Biology 9700/31 — October/November 2019

Background Concept

The candidate must make up a range of ethanol concentrations from a 100% stock (A) and distilled water (W). Proportional (serial) dilution is a standard technique in which a fixed total volume is divided between the stock solution and the diluent. The volume of stock needed for a target percentage $C$ in a total volume $V_T$ is given by

and the diluent makes up the remainder so that $V_{\text{stock}} + V_{\text{diluent}} = V_T$. This guarantees that the moles of ethanol per cm³ scale linearly with the stated percentage.

Understanding the Question

The investigation tests how ethanol concentration affects the release of red pigment from beetroot. The candidate must first decide on a sensible range of concentrations and the volumes required to make each one. Table 1.2 already provides the two end-points (100% and 0%), so three intermediate values are needed to give a smooth trend.

The command word is "complete" — only the missing rows are required. The mark scheme requires at least three further concentrations and that the volumes are correct.

Approach

Pick concentrations that are evenly spaced across the range so that any trend in pigment release will be detectable. The natural choice is 25%, 50% and 75%, dividing the 0–100% interval into four equal steps. Then apply the dilution formula to each.

Step-by-Step Reasoning

• The total volume per tube is $20\ \text{cm}^3$.
• For 75% ethanol: $V_A = 0.75 \times 20 = 15.0\ \text{cm}^3$ of A and $20 - 15 = 5.0\ \text{cm}^3$ of W.
• For 50% ethanol: $V_A = 0.50 \times 20 = 10.0\ \text{cm}^3$ of A and $20 - 10 = 10.0\ \text{cm}^3$ of W.
• For 25% ethanol: $V_A = 0.25 \times 20 = 5.0\ \text{cm}^3$ of A and $20 - 5 = 15.0\ \text{cm}^3$ of W.
• The end-points (100% and 0%) are already in the table, so the final table contains five rows spanning 0–100%.

Key Takeaways

• Proportional dilution gives evenly spaced concentrations; the total volume in each tube must be identical.
• The 100% stock + water diluent method is the standard way to make a dilution series in Paper 3.

Common Mistakes

• Letting $V_A + V_W$ differ between tubes (the dependent variable, the disc, is then exposed to different total volumes).
• Using awkward concentrations (e.g. 33%, 67%) that make volumes awkward to measure.
• Reversing the order so volumes do not decrease monotonically down the table.

Things to Be Careful About

• Volumes should be quoted to one decimal place to match the precision given in the existing rows (20.0, 0.0).
• Ensure the chosen concentrations span the full range, otherwise the trend at one extreme cannot be observed.

Background Concept

A Paper 3 results table has strict conventions: column headings must contain the quantity and any unit (e.g. "colour intensity (key)" — no numerical unit because the scale is categorical), units must NOT be repeated in the body of the table, the independent variable (concentration) comes first, and every row must contain a reading. Because the colour intensity is judged against the key in Fig. 1.1, the data are qualitative but ordered, so they can still reveal a clear trend.

Understanding the Question

The candidate has just performed the practical and is asked to record what they saw. They must:

• design a table from scratch (the question gives none);
• use the + to ++++++ scale shown in Fig. 1.1;
• record all five samples;
• show the trend (intensity falls as concentration falls).

The mark scheme rewards each of these features individually.

Approach

First, decide the headings: percentage concentration of ethanol (no unit, since it is already a percentage), and colour intensity expressed as a symbol or a description such as "++++++". Then fill in the observed + rating for each tube, working from highest to lowest concentration so the trend is obvious. The highest concentration should give the deepest red and the lowest (water only) should give essentially no colour.

Step-by-Step Reasoning

• Heading 1: independent variable, percentage concentration of ethanol (no unit in the body — the % sign is built into the heading).
• Heading 2: dependent variable, colour intensity (key) or simply the symbol observed (++++++, etc.).
• Each row: a single + symbol or short string of +s taken directly from the Fig. 1.1 key.
• Trend: as concentration decreases, the number of +s decreases — most intense at 100% and least intense at 0%.
• The candidate may observe the same intensity (e.g. ++++) in more than one tube; this is acceptable and should be recorded as such.

