Biology 9700/34 — October/November 2022

Background Concept

A serial dilution is a stepwise dilution in which the concentration is reduced by a fixed factor at each step. In a half-strength (1:2) serial dilution the concentration is halved between successive tubes or beakers. Serial dilutions are used to:

• produce a wide range of concentrations from a small volume of stock solution,
• keep all dilutions related to one another by a known factor (so an unknown sample can be read off the scale), and
• avoid the need to weigh out many separate masses of solute.

For a 1:2 dilution, equal volumes of solution and diluent (water) are mixed, so:

If $10\ \text{cm}^3$ of solution is added to $10\ \text{cm}^3$ of water, the total is $20\ \text{cm}^3$ and the concentration is exactly halved. After mixing, $10\ \text{cm}^3$ is taken for the assay and $10\ \text{cm}^3$ is carried forward to the next beaker.

Understanding the Question

The stem tells us we have $20\ \text{cm}^3$ of a $0.1\%$ stock solution, A, and that we must dilute it by half between each successive step to give four further concentrations in addition to A itself. We need $10\ \text{cm}^3$ of each concentration available to use. We have to complete Fig. 1.2 by drawing three more beakers and labelling, for each beaker, the volume of A (or transferred solution) entering it and the volume of W added, plus the resulting concentration beneath.

The command word here is complete — it is essentially a drawing/label task. The marks reward correct concentrations, correct transfer volumes and correct water volumes.

Approach

1. Calculate the four new concentrations by halving $0.1\%$ four times.
2. Recognise the procedure: each new beaker is made by mixing equal volumes ($10\ \text{cm}^3 + 10\ \text{cm}^3$), so the concentration is halved.
3. After mixing, remove $10\ \text{cm}^3$ for use; the remaining $10\ \text{cm}^3$ is carried forward as the source for the next beaker.
4. Draw three additional beakers in a row, with arrows showing the transfer volume and the water volume for each.

Step-by-Step Reasoning

Starting concentration: $0.1\%$.

• Beaker 1: $0.1\%$ — provided. The figure shows $20\ \text{cm}^3$ of A entering, $0\ \text{cm}^3$ of W added, and $10\ \text{cm}^3$ is removed "to use".
• Beaker 2: $10\ \text{cm}^3$ of $0.1\%$ from beaker 1 is transferred in, $10\ \text{cm}^3$ of W is added. The new concentration is $0.1\% / 2 = 0.05\%$.
• Beaker 3: $10\ \text{cm}^3$ of $0.05\%$ from beaker 2 is transferred in, $10\ \text{cm}^3$ of W is added. The new concentration is $0.05\% / 2 = 0.025\%$.
• Beaker 4: $10\ \text{cm}^3$ of $0.025\%$ from beaker 3 is transferred in, $10\ \text{cm}^3$ of W is added. The new concentration is $0.025\% / 2 = 0.0125\%$.
• Beaker 5: $10\ \text{cm}^3$ of $0.0125\%$ from beaker 4 is transferred in, $10\ \text{cm}^3$ of W is added. The new concentration is $0.0125\% / 2 = 0.00625\%$.

So the four new concentrations in order are: $0.05\%$, $0.025\%$, $0.0125\%$ and $0.00625\%$. For each new beaker the arrows must clearly show $10\ \text{cm}^3$ of the previous solution entering and $10\ \text{cm}^3$ of W entering. The percentage label goes under each beaker.

Key Takeaways

• A 1:2 serial dilution halves the concentration at every step; successive concentrations follow a geometric series.
• When equal volumes are mixed, the new concentration is exactly the old concentration divided by 2.
• The "$10\ \text{cm}^3$ to use" message tells you how much is removed before the next transfer — always keep the carried-forward volume the same so each step halves the concentration.

