Biology 9700/52 — May/June 2018

Background Concept

The independent variable is what the experimenter deliberately changes. The dependent variable is what is measured to see the effect of that change. Variables that could affect the result but are kept constant are control variables.

Understanding the Question

The student is going to test the hypothesis: "As light intensity increases, the rate of the light-dependent stage of photosynthesis will increase." We must identify what will be changed (the independent variable) and what will be measured (the dependent variable).

Approach

Read the hypothesis: the rate of the light-dependent reaction is the outcome; light intensity is the factor being changed. Decide how each can be measured in this set-up.

Step-by-Step Reasoning

• Light intensity is the factor being changed. In this apparatus, light intensity is varied by moving the lamp closer to or further from the tube. The formula $I = 1/\text{distance}^2$ relates the two, but the experimenter simply changes the distance.
• Rate of the light-dependent reaction is the outcome. The student will measure how long it takes for the DCPIP to be reduced (i.e. for the blue colour to disappear / the mixture to turn green). Time is therefore the dependent variable, since rate = $1/\text{time}$ when the amount of DCPIP is constant.

Key Takeaways

• The variable that the experimenter deliberately changes is independent.
• The variable measured to record the effect is dependent.
• "Time to decolourise" is a valid dependent variable when the amount of DCPIP is constant, because rate is inversely related to that time.

Common Mistakes

• Stating "rate of photosynthesis" as the dependent variable — the student cannot measure rate directly; they measure the time for the colour change, and infer rate from it.
• Confusing distance and light intensity — the experimenter moves the lamp, but the variable being changed is light intensity, with distance as a means to vary it.

Things to Be Careful About

Either "light intensity" or "distance of the lamp from the tube" is accepted as the independent variable; they are linked by the formula in Fig. 1.4.

Background Concept

To test the effect of light intensity on the rate of the light-dependent reaction, the experimenter varies the light falling on a constant chloroplast + DCPIP mixture and records the time taken for the DCPIP to be reduced. Because the light intensity from a point source falls off as $1/\text{distance}^2$ (the inverse-square law), moving the lamp is the standard way of changing intensity. To make the test fair, every other factor that could affect the reaction (volumes, concentrations, temperature, water-tank position) must be kept constant.

Understanding the Question

The student is going to use the apparatus in Fig. 1.3 to test: "As light intensity increases, the rate of the light-dependent stage of photosynthesis will increase." We need a method detailed enough for another person to follow, in a logical order, that will generate the data required to test the hypothesis.

Approach

1. Decide how to vary the independent variable (light intensity) — by moving the lamp.
2. Decide what to measure (time for DCPIP to decolourise) and how to judge the end-point (compare with a tube of chloroplasts without DCPIP, which stays green).
3. List the variables to keep constant and how to control them.
4. Plan replication and a mean, and the safety precautions.

Step-by-Step Reasoning

• Varying the intensity. Because $I = 1/\text{distance}^2$, moving the lamp changes the intensity. The mark scheme requires at least five distances in the range 10–200 cm, so plan e.g. 20, 40, 60, 80 and 100 cm.
• The measurement. The reaction is followed by the colour change of DCPIP (blue → colourless/green). The time from exposure to full decolourisation is recorded; rate is then $1/\text{time}$ (constant amount of DCPIP). To judge the end-point reliably, the student should use a colour comparator — a separate tube containing the chloroplast suspension plus distilled water (i.e. tube 2 from the original investigation), which is the colour the decolourised mixture will become. Match the test tube to this comparator by eye.
• Controls (standardised variables). The same volume of chloroplast suspension, the same volume of DCPIP, and the same concentration of DCPIP must be used at every distance. The water tank must stay at the same distance from the tube/light source at every trial, otherwise it is filtering a different fraction of the beam.
• Reliability. Repeat each distance at least twice and calculate a mean. Anomalous readings can be identified and discarded before averaging.
• End-point timing. Start the timer the moment the tube is exposed to light. Stop the timer the moment the colour matches the comparator. Recording the time taken (rather than the colour at a fixed time) makes the rate calculation straightforward.
• Safety. The lamp becomes hot during use. Do not touch the bulb or its housing; allow it to cool before handling. The chloroplast suspension is harmless but care should be taken to avoid spillages on the bench.

