Biology 9700/42 — October/November 2024

Background Concept

A population is a group of organisms of the same species living in the same area and capable of interbreeding. Each population carries a gene pool — the complete set of alleles of every gene in that population — and the relative frequencies of those alleles are called the allele frequencies.

These frequencies are not static. Several processes can change them, generating or removing genetic variation and shifting the population away from (or back towards) Hardy–Weinberg equilibrium. Because the question specifies a wildlife population, artificial selection is not relevant and is explicitly rejected by the mark scheme.

The processes fall into three broad categories:

• Mechanisms that create new variation — mutation, genetic recombination.
• Mechanisms that change frequencies randomly (no selection) — genetic drift, founder effect, bottleneck effect.
• Mechanisms that change frequencies non-randomly (because of selection or movement) — natural selection, migration / gene flow.

Understanding the Question

The command word is "outline" — give the main features briefly, without going into full detail. You need to identify six processes that may change allele frequencies in a wildlife population, and give a short one-clause description of each. The mark scheme lists eight creditable points, and any six earn full marks, so you have some flexibility about which six you include.

Be precise with terminology: the mark scheme explicitly rejects "artificial selection" and "stabilising selection" for this question. If you mention selection, it must be directional or disruptive.

Approach

1. List the processes you can confidently describe.
2. For each, write a single short sentence that captures the mechanism and how it changes allele frequencies.
3. Choose six that together cover the range of mechanisms (variation-generating, random, non-random) — this shows the examiner that you understand the different ways allele frequencies can shift.

Step-by-Step Reasoning

1. Mutation. A mutation is a change in the base sequence of DNA. It can create a new allele where none existed before, and so changes the allele frequencies of the gene pool. Mutation is the ultimate source of all genetic variation on which the other processes act.

2. Genetic drift. Random change in allele frequency that occurs because the individuals that happen to survive and reproduce do not pass on a perfectly representative sample of the gene pool. The smaller the population, the more pronounced the effect — some alleles may be lost entirely, others may reach fixation, purely by chance.

3. Founder effect. A special case of genetic drift: a small number of individuals colonise a new area and found a new population. The allele frequencies in that new population are determined by the (non-representative) genotypes of the founders, and can differ markedly from the parent population.

4. Bottleneck effect. Another special case of genetic drift: a large, sudden fall in population size (e.g. disease, volcanic eruption, severe winter) leaves only a small, non-representative sample of the original gene pool. Allele frequencies in the survivors — and in the population as it rebuilds — are shifted by chance.

5. Natural selection. The non-random differential survival and reproduction of individuals because their phenotypes are better adapted to the environment. Directional selection shifts the mean phenotype (and the underlying allele frequencies) in one direction. Disruptive selection favours both extremes of a range at the expense of intermediates, so two divergent allele sets are maintained. The mark scheme ignores artificial and stabilising selection here.

6. Migration / gene flow. When individuals (or their gametes, in plants) move between populations and interbreed, they carry their alleles with them. The donor population loses those alleles (or some of them), and the recipient gains them, so the allele frequencies of both populations shift.

7. Genetic recombination. During meiosis, independent assortment of homologous chromosomes and crossing over between non-sister chromatids of homologues produce new combinations of alleles in the gametes. Different combinations of alleles appear in the offspring, and this is the genetic variation on which the other processes can act. (Note: recombination does not create new alleles — it merely reshuffles existing ones.)

Key Takeaways

• Allele frequencies are not static; they can be changed by mutation, genetic drift (including founder and bottleneck effects), natural selection, migration / gene flow and genetic recombination.
• Genetic drift, founder effect and bottleneck effect are random processes — they change allele frequencies with no reference to whether the alleles are advantageous.
• Natural selection is non-random — allele frequencies change because particular alleles give a survival/reproductive advantage.
• The smaller the population, the faster allele frequencies can be moved by random (non-adaptive) processes.

Common Mistakes

• Writing artificial selection or stabilising selection — both are explicitly rejected by the mark scheme here, because the question specifies wildlife populations.
• Confusing founder effect and bottleneck effect — both are forms of genetic drift, but the founder effect involves a small number of individuals starting a new population, whereas the bottleneck effect involves a sudden reduction in size of an existing population.
• Describing the processes in general terms ("evolution happens", "natural selection acts on variation") rather than linking them directly to allele frequencies.
• Confusing mutation with genetic recombination — mutation creates new alleles; recombination only reshuffles existing ones.

Things to Be Careful About

• Use the technical term ("genetic drift", "founder effect", "bottleneck effect", "gene flow") rather than just describing the situation, because the terms themselves are usually the mark.
• For natural selection, the mark scheme accepts directional or disruptive but not stabilising or artificial.
• Mutation is the only mechanism that creates new alleles — recombination only rearranges existing ones, but in doing so it produces new genotypes.
• You only need six points for six marks; do not pad your answer with weak or incorrect extra material that the examiner will then not credit.

Background Concept

Selective breeding (also called artificial selection) is the deliberate choice by humans of which individuals in a population are allowed to reproduce, in order to shift the population's gene pool towards a desired combination of traits. Two genetic conditions are central:

• Homozygosity — both alleles at a locus are identical. A homozygous plant breeds true: all its gametes carry the same allele at that locus, so all its offspring are genetically identical at that locus.
• Heterozygosity — the two alleles at a locus are different. Heterozygotes often show hybrid vigour (also called heterosis): they grow faster, are larger, and are more robust than either parental line, because the masking of deleterious recessive alleles and the complementary action of different alleles both contribute.

Understanding the Question

The question has a deliberate tension. Uniform height requires genetic similarity between plants — they must all carry the same alleles for the height-controlling genes. But vigour requires genetic diversity — at least some loci must be heterozygous, because the F1 hybrid of two different inbred lines is what produces hybrid vigour. The candidate has to explain how a single breeding programme satisfies both requirements.

The mark scheme rewards the linked ideas: inbreeding → homozygosity → uniformity, and outbreeding / hybridisation → F1 heterozygosity → vigour.

Approach

• Start with the trait that needs uniformity (height) and identify the genetic state that gives it: homozygosity.
• Identify the breeding method that fixes homozygosity: inbreeding.
• Then identify what is needed for vigour: heterozygosity, via crossing two different homozygous lines.
• Bring the two together: inbreed to make pure-breeding parental lines, then cross the lines to get uniform-but-vigorous F1 hybrid seed.

Step-by-Step Reasoning

1. Inbreeding. Inbreeding is the crossing of closely related individuals — in maize this is done by self-pollination (transferring pollen from a flower to the silks of the same plant, or to a genetically very similar plant). Generation after generation, this drives the offspring towards homozygosity at all loci, because each parent can only pass on one of its two alleles and similar parents pass on similar alleles. After several generations, each plant is essentially homozygous at every locus.

2. Homozygous lines are uniform. A homozygous plant is genetically identical (at every locus) to every other plant in the same inbred line, so all the plants in that line have the same height-controlling alleles and are of uniform height.

3. Outbreeding / hybridisation. Two different inbred lines (each homozygous and uniform but for different alleles) are then crossed by deliberate hybridisation — the breeder transfers pollen from one inbred line to the silks of the other. All the F1 offspring receive one allele from line A and one from line B at every locus, so they are all heterozygous at every locus.

4. F1 hybrid vigour. Because the F1 plants are all heterozygous (and genetically identical to one another), they are both uniform and vigorous. The heterozygosity produces hybrid vigour: the F1 plants grow faster, larger and more robustly than either of the parental inbred lines. This is why farmers buy new F1 hybrid maize seed every year rather than saving seed from their own crop — the F2 generation would segregate (different F2 plants would inherit different combinations of the parental alleles), losing both uniformity and vigour.

