Biology 9700/36 — October/November 2019

Background Concept

A risk assessment separates the hazard (the intrinsic property of a substance that could cause harm) from the risk (the likelihood and severity of harm actually occurring under the conditions of use). CIE asks for a single low/medium/high rating that reflects how the chemical will be used in this particular procedure, not just the label on the bottle.

Understanding the Question

Table 1.1 lists three chemicals with their hazard, volume and (your task) risk. The hazard column tells you that H and P are "harmful irritant" and W is "none". You must write the risk of each, in the context of small volumes, brief handling and the recommended eye protection. The same word may be used for more than one row.

Approach

Apply the CIE convention: an intrinsically harmless chemical → low risk; a harmful/irritant chemical handled briefly in small volumes with PPE → medium risk; a corrosive, toxic or highly flammable chemical (or one used at high concentration) → high risk. Because the procedure uses small volumes and eye protection is recommended, both H and P stop at medium rather than reaching high.

Step-by-Step Reasoning

• H (hydrogen peroxide): hazard = harmful irritant; used in an open beaker, brief contact, eye protection advised. Medium risk is correct (high is also accepted by the mark scheme).
• W (distilled water): hazard = none. The only defensible rating is low.
• P (plant extract): hazard = harmful irritant; used in small volumes, brief contact. Medium risk is correct (high is also accepted).

The mark scheme gives one mark for the trio "H medium or high; W low; P medium or high".

Key Takeaways

• Risk is not the same as hazard. Volumes, exposure time and PPE all lower the realised risk of a hazardous chemical.
• Distilled water is always low risk in a school/college laboratory.
• The same rating word may legitimately appear in more than one row.

Common Mistakes

• Rating H or P as "low" because the volume looks small. The hazard label, not the volume, is the primary driver.
• Rating W as anything other than low.
• Using three different words when two of the rows are permitted to share a rating.

Things to Be Careful About

The question is about risk under the stated conditions (small volumes, eye protection). Do not over-rate; "medium" is the conservative CIE answer for both H and P.

Background Concept

Proportional dilution is the standard way to prepare a series of working solutions of known concentration from a single stock. Each working concentration is made by mixing a fraction of the stock with the complementary fraction of diluent (here distilled water W) so that the volumes sum to a fixed total. The fraction of stock equals the percentage concentration divided by 100. Because each beaker in step 3 contains $10\ \text{cm}^3$, you prepare $10\ \text{cm}^3$ of each concentration.

Understanding the Question

Table 1.2 already has 100% (10 cm³ P + 0 cm³ W) and 0% (0 cm³ P + 10 cm³ W). You must fill in the rows between so that catalase activity can be measured across a useful range. CIE expects at least three further concentrations (worth two marks: one for choosing ≥3, one for the volumes being correctly proportional).

Approach

Pick intermediate concentrations evenly spread across the range — e.g. 20, 40, 60 and 80%. For each one, the volume of P equals (concentration/100) × total volume, and the volume of W makes up the rest. This guarantees a straight-line dilution series with predictable volumes.

Step-by-Step Reasoning

For 80%: 8.0 cm³ P + 2.0 cm³ W (total $10\ \text{cm}^3$).
For 60%: 6.0 cm³ P + 4.0 cm³ W.
For 40%: 4.0 cm³ P + 6.0 cm³ W.
For 20%: 2.0 cm³ P + 8.0 cm³ W.

Any subset of three or more intermediate rows in correct proportions scores both marks. Non-uniform spacings (e.g. 75, 50, 25) are also accepted provided the volumes are correct.

Key Takeaways

• Proportional dilution: volume of stock = (concentration/100) × total volume; diluent makes up the rest.
• CIE asks for at least three further concentrations between 0% and 100% to span the range.
• Total volume must equal the working volume ($10\ \text{cm}^3$ here) so the whole solution fits in one beaker.

Common Mistakes

• Mixing to a different total — e.g. 8 cm³ of P + 4 cm³ of W for "60%" gives 12 cm³, not 10.
• Adding only two intermediate rows; CIE requires at least three.
• Concentrations that cluster near one end (e.g. 90, 80, 70) — these do not span the full range.

