Biology HL

1. 1.

   Outline the role of the enzyme DNA helicase and the enzyme DNA polymerase during DNA replication.

   [3 marks] · no calculator

   Worked explanation

   1. Helicase separates paired DNA strands by disrupting hydrogen bonding, exposing two templates rather than cutting the covalent backbone.
   2. DNA polymerase extends a primer by adding complementary nucleotides to the new strand's 3' end, so synthesis proceeds 5' to 3'.
   3. Because templates are antiparallel, one new strand can follow the fork continuously while the other is made in separately initiated Okazaki fragments.

   Common mistake: Polymerase cannot solve antiparallel geometry by synthesising one new strand 3' to 5'; both new strands grow 5' to 3'.

   Marking points

   • States that DNA helicase unwinds the double helix and separates the two strands by breaking the hydrogen bonds between complementary base pairs.
   • States that DNA polymerase synthesizes a new complementary strand by adding free nucleotides that pair with the exposed bases on the template strand.
   • States that DNA polymerase can only add nucleotides in the 5′ to 3′ direction, which is why replication proceeds continuously on one strand and discontinuously (in fragments) on the other.

   Examiner tip: The 5′-to-3′ restriction on DNA polymerase is the key reason one new strand (leading) is synthesized continuously and the other (lagging) is synthesized as Okazaki fragments.

2. 2.

   Marking analysis: A learner attempts the following task: “Outline the role of the enzyme DNA helicase and the enzyme DNA polymerase during DNA replication.” Their response addresses only this point: “States that DNA helicase unwinds the double helix and separates the two strands by breaking the hydrogen bonds between complementary base pairs.” Evaluate the response against the complete 3-mark task. Identify what earns credit and state every additional requirement needed for full marks.

   [3 marks] · no calculator

   Marking points

   • Recognises credit for the stated point: States that DNA helicase unwinds the double helix and separates the two strands by breaking the hydrogen bonds between complementary base pairs.
   • Checks this criterion, crediting equivalent work already shown or identifying what is missing: States that DNA polymerase synthesizes a new complementary strand by adding free nucleotides that pair with the exposed bases on the template strand.
   • Checks this criterion, crediting equivalent work already shown or identifying what is missing: States that DNA polymerase can only add nucleotides in the 5′ to 3′ direction, which is why replication proceeds continuously on one strand and discontinuously (in fragments) on the other.

   Examiner tip: Check each criterion, including equivalent work already shown. Do not demand an extra step merely to rewrite an equivalent answer. Do not award the same idea twice or infer work the learner did not show.

3. 3.

   Explain the role of DNA ligase in joining Okazaki fragments on the lagging strand during replication.

   [2 marks] · no calculator

   Worked explanation

   1. Repeated primers initiate short stretches of lagging-strand synthesis because polymerase can only extend 5' to 3'.
   2. After primer removal and replacement, adjacent DNA stretches still have backbone nicks. DNA ligase seals them by forming covalent phosphodiester links, producing a continuous strand.

   Common mistake: Ligase seals sugar-phosphate-backbone nicks; it does not join complementary strands by adding hydrogen bonds between bases.

   Marking points

   • States that the lagging strand is synthesized discontinuously as a series of short Okazaki fragments, each begun from a separate primer.
   • States that DNA ligase joins (seals) the gaps between adjacent Okazaki fragments, forming one continuous strand.

   Examiner tip: DNA ligase is also the enzyme used in genetic engineering to join fragments of DNA from different sources, making use of its natural strand-joining function.

4. 4.

   Marking analysis: A learner attempts the following task: “Explain the role of DNA ligase in joining Okazaki fragments on the lagging strand during replication.” Their response addresses only this point: “States that the lagging strand is synthesized discontinuously as a series of short Okazaki fragments, each begun from a separate primer.” Evaluate the response against the complete 2-mark task. Identify what earns credit and state every additional requirement needed for full marks.

   [2 marks] · no calculator

   Marking points

   • Recognises credit for the stated point: States that the lagging strand is synthesized discontinuously as a series of short Okazaki fragments, each begun from a separate primer.
   • Checks this criterion, crediting equivalent work already shown or identifying what is missing: States that DNA ligase joins (seals) the gaps between adjacent Okazaki fragments, forming one continuous strand.

   Examiner tip: Check each criterion, including equivalent work already shown. Do not demand an extra step merely to rewrite an equivalent answer. Do not award the same idea twice or infer work the learner did not show.

5. 5.

   Outline the process of post-transcriptional modification of pre-mRNA in eukaryotic cells, including splicing.

   [3 marks] · no calculator

   Worked explanation

   1. A eukaryotic primary transcript can contain introns between exons. Splicing removes introns and joins retained exon sequences.
   2. A 5' cap and poly-A tail help transcript stability, export and translation, producing a processed mRNA rather than leaving the raw transcript unchanged.
   3. Exons include retained untranslated regions as well as coding sequence, so retained does not mean every exon base specifies an amino acid.

