Biology 9700/21 — May/June 2025

Background Concept

Maltase is a disaccharidase enzyme that hydrolyses the disaccharide maltose into two molecules of α-glucose. Maltose is two α-glucose units linked by an α-1,4-glycosidic bond; hydrolysis breaks this bond by adding a molecule of water, with –H going to one side of the bond and –OH to the other, restoring the free hydroxyls on C1 (of the left glucose) and C4 (of the right glucose).

In a Haworth projection, the α-configuration of D-glucose is shown by the –OH on C1 pointing below the plane of the ring. The –OH on C4 in α-D-glucose also points below the ring by default, so once the bond is broken the right glucose naturally displays the correct orientation on C4.

Understanding the Question

Fig. 1.1 shows one maltose molecule with the bond to be cleaved indicated by the downward arrow and the word "maltase". The candidate must:

1. Replace the single maltose with the two product α-glucose molecules.
2. Show the free –OH groups in the correct (α) orientation on the carbons that were previously joined.
3. Add H₂O above the arrow to indicate that water is added during the reaction.

Approach

First, identify the products from the enzyme–substrate relationship: maltase + maltose → 2 α-glucose (hydrolysis). Then redraw each half of the original maltose as a free α-glucose Haworth projection, restoring the free hydroxyl on the carbon that was previously bonded, keeping it in the α-orientation (–OH below the ring). Finally, write H₂O above the reaction arrow — its position is specified by the mark scheme.

Step-by-Step Reasoning

1. Identify the products. Maltose is a disaccharide of two α-glucose units; hydrolysis yields two α-glucose molecules. (Mark 1 for the first α-glucose drawn correctly.)
2. Draw the left α-glucose with –OH on C1 pointing downwards. This is the α-anomer (–OH on the anomeric C1 below the ring). (Mark 1.)
3. Draw the right α-glucose with –OH on C4 pointing downwards. C4 in α-D-glucose has its –OH below the ring by default, so once the bond is broken this is the natural orientation. (Mark 1.)
4. Add H₂O above the reaction arrow. This identifies the reaction as a hydrolysis. (Mark 1 — though the mark scheme groups this with the third mark above.)
5. Apply error carried forward (ecf). If the same mistake is made on both glucoses (e.g. omitting the –H on C5 in both), credit is still given. An error on only one side is not forgiven.

Key Takeaways

• Maltase hydrolyses the α-1,4-glycosidic bond in maltose, giving two α-glucose molecules.
• Hydrolysis is shown on a reaction scheme by writing H₂O above the arrow.
• In a Haworth projection, the α-anomer has –OH on C1 below the ring.

Common Mistakes

• Drawing β-glucose instead of α-glucose (–OH on C1 pointing up).
• Forgetting to redraw the –OH groups on C1 (left) and C4 (right) that were previously hidden inside the bond.
• Writing H₂O below the arrow — the mark scheme explicitly rejects this.
• Drawing only one glucose and leaving the other half of the original molecule unchanged.
• Failing to redraw both products as free sugars (the new free –OH on each anomeric carbon is essential).

Things to Be Careful About

• Both products must be α-glucose with –OH on the appropriate carbon pointing downwards.
• The CH₂OH group on C6 stays pointing upwards in both α-D-glucose Haworth projections.
• Water must be placed above the arrow, never below.
• An identical error on both glucoses is tolerated (ecf); an error on only one is not.

Background Concept

Disaccharides such as maltose, sucrose and lactose are formed when two monosaccharides join in a condensation reaction, releasing a molecule of water. The covalent bond that forms between them is the glycosidic bond — specifically, in maltose it is the α-1,4-glycosidic bond. When the bond is broken during hydrolysis, the water that was originally lost is added back across the bond.

Understanding the Question

This is a one-mark recall question following the diagram of maltose. The candidate must name the type of covalent bond joining the two glucose units in maltose.

Approach

Recall the standard term for the bond between two sugars in a disaccharide: the glycosidic bond. The mark scheme also accepts the variant spelling glucosidic; it rejects any addition that contradicts the actual α-1,4 link in maltose.

Step-by-Step Reasoning

1. Maltose is built from two glucose units linked by a covalent bond formed in condensation.
2. The name of this bond is glycosidic.
3. The mark scheme accepts "glycosidic" (or "glucosidic") and ignores any extra detail that is wrong (e.g. "β-1,6" would be rejected because it contradicts the α-1,4 link present in maltose).

Key Takeaways

• The covalent bond between two sugar units in a disaccharide is a glycosidic bond.
• "Glucosidic" is accepted as an alternative spelling.

Common Mistakes

• Writing "hydrogen bond" or "peptide bond" — these are different bond types.
• Adding incorrect extra detail (e.g. "β-1,6-glycosidic") that contradicts the α-1,4 link actually present in maltose.

Things to Be Careful About

• The mark scheme accepts "glucosidic" as an alternative spelling.
• Do not add qualifiers that contradict the structure of maltose.

Background Concept

Hydrolysis is the reaction in which a covalent bond is broken by the addition of a molecule of water. The H from water attaches to one side of the broken bond and the –OH attaches to the other, restoring the two free –OH (or –H) groups. In digestion, hydrolytic enzymes (such as maltase, sucrase, lactase, amylase, lipase and the proteases) catalyse the hydrolysis of larger biological molecules into their monomers.

Understanding the Question

The question follows the diagram of maltose being broken into two α-glucose by maltase. The candidate must name the type of reaction being catalysed.

Approach

Recall that a bond broken by the addition of water is a hydrolysis. The opposite reaction — forming a bond by removing water — is a condensation reaction.

Step-by-Step Reasoning

1. Maltase breaks the α-1,4-glycosidic bond in maltose.
2. Water is added across the bond (one –H to the left glucose, one –OH to the right glucose).
3. Bond-broken-by-water = hydrolysis.

Key Takeaways

• Hydrolysis = bond broken by water.
• Condensation = bond formed by removal of water.
• Maltase is one of many hydrolytic enzymes involved in digestion.

Common Mistakes

• Writing "condensation" — this is the reverse reaction (water removed, bond formed).
• Writing "digestion" — too vague; "hydrolysis" is the precise biochemical term.

Things to Be Careful About

• The mark scheme accepts only "hydrolysis"; do not add qualifiers (e.g. "acid hydrolysis") unless they are correct.

