Biology 9700/35 — May/June 2020

Background Concept

Risk assessment distinguishes hazard (the intrinsic property of a substance that could cause harm — toxic, irritant, flammable, etc.) from risk (the likelihood and severity of that harm actually occurring during the procedure). CIE practical papers express risk using the words low, medium or high. A substance can have a recognised hazard but still pose a low risk if it is used in tiny quantities with appropriate controls; conversely, even dilute irritants can present medium risk if skin or eye contact is plausible.

In this investigation the four solutions are:

• Y — yeast suspension (biological, not classed as hazardous).
• H — hydrogen peroxide solution, labelled harmful irritant.
• C — 1.0% copper sulfate solution, labelled harmful irritant.
• W — distilled water (control), already completed as low.

Understanding the Question

Table 1.1 lists each solution with its hazard and (for W) a sample risk. You must complete the risk column for Y, H and C. Each word — low / medium / high — may be used once, more than once, or not at all.

Approach

Match the hazard wording to the appropriate risk band:

• "none" → low
• "harmful / irritant" → medium
• "very harmful / corrosive / toxic / flammable" → high

Step-by-Step Reasoning

1. Y (yeast suspension): hazard is "none". With no hazardous property, the chance of harm is negligible → low risk.
2. H (hydrogen peroxide solution): hazard is "harmful irritant". At the working concentration, contact with skin or eyes would cause irritation, so the risk is meaningful → medium risk.
3. C (1.0% copper sulfate solution): also "harmful irritant". Same logic as H → medium risk.
4. W (distilled water): already filled in as "low" — water is essentially non-hazardous.

The mark scheme accepts medium or high for H and C but explicitly rejects low because both carry an irritant warning.

Key Takeaways

• Risk = likelihood × severity, judged against how the chemical is actually used.
• "Harmful irritant" maps to at least medium risk, never low.
• Risk is relative to the procedure and quantity, not an absolute property of the chemical.

Common Mistakes

• Writing "low" for H or C because the volumes are small — risk considers the consequences of contact, not the quantity handled.
• Writing "high" for H or C without justification; "harmful irritant" corresponds to medium in CIE standard wording.
• Stating the hazard ("harmful", "irritant") instead of the risk category.

Things to Be Careful About

• The note "you may use each word once, more than once or not at all" is a hint that the three answers are not all distinct — here "medium" appears twice.
• Even though H and C are handled briefly, skin contact is plausible, so medium risk is the honest assessment.

Background Concept

A serial dilution is a stepwise dilution in which each successive tube/beaker contains a known fraction of the previous concentration, made up to the same total volume with diluent (here distilled water, W). A halving serial dilution halves the concentration at every step. Because the same fixed volume is transferred each time and topped up with the same fixed volume of diluent, the dilution factor is exactly ½ per step.

The mathematical relationship is:

where $c_1$ is the starting concentration and $c_n$ is the concentration in beaker $n$.

Understanding the Question

You must extend Fig. 1.1 (which already shows beaker 1 at 1.0% CuSO₄ with the first transfer arrow drawn) by drawing extra beakers, labelling the concentration available to use below each beaker, and adding arrows above each beaker showing the volume of C transferred in and the volume of W added. The dilution must halve the concentration between each successive beaker.

Approach

Work out the concentrations first, then add the arrows in the right places:

• Start with 20 cm³ of 1.0% CuSO₄ in beaker 1.
• Take 10 cm³ from beaker 1 into beaker 2, then add 10 cm³ of W to beaker 2 → beaker 2 now holds 20 cm³ at 0.5%.
• Take 10 cm³ from beaker 2 into beaker 3, add 10 cm³ of W → beaker 3 holds 20 cm³ at 0.25%.
• Continue halving for beakers 4 and 5.

Step-by-Step Reasoning

Starting concentration $c_1 = 1.0\%$:

For each beaker after the first, 10 cm³ of the previous (more concentrated) solution is transferred in, and 10 cm³ of distilled water W is added to restore the total volume to 20 cm³. The final beaker receives its 10 cm³ from beaker 4 plus 10 cm³ of W but transfers nothing onwards (it is the lowest concentration to use). Each label below a beaker states the volume and concentration available — "10 cm³ of [X]% copper sulfate solution to use" — matching the wording already given under beaker 1.

