Biology 9700/42 — May/June 2024

Background Concept

Phenotype is the observable characteristic of an organism (its morphology, physiology, biochemistry or behaviour). It is not determined by genotype alone; the environment also acts on the developing and adult organism to produce the final phenotype. For this reason, phenotypic variation in a population arises from three distinct sources, and the mark scheme tests whether the candidate can name each one and supply a credible example.

The three sources are:

1. Genetic (genotypic) variation – different alleles or different combinations of alleles at the same loci. The ultimate source of new alleles is mutation (gene mutations and chromosomal mutations). During meiosis, crossing over between non-sister chromatids of homologous chromosomes and the independent assortment of homologous chromosomes (and of maternal/paternal chromosome pairs at metaphase I) reshuffle existing alleles into new combinations. Random fertilisation and random mating then determine which combinations actually come together in a zygote.
2. Environmental variation – abiotic and biotic factors in the organism's surroundings act on individuals of the same genotype to produce different phenotypes. Common examples the mark scheme credits include climate, disease, food availability, soil pH and selection pressures.
3. Gene–environment interaction – many genes are expressed only under particular environmental conditions. The same genotype can produce different phenotypes in different environments (e.g. the Agouti mouse coat-colour gene produces yellow fur only on a high-vitamin diet), and the same environment can be read differently by different genotypes. The mark scheme explicitly credits the wording "gene (expression) is modified by environment".

Understanding the Question

Part (a) is a 3-mark "describe" item worth three independent credit points. The candidate must name one cause of variation and supply an example for each. The mark scheme allows either a category label (e.g. "genetic") or a specific example (e.g. "mutation") to score the point, but a generic statement such as "variation is caused by genes" without an example does not unlock the mark for the example strand.

Approach

Treat the answer as three short, parallel statements: genetic cause + example, environmental cause + example, interaction of the two + example. Use precise CIE terminology — "alleles", "mutation", "crossing over", "independent assortment" — rather than colloquial substitutes ("genes mix up").

Step-by-Step Reasoning

• Mark 1 (genetic): Name genetic variation and give one process that creates it. Mutation is the source of all new alleles, so it is the strongest example; crossing over, independent assortment, random fertilisation and random mating are also credited.
• Mark 2 (environmental): Name environmental variation and give one factor. Any of climate, disease, food availability, soil pH or selection pressure is credited. "Diet" alone is usually insufficient without the example being clearly environmental.
• Mark 3 (interaction): State that phenotype results from genes and environment acting together. The mark scheme accepts either "combination / interaction, of genes and environment" or the example "gene (expression) is modified by environment".

Key Takeaways

• Phenotype = genotype + environment + their interaction.
• Mutation is the only source of new alleles; the other genetic mechanisms only reshuffle existing alleles.
• Continuous variation is usually polygenic and strongly affected by the environment; discontinuous variation is usually controlled by one or few genes with little environmental influence.

Common Mistakes

• Saying "genes" without naming alleles, genotypes or a process — this loses the example mark.
• Confusing discontinuous and continuous variation with their causes; the question asks about causes, not patterns.
• Omitting the gene–environment interaction. Many candidates list only genetic and environmental causes and forget that the two interact, so they score only 2 of the 3 marks.
• Writing "natural selection causes variation" — natural selection acts on variation; it does not create it.

Things to Be Careful About

• The mark scheme credits either a category label or an example for each point, not both required.
• "Selection pressure" is listed as an environmental example; do not write "selection causes variation".
• For the interaction mark, the wording "genes and environment work together" is too vague to score — explicitly state that gene expression is modified by the environment, or use the mark scheme's wording.

Background Concept

A gene is a section of DNA occupying a fixed position (locus) on a chromosome. Within a population, different versions of the same gene are called alleles. Any individual diploid organism carries only two alleles at a locus (one on each chromosome of a homologous pair), but a population as a whole can carry many different alleles at that single locus — this is what is meant by multiple alleles (sometimes called an allelic series).

A classic example is the human ABO blood group gene, which has three alleles ($I^A$, $I^B$ and $I^O$); the tortoise beetle gene here is even more extreme, with four alleles at the same locus.

Understanding the Question

The question simply asks the candidate to justify the use of the term "multiple alleles" for the colour-pattern gene in this beetle. The information required is given in the stem: gene $L$ has four different alleles, $L^V$, $L^T$, $L^R$ and $L^r$.

The command word is "explain why… can be described as" — so the answer must make the link between the fact (four alleles) and the term (multiple alleles).

Approach

State that the gene has more than two alleles and connect that to the term "multiple alleles". A single sentence is enough for this 1-mark part.

Step-by-Step Reasoning

• Look at the information provided about gene $L$: four different alleles are listed.
• Recall the definition: multiple alleles describes a gene that has more than two different alleles within a population.
• Connect the two: 4 > 2, so this is a multiple-allele system.

Key Takeaways

• Multiple alleles = more than two alleles at one locus in the population.
• An individual still only has two of these alleles (one on each chromosome of the homologous pair).
• Each different combination of two alleles gives a (potentially) different phenotype.

Common Mistakes

• Saying the beetle has four alleles — the individual beetle only has two; it is the gene in the population that has four alleles.
• Confusing "multiple alleles" with "multiple genes" — they are not the same thing.

Things to Be Careful About

Just stating "there are four alleles" alone may not be enough on its own; link it explicitly to the idea that more than two alleles means multiple alleles. The mark scheme accepts "a gene which has more than two alleles".

Background Concept

A monohybrid cross follows the inheritance of a single gene. The expected offspring of a cross are predicted by:

1. identifying the parental genotypes from their phenotypes;
2. writing out the gametes each parent can produce (one allele per gamete for a diploid);
3. combining the gametes in a Punnett square to give the offspring genotypes;
4. translating each offspring genotype into a phenotype using a key such as Table 3.1.

Where multiple alleles exist, each individual still only carries two of them, so the cross is worked out exactly as for any monohybrid cross.

Understanding the Question

A dfm-b beetle is crossed with another dfm-b beetle. Table 3.1 shows that the dfm-b phenotype has only one possible genotype: $L^T L^R$. The question asks for a complete genetic diagram, with parents, gametes, offspring genotypes, offspring phenotypes and a phenotypic ratio — six labelled sections, each worth a mark in the mark scheme.

Approach

• Read off the only genotype for dfm-b from Table 3.1.
• Show both parents as $L^T L^R$.
• A heterozygous $L^T L^R$ parent produces two kinds of gamete, $L^T$ and $L^R$.
• Combine the gametes in a 2 × 2 Punnett square.
• Read off offspring genotypes and convert each to a phenotype using Table 3.1.
• State the ratio clearly.

Step-by-Step Reasoning

1. Parental genotypes. Table 3.1 lists dfm-b with a single genotype: $L^T L^R$. Therefore both parents must be $L^T L^R$.
2. Gametes. Each $L^T L^R$ parent can pass on either $L^T$ or $L^R$, so each parent produces two kinds of gamete in equal proportions: $L^T$ and $L^R$.
3. Punnett square. A standard 2 × 2 grid gives four boxes:
   • $L^T L^T$ (top-left)
   • $L^T L^R$ (top-right)
   • $L^T L^R$ (bottom-left)
   • $L^R L^R$ (bottom-right)
4. Offspring genotypes. Genotype ratio 1 $L^T L^T$ : 2 $L^T L^R$ : 1 $L^R L^R$.
5. Offspring phenotypes (read off Table 3.1):
   • $L^T L^T$ → dfm-a (darien f. militaris-a)
   • $L^T L^R$ → dfm-b
   • $L^R L^R$ → rufipennis
6. Phenotype ratio: 1 dfm-a : 2 dfm-b : 1 rufipennis.