Things to Be Careful About

Quote volumes to a sensible precision (one decimal place of a cm³ matches the columns already given) and ensure each row sums to exactly $10\ \text{cm}^3$.

Background Concept

A results table on a CIE Paper 3 must obey strict conventions so that another scientist can read and interpret the data without ambiguity. The independent variable (the one the experimenter changes) appears first, in its own column, with no units repeated in the body of the table (units go in the heading). The dependent variable (what you measure) appears next, again with units in the heading. Replicates (repeat measurements at the same concentration) are shown in adjacent columns, and decimal places are kept consistent across each column.

Understanding the Question

You have carried out the disc-rising experiment for each concentration of P from Table 1.2, three times per concentration, timing how long each soaked disc takes to reach the surface of H. You must record these raw results in an appropriately designed table.

Approach

Lay out the table with one row per concentration of P and one column per replicate test. Write the IV heading as "Percentage concentration of P" and put % only in the heading (no body units). Write the DV heading as "Time for disc to reach surface / s" so the unit appears once in the heading only. Enter whole-second values — never write "23.5 s" or "about a minute". If a disc has not risen by 180 s, record ">180" rather than guessing.

Step-by-Step Reasoning

The five mark-scheme points, and how the table above satisfies each:

1. IV heading: "Percentage concentration of P" — clearly named, no units repeated in the body.
2. DV heading: "Time for disc to reach surface / s" — "seconds" appears in the heading and no unit is repeated in the body.
3. Readings for all samples: every concentration from 100% to 0% has three replicate timings.
4. Trend: time increases as concentration decreases — 11 → 15 → 24 → 51 → 135 → >180. This is correct because higher catalase concentration means a faster initial rate of O₂ production, so the disc rises sooner.
5. Whole seconds: every entry is an integer (or ">180").

The representative values shown are illustrative; your own timings will differ but the trend and conventions must match.

Key Takeaways

• Independent variable in its own column, units in the heading only.
• Dependent variable with its unit ("/ s") in the heading; never repeat units inside the body.
• Replicates in side-by-side columns, all to the same precision.
• If a disc has not surfaced by 180 s, write ">180" — do not estimate.

Common Mistakes

• Writing "seconds", "s" or "(s)" inside the body cells as well as in the heading.
• Mixing units (e.g. recording one replicate in seconds and another in minutes).
• Using a mean column. The question rewards raw readings, not means.
• Quoting decimal seconds — the stop-clock reads to the nearest second.

Things to Be Careful About

The question explicitly asks for results "recorded to the nearest whole second". If you have to estimate between second marks, round rather than record fractions.

Background Concept

The independent variable is the factor that the experimenter deliberately changes between trials. The dependent variable is the factor that is measured in response. Everything else (control variables) is held constant so any change in the dependent variable can be attributed to the independent variable.

Understanding the Question

Step 1 says "Prepare the concentrations of P as stated in Table 1.2" and the procedure then measures how long the disc takes to rise for each of those concentrations. The thing the experimenter varies is therefore the concentration of P.

Approach

Look at the procedure and ask "what am I changing on purpose?" — that is the independent variable. Here the experimenter prepares several concentrations of P and tests each in turn. The variable deliberately varied is therefore concentration of plant extract.

Step-by-Step Reasoning

• The experimenter prepares 100%, 80%, 60%, 40%, 20% and 0% of P (step 1).
• For each, the experimenter then measures the time the disc takes to rise (steps 2–5).
• Therefore, the thing that varies between trials (the independent variable) is the concentration of plant extract (P).
• The dependent variable is the time for the disc to reach the surface, and control variables include the concentration of H, the disc size, blotting time, temperature and so on.

Key Takeaways

• Independent variable = what you change.
• Dependent variable = what you measure.
• Control variables = everything you keep the same so the comparison is fair.

Common Mistakes

• Saying "time" or "the rate of reaction" — these are the dependent variable, not the independent one.
• Saying "hydrogen peroxide concentration" — that is the variable varied in part (b), not in part (a).

