Biology 9700/54 — May/June 2025

Background Concept

A solution of stated molarity contains a fixed number of moles of solute per cubic decimetre (dm³) of solution. By definition, 1 mol dm⁻³ requires one mole (in grams equal to the molecular mass) dissolved in 1 dm³. A 1 × 10⁻³ mol dm⁻³ solution therefore requires 1/1000 of that mass in 1 dm³. The diluent must be pure (distilled or deionised) water because tap-water salts and chlorine could interfere with the grains' germination.

Understanding the Question

You are told the molecular mass of ABA is 264 g and that the student must prepare a stock solution of concentration 1 × 10⁻³ mol dm⁻³. The mark scheme credits two points: the correct diluent and the correct mass in 1 dm³.

Approach

Use
$\text{mass} = \text{molarity} \times \text{molecular mass} \times \text{volume (in dm}^3\text{)}$
For 1 dm³ of 1 × 10⁻³ mol dm⁻³ ABA: 1 × 10⁻³ × 264 × 1 = 0.264 g. State the diluent, the mass and the final volume.

Step-by-Step Reasoning

1. 1 dm³ of a 1 × 10⁻³ mol dm⁻³ solution must contain 1 × 10⁻³ mol of ABA.
2. The molecular mass of ABA is 264 g mol⁻¹, so 1 × 10⁻³ mol weighs 0.264 g.
3. Dissolve 0.264 g of ABA in distilled (deionised) water.
4. Make the solution up to exactly 1 dm³ in a volumetric flask.
5. The diluent must be distilled/deionised water so contaminants do not affect germination.

Key Takeaways

• Molarity (mol dm⁻³) × molecular mass (g) × volume (dm³) gives the mass of solute.
• A stock solution is the concentrated solution from which all working dilutions are subsequently prepared.
• Distilled/deionised water is required to avoid confounding variables.

Common Mistakes

• Stating 264 g (using the molecular mass directly, ignoring the 1 × 10⁻³ factor).
• Naming the volume but not the mass, or the mass but not the diluent.
• Saying "water" without qualifying it as distilled/deionised.

Things to Be Careful About

• "Stock solution" means concentrated; this is then diluted in part (a)(ii).
• The mass should be measured on a balance, not assumed, and the final volume should be made up precisely in a volumetric flask.

Background Concept

A serial dilution is a sequence of ten-fold (or other factor) dilutions, each made by diluting a fixed fraction of the previous solution into the next volume. It is much faster and more accurate than weighing out four separate masses. Each step reduces the concentration by the dilution factor (here 10), and the concentration is calculated by:

where V₁ is the volume transferred and V₂ is the total volume after the diluent is added.

Understanding the Question

The student has a 1 × 10⁻³ mol dm⁻³ stock and needs four working concentrations (1 × 10⁻⁴, 1 × 10⁻⁵, 1 × 10⁻⁶ and 1 × 10⁻⁷ mol dm⁻³), each made up to a final volume of 100 cm³. You must describe how to make these dilutions, but should not repeat the procedure for making the stock itself (that was part (a)(i)).

Approach

Recognise that consecutive ten-fold dilutions are the most efficient method. Show how the first dilution is performed (10 cm³ stock into 90 cm³ water to give 100 cm³ of 1 × 10⁻⁴), then describe one subsequent step (10 cm³ of that into 90 cm³ water to give 1 × 10⁻⁵, and so on).

Step-by-Step Reasoning

1. First dilution: take 10 cm³ of the 1 × 10⁻³ mol dm⁻³ stock and add it to 90 cm³ of distilled water. The total volume is 100 cm³ and the concentration has fallen by a factor of 10, giving 1 × 10⁻⁴ mol dm⁻³ ABA.
2. Second dilution: take 10 cm³ of this 1 × 10⁻⁴ solution and add to 90 cm³ of distilled water → 1 × 10⁻⁵ mol dm⁻³ ABA.
3. Third dilution: take 10 cm³ of the 1 × 10⁻⁵ solution and add to 90 cm³ of distilled water → 1 × 10⁻⁶ mol dm⁻³ ABA.
4. Fourth dilution: take 10 cm³ of the 1 × 10⁻⁶ solution and add to 90 cm³ of distilled water → 1 × 10⁻⁷ mol dm⁻³ ABA.
5. Each dilution should be mixed thoroughly (e.g. using a separate glass rod or by shaking a stoppered flask) before the next transfer.