Key Takeaways

• Inbreeding → homozygosity → genetic uniformity → uniform height.
• Outbreeding / hybridisation of two inbred lines → F1 heterozygosity → hybrid vigour.
• The two processes are used in sequence in the same breeding programme: first inbreed to make pure lines, then cross the lines to make F1 hybrid seed.
• F1 hybrid vigour is lost in the F2 generation, so commercial hybrid seed must be re-produced every year by re-crossing the parental inbred lines.

Common Mistakes

• Describing only inbreeding — this gives uniformity but not vigour. The question requires both targets, and the marks are split across both methods.
• Describing only outbreeding — this can give vigour, but without first inbreeding the parental lines, the F1 offspring will not be uniform.
• Getting the genetic relationship the wrong way round — writing "outbreeding produces homozygosity" or "inbreeding produces vigour". The relationships go the opposite way.
• Mentioning stabilising selection or natural selection — this is selective breeding (artificial), and the breeding process is inbreeding followed by outbreeding, not natural selection.
• Failing to identify the F1 generation specifically — the mark scheme gives credit for linking the F1 to heterozygosity and to vigour.

Things to Be Careful About

• Keep the genetic terms straight: homozygous (identical alleles) versus heterozygous (different alleles).
• Recognise that uniformity and vigour are not contradictory in the F1 — uniformity comes from the F1 plants being genetically identical to one another, while vigour comes from them being heterozygous at each locus. Both conditions hold simultaneously.
• The mark scheme does not require the technical term "heterosis" — it accepts "hybrid vigour" or a simple description — but using "hybrid vigour" makes the link to heterozygosity explicit and shows clear understanding.
• Note that F1 hybrid seed is uniform only as a generation; the individual plants in the F1 generation are genetically identical to one another but genetically different from both parental inbred lines.

Background Concept

Genetic engineering (recombinant DNA technology) is the deliberate alteration of the genetic makeup of an organism by introducing, removing or modifying DNA. Three ideas must all be present: something is done to the DNA itself (manipulation); that DNA is moved into a host cell or organism; and once inside, it is expressed (transcribed and translated) so that the host gains a new characteristic or makes a new protein. Common laboratory steps include cutting DNA with restriction endonucleases, joining fragments with DNA ligase, and using a vector (typically a plasmid or a virus) to carry the recombinant DNA into the recipient. The result is often described as a genetically modified organism (GMO), and the field now also includes gene editing (e.g. CRISPR/Cas9) where the existing DNA sequence is changed in situ rather than a new gene being added.

Understanding the Question

Part (a) is a 3-mark "explain" command-word item. The mark scheme asks for any three of: manipulation of DNA, transfer into a cell/organism, expression, modified phenotype / new protein, or an additional valid point (such as the possibility of producing a GMO). The candidate must say enough to convince the examiner that the technique is intentional, that DNA is the material being worked on, and that a tangible biological outcome follows.

Approach

Write a short, single-sentence (or compact) definition that explicitly mentions the three linked ideas: (1) the material acted on (DNA/gene/allele), (2) what is done to it (manipulation/transfer), and (3) what happens as a result (expression, leading to a new protein or phenotype change). Mirror the language of the mark scheme.

Step-by-Step Reasoning

• Point 1 – Manipulation: genetic engineering involves deliberately changing (manipulating) genetic material – gene, allele or DNA. This is the defining feature that separates it from selective breeding.
• Point 2 – Transfer: the manipulated gene/allele is moved into a recipient cell or organism, typically using a vector (plasmid or virus) or physical methods.
• Point 3 – Expression: once inside, the gene is transcribed and translated; this is the key step that produces the biological effect.
• Point 4 – Outcome: the cell/organism now produces a (new) protein or shows a modified phenotype – e.g. a bacterium making human insulin, a crop resistant to herbicide, or – relevant to this paper – production of a missing functional protein in a human disease.

Key Takeaways

• Genetic engineering = deliberate change to an organism's DNA, expressed as a new protein or phenotype.
• It is distinct from selective breeding (no DNA is moved) and from gene therapy (where the new DNA is added to a patient for medical benefit).
• Modern genetic technology also includes gene editing – changing the existing DNA in place, rather than adding foreign DNA.

Common Mistakes

• Saying only that "DNA is changed" without mentioning transfer into another cell or expression – loses 2 of the 3 marks.
• Confusing genetic engineering with cloning (cloning copies an entire genome; engineering changes specific DNA).
• Writing "modification of genes" alone – the mark scheme requires that the result is a transferred, expressed gene.

Things to Be Careful About

• Use the precise term "genetic material / gene / allele / DNA" – avoid vague phrases like "the genetics of an organism".
• For full marks, the definition must link manipulation + transfer + expression in a single, integrated answer.

Background Concept

In Mendelian inheritance, each gene exists as two alleles (alternative forms) on homologous chromosomes. For a single gene with two alleles, an organism can be homozygous dominant (e.g. $RR$), homozygous recessive (e.g. $rr$), or heterozygous (e.g. $Rr$). The dominant allele is expressed in the phenotype whenever it is present, so $RR$ and $Rr$ individuals look the same; the recessive allele is only expressed when no dominant allele is present (i.e. $rr$).

A monohybrid cross follows the inheritance of one gene. Mendel's first law (the law of segregation) states that the two alleles at a locus separate (segregate) during meiosis, so each gamete carries only one allele of each gene. When a heterozygote is self-crossed ($Rr \times Rr$), the gametes ($R$ and $r$ in equal proportions) fuse at random, giving a genotypic ratio of $1\ RR : 2\ Rr : 1\ rr$ and a phenotypic ratio of $3$ dominant : $1$ recessive.

Understanding the Question

The question describes a tomato cross: pure-breeding red ($\times$ homozygous $RR$) crossed with pure-breeding yellow ($\times$ homozygous $rr$) gives an F1 in which all fruits are red — this tells you that red is dominant to yellow and that the F1 plants are all $Rr$. The question then asks for a genetic diagram of the F1 $\times$ F1 cross that produces the F2 generation. You are told to use the symbols R and r for the alleles.

The command word is construct a genetic diagram, so a clear Punnett square showing parents, gametes, offspring genotypes and phenotypes is the expected answer. Three marks are available: one for the parental genotypes and gametes, one for the offspring genotypes, and one for the offspring phenotypes.

Approach

1. Identify the parental generation: the F1 plants, all of which are $Rr$ (because red is dominant).
2. List the gametes each parent can produce: $R$ or $r$ (with equal probability).
3. Set out a Punnett square (a 2 × 2 grid) combining the gametes from each parent.
4. Read off the four offspring boxes, give each genotype, then assign the phenotype (red or yellow) to each.
5. Tally the phenotypes to give the ratio.

Step-by-Step Reasoning

• Parental genotypes ($Rr \times Rr$): The F1 individuals came from a cross $RR \times rr$, so every F1 is $Rr$. When the F1 are crossed with each other, both parents are $Rr$ — this is the parental genotype line that earns the first mark.
• Gametes ($R$ and $r$ from each parent): According to the law of segregation, the homologous chromosomes carrying $R$ and $r$ separate at meiosis I, so each parent produces gametes with $R$ and gametes with $r$ in equal numbers. The mark scheme requires the gametes to be written out — typically above and to the side of the Punnett square.
• Offspring genotypes ($RR$, $Rr$, $Rr$, $rr$): Filling in the four cells of the Punnett square gives $RR$ (top left), $Rr$ (top right and bottom left), and $rr$ (bottom right). The two $Rr$ cells must both be written for the second mark.
• Phenotypes (red, red, red, yellow): Because $R$ is dominant, $RR$ and $Rr$ both produce red fruits; only $rr$ produces yellow fruits. The third mark is awarded for writing the phenotype beneath or alongside each genotype.
• Ratio: Summing the phenotypes gives 3 red : 1 yellow.

