Biology 9700/53 — May/June 2018

Background Concept

The independent variable is what the experimenter deliberately changes between trials. The dependent variable is what is measured to record the response. The relationship is summarised as: "As the independent variable is changed, the dependent variable is measured." All other factors that could influence the result must be kept constant (controlled variables).

Understanding the Question

The students are testing the hypothesis "as light intensity increases the rate of photosynthesis increases". They are using a lamp whose distance from the container can be changed, and they are reading the colour of the hydrogencarbonate indicator as a proxy for CO₂ uptake (and hence for photosynthetic rate).

Approach

The variable the experimenter changes is the independent variable. The variable the experimenter measures (or reads off a colour standard) is the dependent variable.

Step-by-Step Reasoning

• Independent: the students move the lamp (or the container) to different distances. Distance is being changed, so the directly-changed physical quantity is distance — but the biological quantity of interest is light intensity, which depends on distance. The mark scheme accepts either, since the two are linked by $I = 1/\text{distance}^2$.
• Dependent: the rate of photosynthesis is being measured indirectly via the colour of the indicator. Comparing the experimental tube's colour to Table 1.1 yields a pH, so the colour itself (or the pH read off the standard) is the dependent variable.

Key Takeaways

• Independent variable: light intensity / distance.
• Dependent variable: colour / pH.
• "Rate of photosynthesis" is the biological quantity the students are ultimately interested in, but it is not measured directly; it is inferred from colour/pH.

Common Mistakes

• Writing "rate of photosynthesis" as the dependent variable. The rate is INFERRED from the colour, but the actual measurement is the colour or pH.
• Reversing the two variables.

Things to Be Careful About

The mark scheme allows either form (light intensity OR distance) for the independent variable. Always state the variable the experimenter directly manipulates, but recognise that for this apparatus the physical manipulation is the distance and the biological variable is the light intensity.

Background Concept

A good experimental method must (i) vary the independent variable across a sensible range, (ii) keep all other variables constant, (iii) include a control to show that any change is due to the treatment, (iv) be replicated so that anomalies can be identified and a mean calculated, and (v) identify a relevant hazard and how to deal with it.

Understanding the Question

The students want to test whether photosynthetic rate rises with light intensity. Their indicator responds to pH (and therefore to CO₂): if photosynthesis exceeds respiration, CO₂ falls, pH rises, the colour moves up the table towards purple. They have a single container and a lamp whose distance from the container can be changed.

Approach

Pick five points from the nine possible marking points, choosing the most decisive features of a strong method.

Step-by-Step Reasoning

• Vary the light intensity: move the lamp, or move the container, to different distances from each other. The mark scheme also accepts using neutral density filters, different wattage bulbs, or a dimmer switch.
• Range of values: state five specific distances, all in the range 10–200 cm. Five values are needed to draw a meaningful graph of rate vs. intensity.
• Exclude other light: wrap the tube/container in metal foil, or work in a dark room with the lamp as the only source. Without this, ambient daylight would swamp the variable the students are trying to change.
• Standardise: use the same volume of indicator and the same number (or mass, or volume) of algal balls at every distance. Without this, the rate per algal ball cannot be compared.
• Fixed time: expose each container for the same length of time. Comparing final colour readings only makes sense if the time is identical.
• Control: set up an identical tube but with no photosynthesis occurring — replace the living algal balls with glass beads, or with algal balls that have been killed (e.g. by boiling). The control should show little or no colour change, confirming that the change seen in the live-algae tube is due to photosynthesis.
• Replication: do each distance at least twice; calculate a mean. This is essential because judging the colour change is subjective and any single reading is unreliable.
• Safety: hydrogencarbonate indicator is an irritant; wear gloves and goggles. (Other accepted hazards include allergy to alginate/algae.)

Key Takeaways

• Five mark-scheme points are needed: a method to vary the IV, five distances in 10–200 cm, control of light, a fixed recording time, and three further points from {control, standardised indicator, standardised algal balls, replication/mean, safety}.
• A clean method lists the variables (varied, measured, controlled) explicitly and gives numerical details (distances, time) so that another student could repeat the work exactly.