Common Mistakes

• Halving only once and giving, e.g., $0.05\%$ and $0.025\%$ only: the question asks for four additional concentrations.
• Adding the wrong volume of water (e.g. $5\ \text{cm}^3$) — to halve the concentration by mixing, the volumes added must be equal.
• Forgetting to write the concentration below each beaker.
• Using units of $\text{cm}^3$ for concentration (concentration is given as a percentage here).

Things to Be Careful About

• The first beaker is A at $0.1\%$ — it is supplied, not made by dilution. Your arrows for beaker 1 should show $0\ \text{cm}^3$ of W.
• The volumes in the diagram are $10\ \text{cm}^3$ transferred and $10\ \text{cm}^3$ of water added at every new beaker, not $20\ \text{cm}^3$ of the previous solution.
• $0.00625\%$ is a small number but it is correct; don't round it to $0.006\%$ or $0.01\%$ — the mark scheme requires the exact value.

Background Concept

In any practical investigation the independent variable is the one the experimenter deliberately changes, and the dependent variable is the one measured. CIE expects results to be recorded in a table that:

1. has a clear heading for each column that names the quantity and (where appropriate) gives the unit,
2. puts the independent variable as the left-hand column,
3. puts units only in the heading, not repeated against every value,
4. uses a consistent number of decimal places down each column,
5. has a ruled border and clear rows/columns.

The biological principle behind this titration: iodine ($I_2$) is reduced to colourless iodide ($I^-$) by ascorbic acid; once all the ascorbic acid has been oxidised, further drops of iodine react with starch to give the blue-black starch–iodine complex. The end-point is the volume at which the blue colour persists for $10\ \text{s}$. The higher the ascorbic acid concentration, the more iodine is needed to reach the end-point.

Understanding the Question

The candidate has just carried out the titration for each concentration of ascorbic acid prepared in (a)(i). They must now record their own results in an appropriate table. The marks reward:

1. Correct headings (independent variable first, no units in the body).
2. Results for all concentrations.
3. Decimal places consistent and at least 1 d.p.
4. Correct trend: more iodine needed for higher concentrations of ascorbic acid.

Approach

1. Decide the layout: two columns — ascorbic acid concentration (independent) on the left, volume of iodine (dependent) on the right.
2. Decide decimal places. Volumes from a $1.0\ \text{cm}^3$ syringe read to $0.05\ \text{cm}^3$ (one drop ≈ $0.05\ \text{cm}^3$) are typically recorded to one decimal place, e.g. $0.4$, $0.8$, $1.0$, $1.5$, $2.2$.
3. Order the rows by concentration (usually descending, so the largest value is at the top).
4. Check the trend: the row with the highest ascorbic acid concentration should have the largest volume of iodine.

Step-by-Step Reasoning

Because the candidate performs the experiment, I cannot know their exact values. Representative values consistent with the trend expected by the mark scheme (and using an evenly-stepped dilution) are:

What matters for the marks:

• The values decrease down the second column as the concentration decreases (or equivalently, increase up the table as concentration increases).
• All five concentrations are present, with results to 1 d.p. (or more).
• The headings are correct and units are in the headings only.

Key Takeaways

• Independent variable → first column; dependent variable → subsequent columns.
• Units in the heading, not in the body.
• A consistent number of decimal places down each column.
• In an ascorbic-acid/iodine titration, more ascorbic acid means more iodine is needed to reach the end-point.

Common Mistakes

• Writing the units (e.g. $\%$, $\text{cm}^3$) inside the data cells rather than only in the headings.
• Putting the dependent variable in the first column.
• Inconsistent decimal places (e.g. $0.5$ in one row, $1.25$ in another).
• Reversing the trend: writing a larger volume of iodine for a lower ascorbic acid concentration.

Things to Be Careful About

• "At least one decimal place" means $1.0$ is fine, but $1$ alone is not.
• The trend check is the most reliable way to catch errors: if the highest concentration has the smallest iodine volume, something is wrong.
• If a sample did not turn blue after $5.0\ \text{cm}^3$, record "more than $5.0$" — do not write $5.0$ exactly.