   Common mistake: Calling every exon coding and every non-coding base an intron misclassifies untranslated exonic regions.

   Marking points

   • States that many eukaryotic pre-mRNAs contain introns and exons; exons are retained in mature mRNA and can include untranslated regions as well as protein-coding sequence.
   • States that during splicing, introns are removed and exons are joined together to form the mature mRNA.
   • States that a 5′ cap and a poly-A tail are also added to the pre-mRNA, which help protect the mRNA and assist in its export from the nucleus and translation.

   Examiner tip: Only mature mRNA (after splicing and processing) leaves the nucleus to be translated at the ribosome — the primary transcript itself is never translated directly.

6. 6.

   Marking analysis: A learner attempts the following task: “Outline the process of post-transcriptional modification of pre-mRNA in eukaryotic cells, including splicing.” Their response addresses only this point: “States that many eukaryotic pre-mRNAs contain introns and exons; exons are retained in mature mRNA and can include untranslated regions as well as protein-coding sequence.” Evaluate the response against the complete 3-mark task. Identify what earns credit and state every additional requirement needed for full marks.

   [3 marks] · no calculator

   Marking points

   • Recognises credit for the stated point: States that many eukaryotic pre-mRNAs contain introns and exons; exons are retained in mature mRNA and can include untranslated regions as well as protein-coding sequence.
   • Checks this criterion, crediting equivalent work already shown or identifying what is missing: States that during splicing, introns are removed and exons are joined together to form the mature mRNA.
   • Checks this criterion, crediting equivalent work already shown or identifying what is missing: States that a 5′ cap and a poly-A tail are also added to the pre-mRNA, which help protect the mRNA and assist in its export from the nucleus and translation.

   Examiner tip: Check each criterion, including equivalent work already shown. Do not demand an extra step merely to rewrite an equivalent answer. Do not award the same idea twice or infer work the learner did not show.

7. 7.

   Explain how alternative splicing of the same pre-mRNA transcript can produce different protein products.

   [3 marks] · no calculator

   Worked explanation

   1. One pre-mRNA can be spliced into different retained exon combinations, for example 1-2-4 versus 1-3-4 when those alternatives are compatible.
   2. These mature RNAs carry different coding sequences without requiring a change to the original DNA gene.
   3. Translation can produce proteins with different amino-acid segments and functions, explaining how one gene can yield multiple related products. Not every hypothetical exon combination necessarily gives a functional protein.

   Common mistake: Alternative splicing rearranges or selects RNA segments; it is not a new DNA mutation in each cell.

   Marking points

   • States that alternative splicing means different combinations of exons can be joined together (or some exons excluded) from the same pre-mRNA transcript.
   • States that each different combination of exons produces a different mature mRNA sequence.
   • States that translation of these different mRNA sequences produces different (though related) protein products from a single gene.

   Examiner tip: Alternative splicing explains how the human genome, with a relatively limited number of genes, can produce a much larger number of distinct proteins.

8. 8.

   Marking analysis: A learner attempts the following task: “Explain how alternative splicing of the same pre-mRNA transcript can produce different protein products.” Their response addresses only this point: “States that alternative splicing means different combinations of exons can be joined together (or some exons excluded) from the same pre-mRNA transcript.” Evaluate the response against the complete 3-mark task. Identify what earns credit and state every additional requirement needed for full marks.

   [3 marks] · no calculator

   Marking points

   • Recognises credit for the stated point: States that alternative splicing means different combinations of exons can be joined together (or some exons excluded) from the same pre-mRNA transcript.
   • Checks this criterion, crediting equivalent work already shown or identifying what is missing: States that each different combination of exons produces a different mature mRNA sequence.
   • Checks this criterion, crediting equivalent work already shown or identifying what is missing: States that translation of these different mRNA sequences produces different (though related) protein products from a single gene.

   Examiner tip: Check each criterion, including equivalent work already shown. Do not demand an extra step merely to rewrite an equivalent answer. Do not award the same idea twice or infer work the learner did not show.

9. 9.

   Outline the role of transcription factors in controlling gene expression in eukaryotic cells.

   [3 marks] · no calculator

   Worked explanation

   1. A transcription factor binds a specific regulatory DNA sequence, such as a promoter-associated site or enhancer.
   2. It can help recruit transcription machinery or inhibit it, changing the rate at which RNA is made rather than directly translating protein.
   3. Different cells contain different factors and signals. Consequently a liver cell and neuron can express different genes despite broadly sharing the same genome.

   Common mistake: An enhancer can act from a distant DNA site; regulatory binding need not mean the factor physically sits in the coding sequence.

   Marking points

   • States that transcription factors are proteins that bind to specific regulatory DNA sequences (e.g. promoters or enhancers) near a gene.
   • States that binding of a transcription factor can either promote or inhibit the binding of RNA polymerase to the gene, thereby increasing or decreasing the rate of transcription.
   • States that different cell types express different sets of transcription factors, which is one mechanism explaining how cells with identical DNA can express different genes and differentiate.