Common Mistakes

• Saying "amount" instead of "volume" or "concentration" for the indicator — the mark scheme specifically rejects "amount" here.
• Recording the time taken to reach a particular colour instead of recording the colour after a fixed time — both are accepted by the mark scheme, but only one needs to be described.
• Forgetting the control, or setting up a control that does not test what the experiment claims (e.g. a tube with indicator but no algal balls AND no beads — this does not show whether the beads themselves are inert).
• Skipping replication, or doing a single replicate at each distance.

Things to Be Careful About

The mark scheme offers 9 possible points but only 5 are needed. Choose the strongest, most specific 5: the cleanest answers in the order above are usually the most reliable for credit.

Background Concept

A tungsten-filament lamp emits substantial infrared radiation alongside visible light. If the lamp is close to the container, this infrared warms the indicator solution. Temperature affects the rate of photosynthesis (and the rate of CO₂ dissolution), so any warming would be a confounding variable. A heat filter absorbs the infrared but transmits visible light, decoupling the two effects.

Understanding the Question

The students want to know what practical step would stop the lamp from warming the solution, so that the only thing varying between trials is the visible light intensity.

Approach

Recall that "heat filter" is the standard CIE term. The simplest practical filter is a flat-sided glass tank of water (water absorbs infrared strongly) or a sheet of heat-absorbing glass placed between lamp and container.

Step-by-Step Reasoning

• The lamp emits visible light AND infrared. The visible light is the experimental variable; the infrared is an unwanted confounder.
• Place a clear heat filter in the path of the light. Water in a glass tank is the cheapest effective filter; a sheet of double- or triple-glazed glass works similarly.
• Modern LED or halogen-cold-light sources emit very little infrared and are accepted by the mark scheme as an alternative.

Key Takeaways

• A heat filter is the standard way to control temperature when using a filament lamp as a light source in photosynthesis investigations.
• A water-filled glass tank is a cheap, accessible heat filter.

Common Mistakes

• Suggesting "move the lamp further away" — this changes the light intensity (the IV), defeating the purpose of the experiment.
• Suggesting "use a lower-wattage bulb" — this also changes the light intensity.
• Saying "place a thermometer in the solution" — that monitors temperature but does not control it.

Things to Be Careful About

The mark scheme specifically accepts "description of heat filter e.g. container of water / double glazed glass sheet" or "cold light source e.g. LED lights". Vague answers such as "use a filter" do not earn the mark — name what the filter is.

Background Concept

Every measurement has an associated uncertainty. Subjective comparisons (matching a colour by eye) carry a large human uncertainty: two observers can legitimately place the same solution at different points on the colour standard. A pH probe converts the same chemical property into a voltage that the instrument displays as a number; this is an objective measurement whose uncertainty is set by the probe's calibration, not by the observer's eyes.

Understanding the Question

The original method relied on matching a colour to a printed scale (Table 1.1). The students want to know why a pH probe would give better data.

Approach

Identify what is wrong with the original colour-comparison method and how a probe fixes it.

Step-by-Step Reasoning

• Subjectivity of colour matching: the eye cannot reliably discriminate between adjacent colours on the scale (e.g. between dark orange and red, or between light magenta and dark magenta). Two students comparing the same tube could legitimately record different pH values.
• Probe advantages: a pH probe gives a numerical reading (e.g. 8.7) to one or two decimal places, is not influenced by the observer, and detects smaller differences in pH than the eye can resolve between adjacent colour steps. The mark scheme also accepts "more sensitive" or "gives a quantitative result".

Key Takeaways

Subjective measurements are less accurate and less reproducible than objective ones. A pH probe is more accurate because it removes human judgement and resolves smaller pH differences.

Common Mistakes

• Saying "more accurate" without explaining why. The mark scheme requires either the subjectivity of colour, the difference between observers, or the higher sensitivity of the probe.
• Confusing accuracy with precision. The probe is also more precise, but the mark-scheme answers focus on the subjectivity and the quantitative nature of the reading.