Key Takeaways

• Headings carry the unit; the body of the table carries only numbers or symbols.
• The independent variable (concentration) heading is written first, the dependent variable (intensity) second.
• Use the symbol key given in the question — do not invent your own scale.

Common Mistakes

• Putting "%" in the body of the table (it belongs only in the heading).
• Using vague descriptions such as "pale red" instead of the + key, when a key has been provided.
• Leaving the 0% row blank because the solution is clear — it still counts as a reading and shows the lower limit of the trend.

Things to Be Careful About

• The mark scheme is explicit that the IV heading comes before the DV heading; reversing them loses a mark.
• The symbol key is + (no colour) up to ++++++ (deep red) — five +'s between the extremes, not six.

Background Concept

In any experiment the independent variable is what the experimenter deliberately changes between trials; the dependent variable is what is measured as a result. All other factors must be kept constant (controlled variables).

Understanding the Question

The investigation is described in the question stem: the candidate prepares different concentrations of ethanol, places beetroot discs into each, and records the colour intensity. The thing that is changed on purpose from tube to tube is the concentration of ethanol.

Approach

Read the procedure and ask: "What did I vary from one tube to the next?" — the answer is the concentration of ethanol. The colour intensity is the dependent variable; everything else (volume, disc number, time, temperature) is controlled.

Step-by-Step Reasoning

• Step 1 of the procedure makes different ethanol concentrations: this is the manipulation.
• Step 13 records the colour: this is the measurement.
• Therefore the concentration of ethanol is the independent variable and colour intensity is the dependent variable.

Key Takeaways

• Independent variable = what is changed on purpose.
• Dependent variable = what is measured.
• Controlled variables = everything kept the same.

Common Mistakes

• Naming "amount of pigment released" or "colour of the solution" — these are the dependent variable, not the independent one.
• Naming the type of tissue (beetroot) — that is a constant, not a variable.

Things to Be Careful About

• The variable must be stated precisely: "concentration of ethanol" (with units understood) is required; vague phrases such as "the amount of ethanol" are not credited.

Background Concept

The cell-surface membrane is a fluid phospholipid bilayer with embedded proteins. Its integrity depends on the hydrophobic interactions between the fatty-acid tails of the phospholipids and on the hydrogen bonds and weak interactions that hold each protein in its correct 3-D shape. Anything that disrupts these interactions makes the membrane "leaky".

Ethanol is a small, non-polar (or amphipathic) molecule that:

• disrupts hydrophobic interactions between phospholipid tails, so the bilayer becomes more fluid and phospholipids can be solubilised (washed out of the membrane);
• interferes with hydrogen bonding and hydrophobic interactions in membrane proteins, so they denature and lose their function;
• the combined effect is an increase in membrane permeability, allowing the red pigment (betalain) held inside the cell vacuole to leak out by diffusion down its concentration gradient.

The more ethanol present, the more phospholipid and protein is disrupted per unit time, so the higher the concentration the greater the pigment release.

Understanding the Question

The candidate has observed a clear trend: the more concentrated the ethanol, the deeper the red colour of the surrounding solution. The question asks them to explain this trend, which means giving the biological reasons (membrane disruption → increased permeability → more diffusion), not just restating the observation.

Approach

Work from the cause to the effect: start with what ethanol does to membrane molecules, then state the consequence for membrane function, then link that consequence to the observed colour change.

Step-by-Step Reasoning

• Cause 1 — proteins denatured: ethanol disrupts the bonds holding membrane proteins in shape, so they unfold and can no longer function as channels/barriers.
• Cause 2 — phospholipids dissolved: the hydrophobic tail region of the bilayer is destabilised by ethanol, so phospholipid molecules are removed from the membrane, leaving gaps.
• Consequence — increased permeability: the membrane becomes more permeable to solutes that would not normally cross it.
• Link to observation — more diffusion: betalain pigment inside the vacuole diffuses out through the leaky membrane; the higher the ethanol concentration, the greater the membrane damage and the more pigment released, giving a deeper red colour.