Key Takeaways

• A good plan names the independent variable, how it is varied, the dependent variable and how it is measured, the variables that must be standardised, replication, and a safety point.
• Use a colour comparator to judge an end-point reliably by eye.
• Vary the lamp–tube distance over at least five values across a wide range.

Common Mistakes

• Stating only one or two distances — the mark scheme requires at least five in the range 10–200 cm.
• Forgetting to use a comparator (e.g. tube 2) to decide when to stop timing.
• Varying the concentration of DCPIP between trials — that would change the dependent variable's scale.
• Failing to start the timer the moment the tube is illuminated.
• Omitting a safety point.

Things to Be Careful About

• "Time for the DCPIP to decolourise" is the measurement; "rate of the light-dependent reaction" is what is inferred from it.
• The water tank must be at the same distance from the tube (or lamp) at every trial.
• Distances must be in the range 10–200 cm; placing the lamp right next to the tube (a few cm) may overheat the sample, while very large distances give very low intensities and ambiguous end-points.

Background Concept

A conclusion about when a reaction finishes can only be supported by data that show the reaction is still proceeding (absorption still falling) just before the supposed end-point, and that the reading has stopped changing just after. With data every minute, an end-point can only be pinned to the nearest whole minute.

Understanding the Question

The student concluded that the reaction in Fig. 1.5 was complete at 8 minutes because the curve appears to plateau from 8 min onwards. We are asked why the data on the graph do not actually support that precise conclusion.

Approach

Look at the spacing of the data points. The student took readings every minute, so the only readings are at 1, 2, 3, … 7, 8, 9, 10, 11, 12 min. There is no reading between 7 and 8 min.

Step-by-Step Reasoning

Because no absorption value is recorded between 7 and 8 minutes, the student cannot tell whether the reaction was still going at, say, 7 min 30 s and only just completed by 8 min, or whether it had already finished by 7 min 30 s. The data only allow the conclusion that the reaction had completed by 8 min, not that it completed at 8 min.

Key Takeaways

• A conclusion about timing must be backed up by data at the right resolution.
• "Completed at 8 min" and "completed by 8 min" are different claims.

Common Mistakes

• Saying the data "don't show a plateau" — they do; the issue is the spacing of the points, not the existence of a plateau.
• Saying the data are inaccurate.

Things to Be Careful About

A reading taken every minute can only pin an end-point to the nearest minute, so 8 min is the resolution limit of the data.

Background Concept

The rate of change of one variable with respect to another at a particular point on a curve is given by the gradient of the tangent at that point. The tangent is a straight line that just touches the curve at the point of interest and has the same slope as the curve at that point. A large triangle (i.e. one that uses well-separated points on the tangent) gives a more accurate estimate of the gradient than a small one.

Understanding the Question

The student needs to find the instantaneous rate at which absorption is changing at the 6-minute mark on the curve. The curve is decreasing through $(6,\ 0.9)$, so the rate will be negative. The mark scheme accepts values in the range $0.2$ to $0.4$ in magnitude.

Approach

1. Place a ruler on the curve so that it just touches the curve at $t = 6\ \text{min}$ and has the same gradient there — i.e. it follows the curve's direction at that point without cutting it.
2. Choose two well-separated points on the tangent (well clear of the curve's bends).
3. Read off their $(x, y)$ coordinates and use $\Delta y / \Delta x$.

Step-by-Step Reasoning

A typical tangent at $(6,\ 0.9)$ on this curve passes through approximately $(3,\ 2.3)$ and $(8,\ 0.6)$:

The negative sign indicates the absorption is still decreasing. The mark scheme's accepted range is $0.2$–$0.4$ in magnitude, so a value of about $-0.3$ is appropriate. Different reasonable tangents will give slightly different values in this range.

Key Takeaways

• The rate of change at a point is the gradient of the tangent at that point.
• A large triangle gives a more accurate gradient than a small one.
• The mark scheme accepts a magnitude in $0.2$–$0.4$, so the answer is not unique.