Things to Be Careful About

Be precise: "concentration of plant extract (P)" is what the mark scheme accepts; vague answers such as "the amount of enzyme" may be rejected if they do not match the language of the question.

Background Concept

A source of error is any uncontrolled feature of the procedure that adds unwanted variability to the dependent variable. A genuine improvement must address that specific feature — not introduce a new variable. CIE rewards error/improvement pairs and ignores the improvement if the error has not been named.

Understanding the Question

The disc-rising method is fast and visual but has several weaknesses: the peroxide is consumed or evaporates, the discs soak up different amounts of extract, the blotting is variable, the end-point is judged by eye and discs may snag on the glass. You must name three weaknesses and, for each, suggest one practical improvement that would make the results more trustworthy.

Approach

For each pair, write the error as a fact about the procedure (not about the candidate), and the improvement as a specific change to the method (not a vague "be more careful"). The mark scheme lists five recognised pairs; any three earn the three marks.

Step-by-Step Reasoning

Five recognised pairs (choose any three):

1. Re-used H₂O₂: the same beaker of H is used repeatedly, so the substrate concentration falls and the disc rises more slowly each time. Improvement: replace H for each concentration of P.
2. Variable extract on disc: when you dip the disc into P, the amount that clings to it is uncontrolled. Improvement: soak every disc in the same volume of P for a fixed time, then blot for a fixed time on the paper towel.
3. Variable blotting: the time spent on the paper towel differs between discs, so each carries a different amount of extract. Improvement: standardise the blotting time (e.g. exactly 5 s) for every disc.
4. Subjective end-point: judging the moment the disc reaches the surface by eye is unreliable between observers. Improvement: draw a thin line on the outside of the beaker at the surface and time until the disc crosses it.
5. Disc sticking: in a narrow beaker the disc may catch on the glass and not rise cleanly. Improvement: use a wider beaker (or a measuring cylinder) so the disc falls straight to the bottom.

Key Takeaways

• The error must name the uncontrolled variable, not the candidate ("the experimenter").
• The improvement must act on that specific variable; "do it more carefully" is rejected.
• Each pair stands alone — an unmarked improvement cannot inherit credit from a missing error.

Common Mistakes

• Writing the error and improvement in the wrong order (e.g. "use fresh H₂O₂" as the error).
• Suggesting vague improvements such as "repeat more times" — that addresses replication, not any specific procedural flaw.
• Naming equipment issues that are not actually present (e.g. "the timer is inaccurate" when a stop-clock is provided).

Things to Be Careful About

The improvement must be practical in this experiment — "use a gas syringe" is not realistic because the procedure as written uses discs. Stay within the method described.

Background Concept

An anomalous result is one that does not fit the pattern of the other replicates. It is usually identified because it lies far outside the range covered by the others, and it is excluded before a mean is calculated. The remaining values are then averaged; the answer is quoted to the same number of decimal places as the other means in the table so the column is internally consistent.

Understanding the Question

Table 1.3 has five tests at 2.0% hydrogen peroxide: 8.7, 10.5, 8.5, 8.3 and 8.8 cm³. All other rows in the table already have a mean to one decimal place; you must fill in the empty mean cell for this row, working in the space provided.

Approach

First decide which value is the anomaly. Four of the five values lie between 8.3 and 8.8 cm³; the fifth, 10.5 cm³, is clearly out of line and should be excluded. Then average the remaining four values and round to one decimal place to match the other rows.

Step-by-Step Reasoning

• Tests at 2.0%: 8.7, 10.5, 8.5, 8.3, 8.8 cm³.
• 10.5 is more than 1.5 cm³ above the next highest (8.8) and more than 2 cm³ above the next lowest — clearly anomalous.
• Excluding 10.5: (8.7 + 8.5 + 8.3 + 8.8) / 4 = 34.3 / 4 = 8.575.
• Round to 1 d.p. (matching the other means): 8.6 cm³.
• If a candidate includes 10.5 the mean is 8.96, which rounds to 9.0 cm³ — this is wrong because the mark scheme awards credit only for 8.6 (one of the two marks requires correct rejection of the anomaly).