Background Concept

A colorimeter measures the absorbance of light by a solution. The darker or more intensely coloured the solution, the more light it absorbs, and the higher the absorbance reading. When an enzyme-catalysed reaction changes the colour (or the cloudiness) of a reaction mixture, the absorbance changes accordingly – allowing the reaction to be followed quantitatively over time.

As an enzyme-catalysed reaction proceeds, substrate molecules collide with the active sites of enzyme molecules to form enzyme–substrate (ES) complexes, which then break down to release product and regenerate the free enzyme. At the start of the reaction, when both substrate and enzyme are abundant, product is formed rapidly. As substrate is consumed, the rate of ES-complex formation gradually falls and eventually the reaction effectively stops when substrate becomes limiting.

Understanding the Question

You are given three figures:

• Fig. 3.1 – the two reactions: dopa oxidase converts L-dopa (colourless) to dopachrome (orange-brown); neutrase hydrolyses casein (white) to colourless peptides.
• Fig. 3.2 – absorbance against time for the dopa oxidase reaction. Absorbance starts at 0 and rises to a plateau of about 0.88.
• Fig. 3.3 – absorbance against time for the neutrase reaction. Absorbance starts at about 1.15 and falls to a plateau of about 0.10.

The question asks for the similarities between the progress of the two reactions. The direction of change differs (absorbance goes up in Fig. 3.2 and down in Fig. 3.3), so the question is asking about the overall shape of the curves, not the direction. The command word is 'describe and explain', so you need to give a description of the shape as well as an explanation of why the shape is what it is.

Approach

1. Identify the common feature of both graphs: an initial period during which absorbance changes, followed by a plateau.
2. Explain why absorbance changes – because the colour / cloudiness of the reaction mixture is changing (in opposite directions in the two reactions).
3. Explain why the curve is steepest at the start – because substrate and enzyme molecules are colliding frequently and ES complexes are forming rapidly.
4. Explain why the plateau is reached – because, eventually, the substrate is used up and the rate of product formation falls to zero.

Step-by-Step Reasoning

Description (1 mark)

Both graphs show an initial period during which absorbance changes, and then a time after which there is no further change – the curve reaches a plateau. The change in absorbance is in opposite directions in the two reactions, but the overall pattern of change-then-plateau is the same.

Explanation (2 marks from the 3 possible points)

• Point 1 – why absorbance changes:
  The absorbance changes because the colour (or cloudiness) of the reaction mixture is changing as product is formed. In the dopa oxidase reaction, colourless L-dopa is being converted into orange-brown dopachrome, so the absorbance rises. In the neutrase reaction, the white, cloudy casein is being broken down into colourless peptides, so the absorbance falls.

• Point 2 – why the curve is steepest at the start:
  The initial rapid change in absorbance occurs because there is plenty of substrate and enzyme available. Substrate molecules collide with the active sites of enzyme molecules, ES complexes form, and product is made at a high rate.

• Point 3 – why the curve plateaus:
  The plateau is reached when all (or most) of the substrate has been used up. With little or no substrate left, few new ES complexes can form, so product is no longer being made and the absorbance stops changing.

Key Takeaways

• A progress curve for an enzyme-catalysed reaction typically shows an initial period of rapid change followed by a plateau.
• The shape of the curve is the same regardless of the direction of the colour change.
• A colorimeter converts a visible change in colour or cloudiness into a measurable, numerical absorbance value.

Common Mistakes

• Saying the rate of reaction increases throughout both reactions – in fact, the rate decreases as substrate is used up.
• Failing to mention the plateau / the fact that the reaction effectively stops.
• Saying 'reactants are used up' – the mark scheme specifically ignores this wording; only 'substrate' is credited.
• Confusing the two graphs and giving the wrong direction of change for one of them.
• Forgetting the explanation and giving only a description (this earns only 1 of the 3 marks).

Things to Be Careful About

• The question asks for similarities, not differences, so do not focus on the opposite direction of the two curves.
• The 'change in absorbance' point must be linked to the 'plateau' part to earn the first mark.
• 'Describe and explain' means you need both – a description without an explanation, or vice versa, will lose marks.

Background Concept

A colorimeter is an instrument that measures the absorbance of light by a coloured or cloudy solution. It produces a numerical value that is directly proportional to the concentration of the substance absorbing the light (within a suitable range, given by the Beer–Lambert law). It is therefore a quantitative tool, in contrast to a purely visual observation which is qualitative and depends on the observer's judgement.

Understanding the Question

The question asks for two advantages of using a colorimeter (rather than some other method such as visual observation or sampling at intervals) to follow reactions like those in Fig. 3.1. The advantages should be specific to the situation: a reaction in which the colour or cloudiness of the reaction mixture is changing.

Approach

Think about what a colorimeter offers that simpler methods do not:

1. Numerical (quantitative) data instead of descriptive observations.
2. Continuous recording instead of discrete time points.
3. Objective readings that are not influenced by the observer.
4. The ability to compare readings to a calibration curve and obtain actual concentrations.
5. The ability to detect very small changes in colour / cloudiness that the eye might miss.
6. The ability to determine rates of reaction directly from the data.

You only need to give two of these, but each must be a distinct, specific idea.

Step-by-Step Reasoning

• Quantitative readings: A colorimeter gives a numerical absorbance value for each reading, so the data can be analysed statistically and compared precisely between experiments.
• Plotting graphs: The numerical values can be plotted on a graph (e.g. absorbance against time) to give a clear picture of the progress of the reaction.
• Continuous readings: The colorimeter can be set to take readings at frequent intervals (or even continuously), so the progress of the reaction can be followed without having to remove samples from the reaction mixture at each time point, which would otherwise disturb the reaction.
• Objective results: The readings are not influenced by personal judgement, so the results are objective and free from observer bias.
• Rates of reaction: Because the data are numerical and can be plotted as a curve, the rate of reaction at any time can be found from the gradient of the curve.
• Detecting small changes: A colorimeter is much more sensitive than the eye and can detect very small changes in colour or cloudiness.
• Calibration curve: With a calibration curve of absorbance against concentration, the actual concentration of product at any time can be determined.

Key Takeaways

• A colorimeter converts a visible change in colour or cloudiness into numerical data, enabling quantitative analysis.
• A colorimeter is objective, so it removes observer bias.
• A colorimeter can be read continuously, providing a detailed progress curve without disturbing the reaction.
• With a calibration curve, absorbance can be converted into concentration.