   Examiner tip: Transcription factor control is a key mechanism of differential gene expression — it explains how a liver cell and a neuron, with identical DNA, end up with very different sets of active proteins.

10. 10.

    Marking analysis: A learner attempts the following task: “Outline the role of transcription factors in controlling gene expression in eukaryotic cells.” Their response addresses only this point: “States that transcription factors are proteins that bind to specific regulatory DNA sequences (e.g. promoters or enhancers) near a gene.” Evaluate the response against the complete 3-mark task. Identify what earns credit and state every additional requirement needed for full marks.

    [3 marks] · no calculator

    Marking points

    • Recognises credit for the stated point: States that transcription factors are proteins that bind to specific regulatory DNA sequences (e.g. promoters or enhancers) near a gene.
    • Checks this criterion, crediting equivalent work already shown or identifying what is missing: States that binding of a transcription factor can either promote or inhibit the binding of RNA polymerase to the gene, thereby increasing or decreasing the rate of transcription.
    • Checks this criterion, crediting equivalent work already shown or identifying what is missing: States that different cell types express different sets of transcription factors, which is one mechanism explaining how cells with identical DNA can express different genes and differentiate.

    Examiner tip: Check each criterion, including equivalent work already shown. Do not demand an extra step merely to rewrite an equivalent answer. Do not award the same idea twice or infer work the learner did not show.

11. 11.

    Outline how DNA replication is described as a semi-conservative and bidirectional process at a replication origin.

    [2 marks] · no calculator

    Course scope: Optional enrichment

    Syllabus placement evidence: D1.1.1-D1.1.9, printed pp.85-86. Semi-conservative replication and strand directionality are prescribed, but the required origin/bubble/bidirectional-fork mechanism is not listed. The whole mixed task is enrichment because the extra mechanism is required.

    Worked explanation

    1. An origin initiates a replication bubble with forks moving away in opposite directions; bidirectional describes fork movement.
    2. Each daughter double helix retains one parental strand and receives one newly synthesised partner; semi-conservative describes strand composition.
    3. These are independent descriptions: multiple origins and two outward forks can speed chromosome copying while every newly extended strand still grows 5' to 3'.

    Common mistake: Bidirectional replication does not mean DNA polymerase synthesises in both chemical directions.

    Marking points

    • Explains that semi-conservative replication produces daughter DNA molecules each containing one parental strand and one newly synthesised strand; replication is initiated at an origin.
    • States that replication proceeds in both directions away from the origin simultaneously (bidirectionally), speeding up the overall replication of the DNA molecule.

    Examiner tip: Bidirectional replication from multiple origins along a eukaryotic chromosome is essential — a single origin replicating in only one direction would take far too long to copy an entire chromosome.

12. 12.

    Marking analysis: A learner attempts the following task: “Outline how DNA replication is described as a semi-conservative and bidirectional process at a replication origin.” Their response addresses only this point: “Explains that semi-conservative replication produces daughter DNA molecules each containing one parental strand and one newly synthesised strand; replication is initiated at an origin.” Evaluate the response against the complete 2-mark task. Identify what earns credit and state every additional requirement needed for full marks.

    [2 marks] · no calculator

    Marking points

    • Recognises credit for the stated point: Explains that semi-conservative replication produces daughter DNA molecules each containing one parental strand and one newly synthesised strand; replication is initiated at an origin.
    • Checks this criterion, crediting equivalent work already shown or identifying what is missing: States that replication proceeds in both directions away from the origin simultaneously (bidirectionally), speeding up the overall replication of the DNA molecule.

    Examiner tip: Check each criterion, including equivalent work already shown. Do not demand an extra step merely to rewrite an equivalent answer. Do not award the same idea twice or infer work the learner did not show.

13. 13.

    Explain why a single-base substitution within an intron, away from splice sites and known regulatory sequences, may be less likely to alter a protein's amino acid sequence than a substitution in its protein-coding sequence.

    [3 marks] · no calculator

    Worked explanation

    1. Under the specified normal-splicing assumption, the intronic base is removed from pre-mRNA, so it is not directly read as part of a translated codon.
    2. A coding substitution remains in mRNA and can change an amino acid or create a stop codon, although degeneracy makes some substitutions synonymous.
    3. The comparison excludes splice sites and known regulatory sequences deliberately: mutations there can strongly alter expression, so non-coding is not synonymous with unimportant.

    Common mistake: An intronic position does not guarantee no effect; the prompt's location and splicing assumptions are necessary to the comparison.

    Marking points

    • States that introns are removed from the pre-mRNA during splicing and are not present in the mature mRNA that is translated.
    • States that the protein-coding sequence retained in mature mRNA determines amino acid sequence, so substitutions there can be missense or nonsense, although some are synonymous.
    • Explains that an intronic substitution removed by normal splicing does not directly change translated codons, whereas some coding substitutions do; neither position alone guarantees a phenotypic effect.