Background Concept

A "source of error" in practical work is anything that introduces uncertainty into the measurement. For titrations the dominant errors are usually end-point detection (when does the indicator change?) and volume reading (how precisely can the syringe or burette be read?). Both are made worse if the solution is dark or intensely coloured.

In this experiment the end-point is the appearance of a blue colour that persists for at least $10\ \text{s}$. This requires a subjective judgement — the colour change is gradual, and a candidate may declare the end-point too early (when the blue colour is transient) or too late (after adding more iodine than was strictly needed).

Understanding the Question

The question asks for one significant source of error in steps 8 to 17 — that is, in drawing up the iodine, releasing it drop by drop, judging the end-point, and reading the syringe. The mark scheme gives two credited errors: difficulty judging the end-point, and difficulty reading the syringe because the iodine solution is too dark.

Approach

1. Identify where the result is most uncertain.
2. Match it to a specific feature of the procedure (end-point, syringe reading, drop size, mixing).
3. State the error clearly and concisely.

Step-by-Step Reasoning

• The end-point requires a subjective "blue for at least $10\ \text{s}$" judgement. Different candidates will stop at different points. This is a recognised, significant error.
• The iodine solution is a dark brown liquid, so the markings on the $1.0\ \text{cm}^3$ syringe barrel are hard to read through it. The reading is therefore imprecise. This is also a recognised, significant error.

Either answer is sufficient. The mark scheme does not credit vague answers like "human error" or "parallax error".

Key Takeaways

• For a colour-change titration, the end-point is usually the dominant source of error.
• For coloured solutions, reading the syringe or burette is the second-biggest source of error.
• Vague answers ("human error") are not credited; be specific about what was hard to do and why.

Common Mistakes

• Writing "human error" — too vague, not credited.
• Writing "parallax error" — the syringe is vertical and the candidate looks straight down at it, so parallax is not the issue.
• Naming an error that is not actually significant, e.g. "the test-tube might fall over".

Things to Be Careful About

• "Significant" means the error noticeably affects the recorded volume. The two end-point / reading errors are the ones that change the result by a measurable amount.
• If the candidate is asked for one error, do not list several — credit is only for the first error and the rest are ignored.

Background Concept

The investigation is comparing two treatments of the same vegetable extract: U (uncooked) and C (cooked for $60\ \text{min}$). Because both are run through the same procedure as the standard concentrations, the volume of iodine required gives a relative measure of their ascorbic acid content. The biological expectation is that heating decreases the ascorbic acid concentration (ascorbic acid is heat-labile and oxidises on heating), so the cooked extract C should require less iodine than the uncooked extract U.

Understanding the Question

The candidate has just carried out steps 18–21: titrate U against iodine, record the volume; then titrate C against iodine, record the volume. The marks reward both volumes being recorded and the volume for U being greater than the volume for C.

The actual numerical values depend on the candidate's own experiment, so the mark scheme cannot pin down a single "correct" answer — it just requires the qualitative ordering to be correct.

Approach

1. Read the syringe to the same precision as in (a)(ii).
2. If the blue colour did not persist after $5.0\ \text{cm}^3$ of iodine, write "more than $5.0$".
3. Confirm the expected trend: $\text{volume for } \mathbf{U} > \text{volume for } \mathbf{C}$.

Step-by-Step Reasoning

Using the representative data above: $1.6\ \text{cm}^3$ of iodine was needed for U and $0.8\ \text{cm}^3$ for C. Because $1.6 > 0.8$, the cooked extract has roughly half the ascorbic acid of the uncooked extract, which is consistent with the prediction. The candidate's own values may differ but the ordering U > C is what the mark scheme checks.

Key Takeaways

• An unknown sample can be compared to a calibration series by the volume of titrant needed to reach the end-point.
• Heating destroys ascorbic acid, so the cooked extract should always require less iodine than the uncooked one.
• Always record to the same precision across all measurements in an experiment.