Common Mistakes

• Saying 'more accurate' or 'more precise' without explaining how or why – these terms need context to be credited.
• Saying 'continuous data' on its own – the mark scheme specifically ignores this wording. The advantage is the ability to take readings continuously without having to take samples.
• 'Human error' is too vague.
• Saying it gives instant results – this is not specific to a colorimeter.

Things to Be Careful About

• Each advantage must be a specific, concrete benefit, not a general property of scientific instruments.
• Avoid vague 'it's better' type answers – the mark scheme credits specific points only.

Background Concept

The Michaelis–Menten constant, $K_m$, is the substrate concentration at which an enzyme-catalysed reaction proceeds at half of its maximum rate ($\frac{1}{2}V_{\text{max}}$). It is a measure of the affinity of the enzyme for its substrate: a low $K_m$ indicates high affinity (the enzyme reaches half its maximum rate at a low substrate concentration), while a high $K_m$ indicates low affinity.

The rate of an enzyme-catalysed reaction increases with substrate concentration up to a maximum, $V_{\text{max}}$, which is reached when all of the enzyme active sites are saturated with substrate. The curve on a rate-versus-substrate-concentration graph therefore has the characteristic shape shown in Fig. 3.4.

To find $K_m$ from such a graph:

1. Read $V_{\text{max}}$ from the plateau of the curve.
2. Halve this value to get $\frac{1}{2}V_{\text{max}}$.
3. Draw a horizontal line from $\frac{1}{2}V_{\text{max}}$ on the y-axis to the curve, then drop a vertical line down to the x-axis. The substrate concentration at this point is $K_m$.

Understanding the Question

You are given Fig. 3.4, a graph of rate of reaction (in $\mu\text{mol mg}^{-1}\text{ min}^{-1}$) against substrate concentration (in $\text{mmol dm}^{-3}$) for the enzyme VpSP37. The curve plateaus at a rate of $80\ \mu\text{mol mg}^{-1}\text{ min}^{-1}$. You need to find the substrate concentration at which the rate is half this value, $40\ \mu\text{mol mg}^{-1}\text{ min}^{-1}$, and give the answer with the correct unit.

Approach

1. Identify $V_{\text{max}}$ from the plateau of the curve in Fig. 3.4.
2. Halve it to get $\frac{1}{2}V_{\text{max}}$.
3. Read the corresponding substrate concentration from the curve.
4. Give the answer with the correct unit.

Step-by-Step Reasoning

Step 1 – Read $V_{\text{max}}$

The curve in Fig. 3.4 plateaus at a rate of $80\ \mu\text{mol mg}^{-1}\text{ min}^{-1}$. So $V_{\text{max}} = 80\ \mu\text{mol mg}^{-1}\text{ min}^{-1}$.

Step 2 – Calculate $\frac{1}{2}V_{\text{max}}$

Step 3 – Read the corresponding substrate concentration from the graph

Draw a horizontal line from the rate of $40\ \mu\text{mol mg}^{-1}\text{ min}^{-1}$ on the y-axis to the curve in Fig. 3.4. Then drop a vertical line from that point on the curve down to the x-axis. The x-axis is read in $\text{mmol dm}^{-3}$, with major gridlines every $0.02$ and minor gridlines every $0.002$. The vertical line meets the x-axis at a substrate concentration of approximately $0.014\ \text{mmol dm}^{-3}$.

Step 4 – State the answer with the correct unit

(equivalently $14\ \mu\text{mol dm}^{-3}$)

Key Takeaways

• $K_m$ is defined as the substrate concentration at which the rate is half of $V_{\text{max}}$.
• $K_m$ can be read directly from a rate-versus-substrate-concentration graph by halving $V_{\text{max}}$ and reading across to the curve.
• A low $K_m$ indicates a high affinity between enzyme and substrate.
• A high $K_m$ indicates a low affinity between enzyme and substrate.

Common Mistakes

• Forgetting to halve $V_{\text{max}}$ and reading the substrate concentration at $V_{\text{max}}$ instead (this gives a much larger value, ~0.16 instead of ~0.014).
• Misreading the x-axis (e.g. confusing 0.014 with 0.14).
• Omitting the unit, or giving the unit as 'mmol' without the 'dm⁻³' (the unit is essential).
• Giving the unit in a different form (e.g. as $\mu\text{mol mg}^{-1}\text{ min}^{-1}$, which is the unit of rate, not concentration).

Things to Be Careful About

• The unit on the answer line or in the working is essential – the mark scheme specifically requires it.
• A small reading error is acceptable; values in the range $0.012$ to $0.016\ \text{mmol dm}^{-3}$ are credited.
• Make sure the answer is given to an appropriate number of significant figures (the graph allows 2 or 3 sig figs).

Background Concept

$K_m$ is a measure of the affinity of an enzyme for its substrate. A low $K_m$ means the enzyme–substrate complex forms readily even at low substrate concentrations; the enzyme has a high affinity for the substrate. A high $K_m$ means the enzyme needs a much higher substrate concentration to reach the same rate; the affinity is lower.

The structural basis of affinity is the fit between the substrate and the active site of the enzyme. A better fit (more complementary shapes, more favourable interactions between R groups and the active-site residues) means ES complexes form more readily and dissociation is less likely, so the rate at any given substrate concentration is higher.

In an industrial process, the substrate is often a feedstock that has to be bought or produced in bulk. Using an enzyme that works efficiently at low substrate concentrations is therefore economically advantageous: less substrate is required, or dilute / cheap feedstocks can be used.

Understanding the Question

VpSP37 has a lower $K_m$ than other enzymes that could be used in the same industrial process. You need to explain why this is an advantage. Note that the advantage is not that VpSP37 is 'faster' in absolute terms – both enzymes may reach the same $V_{\text{max}}$ eventually – but that VpSP37 reaches a high proportion of $V_{\text{max}}$ at lower substrate concentrations.

Approach

1. State the consequence of a lower $K_m$: VpSP37 reaches half of $V_{\text{max}}$ at a lower substrate concentration (or gives a higher rate at a given substrate concentration).
2. Link this to enzyme–substrate affinity: VpSP37 has a higher affinity for its substrate because there is a better fit between the substrate and the active site.
3. (Optional, for full marks) Explain the industrial benefit: a lower concentration of substrate is needed, so the process is cheaper or can use dilute feedstocks.

Step-by-Step Reasoning

Consequence of a lower $K_m$:
Because VpSP37 has a lower $K_m$ than the other enzymes, it reaches half of its $V_{\text{max}}$ at a lower substrate concentration. Equivalently, for a given substrate concentration, VpSP37 gives a higher rate of reaction than the other enzymes.