    Examiner tip: Non-coding does not mean non-functional. Regulatory or splice-site mutations can strongly affect expression, and exons can contain untranslated regions. The comparison assumes normal splicing and excludes those regulatory positions.

14. 14.

    Marking analysis: A learner attempts the following task: “Explain why a single-base substitution within an intron, away from splice sites and known regulatory sequences, may be less likely to alter a protein's amino acid sequence than a substitution in its protein-coding sequence.” Their response addresses only this point: “States that introns are removed from the pre-mRNA during splicing and are not present in the mature mRNA that is translated.” Evaluate the response against the complete 3-mark task. Identify what earns credit and state every additional requirement needed for full marks.

    [3 marks] · no calculator

    Marking points

    • Recognises credit for the stated point: States that introns are removed from the pre-mRNA during splicing and are not present in the mature mRNA that is translated.
    • Checks this criterion, crediting equivalent work already shown or identifying what is missing: States that the protein-coding sequence retained in mature mRNA determines amino acid sequence, so substitutions there can be missense or nonsense, although some are synonymous.
    • Checks this criterion, crediting equivalent work already shown or identifying what is missing: Explains that an intronic substitution removed by normal splicing does not directly change translated codons, whereas some coding substitutions do; neither position alone guarantees a phenotypic effect.

    Examiner tip: Check each criterion, including equivalent work already shown. Do not demand an extra step merely to rewrite an equivalent answer. Do not award the same idea twice or infer work the learner did not show.

15. 15.

    Outline the role of proofreading by DNA polymerase in maintaining the accuracy of DNA replication.

    [2 marks] · no calculator

    Syllabus placement evidence: D1.1.9 (AHL), printed p.86. DNA polymerase III proofreading, removal of a terminal mismatched nucleotide and correct replacement are explicitly prescribed at HL; no separate detailed repair pathway is required here.

    Worked explanation

    1. A mismatched nucleotide at the newly extended end distorts normal pairing and can be detected by a proofreading polymerase.
    2. Its proofreading activity removes the incorrect nucleotide, permitting correct incorporation and resumed extension.
    3. This lowers copying error frequency but does not eliminate all mutations; separate repair processes also contribute to genome fidelity.

    Common mistake: Proofreading is not recognition that a mutation would be harmful; it responds to molecular mismatches rather than an organism's future needs.

    Marking points

    • States that DNA polymerase can detect when an incorrectly paired (mismatched) nucleotide has been added to the new strand.
    • States that DNA polymerase removes the incorrect nucleotide and replaces it with the correct one, greatly reducing the error rate of DNA replication.

    Examiner tip: Proofreading is one of several mechanisms (along with separate DNA repair systems) that keep the overall mutation rate during replication extremely low.

16. 16.

    Marking analysis: A learner attempts the following task: “Outline the role of proofreading by DNA polymerase in maintaining the accuracy of DNA replication.” Their response addresses only this point: “States that DNA polymerase can detect when an incorrectly paired (mismatched) nucleotide has been added to the new strand.” Evaluate the response against the complete 2-mark task. Identify what earns credit and state every additional requirement needed for full marks.

    [2 marks] · no calculator

    Marking points

    • Recognises credit for the stated point: States that DNA polymerase can detect when an incorrectly paired (mismatched) nucleotide has been added to the new strand.
    • Checks this criterion, crediting equivalent work already shown or identifying what is missing: States that DNA polymerase removes the incorrect nucleotide and replaces it with the correct one, greatly reducing the error rate of DNA replication.

    Examiner tip: Check each criterion, including equivalent work already shown. Do not demand an extra step merely to rewrite an equivalent answer. Do not award the same idea twice or infer work the learner did not show.

17. 17.

    Outline how the enzyme reverse transcriptase is used in genetic engineering to produce complementary DNA (cDNA) from an mRNA template.

    [3 marks] · no calculator

    Course scope: Optional enrichment

    Syllabus placement evidence: D1.2.14-D1.2.16, printed p.88; D1.3.8-D1.3.10, p.90. Introns, splicing and gene editing are prescribed, but producing cDNA from mature mRNA with reverse transcriptase is not among those methods. Required reverse-transcription workflow exceeds merely applying intron removal.

    Worked explanation

    1. Isolate mature mRNA for the expressed gene and use reverse transcriptase to make complementary DNA from that RNA template.
    2. Generate a second DNA strand to obtain double-stranded cDNA. Because the source mRNA was spliced, its cDNA lacks the corresponding introns.
    3. With suitable expression controls, this provides coding sequence a bacterium can use without having to splice human introns. Other engineering steps and protein processing may still be needed.

    Common mistake: Copying genomic human DNA directly does not remove introns; mature-mRNA templating is what gives cDNA this advantage.

    Marking points

    • States that reverse transcriptase is an enzyme that synthesizes a strand of DNA using an RNA molecule (e.g. mRNA) as its template.
    • States that the resulting single-stranded DNA is complementary to the mRNA template, and is then made double-stranded (by DNA polymerase) to form cDNA.
    • States that because cDNA is made from mature mRNA, it lacks introns, making it useful for expressing eukaryotic genes in prokaryotic organisms (which cannot splice introns) during genetic engineering.