Key Takeaways

• A halving serial dilution gives concentrations $1.0, 0.5, 0.25, 0.125, 0.0625\%$ — five useful concentrations plus the W control.
• Each step requires both a transfer of the previous solution and an equal-volume addition of diluent to keep total volume constant.
• The mark scheme requires the concentration sequence, all transfer arrows (10 cm³ each), all water additions (10 cm³ of W each), and the volume/concentration labels under each beaker.

Common Mistakes

• Forgetting to add water back to the beaker — this leaves only 10 cm³ in the beaker and breaks the dilution factor.
• Labelling the wrong concentration under a beaker (e.g. putting 0.5% under the third beaker).
• Using different volumes for the water addition (e.g. adding 20 cm³ of W instead of 10 cm³).
• Putting the % sign inconsistently; the mark scheme requires % stated at least once.

Things to Be Careful About

• The volume available to use is 10 cm³ from each beaker (20 cm³ in beaker 1 is split 10 cm³ for transfer + 10 cm³ to use).
• Beaker 5 has no outgoing transfer arrow — only an incoming transfer from beaker 4 and an incoming water addition.
• The label must include both volume and concentration of solution to use, matching the wording for beaker 1.

Background Concept

Results tables in CIE Paper 3 follow strict conventions. Each column must have a heading that includes both a quantity and a unit. The independent variable (the one deliberately changed by the experimenter) is placed in the left-hand column and the dependent variable (the one measured) on the right. Decimal places must be consistent down a column. Where the dependent variable is a time measured by stop-clock, the value should be recorded in whole seconds (rounded, not given with spurious decimal precision).

In this experiment yeast catalase decomposes hydrogen peroxide:

Copper sulfate is a non-competitive inhibitor of catalase (it binds away from the active site and distorts the enzyme's tertiary structure). Higher inhibitor concentrations therefore slow the reaction more, so the time taken to collect a fixed volume of oxygen increases with [CuSO₄].

Understanding the Question

After performing steps 1–14 the candidate times how long it takes the water level inside the inverted small test-tube to fall to the 8 cm mark as oxygen displaces water. The candidate records their own times for each of the six conditions: 0% (W control), 0.0625%, 0.125%, 0.25%, 0.5%, 1.0%.

Approach

Build a clean two-column table with the IV (% CuSO₄) on the left, headed by a quantity-plus-unit label, and the DV (time) on the right, also with a unit. Record each time as a whole number of seconds. The expected trend is monotonic: time increases with concentration.

Step-by-Step Reasoning

1. Heading for the independent variable — "Percentage concentration of copper sulfate solution (%)". The unit % is in the heading (not in the body). The mark scheme explicitly requires the IV to be on the left.
2. Heading for the dependent variable — "Time / s". Unit s placed in the heading, not the body.
3. Rows — six rows, one for each concentration tested, in ascending concentration order so the trend reads from top to bottom.
4. Data — each value is the time in whole seconds taken for the water to fall to the line. If the reaction had not finished within 180 s the recording must be the literal phrase 'more than 180'.
5. Expected trend — the longest time must be at the highest percentage of CuSO₄ (1.0%), because more inhibitor slows catalase more. The control (0%, W) should give the shortest time.

Representative (illustrative) times that demonstrate the trend:

These are example values only — the candidate's actual times depend on the activity of their yeast suspension and the temperature of the water bath, but the order must be monotonic.

Key Takeaways

• IV on the left, DV on the right; both headed by quantity + unit.
• Whole-second precision for stop-clock times; 'more than 180' is the correct way to record a reaction that did not complete.
• The trend must be longest time = highest [CuSO₄], because higher inhibitor concentration slows the enzyme.