Key Takeaways

• A pure-breeding red crossed with a pure-breeding yellow gives a uniform F1 (here, all red) — the parental phenotype seen in the F1 identifies which trait is dominant.
• An F1 $\times$ F1 monohybrid cross gives a 3:1 phenotypic ratio (and 1:2:1 genotypic ratio) in the F2.
• A genetic diagram should always show, in order: parental phenotypes, parental genotypes, gametes, offspring genotypes, and offspring phenotypes. This is the conventional layout credited by the mark scheme.

Common Mistakes

• Writing the F1 parental genotype as $RR$ (forgetting that pure breeding was the P generation, not the F1).
• Listing only one gamete from each parent (e.g. only $R$) instead of both $R$ and $r$ — the mark scheme requires the two gamete types.
• Writing offspring as $RRr$ or $RRRr$ instead of separating them into the four boxes.
• Labelling $Rr$ offspring as yellow because they "carry the recessive allele" — dominance means the phenotype is determined by the allele that is present, not the ones that are absent.
• Forgetting the parental phenotype line — a complete genetic diagram starts with phenotypes, not just genotypes.

Things to Be Careful About

• Use the symbols exactly as instructed: R and r, with the dominant allele in upper case and the recessive allele in lower case.
• The mark scheme credits the answer only if parents, gametes, offspring genotypes and phenotypes are all present and in the right order. A diagram with a single line of "$Rr \times Rr \rightarrow 3$ red : $1$ yellow" will not earn full marks.
• "Pure breeding" means homozygous — for this question, the pure-breeding red parent is $RR$ and the pure-breeding yellow parent is $rr$. This is P generation context; the question only asks for the F1 $\times$ F1 cross.
• Do not introduce extra alleles (e.g. $R'$ or $r_y$); the question restricts you to $R$ and $r$.

Background Concept

A dihybrid cross follows the inheritance of two different genes at the same time. When the two genes are located on different autosomes (i.e. on different pairs of homologous chromosomes), the alleles of one gene assort independently of the alleles of the other gene during meiosis. This is Mendel's second law, the law of independent assortment.

A cross between a parent homozygous dominant for both genes (e.g. $AABB$) and a parent homozygous recessive for both genes (e.g. $aabb$) produces F1 individuals that are heterozygous at both loci ($AaBb$). The F1 expresses only the two dominant phenotypes because $A$ and $B$ are dominant. When two F1 individuals are crossed, the gametes they produce are $AB$, $Ab$, $aB$ and $ab$ in equal proportions (because of independent assortment), and the random fusion of these gametes produces the classic dihybrid phenotypic ratio of $9 : 3 : 3 : 1$ in the F2. Of these 16 parts, 9 show both dominant phenotypes, 3 show one dominant + the other recessive, 3 show the other dominant + the one recessive, and 1 shows both recessive phenotypes — the last category (and the two 3s in part) is "recombinant" relative to the original parents.

Understanding the Question

This part sets up a theoretical dihybrid cross with two unlinked autosomal genes (i.e. genes on different chromosomes). You are asked to do two things: (1) state the F2 phenotypic ratio, and (2) explain why some offspring in the F2 have phenotypes that were not present in either of the original (P generation) parents. The P generation shows only the two parental phenotypes (both dominant, and both recessive); the F2 contains four phenotypes, two of which mix a dominant trait of one gene with a recessive trait of the other — these "new" phenotypes are what the question wants explained.

The command words are state (the ratio — just a number) and explain (the reason for the recombinant phenotypes — must give the mechanism, not just restate the ratio). Three marks are available: one for the ratio and two for the explanation.

Approach

1. Recognise the cross as a standard dihybrid cross between two double-heterozygous F1 individuals.
2. State the classic F2 ratio: 9:3:3:1.
3. Identify the source of the "new" phenotypes: independent (random) assortment of homologous chromosomes / bivalents at a specific stage of meiosis (metaphase I, where the orientation of each bivalent on the equator is random, or metaphase II for sister chromatids).
4. Link the chromosome behaviour to the four types of gamete, and therefore to the appearance of recombinant phenotypes in the F2.

Step-by-Step Reasoning

• Ratio (mark 1): A dihybrid F2 from a $AaBb \times AaBb$ cross always shows the $9:3:3:1$ phenotypic ratio when both genes show complete dominance and are unlinked. The 9 represents the double-dominant phenotype, the two 3s each represent one dominant + one recessive phenotype, and the 1 is the double-recessive phenotype. Two of these four phenotypes (the two 3s) were not seen in the P generation.
• Independent assortment (mark 2): During meiosis, the two pairs of homologous chromosomes (one carrying $A/a$ and the other carrying $B/b$) line up independently of one another. This is the source of the four kinds of gamete.
• Stage of meiosis (mark 3): The mark scheme specifically requires a named stage — metaphase I is the canonical answer (bivalents align on the equator; the orientation of each bivalent is random, so the maternal/paternal chromosome of one pair can be combined with either chromosome of the other pair). Metaphase II is also accepted. Note that crossing over is ignored by the mark scheme for this question, so it should not be given as the explanation.
• Link to phenotype: The independent orientation of bivalents means that the $A$ allele can end up in a gamete with either $B$ or $b$, and likewise for $a$. F1 gametes are therefore $AB$, $Ab$, $aB$ and $ab$ in equal proportions, and when these fuse at random, the $Ab$ and $aB$ combinations produce phenotypes that combine a dominant trait of one gene with a recessive trait of the other — phenotypes that were not present in the original parents.

Key Takeaways

• A standard dihybrid F2 (from a cross of two F1 double heterozygotes) has a phenotypic ratio of $9 : 3 : 3 : 1$.
• The new (recombinant) phenotypes in the F2 are produced by independent assortment of the two pairs of homologous chromosomes during meiosis.
• The classic meiotic stage to cite is metaphase I, where random orientation of bivalents generates the four gamete types.

Common Mistakes

• Writing the ratio as 1:2:1:2:4:2:1:2:1 or another genotypic-style ratio instead of the 9:3:3:1 phenotypic ratio.
• Saying that the new phenotypes arise "because of meiosis" without specifying what happens in meiosis — the mark scheme requires independent assortment AND a named stage (e.g. metaphase I).
• Attributing the recombinant phenotypes to crossing over — the mark scheme explicitly says to ignore references to crossing over for this question.
• Citing "random mating" or "random fertilisation" as the cause — these are true but they do not, by themselves, generate new combinations of alleles; it is the assortment of chromosomes into gametes that creates the new allele combinations in the first place.
• Confusing the F2 with the F1: the F1 is uniform, and the F2 is where the 9:3:3:1 ratio appears.

Things to Be Careful About

• State the ratio precisely in the form "9:3:3:1" — variants such as "9-3-3-1" or "9, 3, 3, 1" may not be credited.
• For the explanation, two separate ideas are needed for the two marks: (i) independent / random assortment, and (ii) of homologous chromosomes / bivalents (or sister chromatids) at a named stage of meiosis. A single sentence that lumps all of this together may miss a mark; a candidate who only says "independent assortment" without naming the stage will not earn the third mark.
• Do not describe crossing over as the source of the new phenotypes for this question — the mark scheme instructs the examiner to ignore it.
• Keep the explanation focused on the mechanism (chromosome behaviour) rather than the outcome (the 9:3:3:1 ratio). The mark scheme is rewarding the mechanism, not the result.