Key Takeaways

• A phenotype produced by a unique genotype in a multiple-allele system makes deducing the cross straightforward.
• The cross is still a standard monohybrid cross (1:2:1 genotypic and phenotypic in this case).
• A Punnett square must be drawn clearly with gametes labelled and genotypes linked to phenotypes.

Common Mistakes

• Assuming dfm-b must be a heterozygote of two different alleles from outside the table — Table 3.1 explicitly gives the only genotype, $L^T L^R$.
• Forgetting to link each offspring genotype to its phenotype in the diagram (the mark scheme requires the phenotypes labelled next to the genotypes).
• Writing the ratio in the wrong direction (phenotype order does not matter as long as it is consistent).
• Calling the offspring genotype $L^R L^T$ instead of $L^T L^R$ — the order does not matter for a heterozygote, but be consistent with the allele order used in the question.

Things to Be Careful About

• The Punnett square must show all four boxes filled in.
• Gametes should be written as single alleles (one allele per gamete, e.g. $L^T$ and $L^R$).
• Offspring phenotype names should match exactly those used in Table 3.1: dfm-a, dfm-b, rufipennis.

Background Concept

When two alleles are both fully expressed in the heterozygote (so the heterozygote shows features of both homozygotes) they are said to be codominant. This is different from:

• Dominant/recessive — the dominant allele is fully expressed and the recessive one is hidden in the heterozygote (e.g. $L^V$ over $L^r$, where $L^V L^r$ is veraguensis, indistinguishable from $L^V L^V$).
• Incomplete dominance — the heterozygote has an intermediate phenotype (not the case here, since the heterozygote clearly shows both parental patterns, not a blend).

Where a gene has several alleles they can be arranged in a dominance hierarchy in which some alleles are dominant over others, while two particular alleles may be codominant with each other.

Understanding the Question

The question asks two things:

1. Identify the codominant pair from the data in Table 3.1.
2. List the dominance hierarchy with alleles from most to least dominant.

The candidate must use Table 3.1 to compare heterozygotes with homozygotes.

Approach

• Look at every heterozygote in Table 3.1.
• For each heterozygote, decide whether the phenotype shows only one of the two homozygote patterns (dominance/recessiveness) or shows both of them (codominance).
• Build the hierarchy from those observations.

Step-by-Step Reasoning

Codominance check. Compare the heterozygote $L^T L^R$ (dfm-b) with the two homozygotes $L^T L^T$ (dfm-a, all black and red striped) and $L^R L^R$ (rufipennis, black pronotum and red elytra):

• The pronotum of dfm-b is black — exactly as in rufipennis, the $L^R L^R$ pattern (i.e. the $L^R$ contribution is visible).
• The elytra of dfm-b are black and red striped — exactly as in dfm-a, the $L^T L^T$ pattern (i.e. the $L^T$ contribution is visible).
• Both parental phenotypes are clearly visible side-by-side, so $L^T$ and $L^R$ are codominant.

Dominance hierarchy. Look at the remaining heterozygotes:

• $L^V L^R$ → rufipennis (the rufipennis $L^R L^R$ phenotype). So $L^R$ is dominant over $L^V$.
• $L^V L^T$ → dfm-a (the $L^T L^T$ phenotype). So $L^T$ is dominant over $L^V$.
• $L^T L^r$ → dfm-a (the $L^T L^T$ phenotype). So $L^T$ is dominant over $L^r$.
• $L^R L^r$ → rufipennis (the $L^R L^R$ phenotype). So $L^R$ is dominant over $L^r$.
• $L^V L^r$ → veraguensis (the $L^V L^V$ phenotype). So $L^V$ is dominant over $L^r$.

Combining: $L^T$ and $L^R$ are at the top (codominant with each other and each dominant over $L^V$); $L^V$ is dominant over $L^r$, which is the least dominant.

Key Takeaways

• Codominance is recognised when the heterozygote shows BOTH homozygote phenotypes at once (in different parts of the body, here the pronotum vs. the elytra).
• A dominance hierarchy arranges multiple alleles from most to least dominant; codominant alleles sit at the same level of the hierarchy.
• For each heterozygous pair, compare the phenotype with both corresponding homozygotes to decide the relationship.

Common Mistakes

• Confusing codominance with incomplete dominance. Incomplete dominance would give a blend (e.g. pink from red × white), but here the patterns are spatially separate (black pronotum from $L^R$, striped elytra from $L^T$) — both parental patterns are visible intact.
• Saying that $L^T$ and $L^R$ are simply both "dominant" and missing the codominance — the mark scheme specifically requires the recognition that they are codominant.
• Putting $L^V$ above $L^T$ or $L^R$ in the hierarchy — Table 3.1 shows $L^V L^T$ and $L^V L^R$ heterozygotes both look like the $L^T L^T$ and $L^R L^R$ homozygotes respectively, confirming that $L^V$ is recessive to both.

Things to Be Careful About

• Read the table carefully: dfm-b has genotype $L^T L^R$ — there is no other option, and the pattern in dfm-b is not a blend but a combination of the two homozygote patterns, which is the hallmark of codominance.
• When writing the hierarchy, the symbol used is often "=" or putting both alleles at the top (e.g. "$L^T = L^R > L^V > L^r$"). The mark scheme accepts "$L^T$ and $L^R > L^V > L^r$".

Background Concept

The central dogma of molecular biology describes how genetic information flows from DNA to protein to phenotype. A gene is a specific DNA sequence that codes for a protein (or a functional RNA). The protein then performs a function in the cell or body that contributes to the organism's observable characteristics — its phenotype.

Haemophilia is a classic Cambridge example used to illustrate this gene → protein → phenotype relationship. It results from mutations in genes coding for blood-clotting factors. Haemophilia A (the more common form) is caused by mutations in the F8 gene, which sits on the X chromosome and codes for clotting factor VIII. Haemophilia B involves F9 and factor IX.

Because F8 is on the X chromosome, inheritance is sex-linked. Males (XY) have only one X, so a single recessive allele produces the affected phenotype — males are hemizygous for X-linked genes. Females (XX) need two recessive alleles to be affected; heterozygous females are phenotypically normal carriers.

Understanding the Question

The question asks you to explain the gene → protein → phenotype relationship, using haemophilia as a worked example. The marks are spread across:

1. The general principle that a gene codes for a protein that determines the phenotype.
2. Naming the specific gene (F8) and the protein (factor VIII).
3. Stating that the gene is on the X chromosome (sex-linked).
4. Stating that the recessive allele codes for non-functioning / less / no factor VIII.
5. Explaining the resulting phenotype (blood does not clot / excessive bleeding).
6. Linking at least some genotypes to phenotypes.

