Biology 9700/51 — October/November 2025

Background Concept

A dilution is calculated using $C_1 V_1 = C_2 V_2$, where $C$ is concentration and $V$ is volume. To make a working solution from a more concentrated stock, a small volume of stock is added to a larger volume of diluent (the solvent mixture here) so that the final volume and concentration match the target. The dilution factor ($C_\text{stock} / C_\text{final}$) tells you the ratio of stock to total final volume.

Understanding the Question

The student has $0.1\ \text{mol dm}^{-3}$ juglone stock and needs to make $500\ \text{cm}^3$ of $1.0 \times 10^{-3}\ \text{mol dm}^{-3}$ juglone solution in a single dilution, using measuring cylinders. The question asks for the volumes used.

Approach

Apply $C_1 V_1 = C_2 V_2$:

• $C_1 = 0.1\ \text{mol dm}^{-3}$ (stock)
• $C_2 = 1.0 \times 10^{-3}\ \text{mol dm}^{-3}$ (final)
• $V_2 = 500\ \text{cm}^3$ (final volume)
  Solve for $V_1$. The remainder of the $500\ \text{cm}^3$ is solvent.

Step-by-Step Reasoning

So $5\ \text{cm}^3$ of the $0.1\ \text{mol dm}^{-3}$ juglone stock is required, plus $500 - 5 = 495\ \text{cm}^3$ of the solvent mixture, to make the final volume up to $500\ \text{cm}^3$ at the target concentration.

Key Takeaways

• The dilution equation $C_1 V_1 = C_2 V_2$ is the standard tool for working out dilution volumes.
• The dilution factor (here 100-fold) tells you the ratio of stock to total final volume.
• The diluent (solvent) volume = final volume − stock volume.

Common Mistakes

• Forgetting that the final volume is fixed at $500\ \text{cm}^3$ and adding the stock ON TOP of the solvent.
• Using the wrong dilution factor (e.g. 10-fold instead of 100-fold, because of misreading the powers of ten).
• Misplacing decimal points on the concentrations.

Things to Be Careful About

State both volumes clearly with correct units and the qualifier ($0.1\ \text{mol dm}^{-3}$ juglone / solvent mixture) so that the two volumes cannot be confused.

Background Concept

Percentage error is inversely related to the precision of the apparatus used. Percentage error = (absolute apparatus error ÷ measured value) × 100. Measuring cylinders typically have an absolute error of ±0.5 to ±1 cm³, whereas a graduated pipette, burette, or volumetric glassware has a much smaller absolute error. Because the same absolute error is a smaller proportion of a larger volume, simply using larger volumes also reduces the percentage error.

Understanding the Question

The stem states that the student used measuring cylinders to perform the dilution. The question asks for ONE improvement that would reduce the percentage error.

Approach

Either: switch to more precise apparatus (a graduated pipette, syringe, micropipette or burette to measure the small juglone volume; a volumetric flask to make up to the mark), OR scale up both volumes so that the apparatus error is a smaller fraction of the measured value.

Step-by-Step Reasoning

A measuring cylinder measuring 5 cm³ has a typical error of ±0.5 cm³, i.e. about 10% error. A graduated pipette or burette can measure 5 cm³ with an error of ±0.05 cm³, i.e. about 1% error. Alternatively, doubling the scale (preparing 1000 cm³ instead of 500 cm³) halves the percentage error from the apparatus.

Key Takeaways

• Smaller-volume measurements need more precise apparatus.
• Percentage error = (absolute error ÷ measured value) × 100.
• Volumetric glassware is more accurate than measuring cylinders.

Common Mistakes

• Suggesting "use a measuring cylinder" (the apparatus already used).
• Naming apparatus without any link to reducing percentage error.
• Listing several improvements when the question asks for ONE.

Things to Be Careful About

The question asks for one improvement only, not many. State it clearly and identify which stage of the dilution (measurement of stock, or making up to volume) it improves.

Background Concept

A control is a treatment that lacks the factor under investigation but is otherwise identical to the experimental treatment. It is run alongside the experimental group so that the effect of the independent variable can be isolated from confounding factors such as the solvent, the apparatus, or the environmental conditions.