Background Concept

Variation between individuals in a population can be classified in two main ways:

• Discontinuous variation is qualitative and produces a small number of distinct, non-overlapping categories (e.g. ABO blood group, tongue-rolling, seed shape in Mendel's peas). It is usually determined by one gene (or a very small number of genes) with a small number of alleles, so the environment has little effect on which category an individual falls into.
• Continuous variation is quantitative and shows a smooth gradation between two extremes (e.g. height, mass, skin colour, milk yield). It is usually determined by many genes (polygenes), each contributing a small additive effect to the phenotype. Environmental factors (nutrition, temperature, exercise, sunlight exposure) also influence where on the scale an individual lies, so the phenotype is the result of genotype + environment.

The key distinctions tested by Table 5.1 are:

Understanding the Question

The question provides four statements and asks which apply to discontinuous variation, which to continuous variation, and which to both. A tick is placed in a box only if the statement applies to that type.

Approach

For each statement, recall the defining features of discontinuous and continuous variation and decide which column it fits.

Step-by-Step Reasoning

1. "often involves one gene only" — Discontinuous variation is typically controlled by a single gene (or very few) with discrete alleles. ✓ for discontinuous; ✗ for continuous (which involves many genes). → Tick discontinuous.
2. "environmental factors may affect gene expression" — Continuous variation is strongly influenced by environment (e.g. nutrition affects height, sunlight affects skin colour). The mark scheme ticks this only for continuous. → Tick continuous.
3. "there is an additive effect of genes that contributes to the phenotype" — Continuous variation results from many genes each contributing a small additive effect (polygenes). → Tick continuous.
4. "there are distinct differences between the various forms of a characteristic" — Discontinuous variation has discrete, non-overlapping categories. → Tick discontinuous.

Key Takeaways

• Discontinuous variation = few genes, no overlap, environment has little effect.
• Continuous variation = many genes (polygenes) with additive effects, strongly influenced by environment, no distinct categories.

Common Mistakes

• Ticking "environmental factors may affect gene expression" for both columns. The mark scheme credits this only for continuous, because although environment can also affect discontinuous traits, the statement is characteristic of continuous variation in the context of the syllabus.
• Ticking "additive effect of genes" for discontinuous. This is the hallmark of polygenic (continuous) variation.

Things to Be Careful About

• Read each statement carefully and match it to the type of variation whose defining features it describes, not just traits that might apply in unusual cases.
• A tick goes in a box only if the statement is a defining feature of that type of variation.

Background Concept

Paper chromatography is a separation technique that exploits the different solubilities of mixture components in a solvent and their different affinities for the stationary phase (the chromatography paper). When a piece of paper with a spot of pigment mixture is placed in a solvent, the solvent moves up the paper by capillary action, carrying the pigments with it. Pigments that are more soluble in the solvent (and less strongly adsorbed to the paper) travel further.

Each pigment, under given conditions (solvent, temperature, paper type), has a characteristic retardation factor ($R_f$) value, defined as:

The $R_f$ value is a ratio and therefore has no units. It always lies between 0 (pigment did not move) and 1 (pigment moved as far as the solvent front).

Understanding the Question

The question asks how the results of paper chromatography can be used to identify which chloroplast pigments are present. The "results" here are the chromatogram showing separated pigment spots at different positions on the paper. The task is to convert those positions into identifications.

Approach

The approach is in three steps: (1) measure the distances travelled by the solvent and by each pigment from the baseline; (2) calculate an $R_f$ value for each pigment spot; and (3) compare each $R_f$ value with a known reference (e.g. published tables or values obtained for pure pigment standards run under the same conditions).

Step-by-Step Reasoning

1. Measure the distances. With the chromatogram dry, mark the original baseline (where the pigment was applied) and the solvent front (the highest point reached by the solvent). For each coloured spot, measure from the baseline to the centre of the spot (distance moved by the pigment) and from the baseline to the solvent front (distance moved by the solvent).
2. Calculate the $R_f$ value. Divide the distance moved by the pigment by the distance moved by the solvent. The result is a number between 0 and 1, characteristic of that pigment under the conditions used.
3. Compare with known values. $R_f$ values for the common chloroplast pigments (chlorophyll a, chlorophyll b, carotene, xanthophyll etc.) are tabulated for standard solvent systems. By matching the calculated $R_f$ values with these reference values, each spot can be identified.

Key Takeaways

• $R_f$ is a ratio (no units) and is specific to a given pigment, solvent and paper type.
• Identification by chromatography relies on $R_f$ values being reproducible under fixed conditions.
• Always compare like with like: an $R_f$ value is only meaningful if the solvent and conditions match those used to generate the reference table.

Common Mistakes

• Measuring from the wrong baseline (e.g. the bottom edge of the paper rather than the line where pigment was applied).
• Dividing the wrong way round — always distance moved by pigment ÷ distance moved by solvent.
• Quoting an $R_f$ greater than 1, which signals a measurement error.
• Forgetting that comparison must be with values obtained in the same solvent system, since $R_f$ depends on the solvent.

Things to Be Careful About

• Measure to the centre of each spot, not the leading edge, especially if spots are diffuse.
• Make sure the chromatogram is fully dry before measuring.
• Quote $R_f$ to two decimal places for the level of precision the mark scheme expects.

Background Concept

Chlorophyll a is the primary pigment in the reaction centres of photosystems I and II. It is the only pigment that can directly convert absorbed light energy into chemical energy (as excited electrons that enter the electron transport chain). However, chlorophyll a does not absorb all wavelengths of visible light equally well — for example, it absorbs red and blue light strongly but absorbs green light poorly.

Plants also contain a range of accessory pigments, principally:

• Carotenes (e.g. β-carotene) — orange-yellow pigments.
• Xanthophylls — yellow pigments (oxygenated carotenes).
• Chlorophyll b — a yellow-green pigment.

These accessory pigments have absorption spectra that complement chlorophyll a, so together they harvest more of the incoming light energy.

Understanding the Question

The question asks specifically about the role of carotene and xanthophyll in photosynthesis. The mark scheme wants you to say what these pigments do — not what they are or where they are found.

Approach

The key idea is that these are accessory pigments: they assist the primary pigment (chlorophyll a) by absorbing additional wavelengths of light and then transferring that energy onwards. You need to make three things clear: (1) what they absorb that chlorophyll a does not, (2) the consequence of this (extending the wavelength range usable by the plant), and (3) what happens to the energy they capture (it is passed to the reaction centre).

Step-by-Step Reasoning

1. Wavelengths absorbed. Carotene and xanthophyll absorb wavelengths of light (such as blue-green and other regions) that are not absorbed efficiently by chlorophyll a. In other words, they fill in the "gaps" in chlorophyll a's absorption spectrum.
2. Extended range. Because of this complementary absorption, the overall range of wavelengths of light that the plant can use for photosynthesis is wider than it would be with chlorophyll a alone. More of the incoming solar energy is therefore harvested.
3. Energy transfer. The energy absorbed by carotene and xanthophyll is passed on (by resonance transfer) to the reaction centre / primary pigment / chlorophyll a within the photosystem. From there, the energy can drive the photochemical reactions of the light-dependent stage.

Carotene and xanthophyll do not have their own reaction centres and do not directly emit electrons or reduce electron carriers; they are simply energy funnels.

Key Takeaways

• Accessory pigments broaden the absorption spectrum of the plant.
• The energy they capture is transferred to chlorophyll a — it is not used directly.
• This is why the action spectrum (rate of photosynthesis vs wavelength) does not perfectly match the absorption spectrum of chlorophyll a alone.