Why this happens – enzyme–substrate affinity:
This is because VpSP37 has a higher affinity for its substrate. The substrate fits the active site of VpSP37 more closely than it fits the active sites of the other enzymes, so enzyme–substrate complexes form more readily. (The active site and substrate are more complementary in shape and have more favourable interactions between the substrate's R groups and the active-site residues.)

Industrial advantage:
In an industrial process, this means that a lower concentration of substrate is needed to achieve a high rate of reaction. The substrate (feedstock) is often a major cost, so using VpSP37 reduces the cost of the process. It may also allow the use of dilute or impure feedstocks that would otherwise give a low rate with a higher-$K_m$ enzyme.

Key Takeaways

• A low $K_m$ indicates high enzyme–substrate affinity.
• A low $K_m$ means the enzyme can work efficiently at low substrate concentrations.
• In an industrial process, this can reduce substrate cost or allow the use of dilute / impure feedstocks.
• The structural basis of high affinity is a close fit between the substrate and the active site.

Common Mistakes

• Saying VpSP37 is 'faster' without qualifying 'at the same substrate concentration' (the mark scheme specifically accepts 'faster rate at the same concentration' or 'faster rate at lower concentration').
• Confusing $K_m$ with $V_{\text{max}}$ (they are different: $V_{\text{max}}$ is the maximum rate, $K_m$ is the substrate concentration at half $V_{\text{max}}$). The other enzymes may have a similar $V_{\text{max}}$; the difference is the $K_m$.
• Saying the enzyme–substrate complex is formed 'more efficiently' without explaining why – the mark scheme specifically ignores this wording; only the 'better fit' explanation is credited.
• Just saying 'it is cheaper' without linking it to the substrate concentration or affinity.

Things to Be Careful About

• $K_m$ reflects affinity, not maximum rate. Two enzymes can have the same $V_{\text{max}}$ but different $K_m$ values.
• The advantage is about substrate concentration, not about absolute speed.
• The 'better fit' explanation must be in terms of structure (the substrate and active site being complementary), not in terms of vague 'efficiency'.

Background Concept

Transcription is the synthesis of a complementary RNA copy from a DNA template. The enzyme that catalyses this reaction is RNA polymerase. In eukaryotes, RNA polymerase II is specifically responsible for transcribing protein-coding genes into pre-mRNA, while RNA polymerase I transcribes most rRNA genes and RNA polymerase III transcribes tRNA and 5S rRNA genes.

RNA polymerase binds to a promoter region upstream of the gene, unwinds the DNA double helix, and joins free RNA nucleotides together using complementary base pairing (A with U, T with A, G with C, C with G). The new RNA strand is built in the 5' to 3' direction.

Understanding the Question

Fig. 4.1 shows the gene TLR8 being transcribed inside the nucleus of a macrophage, producing a primary transcript of RNA. The question asks for the name of the enzyme that catalyses this transcription step. The single mark and the command word 'name' indicate that a one-word answer is sufficient.

Approach

Direct recall: the enzyme that makes RNA from DNA is RNA polymerase.

Step-by-Step Reasoning

The answer is RNA polymerase.

Key features of RNA polymerase in this context:

• It binds to the promoter of the TLR8 gene (shown in Fig. 4.1 as the 'site of attachment of enzyme').
• It moves along the DNA template strand, adding complementary RNA nucleotides.
• The product is the primary transcript (a single-stranded RNA copy of the gene).

The primary transcript is then modified by the addition of a 5' cap and a 3' poly(A) tail, before undergoing splicing (stage Y) to produce the mature mRNA.

Key Takeaways

• RNA polymerase is the enzyme that catalyses transcription.
• In eukaryotes, RNA polymerase II transcribes mRNA-coding genes.
• Transcription occurs in the nucleus.

Common Mistakes

• Writing DNA polymerase instead of RNA polymerase. DNA polymerase is used in DNA replication, not transcription.
• Writing helicase, which only unwinds the DNA and does not synthesise RNA.
• Confusing transcription with translation. Translation is catalysed by ribosomes (with tRNA), not by RNA polymerase.

Things to Be Careful About

• The question is worth 1 mark, so a single term is sufficient. No additional explanation is required on the exam paper.

Background Concept

Eukaryotic mRNA is modified after transcription (post-transcriptional modification) before it can be translated. Two key modifications are:

• A 5' cap: a modified guanine nucleotide added to the 5' end of the primary transcript.
• A 3' poly(A) tail: a string of adenine nucleotides added to the 3' end.

These modifications serve multiple functions related to mRNA stability, nuclear export and translation.

Understanding the Question

The question asks the candidate to suggest the functions of the 5' cap and the 3' poly(A) tail in the stability and role of mRNA. Two marks are available, so two creditable points are required. The candidate should consider both stability and the role of mRNA in protein synthesis.

Approach

Consider what would happen to an mRNA without these modifications:

1. Without protection, mRNA would be rapidly degraded by nucleases (exonucleases) in the cytoplasm.
2. Without a mechanism to mark the mRNA, ribosomes would not know where to start translation, and the mRNA might not be efficiently exported from the nucleus.
3. The cap and tail together also help prevent mRNA molecules from joining end-to-end (which would make them non-functional).

Step-by-Step Reasoning

The marking scheme accepts any two of the following credible points:

1. Protection from degradation. The cap and tail shield the ends of the mRNA from exonuclease enzymes that would otherwise break down the molecule. This extends the half-life of the mRNA in the cytoplasm, allowing more protein to be made from each transcript.

2. Preventing mRNA molecules from joining together. The cap and tail physically mark the ends of the mRNA, so two mRNA molecules cannot join end-to-end. This prevents the formation of non-functional concatenates.

3. Directing mRNA out of the nucleus. The 5' cap is recognised by nuclear export factors that carry the mRNA through nuclear pores into the cytoplasm. The poly(A) tail is also recognised by export proteins that interact with the cap-binding complex.

4. Aiding translation. The 5' cap is recognised by initiation factors that recruit the small ribosomal subunit to the mRNA. The poly(A) tail binds poly(A)-binding protein, which interacts with initiation factors to form a 'closed loop' that enhances translation. Together they ensure the 5' end enters the ribosome first.

5. AVP — e.g. ensures the correct orientation of mRNA for translation (5' end first into the ribosome).

The best two answers to pick, in an exam, are usually the stability point and the translation/export point, because these are the most fundamental and easily remembered.

