / misread OR that portion of peat contained less organic matter / more water / stones OR the incubator was not at 15 °C, OWTTE;

• to prevent photosynthesis by any mosses / algae in the peat, which would take up carbon dioxide;
• (so) the measured rate would be lower than the true rate of release by decomposition, OWTTE;

• water content / moisture of the peat;
• volume of the jar / volume of air above the peat;
• the place in the bog / the core that the peat came from (or peat mixed before dividing it);
• the time and conditions of storage between collecting the peat and sealing the jars;
• the time the peat was left to reach the incubator temperature before the first reading;

Mark the first two only. Do not accept mass of peat, depth of collection, light, temperature or the 24 hours, which are already described.

• in the bog the peat is waterlogged so little oxygen reaches it, whereas the jars held air, so decomposition in the jars may be faster than in the bog, OWTTE;
• only carbon dioxide was measured, but waterlogged peat also releases carbon as methane;
• in the bog the temperature varies from day to night and through the year rather than staying constant;
• 24 hours is a short time / disturbing the peat may give a brief burst of decomposition that does not last;
• peat from one depth at one site may not represent the whole bog / other bogs;
• in the bog living plants (e.g. *Sphagnum*) also take up carbon dioxide, so the net loss depends on photosynthesis as well as decomposition;

• vary the water content of the peat, e.g. peat kept covered with water in some jars and allowed to drain in others / several levels of water content;
• keep the temperature the same in every jar (e.g. 15 °C), with the other variables controlled as before;
• measure the rate of release of carbon dioxide (and methane) in the same way, with at least three jars for each treatment;
• compare the mean rates: a higher rate in drained peat would support the idea, OWTTE;

• 1.5 µm⁻¹ (surface area 50.3 µm² ÷ volume 33.5 µm³);

Units required. Accept 1.50 µm⁻¹. Accept 1.5 : 1 with the unit µm⁻¹ stated.

• the surface area-to-volume ratio of a 40 µm cell is 0.15 µm⁻¹ / one tenth of that of a 4 µm cell (ECF from (a));
• uptake per unit volume falls from about 3.2 at 4 µm to about 0.3 at 40 µm / by a factor of about 10, the same factor as the ratio, so the data support the claim;
• other pairs of species fit as well, e.g. from 2 µm to 20 µm the ratio falls tenfold and uptake falls from about 5.6 to about 0.6 / uptake ÷ surface area-to-volume ratio is about 2 for every species;
• but the species differ in other ways (e.g. rate of growth / uptake capacity of each unit of membrane), so the correlation does not show that the ratio is the cause, OWTTE;
• the cells are only roughly spherical, so the ratios calculated from the diameter are estimates;
• only eight species / one value per species / no replicates or error bars are shown, so the strength of the relationship is uncertain;

Accept readings of 3.1 to 3.3 at 4 µm and 0.2 to 0.4 at 40 µm, and a factor of about 8 to 16; the 2 µm and 20 µm points (5.5 to 5.7 and 0.55 to 0.65) can be read more closely. For 3 marks the answer must include both support from the data and at least one limitation.

• small cells have a larger surface area-to-volume ratio, so they take up more nitrate per unit volume (e.g. about 5.6 at 2 µm against about 0.3 at 40 µm);
• where nitrate is scarce, small cells can still obtain enough nitrate for their volume to grow and divide, whereas large cells grow more slowly / are outcompeted, OWTTE;
• where nitrate is plentiful, uptake across the surface can meet the needs of a larger volume, so large cells are no longer at a disadvantage, OWTTE;

• 1.19 µm⁻¹ (volume 2 × 33.5 = 67.0 µm³, so diameter = 4 × ∛2 = 5.04 µm; surface area 79.8 µm² ÷ 67.0 µm³, or 6 ÷ 5.04);

Units required. Accept 1.2 µm⁻¹. ECF from the volume calculated in (a). Do not accept 0.75 µm⁻¹ (the diameter doubled, or the volume doubled with the surface area unchanged).

• as the cell grows its surface area-to-volume ratio falls from 1.5 to 1.19 µm⁻¹ / by about a fifth, because the volume rises faster than the surface area (ECF from (a) and (d));
• the need for nitrate depends on the volume / amount of cytoplasm, but uptake depends on the area of plasma membrane, so uptake per unit volume falls as the cell grows, OWTTE;
• a cell that went on growing would have an ever lower ratio, until uptake across its surface could not supply its cytoplasm;
• division restores the ratio of each new cell to 1.5 µm⁻¹ / to its value after the previous division, OWTTE;

• measures cell width and scale bar on the drawing and divides (cell width ÷ scale bar length) / working shows the ratio of the two measurements;
• 60 (μm);

Accept 55–65 μm. Units required for the second mark. The ratio method scores the first mark even if the arithmetic then slips.

• II: chloroplast;
• III: (large central) vacuole;

Do not accept 'vesicle' for III.

• plant (cell);
• because it has a cell wall (I) OR chloroplasts (II) OR a large central vacuole (III), none of which occur in animal cells;

The reason mark requires a named structure from the drawing. No mark for the reason if the deduction is wrong.

• (nucleus:) contains the chromosomes/DNA, whose genes control the activities of the cell, OWTTE;

Accept: site of DNA replication / transcription. No mark for naming the structure without a function.