    Examiner tip: Reverse transcriptase copies RNA into DNA, enabling cDNA preparation and forming part of retroviral replication. It reverses transcription, not the central dogma's prohibition on transfer of sequence information out of protein; most RNA viruses do not use reverse transcription.

18. 18.

    Marking analysis: A learner attempts the following task: “Outline how the enzyme reverse transcriptase is used in genetic engineering to produce complementary DNA (cDNA) from an mRNA template.” Their response addresses only this point: “States that reverse transcriptase is an enzyme that synthesizes a strand of DNA using an RNA molecule (e.g. mRNA) as its template.” Evaluate the response against the complete 3-mark task. Identify what earns credit and state every additional requirement needed for full marks.

    [3 marks] · no calculator

    Marking points

    • Recognises credit for the stated point: States that reverse transcriptase is an enzyme that synthesizes a strand of DNA using an RNA molecule (e.g. mRNA) as its template.
    • Checks this criterion, crediting equivalent work already shown or identifying what is missing: States that the resulting single-stranded DNA is complementary to the mRNA template, and is then made double-stranded (by DNA polymerase) to form cDNA.
    • Checks this criterion, crediting equivalent work already shown or identifying what is missing: States that because cDNA is made from mature mRNA, it lacks introns, making it useful for expressing eukaryotic genes in prokaryotic organisms (which cannot splice introns) during genetic engineering.

    Examiner tip: Check each criterion, including equivalent work already shown. Do not demand an extra step merely to rewrite an equivalent answer. Do not award the same idea twice or infer work the learner did not show.

19. 19.

    Distinguish between the structure and role of stamens and carpels in a flowering plant.

    [4 marks] · no calculator

    Worked explanation

    1. A stamen has an anther supported by a filament. The anther produces pollen, whose male gametes contribute to fertilisation.
    2. A carpel includes stigma, style and ovary: the stigma receives pollen, the style provides a pollen-tube route and ovules in the ovary contain the female gametes.
    3. Pollination is pollen transfer, while fertilisation is gamete fusion; afterwards ovules develop into seeds and the ovary can develop into fruit.

    Common mistake: Pollen landing on the stigma is not already fertilisation; the male gamete must reach the ovule.

    Marking points

    • States that a stamen consists of an anther (which produces pollen containing the male gametes) supported on a filament.
    • States that a carpel consists of a stigma, style and ovary; the stigma receives pollen and the ovary contains ovules with the female gametes.
    • States that the stamen is the male reproductive structure, producing and releasing pollen.
    • States that the carpel is the female reproductive structure, where fertilisation and seed development occur after pollination.

    Examiner tip: Structure and function must be linked for each organ, not listed separately — state what each part is made of and what it does.

20. 20.

    Marking analysis: A learner attempts the following task: “Distinguish between the structure and role of stamens and carpels in a flowering plant.” Their response addresses only this point: “States that a stamen consists of an anther (which produces pollen containing the male gametes) supported on a filament.” Evaluate the response against the complete 4-mark task. Identify what earns credit and state every additional requirement needed for full marks.

    [4 marks] · no calculator

    Marking points

    • Recognises credit for the stated point: States that a stamen consists of an anther (which produces pollen containing the male gametes) supported on a filament.
    • Checks this criterion, crediting equivalent work already shown or identifying what is missing: States that a carpel consists of a stigma, style and ovary; the stigma receives pollen and the ovary contains ovules with the female gametes.
    • Checks this criterion, crediting equivalent work already shown or identifying what is missing: States that the stamen is the male reproductive structure, producing and releasing pollen.
    • Checks this criterion, crediting equivalent work already shown or identifying what is missing: States that the carpel is the female reproductive structure, where fertilisation and seed development occur after pollination.

    Examiner tip: Check each criterion, including equivalent work already shown. Do not demand an extra step merely to rewrite an equivalent answer. Do not award the same idea twice or infer work the learner did not show.

21. 21.

    Explain two ways in which meiosis generates genetic variation among gametes.

    [4 marks] · no calculator

    Worked explanation

    1. In prophase I, corresponding segments can cross between non-sister chromatids of homologous chromosomes, producing new linked allele combinations along chromatids.
    2. At metaphase I, each homologous pair's orientation is random relative to other pairs, so segregation generates different combinations of maternal and paternal chromosomes.
    3. The first mechanism recombines segments within chromosomes; the second reshuffles chromosome combinations. Neither requires creating new alleles by mutation.

    Common mistake: Crossing over normally concerns non-sister chromatids of homologues, not arbitrary non-homologous chromosomes.

    Marking points

    • States that crossing over exchanges corresponding DNA segments between non-sister chromatids of homologous chromosomes during prophase I.
    • Explains that chromatids participating in crossing over can acquire new combinations of maternal and paternal alleles along a chromosome.
    • States that independent assortment occurs when homologous pairs align randomly at the metaphase I plate, so each pair segregates independently of other pairs.
    • States that this produces many different combinations of maternal and paternal chromosomes in the resulting gametes.