Common Mistakes

• Writing the volumes with different decimal places (e.g. $1.6$ for U and $0.85$ for C).
• Reversing the trend (writing C > U) — this contradicts the biology and forfeits the mark.
• Writing "more than $5.0$" for C when the blue colour did in fact appear within $5.0\ \text{cm}^3$.

Things to Be Careful About

• The result is student-dependent; the mark is awarded for the correct ordering and a numerical value with units, not for a specific value.
• Be consistent with the precision of your earlier results in (a)(ii).

Background Concept

A scale bar (or number line) is a way of representing a continuous range of values when you have a calibration series. Once the positions of the standards are marked, any unknown can be placed on the same line by eye and its value read off.

The serial dilution in (a)(i) was a 1:2 (half-concentration) series. The five standard concentrations are $0.10\%$, $0.05\%$, $0.025\%$, $0.0125\%$, $0.00625\%$. On a linear scale, however, halving the concentration does not mean equal spacing on the bar — the spacing is proportional to the absolute concentration, not the log. So a $0.10\%$ mark is twice as far from $0$ as a $0.05\%$ mark, and the smallest concentrations ($0.0125\%$ and $0.00625\%$) cluster very close to the left end.

Understanding the Question

The question shows a horizontal line with the two ends labelled $0.00\%$ and $0.10\%$, and a single tick already marked at the midpoint. The candidate must label the scale bar with the five standard concentrations used in the experiment, in the correct positions.

The mark scheme requires the scale to be correctly labelled with the different concentrations.

Approach

1. Mark $0$ at the left end and $0.10\%$ at the right end.
2. Place each of the five known concentrations at a position proportional to its value:
   • $0.05\%$ is exactly halfway along the bar (this is the tick already drawn).
   • $0.025\%$ is at $1/4$ of the way along.
   • $0.0125\%$ is at $1/8$ of the way along.
   • $0.00625\%$ is at $1/16$ of the way along.
3. Write each label clearly above or below the tick mark.

Step-by-Step Reasoning

The mid-tick already shown is at the $0.05\%$ position (half of $0.10\%$). The other labels are placed at the proportional positions: $0.025\%$ at $1/4$ of the total length, $0.0125\%$ at $1/8$, and $0.00625\%$ at $1/16$. The candidate's own data from (a)(ii) has already established that these concentrations are valid, and they should now be placed in proportion on the bar.

Key Takeaways

• A linear scale bar places values at positions proportional to their magnitude, not logarithmically.
• A $1\!:\!2$ dilution series gives values that compress near zero on a linear scale — most of the bar's length is taken up by the higher concentrations.

Common Mistakes

• Spacing the values evenly (e.g. equal gaps between $0.00625\%$, $0.0125\%$, $0.025\%$, $0.05\%$, $0.10\%$) — this is wrong because the gaps on a linear scale should be proportional to the values, not equal.
• Forgetting the smallest concentration $0.00625\%$, which sits very close to the left end and is easy to miss.

Things to Be Careful About

• The midpoint tick is at $0.05\%$, not at $0.05\% \times 0.5$.
• The scale bar is linear, not logarithmic — the spacing must be proportional.

Background Concept

This part uses the calibration scale built up in (a)(v). The volume of iodine is directly proportional to the ascorbic acid concentration (more ascorbic acid = more iodine required). A standard curve or scale bar therefore lets us read the concentration of an unknown by locating the position on the bar where the iodine volume matches.

Because the serial dilution halves concentration at each step, equal volumes of iodine map to concentrations in a simple ratio. For example, if a $0.05\%$ standard needs $1.4\ \text{cm}^3$ of iodine and a $0.025\%$ standard needs $0.9\ \text{cm}^3$, then an extract needing $1.6\ \text{cm}^3$ of iodine has a concentration slightly higher than $0.05\%$ (interpolated between $0.05\%$ and the next-higher standard, $0.10\%$, which needs $2.2\ \text{cm}^3$).