Key Takeaways

• A serial dilution multiplies small errors if the same pipette or rod is reused without rinsing — use clean, rinsed glassware between steps.
• Distilled/deionised water is the diluent of choice.
• Each working concentration must be made up to exactly 100 cm³ so that subsequent additions to the grains are equal in volume.

Common Mistakes

• Adding 1 cm³ to 99 cm³ (which is also correct, but the mark scheme prefers 10 + 90) — either is acceptable.
• Stating the final concentration but not the volumes used.
• Repeating the stock-solution preparation in this part — the question explicitly forbids it.
• Forgetting to mix between dilutions.

Things to Be Careful About

• Volumes must sum to 100 cm³ for each working solution.
• A separate, clean pipette (or rinsed one) should be used at each step to avoid carry-over contamination.

Background Concept

The independent variable (IV) is the one the experimenter deliberately changes; the dependent variable (DV) is what is measured to record the effect of that change. Anything else that could influence the result must be kept constant — these are the controlled variables. Confusing the IV with the DV is the single most common mistake in planning questions.

Understanding the Question

The student has soaked the grains in a fixed concentration of GA and is now adding different concentrations of ABA. The IV must therefore be the concentration of ABA. The DV is the response of the wheat grains — how many germinate, or alternatively how much the shoot grows. The mark scheme accepts either answer for the DV.

Approach

Read the stem carefully: the student varies the ABA concentration and observes germination (and, from part (c), shoots are also measured). The variable deliberately altered is the IV; the outcome measured is the DV.

Step-by-Step Reasoning

1. The student makes several different ABA solutions (1 × 10⁻⁴, 1 × 10⁻⁵, 1 × 10⁻⁶ and 1 × 10⁻⁷ mol dm⁻³) and adds each to a different group of 50 grains.
2. The variable deliberately changed from one group to the next is the concentration of ABA — so this is the IV.
3. The variable measured at the end of the experiment — how many grains have germinated, or how long the shoots are — is the DV.
4. Everything else (GA pre-soak time, GA concentration, temperature, light, time of measurement, number of grains per group) must be standardised as controlled variables.

Key Takeaways

• IV = what you change; DV = what you measure.
• The mark scheme accepts "number of grains germinated" OR "length of shoot" as the DV.
• All other conditions must be controlled to make a fair test.

Common Mistakes

• Saying "amount of ABA" rather than "concentration of ABA" — concentration is the precise scientific term.
• Confusing IV and DV — e.g. writing the number germinated as the IV.
• Adding "time" as the IV when the question is about different concentrations.

Things to Be Careful About

• "Concentration" must be used; "amount" is too vague.
• The independent variable is what is varied, not what is added (even though ABA is added).

Background Concept

A valid experiment must include (i) a control that receives every treatment except the IV, (ii) standardisation of every other variable, (iii) replication so a mean can be calculated, (iv) a clear, quantitative measurement of the DV at a fixed time, and (v) appropriate safety. Each of these earns marks under the mark scheme's "quality of results" category.

The biology: ABA inhibits germination; without light and at the right temperature (10–25 °C), wheat germinates best. So the experiment must be conducted in the dark, at a controlled temperature within that range, and germination must be assessed after a known period.

Understanding the Question

You are asked to write a procedure (NOT including how to dilute the stock — that was part (a)(ii)) that another person could follow. Five marks are available from a list of seven creditable points in the mark scheme. Cover as many as you can; aim for at least five distinct, well-stated points.