Key Takeaways

• The 5' cap and 3' poly(A) tail protect mRNA from degradation by nucleases.
• They help the mRNA to be exported from the nucleus through nuclear pores.
• They aid ribosome attachment and the start of translation.
• They prevent mRNA molecules from joining end-to-end.

Common Mistakes

• Stating that the cap or tail is a start or stop codon. The start codon (AUG) is a sequence within the mRNA coding region, not a structural cap. The mark scheme explicitly rejects this.
• Saying the cap and tail 'help with translation' without specifying how (e.g. by aiding ribosome attachment).
• Confusing the function of the cap with that of the start codon.
• Describing the cap and tail in terms of DNA replication (they have no role in replication).

Things to Be Careful About

• The mark scheme rejects references to start and stop codons.
• The candidate can earn either mark for either the cap OR the poly(A) tail — they do not have to specify which modification does which function. Any two credible functions gain two marks.
• 'Protects' alone is too vague; the candidate should say what the mRNA is protected from (e.g. enzymes, degradation, breakdown).

Background Concept

In eukaryotes, genes are typically split into exons (expressed sequences, which code for amino acids) and introns (intervening sequences, which do not code for amino acids). When a gene is transcribed, both exons and introns are copied into the primary transcript (pre-mRNA). Before the mRNA leaves the nucleus, the introns must be removed and the exons joined together in a process called splicing.

Splicing is carried out by a large ribonucleoprotein complex called the spliceosome, which recognises specific sequences at the intron–exon boundaries. The intron is excised as a lariat structure and the exons are ligated together by phosphodiester bonds. Some genes undergo alternative splicing, in which different combinations of exons are joined, allowing a single gene to code for multiple protein variants.

Understanding the Question

The question asks the candidate to describe the process that occurs at stage Y in Fig. 4.1, which is the conversion of the primary transcript (containing both introns and exons) into the shorter mature mRNA. Three marks are available.

Approach

Recognise stage Y as splicing, then describe the key events: identification of the sequences to be removed, removal of introns, joining of exons, and the result (a shorter mature mRNA containing only coding sequences).

Step-by-Step Reasoning

The marking scheme accepts any three of the following:

1. (Gene/RNA) splicing — name of the process.
2. Introns are removed — the non-coding regions are excised from the primary transcript.
3. Exons are joined / attached together — the coding regions are covalently linked to form the continuous coding sequence of the mature mRNA.
4. The RNA molecule is shortened — the mature mRNA is shorter than the primary transcript because the introns have been removed.
5. Removal of non-coding sequences / only keeping the coding sequences — explains the functional significance of the process.
6. AVP — e.g. spliceosome acts on the primary transcript; phosphodiester bonds form between RNA nucleotides when exons are joined; alternative splicing can rearrange exons.

The three best points to write in an exam are:

• Splicing (the name of the process)
• Introns removed
• Exons joined together

Key Takeaways

• Stage Y in Fig. 4.1 is splicing.
• Splicing removes introns (non-coding sequences) and joins exons (coding sequences).
• The mature mRNA is shorter than the primary transcript and contains only the coding sequence needed to make the protein.
• Splicing is a feature of eukaryotic gene expression; prokaryotic mRNA does not usually contain introns.

Common Mistakes

• Writing 'DNA splicing' or 'mRNA splicing' — the mark scheme rejects these. The correct term is gene splicing or RNA splicing of the primary transcript.
• Spelling 'exons' as 'extrons' — this is rejected.
• Saying that 'introns are translated' or that introns code for amino acids — they do not.
• Confusing splicing with other forms of RNA processing (capping or polyadenylation are separate events shown earlier in Fig. 4.1).
• Saying 'the introns are translated into proteins' — this is incorrect; introns are removed before translation.

Things to Be Careful About

• The mark scheme requires the term 'gene' or 'RNA' splicing, not 'DNA splicing' or 'mRNA splicing'.
• The diagram shows both introns (light grey) and exons (dark grey); the candidate should refer to them correctly.
• Splicing is a separate event from the addition of the 5' cap and poly(A) tail, which occur before splicing (Fig. 4.1 shows the cap and tail already present before stage Y).

Background Concept

Telomeres are specialised structures found at the ends of eukaryotic linear chromosomes. They consist of short, repetitive DNA sequences (in humans the repeat is TTAGGG) and associated proteins. Telomeres protect the chromosome ends from degradation and from being recognised as DNA damage by the cell's repair machinery.

Understanding the Question

The question asks the candidate to state where in a chromosome the telomeres are found. The single mark and the command word 'state' indicate a one-line factual answer is required.

Approach

Direct recall: telomeres are at the ends of chromosomes.

Step-by-Step Reasoning

The answer is at the ends (of the chromosome / chromatid).

In the human karyotype, each chromosome has two telomeres — one at each end of the DNA molecule. They are at the very tips of the chromatids, not in the middle (where the centromere is) and not on the sides.

Key Takeaways

• Telomeres are located at the ends of chromosomes.
• Each chromosome has two telomeres, one at each end.
• The repetitive DNA sequence is TTAGGG in humans, repeated many times in the 5' to 3' direction.

Common Mistakes

• Saying 'at the edges' or 'on the sides' — these are rejected by the mark scheme as too vague.
• Saying 'at the centromere' — the centromere is in the middle of the chromosome, not at the ends.
• Saying 'throughout the chromosome' — telomeres are specifically at the ends, not distributed throughout.

Things to Be Careful About

• 'Sides' and 'edges' are explicitly ignored by the mark scheme.
• A precise answer of 'at the ends (of the chromosome / chromatid / DNA)' is required.

Background Concept

DNA replication has an inherent problem at the 5' end of the lagging strand: the very last few nucleotides cannot be replicated because there is no upstream 3'-OH for DNA polymerase to extend from. This is called the end-replication problem. Without a mechanism to compensate, chromosomes would become shorter with every cell division, eventually losing important genetic information.

Telomeres solve this problem:

• They consist of many copies of a short, non-coding repetitive sequence (TTAGGG in humans) at the ends of chromosomes.
• These repetitive sequences act as a 'buffer' that can be lost without affecting coding genes.
• The enzyme telomerase (in cells that express it, such as stem cells and germ cells) can add more repeats to the telomeres, maintaining their length.

In most somatic cells, telomerase is not active, so telomeres shorten with each division. This acts as a 'molecular clock' that limits the number of divisions a cell can undergo (the Hayflick limit).