Approach

Lay out the causal chain first — gene → protein → protein's function lost → phenotype — then add concrete genotype–phenotype examples so the sex-linked nature of the inheritance is obvious. Use italics for the gene name (F8) and roman text for the protein (factor VIII) — Cambridge expects this typographical distinction.

Step-by-Step Reasoning

• General principle: A gene codes for a protein, and the protein's activity in the cell or body determines the phenotype.
• Specific gene and protein: The F8 gene is located on the X chromosome (sex-linked) and codes for the protein factor VIII, one of the proteins in the blood-clotting cascade.
• Recessive allele effect: The recessive allele codes for a non-functioning version of factor VIII. The mark scheme accepts "non-functioning", "less", or "no" factor VIII as alternatives — any of these earns the mark.
• Phenotype: Factor VIII normally helps activate factor X in the clotting cascade, which eventually leads to fibrinogen being converted to fibrin and a blood clot forming. Without functional factor VIII, the clot forms too slowly, so bleeding after injury is prolonged. This is haemophilia.
• Genotype–phenotype links:
  • $X^F X^F$ — homozygous dominant female, unaffected.
  • $X^F X^f$ — heterozygous female, unaffected carrier (because F is dominant over f).
  • $X^f X^f$ — homozygous recessive female, affected.
  • $X^F Y$ — male carrying the dominant allele, unaffected.
  • $X^f Y$ — male carrying the recessive allele, affected.
• Because a single recessive allele on the X chromosome is enough to produce the phenotype in males, haemophilia is much more common in males than in females.

Key Takeaways

• The gene → protein → phenotype chain: a gene codes for a protein, and the protein's function in the body determines the phenotype.
• Haemophilia A is caused by recessive mutations in F8, the X-linked gene coding for factor VIII.
• The recessive allele produces non-functioning factor VIII, impairing blood clotting.
• Sex-linked inheritance means recessive X-linked alleles express in males (hemizygous) more readily than in females.

Common Mistakes

• Vague phrasing such as "a gene causes haemophilia" without specifying F8 and factor VIII — this misses the protein link in the middle of the chain.
• Saying F8 is on an autosome — it is on the X chromosome.
• Saying haemophilia is dominant — it is recessive.
• Failing to link genotype to phenotype with concrete examples for both sexes.
• Confusing the gene (italic) with the protein (roman) — Cambridge expects F8 and factor VIII.

Things to Be Careful About

• Use italic for gene names (F8) and roman for protein names (factor VIII).
• Genotypes for X-linked genes in males must include the Y — e.g. $X^F Y$, not just $X^F$.
• A heterozygous female is a carrier — phenotypically normal but able to pass the recessive allele to her offspring.
• The mark scheme accepts "no", "less", or "non-functioning" factor VIII as alternatives — any of these earns the mark, but the protein must be named.

Background Concept

Plant height is largely controlled by the hormone gibberellin, which promotes cell elongation in stem internodes. Gibberellins are a family of related plant hormones; the form most active in stem elongation is GA$_1$. Active GA$_1$ is synthesised from precursors by a multi-step pathway; the final step is catalysed by the enzyme GA 3β-hydroxylase (also called GA3-oxidase), which converts inactive gibberellin precursors into active GA$_1$.

In pea plants (Pisum sativum), the gene coding for this final enzyme is the Le gene. The dominant allele Le codes for a functional version of the enzyme; the recessive allele le codes for a non-functional version. In fact, the le allele carries a single base substitution that replaces a threonine with an alanine in the enzyme, inactivating it.

Understanding the Question

The question asks you to connect the lele genotype in pea plants to the dwarf phenotype via the gene product and its effect on gibberellin and cell elongation. The causal chain is:

lele → non-functional GA 3β-hydroxylase → no active GA$_1$ → no cell elongation → dwarf plant.

Approach

Walk through the chain one step at a time, making every link explicit. Mention the specific enzyme and the specific active form of gibberellin (GA$_1$) so the answer is precise rather than vague. This question is essentially the gene → protein → phenotype chain applied to an enzyme in a hormone-biosynthesis pathway.

Step-by-Step Reasoning

• Genotype: The plant is lele — homozygous recessive, carrying two copies of the recessive le allele and no dominant Le allele.
• Gene product: The le allele codes for a non-functional version of the enzyme GA 3β-hydroxylase, the final enzyme in the gibberellin biosynthesis pathway.
• Gibberellin: Because the enzyme does not work, the inactive precursor cannot be converted into active GA$_1$. The plant therefore has no (or much less) active gibberellin.
• Cellular effect: Active GA$_1$ normally stimulates cell elongation in stem internodes. Without it, cells in the internodes do not elongate, so the internodes remain short.
• Phenotype: A pea plant with short internodes has a short stem overall — the dwarf phenotype.

This is essentially the same gene → protein → phenotype chain as in part (a), but the "protein" here is an enzyme in a metabolic pathway rather than a clotting factor.

Key Takeaways

• The Le/le gene in pea plants codes for GA 3β-hydroxylase, the final enzyme in gibberellin synthesis.
• The recessive le allele codes for a non-functional enzyme, so no active GA$_1$ is made.
• Without active GA$_1$, stem cells do not elongate and the plant is dwarf.
• This is a useful example of how a gene can affect a phenotype indirectly — through its role in synthesising a hormone — rather than directly through a structural protein.

Common Mistakes

• Saying lele "does not code for anything" — it codes for a protein, but that protein is non-functional.
• Saying the le allele "stops gibberellin being made" without specifying the enzyme and the active form GA$_1$.
• Confusing gibberellin with auxin — they are different plant hormones with different roles.
• Saying the dwarf plant has shorter cells because the cells "do not divide" — gibberellin mainly affects cell elongation, not cell division.
• Vague statements like "the le allele prevents growth" — the answer must go through the enzyme and the active gibberellin, not jump straight from allele to dwarf.

Things to Be Careful About

• Use italic for gene names (Le, le) and roman for proteins and hormones (GA 3β-hydroxylase, GA$_1$).
• It is the active gibberellin GA$_1$ that is missing, not gibberellin in general.
• The phenotype is "short stems" or "dwarf" — be specific about which part of the plant is affected.
• The mark scheme credits the specific molecular change (alanine instead of threonine in the enzyme) as an extra point if mentioned, but this is not required.

Background Concept

Gene expression in eukaryotes is regulated at many levels, but the first decision is usually whether transcription occurs. For RNA polymerase to begin transcribing a gene, it must bind to the gene's promoter. Often, RNA polymerase can only bind efficiently with the help of activator proteins called transcription factors, which bind to specific DNA sequences near the promoter.

In plants, growth-related genes (such as those involved in cell elongation) are activated by a family of transcription factors called PIFs (Phytochrome-Interacting Factors). PIFs bind to the promoters of these genes and stimulate RNA polymerase to begin transcription.

DELLA proteins are a family of regulatory proteins that act as repressors of these growth genes. They do this by binding to PIFs and physically preventing PIFs from binding to DNA. When DELLA is bound to PIF, the growth genes cannot be transcribed.

Understanding the Question

The question asks you to explain the molecular mechanism by which DELLA proteins repress gene expression. The chain of events is:

DELLA binds PIF → PIF cannot bind DNA → RNA polymerase cannot bind → no transcription.

The mark scheme rewards each step in the chain, so the answer must include all four steps.