Approach

Structure the procedure logically:

• Control
• Set-up of each group
• Standardisation of environmental variables (dark, temperature, volume)
• Measurement at a fixed time
• Safety

Step-by-Step Reasoning

1. Control. Set up one extra group of 50 GA-soaked grains and add the same volume of distilled water instead of any ABA solution. This shows the effect of GA alone, against which ABA treatments can be compared.
2. Substrate. Place the grains on moist filter paper, cotton wool or soil in a labelled container (e.g. Petri dish) — the substrate must support germination.
3. Dark. Place all containers in the dark (e.g. in a cupboard or wrapped in black paper / foil) because wheat germinates best without light.
4. Constant temperature. Use an incubator or a temperature-controlled room set to a constant value within 10–25 °C (e.g. 20 °C). This rules out temperature as a confounding factor.
5. Equal volumes. Add the same stated volume (e.g. 20 cm³) of each ABA solution to each container so that any difference is due to concentration, not volume.
6. Measurement at a fixed time. After a stated, identical time period (e.g. 7 days) for every group, count the number of grains that have germinated (those with an emerged shoot and root), or measure shoot length.
7. Safety. ABA is an irritant / allergen. Hazard = ABA; risk = skin/eye irritation or allergic reaction; precaution = wear gloves and a face mask (PPE). Wheat grains themselves may also cause allergies — same PPE applies.

Key Takeaways

• A control is essential for a fair test.
• All variables other than the IV must be standardised.
• The DV must be measured quantitatively at a single, fixed time.
• Safety must state hazard + risk + precaution.
• Procedure should be written so that another person could replicate it without further instruction.

Common Mistakes

• Omitting the control.
• Failing to keep temperature constant or to specify a temperature within 10–25 °C.
• Saying "leave in a warm place" instead of naming a controlled temperature.
• Vague safety — "be careful" or "human error" scores nothing; you must name a hazard, a risk and a precaution.
• Forgetting to state a time for measurement.

Things to Be Careful About

• "Constant temperature" or "stated temperature" within the 10–25 °C window is required.
• The control is the GA-only group (no ABA), not "no GA".
• Use precise volumes (cm³) and times (days).

Background Concept

Percentage germination standardises the count to a 0–100 scale, allowing fair comparison between groups that started with the same number of grains (here, 50 per group). It expresses the proportion that succeeded as a fraction of 100.

Understanding the Question

You are told the student divided the grains into groups of 50 and treated each with a different ABA concentration. You must state the formula used to convert the raw count of germinated grains into a percentage.

Approach

Use the universal percentage formula:

Apply it to germination: part = grains that germinated, whole = total grains in the group.

Step-by-Step Reasoning

1. After the experiment, the student counts the grains in each group that have germinated (e.g. shoot and root emerged).
2. The total per group is 50 (given in the stem).
3. The percentage germination is therefore:

1. This yields a value between 0 % and 100 % for each ABA concentration.

Key Takeaways

• Percentage = (part / whole) × 100.
• Standardising the starting number (here 50 per group) is what makes the comparison valid.

Common Mistakes

• Writing "divide by 50" without multiplying by 100.
• Using the wrong denominator (e.g. dividing by number germinated).
• Forgetting the ×100 step.

Things to Be Careful About

• Mark scheme requires the formula to be stated as "number germinated ÷ total × 100" — not just "mean of 50".

Background Concept

When asked to predict a graph, you must use information already given. Here the stem tells you ABA inhibits germination and that it is active in plant tissues between 1 × 10⁻⁶ and 1 × 10⁻⁴ mol dm⁻³, with the concentrations tested ranging from 1 × 10⁻⁷ to 1 × 10⁻⁴ mol dm⁻³. Therefore as the concentration of ABA rises, the percentage germination should fall — the highest concentrations prevent germination entirely, while very low concentrations (1 × 10⁻⁷) have negligible effect and germination is essentially 100 %.

Understanding the Question

You are asked to complete the blank axes of Fig. 1.2 by sketching the graph you would expect. Three marks: correctly oriented axes with labels and units (1), correct values on the x-axis (1), correct trend line (1).

Approach

• The IV (concentration of ABA) goes on the x-axis.
• The DV (percentage germination) goes on the y-axis, scaled 0–100.
• The x-axis values must match the range of ABA concentrations used in the experiment (1 × 10⁻⁷ to 1 × 10⁻⁴ mol dm⁻³), and the points can be marked at 1 × 10⁻⁷ and at 1 × 10⁻⁴ (or 1 × 10⁻³).
• The line must show percentage germination falling as ABA concentration rises.