Understanding the Question

The student compares cucumber seeds in juglone solution against a "control treatment". The question asks what that control should be and why it is included.

Approach

The control must differ from the juglone treatment by ONLY the juglone. Because juglone is dissolved in a solvent mixture, the solvent mixture alone (with no juglone) is the correct control. The purpose of the control is to confirm that any difference in outcome is caused specifically by the juglone.

Step-by-Step Reasoning

Running the solvent mixture alone alongside the juglone treatment allows the student to confirm that any differences in germination percentage or seedling growth are caused by juglone and not by the solvent, the volume of liquid added, the soil, or the environmental conditions. Without this comparison, any observed difference could not be confidently attributed to juglone.

Key Takeaways

• A control must differ from the experimental treatment by ONLY the independent variable.
• The control provides the baseline against which the experimental treatment is compared.
• In this case, the correct control is the solvent mixture alone - NOT pure water, because the solvent itself could affect germination.

Common Mistakes

• Suggesting "water" or "no treatment" rather than the solvent mixture (since juglone is dissolved in a solvent mixture, the solvent alone is the correct control).
• Stating the reason vaguely as "to compare" without explaining that the control isolates juglone as the cause.

Things to Be Careful About

The control must include the same solvent mixture (without juglone), not pure water, because the solvent itself could influence germination. The control volume and conditions should also be matched to those of the juglone treatment.

Background Concept

A valid comparison requires standardisation of all variables except the independent variable. Here there are two measurable outcomes (percentage germination and seedling growth) and the method must include replicates, controlled conditions, defined endpoints, and safety considerations for any chemical hazard.

Understanding the Question

The student needs a detailed method, set out logically and in enough detail for another person to follow, that tests the effect of juglone on (i) the percentage of cucumber seeds that germinate and (ii) the growth of each seedling 10 days after the seedling appears above the soil. Marks are available for standardisation, control of variables, replication, the measurement procedure, and a hazard/risk/precaution safety statement.

Approach

Structure the method in the order in which it would be carried out:

1. Preparation of the treatment trays.
2. Standardisation (same seeds, same volume/concentration of solutions, same soil mass, same conditions).
3. Maintenance (reapply juglone so exposure continues).
4. Measurement of germination percentage.
5. Measurement of growth at 10 days.
6. Replication and means.
7. Safety.

Step-by-Step Reasoning

• Standardising exposure: Use the same mass/volume of the same soil in identical trays; sow the same number of cucumber seeds of the same variety and similar initial size; add the same volume ($500\ \text{cm}^3$) of solution to each tray. This ensures both groups start from the same baseline.
• Same seeds: Use the same species/variety, same age, same size of cucumber seeds in both trays so that any difference in outcome cannot be due to seed quality.
• Constant temperature: Place both trays in the same environment at the same temperature, ideally in an incubator or a controlled-temperature room (or at least in the same room, monitored with a thermometer).
• Constant light intensity: Place both trays under the same light source for the same photoperiod, so photosynthesis and growth are not differentially affected.
• Another standardised variable: Equal watering (same volume at the same intervals), same planting depth, same spacing of seeds.
• Reapply juglone after germination: The original $500\ \text{cm}^3$ will be exhausted as the plant grows. To maintain the experimental exposure, reapply the same volume of the appropriate solution (juglone to the experimental tray, solvent to the control) at stated intervals during the 10 days.
• Counting germination: After a fixed/stated time, count the number of seeds that have germinated (cotyledons visible above soil) out of the total sown in each tray, and calculate percentage germination.
• Measuring growth at 10 days: Use a ruler to measure the height of each seedling from soil level to the tip of the longest leaf, 10 days after that seedling emerged above the soil. Record measurements (cm) and calculate the mean for each treatment.
• Large sample + mean: Use a large number of seeds in each tray (e.g. 50) so that a meaningful mean can be calculated and anomalies are detectable. Repeat the whole experiment at least once more with fresh seeds to obtain replicate means.
• Safety: Juglone and the solvent mixture are irritant/toxic; wear gloves and a mask when preparing and applying solutions; soil may contain pathogens - wear gloves and wash hands afterwards.