    Examiner tip: Crossing over creates new allele combinations within a single chromosome; independent assortment creates new combinations of whole chromosomes — keep these two mechanisms clearly separated.

22. 22.

    Marking analysis: A learner attempts the following task: “Explain two ways in which meiosis generates genetic variation among gametes.” Their response addresses only this point: “States that crossing over exchanges corresponding DNA segments between non-sister chromatids of homologous chromosomes during prophase I.” Evaluate the response against the complete 4-mark task. Identify what earns credit and state every additional requirement needed for full marks.

    [4 marks] · no calculator

    Marking points

    • Recognises credit for the stated point: States that crossing over exchanges corresponding DNA segments between non-sister chromatids of homologous chromosomes during prophase I.
    • Checks this criterion, crediting equivalent work already shown or identifying what is missing: Explains that chromatids participating in crossing over can acquire new combinations of maternal and paternal alleles along a chromosome.
    • Checks this criterion, crediting equivalent work already shown or identifying what is missing: States that independent assortment occurs when homologous pairs align randomly at the metaphase I plate, so each pair segregates independently of other pairs.
    • Checks this criterion, crediting equivalent work already shown or identifying what is missing: States that this produces many different combinations of maternal and paternal chromosomes in the resulting gametes.

    Examiner tip: Check each criterion, including equivalent work already shown. Do not demand an extra step merely to rewrite an equivalent answer. Do not award the same idea twice or infer work the learner did not show.

23. 23.

    In a hypothetical sweet-pea cross, purple flower colour P is completely dominant to red p, and long pollen L to round l. The loci are linked. Assume a 10% recombination frequency, approximately 10 map units. A PL/pl plant is test-crossed with pl/pl, with equal viability of offspring classes. State the expected phenotype proportions and explain why they differ from the 1:1:1:1 ratio for an independently assorting dihybrid test cross.

    [5 marks]

    Worked explanation

    1. The tester pl/pl contributes only pl, so offspring phenotypes reveal gametes made by the PL/pl parent.
    2. A 10% recombinant fraction leaves 90% parental gametes: PL and pl each 45%; recombinant Pl and pL each 5%, assuming equal class viability.
    3. Offspring are therefore 45% purple-long, 45% red-round, 5% purple-round and 5% red-long.
    4. Unlinked loci would give four 25% classes in a test cross, not 9:3:3:1. The map-unit interpretation is approximate; the supplied recombination fraction is the calculation's direct input.

    Common mistake: A 10% recombinant fraction is shared across both recombinant classes, not 10% for each.

    Marking points

    • States that the loci are linked and do not show the independent assortment that would give 1:1:1:1 in this test cross.
    • States that most gametes are parental types: PL and pl, each at a frequency close to 45% (since 10% total are recombinants, split between the two recombinant classes).
    • States that a minority of gametes are recombinant types: Pl and pL, each at a frequency of about 5% (half of the 10% map distance).
    • States the expected offspring ratio as approximately 45% purple long : 45% red round : 5% purple round : 5% red long.
    • Explains that only 10% of the heterozygote's gametes are recombinant under the stated assumption, so parental offspring classes predominate rather than all four classes being equally frequent.

    Examiner tip: A test cross is not an F1 dihybrid self-cross: 9:3:3:1 is not its unlinked baseline. Recombination percentage approximates map distance over short intervals; undetected multiple crossovers make the equality unreliable over longer intervals.

24. 24.

    Marking analysis: A learner attempts the following task: “In a hypothetical sweet-pea cross, purple flower colour P is completely dominant to red p, and long pollen L to round l. The loci are linked. Assume a 10% recombination frequency, approximately 10 map units. A PL/pl plant is test-crossed with pl/pl, with equal viability of offspring classes. State the expected phenotype proportions and explain why they differ from the 1:1:1:1 ratio for an independently assorting dihybrid test cross.” Their response addresses only this point: “States that the loci are linked and do not show the independent assortment that would give 1:1:1:1 in this test cross.” Evaluate the response against the complete 5-mark task. Identify what earns credit and state every additional requirement needed for full marks.

    [5 marks]

    Marking points

    • Recognises credit for the stated point: States that the loci are linked and do not show the independent assortment that would give 1:1:1:1 in this test cross.
    • Checks this criterion, crediting equivalent work already shown or identifying what is missing: States that most gametes are parental types: PL and pl, each at a frequency close to 45% (since 10% total are recombinants, split between the two recombinant classes).
    • Checks this criterion, crediting equivalent work already shown or identifying what is missing: States that a minority of gametes are recombinant types: Pl and pL, each at a frequency of about 5% (half of the 10% map distance).
    • Checks this criterion, crediting equivalent work already shown or identifying what is missing: States the expected offspring ratio as approximately 45% purple long : 45% red round : 5% purple round : 5% red long.
    • Checks this criterion, crediting equivalent work already shown or identifying what is missing: Explains that only 10% of the heterozygote's gametes are recombinant under the stated assumption, so parental offspring classes predominate rather than all four classes being equally frequent.