Common Mistakes

• Putting the DV (time) on the left of the table — this loses the IV-heading mark.
• Recording times like "0:35" or "35.5 s" or "35 s (approx.)" — whole seconds with no other formatting.
• Writing the unit inside the body cells ("30 seconds", "40 s") instead of in the heading.
• Reversing the trend or showing non-monotonic data — likely to lose the trend mark unless justified.

Things to Be Careful About

• Six rows are expected (0%, 0.0625%, 0.125%, 0.25%, 0.5%, 1.0%); the mark scheme requires at least five.
• 'more than 180' is acceptable as a recording if the reaction was very slow.
• The expected trend (highest % = longest time) is the fifth marking point and must be visible from the recorded values.

Background Concept

In any controlled experiment the independent variable (IV) is the factor the experimenter deliberately changes between trials, the dependent variable (DV) is the factor measured as the outcome, and all other factors are controlled (kept constant). Identifying these correctly is the first step in designing or interpreting an experiment.

Understanding the Question

The candidate has just carried out an investigation in which several different concentrations of copper sulfate were tested. The question asks which factor was the IV — i.e. what was being deliberately varied.

Approach

Look at the method: a series of different concentrations of CuSO₄ were prepared by serial dilution, and the time for a fixed volume of oxygen to be collected was measured for each. The thing being changed from one trial to the next is therefore the CuSO₄ concentration; the thing being measured is the time.

Step-by-Step Reasoning

1. The serial dilution prepared six different CuSO₄ concentrations: 0% (W control), 0.0625%, 0.125%, 0.25%, 0.5%, 1.0%.
2. For each of these, the time taken for the water level to fall to the 8 cm line was measured.
3. What is varied = concentration of copper sulfate solution.
4. What is measured = time for oxygen collection.

So the IV is the concentration of copper sulfate solution.

Key Takeaways

• IV is what the experimenter changes deliberately (the serial dilution provides a series of values).
• DV is what is measured (time here).
• All other variables (volume of yeast, volume of H₂O₂, temperature, total reaction volume) are controlled.

Common Mistakes

• Saying "copper sulfate" without specifying concentration — this conflates the substance with the variable.
• Stating the DV (time, or volume of oxygen) instead of the IV.
• Confusing "independent" with "controlled" variables.

Things to be Careful About

• The mark scheme accepts '(percentage) concentration of copper sulfate (solution)' but rejects "copper sulfate" alone.
• Include both "percentage" and "concentration" for clarity, plus the word solution or CuSO₄.

Background Concept

Enzyme activity is strongly affected by temperature. As temperature rises, kinetic energy increases and the rate of successful enzyme–substrate collisions rises, but beyond the optimum temperature the enzyme denatures and activity falls sharply. To investigate the effect of temperature you must hold all other variables (pH, substrate concentration, enzyme concentration, inhibitor concentration) constant and vary only the temperature. A water-bath is the standard tool for maintaining a reaction mixture at a chosen temperature.

Understanding the Question

You are asked to modify the existing CuSO₄-inhibition experiment so that it instead investigates the effect of temperature on the breakdown of hydrogen peroxide. Specifically, you must describe how the independent variable will be changed. (How the temperature will be measured / controlled and what range of values to use.)

Approach

The existing experiment changes concentration of CuSO₄ between trials. To investigate temperature instead:

1. Hold CuSO₄ concentration at a fixed value (e.g. none / W, or a constant low concentration).
2. Replace concentration variation with temperature variation.
3. Use multiple water-baths at chosen temperatures.
4. Allow the reactants to equilibrate to the target temperature before mixing, so the reaction begins at that temperature.

Step-by-Step Reasoning

1. Apparatus: several thermostatically-controlled water-baths (one per temperature tested).
2. Range and interval: at least five different temperatures spaced across a sensible range. CIE typical practice is $0\ ^{\circ}\text{C}$ to $100\ ^{\circ}\text{C}$; representative choices are $20, 30, 40, 50, 60\ ^{\circ}\text{C}$ (or wider: $10, 25, 40, 55, 70\ ^{\circ}\text{C}$).
3. Procedure: place each reaction tube in its water-bath and leave it to equilibrate for a fixed time (e.g. 2 minutes) so that the contents reach the bath temperature before mixing yeast and hydrogen peroxide.
4. Start timing when the H₂O₂ is added and measure the same outcome (time to collect oxygen to the 8 cm line, or volume of oxygen in a fixed time).
5. Keep all other variables constant: same volume of Y (5 cm³), same volume of H (2 cm³), same concentration of CuSO₄ (e.g. none / W), same total volume, same small test-tube.