Approach

Walk through the repression step-by-step. The key insight — and the most common place students slip up — is that DELLA does NOT bind to DNA directly. It acts by sequestering the transcription factor (PIF) that would otherwise activate the gene. From there, the rest of the chain (PIF off DNA → RNA polymerase off promoter → no transcription) follows.

Step-by-Step Reasoning

• In the absence of active gibberellin, DELLA proteins are stable in the cell and bind to PIF transcription factors.
• With DELLA attached, PIF cannot bind to the promoter region of growth-related genes.
• Without PIF on the promoter, RNA polymerase cannot bind to the promoter either — PIF is needed to help recruit or stabilise RNA polymerase at the promoter.
• Therefore transcription does not occur, and no mRNA is made from the growth-related genes.
• The growth-related genes are not expressed, so the plant remains dwarf.

For completeness, when active gibberellin is present, it binds to its receptor GID1. The DELLA–GID1–gibberellin complex is then tagged with ubiquitin and degraded by the proteasome. This releases PIF to bind DNA and activate transcription, lifting the repression — but this lifting mechanism is not what the question asks about.

Key Takeaways

• DELLA proteins repress transcription by binding to PIF transcription factors, not by binding DNA themselves.
• When PIF is sequestered by DELLA, PIF cannot bind the promoter.
• Without PIF on the promoter, RNA polymerase cannot bind and transcription does not occur.
• This is a classic example of a transcriptional repressor working by inhibiting an activator.

Common Mistakes

• Saying DELLA binds directly to DNA — it does not; it binds the transcription factor.
• Saying DELLA "stops translation" — DELLA affects transcription (mRNA synthesis), not translation (protein synthesis). This is one of the most common errors on this question.
• Confusing DELLA with the gibberellin receptor (GID1) or with gibberellin itself.
• Skipping the RNA polymerase step — the mark scheme specifically credits "RNA polymerase cannot bind to promoter", so this step must appear in the answer.

Things to Be Careful About

• Transcription = making mRNA from DNA. Translation = making protein from mRNA. Do not confuse them.
• The order of events matters: PIF is blocked from binding DNA BEFORE RNA polymerase can bind.
• DELLA acts in the absence of active gibberellin. When gibberellin is present, DELLA is degraded and repression is lifted — so DELLA repression and gibberellin activation are two sides of the same control switch.

Background Concept

BRCA1 codes for a tumour-suppressor protein. Tumour-suppressor proteins normally stop a cell becoming cancerous in two ways: by repairing damage to its DNA, and by triggering apoptosis (programmed cell death) when the damage is too severe to repair. A mutation in BRCA1 is therefore a loss-of-function mutation — the protective protein either is not made or does not work. Because the cell can no longer police its own DNA, further mutations can build up unchecked, and a cell that should have died can keep dividing. The end result is uncontrolled mitosis in breast tissue and (because the gene is also expressed elsewhere) an increased risk of ovarian cancer.

Proteins only work when they are folded into a specific tertiary shape, which is held together by hydrogen bonds, ionic bonds, disulfide bridges and hydrophobic interactions between specific R groups. The shape is determined by the order of amino acids (the primary structure), which is set by the base sequence of the gene. A single-base change (a point mutation) can swap one amino acid for another, and if that amino acid is in a key position the protein can no longer fold correctly, losing its function.

Understanding the Question

The command word is "state … and explain". The candidate must (i) give a single clinical reason a doctor would offer for testing, and (ii) follow the cause-and-effect chain from the mutated gene to cancer. The mark scheme splits this into one mark for the reason and up to three marks drawn from a list of six creditable points in the explanation.

Approach

For the reason, the most common and accepted answer is a family history of breast or ovarian cancer, because the mutation is heritable (autosomal dominant) and the cancers cluster in families. For the explanation, walk down the chain in this order so that each point follows logically from the one before:

1. The mutation changes the base sequence → primary structure of the protein.
2. The wrong R groups disrupt folding → tertiary (3-D) shape changes.
3. The protein becomes non-functional.
4. Without a working tumour suppressor, DNA damage is not repaired and damaged cells are not killed.
5. Mutations accumulate over time.
6. Some of those further mutations affect proto-oncogenes / tumour-suppressor genes controlling the cell cycle, so breast cells undergo uncontrolled mitosis → cancer.

Step-by-Step Reasoning

Mark 1 — reason for testing. BRCA1 mutations are heritable (autosomal dominant). If a close relative — mother, sister, aunt — has had breast or ovarian cancer, the patient's own risk of carrying the mutation is increased, so a positive test would change the screening and preventive options offered.

Marks 2–4 — explanation. Pick any three from the list:

• "Change in the primary structure" — this is the direct effect of the base change; a codon now codes for a different amino acid.
• "Change in the folding / 3-D / tertiary structure" — the new R groups cannot form the original bonds, so the protein misfolds.
• "Non-functional protein" — the consequence of misfolding; this is the key biological point.
• "DNA is not repaired / cells with damaged DNA do not die" — the two cellular jobs of BRCA1 are lost.
• "Mutations accumulate" — because there is no DNA repair and no apoptosis of badly damaged cells.
• "Uncontrollable cell division / mitosis of breast cells" — the proximate cause of a tumour.

A short narrative that joins three of these points will earn the marks. There is no need to write all six; any three accepted points score the three explanation marks.

Key Takeaways

• BRCA1 mutations are loss-of-function mutations in a tumour-suppressor gene; loss of function removes DNA-repair and apoptosis checkpoints.
• The clinical reason to test is a relevant family history, because the mutation is inherited in an autosomal-dominant pattern.
• A protein's function depends on its tertiary structure, which is set by its primary structure, which is set by the gene's base sequence — so a point mutation can abolish function.

Common Mistakes

• Vague answers such as "because it runs in the family" (without naming breast / ovarian cancer).
• Jumping straight to "mutations accumulate" without explaining why they accumulate (because the tumour suppressor is non-functional).
• Saying "the gene causes cancer" rather than the protein, or saying "the cell mutates" rather than the DNA mutates.
• Confusing the role of BRCA1 with that of an oncogene — BRCA1 is a tumour suppressor, not a cancer-promoting gene.

Things to Be Careful About

• The mark scheme is explicit that "change in primary structure" and "change in tertiary structure" are two separate marks, so be careful to state both if you need both.
• "DNA is not repaired" is the first cellular consequence; "mutations accumulate" is the second. Mark scheme awards marks for either or both.
• The end of the chain — uncontrolled mitosis — must refer specifically to breast (or ovarian) cells, because the gene is only expressed in those tissues and the cancer risk is tissue-specific.

Background Concept

Polymerase chain reaction (PCR) is a way of making many millions of copies of a specific DNA sequence in vitro. The reaction needs: a DNA template, two short primers that flank the region of interest, a thermostable DNA polymerase (Taq polymerase), and a supply of the four deoxyribonucleotide triphosphates (dNTPs). Each cycle of denaturation, annealing and extension doubles the number of target DNA molecules, so 30 cycles produce approximately $2^{30} \approx 10^9$ copies.

DNA sequencing, by contrast, needs a relatively large amount of pure, single-stranded DNA of known sequence at the region of interest. A clinical sample (a cheek swab, a blood sample) yields only nanograms of DNA, and the BRCA1 gene makes up only a tiny fraction of that.