Common Mistakes

• Saying that carotene and xanthophyll carry out the light-dependent reactions — they do not; they only pass energy to chlorophyll a.
• Saying they "absorb all wavelengths" or "trap light" without specifying which wavelengths or what happens next.
• Confusing accessory pigments with primary pigments and saying chlorophyll a is the only pigment — chlorophyll a is the only primary pigment.
• Omitting any mention of energy transfer.

Things to Be Careful About

• Use the precise terms "reaction centre", "primary pigment" or "chlorophyll a" — these are the terms the mark scheme accepts for where the energy is passed to.
• "Accessory pigment" is the umbrella term the mark scheme wants.
• Note that the energy transfer is to chlorophyll a in the reaction centre, not to any chlorophyll a molecule in the antenna complex.

Background Concept

The light-independent stage of photosynthesis (also called the Calvin cycle, after its discoverer Melvin Calvin) is a cyclical series of enzyme-controlled reactions that takes place in the stroma of the chloroplast. It does not require light directly, but it depends on the products of the light-dependent stage: ATP (an energy source) and reduced NADP (a source of reducing power / hydrogen).

The overall purpose of the cycle is to fix inorganic carbon dioxide into organic molecules — ultimately carbohydrate — using the energy and reducing power supplied by the light-dependent stage. Each turn of the cycle fixes one molecule of CO₂; three turns fix three molecules of CO₂ and yield one molecule of triose phosphate (a 3-carbon sugar) that can leave the cycle.

Understanding the Question

The stem tells you that ATP and reduced NADP from the light-dependent stage are used in the light-independent stage. Part (c) asks you to describe the light-independent stage — that is, the Calvin cycle. Seven marks are available, so the answer should be a thorough, step-by-step account covering: the entry of CO₂, the role of rubisco and RuBP, the formation of GP and TP, the regeneration of RuBP, and what happens to the intermediates that leave the cycle.

Approach

Work systematically through the cycle in the order the reactions occur:

1. CO₂ entry and fixation (combining with RuBP).
2. The unstable 6C intermediate and its breakdown into GP.
3. Reduction of GP to TP (using ATP and reduced NADP).
4. Regeneration of RuBP from TP (using further ATP).
5. Fate of the TP and GP that leave the cycle (the products of photosynthesis).

Keep the energy/reducing-power link to the light-dependent stage in mind throughout — that is what makes the cycle "light-independent but light-dependent in its products".

Step-by-Step Reasoning

1. Carbon dioxide fixation. CO₂ combines with the 5-carbon acceptor molecule ribulose bisphosphate (RuBP). This reaction is catalysed by the enzyme rubisco (ribulose bisphosphate carboxylase/oxygenase), which is the most abundant protein on Earth. The combination of CO₂ with RuBP is called carbon dioxide fixation.
2. Unstable 6C intermediate and GP formation. The product of CO₂ + RuBP is an unstable 6-carbon compound that has only a transient existence. It is immediately hydrolysed into two molecules of glycerate 3-phosphate (GP), a 3-carbon compound. (GP is also sometimes called 3-phosphoglycerate, 3-PGA.)
3. Reduction of GP to TP. Each molecule of GP is then reduced (gains hydrogen) to triose phosphate (TP). The reducing agent is reduced NADP, and the energy required comes from ATP. Both of these were produced by the light-dependent stage. NADP is re-oxidised to NADP⁺ (which returns to the light-dependent stage to be reduced again), and ADP + Pi is released (which is re-phosphorylated to ATP in the light-dependent stage).
4. Regeneration of RuBP. For every 6 molecules of TP produced, 5 are used to regenerate RuBP so that the cycle can continue. This regeneration step also uses ATP.
5. Fate of the remaining TP. The 1 molecule of TP in 6 that does not get used to regenerate RuBP is a net product of the cycle. From this TP, the plant can make a wide range of organic molecules:
   • Hexose sugars (glucose, fructose), which can be combined to form disaccharides such as sucrose and maltose, or joined to form polysaccharides such as starch (storage) and cellulose (cell walls).
   • Lipids — the glycerol backbone and the carbon skeletons of fatty acids are made from TP.
   • Amino acids — by adding amine groups, GP and TP can be converted to amino acids and hence to proteins.
6. The whole cyclical pathway is known as the Calvin cycle. It takes place in the stroma of the chloroplast, where the relevant enzymes (including rubisco) are located.

Key Takeaways

• The Calvin cycle is the carbon-fixation pathway of photosynthesis.
• It needs ATP and reduced NADP from the light-dependent stage to run.
• CO₂ is fixed by combining with RuBP in a rubisco-catalysed reaction.
• The first stable product is GP (3C); this is reduced to TP (3C) using the products of the light-dependent stage.
• Most TP is recycled to RuBP; some TP leaves the cycle to form sugars, polysaccharides, lipids and amino acids.
• Six turns of the cycle (fixing 6 CO₂) give a net output of 2 TP molecules (one hexose equivalent).

Common Mistakes

• Saying the 6C compound is stable — it is unstable and breaks down almost immediately into two GP molecules.
• Forgetting the role of rubisco — this enzyme is the gateway to the cycle and is credit-worthy.
• Saying GP is oxidised to TP — the reverse is true; GP is reduced to TP.
• Confusing the direction of the cycle: GP → TP → RuBP, not the other way round.
• Saying the cycle produces glucose directly — glucose formation requires two TP molecules to combine, which is a separate step and not a single turn of the cycle.
• Stating the cycle takes place in the grana / thylakoids — it takes place in the stroma.

Things to Be Careful About

• Keep the role of ATP and reduced NADP clear: GP → TP uses both, RuBP regeneration uses ATP.
• Specify that the unstable 6C compound splits into two GP — do not say "into GP" or "into one GP".
• "Carbon dioxide fixation" is the technical term the mark scheme wants for the addition of CO₂ to RuBP.
• The use of "Calvin cycle" is an acceptable term for the whole pathway.
• "Hexose / sucrose / maltose / starch / cellulose / glycerol / lipids / amino acids" — the mark scheme accepts any of these as a credible product of TP, so you only need to name one or two to earn the mark.

Background Concept

Excitation–contraction coupling links the arrival of an action potential at the muscle surface to shortening of the sarcomere. The sequence has three broad phases:

1. Electrical excitation — depolarisation spreads across the sarcolemma and down into the T-tubules.
2. Ca²⁺ release and tropomyosin shift — $\text{Ca}^{2+}$ exits the sarcoplasmic reticulum (SR), binds troponin, and pulls tropomyosin off the actin's myosin-binding sites.
3. Cross-bridge cycling and shortening — myosin heads bind actin, tilt (power stroke), release and rebind, dragging the thin filaments over the thick filaments and shortening the sarcomere.

Understanding the Question

Table 8.1 lists ten events (A–J) that occur during muscle contraction but in the wrong order. You must slot the correct letter into each row 1–10 of Table 8.2. Row 5 is already filled with J (troponin changes shape) and row 10 is already filled with D (sarcomere shortens).

Approach

Trace the cause-and-effect chain forwards from the start (sarcolemma depolarised) and backwards from the end (sarcomere shortens). Insert each event at the point where its cause is satisfied and its consequence makes sense.