Common Mistakes

• Drawing the tangent at the wrong time (e.g. at 4 min or 8 min).
• Using two points on the curve instead of on the tangent — that gives the average gradient over an interval, not the instantaneous rate.
• Quoting the units of the answer as just "arbitrary units" rather than "arbitrary units per minute".

Things to Be Careful About

The units of rate are arbitrary units per minute. The mark scheme's "0.2 to 0.4" is the magnitude; including the negative sign is correct because the curve is decreasing.

Background Concept

A lower light intensity means fewer photons reach the chloroplasts, so the rate of the light-dependent reaction is lower, and DCPIP is reduced more slowly. The same total amount of DCPIP is present in the tube, so the final absorption value reached (when all the DCPIP is reduced) is the same as in the high-intensity tube — only the time taken to reach it changes.

Understanding the Question

We need to sketch on Fig. 1.5 the curve the student would expect for a tube exposed to a low light intensity. The original curve was drawn at high light intensity.

Approach

Predict how the rate changes (lower → slower decrease) and how the end-point changes (same total DCPIP → same final absorption, but reached later).

Step-by-Step Reasoning

• Slower decrease: the curve drops less steeply than the high-intensity curve.
• Same final value: both tubes contain the same amount of DCPIP, so both must end at the same absorption ($\approx 0.6$).
• Reached later: the low-intensity curve will reach the plateau at a time $> 8$ min.
• The curve must therefore be to the right of the original at every time before the plateau, and it should be drawn across the whole 1–12 min range of the x-axis.

Key Takeaways

• A change in an independent variable that slows the reaction produces a curve that descends more slowly but reaches the same end-point.
• The final value is set by the amount of substrate (DCPIP), not by the rate.

Common Mistakes

• Drawing the curve to the left of the original — this would imply a higher rate, the opposite of what low light intensity gives.
• Drawing the curve plateauing at a higher absorption — that would imply less DCPIP was reduced.
• Drawing only part of the curve (e.g. not extending it across the full 1–12 min range).

Things to Be Careful About

The mark scheme requires the curve to be to the right of the original and to span the full 1–12 min range.

Background Concept

A null hypothesis ($H_0$) is the statistical statement that there is no significant difference (or no significant effect/association) between the populations from which the samples are drawn. It is the hypothesis that the t-test evaluates directly; the alternative hypothesis ($H_1$) is the opposite — that a significant difference does exist. If the calculated $t$ exceeds the critical value at $p < 0.05$, $H_0$ is rejected.

Understanding the Question

The student is going to use a t-test to compare tubes 1 and 2 from Table 1.3 — i.e. chloroplasts + DCPIP with no weed killer (tube 1) and chloroplasts + DCPIP with weed killer X (tube 2). We must state the null hypothesis for that comparison.

Approach

State, in words, that the means of the two populations from which the samples are drawn are equal, naming the two tubes.

Step-by-Step Reasoning

The t-test is comparing the mean absorption values of the contents of tube 1 and tube 2. The null hypothesis is therefore that these two means are not significantly different — i.e. that any observed difference between the sample means is due to chance variation between replicates, not to the weed killer.

Key Takeaways

• A null hypothesis is always a statement of "no significant effect" or "no significant difference".
• It must refer to the specific quantities being compared.

Common Mistakes

• Phrasing the null hypothesis as a research hypothesis ("weed killer X changes the rate") — the research hypothesis is the alternative, not the null.
• Omitting the word "significant" — without it, the null is trivially false because the means are unlikely to be exactly equal.

Things to Be Careful About

The null hypothesis always contains "no significant difference" (or equivalent), and refers to the populations, not just the samples.

Background Concept

The t-test makes assumptions about the data:

1. The data are continuous (so a mean is meaningful).
2. The data are normally distributed in each population (or the sample is large enough for the central limit theorem to make the sampling distribution of the mean approximately normal — the t-test is robust to mild departures when sample sizes are similar).
3. The two populations have similar standard deviations (homogeneity of variance).
4. The samples are small (typically $n < 30$) — the t-test is the correct test for small samples; for large samples the z-test would be used instead, but the two converge.
5. The samples are independent.