Key Takeaways

• A valid investigation of temperature requires varying temperature while holding every other variable constant.
• Equilibration before mixing is essential — otherwise the reaction starts at room temperature and the recorded effect of the bath is partly lost.
• A range of at least five temperatures is required to identify a trend and any optimum.

Common Mistakes

• Saying "change the temperature" without specifying a range or number of values.
• Failing to equilibrate the reactants — reaction starts at a temperature different from that of the bath.
• Changing more than one variable at once (e.g. varying CuSO₄ and temperature).
• Using the word "heat" vaguely instead of "water-bath at a controlled temperature".

Things to be Careful About

• A water-bath is the standard controlled-temperature device; an open Bunsen flame is not a controlled variable.
• Above ~$60\ ^{\circ}\text{C}$ yeast catalase denatures, so including values up to $100\ ^{\circ}\text{C}$ captures the fall in activity as well as the rise — useful for showing an optimum.
• The mark scheme requires both the water-bath point and at least five different temperatures OR explicit mention of equilibrating before mixing.

Background Concept

When a reaction produces a gas that escapes from an open container, the mass of the system falls. The fall in mass equals the mass of gas that has left. The percentage change expresses that loss relative to the starting mass and lets results from reactions of different sizes be compared.

The two formulas are:

The percentage change carries a negative sign when the final mass is less than the initial mass — this is not optional, it tells the reader the sample lost mass.

Understanding the Question

Table 1.2 gives the initial mass (249.55 g) and final mass (243.31 g) of the reaction mixture in the student's conical flask. You must calculate the change in mass and then the percentage change in mass, and write both into the empty cells of Table 1.2.

Approach

Apply the two formulas in order. Subtract the initial mass from the final mass to get the change (negative because mass has been lost as oxygen). Divide by the initial mass and multiply by 100 for the percentage change. Round the percentage change to an appropriate number of significant figures (the input masses are given to 4 sig figs, so the percentage change is quoted to 3 sig figs).

Step-by-Step Reasoning

1. Change in mass:

1. Percentage change:

Carry out the division:

Multiply by 100:

1. Significant figures: the initial and final masses are given to 4 sig figs (249.55, 243.31); the answer should be quoted to 3 sig figs (matching the precision of the change in mass which has only 3 sig figs). $-2.50\%$ is acceptable; $-2.5\%$ is acceptable but loses a sig fig.

Key Takeaways

• The change in mass is final − initial, and is negative when mass is lost.
• Percentage change = (change ÷ initial) × 100; the sign is preserved.
• Match the precision of the answer to the precision of the data given.

Common Mistakes

• Subtracting the wrong way round (initial − final) and getting a positive number.
• Forgetting the negative sign on the percentage change — losing the meaning of "loss of mass".
• Quoting $-2.5005\%$ without rounding, or rounding to $-3\%$, which loses precision.
• Using $249.55$ as the denominator and then dividing by $6.24$ instead of $-6.24$, dropping the sign.

Things to be Careful About

• The mark scheme accepts $-2.5\%$ as well as $-2.50\%$, but rejects $-2\%$ (too few sig figs).
• The units for the change in mass are grams (g); the percentage change has no units (it is dimensionless).
• Write the percentage change with its sign in the table.

Background Concept

The reliability of a quantitative result is judged by how close repeated measurements are to one another (and to the true value). Random errors (e.g. slight variations in starting mass, timing, room temperature, exact volume of H₂O₂) affect each individual reading; their effect is reduced by taking replicates and using their mean as the best estimate of the true value. The more independent repeats, the smaller the standard deviation around the mean and the more confidence one can place in the result.