Key Takeaways

• An "explain" answer needs the mechanism (what is disrupted) AND the consequence (what changes as a result).
• Non-polar solvents increase membrane permeability by dissolving lipids and denaturing proteins.
• The betalain pigment is too large to cross an intact membrane, so its appearance in the solution is direct evidence of membrane damage.

Common Mistakes

• Saying only that "ethanol damages the membrane" without specifying proteins and phospholipids.
• Confusing the effect of ethanol with that of temperature (heat) — heat denatures proteins but does not dissolve phospholipids.
• Stating the observation (more colour) without explaining it in terms of membrane permeability.
• Saying ethanol "kills the cell" — the cells may still be alive; what matters is membrane integrity, not viability.

Things to Be Careful About

• The fourth mark point in the scheme requires BOTH "higher concentrations of ethanol" AND "more diffusion of pigment" — state both, not just one.

Background Concept

A source of error is any feature of the procedure that introduces uncontrolled variation into the results. The mark scheme accepts only those that are (a) genuinely present in the procedure given, (b) described specifically (not "human error"), and (c) paired with an improvement that directly addresses the named problem.

In this experiment the four main weaknesses of the procedure as written are:

1. the colour key is judged by eye → subjective;
2. discs are cut freehand with a blade → variable thickness;
3. all tubes are prepared in sequence → time-lag between the first and last immersion;
4. discs are washed only once → variable surface pigment remaining on each disc.

Understanding the Question

The candidate must list two distinct sources of error and a matching improvement for each. Each pair should be specific and biologically/practically sensible.

Approach

Walk mentally through the procedure and ask at each step: "What could go wrong here, and how could I fix it?" Pick the two issues that are most likely to have caused genuine variation in your own results.

Step-by-Step Reasoning

Pair 1 — colour judgement:

• Error: the + key in Fig. 1.1 is a subjective scale — observer bias, lighting and the position of the test-tube all affect the reading.
• Improvement: a colorimeter measures absorbance of the solution at a fixed wavelength (e.g. 530 nm for red) and gives a numerical value independent of the observer; a colour chart with a defined red scale is a cheaper alternative.

Pair 2 — disc thickness:

• Error: the procedure says "cut discs with a single-edged blade" but does not control thickness; discs of 1, 2 or 3 mm would have very different surface-area-to-volume ratios and different total pigment contents.
• Improvement: a cutting template (a series of slits 2 mm apart) or a mechanical chopper with an adjustable spacer gives identical discs every time.

Pair 3 — time lag (alternative):

• Error: the tubes are set up one after the other, so the first tube is in ethanol for several minutes longer than the last by the time they are all read.
• Improvement: use a staggered start — add the discs to each tube at timed intervals so that all tubes have been in ethanol for the same total time (5 minutes) when read.

Pair 4 — single wash (alternative):

• Error: step 5–7 wash the discs in water only once; pigment released by cutting remains on the surface and adds to the colour seen in the test-tube.
• Improvement: standardise the wash — wash all discs in the same volume of water for the same time, and repeat the wash until the water runs clear, so any surface pigment is fully removed before exposure to ethanol.

Key Takeaways

• Each error must have a paired improvement; you cannot list two errors and one improvement.
• "Human error" is too vague — say what the human was doing and why it is a problem.
• Improvements should be practical in a school lab: a colorimeter or colour chart is fine; a mass spectrometer is not.

Common Mistakes

• Vague errors such as "the experiment was inaccurate" — name the procedure step that was inaccurate.
• Improvements that do not address the named error (e.g. "use a more accurate balance" when the error was about colour judgement).
• Listing the same error twice with two different wordings — it only counts once.

Things to Be Careful About

• The mark scheme explicitly rejects "use more repeats" as an improvement for these particular errors — the issue is the measurement or the method, not the sample size.
• Keep each pair short and on the same line so the examiner can pair them up easily.