    Examiner tip: Check each criterion, including equivalent work already shown. Do not demand an extra step merely to rewrite an equivalent answer. Do not award the same idea twice or infer work the learner did not show.

25. 25.

    Distinguish between polygenic inheritance and epistasis, each with a brief example.

    [4 marks] · no calculator

    Course scope: Optional enrichment

    Syllabus placement evidence: D3.2.14 and D3.2.16-D3.2.20, printed pp.99-100. Polygenic variation, unlinked crosses and linkage are prescribed. Required epistatic masking is a different between-locus interaction not listed among those outcomes, making the mixed comparison enrichment.

    Worked explanation

    1. Polygenic inheritance means several loci contribute to a trait, such as human height; environmental effects can add further continuous variation.
    2. Epistasis concerns interaction: one locus changes the phenotypic expression of another. In Labradors, an ee genotype prevents dark coat-pigment deposition regardless of the B-locus brown/black variant.
    3. The terms describe different properties, number of loci versus their interaction, so a polygenic trait can also involve epistasis.

    Common mistake: Epistasis is between loci, unlike simple dominance between alleles at one locus; it is not the opposite of polygenic inheritance.

    Marking points

    • States that polygenic inheritance involves two or more loci contributing to one trait, often modelled with additive effects, although interactions between loci can also occur.
    • Gives a valid example of polygenic inheritance, such as human skin colour or height, which show continuous variation.
    • States that epistasis is when the allele of one gene masks or modifies the phenotypic expression of a different gene at another locus.
    • Gives a valid example of epistasis, such as coat colour in Labrador retrievers, where the E gene determines whether the B (colour) gene's effect is expressed at all.

    Examiner tip: Polygenic describes how many loci contribute to a trait; epistasis describes interaction between loci. These are not mutually exclusive: a polygenic trait can include both additive and epistatic effects.

26. 26.

    Marking analysis: A learner attempts the following task: “Distinguish between polygenic inheritance and epistasis, each with a brief example.” Their response addresses only this point: “States that polygenic inheritance involves two or more loci contributing to one trait, often modelled with additive effects, although interactions between loci can also occur.” Evaluate the response against the complete 4-mark task. Identify what earns credit and state every additional requirement needed for full marks.

    [4 marks] · no calculator

    Marking points

    • Recognises credit for the stated point: States that polygenic inheritance involves two or more loci contributing to one trait, often modelled with additive effects, although interactions between loci can also occur.
    • Checks this criterion, crediting equivalent work already shown or identifying what is missing: Gives a valid example of polygenic inheritance, such as human skin colour or height, which show continuous variation.
    • Checks this criterion, crediting equivalent work already shown or identifying what is missing: States that epistasis is when the allele of one gene masks or modifies the phenotypic expression of a different gene at another locus.
    • Checks this criterion, crediting equivalent work already shown or identifying what is missing: Gives a valid example of epistasis, such as coat colour in Labrador retrievers, where the E gene determines whether the B (colour) gene's effect is expressed at all.

    Examiner tip: Check each criterion, including equivalent work already shown. Do not demand an extra step merely to rewrite an equivalent answer. Do not award the same idea twice or infer work the learner did not show.

27. 27.

    A population of 500 beetles is in Hardy-Weinberg equilibrium for a gene with two alleles, A (dominant) and a (recessive). 180 beetles show the recessive phenotype. Calculate the allele frequencies of A and a, and the number of heterozygous beetles.

    [5 marks]

    Worked explanation

    1. Under complete dominance, recessive beetles are aa: q squared = 180/500 = 0.36.
    2. Take the square root: q = 0.60, so p = 1 - q = 0.40.
    3. Expected heterozygote frequency = 2pq = 2(0.40)(0.60) = 0.48. Multiply by population size: 0.48 x 500 = 240 beetles.
    4. Check AA = 0.16 x 500 = 80; 80 + 240 + 180 = 500. The calculation relies on the stated equilibrium and two-allele model.

    Common mistake: The 320 dominant-phenotype beetles include both AA and Aa; they cannot all be counted as heterozygotes.

    Marking points

    • Calculates q² = 180/500 = 0.36.
    • Calculates q = √0.36 = 0.6, so the frequency of allele a is 0.6.
    • Calculates p = 1 − q = 0.4, so the frequency of allele A is 0.4.
    • Calculates the heterozygous frequency as 2pq = 2(0.4)(0.6) = 0.48.
    • Calculates the number of heterozygous beetles as 0.48 × 500 = 240.

    Examiner tip: Always start from the recessive phenotype frequency, since it is the only phenotype that maps to a single, unambiguous genotype (aa).

28. 28.