Step-by-Step Reasoning

Working from the start:

1. C — sarcolemma depolarised: the action potential arrives at the muscle surface.
2. F — T-tubule system membranes depolarised: the depolarisation is conducted deep into the fibre via the T-tubules, which are continuous with the sarcolemma.
3. A — calcium ions diffuse out of sarcoplasmic reticulum: depolarisation of the T-tubules opens voltage-gated $\text{Ca}^{2+}$-release channels (ryanodine receptors) in the SR, so $\text{Ca}^{2+}$ flows down its gradient into the sarcoplasm.
4. E — calcium ions bind to troponin: troponin has $\text{Ca}^{2+}$-binding sites; once $\text{Ca}^{2+}$ is in the sarcoplasm it binds.
5. J — troponin changes shape (given): $\text{Ca}^{2+}$ binding causes a conformational change in troponin.
6. I — tropomyosin moves: troponin's change of shape pulls tropomyosin away from the myosin-binding sites on actin.
7. G — binding sites on actin exposed: with tropomyosin shifted, the myosin-binding sites are now accessible.
8. B — myosin heads bind to actin: cross-bridges form.
9. H — myosin heads tilt: the power stroke — ADP + Pi are released and the head pivots, sliding the thin filament.
10. D — sarcomere shortens (given): repeated cross-bridge cycles pull the Z-lines closer together.

Key Takeaways

• Excitation–contraction coupling flows: electrical signal → $\text{Ca}^{2+}$ release → troponin/tropomyosin shift → cross-bridge formation → power stroke → sarcomere shortening.
• The T-tubule system is essential: it carries the action potential deep into the fibre so that all myofibrils depolarise simultaneously.
• Troponin and tropomyosin are the regulatory proteins; $\text{Ca}^{2+}$ is the trigger that unmasks actin's binding sites.

Common Mistakes

• Putting "myosin heads bind to actin" (B) before "binding sites on actin exposed" (G) — without exposed sites, myosin cannot bind.
• Putting $\text{Ca}^{2+}$ release (A) after $\text{Ca}^{2+}$ binding to troponin (E) — release must precede binding.
• Forgetting that T-tubule depolarisation (F) is the step between sarcolemma depolarisation (C) and SR $\text{Ca}^{2+}$ release (A).
• Putting tropomyosin movement (I) before troponin change of shape (J) — J drives I.

Things to Be Careful About

The order tests logical dependence: each event must occur before its consequence. Work forwards from the depolarisation and backwards from the sarcomere shortening, and meet in the middle.

Background Concept

At the neuromuscular junction the motor neurone's synaptic knob releases the neurotransmitter acetylcholine ($\text{ACh}$) onto the muscle fibre's motor end plate. The sequence is:

1. An action potential arrives at the synaptic knob.
2. Voltage-gated $\text{Ca}^{2+}$ channels open and $\text{Ca}^{2+}$ flows into the knob down its electrochemical gradient.
3. The $\text{Ca}^{2+}$ causes vesicles containing $\text{ACh}$ to fuse with the presynaptic membrane, releasing $\text{ACh}$ into the synaptic cleft (exocytosis).
4. $\text{ACh}$ diffuses across the cleft and binds to nicotinic receptors on the motor end plate (sarcolemma).
5. The receptors open, allowing $\text{Na}^{+}$ (and some $\text{K}^{+}$) to flow; the net influx of $\text{Na}^{+}$ depolarises the sarcolemma.
6. If threshold is reached, an action potential fires and propagates along the muscle fibre, triggering excitation–contraction coupling (the events ordered in part (b)) and contraction.

The strength of muscle contraction is therefore set by the amount of $\text{ACh}$ released, which depends on how much $\text{Ca}^{2+}$ enters the presynaptic knob.

Understanding the Question

In Lambert–Eaton myasthenic syndrome (LEMS), autoantibodies bind to and disable the voltage-gated $\text{Ca}^{2+}$ channels on the presynaptic knob of motor neurones. The question asks you to suggest and explain how this leads to weaker muscle contraction — i.e. trace the consequences through the neuromuscular junction to the contraction itself.

Approach

Follow the synaptic-transmission chain step by step. At each stage, state how a smaller $\text{Ca}^{2+}$ influx propagates to a weaker downstream effect, and end with the consequence for muscle contraction.

Step-by-Step Reasoning

1. Ca²⁺ entry reduced. Antibodies bind to the voltage-gated $\text{Ca}^{2+}$ channels on the presynaptic knob, so fewer channels are functional. When the action potential arrives, less $\text{Ca}^{2+}$ enters the knob.
2. Vesicle fusion reduced. $\text{Ca}^{2+}$ is the trigger for vesicle fusion with the presynaptic membrane. With less $\text{Ca}^{2+}$ inside, fewer vesicles fuse, so less $\text{ACh}$ is released into the synaptic cleft by exocytosis.
3. Receptor activation reduced. Less $\text{ACh}$ in the cleft means fewer $\text{ACh}$ molecules bind to receptors on the motor end plate.
4. Na⁺ entry reduced. Each $\text{ACh}$-bound receptor opens to allow $\text{Na}^{+}$ (and some $\text{K}^{+}$) through. With fewer receptors activated, fewer $\text{Na}^{+}$ channels are open and less $\text{Na}^{+}$ enters the muscle fibre.
5. Depolarisation reduced. The end-plate potential is therefore smaller. It may not reach threshold, so fewer (or no) action potentials are generated, and excitation–contraction coupling is initiated less often / less strongly.
6. Weaker contraction. Each of the events described in part (b) — $\text{Ca}^{2+}$ release from the SR, troponin/tropomyosin shift, cross-bridge cycling — occurs less often or less synchronously, so the muscle contraction is weaker.

Any four of these links earns full credit; the mark scheme lists seven alternative or additional points.

Key Takeaways

• $\text{Ca}^{2+}$ entry into the presynaptic knob is the trigger that couples the action potential to neurotransmitter release.
• Anything that reduces $\text{Ca}^{2+}$ entry (in this case, autoantibodies against the channels) reduces $\text{ACh}$ release and therefore reduces muscle activation.
• LEMS is a useful contrast with myasthenia gravis, where the autoantibodies target the $\text{ACh}$ receptors on the muscle side rather than the $\text{Ca}^{2+}$ channels on the neurone side — the clinical picture (weak muscles) is similar but the mechanism is different.

Common Mistakes

• Saying "less $\text{ACh}$ binds to receptors" without explaining why (because less $\text{ACh}$ was released, because less $\text{Ca}^{2+}$ entered). Each link in the chain should be connected.
• Writing "less $\text{Ca}^{2+}$ released from the SR" instead of "less $\text{Ca}^{2+}$ enters the presynaptic knob" — the $\text{Ca}^{2+}$ channels mentioned in the stem are on the presynaptic membrane, not on the SR.
• Saying "no muscle contraction" or "muscle cannot contract at all" — the contraction is weaker, not abolished, because the channels are not completely blocked and the effect is partial.
• Discussing the postsynaptic membrane instead of the presynaptic knob — the antibodies act on the knob.

Things to Be Careful About

• Mark scheme point 7 mentions "less $\text{Ca}^{2+}$ released from the SR / less binding of $\text{Ca}^{2+}$ to troponin / fewer sites exposed / fewer cross bridges" — this is an optional downstream point linking the synaptic effect back to the contraction events in part (b). It is acceptable but not required.
• Use precise terminology: "sarcolemma", not "postsynaptic membrane"; "presynaptic knob", not "axon terminal" alone; "exocytosis" for vesicle release.
• "Ignore postsynaptic membrane" in the mark scheme — the antibodies do not bind there.

Background Concept

The IUCN Red List categorises species by extinction risk. "Critically endangered" sits just below "extinct in the wild" — these populations face an extremely high risk of extinction in the immediate future. Species become critically endangered when one or more of their vital rates (births, deaths, dispersal) are pushed out of balance, typically by human activity.

The red ruffed lemur (Varecia rubra) is endemic to the Masoala peninsula of north-east Madagascar. Madagascar has been isolated for ~150 million years, so its lemurs have evolved without close relatives elsewhere and many are highly specialised to particular forest niches. Endemic species with small ranges are especially vulnerable because they have nowhere else to go when their habitat is altered.