Understanding the Question

The command word is "suggest why", which means the answer should explain the practical reason PCR is run before sequencing in this test. The mark scheme gives two complementary creditable points: the size of the starting sample, and the purpose of the step (amplification).

Approach

State that the clinical sample is small, then state that PCR makes many copies of the target sequence. That is the whole logic — the test cannot proceed without enough DNA to read.

Step-by-Step Reasoning

Mark 1. "Only a small amount of DNA is available" — the mark scheme explicitly accepts "small amount of DNA" or, by alternative wording (A), "gene". A blood sample or saliva sample contains only a few micrograms of total DNA, of which BRCA1 is a tiny fraction.

Mark 2. "Amplify, DNA / (BRCA1) gene" — the function of PCR. After amplification there is enough identical DNA for the sequencer to read the order of bases, and therefore to spot any mutation.

Key Takeaways

• PCR's role in diagnostic testing is to convert a tiny clinical sample into a usable quantity of a specific sequence.
• Without PCR, the DNA available would be too little for a sequencer to read reliably.

Common Mistakes

• Writing vaguely that PCR "makes more DNA" without explaining why more DNA is needed (i.e. the starting amount is too small to sequence).
• Describing the stages of PCR (denaturation, annealing, extension) — the question asks why it is used, not how it works.
• Saying PCR "finds the mutation" — it does not; sequencing (or another detection method) does. PCR only copies the DNA so the mutation can be detected.

Things to Be Careful About

• The mark scheme will accept "amplify the gene" as well as "amplify the DNA" — both score.
• Do not describe the components of the PCR mixture in detail; the question is not about the method itself.

Background Concept

This is a population-frequency calculation of the type CIE sets regularly under the heading "mathematical requirements". Two pieces of information are given: a frequency of the mutation in the population (1 in 400 females) and a conditional probability of developing the disease, given the genotype (70 %). The two must be combined by multiplication, after first converting the "1 in 400" statement into a count of people.

Standard form expresses a number as $a \times 10^{n}$ where $1 \leq a < 10$ and $n$ is an integer. The final answer should be quoted in standard form because the population is large.

Understanding the Question

The question gives three numbers — the total female population ($3.5 \times 10^{7}$), the carrier frequency (1/400), and the probability of developing cancer given carrier status (70 %) — and asks for an estimated count in standard form, with working shown.

Approach

• Step 1: divide the total female population by 400 to estimate the number of females with at least one mutated BRCA1 allele.
• Step 2: multiply that count by 0.70 to estimate the number of those who will develop breast cancer by age 80.
• Step 3: write the result in standard form.

Step-by-Step Reasoning

Step 1 — number of females with the mutated allele.

So 87 500 females are estimated to carry a mutated BRCA1 allele.

Step 2 — 70 % of those will develop breast cancer by age 80.

Step 3 — standard form. The mantissa 6.125 already lies in $[1, 10)$, so the answer in standard form is $6.125 \times 10^{4}$. The mark scheme accepts rounding to $6.1 \times 10^{4}$ for full credit, with $61\ 250$ (i.e. not in standard form) scoring one mark, and a correct $8.75 \times 10^{4}$ or $87\ 500$ at step 1 also scoring one mark.

Key Takeaways

• "1 in 400" must be applied as a division of the total, not as a multiplication.
• Percentages are applied as a multiplication by the decimal form (70 % = 0.70).
• Standard form requires the mantissa to be in $[1, 10)$; convert by shifting the decimal point and changing the power of ten.

Common Mistakes

• Multiplying by 400 instead of dividing (treating "1 in 400" as a multiplication factor).
• Forgetting to convert 70 % to 0.70, so leaving the answer as $8.75 \times 10^{4}$.
• Writing the final answer as 61 250 without converting to standard form — this still gets one mark but loses the standard-form mark.
• Carrying too few significant figures (e.g. quoting $6 \times 10^{4}$) and losing the second mark.

Things to Be Careful About

• Show every step of the working; the mark scheme awards one mark for reaching $8.75 \times 10^{4}$ (or $87\ 500$) even if the second step is wrong, via error-carried-forward.
• The final answer must be in standard form, written as $6.1 \times 10^{4}$ (or $6.125 \times 10^{4}$ for two marks).

Background Concept

Survival curves plot the probability that a subject is still alive at each time point after a treatment. A curve that stays high (close to 1.0) means most subjects are still alive; a curve that drops quickly means many have died. Comparing two curves tells you which group is doing better after the same treatment.

A useful rule for evaluating a hypothesis from a graph: look at (1) the direction of the difference (is the hypothesised group higher or lower?), (2) the magnitude (how big is the gap?), and (3) the duration (does the difference persist for the whole follow-up?). A common weakness in many biology studies is the absence of a statistical test such as a log-rank test for survival data, which would tell you whether the difference is significant or could have arisen by chance.

Understanding the Question

Part (d) gives a hypothesis from the researchers — that the DNA-damaging drug is more effective in women with a BRCA1 mutation — and a survival graph for two cohorts. The command word is "discuss", which means the candidate should make judgements both for and against the hypothesis, supporting points with evidence read off the graph where possible. The mark scheme credits up to two points drawn from a list that includes both supporting and limiting evidence.

Approach

Read the graph and identify:

• which curve is higher at any given time;
• how the gap between the curves changes with time;
• a numerical comparison at a specific time point;
• any limitation of the data (e.g. absence of a statistical test or sample size).

Step-by-Step Reasoning

Supporting evidence:

• The BRCA1 mutation curve is above the no-mutation curve at every time point, so women with the mutation have a higher probability of survival.
• The gap between the two curves widens with time: by 180 months the mutation group is at ~0.70 while the no-mutation group has fallen to ~0.47. This pattern is exactly what the hypothesis predicts — the drug is more effective in the mutation group, so they keep living while the non-mutation group continues to die.
• The follow-up is 15 years (180 months), which is long enough for a meaningful comparison in breast cancer.

Limiting evidence:

• The graph is just a description of the two groups. Without a statistical test (e.g. a log-rank test, a chi-squared test on the numbers surviving at 180 months) and without knowing the sample sizes, the difference cannot be confirmed as significant rather than a chance result of small numbers.

Two creditable points — a comparison of the curves, a data quote, and/or a study limitation — score the two marks.

Key Takeaways

• Reading a survival curve: look at the y-values at a given x, and at how the curves diverge or converge over time.
• A numerical data quote (with units) is much stronger support than a vague "the curve is higher".
• "Discuss" questions almost always expect both supporting and limiting evidence, with the limitation often being the lack of a statistical test.

Common Mistakes

• Vague claims such as "the drug works better" without quoting a specific value from the graph.
• Forgetting the negative side of the discussion — a "discuss" answer that only supports the hypothesis cannot earn the second mark if the mark scheme expects a limitation.
• Misreading the graph — the BRCA1 mutation group is the higher curve, because the drug (which damages DNA) is more lethal to tumour cells that already have a faulty DNA-repair pathway, so their tumour cells die and the woman lives longer.
• Quoting figures without units (months).