The CIE mark scheme accepts any one of: continuous; normally distributed; similar standard deviations; sample size less than 30.

Understanding the Question

The student is going to use a t-test on the absorption data from Table 1.3. We are asked to state one feature of the data that allows the t-test to be used.

Approach

Pick one of the assumptions above that is met by these data.

Step-by-Step Reasoning

• Continuous data: absorption values from a colorimeter are measured on a continuous scale, so a mean is appropriate.
• Normally distributed: the population of absorption values for each tube is assumed to be approximately normal — reasonable for replicated colorimeter readings.
• Similar standard deviations: the variation between replicates of one tube should be similar to the variation between replicates of another.
• Small sample size: the mark scheme for part (iii) shows $n = 12$ per tube, which is less than 30, so the t-test is preferred over the z-test.

Any one of these is sufficient.

Key Takeaways

• The t-test is the appropriate test for comparing two means from small, continuous, approximately normal samples with similar variances.
• For larger samples the z-test is used; the two converge as $n$ grows.

Common Mistakes

• Listing several assumptions and getting them confused.
• Stating an assumption that is not a feature of the data (e.g. "the samples are random") without justification.

Things to Be Careful About

Any one of the four features is accepted; pick the one you can justify most clearly.

Background Concept

For a two-sample (unpaired) t-test the degrees of freedom are $n_1 + n_2 - 2$, where $n_1$ and $n_2$ are the sample sizes of the two groups being compared. Each observation "uses up" one degree of freedom for the mean; one further degree of freedom is used for the difference between the two means; the remaining observations are free to vary.

Understanding the Question

Each t-test in Table 1.4 has 22 degrees of freedom, and the two tubes being compared in each test had an equal number of replicates. We need to find the number of replicates per tube.

Approach

Set $n_1 = n_2 = n$ and use $\text{df} = 2n - 2 = 22$ to solve for $n$.

Step-by-Step Reasoning

So 12 replicates were carried out for each tube.

Key Takeaways

• $\text{df} = n_1 + n_2 - 2$ for a two-sample t-test.
• If $n_1 = n_2$, then $\text{df} = 2n - 2$.

Common Mistakes

• Using $\text{df} = n - 1$ (which is the formula for a one-sample t-test, not a two-sample one).
• Using $\text{df} = n_1 + n_2$ (forgetting to subtract 2).
• Halving 22 to get 11 (the formula is $n_1 + n_2 - 2$, not $(n_1 + n_2)/2$).

Things to Be Careful About

The t-test here compares two tubes, so the two-sample formula applies.

Background Concept

A t-test is significant at $p < 0.05$ when the calculated $t$ value exceeds the critical value. The critical value is read from a t-table at the given degrees of freedom and the chosen probability level; here the critical value is given as 2.07. If calculated $t > 2.07$, the null hypothesis (no significant difference) is rejected.

Understanding the Question

Table 1.4 shows the result of the t-test comparing weed killer X with weed killer Y is significant. We must state what this tells us about the calculated $t$ value.

Approach

A "significant" result at $p < 0.05$ means the calculated $t$ value exceeded the critical value for that probability level.

Step-by-Step Reasoning

The critical value for $t$ at $p < 0.05$ with 22 df is 2.07. Because the result in Table 1.4 for X vs Y is "significant", the student's calculated $t$ for that comparison must have been greater than 2.07. This means the difference between the mean absorption values for tubes 2 and 3 is unlikely to be due to chance variation between replicates — i.e. weed killer X and weed killer Y have significantly different effects on the light-dependent reaction.

Key Takeaways

• "Significant" in a t-test summary means calculated $t$ > critical value.
• The exact calculated $t$ is not given; only the significance is reported.

Common Mistakes

• Stating that calculated $t$ is equal to 2.07 — the table does not give the exact value, only that it exceeds the critical value.
• Confusing "significant" with "large"; the word has a specific statistical meaning here.

Things to Be Careful About

The wording in the mark scheme: "greater than 2.07" or "2.07 is less than the calculated value". Either is acceptable; do not give the calculated value as 2.07 itself.