Biology 9700/53 — May/June 2021

Background Concept

When an enzyme reaction produces a coloured product, the rate of reaction can be followed by measuring how quickly the colour develops. There are two broad approaches:

1. Qualitative — describing the colour by eye ("light brown", "dark brown"). This depends on the observer and cannot easily be compared between samples.
2. Quantitative — measuring absorbance with a colorimeter (or transmission with a spectrophotometer). The absorbance is directly proportional to the concentration of the coloured product (Beer–Lambert law) and gives a numerical value that can be plotted, compared and analysed statistically.

Understanding the Question

The students need to react catechol oxidase with catechol and judge how active the enzyme is. The brown product melanin is coloured, but its shade is hard to grade visually. The question asks why a colorimeter is preferable to judging the colour by eye.

Approach

Identify the key benefit of colorimetry: it produces a quantitative, objective, numerical reading that is independent of who is looking at the tube. The mark scheme rewards the words "quantitative" or equivalent (e.g. "numerical").

Step-by-Step Reasoning

• Different students would disagree about whether a brown solution is "moderately" or "strongly" coloured.
• A colorimeter measures the absorbance of light of a chosen wavelength through the solution; the reading is a number that does not depend on the observer.
• This numerical value can be compared between tubes, plotted on a graph and used to calculate a rate.

Key Takeaways

• Colorimetry converts a subjective colour judgement into an objective numerical value.
• A quantitative result allows comparison between treatments and statistical analysis.

Common Mistakes

• Saying the colorimeter is "more accurate" without explaining why — the key point is that the result is quantitative/objective.
• Saying it is "easier" — irrelevant; the mark is for the nature of the data produced.

Things to Be Careful About

Do not write vague statements such as "more reliable" without specifying the comparison (subjective judgement by eye vs. objective numerical reading).

Background Concept

In any controlled investigation three kinds of variable are recognised:

• Independent variable (IV) — what the experimenter deliberately changes.
• Dependent variable (DV) — what is measured to see the effect of changing the IV.
• Standardised (controlled) variables — everything kept the same so that any change in the DV can be attributed to the IV.

Understanding the Question

The procedure (step 3) tells us that the students vary the concentration of the catechol solution and measure the absorbance with a colorimeter. Therefore the concentration of catechol must be the IV and the absorbance must be the DV.

Approach

Read the procedure carefully and pick out the verb "repeated using different concentrations of catechol solution" (the thing changed) and the verb "a colorimeter was used to measure the absorbance" (the thing measured).

Step-by-Step Reasoning

• "Different concentrations of catechol solution" = the variable the students deliberately alter → independent variable.
• "Absorbance of this mixture" measured with a colorimeter = the response being recorded → dependent variable.

Key Takeaways

• IV = what you change; DV = what you measure.
• The DV should ideally be linked to the rate of the enzyme-catalysed reaction; here absorbance is proportional to melanin concentration, so absorbance after a fixed time is proportional to the initial rate.

Common Mistakes

• Writing "amount of catechol" rather than "concentration of catechol" — concentration is the correct scientific term for an aqueous solution.
• Writing "rate of reaction" as the DV — the students are not measuring rate directly; they are measuring absorbance after a fixed time, which is then used as the initial rate.

Things to Be Careful About

The mark scheme requires the exact terms: concentration of catechol / substrate and absorbance. Do not paraphrase as "colour intensity" or "amount of melanin".

Background Concept

This question tests the planning of a controlled enzyme investigation. Catechol oxidase catalyses the conversion of catechol (pale pink) to melanin (brown). The brown colour can be quantified by absorbance at a chosen wavelength using a colorimeter; absorbance after a fixed short time is proportional to the initial rate of reaction. An inhibitor that lowers the rate can be characterised by the change it produces on the Michaelis–Menten curve (lower $V_{\text{max}}$, unchanged $K_m$ for a non-competitive inhibitor).

A good plan must:

• state a clear hypothesis;
• identify IV, DV and standardised variables;
• describe a workable, ordered procedure that another person could follow;
• include appropriate controls (reaction without inhibitor);
• include replication (≥3 repeats) and the calculation of means;
• address a relevant safety hazard.

Understanding the Question

The students already have the procedure for an enzyme-only investigation (steps 1–3 in the question stem). They now need to extend that procedure so that they can test the effect of inhibitor Y on the same reaction, using a range of catechol concentrations. The question says "the description of your method should be set out in a logical way and be detailed enough for another person to follow" and explicitly tells you not to repeat the enzyme-extraction details from part (a).

Approach

Adopt the standard structure for a Paper 5 plan:

1. Hypothesis.
2. Variables — IV, DV, standardised variables.
3. Method — dilution series, mixing protocol, measurement, control.
4. Replication and analysis.
5. Safety.

Hit each of the 13 possible marking points the examiner has identified so that you cover the 8 marks on offer.

Step-by-Step Reasoning

Hypothesis (covers the scientific rationale): The simplest statement — adding inhibitor Y will lower the rate of reaction compared with the uninhibited reaction — is sufficient.