Step-by-Step Reasoning

1. y-axis: label "percentage germination" (no units — it is a %). Scale from 0 to 100.
2. x-axis: label "concentration of ABA / mol dm⁻³". Mark 1 × 10⁻⁷ and (at least) 1 × 10⁻⁴ — these are the extreme concentrations the student actually tested.
3. Sketch a smooth curve starting near 100 % at the lowest ABA concentration (1 × 10⁻⁷ mol dm⁻³ — well below the active range, so GA still promotes germination) and decreasing to near 0 % at the highest ABA concentration the student tested (1 × 10⁻⁴ mol dm⁻³ — within the active range, so germination is strongly inhibited).
4. The curve should be concave-up (steeper fall at the higher end), reflecting the biological fact that inhibition increases with concentration up to a maximum.

Key Takeaways

• Always place the IV on the x-axis and the DV on the y-axis.
• Label each axis with both quantity AND unit.
• The predicted trend must be biologically justified — here, ABA inhibits germination, so high [ABA] means low germination.

Common Mistakes

• Swapping the axes (concentration on y, germination on x).
• Omitting units (especially mol dm⁻³).
• Drawing a straight line instead of a curve.
• Drawing the curve the wrong way (rising) because of confusion with GA.
• Marking x-axis values that do not match the actual concentrations used.

Things to Be Careful About

• The x-axis range must include the concentrations used in the experiment.
• The curve does not have to pass through specific numerical points — it is a sketch showing trend.

Background Concept

The standard error of the mean (SEM) estimates how far the sample mean is likely to lie from the true population mean. It is calculated from the standard deviation divided by √n. A small SE means the sample mean is a precise estimate; a large SE means it could be quite far from the true value. Means with non-overlapping SEs are likely to be significantly different; means whose SEs overlap substantially are unlikely to be so.

Understanding the Question

You are given Table 1.1 with five mean shoot lengths, each ± a standard error. You must explain what the SE tells you about the reliability of the means, ideally with reference to actual values in the table.

Approach

State the general principle (smaller SE = more reliable mean), then give a worked example from the table (e.g. compare the SEs of two concentrations and comment on overlap / non-overlap).

Step-by-Step Reasoning

1. The SE is a measure of how precisely the sample mean estimates the true mean of the population.
2. A smaller SE means that the mean is more reliable — repeated samples would give similar values.
3. Equivalently, a smaller SE implies the sample mean is closer to the true population mean.
4. From Table 1.1, the smallest SE is ±5 mm (at 1 × 10⁻⁷ mol dm⁻³), so that mean (150 mm) is the most reliable.
5. Comparing 1 × 10⁻⁷ mol dm⁻³ (150 ± 5 mm) and 1 × 10⁻⁶ mol dm⁻³ (175 ± 15 mm): the SEs do not overlap (150 ± 5 = 145–155; 175 ± 15 = 160–190), so the difference between these two means is likely to be statistically significant.
6. Conversely, 1 × 10⁻⁶ mol dm⁻³ (175 ± 15 mm) and 1 × 10⁻⁴ mol dm⁻³ (180 ± 8 mm) have SEs that overlap (175 ± 15 = 160–190; 180 ± 8 = 172–188), so those means are unlikely to differ significantly.

Key Takeaways

• SE indicates precision of the mean estimate; the smaller, the more reliable.
• Non-overlapping SEs are a useful (though not strict) indication of a significant difference.
• Always quote specific values from the table to support your interpretation.

Common Mistakes

• Defining SE without relating it to the mean's reliability.
• Failing to quote values from Table 1.1.
• Confusing standard error with standard deviation (SD measures spread of individuals; SE measures precision of the mean).

Things to Be Careful About

• The mark scheme requires both the principle AND a numerical example.
• Do not overstate: SE overlap is suggestive but the proper test is the t-test (part d iii).

Background Concept

A null hypothesis (H₀) is a statement of "no effect" or "no difference" between the groups being compared. It is what a statistical test either rejects or fails to reject. For a t-test comparing two means, the null hypothesis is that the two population means are equal. The alternative hypothesis (H₁) is the converse, that there IS a difference.

Understanding the Question

You are told the student ran a t-test comparing the no-GA group with the 1 × 10⁻⁵ mol dm⁻³ GA group. From Table 1.1 the means are 120 mm (no GA) and 210 mm (1 × 10⁻⁵ mol dm⁻³). You must state the null hypothesis for this comparison.