Understanding the Question

The candidate must place the letters U and C on the scale bar in positions consistent with the volumes of iodine recorded in (a)(iv). The mark scheme requires both to be in the correct position.

Approach

1. Compare the volume of iodine for U with the volumes for the standards (from (a)(ii)).
2. Locate the standard whose iodine volume is the same (or interpolate between the two nearest standards).
3. Repeat for C.
4. Place U and C on the bar.

Step-by-Step Reasoning

Using the representative data:

So U is plotted a little to the right of the $0.05\%$ mark, and C is plotted about halfway between $0.025\%$ and $0.05\%$ (slightly to the right of $0.025\%$). Importantly, U must be to the right of C on the bar.

Key Takeaways

• A standard series converts a measured quantity (here, volume of titrant) into the concentration of an unknown by interpolation.
• The accuracy of the estimate is limited by the spread of the standards: with five standards covering $0.00625\%\!-\!0.10\%$, the resolution is about $0.006\%$ near zero and much coarser near the high end.

Common Mistakes

• Reversing the order (placing C to the right of U) — this contradicts both the recorded volumes and the biology.
• Placing U and C at the same position, which implies equal ascorbic acid concentrations in cooked and uncooked extract.
• Forgetting to write the letters at all.

Things to Be Careful About

• The scale bar is a calibration curve, not a bar chart. Each unknown is placed on the same axis as the standards, not on a separate axis.
• The position of the letters should be consistent with the candidate's own volumes from (a)(iv), even if those volumes are unusual.

Background Concept

"Improving" a procedure means reducing the dominant sources of uncertainty. For this titration there are two big ones: the end-point is subjective, and the concentration scale is coarse (only five standards spanning $0.00625\%$ to $0.10\%$). Improvements therefore target one of those two issues, or replace the drop-by-drop method with a more precise dispenser.

Understanding the Question

The question asks for one modification that would give a more accurate estimate of the concentration of ascorbic acid in U and C, without using a colorimeter. The mark scheme accepts three categories: (1) a narrower range of standards, (2) repeats and a mean, (3) a more precise dispenser (micropipette).

Approach

1. Identify the dominant source of error (end-point, coarse scale, or coarse drop size).
2. Choose a modification that directly addresses it.
3. State the modification clearly and specifically.

Step-by-Step Reasoning

• Narrower range: with standards clustered around the unknown's value, the interpolation is more accurate. For example, if U is near $0.04\%$, prepare standards at $0.03\%$, $0.035\%$, $0.04\%$, $0.045\%$ and $0.05\%$ — much finer resolution than the original $1\!:\!2$ series.
• Repeat and mean: doing each titration three times and averaging smooths out the random end-point error. The improvement is paired with the error identified in (a)(iii).
• Micropipette: a micropipette delivers a fixed, smaller volume per "drop" (typically microlitres), so the resolution around the end-point is finer — you can stop closer to the true equivalence point.

Any one of these is accepted.

Key Takeaways

• Improvements should target the dominant source of error identified earlier in the question.
• Repeats and a mean reduce random error; finer standards reduce systematic interpolation error; finer dispensers reduce end-point error.

Common Mistakes

• Vague answers like "be more careful" or "do it more accurately" — not credited.
• "Use a colorimeter" — explicitly excluded by the question.
• Suggesting changes that don't affect the precision, e.g. "use a different vegetable".

Things to Be Careful About

• The mark scheme credits only one improvement; if you list several, only the first is read.
• The improvement must be specific and actionable (e.g. "repeat three times and take a mean" not just "repeat").