Understanding the Question

The stem establishes that Varecia rubra is found only in the Masoala rainforests and is listed as critically endangered. You are asked to suggest reasons — that is, to come up with credible biological explanations that could account for the decline. You need three distinct points.

Approach

Think of the major categories of threat that drive species towards extinction, and apply each to a small-range rainforest mammal in Madagascar:

• Habitat-related threats: deforestation, fragmentation, degradation
• Direct exploitation: hunting, poaching, illegal trade
• Environmental change: climate change, altered rainfall
• Biological threats: new diseases, invasive predators/competitors

Pick the three you can support with the most specific detail.

Step-by-Step Reasoning

1. Habitat loss — Madagascar has lost a large proportion of its primary forest to slash-and-burn agriculture ("tavy"), commercial logging, charcoal production and mining. The Masoala region has experienced significant deforestation, which removes the fruiting trees and canopy the lemurs depend on for food and shelter.
2. Hunting / poaching — Lemurs are hunted for bushmeat in parts of Madagascar, and live animals are illegally captured for the exotic pet trade. Distinctive, brightly coloured primates like V. rubra are particularly sought-after.
3. Climate change — Madagascar's climate is becoming drier and cyclones are becoming more frequent and severe. Both reduce forest productivity and fruiting, causing food shortages.
4. Competition for food / resources — As human populations expand into forest margins, they harvest the same wild fruits the lemurs depend on, depleting food supply.
5. New disease — Contact with humans, domestic animals or invasive species exposes lemurs to pathogens they have not co-evolved with, against which they have little immunity.
6. Illegal pet / fur trade — Specific to attractive species like V. rubra; wild-caught individuals are sold on international markets.

Any three of these earn full marks.

Key Takeaways

• Critically endangered species usually face multiple, interacting threats — not just one.
• Endemic species with restricted ranges are the most vulnerable because they cannot migrate to alternative habitats.
• Madagascar's lemurs are the most threatened mammal group on Earth; habitat loss and hunting are the dominant drivers.

Common Mistakes

• Writing vague answers such as "human activity" without specifying which activity.
• Repeating the same idea in different words (e.g. "deforestation" and "habitat loss" together) — only one would be credited.
• Failing to make answers specific to a rainforest mammal in Madagascar.

Things to Be Careful About

• "Suggest" questions accept any biologically credible answer; the mark scheme lists the most common ones but is not exhaustive.
• Each marking point must be a distinct idea, not a paraphrase of another.
• Generic phrases such as "loss of biodiversity" or "extinction risk" earn no credit — name the mechanism.

Background Concept

Captive breeding programmes are run by zoos worldwide as insurance against extinction in the wild, and as a source of individuals for reintroduction programmes. Studbooks, coordinated by bodies such as WAZA (World Association of Zoos and Aquariums) and the regional zoo associations, track the pedigree of every individual in a managed population and recommend pairings to maintain genetic diversity.

Graphs of population data are a standard way to monitor programme success over time. When describing such a graph you should describe:

1. the overall trend (general direction of change);
2. any key transitions (sharp rises/falls, plateaus);
3. a specific data quote to support your description.

Understanding the Question

Fig. 9.2 shows a line graph of the number of captive-born red ruffed lemurs held in North American zoos from 1970 to 2020. You are asked to describe the results — i.e. state what the graph shows, supported with data.

Approach

Identify the shape of the curve segment by segment:

• 1970 → 1980: small, slow rise (from 5 to about 32)
• 1980 → 1990: very steep rise (from 32 to 175) — the most prominent feature
• 1990 → 2010: roughly flat with a small rise to a peak around 2010
• 2010 → 2020: small decrease

Then select a specific data quote that supports one of these features.

Step-by-Step Reasoning

The most descriptive points are:

1. Trend 1 (the dominant feature): The number of captive-born lemurs increased very steeply/rapidly between 1980 and 1990.

2. Trend 2 (the later phase): From 1990 onwards the number remained roughly constant / plateaued, with only small fluctuations — a slight rise to a peak around 2010, then a small decrease by 2020.

3. Data quote — must give two numbers with their years, e.g.:

   • "In 1980 there were about 32 captive-born lemurs; by 1990 this had risen to 175."
   • "In 1990 there were 175 lemurs; in 2000 the figure was unchanged at 175."
   • "At the 2010 peak there were about 185; by 2020 this had fallen slightly to 170."

Each marking point requires its own clear, separate statement.

Key Takeaways

• For "describe a graph" questions: trend → key transition → data quote is the reliable structure for full marks.
• Always pair each number with its year and the unit ("number of captive-born lemurs").
• Comparative language ("increased", "plateaued", "fluctuated", "decreased") is more informative than vague terms like "went up a lot".

Common Mistakes

• Quoting only one number — the mark scheme requires two numbers and two years.
• Only mentioning the start and end values without describing the shape in between.
• Confusing this captive-born dataset with the wild population — the graph is specifically captive-born individuals.

Things to Be Careful About

• The data values from the mark scheme are 1970 = 5, 1980 = 32/33, 1990 = 175, 2000 = 175, 2010 = 184/185, 2020 = 170.
• "Plateau" or "remains constant" is the technical term for the flat region — using it scores the second mark more clearly than "stays the same".

Background Concept

Homeostasis is the maintenance of a constant internal environment, even when the external environment changes. For any factor that is regulated (e.g. blood glucose concentration, body temperature, water potential), the body uses the same basic pathway:

• A receptor detects a deviation from the set point (the norm).
• A coordinator (usually part of the central nervous system — the brain, or specifically the hypothalamus for many factors) receives the information and decides on a response.
• A signal is sent to an effector (a muscle or a gland), which carries out a response.
• The response returns the factor to its set point through negative feedback — i.e. the response opposes the original deviation, so the system is switched off once the factor is back to normal.

Negative feedback is the key principle: any rise triggers mechanisms to bring the factor down; any fall triggers mechanisms to bring it up. The factor therefore oscillates around the set point rather than staying perfectly constant.

Understanding the Question

This is a generic 'principles of homeostasis' question. The candidate must give the standard pathway and the idea of negative feedback. Four marks are available, so four clear points are needed. The question stem specifically mentions blood glucose, but the marks are awarded for the general principle, not the glucose-specific example.

Approach

State the pathway in order: stimulus → receptor → coordinator → signal → effector → response. Then add the negative feedback idea — that the response returns the factor to the set point and is then switched off.

Step-by-Step Reasoning

1. Receptor — a change in the internal factor (e.g. a rise in blood glucose) is detected by a specific receptor cell.
2. Coordinator — information is sent to a coordinator, most commonly the brain / CNS / hypothalamus. This decides on an appropriate response.
3. Signal to effector — impulses (along neurones) and/or a hormone (in the blood) are sent to the effector.
4. Effector response — the effector (a muscle or a gland) carries out a response that opposes the original change.
5. Return to set point — the factor is returned to its set point / norm.
6. Negative feedback — because the response opposes the deviation, the system is self-correcting: once the factor is back to normal, the response is switched off. This is what makes homeostatic control stable.

Key Takeaways

• Homeostasis is the maintenance of a constant internal environment.
• The general pathway is: receptor → coordinator → effector.
• Negative feedback returns the factor to its set point and is the key principle that makes the system stable.

Common Mistakes

• Confusing receptor with effector — a receptor detects; an effector acts.
• Saying 'returns the factor to normal' without naming the mechanism (negative feedback) — both points are needed.
• Forgetting to mention the coordinator — the pathway is not receptor → effector directly; information must be processed.
• Writing 'the body responds' without specifying the type of response or the effector — a vague statement does not earn the mark.