Approach

A null hypothesis always specifies:

• The two groups being compared
• The variable measured
• A statement that there is NO difference between them.

Step-by-Step Reasoning

1. The two groups are: (a) wheat grains with no GA treatment, and (b) wheat grains treated with 1 × 10⁻⁵ mol dm⁻³ GA.
2. The variable measured is the mean shoot length.
3. Therefore the null hypothesis is: there is no significant difference between the mean shoot length of these two groups.
4. The wording must include both groups and the variable; do not simply say "GA has no effect".

Key Takeaways

• Null hypothesis = statement of no difference.
• It must mention both groups and the variable being compared.
• It is what the t-test either rejects (significant difference) or fails to reject (no significant difference).

Common Mistakes

• Phrasing it as a question ("is there a difference ...?") — a hypothesis is a statement.
• Omitting one of the two groups.
• Saying "no difference" without saying "no significant difference" (though "no difference" alone is usually accepted by CIE).

Things to Be Careful About

• The mark scheme requires the wording to mention both treatments and the shoot length.

Background Concept

The t-test assesses whether the difference between two sample means is statistically significant. The result is a calculated t-value; this is compared against a critical value (looked up in a t-table) at a chosen probability (p = 0.05 is conventional) and the appropriate degrees of freedom.

Degrees of freedom for a two-sample (unpaired) t-test:

where n₁ and n₂ are the sample sizes of the two groups. The critical value decreases as df increases.

The decision rule:

• If |calculated t| > critical t → reject H₀ (significant difference).
• If |calculated t| ≤ critical t → accept H₀ (no significant difference).

Understanding the Question

Each group of wheat grains contained 20 grains (n = 20), so each t-test comparison has df = 20 + 20 − 2 = 38. You must explain the three steps to determine whether the difference is significant.

Approach

Lay out the procedure in order:

1. Calculate df from sample sizes.
2. Look up the critical t-value at p = 0.05 for that df.
3. Compare the calculated t with the critical t and state the decision.

Step-by-Step Reasoning

1. Sample sizes: each group has 20 wheat grains, so n₁ = 20 and n₂ = 20.
2. Degrees of freedom:

1. From a t-table, the critical value at p = 0.05 and df = 38 is approximately 2.024 (one-tailed) or 2.024 for a two-tailed test (the mark scheme accepts either orientation).
2. The student calculates a t-value for the two groups from their means and SEs.
3. If the calculated t-value exceeds 2.024, the difference is significant at the 5 % level — reject the null hypothesis.
4. If the calculated t-value is less than or equal to 2.024, accept (fail to reject) the null hypothesis — no significant difference.

In this investigation, the means are very different (120 vs 210 mm at the peak), so the calculated t is large and H₀ is rejected — confirming that GA significantly increases shoot length at 1 × 10⁻⁵ mol dm⁻³.

Key Takeaways

• df = n₁ + n₂ − 2 for an unpaired two-sample t-test.
• Compare calculated t with the critical value at p = 0.05 for that df.
• Reject H₀ if calculated t > critical t; otherwise accept it.

Common Mistakes

• Using the wrong df formula (e.g. n₁ × n₂ or n₁ + n₂).
• Comparing the calculated t with a critical value at the wrong probability (e.g. p = 0.01).
• Stating "accept" rather than "fail to reject" — both are accepted by CIE, but be clear about the logic.

Things to Be Careful About

• The mark scheme requires three linked points: df, comparison with critical value at p = 0.05, and the reject/accept decision.
• Critical value depends on df; do not quote a single fixed number like 1.96 unless df is appropriate.

Background Concept

The arithmetic mean summarises a set of repeated measurements. For $n$ measurements $x_1, x_2, \dots, x_n$, $\bar{x} = \dfrac{\sum x_i}{n}$. Quoting the mean with the same units and decimal places as the original readings keeps the table consistent and shows the reader the precision of the data.

Understanding the Question

The four trial values for 'mitochondria with succinate' are 50.22, 49.10, 48.53 and 50.15 (all in $\text{nmol min}^{-1}$). The mean cell in Table 2.1 is blank and needs filling.