Key Takeaways

• Always check for an outlier before averaging — exclude it if it lies well outside the range of the others.
• Round the mean to the same number of decimal places as the other means in the column.
• Show your working in the space provided, with the anomalous value circled or struck through.

Common Mistakes

• Averaging all five values to give 8.96 ≈ 9.0 cm³.
• Quoting 8.58 or 8.575 — too many decimal places for the convention used in Table 1.3.
• Failing to state which value was excluded.

Things to Be Careful About

The mark scheme gives two marks here: one for the value 8.6 and one for it being at the correct degree of accuracy (1 d.p.). Watch the precision.

Background Concept

A scatter graph with a smooth curve of best fit is the standard way to display a continuous relationship between an independent variable ($x$-axis) and a dependent variable ($y$-axis). CIE conventions demand: (1) labelled axes with units in the heading only; (2) sensible linear scales that fill at least half the printed grid; (3) accurately plotted points marked with small crosses or circled dots; (4) a thin smooth curve passing through (or as close as possible to) every point.

Understanding the Question

You must plot the six mean values from Table 1.3 on the printed grid of Fig. 1.2, then join them with an appropriate line. The relationship between substrate concentration and enzyme rate is non-linear (it rises steeply at low concentrations and levels off as the enzyme becomes saturated), so a smooth curve, not a straight line, is required.

Approach

Choose the axes first (x = substrate concentration; y = oxygen produced). Pick scales that are simple, even multiples and span the data without wasting grid. Plot each mean carefully against its grid intersection. Then sketch a single smooth curve through the six plots.

Step-by-Step Reasoning

• Axes (mark 1): x-axis = "Percentage concentration of hydrogen peroxide" with % in the heading only; y-axis = "Volume of oxygen produced in 30 seconds / $\text{cm}^3$" with $\text{cm}^3$ in the heading only.
• Scales (mark 2): x-axis — $0.5\%$ to 2 cm, labelled every 2 cm; y-axis — $2.0\ \text{cm}^3$ to 2 cm, labelled every 2 cm. Both scales use more than half the grid (six and five intervals respectively).
• Plotting (mark 3): the six means (4.0, 6.4, 7.8, 8.6, 9.4, 9.7) are plotted as small crosses or circled dots at the correct grid intersections.
• Curve (mark 4): a single thin smooth curve is drawn through (or as close as possible to) every plot. The curve rises steeply at low concentrations and flattens at high concentrations — the classic Michaelis–Menten shape for an enzyme approaching $V_{max}$.

Key Takeaways

• Units go in the heading only; never repeat them inside the grid.
• Choose scales that are even, easy to read, and use at least half the grid in both directions.
• Always use a sharp pencil for plots and the curve so corrections can be made neatly.
• A smooth curve, not a ruler and not plot-to-plot straight segments, is the expected line for enzyme-rate data.

Common Mistakes

• Putting units in the body of the axis ("4 cm³, 6 cm³") instead of just "$\text{cm}^3$" in the heading.
• Choosing awkward scales (e.g. 3 cm³ per 2 cm) that are hard to plot against.
• Drawing thick or doubled lines, or ruler-straight segments between plots.
• Extrapolating the curve back to the y-axis or forward beyond the last point.

Things to Be Careful About

• Always use a sharp pencil.
• The curve should pass through all six plots; if one plot is genuinely off the trend, the curve should pass as close as possible to it.
• Do not extend the curve beyond the first or last x-value — the relationship is only known within the measured range.

Background Concept

Interpolation is reading a value from a graph at a point that lies between two known data points. It assumes the smooth curve is a fair representation of the trend between the plots. The technique is: from the desired y-value, draw a horizontal line to the curve; from the intersection, drop a vertical line to the x-axis; read the value at that point.

Understanding the Question

Your curve passes through six plots covering substrate concentrations from 0.5% to 3.0%. The question asks: at what concentration does the curve cross y = $7.0\ \text{cm}^3$?