Things to Be Careful About

• The mark scheme accepts values within a small range, so a quote such as "at 180 months, ~0.70 compared with ~0.47" scores even though the precise values on the grid are slightly different.
• The y-axis is probability of survival, not percentage — but the mark scheme accepts both, because probability 0.70 is the same as 70 %.
• "No statistical test carried out" and "sample size not stated" are alternative ways of saying the same limitation; either scores one mark.

Background Concept

A mitochondrion has a double membrane: a smooth outer membrane and a highly folded inner membrane (folded into cristae to give a large surface area). The inner membrane is unusually rich in protein and almost impermeable to most ions and small polar molecules; it is crossed only via specific transport proteins. Between the two membranes is the intermembrane space, and enclosed by the inner membrane is the matrix, where the Krebs cycle takes place.

Membrane permeability depends on the nature of the molecule. Small, non-polar molecules such as $\text{O}_2$, $\text{CO}_2$ and steroid hormones pass through membranes freely by simple diffusion, dissolving in and crossing the phospholipid bilayer directly. They do NOT need protein channels or carriers, and the cell cannot regulate their movement. By contrast, polar molecules and ions (e.g. $\text{H}^+$, glucose, $\text{Na}^+$, $\text{K}^+$) cannot cross the phospholipid bilayer unaided and must move through specific transport proteins (by facilitated diffusion or active transport).

Oxygen is required in the matrix because it is the final electron acceptor of the electron transport chain (ETC) at the inner mitochondrial membrane, where it is reduced to water. It must therefore diffuse from the cytoplasm, across both mitochondrial membranes, into the matrix.

Understanding the Question

Fig. 6.1 is a schematic cross-section of part of a mitochondrion, with proteins A, B and C (ETC components) and D (ATP synthase) embedded in the inner membrane, plus channel proteins in the outer membrane. The question asks the candidate to use arrows to show the route $\text{O}_2$ takes from the cytoplasm to the matrix. The mark scheme explicitly REJECTS arrows drawn through any protein (channel proteins, A, B, C or D), so the test is essentially: do you know that $\text{O}_2$ crosses membranes through the phospholipid bilayer and not through proteins?

Approach

Identify the two membranes $\text{O}_2$ must cross (outer and inner), remember that $\text{O}_2$ is small and non-polar, and decide which part of each membrane the arrow should pass through: through the phospholipid bilayer (the gap between the protein complexes), not through any of the labelled or unlabelled proteins shown.

Step-by-Step Reasoning

1. $\text{O}_2$ is needed in the matrix because it is the terminal electron acceptor of oxidative phosphorylation. The matrix is enclosed by the inner mitochondrial membrane; to reach it from the cytoplasm, $\text{O}_2$ must cross both the outer and the inner mitochondrial membranes.
2. Because $\text{O}_2$ is a small, non-polar molecule, it crosses biological membranes by simple diffusion directly through the phospholipid bilayer. It does not use transport proteins.
3. Therefore the arrow through the outer membrane must pass through the phospholipid layer (the lipid bilayer drawn in the figure), not through any of the channel proteins embedded in it.
4. Similarly, the arrow through the inner membrane must pass through the phospholipid layer between protein complexes A, B, C and D, not through any of those proteins.
5. The route is: cytoplasm → (through outer-membrane phospholipid layer) → intermembrane space → (through inner-membrane phospholipid layer) → mitochondrial matrix.

Key Takeaways

• The phospholipid bilayer is permeable to small, non-polar molecules ($\text{O}_2$, $\text{CO}_2$) and impermeable to ions and most polar molecules.
• The ETC and ATP synthase are embedded in the inner mitochondrial membrane and participate in oxidative phosphorylation, NOT in moving $\text{O}_2$ into the matrix.
• Mark-scheme trap: arrows drawn through the proteins shown will not earn the mark, even if the route is correct.

Common Mistakes

• Drawing arrows through the channel proteins of the outer membrane or through proteins A, B, C or D of the inner membrane. The mark scheme rejects this. The mistake shows a failure to understand that $\text{O}_2$ is small and non-polar and crosses the bilayer directly.
• Drawing only one arrow (e.g. only across the outer membrane). Both membranes must be crossed.
• Drawing the arrow ending in the intermembrane space rather than in the matrix.

Things to Be Careful About

• The mark scheme is precise: arrows must pass through the phospholipid layers, not through proteins.
• Although $\text{O}_2$ is consumed at the ETC, the figure here asks for the route taken to GET to the matrix; the candidate is not asked to show where $\text{O}_2$ is used.
• Read the wording: the final arrow tip should be in the matrix.

Background Concept

Aerobic respiration produces most of the cell's ATP in the mitochondrion. After glycolysis in the cytoplasm and the link reaction and Krebs cycle in the matrix, the reduced coenzymes NADH and FADH2 carry high-energy electrons to the inner mitochondrial membrane. There, the electron transport chain (ETC), a series of protein complexes and electron carriers embedded in the inner membrane, uses the energy from these electrons to pump protons ($\text{H}^+$) from the matrix into the intermembrane space. This creates an electrochemical gradient (a much higher $\text{H}^+$ concentration in the intermembrane space than in the matrix) — the proton-motive force.

The protons can flow back down their electrochemical gradient only through ATP synthase (also called ATP synthetase), a large stalked protein complex embedded in the inner membrane. As protons flow through ATP synthase, the potential energy they release drives the phosphorylation of ADP + Pi to form ATP. This coupling of electron transport (which builds the gradient) to ATP synthesis (which uses the gradient) is called chemiosmosis (proposed by Peter Mitchell in 1961), and it is the major source of ATP in aerobic cells. About 2.5 ATP are made per NADH and 1.5 ATP per FADH2 by this mechanism.

Understanding the Question

Part (b) is worth 7 marks. It asks two things:

1. Identify protein D and the group of proteins A, B, C (1 mark each).
2. Describe the roles of these proteins in oxidative phosphorylation (up to 5 further marks).

The mark scheme rewards the role of ETC proteins (electrons travel along the chain, energy released, $\text{H}^+$ pumped into intermembrane space, proton gradient created) and the role of ATP synthase ($\text{H}^+$ diffuse through, ATP made from ADP + Pi, term 'chemiosmosis' used). Both halves must be clearly described to earn full marks.

Approach

Name D and A/B/C first, then separate the description into two halves — what the ETC does, then what ATP synthase does. The proton gradient is the bridge between the two halves. End by naming the process 'chemiosmosis'.