Understanding the Question

The student has carried out the experiment once and obtained a single percentage change. The question asks how they could increase confidence in that specific result.

Approach

Identify the source of uncertainty: a single measurement. The remedy is replication combined with averaging — the standard technique for improving reliability in any quantitative experiment.

Step-by-Step Reasoning

1. The single percentage change ($−2.50\%$) is just one value. It could be $1\%$ too high or too low due to random error in weighing, timing, or reactant volumes.
2. To check whether the value is reproducible, repeat the experiment several times under the same conditions.
3. Calculate the mean of the resulting percentage changes. The mean is less affected by random fluctuations in any one trial.
4. (Optional, not required for the mark) calculate the standard deviation or 95% confidence interval to quantify the spread.

Key Takeaways

• Confidence in a result comes from showing that it can be reproduced.
• The mean of several replicates is a more reliable estimate than any single reading.
• This is true for any quantitative measurement, not just mass-loss experiments.

Common Mistakes

• Suggesting vague "be more careful" or "human error" answers — these do not address the specific issue of a single reading.
• Suggesting a control experiment — that addresses validity, not confidence in this particular value.
• Saying "repeat and take an average" without explicitly mentioning what is being averaged.

Things to be Careful About

• The mark scheme credits only repeat + calculate a mean; one without the other is incomplete.
• This question is about confidence in this result, not about improving the method overall — so changing apparatus or technique is not relevant.

Background Concept

A bar chart is the correct display for a discrete (categorical) independent variable. Each bar represents one category, and the bar's height encodes the value of the dependent variable for that category. CIE conventions for bar charts:

• x-axis carries the discrete categories (here the letters N, P, Q, R, S);
• y-axis is a continuous scale of the dependent variable, beginning at 0;
• bars are of equal width, with gaps between them;
• the y-scale should be linear and use at least half the grid;
• bars should be drawn with thin continuous lines, with a clear horizontal top edge.

If the IV were a continuous variable (e.g. concentration on the x-axis), a line graph would be appropriate instead.

Understanding the Question

Table 1.3 lists five different inhibitors (N, P, Q, R, S) and the volume of oxygen collected from each reaction after 5 minutes. The IV is categorical (the identity of the inhibitor) and the DV is continuous (volume of oxygen in cm³). The grid in Fig. 1.6 is supplied blank. You must draw a bar chart of this data, following the conventions above.

Approach

1. Choose the y-scale so that the tallest bar fits comfortably and the scale uses at least half the grid. The largest value is 8.6 cm³, so a scale of 0–10 cm³ with major gridlines every 2 cm³ is appropriate.
2. Label both axes (quantity + unit on y, category on x).
3. Draw five bars of equal width, separated by gaps, with one bar above each inhibitor label.
4. Mark each bar's top with a thin horizontal line at the correct height (to the nearest mm on the grid).

Step-by-Step Reasoning

The five data points to plot are:

The range of values is 3.9 to 8.6 cm³. With 2 cm = 2 cm³, each bar top can be located at:

• N: $4.3 \times 2 = 8.6$ cm above the x-axis (bar reaches the top label "8") — actually $8.6\ \text{cm}^3$ at gridline "8" plus three small squares (since $1\ \text{cm}^3 = 1$ small square on this scale).
• P: $8.2\ \text{cm}^3$, just below N's bar by $0.4$ cm on the grid.
• Q: $4.3\ \text{cm}^3$, just above the "4" gridline by $0.3$ cm.
• R: $7.8\ \text{cm}^3$, just below the "8" gridline by $0.2$ cm.
• S: $3.9\ \text{cm}^3$, just below the "4" gridline by $0.1$ cm.

Key Takeaways

• Use a bar chart for a categorical IV and a line graph for a continuous IV.
• y-axis must begin at 0 and use a sensible scale that occupies at least half the grid.
• Bars are of equal width with gaps, drawn with a sharp pencil and a clear flat top line.
• Both axes must be labelled with quantity + unit.