Things to Be Careful About

• The mark scheme credits the idea that the response returns the factor to the set point, not just that a response occurs.
• 'Negative feedback' is a specific term and must be used (or described) for the final mark.
• A well-prepared answer mentions both a nervous and a hormonal signal as possible links between coordinator and effector — this matches the 'or (named) hormone' alternative in the mark scheme.

Background Concept

Glucagon raises blood glucose by binding to a receptor on the surface of liver cells and triggering an intracellular second-messenger cascade:

1. Glucagon binds its transmembrane receptor.
2. A G protein on the inner face of the membrane is activated.
3. G protein activates adenylyl cyclase, which converts ATP to cyclic AMP (cAMP).
4. cAMP activates protein kinase A (PKA).
5. PKA begins a cascade that ultimately activates glycogen phosphorylase.
6. Active glycogen phosphorylase breaks down glycogen to glucose 1-phosphate (glycogenolysis), raising blood glucose.

The PYGL gene codes for glycogen phosphorylase itself. A mutation can change amino acid sequence, which alters the protein's tertiary structure and therefore the shape of its active site, so the enzyme can no longer bind / catalyse its substrate efficiently. This is GSDVI.

Understanding the Question

The candidate is told that the PYGL gene is mutated, so the enzyme it codes for is defective. The question is asking why cell signalling by glucagon is 'likely to be affected'. The key reasoning is: the early signalling steps (receptor, G protein, adenylyl cyclase, cAMP, PKA) are themselves proteins coded by other genes, so they are still functional. What is missing or non-functional is the final target of the cascade — glycogen phosphorylase. The cell-signalling events still happen, but the metabolic end-point is lost.

Approach

First say the cascade still occurs (name two steps to earn the marks). Then state the consequence of the PYGL mutation: glycogen phosphorylase is absent or non-functional, with a change in active site. End with the metabolic outcome.

Step-by-Step Reasoning

1. The cascade still happens — glucagon binds to its receptor on the liver cell, activates the G protein, which activates adenylyl cyclase, which converts ATP to cAMP, which activates protein kinase A. The mark scheme accepts any two of these details.
2. The mutation affects the target enzyme — because PYGL is mutated, the liver cell produces no / less (functioning) glycogen phosphorylase, OR the glycogen phosphorylase produced is non-functioning.
3. Why it is non-functioning — the mutation changes the tertiary structure of the enzyme, so the shape of its active site is altered and it cannot bind glycogen / catalyse glycogenolysis.
4. Consequence — little or no glycogen is converted to glucose, so glycogenolysis is impaired and blood glucose cannot be raised by this route.

Key Takeaways

• A gene mutation affects the protein that gene codes for, not the unrelated proteins upstream in a signalling cascade.
• The glucagon cAMP cascade itself is intact in GSDVI; the target enzyme is the one that is defective.
• A change in amino acid sequence alters tertiary structure and the active site, abolishing enzyme activity.

Common Mistakes

• Stating that glucagon cannot bind its receptor — incorrect, because PYGL codes for glycogen phosphorylase, not the glucagon receptor.
• Saying the cascade 'cannot happen' — wrong; the cascade proteins are coded by other genes and are unaffected.
• Forgetting to link the mutation to a change in tertiary structure / active site — this is the underlying reason the enzyme is non-functional.
• Writing vaguely that 'glucagon signalling is reduced' without specifying which step is affected and which is not.

Things to Be Careful About

• The mark scheme explicitly requires naming the consequence of the mutation (no / less / non-functioning glycogen phosphorylase) and the structural reason (changed tertiary structure / active site).
• Note that 'cell signalling' is technically still occurring — the question's wording is a prompt to recognise that the defect is in the downstream enzyme, not in the cascade itself.
• The mark scheme allows credit for the alternative that the enzyme is not activated by the cascade, even if it is structurally intact — both interpretations are acceptable.

Background Concept

After a carbohydrate-rich meal, blood glucose rises. The normal response:

1. The rise is detected by β cells of the islets of Langerhans in the pancreas.
2. Insulin is released.
3. Insulin stimulates glucose uptake by muscle and adipose tissue, and in the liver it activates glycogen synthase, which converts glucose to glycogen for storage (glycogenesis).
4. As blood glucose returns to normal, insulin release falls.
5. The other key regulatory hormone is glucagon, released by α cells when blood glucose is low; glucagon raises blood glucose via glycogenolysis and gluconeogenesis.

Glycogen synthase is coded by GYS2. A mutation in GYS2 means the liver cannot efficiently convert glucose to glycogen, so the body's main short-term storage route for excess glucose is impaired (GSD0).

Understanding the Question

The question is 'suggest' — so the candidate is expected to reason through the consequences of the enzyme defect, not just state one. Three marks require three distinct, sensible points about what would happen if a person with GSD0 ate a glucose-rich meal.

Approach

Start with the immediate effect (glycogen cannot be stored), then think about where the excess glucose goes: it stays in the blood, spills into the urine, or is shunted into other pathways such as lipid synthesis. Finally, mention the hormonal feedback consequence (glucagon release is inhibited by high blood glucose).

Step-by-Step Reasoning

1. The defect — GYS2 is mutated, so the liver produces no / less (functioning) glycogen synthase, or the enzyme produced is non-functional. Glycogen cannot be made from the excess glucose.
2. Blood glucose remains high — without the glycogen buffer, blood glucose concentration stays elevated after the meal.
3. Renal threshold exceeded — once blood glucose exceeds the kidney's transport maximum (~10 mmol dm⁻³), glucose appears in the urine (glycosuria), drawing water with it by osmosis (polyuria, dehydration, thirst).
4. Lipid synthesis — excess glucose is diverted into lipogenesis (conversion to fatty acids and then triglycerides in the liver and adipose tissue), potentially leading to raised blood lipid levels and fat deposition.
5. Hormonal feedback — the high blood glucose inhibits glucagon release from α cells (negative feedback). Because both glycogenolysis (in part b reasoning) and glycogenesis are impaired, the person has limited ability to control blood glucose either up or down by these routes.
6. Other consequences — the mark scheme accepts an AVP such as: affects blood water potential (leading to dehydration, thirst, raised blood pressure), or produces long-term effects (tiredness, coma in severe cases, vascular damage similar to diabetes mellitus).

Key Takeaways

• Glycogen is the body's main short-term store of excess glucose from meals.
• If glycogen synthase is defective, the meal-derived glucose cannot be buffered by the liver.
• The consequences mirror several features of diabetes mellitus: hyperglycaemia, glycosuria, polyuria, raised lipid synthesis, dehydration.
• Loss of the glycogen storage route does not lower blood glucose on its own — it just removes the buffering capacity.

Common Mistakes

• Writing 'blood glucose will decrease' — wrong; with no glycogen storage, the excess glucose stays in the blood.
• Saying 'glucose will not be absorbed' — wrong; absorption in the gut is independent of the liver's storage enzyme.
• Missing the link to lipid synthesis or urine — the question asks for consequences, so several distinct points are needed, not just the hyperglycaemia.
• Confusing the direction of glucagon signalling — high blood glucose inhibits glucagon release (negative feedback); do not write the opposite.

Things to Be Careful About

• 'Suggest' questions require plausible reasoning, not absolute certainty — but the reasoning must still be biologically sound (e.g. linking hyperglycaemia to glycosuria is acceptable; linking it to sudden weight loss is not, since lipid synthesis would tend to increase mass).
• The mark scheme accepts AVP answers such as 'dehydration', 'thirst', 'raised blood pressure' or 'affects blood water potential' — any of these can be used as a third mark if the first three are stronger.
• The answer should be specifically about a person with GSD0, not a generic description of diabetes — phrase each point in the context of the enzyme defect.