Things to Be Careful About

The mark scheme accepts any of: judging colour is subjective; different people see colours differently; pH probe is more sensitive; pH probe gives a numerical/quantitative result. Give the one that flows most naturally in your sentence.

Background Concept

The pH rises because photosynthesis removes CO₂ from the indicator. The rate of pH rise is therefore a measure of the rate of photosynthesis. A lower light intensity gives a lower rate of photosynthesis (within the light-limiting range), so pH rises more slowly, and the plateau (where photosynthesis equals the rate at which CO₂ is supplied from the air / hydrogencarbonate equilibrium) is reached later, if at all within the 10 minutes shown.

Understanding the Question

The students already have a curve for a high light intensity. They want to know what to draw for a low light intensity. The student must add a curve to the printed Fig. 1.3, not just describe it in words.

Approach

Reduce the gradient of the curve. Keep the same starting pH. Keep the curve to the right of the original across the entire 0–10 min range.

Step-by-Step Reasoning

• The starting pH is the same (8.4) because the indicator is freshly mixed.
• The rate of pH rise is lower, so the slope is shallower.
• The new curve must lie to the right of (or below) the original at every time point from 0 to 10 minutes.
• The plateau, if reached, is reached later than 5 minutes; it may not be reached within the 10-minute window at all.
• The mark scheme requires the curve to be drawn from 0 to 10 minutes — do not stop the new curve early.

Key Takeaways

• Lower IV → shallower gradient on a rate graph.
• The plateau represents the system reaching a steady state (rate of CO₂ uptake = rate of CO₂ supply from buffer equilibrium). At lower light intensity this steady state is reached later, if at all.
• When sketching on a printed figure, the curve must be drawn across the full axis range, not truncated.

Common Mistakes

• Drawing a curve that starts at a different pH — the start pH is determined by the indicator mix, not by light intensity.
• Drawing the new curve steeper than the original, or to the left of it.
• Stopping the curve before 10 minutes.
• Drawing a curve that plateaus at the same time as the original — the plateau must be later, because the steady state is reached later.

Things to Be Careful About

The mark scheme specifies two features: the curve must be to the right of the original AND it must span the full 0–10 minute range. Either of these missing loses the mark.

Background Concept

A limiting factor is the resource in shortest supply that caps the rate of a process. In photosynthesis the classic limiting factors are light intensity, CO₂ concentration and temperature. In this experiment the indicator solution is essentially a concentrated hydrogencarbonate buffer; it holds a large reservoir of dissolved inorganic carbon (as HCO₃⁻) that can be drawn upon by the algae throughout the experiment.

Understanding the Question

The students used only one light intensity and recorded pH over time. They must justify that the resulting rate is not limited by CO₂ availability, otherwise the rate measured is not the maximum possible rate at that light intensity.

Approach

Identify the chemical role of the indicator solution: it is a hydrogencarbonate buffer, and the equilibrium

$\text{HCO}_3^- \rightleftharpoons \text{CO}_2 + \text{OH}^-$

continuously replenishes CO₂ as the algae use it. So CO₂ is plentiful, and the only thing changing the rate is the light intensity.

Step-by-Step Reasoning

• The indicator solution is made from sodium hydrogencarbonate, which dissociates to give hydrogencarbonate ions (HCO₃⁻).
• The hydrogencarbonate ions are in equilibrium with dissolved CO₂.
• As algal photosynthesis removes CO₂, more CO₂ is released from the hydrogencarbonate reservoir, so the CO₂ concentration is effectively held high.
• Because CO₂ is supplied in excess, it is not the factor limiting the rate of photosynthesis in this investigation.

Key Takeaways

• A buffer can act as a reservoir of the very ion it buffers, providing a continuous supply of substrate to a reaction that consumes the substrate.
• "Not limiting" means the resource is present in excess relative to the demand.

Common Mistakes

• Saying "CO₂ is provided by the air" — this is true but not what the indicator is doing; the question is specifically about the indicator solution.
• Saying "the algae make their own CO₂" — the algae use CO₂, not make it (in this context).
• Vague answers such as "there is enough CO₂" without saying where it comes from.