Common Mistakes

• Drawing a line graph instead of a bar chart (this loses marks because the IV is categorical).
• Leaving the y-axis without a unit (the unit cm³ must be in the heading).
• Using unequal bar widths or omitting the gaps between bars.
• Choosing a scale that does not use at least half the grid (e.g. 0–100 when all values are < 10).
• Plotted bar tops that are visibly off the correct gridline (mark scheme requires plotting accurately).

Things to be Careful About

• The y-axis label must include both the quantity ("volume of oxygen after 5 minutes") and the unit ($\text{cm}^3$ in the heading).
• The x-axis label is the word inhibitor — the letters N, P, Q, R, S go under the bars, not as the axis title.
• 2 cm on the grid = 2 cm³ (i.e. 1 cm = 1 cm³) is a natural and recommended scale.

Background Concept

Enzymes are biological catalysts with an active site — a small region whose three-dimensional shape is complementary to the substrate. The substrate binds to the active site to form an enzyme–substrate complex (ESC); the enzyme then catalyses the conversion of substrate to product.

Enzyme inhibitors reduce the rate of an enzyme-controlled reaction. There are two main classes:

• Competitive inhibitors have a shape similar to the substrate and bind reversibly to the active site. They compete with the substrate for the active site, but inhibition can be overcome by adding more substrate.
• Non-competitive inhibitors bind to a site other than the active site (an allosteric site), changing the enzyme's shape so the substrate can no longer bind. Inhibition cannot be overcome by adding more substrate.

Understanding the Question

Table 1.3 shows that different inhibitors give very different rates of oxygen production, suggesting they inhibit the catalase enzyme to different extents. The question states that one of these inhibitors is competitive and asks you to explain how competitive inhibitors affect an enzyme-controlled reaction in general.

Approach

The explanation should connect four ideas in order: shape similarity to substrate → fits the active site → competes with substrate → fewer ESCs form → reaction slows. Each link in the chain is a mark.

Step-by-Step Reasoning

1. Shape similarity: the inhibitor's overall three-dimensional shape resembles that of the substrate (the natural substrate of catalase is hydrogen peroxide, H₂O₂). This is why the inhibitor is recognised by the enzyme's active site at all.
2. Complementary to active site: because the inhibitor is similar in shape to the substrate, it has a complementary shape to the active site of the enzyme. It therefore fits into the active site where the substrate would normally bind.
3. Competition: the inhibitor binds to the active site, occupying it in place of the substrate. The substrate and inhibitor compete for the same site. This is reversible — the inhibitor can leave and the substrate can then bind.
4. Consequence: because some active sites are occupied by inhibitor instead of substrate, the number of enzyme-substrate complexes (ESCs) formed per unit time falls. The reaction rate therefore decreases. (Note that adding more substrate can out-compete the inhibitor and restore the rate.)

In the data of Table 1.3, a competitive inhibitor would be the one with the smallest volume of oxygen unless substrate concentration is saturating; a non-competitive inhibitor would still produce a low volume of oxygen even at very high substrate concentration.

Key Takeaways

• Competitive inhibition depends on shape mimicry of the substrate.
• The inhibitor and substrate compete for the same active site.
• The inhibition is reversible and is overcome by high substrate concentrations.
• The consequence is fewer ESCs and a lower reaction rate.

Common Mistakes

• Saying the inhibitor "blocks the active site" without mentioning its shape similarity to the substrate — this misses the mechanistic reason why the inhibitor can fit the site.
• Confusing competitive with non-competitive inhibition (e.g. saying the inhibitor binds to a different site).
• Saying the inhibitor "destroys" or "denatures" the enzyme — competitive inhibitors do not alter the enzyme's tertiary structure.
• Omitting the consequence (fewer ESCs, slower rate).

Things to be Careful About

• The mark scheme awards any three of the four listed points; including all four is safest.
• The term enzyme-substrate complex (ESC) is required — "enzyme-substrate" alone is too vague.
• Avoid using the word "competitive" as the explanation (it is the term being defined).