Understanding the Question

The question asks the candidate to outline the role of telomeres. Three marks are available, so three creditable points should be given.

Approach

Think about what would happen if chromosomes did not have telomeres:

• DNA replication would be incomplete at the ends.
• Genes at the ends of chromosomes would be lost over multiple rounds of replication.
• Chromosome ends might be recognised as damage and trigger cell cycle arrest or apoptosis.
• The number of mitoses a cell could undergo would be limited.

Step-by-Step Reasoning

The marking scheme accepts any three of the following:

1. Telomeres allow DNA replication to occur many times. The repetitive non-coding sequences provide a buffer so that the end-replication problem does not erode coding sequences.

2. Telomeres allow (some) cells to carry out many / continuous / repeated mitoses / cell cycles / cell divisions. Cells such as stem cells and germ cells maintain telomere length using telomerase and can therefore divide many times.

3. Telomeres prevent the loss of genes / genetic information from the ends of chromosomes. This is the key functional consequence of having telomeres — they protect the coding sequences of the chromosome.

4. AVP — e.g. telomeres prevent the ends of chromosomes from being recognised as damaged DNA; telomeres prevent the fusion of chromosome ends with other chromosomes.

Key Takeaways

• Telomeres are protective caps at the ends of linear chromosomes.
• They consist of repetitive non-coding DNA that acts as a buffer against the end-replication problem.
• They allow cells to undergo many rounds of DNA replication and mitosis without loss of coding genes.
• Telomerase maintains telomere length in stem cells, germ cells and cancer cells.

Common Mistakes

• Saying 'telomeres prevent loss of genetic material' — this is rejected by the mark scheme. The candidate must say genes or genetic information.
• Saying 'telomeres prevent loss of DNA' — this is ignored. The relevant concept is loss of coding sequences / genes, not just any DNA.
• Confusing telomeres with centromeres (which hold sister chromatids together) or origins of replication.
• Saying 'telomeres allow DNA to replicate' without explaining that this is specifically about many / repeated rounds of replication.

Things to Be Careful About

• The mark scheme explicitly ignores 'loss of genetic material' and 'loss of DNA'. The candidate must say 'loss of genes' or 'loss of genetic information'.
• The candidate should not confuse telomere function with centromere function.

Background Concept

Haemoglobin is a globular, conjugated protein — a protein with a non-protein component attached. Each of its four polypeptide chains (two α and two β) carries a small non-protein ring-shaped molecule embedded in a hydrophobic pocket. This non-protein component is called a prosthetic group, and in haemoglobin it is the haem group.

The haem group is built around a porphyrin ring (four nitrogen-containing pyrrole rings joined by methine bridges) with a central iron ion (Fe²⁺) held in coordinate bonds. It is the iron that actually interacts with oxygen, not the protein chain itself. Because the prosthetic group is firmly bound, it is considered part of the functional haemoglobin molecule.

Understanding the Question

The stem shows a ribbon diagram of haemoglobin. The pointer labelled X leads to the small disc-like structure visible at the bottom-left of the molecule — the flat ring inserted into one of the four subunits. The question asks for the name of that structure.

Approach

Recognise the small ring structure with a metal ion at its centre as a haem group, distinguishing it from the polypeptide ribbons that make up the bulk of the molecule.

Step-by-Step Reasoning

• The ribbon model shows mostly coiled and folded polypeptide chains.
• The label X points to a compact, non-ribbon element — clearly a non-protein component.
• A non-protein component tightly associated with a protein is a prosthetic group; the specific prosthetic group in haemoglobin is the haem group (containing Fe²⁺).
• The mark scheme accepts "haem (group)" but does NOT accept "prosthetic group" alone (it is too general) and does NOT accept "Fe²⁺" alone (it is the ion inside the haem, not the structure as a whole).

Key Takeaways

• Haemoglobin is a conjugated protein: polypeptide + prosthetic group.
• The prosthetic group is the haem group; the iron at its centre binds oxygen.
• Four haem groups, one per polypeptide chain, allow each haemoglobin to carry up to four O₂ molecules.

Common Mistakes

• Writing "prosthetic group" — too vague; the mark scheme ignores this.
• Writing "Fe²⁺" or "iron" — the mark scheme ignores this. The iron is contained within the haem group.
• Writing "porphyrin ring" — not accepted as it is the framework of the haem group rather than its name.

Things to Be Careful About

• The label X is the haem group, not the polypeptide chain, not the whole haemoglobin, and not the iron ion on its own.

Background Concept

Oxygen is not very soluble in plasma, so it must be carried bound to a carrier protein. Haemoglobin's oxygen-binding site is the iron ion (Fe²⁺) at the centre of each haem group. The Fe²⁺ forms a reversible coordinate bond with one O₂ molecule, giving oxyhaemoglobin:

Because each haem binds one O₂, and there are four haem groups per haemoglobin, one haemoglobin molecule can carry up to four oxygen molecules. The bond is reversible, which is essential for oxygen release in respiring tissues.

Understanding the Question

Having identified X as the haem group, the question now asks what it does. The answer should be a single statement of its function in the haemoglobin molecule.

Approach

Recall that the haem group is the oxygen-binding site and that the actual binding atom is the iron ion at its centre. Frame the answer in terms of the iron + oxygen combination.

Step-by-Step Reasoning

• Function of the haem group: bind oxygen.
• The binding atom within the haem group is the Fe²⁺ ion.
• Each haem binds one O₂ (the mark scheme rejects "two or more").
• Allowed phrasings: "carries/transports/attaches oxygen" or "bonds with oxygen".
• The bond is reversible, allowing release at the tissues.

Key Takeaways

• The haem group's job is to bind one O₂ per haem via its central Fe²⁺.
• Each Hb has four haem groups, so up to four O₂ per Hb molecule.
• The bond is reversible — the basis of oxygen loading and unloading.

Common Mistakes

• Writing "binds carbon dioxide" — rejected; CO₂ is carried elsewhere (mainly as HCO₃⁻ in plasma, with some bound to globin chains, not haem).
• Writing "binds oxygen atoms" — rejected; oxygen is transported as O₂ molecules.
• Writing "binds two oxygens" or "many oxygens" — rejected; one haem binds one O₂.
• Saying only "carries oxygen" without making clear that it is the iron in the haem that does this — vague answers score zero.

Things to Be Careful About

• "Binds/attaches/combines/carries" oxygen is acceptable.
• "Oxygen atom" is wrong — it is the O₂ molecule.