Background Concept

Ascorbic acid (vitamin C) is a small, water-soluble, reducing molecule. It is chemically unstable: it is readily oxidised to dehydroascorbic acid, especially in the presence of heat, oxygen, light and metal ions. Cooking vegetables in water (boiling, simmering) accelerates this oxidation and also leaches ascorbic acid out of the cells into the cooking water, so the vegetable itself loses vitamin C. Both effects reduce the ascorbic acid concentration of the cooked extract relative to the uncooked one.

In this experiment the cooked extract C required less iodine to reach the end-point than the uncooked extract U, indicating a lower concentration of ascorbic acid. The biological explanation is therefore that heating breaks down ascorbic acid.

Understanding the Question

The question asks the candidate to suggest an explanation for the effect of heating on the ascorbic acid concentration. The mark scheme accepts either:

1. Heating breaks down ascorbic acid (the credited biological explanation), or
2. Heating has no effect on ascorbic acid (accepted if the candidate's data happened to show no difference — for example, if the iodine volumes for U and C were equal).

Approach

1. State the expected effect of heating on ascorbic acid.
2. Link it to the observation (less iodine needed for the cooked extract).

Step-by-Step Reasoning

The result expected, and obtained by most candidates, is that U required more iodine than C because ascorbic acid was destroyed by heating. The explanation is therefore: heating breaks down (oxidises) ascorbic acid, so the cooked extract has a lower concentration.

If a candidate's data shows no change, the alternative answer "heating has no effect on ascorbic acid" is also credited.

Key Takeaways

• Ascorbic acid is heat-labile; heating destroys it.
• A lower ascorbic acid concentration in a sample corresponds to a lower volume of iodine at the end-point.

Common Mistakes

• Vague answers like "cooking changes the vegetable" — not specific to ascorbic acid.
• "Heating increases the ascorbic acid concentration" — biologically incorrect; the trend is the opposite.
• Failing to link the explanation to the data (e.g. just writing "ascorbic acid is destroyed by heat" without mentioning the iodine volumes).

Things to Be Careful About

• The mark scheme awards the mark for either accepted explanation, so a candidate with anomalous data (e.g. C > U) is not penalised as long as the explanation is consistent with their data.

Background Concept

This is a Planning sub-question: it asks how to extend the experiment to investigate a different independent variable (heating time). The general structure of a controlled experiment is:

1. Independent variable — what you deliberately change. Here: heating time.
2. Dependent variable — what you measure. Here: volume of iodine needed to reach the end-point, which is then converted to ascorbic acid concentration via the standard curve.
3. Controlled variables — factors that must be kept the same so the change in the dependent variable can be attributed to the independent variable. Here: concentration of extract, temperature, volume of extract, volume of starch, drop size, etc.
4. Range and intervals — a minimum of five values, evenly spaced, to show a trend.
5. Reliability — repeats and means.

Understanding the Question

The candidate is told that the extract C was heated for $60$ minutes. The question asks how to investigate the effect of different heating times on ascorbic acid concentration. The mark scheme awards marks for any three of:

1. Same concentration.
2. Same temperature.
3. Thermostatically controlled water-bath.
4. Five different times.
5. Times evenly spaced.

Approach

1. State the independent variable (heating time) and the range of times (e.g. $0$–$60$ min) — at least five values, evenly spaced.
2. State what must be kept the same (concentration of extract, temperature, volume).
3. State the apparatus used to control temperature (thermostatic water-bath).
4. Describe how the ascorbic acid concentration is measured (same iodine titration as in (a)(ii)).

Step-by-Step Reasoning

• Heating times: pick five values evenly spaced across a sensible range. For example, $0$, $15$, $30$, $45$, $60$ min (interval of $15$ min). Other equally valid choices are $10$, $20$, $30$, $40$, $50$ min or $0$, $20$, $40$, $60$, $80$ min.
• Same concentration: use the same volume of the same batch of vegetable extract for every time point — otherwise any change in ascorbic acid could be due to the starting concentration rather than the heating time.
• Same temperature: heat at a fixed temperature (e.g. $100\,^\circ\text{C}$ boiling or $80\,^\circ\text{C}$ in a water-bath). Without this, you cannot tell whether the effect is due to time or temperature.
• Thermostatically controlled water-bath: this is the standard way to maintain a constant temperature throughout the experiment; a beaker on a hot plate would fluctuate.
• Measurement: after each heating period, the extract is cooled, then titrated against iodine as in (a)(ii), recording the volume of iodine at the end-point. Plot time (x) against volume of iodine (y), or against calculated ascorbic acid concentration (y).