Key Takeaways

• Standardise EVERY variable except the IV.
• Reapply treatment if exposure must be maintained across the experimental period.
• Use a large sample and a mean for both germination percentage and growth.
• Safety requires naming a hazard AND a risk AND a precaution.

Common Mistakes

• Forgetting to reapply juglone (so the treatment disappears once the seed has imbibed the initial solution).
• Saying "measure growth" without specifying how (ruler, from where to where, at what time after emergence).
• Failing to standardise light, temperature, water or seed source.
• Vague safety statements ("be careful") without naming hazard and precaution.
• Using too few seeds and not calculating a mean.

Things to Be Careful About

• The mark scheme wants BOTH a hazard AND a risk AND a precaution for the safety mark.
• Each standardisation must name WHAT is kept constant (temperature, light, water, seed source) and ideally HOW (incubator, controlled-environment chamber, equal-volume watering).
• The 10-day growth measurement is taken from the moment each individual seedling appears, not 10 days after sowing (so the seedlings in each tray may be measured at different calendar days).

Background Concept

The t-test is a parametric statistical test that compares two means. It is appropriate only when its underlying assumptions are satisfied: the data are continuous, the two samples are independent and drawn from populations that are approximately normally distributed, and the variances (and hence standard deviations) in the two groups are similar. If these conditions fail, a non-parametric alternative (e.g. the Mann-Whitney U test) should be used instead.

Understanding the Question

The investigation compares the time taken to germinate for 10 juglone-treated seeds and 10 untreated seeds. The stem has already accepted that the sample size is appropriate. The question asks for TWO other reasons why the t-test fits this investigation.

Approach

List the assumptions of the t-test that the data in this investigation meets. Any two of the four standard assumptions will earn full marks: (i) continuous data, (ii) comparison of two means, (iii) approximate normality, (iv) similar standard deviations.

Step-by-Step Reasoning

1. Continuous data: Time taken for seeds to germinate is measured on a continuous scale (days, with finer subdivisions possible), which is a requirement for the t-test.
2. Comparison of two means: The hypothesis concerns whether the mean germination time differs between the juglone-treated group and the control group - exactly what the t-test is designed to test.
3. Approximate normality: Time-to-germination data from a population of seeds of the same species is approximately normally distributed, satisfying this assumption.
4. Similar standard deviations: Because the seeds are from the same species and grown under similar conditions (apart from the juglone treatment), the variability in germination time within each group should be similar.

Key Takeaways

• t-test requirements: continuous data, two groups, approximate normality, similar variances.
• The t-test is NOT appropriate for categorical data (use chi-squared) or for comparing more than two means (use ANOVA).
• If normality is in doubt with small samples, a non-parametric test such as the Mann-Whitney U test is preferable.

Common Mistakes

• Repeating the sample-size justification already given in the stem.
• Stating vague reasons like "the data is reliable" or "there are two groups" without naming the statistical property (means, continuous, normal, variances).
• Saying the data is "categorical" or "discrete" (it is continuous).

Things to Be Careful About

"Two groups" alone is not sufficient - the test compares the two MEANS of the groups. The reasoning must refer to the statistical property, not just the experimental design.

Background Concept

For an unpaired two-sample t-test on samples of size $n_1$ and $n_2$, the degrees of freedom is $n_1 + n_2 - 2$. The calculated value of $t$ is compared with a critical value from a t-table at the chosen probability level (commonly $p = 0.05$). If the calculated value is less than the critical value, the null hypothesis is accepted at that probability level - meaning the difference observed is not statistically significant.

Understanding the Question

The stem gives a calculated $t = 2.090$ and a critical-values table for df = 17 to 21. Each group has 10 seeds, so df = 18. The question asks the candidate to state and explain the conclusion.

Approach

1. Calculate df.
2. Look up the critical value at df = 18 and p = 0.05.
3. Compare the calculated t with the critical value.
4. State the conclusion about the null hypothesis and what it means biologically.