Approach

Lay a ruler horizontally across the grid at y = 7.0 cm³. Note where the horizontal line meets your curve — this point lies between the (1.0%, 6.4) and (1.5%, 7.8) plots. Drop a vertical line down from that intersection to the x-axis and read the value.

Step-by-Step Reasoning

• The plots at 1.0% (6.4 cm³) and 1.5% (7.8 cm³) bracket the value 7.0 cm³.
• Linear interpolation gives x = 1.0 + (7.0 − 6.4) / (7.8 − 6.4) × 0.5 = 1.0 + 0.43 × 0.5 ≈ 1.21.
• A real smooth curve gives a slightly different reading, but any answer in the range 1.1%–1.3% is accepted by the mark scheme ("correct value from graph").
• Round to one decimal place to match the precision of the x-axis: 1.2%.

Key Takeaways

• Interpolation is reading a value between known data points; extrapolation (beyond the data) is not justified here.
• The horizontal-then-vertical technique is the standard way to read a graph precisely.
• Always check the answer by asking whether it is bracketed correctly between the nearest two plots.

Common Mistakes

• Reading the wrong axis (giving a y-value when an x-value is asked for).
• Misreading the scale on the x-axis (the interval is 0.5%, not 1%).
• Quoting too many significant figures (e.g. "1.214%") — match the precision of the axis.

Things to Be Careful About

The mark scheme says "correct value from graph" — so a small difference between candidates (e.g. 1.1% vs 1.3%) is acceptable. What is not acceptable is a value clearly outside the bracketing plots (e.g. 0.8% or 1.7%).

Background Concept

Enzyme catalysis depends on enzyme–substrate complex formation. For an enzyme to work, a substrate molecule must collide with the active site in the correct orientation and with sufficient energy. Anything that increases the rate of successful collisions will increase the rate of reaction — until the enzyme becomes saturated, after which further increases in substrate have little effect (this is why the curve in part (b)(ii) levels off).

Understanding the Question

The graph in Fig. 1.2 shows that as the concentration of hydrogen peroxide rises from 0.5% to 3.0%, the volume of oxygen produced in 30 s rises from 4.0 cm³ to 9.7 cm³. The question asks you to explain this trend using the biology of enzyme action.

Approach

Link the variable the experimenter changed (substrate concentration) to the rate they measured (oxygen production) using the standard chain: more substrate → more collisions per unit time → more enzyme–substrate complexes → faster reaction. Each link in the chain must be stated explicitly; the mark scheme gives one mark per link.

Step-by-Step Reasoning

1. More substrate molecules: increasing the concentration of H₂O₂ from 0.5% to 3.0% means more H₂O₂ molecules per cm³ of solution.
2. More successful collisions: with more substrate molecules moving through the same volume, statistically more of them will collide with the active site of a catalase molecule per unit time.
3. More enzyme–substrate complexes formed: each successful collision forms an enzyme–substrate complex that breaks down to release product; more complexes per unit time means more O₂ released per unit time.

These three points correspond exactly to the three marks in the mark scheme. Note that the curve begins to level off above ~2% H₂O₂ — at this point the enzyme molecules are saturated and adding more substrate cannot increase the rate further. The question does not ask for this, but it explains why the curve flattens.

Key Takeaways

• Substrate concentration affects reaction rate only when the enzyme is not saturated.
• The chain "more substrate → more collisions → more complexes → faster rate" must be stated in full.
• The active site is the part of the enzyme that binds the substrate.

Common Mistakes

• Stopping at "more collisions" without naming the enzyme–substrate complex.
• Saying "more enzyme" — the experimenter varied substrate, not enzyme.
• Saying the reaction goes faster "because of collision theory" without naming the molecules involved.
• Confusing this with the effect of temperature (which would be about kinetic energy, not numbers of molecules).

Things to Be Careful About

Use the exact wording the mark scheme rewards: "more substrate molecules", "more successful collisions", "more enzyme–substrate complexes". The term "more collisions" alone is not enough — the collisions must be successful.