Any three of these earn the marks.

Key Takeaways

• A controlled experiment varies only the independent variable and keeps all other relevant variables constant.
• A thermostatically controlled water-bath is the standard way to maintain a constant temperature.
• A minimum of five values of the independent variable, evenly spaced, is needed to identify a trend.

Common Mistakes

• Varying the temperature as well as the time — this confounds the experiment.
• Using only two or three time points — too few to identify a trend.
• Suggesting unevenly spaced times (e.g. $5$, $10$, $30$, $60$, $90$ min).
• "Use a Bunsen burner to heat the extract" — open flames are unsuitable for controlled temperature work and a hazard.

Things to Be Careful About

• "Heating time" is the independent variable; the temperature is a controlled variable, not the independent one.
• A thermostatically controlled water-bath is more controllable than a Bunsen burner and is the expected answer.

Background Concept

The data show how the mass of B. subtilis changes as the ascorbic acid concentration is increased. The biology: ascorbic acid is antimicrobial, so increasing its concentration should reduce bacterial growth (lower mass). The graph should show a downward trend that levels off at higher concentrations — the bacteria have a minimum viable mass and cannot be reduced further.

Understanding the Question

The candidate is given five (x, y) pairs and a blank grid. They must plot the data and join the points. The mark scheme rewards four features:

1. Correct axis labels with units.
2. Sensible scales ($10\ \text{mM}$ per $2\ \text{cm}$ on x; $2.0\ \text{mg}$ per $2\ \text{cm}$ on y) and labels at every major tick.
3. All five points plotted accurately, with small crosses or dots in circles.
4. A thin line joining the points.

Approach

1. Choose the axes: ascorbic acid concentration is the independent variable, so it goes on the x-axis; mass of bacteria is the dependent variable, so it goes on the y-axis.
2. Choose scales that use at least half the grid in both directions and make plotting easy (e.g. $10\ \text{mM} = 2\ \text{cm}$, $2.0\ \text{mg} = 2\ \text{cm}$).
3. Plot each point carefully as a small cross or a dot in a circle.
4. Join the points with a thin line — do not extrapolate beyond the first and last points.

Step-by-Step Reasoning

• x-axis: the data range from $2.5$ to $40.0\ \text{mM}$. A scale of $0$ to $50$ with major ticks every $10\ \text{mM}$ uses most of the grid and allows easy plotting. $10\ \text{mM}$ per $2\ \text{cm}$ is the suggested scale.
• y-axis: the data range from $2.4$ to $9.7\ \text{mg}$. A scale of $0$ to $10$ with major ticks every $2.0\ \text{mg}$ works well; $2.0\ \text{mg}$ per $2\ \text{cm}$ is the suggested scale.
• Points: plot each (x, y) pair with a small cross (or a dot inside a small circle). Read the x-value off the x-axis to the nearest $0.5\ \text{mM}$ and the y-value off the y-axis to the nearest $0.1\ \text{mg}$.
• Line: join the points with a single thin continuous line. Because the data show a clear non-linear (curved) trend, the line should be a smooth curve rather than a series of straight segments.

Key Takeaways

• The independent variable goes on the x-axis; the dependent variable on the y-axis.
• Choose scales that use at least half the grid and make plotting straightforward.
• A line graph is appropriate when the independent variable is continuous (concentration); use a bar chart only for discrete categories.
• A smooth curve is appropriate when the trend is non-linear; a straight line of best fit is appropriate when the trend is linear.