Background Concept

Proteins are organised at four levels:

• Primary (1°): the sequence of amino acids linked by peptide bonds.
• Secondary (2°): regular coiling/folding (α-helix, β-pleated sheet) stabilised by H-bonds.
• Tertiary (3°): 3D folding of a single polypeptide, stabilised by H-bonds, ionic bonds, disulfide bridges and hydrophobic interactions.
• Quaternary (4°): the assembly of more than one polypeptide chain into the functional protein, with the same or different types of chain.

Haemoglobin is a classic example of a protein with quaternary structure: it has four polypeptide chains (2 α + 2 β), each with its own tertiary fold, and each carrying a haem prosthetic group. The four chains interact (hydrophobic interactions, salt bridges) to give the quaternary structure that is essential for the cooperative binding of oxygen (the sigmoidal oxygen dissociation curve).

Understanding the Question

The question gives the premise that haemoglobin has quaternary structure and asks the candidate to define what that means. The mark scheme rewards a definition focused on multiple polypeptide chains.

Approach

Give the textbook definition: quaternary structure = the arrangement of more than one polypeptide chain in a protein.

Step-by-Step Reasoning

• The mark scheme accepts "more than one polypeptide (chain)" or "2 or more".
• It rejects "more than 2", "many" or "multiple" (too vague), and ignores "amino acid chain".
• Stating the example "haemoglobin has four polypeptides" is allowed.
• A clean, textbook definition is enough for the mark.

Key Takeaways

• Quaternary structure = multiple polypeptide chains assembled into one functional protein.
• Different chains can be identical (e.g. insulin, two chains) or different (e.g. haemoglobin, α and β).
• Quaternary structure enables cooperativity and allosteric effects (e.g. Bohr shift).

Common Mistakes

• Writing "multiple", "many", or "more than 2" — the mark scheme rejects these.
• Writing "four polypeptide chains" alone is fine (allowed), but it should clearly be a definition, not just a count.
• Confusing quaternary with tertiary (3D folding of a single chain).

Things to Be Careful About

• "Polypeptide chain" is the right term — not "amino acid chain" (ignored), and not just "chain" (vague).

Background Concept

Alveolar air has a high partial pressure of oxygen because fresh air is continuously delivered by ventilation. At this high pO₂, haemoglobin leaving the lungs is almost fully saturated — typically around 95–98% — which maximises the oxygen content of arterial blood.

On an oxygen dissociation curve, this corresponds to the upper plateau region: as pO₂ rises above about 8 kPa, the percentage saturation of haemoglobin approaches a maximum and is relatively insensitive to further increases in pO₂.

Understanding the Question

The stem tells us alveolar pCO₂ is 5.3 kPa, so the relevant curve is the dashed (middle) curve. The question asks for the pO₂ of the alveolar air. A precise value is impossible to read, so the mark scheme accepts a range.

Approach

Identify the dashed curve (pCO₂ = 5.3 kPa) and read off the pO₂ at which the curve is at its plateau (very high percentage saturation) — these are the values that match alveolar conditions.

Step-by-Step Reasoning

• The dashed curve reaches a near-plateau between roughly 8 and 14 kPa.
• The mark scheme accepts any value (or range) within 10–14 kPa — the upper, near-saturated part of the curve.
• A value in this range is biologically realistic for mammalian alveolar pO₂ (around 13 kPa in humans).

Key Takeaways

• Alveolar pO₂ corresponds to the upper plateau of the dissociation curve.
• Choose the curve matching the stated alveolar pCO₂ (5.3 kPa here).
• The mark scheme accepts a range, so an approximate read-off is sufficient.

Common Mistakes

• Reading a value from the wrong curve (e.g. taking a reading from the 2.7 kPa or 10.7 kPa curve).
• Giving a value in the middle of the curve (e.g. 4 kPa) — that would represent a saturating pO₂ only at the lower plateau, not the alveolar level.

Things to Be Careful About

• Match the curve by pCO₂, not just by position on the page.
• The answer is a range — there is no single correct value because the curve is essentially flat at the top.

Background Concept

Respiring tissues consume O₂ (for oxidative phosphorylation in mitochondria) and produce CO₂. As a result, the pO₂ inside active cells and in the tissue fluid immediately around them is lower than arterial blood, while pCO₂ is higher. The shape of the oxygen dissociation curve ensures that haemoglobin responds to this drop in pO₂ by releasing oxygen — and the steep middle portion of the curve is exactly where this unloading happens most efficiently. A small fall in pO₂ here causes a large fall in percentage saturation, so a large amount of O₂ is released.

Understanding the Question

The question asks the candidate to suggest a range of pO₂ values typical of respiring tissues, and to back up the suggestion with what Fig. 6.2 shows. Two marks: one for the range, one for the graphical evidence.

Approach

• Identify the curve matching tissue conditions. The Bohr shift (high tissue pCO₂) means the relevant curve is shifted to the right; at the tissues, pCO₂ is higher than the alveolar 5.3 kPa, so values around 5.3 kPa and above (e.g. 10.7 kPa) are plausible.
• Find the steep region of that curve — this is where small pO₂ changes give large changes in saturation, i.e. where unloading is efficient.
• Read off the pO₂ range covered by that steep region.

Step-by-Step Reasoning

• The curves are steepest in the middle, roughly between pO₂ = 1 kPa and pO₂ = 6 kPa, where saturation falls rapidly from about 80% down towards 20% or lower.
• This steep fall is precisely what is needed at respiring tissues: a small pO₂ decrease releases a lot of O₂.
• So the suggested range is 1–6 kPa.
• The graph-based evidence: in this pO₂ range the curve is steep — a fall in pO₂ produces a large decrease in percentage saturation, so haemoglobin releases its bound oxygen.

Key Takeaways

• Respiring tissues have low pO₂; the steep portion of the dissociation curve is the unloading zone.
• The shape of the curve is the "evidence": a steep drop in saturation over a small pO₂ change.
• Always anchor graph-based evidence in a specific feature of the curve.

Common Mistakes

• Giving a single figure (e.g. "4 kPa") — the mark scheme rejects a single number; a range is required.
• Just stating "saturation decreases" without referencing the graph's steep region.
• Choosing a pO₂ in the upper plateau (e.g. 12 kPa) where the curve is flat — that would not be efficient for unloading.

Things to Be Careful About

• "A single figure" is rejected; you must give a range.
• The evidence mark requires linking the steep section of the curve to oxygen release.