Things to Be Careful About

The mark scheme requires the link to the indicator / hydrogencarbonate. The point is that the indicator itself is the source of CO₂, not the atmosphere.

Background Concept

A null hypothesis ($H_0$) is the default assumption that there is no effect — no difference between the groups being compared. The alternative hypothesis ($H_1$) states that there is a real difference. A statistical test (here, the t-test) calculates the probability of obtaining the observed data, or more extreme data, IF the null hypothesis is true. If that probability (the p-value) is very small, we reject $H_0$ in favour of $H_1$.

Understanding the Question

The students are comparing treatment 2 (weed killer A) with treatment 1 (no weed killer). The question asks for the null hypothesis that will be tested by their t-test.

Approach

The null hypothesis must (i) state a comparison between the two treatments, (ii) use the word "no difference" or "no significant difference", and (iii) refer to the specific measured quantity (mean pH). A common alternative is to write the hypothesis about the populations from which the samples are drawn, but in CIE mark-scheme terms either form is credited.

Step-by-Step Reasoning

• Identify the two samples: treatment 1 (control, distilled water) and treatment 2 (weed killer A).
• Identify the measured variable: mean pH after 30 minutes.
• State the hypothesis: there is no significant difference between the (mean) pH for treatment 1 and the (mean) pH for treatment 2.

Key Takeaways

• A null hypothesis always says "no significant difference" or "no significant effect".
• It must specify the two groups being compared and the variable being measured.

Common Mistakes

• Stating the alternative hypothesis (e.g. "weed killer A reduces the pH") instead of the null.
• Omitting the word "significant" — without it, the null hypothesis would be trivially false because two random samples rarely have IDENTICAL means.
• Omitting "mean" — the test compares sample means, not individual pH values.
• Reversing the two treatments.

Things to Be Careful About

The mark scheme specifically says "no significant difference". The word "significant" is the load-bearing word; a null hypothesis of "no difference" does not earn the mark.

Background Concept

The (Student's) t-test compares two sample means and asks: "If the two samples were drawn from populations with the same true mean, how likely would we be to obtain two sample means at least as far apart as the ones observed?" The test makes a number of assumptions about the data, and the test is only strictly valid when those assumptions hold.

Understanding the Question

The students have used a t-test. The question asks the student to identify ONE feature of the data that permits the use of a t-test.

Approach

Recall the four classical assumptions of the t-test:

1. The data are continuous (not categorical or rank).
2. The data are approximately normally distributed in each population.
3. The two populations have approximately equal standard deviations (variance).
4. The two samples are independent of each other and the sample sizes are reasonably small (typically $n < 30$ each — beyond this, a z-test would be more appropriate).

The mark scheme accepts any one of these, EXCEPT "compare the means" (which the mark scheme specifically ignores, as it is what the t-test does, not a property of the data).

Step-by-Step Reasoning

• The pH readings are continuous (between 8.5 and 9.2), so condition 1 is met.
• The replicates at each treatment are likely to be approximately normally distributed around their mean.
• The mark scheme explicitly REJECTS "compare the means" as a feature of the data, because that is the purpose of the t-test, not a property of the data.

Key Takeaways

• The t-test is for comparing two means of continuous, approximately normal data.
• For larger sample sizes ($n \geq 30$ each), a z-test is generally used; the t-test is preferred for small samples.
• A vague answer such as "compare the means" is not a feature of the data — it is the aim of the test.

Common Mistakes

• Writing "compare the means" — explicitly ignored by the mark scheme.
• Writing "the data are qualitative" — pH is quantitative.
• Writing "there are two treatments" — the number of groups is not an assumption of the t-test; it is what defines which test to use.

Things to Be Careful About

The mark scheme lists four possible answers, of which one is needed. State any one of them clearly.

Background Concept

For a two-sample (independent) t-test, the degrees of freedom is

where $n_1$ and $n_2$ are the sample sizes of the two groups being compared. The df controls the shape of the t-distribution used as the reference distribution. Knowing df and the sample size of one group lets you calculate the other.

Understanding the Question

The students carried out 7 replicates of treatment 1 and 15 degrees of freedom in their t-test. How many replicates did treatment 2 have?