Common Mistakes

• Swapping the axes (mass on x, concentration on y).
• Using awkward scales, e.g. $0$ to $40$ with no labels in between, or $0$ to $100$ which compresses all the data into the bottom-left corner.
• Plotting points as large blobs rather than small crosses or dots in circles.
• Drawing a thick line or extending the line beyond the data range.
• Drawing a straight line through a curved set of points.

Things to Be Careful About

• Use a sharp pencil so points and lines are precise.
• Make sure each point is plotted to the correct precision: x to the nearest $0.5\ \text{mM}$, y to the nearest $0.1\ \text{mg}$.
• The line should not extrapolate; it should stop at the first and last data points.

Background Concept

A percentage decrease measures the size of a fall relative to the original value. The formula is:

The original value is the starting point of the change, not the final value. In this question the starting concentration is $10.0\ \text{mM}$, so the original mass is the one at $10.0\ \text{mM}$. Reading both masses from the graph is therefore essential.

Understanding the Question

The candidate has plotted a graph of ascorbic acid concentration (x) against mass of B. subtilis (y) in (b)(i). They must now use that graph to find the percentage decrease in mass as the ascorbic acid concentration is raised from $10.0\ \text{mM}$ to $35.0\ \text{mM}$. The mark scheme awards:

1. Correct figures for $10.0\ \text{mM}$ and $35.0\ \text{mM}$ from the graph.
2. (Decrease ÷ $10.0\ \text{mM}$ value) × $100$.

Approach

1. Read $y$ at $x = 10.0\ \text{mM}$ — this is the original value.
2. Read $y$ at $x = 35.0\ \text{mM}$ — this is the final value. The candidate may need to interpolate between $30.0$ and $40.0\ \text{mM}$.
3. Subtract the final value from the original to get the decrease.
4. Divide the decrease by the original value and multiply by $100$.

Step-by-Step Reasoning

• At $x = 10.0\ \text{mM}$: the point is plotted at $y = 7.2\ \text{mg}$. This is the original value.
• At $x = 35.0\ \text{mM}$: this lies halfway between the points at $30.0\ \text{mM}$ ($3.1\ \text{mg}$) and $40.0\ \text{mM}$ ($2.4\ \text{mg}$). The line is slightly curved here (it's flattening off), so a careful read at the midpoint gives $y \approx 2.75\ \text{mg}$. The mark scheme allows ecf from a correctly drawn line.
• Decrease: $7.2 - 2.75 = 4.45\ \text{mg}$.
• Percentage decrease: $\frac{4.45}{7.2} \times 100 \approx 61.8\%$.

Key Takeaways

• A percentage change always uses the starting value as the denominator.
• Reading the starting value directly from a plotted point is more reliable than interpolating; the final value at $35.0\ \text{mM}$ must be interpolated from the curve.
• The line drawn in (b)(i) determines the precision of the read-off — a poorly drawn line will lead to a wrong value in (b)(ii).

Common Mistakes

• Dividing by the final value instead of the original (e.g. $\frac{4.45}{2.75} \times 100 \approx 162\%$) — biologically this is wrong and the mark scheme rejects it.
• Reading the value at $35.0\ \text{mM}$ as $3.1$ (using the $30.0\ \text{mM}$ point) — this gives $\frac{7.2 - 3.1}{7.2} \times 100 = 57\%$, a small under-estimate.
• Forgetting to multiply by $100$ — gives a value of $0.62$ rather than $62\%$.
• Forgetting the percentage sign.

Things to Be Careful About

• "Decrease" means the fall, so it is original − final, not final − original.
• The denominator is the original value (at $10.0\ \text{mM}$), not the average or the final value.
• The final value is read from the line drawn in (b)(i), not from the data table — the data table only has values at $30.0$ and $40.0\ \text{mM}$, so the candidate must interpolate.