Step-by-Step Reasoning

From Table 1.1, at 18 df and $p = 0.05$, the critical value of $t$ is $2.101$.
The calculated value ($t = 2.090$) is LESS than the critical value ($2.101$).

Therefore, the null hypothesis is accepted at $p = 0.05$: there is no significant difference (at $p = 0.05$) between the mean time taken for juglone-treated and untreated crimson clover seeds to germinate.

Key Takeaways

• df = $n_1 + n_2 - 2$ for an unpaired two-sample t-test.
• Calculated t must EXCEED the critical value to REJECT the null hypothesis.
• The standard probability level used is $p = 0.05$ (5%).
• "Accepting" the null hypothesis is not the same as "proving" it; we can only say the data do not provide evidence of a significant difference.

Common Mistakes

• Using the wrong df (e.g. 20 or 19, or $n_1 + n_2 - 1$).
• Saying "the result is significant" when the calculated value is LESS than the critical value.
• Confusing "accept" and "reject" the null hypothesis.
• Failing to mention the probability level used.

Things to Be Careful About

Wording is "accept the null hypothesis" (or "fail to reject"), not "prove the null hypothesis". At $p = 0.05$ there is no significant difference; we cannot say there is definitely no effect, only that the data do not provide evidence of one at this significance level. Note that the calculated $t = 2.090$ is greater than the $p = 0.10$ critical value (1.734) at df = 18, so the result IS significant at $p = 0.10$ but not at $p = 0.05$; by convention the more stringent $p = 0.05$ is used unless stated otherwise.

Background Concept

Data are classified by type because different statistical tests and graphical presentations are appropriate for each type:

• Nominal data are categories with no inherent order (e.g. species names, blood groups, eye colour).
• Ordinal data are ordered categories where the gaps between ranks are not necessarily equal (e.g. low / medium / high).
• Discrete data are countable whole numbers (e.g. number of bacteria, number of leaves).
• Continuous data are numerical values that can take any value within a range, including fractions/decimals, and are obtained by measurement (e.g. length, mass, concentration).

Understanding the Question

We have two variables displayed on Fig. 2.1: the species of bacterium on the x-axis (categorical labels) and the lowest nitrite concentration on the y-axis (numerical values in µmol cm⁻³). The question asks for the correct data type for each.

Approach

Ask: is the variable a name/label with no numerical meaning, or a measured number on a scale? That decides between nominal/ordinal and discrete/continuous. Then ask whether the numbers are restricted to whole numbers (discrete) or can take any value (continuous).

Step-by-Step Reasoning

1. Species of bacterium: the names Yersinia enterocolitica, Salmonella enteritidis, etc. are simply labels. There is no numerical ordering — Escherichia coli is not "greater than" Shigella sonnei. These are categories with no rank, so the data type is nominal.
2. Lowest nitrite concentration: this is a measured quantity (µmol cm⁻³) obtained from the experiment. It can take any value within a range, including non-integer values such as 0.7, 1.5, 2.85. There are no restrictions on the values it can take, so the data type is continuous.

Key Takeaways

• Names/labels without numerical order = nominal.
• Measured quantities that can take any value = continuous.
• Identifying the data type is essential for choosing the correct statistical test later (e.g. a t-test requires continuous data; a chi-squared test works on counts/frequencies).

Common Mistakes

• Writing "qualitative" or "categorical" instead of the precise term "nominal".
• Writing "quantitative" or "discrete" instead of "continuous" — discrete would imply only whole-number values, but concentrations can be non-integer.
• Confusing the bars on the chart (which represent continuous values) with the species labels (which are nominal categories).

Things to Be Careful About

• Use the exact statistical terms nominal and continuous; the mark scheme will not accept synonyms.

Background Concept

A percentage difference expresses the size of a change between two values as a percentage of one of the values. There are two common conventions:

• Percentage change = $\frac{\text{new} - \text{old}}{\text{old}} \times 100$ — used when one value is clearly the reference (e.g. before vs after, smaller vs larger).
• Symmetric percentage difference = $\frac{|a - b|}{(a + b)/2} \times 100$ — used when neither value is privileged.