Variables (marks 6, 7, 8):

• IV: concentration of catechol solution.
• DV: absorbance (initial rate).
• Standardised: same volume and concentration of enzyme extract; same volume and concentration of inhibitor Y; same volume of catechol solution; same pH (pH 7.0 buffer); same temperature; same time before reading absorbance (e.g. 2 minutes); same wavelength setting on the colorimeter.

Method (marks 1, 2, 3, 4, 5, 9, 10):

• Prepare at least five dilutions of catechol from the 1.0% stock (e.g. 0.2%, 0.4%, 0.6%, 0.8%, 1.0%) using the pH 7.0 buffer. State the range and include the % units.
• Keep the enzyme extract cold (e.g. in an ice bath) to preserve its activity.
• Equilibrate enzyme and substrate to the same temperature (e.g. 25 °C water bath) before mixing so that mixing does not cause a temperature jump that would alter the rate.
• Maintain the reaction at a constant temperature throughout, e.g. in the water bath.
• Mix a fixed volume of catechol + fixed volume of inhibitor Y + fixed volume of enzyme, start timing, leave for exactly 2 minutes, then read absorbance on the colorimeter.
• Additional colorimeter detail: zero the colorimeter with a buffer blank at the start and use the same wavelength for every reading.

Control (mark 12): Repeat the entire experiment with the same catechol concentrations but with the inhibitor replaced by the same volume of buffer. This gives the data to compare against (Fig. 1.3).

Replication (mark 11): Carry out at least three replicates of every mixture and calculate the mean absorbance for each catechol concentration.

Safety (mark 13): Catechol (and many enzyme substrates) is irritant. Name the hazard (skin/eye irritant), the risk (contact with skin or eyes) and the precaution (wear goggles and gloves; wash spillages immediately).

Key Takeaways

• A complete plan must cover hypothesis, variables, method (with controls and replication) and safety.
• For enzyme assays the rate must be measured under standardised conditions of temperature, pH, enzyme concentration and timing.
• A dilution series of at least five values across the range gives a graph shape from which inhibition type can be judged.

Common Mistakes

• Repeating the enzyme-extraction details from (a) — explicitly told not to.
• Failing to include a control without Y — without it, the experiment has nothing to be compared with.
• Forgetting to keep volumes of enzyme, substrate and inhibitor constant — any change in these would alter the rate independently of substrate concentration.
• Not stating a fixed time before measuring absorbance — the rate can only be inferred if the measurement time is the same for every tube.
• Vague temperature control ("at room temperature") — should specify a value and the method (water bath).

Things to Be Careful About

• Specify concentrations in % to match the question's wording.
• Use absorbance (not "colour") as the dependent variable, because the colorimeter is the chosen instrument.
• Replication means separate, independent repeats — repeating the same reading on the same tube is not a replicate.

Background Concept

In a Michaelis–Menten plot of initial rate against substrate concentration:

• the curve rises steeply at first (rate limited by substrate availability),
• then plateaus as the enzyme becomes saturated with substrate — the maximum rate is $V_{\text{max}}$.

Different inhibitors alter the curve in characteristic ways:

• Competitive inhibitor: raises the apparent $K_m$ (curve shifts to the right at low [S]) but $V_{\text{max}}$ is unchanged at high [S] because the substrate can out-compete the inhibitor.
• Non-competitive inhibitor: lowers $V_{\text{max}}$ (the inhibitor reduces the number of functional active sites, and this cannot be overcome by adding more substrate) but $K_m$ is unchanged because the inhibitor does not affect the affinity of the remaining active sites for substrate. The curve therefore plateaus below the uninhibited curve.

Understanding the Question

The students have plotted the uninhibited reaction on Fig. 1.3 (a curve rising to a plateau labelled "no inhibitor Y"). They claim that Y is a non-competitive inhibitor. You must sketch the curve that would result if they were correct.

Approach

Recall the two diagnostic features of a non-competitive inhibitor on this type of graph:

1. The plateau is lower ($V_{\text{max}}$ is reduced).
2. The shape at low substrate concentrations can differ slightly, but $K_m$ — the [S] at half $V_{\text{max}}$ — is unchanged.

A practical way to draw it is to sketch a curve of the same general shape but levelling off at, say, two-thirds or half of the height of the original plateau, starting a little to the right of the original curve at low concentrations.

Step-by-Step Reasoning

• Mark a horizontal line at half the original $V_{\text{max}}$ (this defines the new $K_m$, which is the same as the original because $K_m$ is unchanged).
• From that point sketch a curve of similar shape to the original that rises and plateaus at a lower maximum.
• The new curve should sit to the right of the original at low catechol concentrations but converge with the original only at the new, lower plateau (it should not cross the original curve at any point above the new plateau).

Key Takeaways

• Non-competitive inhibition reduces $V_{\text{max}}$; it does not change $K_m$.
• The signature shape is the same general Michaelis–Menten curve but with a lower plateau.