Approach

Apply the formula $df = n_1 + n_2 - 2$ and solve for $n_2$.

Step-by-Step Reasoning

• $df = 15$ (given).
• $n_1 = 7$ (replicates of treatment 1).
• $15 = 7 + n_2 - 2$.
• $n_2 = 15 - 7 + 2 = 10$.
• So treatment 2 had 10 replicates.

Key Takeaways

• df is not the same as sample size: for a t-test, $df = n_1 + n_2 - 2$, so the total sample size is $df + 2 = 17$ here.
• The treatment with more replicates has more weight in the comparison — here, treatment 2 with 10 replicates contributes more to the t-statistic than treatment 1 with 7.

Common Mistakes

• Writing $df = n_1 + n_2$ (forgetting the "− 2").
• Confusing df with total sample size and giving $n_2 = 8$ (which would be the answer if you wrongly used $df = n_1 + n_2$).
• Subtracting instead of adding: $15 = 7 - n_2 - 2 \Rightarrow n_2 = -10$ (nonsense).

Things to Be Careful About

Show the rearrangement of the formula explicitly; the mark scheme accepts the answer of 10 but credit is also given for correct working.

Background Concept

A p-value is the probability of obtaining data at least as extreme as the observed data, assuming the null hypothesis is true. By convention, a p-value below 0.05 is taken to mean that the result is "statistically significant" — the data are unlikely to have arisen by chance if there were truly no effect. The smaller the p-value, the stronger the evidence against the null hypothesis.

In this experiment:

• $H_0$ for each comparison: there is no significant difference between treatment 1 (no weed killer) and the weed-killer treatment.
• A small p-value → reject $H_0$ → the weed killer does affect photosynthesis.
• A large p-value (> 0.05) → fail to reject $H_0$ → the weed killer does not significantly affect photosynthesis.

Understanding the Question

The students concluded that the weed killers HAD a significant effect on the rate of photosynthesis. The question asks to what extent the p-values in Table 1.2 actually support that conclusion. Some p-values support it, one does not, and one effect is stronger than another.

Approach

Compare each p-value with the 0.05 threshold. Identify which are below 0.05 (significant) and which are above (not significant). Note the relative magnitudes where two are below the threshold.

Step-by-Step Reasoning

• Weed killer A: p < 0.001. This is well below 0.05; the result is highly significant. There is a real difference between treatments 1 and 2.
• Weed killer B: p > 0.05. This is above 0.05; the result is not significant. Any apparent difference between treatments 1 and 3 could plausibly be due to chance variation. The data do not support the claim that B affects photosynthesis.
• Weed killer C: p < 0.05. This is below 0.05; the result is significant. There is a real difference between treatments 1 and 4.
• Comparing A and C: both are significant, but A's p-value is smaller (< 0.001 vs < 0.05). The smaller p-value indicates stronger evidence against $H_0$, so the effect of A is greater than the effect of C.

Key Takeaways

• A single p-value threshold (0.05) is used to decide significance; results above it are not significant, results below it are.
• Smaller p-values indicate stronger evidence against the null, so they can be used to rank the strength of different effects.
• A conclusion that is true "on average" can still be wrong for specific treatments — the data must be interpreted treatment by treatment.

Common Mistakes

• Saying "all three p-values are significant" — B's p-value (> 0.05) is NOT significant. The mark scheme explicitly splits the answer between yes (A and C) and no (B).
• Saying "all three p-values are not significant" — A's and C's p-values are below 0.05.
• Comparing the p-values to the WRONG threshold (e.g. 0.01 or 0.10) — the conventional CIE threshold is 0.05.
• Failing to compare the effects of A and C.
• Confusing "smaller p-value" with "weaker effect" — it is the opposite: smaller p-value means stronger evidence that the effect is real, and (assuming the direction of effect is consistent) a larger absolute effect size.

Things to Be Careful About

The mark scheme awards 1 mark for the yes/no split, plus two further marks from the three supporting statements (significance criterion for A/C, ranking of A above C, non-significance of B). Cover at least the split and one supporting statement, ideally all three.