In CIE mark schemes, "percentage difference" is usually treated as a percentage change relative to one of the values, and either direction is often accepted.

Understanding the Question

We must compare the lowest nitrite concentration that prevented growth of Salmonella enteritidis with that for Salmonella typhimurium. From Fig. 2.1 we read:

• S. enteritidis bar reaches just below 1.0, at 0.7 µmol cm⁻³.
• S. typhimurium bar reaches 5.0 µmol cm⁻³.

Approach

1. Read both values carefully off the bar chart (use gridlines, not rough estimates).
2. Calculate the difference and divide by an appropriate base value, then multiply by 100.
3. State the final answer as a percentage.

Step-by-Step Reasoning

1. Read values from Fig. 2.1 (mark scheme point 1). The S. enteritidis bar is clearly below the 1.0 gridline at 0.7 µmol cm⁻³; the S. typhimurium bar is on the 5.0 gridline at 5.0 µmol cm⁻³.
2. Set up the calculation (mark scheme point 2). The difference is $5.0 - 0.7 = 4.3\ \mu\text{mol cm}^{-3}$. Using the smaller value as the denominator gives the larger percentage: $\frac{4.3}{0.7} \times 100$.
3. Evaluate and quote with the % sign (mark scheme point 3). $\frac{4.3}{0.7} \times 100 = 614.28...\ \%$, which rounds to 614 %.

The mark scheme also accepts 86 %, which is what you obtain by dividing the same difference by the larger value: $\frac{4.3}{5.0} \times 100 = 86\ \%$. Both are correct because they each express the relative gap between the two results; the convention you pick determines the magnitude.

Key Takeaways

• Always read bar-chart values to the precision shown by the gridlines.
• Show your working — substitution into the formula earns a mark even if the arithmetic slips.
• The mark scheme often accepts both directions of a percentage-difference calculation.

Common Mistakes

• Reading S. enteritidis as 0.5 (the gridline below) or as some other value; the mark scheme is firm at 0.7.
• Dividing by the wrong base value without realising the answer changes.
• Forgetting the % sign on the final answer.
• Writing the difference (4.3) without converting it into a percentage.

Things to Be Careful About

• Use the gridlines to read values; do not guess.
• State the units of the readings (µmol cm⁻³) even though the final answer is a %.
• Round only at the end, not at intermediate steps.

Background Concept

The y-axis of Fig. 2.1 shows "lowest nitrite concentration at which no bacterial population growth occurred after 24 hours". A species that needs only a tiny concentration of nitrite to stop it growing is highly sensitive (most affected); a species that needs a lot of nitrite before growth stops is resistant (least affected). So:

• The smallest bar = the most affected species.
• The tallest bar = the least affected species.

Understanding the Question

The question asks us to identify, from the bar chart, the species that is most affected by nitrite ions and the species that is least affected. The answer is obtained purely by reading the heights of the bars.

Approach

Compare the bar heights in Fig. 2.1 and identify the shortest and tallest. Convert those into the corresponding species names.

Step-by-Step Reasoning

1. From Fig. 2.1, the shortest bar belongs to Yersinia enterocolitica at approximately 0.5 µmol cm⁻³ — this species is stopped from growing by the smallest amount of nitrite, so it is the most affected.
2. The tallest bar belongs to Salmonella typhimurium at 5.0 µmol cm⁻³ — this species requires a much larger concentration of nitrite before its growth stops, so it is the least affected.

Key Takeaways

• In dose–response data, the species that responds at the lowest dose is the most sensitive.
• Always match the wording of the question ("most/least affected") to the corresponding numerical direction (lowest/highest dose).

Common Mistakes

• Reversing the relationship and naming S. typhimurium as most affected (it is the least affected because it is the most resistant).
• Confusing the direction of the bars — reading "most affected" as "the bar that grows the tallest" rather than "the species needing the smallest dose".

Things to Be Careful About

• Italicise the species binomial names, e.g. Yersinia enterocolitica; the mark scheme credits the correct name only.