Background Concept

A line graph is used when both variables are continuous (numerical) and the independent variable is plotted on the x-axis. A graph is read by humans, so conventions matter: each axis must be labelled with the quantity AND the unit, the scale must be sensible (using at least half the grid, with no awkward breaks), each point must be plotted accurately, and a smooth line of best fit (or a join-the-dots line if the trend is to be emphasised) is drawn through the points.

Here the data are all at negative water potentials, so the x-axis is drawn with 0 at the right-hand end and the negative values extending to the left — an unusual but perfectly valid arrangement.

Understanding the Question

The candidate has a table of six paired values and an empty grid. The grid has its x = 0 marker on the right, so the candidate must work with a reversed x-axis. The question awards four marks: axes labelled and scaled; points plotted; line drawn.

Approach

1. Choose the scale on each axis. A simple scale (e.g. 0.25 MPa = 2 cm on x, 20 arbitrary units = 2 cm on y) keeps the points well spread.
2. Mark the scale on the grid, labelling every 2 cm on both axes (and the unit on the y-axis label).
3. Plot each of the six points carefully as a small cross or a dot in a circle.
4. Draw a single thin smooth curve through all six points.

Step-by-Step Reasoning

Scale:

• x-axis: $0.25\ \text{MPa}$ per 2 cm, so the whole $-1.5\ \text{MPa}$ range occupies $12\ \text{cm}$ on the grid. The grid is wide enough to accommodate this with room to spare.
• y-axis: $20$ arbitrary units per 2 cm, so the range $0$–$100$ occupies $10\ \text{cm}$ on the grid.
• Every 2 cm must be labelled on both axes (i.e. $0, 20, 40, 60, 80, 100$ on y; $0, -0.25, -0.5, -0.75, -1.0, -1.25, -1.5$ on x — though only every 4 cm needs a numerical label, the tick marks must appear every 2 cm).

Plotting the points (using the scale above):

• $(-0.3,\ 95)$: 1.2 cm left of x=0, 9.5 cm up the y-axis.
• $(-0.5,\ 76)$: 4 cm left of x=0, 7.6 cm up.
• $(-0.6,\ 68)$: 4.8 cm left of x=0, 6.8 cm up.
• $(-0.9,\ 50)$: 7.2 cm left of x=0, 5.0 cm up.
• $(-1.2,\ 43)$: 9.6 cm left of x=0, 4.3 cm up.
• $(-1.5,\ 24)$: 12.0 cm left of x=0, 2.4 cm up.

Line: the points form a smooth, decreasing curve — falls steeply at first (between $-0.3$ and $-0.6\ \text{MPa}$), then more gently. A thin smooth curve through all six points is the correct choice (not a straight line of best fit, because the relationship is not linear).

Key Takeaways

• All axes need both a quantity and a unit.
• Choose scales that use at least half the grid in both directions and avoid awkward numbers.
• Use small crosses or dots in circles, not large blobs, so the exact position is unambiguous.
• Join the points with a thin line (not freehand dashes) — straight lines between points if the data are noisy, a smooth curve if the trend is clearly continuous.
• A reversed x-axis (0 on the right) is fine; just label the negative values clearly.

Common Mistakes

• Forgetting the unit on the axis label (e.g. writing only "water potential" without "/ MPa").
• Plotting against the wrong axis (e.g. putting acoustic emission on the x-axis because the x-axis has 0 marked).
• Joining the points with straight line segments and calling it a curve.
• Using a scale like "1 MPa = 2 cm" which compresses the data into a tiny corner of the grid.

Things to Be Careful About

• The y-axis sits on the LEFT of the grid (as standard), but the x = 0 is on the RIGHT, so the y-axis does NOT pass through x = 0 — it passes through the most negative x-value shown.
• The line is a smooth curve, not straight segments: the rate of change is clearly different in different parts of the range.