Step-by-Step Reasoning

1. Identification of D: D is ATP synthase (or ATP synthetase). The mark scheme REJECTS 'ATPase' because ATPase suggests an enzyme that breaks down ATP, while ATP synthase synthesises ATP. The stalked particles seen in electron micrographs of mitochondrial inner membranes are ATP synthase.
2. Identification of A, B, C: These are the electron transport chain (ETC) — a series of electron carriers, including cytochromes, that pass electrons from one to the next.
3. Electrons travel along the ETC: Electrons from reduced NAD (NADH) and reduced FAD (FADH2) enter the chain and pass from carrier to carrier. (In real mitochondria, NADH donates electrons to complex I, FADH2 to complex II, and they converge on coenzyme Q then travel through complexes III and IV to oxygen.)
4. Energy is released: As electrons fall to lower and lower energy levels along the chain, energy is released at each step.
5. Energy used to pump $\text{H}^+$ into the intermembrane space: This energy is harnessed by the ETC complexes to actively transport $\text{H}^+$ (protons) from the matrix across the inner membrane into the intermembrane space, against the concentration gradient.
6. A proton gradient is created: The $\text{H}^+$ concentration in the intermembrane space becomes much higher than in the matrix. This electrochemical gradient stores potential energy — the proton-motive force.
7. Protons diffuse through ATP synthase (D): $\text{H}^+$ flows back down its electrochemical gradient, from the intermembrane space to the matrix, but only through ATP synthase — the inner membrane is otherwise impermeable to $\text{H}^+$.
8. ATP synthesis: As protons flow through ATP synthase, the energy released drives the combination of ADP + Pi to make ATP ($\text{ADP} + \text{P}_\text{i} \rightarrow \text{ATP}$).
9. Chemiosmosis: This whole process — electron transport building a proton gradient, and the gradient driving ATP synthesis — is called chemiosmosis.

Key Takeaways

• ATP synthase is the enzyme that MAKES ATP, not the one that breaks it down; the mark scheme rejects 'ATPase'.
• The ETC and ATP synthase are linked by the proton gradient: the ETC creates it, ATP synthase uses it.
• Chemiosmosis is the term for the coupling of electron transport, the proton gradient, and ATP synthesis.
• The inner mitochondrial membrane is otherwise impermeable to $\text{H}^+$; this is what makes the gradient possible.

Common Mistakes

• Calling D 'ATPase' rather than 'ATP synthase'. The mark scheme explicitly rejects this.
• Stating that the ETC makes ATP. It does not. ATP is made by ATP synthase, using the gradient the ETC has built.
• Saying $\text{H}^+$ is 'pumped out of the mitochondrion'. It is pumped from the matrix to the INTERMEMBRANE SPACE, not out of the organelle. The intermembrane space is still inside the mitochondrion.
• Confusing the direction of $\text{H}^+$ flow. ETC proteins pump $\text{H}^+$ OUT of the matrix; ATP synthase allows $\text{H}^+$ to flow IN to the matrix.
• Forgetting to name the process 'chemiosmosis' — this is a specific mark.
• Saying 'electrons' travel through ATP synthase. It is PROTONS ($\text{H}^+$) that flow through ATP synthase, not electrons. Electrons are already on oxygen by the time ATP is made.

Things to Be Careful About

• The mark scheme requires 'ATP synthase' or 'ATP synthetase', not 'ATPase'.
• Use '$\text{H}^+$' or 'protons' consistently; 'hydrogen ions' is also accepted but 'hydrogen' alone is not specific enough.
• The proton gradient is across the INNER mitochondrial membrane, between the intermembrane space and the matrix.
• Each of the five ETC points (electrons travel, energy released, $\text{H}^+$ pumped, gradient created) is a separate mark; one sweeping sentence will not earn all five.

Background Concept

Mitochondria are unusual organelles: they contain their own circular DNA (mtDNA) and their own ribosomes, and they can carry out protein synthesis. The most widely accepted explanation is the endosymbiotic theory, which proposes that mitochondria evolved from aerobic bacteria that were engulfed by an ancestral eukaryotic cell. The 37 human mitochondrial genes are thought to be the remnant of the genome of that original bacterial endosymbiont. The vast majority of mitochondrial proteins (>1000) are, however, encoded by the nuclear genome, synthesised on cytoplasmic ribosomes and imported into the mitochondrion; only 13 are encoded by mtDNA itself in humans.

Mitochondria need proteins for: aerobic respiration (Krebs cycle enzymes, ETC proteins, ATP synthase), membrane transport (since the inner membrane is highly selective), their own protein synthesis (rRNA, tRNA, ribosomal proteins), and structural roles. Anything encoded by mtDNA must therefore be a protein or RNA that operates in the mitochondrion itself.

Understanding the Question

Part (c) is a 3-mark 'suggest' question. It does not require the candidate to know the exact 13 protein-coding genes, but to use understanding of mitochondrial function to suggest what the 37 genes might do. The mark scheme gives a list of acceptable points.

Approach

Think about everything the mitochondrion needs to do or to make: carry out the link reaction, Krebs cycle and oxidative phosphorylation; build and maintain its own membranes; make its own ribosomes and tRNA so it can translate its own proteins.

Step-by-Step Reasoning

1. Enzymes of the Krebs cycle and link reaction: Mitochondria must carry out both stages in the matrix, so they need enzymes such as pyruvate dehydrogenase (link reaction), citrate synthase, isocitrate dehydrogenase, $\alpha$-ketoglutarate dehydrogenase and others. Some of these are encoded by mtDNA. (In humans, subunits of complexes I, III, IV, V and pyruvate dehydrogenase are among the 13 mtDNA-encoded proteins.)
2. ETC proteins (A, B, C): The electron transport chain consists of cytochromes and other electron carriers. These are partly encoded by mtDNA.
3. ATP synthase (D): The stalked particles are partly encoded by mtDNA (subunits of the F0 and F1 parts).
4. Membrane / transport / structural proteins: The outer and inner membranes and cristae require structural proteins and various transporters to move substrates, ADP, Pi, ATP, and to maintain the proton gradient.
5. tRNA, rRNA and ribosomes: Mitochondria have their own ribosomes (70S, similar to bacterial) and need tRNAs to translate their own mRNA. Of the 37 human mitochondrial genes, 22 encode tRNAs and 2 encode rRNAs; the remaining 13 encode proteins.

Key Takeaways

• Mitochondrial DNA encodes both RNAs (tRNA, rRNA) and a small number of proteins.
• The proteins encoded are key subunits of the respiratory machinery, especially components of oxidative phosphorylation.
• Mitochondrial ribosomes are 70S (like bacterial), supporting the endosymbiotic origin.
• The 13 protein-coding genes of human mtDNA encode subunits of complexes I, III, IV and V of the ETC, plus a few other proteins.

Common Mistakes

• Stating that mitochondrial DNA is linear. It is CIRCULAR, like bacterial DNA — a key piece of evidence for the endosymbiotic theory.
• Stating that mitochondria 'do not need the nucleus' for any of their proteins. In fact, most mitochondrial proteins are encoded by nuclear DNA and imported; mtDNA only encodes a small minority.
• Suggesting mtDNA controls traits inherited from the father. Mitochondrial DNA is inherited almost exclusively from the mother (maternal inheritance) because the egg contributes essentially all the mitochondria to the zygote.
• Listing things like 'energy production' or 'ATP synthesis' as functions of the genes. Genes do not make ATP; they code for the proteins and RNAs that enable ATP synthesis.
• Confusing '37 genes' with '37 proteins'. There are 37 genes in total: 13 protein-coding + 22 tRNA + 2 rRNA.

Things to Be Careful About

• The question asks for functions of the genes (what they do), not functions of the mitochondrion as a whole.
• 'Suggest' questions accept a range of valid answers. Any of the five points in the mark scheme would earn a mark; the candidate only needs to give three.
• Avoid vague answers ('they make the mitochondrion work', 'they produce energy'). Be specific: name the type of protein (enzyme, carrier, ribosome component) or RNA.