Common Mistakes

• Drawing the new curve above the original (e.g. drawing an activator's curve) — non-competitive inhibition always reduces the maximum rate.
• Drawing the new curve plateauing at the same height as the original — that is the shape expected for a competitive inhibitor at very high [S], not a non-competitive one.
• Forgetting to indicate that the new curve should be clearly to the right of the original at low substrate concentrations.

Things to Be Careful About

The examiner awards one mark for "curve to the right of the original curve" and one mark for "plateau below the original curve". You must show BOTH features explicitly.

Background Concept

Two parameters describe the kinetic behaviour of an enzyme following Michaelis–Menten kinetics:

• $V_{\text{max}}$ — the maximum initial rate, reached when all enzyme active sites are saturated with substrate. On a rate vs. [substrate] graph it is the height of the plateau.
• $K_m$ — the Michaelis constant. It is the substrate concentration at which the initial rate is exactly half of $V_{\text{max}}$. $K_m$ is also a (inverse) measure of the apparent affinity of the enzyme for its substrate: a low $K_m$ indicates high affinity, a high $K_m$ indicates lower affinity.

Understanding the Question

Fig. 1.3 shows the uninhibited enzyme's rate curve. The student is asked to mark $V_{\text{max}}$ and $K_m$ on this graph for the no-inhibitor situation.

Approach

Use the definitions literally:

• $V_{\text{max}}$ = the plateau height on the y-axis.
• $K_m$ = the [substrate] on the x-axis corresponding to a rate equal to half $V_{\text{max}}$.

Draw construction lines (dashed) so the examiner can see how the values were obtained.

Step-by-Step Reasoning

1. Identify the plateau of the curve. Draw a horizontal dashed line from this height across to the y-axis. The value on the y-axis is $V_{\text{max}}$. Label it $V_{\text{max}}$ on the y-axis.
2. Halve this value on the y-axis. From this halved value, draw a second horizontal dashed line until it meets the curve.
3. From the point where this half-rate line meets the curve, drop a vertical dashed line down to the x-axis. The point at which it meets the x-axis is $K_m$. Label it $K_m$ on the x-axis.

Key Takeaways

• $V_{\text{max}}$ is read off at the plateau; $K_m$ requires a construction at half the plateau height.
• The labels must be on the axes, not in the middle of the graph — the examiner must be able to read numerical values.

Common Mistakes

• Marking $K_m$ at the origin or at any arbitrary x-axis value rather than at the half-$V_{\text{max}}$ point.
• Drawing the construction lines but failing to label them, so the examiner cannot tell which axis intercept is meant.
• Confusing $V_{\text{max}}$ with the rate at any particular substrate concentration — $V_{\text{max}}$ is specifically the maximum (plateau) rate.

Things to Be Careful About

The construction lines should be dashed (or drawn clearly) so that they can be distinguished from the data curve. The labels $V_{\text{max}}$ and $K_m$ should be placed on the relevant axis, not floating in the middle of the graph.

Background Concept

$K_m$ is the substrate concentration at which the rate is half of $V_{\text{max}}$. It is a measure of the affinity of the enzyme's active site for its substrate.

A non-competitive inhibitor binds to a site on the enzyme other than the active site (an allosteric site). It does not compete with the substrate for binding, so it does not change the affinity of the remaining functional active sites. Therefore:

• $V_{\text{max}}$ decreases (because some enzyme molecules are inactivated and cannot be rescued by adding more substrate),
• $K_m$ is unchanged (the uninhibited active sites still bind substrate with the same affinity).

On a Michaelis–Menten plot the inhibited curve therefore reaches half of the new, lower $V_{\text{max}}$ at the same substrate concentration as the uninhibited curve reached half of the original $V_{\text{max}}$.

Understanding the Question

Part (c)(iv) asks specifically about the effect on $K_m$, not on $V_{\text{max}}$. The student must use the curves drawn on Fig. 1.3 to compare the two $K_m$ values.

Approach

Recall the definition of $K_m$ and the defining feature of non-competitive inhibition. State the conclusion in one sentence.

Step-by-Step Reasoning

• The $K_m$ is read off the x-axis where the rate equals half of the (relevant) $V_{\text{max}}$.
• For the uninhibited curve, $K_m$ corresponds to a particular [catechol].
• For the non-competitively inhibited curve, $V_{\text{max}}$ is lower, so half of $V_{\text{max}}$ is also lower — but the [catechol] at which the rate equals this new half-$V_{\text{max}}$ is the same as before.
• Therefore $K_m$ is unchanged.

Key Takeaways

• Non-competitive inhibitors do not change $K_m$.
• Competitive inhibitors, in contrast, raise the apparent $K_m$ (the curve shifts right at low [S]).

Common Mistakes

• Saying $K_m$ increases — this is true of a competitive inhibitor, not a non-competitive one.
• Saying $K_m$ decreases — the inhibitor does not improve the enzyme's affinity for substrate.
• Confusing $K_m$ with $V_{\text{max}}$ — the question asks specifically about $K_m$, not about the maximum rate.

Things to Be Careful About

The mark scheme explicitly states that the expected answer is that $K_m$ is the same with the non-competitive inhibitor. Do not over-explain; one clear statement is enough for the single mark.