Biology 9700/42 — October/November 2015

Background Concept

Percentage change expresses the difference between two values as a fraction of the starting (initial) value, then multiplies by 100 to convert to a percentage. The formula is:

The result is positive for an increase and negative for a decrease. This kind of calculation is common in respiration and photosynthesis data-handling questions where rates of gas exchange or product formation are tracked over time.

Understanding the Question

A line graph (Fig. 1.1) plots ethanol concentration in a yeast-contaminated sugar sample against time, with data points at 15, 30, 45 and 60 hours. The candidate must read the y-values at 15 h and 45 h, then compute the percentage rise in ethanol between those two times. The answer must be expressed as a percentage.

Approach

1. Read the y-value of the data point at 15 h — the lowest "x" mark, just above the x-axis, at approximately $0.25$ arbitrary units.
2. Read the y-value of the data point at 45 h — the highest point of the curve, at $5.2$ arbitrary units.
3. Substitute into the percentage-change formula and evaluate.

Step-by-Step Reasoning

• At 15 h: the data point lies between gridlines 0.2 and 0.3, very close to 0.25. The mark scheme uses $0.25$ as the accepted reading.
• At 45 h: the data point is plotted exactly on the gridline at $5.2$ (between 5.0 and 5.5, on the 0.2 subdivision).
• Difference: $5.2 - 0.25 = 4.95$ arbitrary units.
• Divide by initial value: $4.95 \div 0.25 = 19.8$.
• Multiply by 100: $19.8 \times 100 = 1980$ %.

The mark scheme awards 1 mark for correctly setting up the calculation $\frac{5.2 - 0.25}{0.25} \times 100$ (or equivalently $\frac{4.95}{0.25} \times 100$) and a second mark for the final answer 1980. Error carried forward (ecf) applies if a candidate misreads the 15 h value but then substitutes it consistently.

Key Takeaways

• Always read graph values against the y-axis gridlines carefully — here, the 0.25 is a sub-gridline reading rather than a major line.
• Percentage change uses the original (initial) value as the denominator, not the new value or the mean.
• Always quote the unit (%) with a percentage answer.
• CIE accepts ecf: an incorrect graph read is forgiven if the rest of the working is consistent.

Common Mistakes

• Reading 15 h as 0.2 (the nearest major gridline) and so getting $\frac{5.0}{0.2} \times 100 = 2500$ % — close, but not what the gridline subdivisions give.
• Dividing by the final value (5.2) instead of the initial value (0.25) — a wrong formula.
• Forgetting to multiply by 100, giving the answer as 19.8.
• Not including the % sign.

Things to Be Careful About

• The y-axis is labelled "arbitrary units" — the answer is therefore still expressed as a percentage, because the initial and final values are in the same units and cancel.
• The 15 h data point is slightly above the 0.2 gridline; reading it requires looking at the small subdivisions. A common acceptable reading is 0.2–0.3; the mark scheme uses 0.25 specifically.

Background Concept

Ethanol (CH3CH2OH) is a small, volatile, polar molecule. It has a relatively low boiling point (78 °C) and a high vapour pressure at room temperature, so an open or partially open aqueous solution loses ethanol to the atmosphere by evaporation. Many microorganisms — including some bacteria (e.g. Acetobacter) and certain yeasts themselves — can use ethanol as a respiratory substrate, oxidising it first to acetaldehyde and then to acetate, which enters the Krebs cycle.

Understanding the Question

The graph shows ethanol concentration peaking at 45 h and then falling sharply to 60 h. The candidate must give one plausible biological or chemical reason for this decrease.

Approach

Think of two broad categories of explanation:

1. Physical loss of ethanol from the system (evaporation into the air).
2. Biological consumption of ethanol by another organism in the sample (e.g. ethanol-oxidising bacteria, or other yeasts switching back to aerobic respiration once the sugar runs out).

Either earns the mark. Avoid vague answers such as "the reaction stopped" (which is wrong — the reaction is still occurring, but the rate of loss now exceeds the rate of production) or "the yeast died" (which is too specific without evidence).

Step-by-Step Reasoning

• After 45 h the sugar in the sap is largely depleted, so yeast fermentation slows and ethanol is no longer being made as fast.
• Meanwhile ethanol continues to leave the system: it evaporates because it is volatile, and/or it is consumed by other microbes (e.g. bacteria that can respire ethanol) that were present in the contaminated sample.
• Net effect: the ethanol concentration falls.

Key Takeaways

• A decrease in product concentration on a time graph needs a loss route (physical or biological), not just a halt in production.
• Ethanol is volatile — it can leave an aqueous solution at room temperature.
• Many microorganisms can use ethanol as a respiratory substrate.
• "Suggest" questions accept any biologically plausible reason, but each reason must be specific.

Common Mistakes

• "The yeast died" — possible, but without evidence; the mark scheme prefers an explicit loss mechanism.
• "The reaction stopped" — the reaction is continuing; the concentration is just falling because loss now exceeds production.
• "The ethanol turned back into sugar" — reverse reaction does not occur; the pathway is essentially irreversible under cellular conditions.
• Vague answers such as "because of the time" or "the concentration changed" — these are not biological explanations.

Things to Be Careful About

• Only one mark is available, so the candidate needs just one clear, specific reason. Either "ethanol evaporated" or "another microorganism metabolises ethanol" is sufficient.