Background Concept

The light-dependent reactions of photosynthesis take place on the thylakoid membranes and require two photosystems working in series: photosystem II (P680) and photosystem I (P700). In NON-CYCLIC photophosphorylation, electrons flow from water through PSII, the electron transport chain, PSI and finally to NADP⁺. The electrons lost from PSII are replaced by the splitting of water – this is the only step in photosynthesis that produces O₂.

Photolysis is the splitting of a molecule using light energy. In the thylakoid lumen, water is broken down by the oxygen-evolving complex (OEC, also called the water-splitting complex, WSC), which contains manganese ions that act catalytically. The reaction can be written:

The electrons replenish those lost by the reaction-centre chlorophyll of PSII; the protons contribute to the proton gradient that drives ATP synthesis by chemiosmosis; the oxygen is released as a by-product (or used in cellular respiration).

Understanding the Question

This is a 'describe' question worth 3 marks on the process by which oxygen is produced in non-cyclic photophosphorylation. We need to give the sequence: how light energy is captured, where the water-splitting happens, what catalyses it, and what the products are.

Approach

The mark scheme offers four creditable points; 'any three from' means the candidate needs three, but giving all four is safer. Cover in order: light capture → photolysis → enzyme catalysing it → products (with the equation as a valid alternative for the products point).

Step-by-Step Reasoning

1. Light capture and photoactivation: light is absorbed by the reaction-centre chlorophyll of photosystem II (P680). The energy raises an electron to a higher energy level (photoactivation / photoionisation), leaving the chlorophyll positively charged and creating an electron 'hole' that must be filled.
2. Photolysis of water: to refill this 'hole', water is split. This step is specifically called photolysis (literally 'splitting by light').
3. Enzyme / complex involved: photolysis is catalysed by the water-splitting enzyme, also called the oxygen-evolving complex, located on the inner (lumenal) surface of the thylakoid membrane.
4. Products: water splits into protons (H⁺), electrons (e⁻) and oxygen (O₂). The equation form is also acceptable: $2\text{H}_2\text{O} \rightarrow 4\text{H}^+ + 4\text{e}^- + \text{O}_2$.

Key Takeaways

• Non-cyclic photophosphorylation involves BOTH photosystems and produces ATP, reduced NADP and O₂.
• The O₂ released in photosynthesis comes entirely from water (not from CO₂) – proven by isotope-labelling experiments with ¹⁸O.
• The water-splitting / oxygen-evolving complex is on the lumen side of the thylakoid membrane; the H⁺ it releases contributes to the proton gradient used by ATP synthase.

Common Mistakes

• Confusing PSII and PSI – PSII (P680) comes first in the non-cyclic pathway; light is absorbed there first.
• Writing 'photoionisation' without specifying that chlorophyll is the molecule being photoionised (or omitting 'photosystem II / P680').
• Saying 'water is broken down by light' without naming the enzyme / complex that catalyses it – the mark scheme credits the water-splitting enzyme / oxygen-evolving complex specifically.
• Writing 'water → hydrogen + oxygen' (atomic hydrogen) instead of 'protons, electrons and oxygen'.
• Confusing photophosphorylation (ATP synthesis driven by light) with the Calvin cycle (CO₂ fixation).

Things to Be Careful About

• 'Photolysis' must be the word used, not 'hydrolysis' or 'splitting' alone.
• Specify which photosystem – PSII (P680), not just 'chlorophyll'.
• 'Oxygen evolving complex' and 'water-splitting enzyme' are both accepted; use whichever you remember.
• The question asks specifically about how OXYGEN is produced – keep the answer focused on photolysis, not on the whole electron transport chain or ATP synthesis.

Background Concept

An action potential is a brief reversal of the membrane potential caused by the rapid influx of Na⁺ ions through voltage-gated sodium channels. The action potential is regenerated along the axon by local circuits: positive ions flow from the depolarised region into the next resting region, depolarising the membrane there to threshold and triggering new voltage-gated Na⁺ channels to open.

In an unmyelinated axon, voltage-gated Na⁺ channels are present along the entire length of the axon, so the action potential must be regenerated at every point along the membrane — conduction is slow and continuous.

In a myelinated axon, the membrane is wrapped in Schwann cells. The myelin is a good electrical insulator, so very few ions can cross the membrane in the myelinated regions. Voltage-gated Na⁺ channels are concentrated only at the nodes of Ranvier — the small unmyelinated gaps between adjacent Schwann cells. Because the myelin prevents ion movement, depolarisation can only occur at the nodes, and the action potential effectively jumps from node to node. This mode of conduction is called saltatory conduction (from the Latin saltare — to jump).

Saltatory conduction is much faster than continuous conduction for two reasons:

1. Fewer action potentials need to be generated (only at each node, not at every micrometre).
2. Local circuits in a myelinated axon are very long — they extend between successive nodes (~1 mm apart) rather than over a fraction of a micrometre, so the impulse effectively travels further in each "step".

Understanding the Question

Part (b) is a four-mark "state and explain" question. The command word "state" requires the headline difference; "explain" requires the biological reason for the difference. A good answer will give one statement of the difference (myelinated is faster) and then several explanatory points (insulation, nodes, depolarisation only at nodes, saltatory conduction, longer local circuits).

Approach

• Open with the observable difference: speed of conduction.
• Then explain WHY the myelin sheath changes conduction: it insulates.
• Then describe the role of the nodes of Ranvier as the only place where the action potential can be regenerated.
• Finally, name the mechanism (saltatory conduction) and the consequence for local circuits.

Step-by-Step Reasoning

1. State the difference. Myelinated neurones conduct impulses faster than unmyelinated ones.
2. Insulation. The myelin sheath is an electrical insulator; it prevents ions from moving across the membrane in the wrapped regions.
3. Nodes of Ranvier. Only at the nodes of Ranvier (gaps between adjacent Schwann cells, with no myelin) are ions able to cross the membrane and voltage-gated Na⁺ channels are concentrated.
4. Depolarisation only at nodes. The action potential can therefore only be generated at the nodes, not along the myelinated sections.
5. Saltatory conduction. The action potential jumps from one node to the next — this is saltatory conduction.
6. Local circuits. Because the internodal distance is long, local circuits are correspondingly longer, so the impulse is regenerated less often; this further increases the speed of conduction in myelinated axons.

Key Takeaways

• Myelinated = fast, saltatory, action potentials only at nodes.
• Unmyelinated = slow, continuous, action potentials regenerated along the whole length.
• The myelin sheath acts as an insulator; the nodes of Ranvier concentrate the voltage-gated channels needed to regenerate the action potential.

Common Mistakes

• Stating only that myelinated is faster, without any explanation (gives only 1 of 4 marks).
• Saying that the impulse "travels through the myelin" — incorrect; the impulse is conducted in the cytoplasm and regenerates at the nodes.
• Confusing the role of the myelin sheath (electrical insulation) with that of the nodes of Ranvier (site of ion exchange and action potential regeneration).
• Writing "action potential jumps along the axon" without naming the mechanism as saltatory conduction.

Things to Be Careful About

The mark scheme explicitly notes that "ora" (or reverse argument) is acceptable for each comparative point, so a candidate who writes the points in terms of unmyelinated neurones (e.g. "slower, continuous conduction, action potentials along whole length") can still earn full marks. Aim to give at least four well-distinguished points to be safe.