Background Concept

Interpreting a graph means using a drawn curve to estimate values that were not measured directly. The technique is:

1. Find the known value on one axis (here $80$ on the y-axis).
2. Draw a thin pencil line across (or up) to intersect the curve.
3. From that intersection, drop a perpendicular line to the other axis and read the value.

If the required value lies between two plotted points, the technique is called interpolation; if it lies outside the plotted range, it is extrapolation (and is less reliable).

Understanding the Question

The candidate must use their own graph from (b)(i) to find the water potential that gives an acoustic emission of $80$ arbitrary units. The mark scheme awards one mark for a correct value read from the graph.

Approach

Use the y-axis value $80$ to locate the corresponding point on the curve, then read the x-axis value directly. A linear interpolation between the two nearest plotted points is a good check.

Step-by-Step Reasoning

• The y-value $80$ lies between the two highest points: $(-0.3,\ 95)$ and $(-0.5,\ 76)$.
• Linear interpolation:

• Reading directly from the graph gives a value in the range $-0.4$ to $-0.5\ \text{MPa}$; $-0.45\ \text{MPa}$ is a representative value.

Key Takeaways

• Graph reading is interpolation between the two nearest data points.
• The unit (MPa) must always be quoted with the numerical answer.
• A range of acceptable answers is expected because each candidate's curve will be drawn slightly differently.

Common Mistakes

• Quoting the answer without the unit (the mark scheme expects "MPa").
• Reading to too many significant figures (e.g. $-0.458\ \text{MPa}$ is over-precise given the spread of the data).
• Reading off the wrong axis because the x-axis is reversed.

Things to Be Careful About

• The x-axis has 0 on the right, so the value at acoustic emission 80 is to the LEFT of 0, and the answer must be negative.
• A reasonable range is $-0.40$ to $-0.50\ \text{MPa}$; any value within this range, read from a properly drawn graph, is correct.

Background Concept

The precision with which a value can be read from a graph depends on how steeply the curve is changing in that region and how many data points lie nearby. Between $-0.3$ and $-0.5\ \text{MPa}$ the curve is changing by about $19$ arbitrary units, but the next measurement is $0.3\ \text{MPa}$ further on, so a small change in the candidate's reading on the y-axis translates to a large change on the x-axis.

Understanding the Question

The candidate has just read a value ($-0.45\ \text{MPa}$) from the graph at acoustic emission $80$. The question asks how the experiment itself could be improved so that this value can be obtained more accurately in the future. This is an evaluation of the independent variable (water potential), not the dependent variable or the method.

Approach

The most direct improvement is to use more values of the independent variable in the region of interest, so the curve is well supported by data where it is changing rapidly.

Step-by-Step Reasoning

• The estimate at $80$ arbitrary units is an interpolation between $(-0.3,\ 95)$ and $(-0.5,\ 76)$.
• The two nearest data points are $0.2\ \text{MPa}$ apart, so a $0.1\ \text{MPa}$ uncertainty in reading corresponds to a $0.05\ \text{MPa}$ uncertainty in the estimate.
• Adding measurements at, say, $-0.35,\ -0.40$ and $-0.45\ \text{MPa}$ would tighten the curve in exactly this region, allowing a more accurate read-off at $80$ arbitrary units.

Key Takeaways

• To improve a graph-reading estimate, add more data in the region of interest.
• The improvement must specifically address the independent variable here, not the dependent variable or the apparatus.

Common Mistakes

• Suggesting "use more accurate equipment" — this is too vague and not specific to the independent variable.
• Suggesting repeats of the same measurement — repeats improve reliability but do not help locate the value more precisely.
• Suggesting a different range altogether (e.g. measuring up to $-3.0\ \text{MPa}$) — this changes the experiment but does not improve the estimate at $80$ arbitrary units.

Things to Be Careful About

• The mark scheme explicitly accepts "reduce gap between intervals"; phrasing such as "use more values closer together" or "add more concentrations in the relevant range" are all valid.
• Do not confuse this with a method improvement (e.g. "standardise the compression rate"); the question is about the independent variable.