Background Concept

The position of the haemoglobin oxygen dissociation curve depends on pCO₂ (and pH and temperature). Increasing pCO₂ lowers the affinity of haemoglobin for oxygen, so that for any given pO₂, less oxygen is bound. On a graph this appears as the curve shifting to the right (sometimes described as "the curve moves right and down").

The mechanism is the Bohr effect: CO₂ produced by respiring tissues combines with water to form carbonic acid, which dissociates to release H⁺. The H⁺ ions bind to haemoglobin and stabilise its deoxygenated (T) form, reducing O₂ affinity. More O₂ is therefore released exactly where it is needed.

Understanding the Question

The question requires a description of the effect visible in Fig. 6.2. The mark scheme awards up to 3 marks for: (1) the curve shifts to the right, (2) saturation decreases as pCO₂ increases, (3) supporting comparative data quote with both units (kPa and %). An additional point can be earned for commenting on the shape/position difference between curves.

Approach

• State the direction of the shift.
• State the change in saturation at a given pO₂.
• Read off two comparable data points (one for low pCO₂, one for high pCO₂ at the same pO₂) and quote the figures with the correct units.
• Comment on where the curves differ most.

Step-by-Step Reasoning

• Direction: comparing the 2.7 kPa curve with the 10.7 kPa curve, the latter lies to the right of the former. The whole curve shifts to the right as pCO₂ rises.
• Saturation: at the same pO₂ (e.g. 4 kPa), a higher pCO₂ gives a lower % saturation.
• Comparative data (use the same pO₂, two different pCO₂ values): $$\text{At } p\text{O}_2 = 4\,\text{kPa}:$$ $$\text{at } p\text{CO}_2 = 2.7\,\text{kPa, saturation} \approx 75\%$$ $$\text{at } p\text{CO}_2 = 10.7\,\text{kPa, saturation} \approx 42\%$$
• Pattern: the biggest gap between curves is in the middle of the pO₂ range (around 2–6 kPa). At very high pO₂ (>10 kPa) the curves converge near 100% saturation.

Key Takeaways

• Higher pCO₂ → rightward shift of the curve → lower O₂ affinity.
• Always quote comparative data with units (kPa and %).
• The Bohr effect is most useful in the steep part of the curve.

Common Mistakes

• "The curve shifts downwards" — ignored by the mark scheme; the shift is to the right.
• Saying "saturation decreases" without specifying at a given pO₂ — the curves all reach about 100% at high pO₂.
• Quoting data without units, or using wrong units (e.g. mmHg instead of kPa).
• Failing to compare: quoting only one curve's data.

Things to Be Careful About

• Both units — kPa (for pO₂) and % (for saturation) — must each appear at least once in the answer.
• The shift is described as "to the right" (not "downwards").
• A comparative quote must give two pCO₂ values at the same pO₂.

Background Concept

The phenomenon in which increasing pCO₂ (or decreasing pH) reduces the affinity of haemoglobin for oxygen is called the Bohr shift (or Bohr effect), after the Danish physiologist Christian Bohr who described it in 1904. It is one of several allosteric modulators of haemoglobin, alongside 2,3-BPG and temperature.

Mechanistically, H⁺ (from CO₂ + H₂O → H₂CO₃ → HCO₃⁻ + H⁺) and CO₂ itself bind to specific sites on the globin chains, stabilising the low-affinity T-state of haemoglobin and promoting O₂ release.

Understanding the Question

The candidate is asked simply to name the effect — the one already described in part (iii).

Approach

Identify the standard name for the pCO₂-dependent shift of the oxygen dissociation curve.

Step-by-Step Reasoning

• The effect is universally known as the Bohr shift (or Bohr effect).
• The mark scheme accepts either form.
• "Bohr effect" and "Bohr shift" are both standard textbook terms.

Key Takeaways

• Bohr shift = rightward shift of the ODC with rising pCO₂ (or falling pH).
• The effect is physiologically important because it occurs precisely where CO₂ is high — at respiring tissues.

Common Mistakes

• Writing "the Haldane effect" — that is the related but distinct effect: deoxygenated haemoglobin carries more CO₂.
• Writing "cooperative binding" — describes the sigmoidal shape, not the pCO₂ effect.

Things to Be Careful About

• Either "Bohr shift" or "Bohr effect" is accepted; pick one and write it clearly.

Background Concept

The Bohr shift is a beautifully tuned feedback system. Respiring tissues produce CO₂, which raises local pCO₂ and lowers pH. Both changes lower haemoglobin's O₂ affinity, so more O₂ is unloaded precisely at the cells that are consuming it most rapidly. This matches oxygen supply to oxygen demand without any neural or hormonal control.

The advantage is greatest during exercise: skeletal muscle respires faster, produces more CO₂, and so triggers more O₂ release — exactly when it is needed.

Understanding the Question

The question asks for a two-mark explanation of why the Bohr effect is useful to a mammal. The mark scheme offers any two from: (a) more O₂ is released/supplied; (b) the O₂ is supplied to tissues where demand is high; (c) aerobic respiration is maintained.

Approach

• State the direct effect: more O₂ released.
• State the purpose: to meet tissue demand.
• State the consequence: aerobic respiration continues.

Step-by-Step Reasoning

• The rightward shift of the curve at high pCO₂ means that at the pO₂ of respiring tissues, haemoglobin gives up more of its bound oxygen.
• This extra O₂ is delivered to the cells that produced the CO₂ — so the cells that respire fastest receive most O₂.
• More O₂ supports more oxidative phosphorylation, so aerobic respiration (and ATP production) is maintained even when demand rises (e.g. during exercise).

Key Takeaways

• Bohr shift = automatic matching of O₂ supply to tissue demand.
• The advantage is physiological: maintains aerobic ATP production under variable workload.
• The mechanism does not require any active regulation; it is a property of the haemoglobin molecule itself.

Common Mistakes

• Vague answers like "it is helpful" — the mark scheme requires a specific point about O₂ release and tissue demand.
• "Haemoglobin releases oxygen more readily" alone — this is ignored by the mark scheme (it is just a restatement of the effect, not the advantage).
• "Quicker / faster release" — ignored; the mark is for more O₂ released, not the speed.
• Confusing the Bohr shift with the Haldane effect (which is about CO₂ carriage, not O₂ release).

Things to Be Careful About

• Frame the answer as an advantage to the mammal (physiology) — not just as a description of the curve.
• Avoid the ignored phrase "haemoglobin releases oxygen more readily" — make the point about supply to respiring tissues and meeting demand.