Approach

Add the four values and divide by 4. Give the answer to two decimal places to match the precision of the data.

Step-by-Step Reasoning

$50.22 + 49.10 = 99.32$; $99.32 + 48.53 = 147.85$; $147.85 + 50.15 = 198.00$. Dividing by 4: $198.00 / 4 = 49.50$. Units $\text{nmol min}^{-1}$ as given in the column heading.

Key Takeaways

A mean must be quoted with the units of the original quantity and (here) to 2 d.p. to match the data. A quick reasonableness check: the four values cluster at 49–50, so a mean just under 50 is sensible.

Common Mistakes

Forgetting the unit, rounding incorrectly (e.g. to one significant figure), or dividing by the wrong number of trials.

Things to Be Careful About

Use the exact figures from the table; do not introduce rounding errors at the intermediate step.

Background Concept

An 'anomalous' result is a reading that does not fit the pattern shown by the other replicates — it lies well outside the cluster formed by the other trials. With only four trials, judgement is made from how far a single value deviates from the cluster.

Understanding the Question

Each row of Table 2.1 has four trial values and a mean. Candidates must circle two values that look anomalous (one mark each).

Approach

Look at each row separately and ask which value lies furthest from the cluster of the other three.

Step-by-Step Reasoning

Row 1 (mitochondria alone): 0.03, 0.02, 0.03, 0.01 — all close to 0.02; no clear anomaly.
Row 2 (with succinate): 50.22, 49.10, 48.53, 50.15 — all lie between 48.5 and 50.2; no clear anomaly.
Row 3 (with ADP and succinate): 139.23, 170.10, 142.67, 138.10 — three values cluster at ~139–143 while trial 2 (170.10) sits ~30 units above them. 170.10 is anomalous.
Row 4 (with cyanide, ADP and succinate): 0.00, 0.10, 0.00, 0.01 — three values are essentially 0, while trial 2 (0.10) is an order of magnitude higher. 0.10 is anomalous.

Both anomalies happen to come from trial 2, suggesting a procedural issue specific to that run (e.g. a leaky seal, incomplete cyanide addition, or contamination with an oxidisable substrate).

Key Takeaways

Identifying anomalies is a judgement based on the spread of replicates; in this paper the marks go to the candidate's reasoning, not to a strict statistical test.

Common Mistakes

Circling a value in the 'mitochondria alone' row (those values are all very small and within noise) or circling two values in the same row (only one per row is anomalous).

Things to Be Careful About

CIE normally wants the two values to come from different rows so the candidate has actually inspected every row. One mark per correct circle.

Background Concept

A single mean hides the spread of the underlying data. Two refinements standard in CIE Paper 5 are:
• Excluding anomalies before re-computing the mean, on the grounds that one rogue value can pull the mean well away from the bulk of the readings.
• Reporting the dispersion — standard deviation (sd), standard error of the mean ($\text{SEM} = \text{sd}/\sqrt{n}$), or a 95% confidence interval (mean $\pm$ 2 $\times$ SEM) — so a reader can see whether replicates agree.

Understanding the Question

Part (ii) identified 170.10 and 0.10 as probable anomalies. The question asks for two processing steps that handle these without throwing away all the data.

Approach

One approach acts on the anomalies directly; the other describes the data around them.

Step-by-Step Reasoning

Approach 1 — exclude and re-average. Removing the two circled values and re-averaging the remaining three trials in each row gives a 'cleaner' mean that is not pulled by the rogue reading.
Approach 2 — quantify the spread. Computing a standard deviation (or standard error, or 95% confidence limits) for each row gives a numerical measure of the spread and an indication of which rows are reproducible and which are not. These statistics are valid whether or not the anomalies are kept.

Key Takeaways

Discarding a possible anomaly is not the same as discarding all the data; statistical description of the spread is the complementary habit.

Common Mistakes

Saying 'repeat the experiment' or 'take more readings' — these are improvements to the method, not processing steps applied to the data already collected. Saying 'calculate a percentage error' — not on the mark scheme here.

Things to Be Careful About

The two marks must be distinct, processing-based responses. CIE wording matters: 'standard deviation', 'standard error' and '95% confidence limits' are individually credit-worthy; pairing them in one answer only counts once.