    Marking analysis: A learner attempts the following task: “A population of 500 beetles is in Hardy-Weinberg equilibrium for a gene with two alleles, A (dominant) and a (recessive). 180 beetles show the recessive phenotype. Calculate the allele frequencies of A and a, and the number of heterozygous beetles.” Their response addresses only this point: “Calculates q² = 180/500 = 0.36.” Evaluate the response against the complete 5-mark task. Identify what earns credit and state every additional requirement needed for full marks.

    [5 marks]

    Marking points

    • Recognises credit for the stated point: Calculates q² = 180/500 = 0.36.
    • Checks this criterion, crediting equivalent work already shown or identifying what is missing: Calculates q = √0.36 = 0.6, so the frequency of allele a is 0.6.
    • Checks this criterion, crediting equivalent work already shown or identifying what is missing: Calculates p = 1 − q = 0.4, so the frequency of allele A is 0.4.
    • Checks this criterion, crediting equivalent work already shown or identifying what is missing: Calculates the heterozygous frequency as 2pq = 2(0.4)(0.6) = 0.48.
    • Checks this criterion, crediting equivalent work already shown or identifying what is missing: Calculates the number of heterozygous beetles as 0.48 × 500 = 240.

    Examiner tip: Check each criterion, including equivalent work already shown. Do not demand an extra step merely to rewrite an equivalent answer. Do not award the same idea twice or infer work the learner did not show.

29. 29.

    In fruit flies, the genes for body colour and wing length are found to be linked, with a recombination frequency of 18%. Explain what this recombination frequency indicates about the relative positions of the two genes.

    [3 marks] · no calculator

    Worked explanation

    1. An 18% recombinant fraction means about 18 of every 100 scored informative offspring carry recombinant rather than parental combinations under the cross assumptions.
    2. Using a simple short-interval map approximation gives about 18 map units between loci; farther separation generally permits more crossovers.
    3. This is not 18 base pairs or an exact physical distance. Undetected multiple crossovers and recombination-rate variation limit the relationship, especially over longer intervals.

    Common mistake: Recombination percentage is not a direct ruler in DNA base pairs and cannot rise indefinitely as loci get farther apart.

    Marking points

    • States that 18% recombinant offspring give an approximate two-point map distance of 18 map units under the simple short-interval approximation.
    • Explains that more widely separated loci generally have more opportunities for crossing over between them, while recombination fraction alone is not an exact physical-distance measure.
    • States that if the genes were very close together, the recombination frequency would be much lower, since crossing over between them would be rare.

    Examiner tip: Recombination frequency is a proxy for physical distance between linked genes, not a direct physical measurement — the relationship is approximately linear only over shorter map distances.

30. 30.

    Marking analysis: A learner attempts the following task: “In fruit flies, the genes for body colour and wing length are found to be linked, with a recombination frequency of 18%. Explain what this recombination frequency indicates about the relative positions of the two genes.” Their response addresses only this point: “States that 18% recombinant offspring give an approximate two-point map distance of 18 map units under the simple short-interval approximation.” Evaluate the response against the complete 3-mark task. Identify what earns credit and state every additional requirement needed for full marks.

    [3 marks] · no calculator

    Marking points

    • Recognises credit for the stated point: States that 18% recombinant offspring give an approximate two-point map distance of 18 map units under the simple short-interval approximation.
    • Checks this criterion, crediting equivalent work already shown or identifying what is missing: Explains that more widely separated loci generally have more opportunities for crossing over between them, while recombination fraction alone is not an exact physical-distance measure.
    • Checks this criterion, crediting equivalent work already shown or identifying what is missing: States that if the genes were very close together, the recombination frequency would be much lower, since crossing over between them would be rare.

    Examiner tip: Check each criterion, including equivalent work already shown. Do not demand an extra step merely to rewrite an equivalent answer. Do not award the same idea twice or infer work the learner did not show.

31. 31.

    Explain the role of non-disjunction during meiosis I in causing Down syndrome (trisomy 21).

    [4 marks] · no calculator

    Worked explanation

    1. During anaphase I, the two homologous chromosome-21 copies fail to separate, so one daughter receives both and the other receives none.
    2. After meiosis II, a gamete can contain two chromosome-21 copies. Fusion with a normal gamete contributing one yields three copies in the zygote.
    3. That meiotic pathway produces trisomy 21, whose extra gene dosage causes Down syndrome. Other pathways, including mosaicism and translocations, exist but are not the pathway requested.

    Common mistake: Meiosis-I non-disjunction concerns homologues; failure of sister chromatids to separate is the meiosis-II version.

    Marking points

    • States that non-disjunction is the failure of homologous chromosomes (chromosome 21) to separate properly during anaphase I.
    • States that this results in one secondary oocyte (or spermatocyte) receiving both copies of chromosome 21, and the other receiving none.
    • States that if a gamete with two copies of chromosome 21 is fertilised by a normal gamete, the resulting zygote has three copies of chromosome 21 (trisomy 21) instead of the normal two.
    • States that this individual develops with Down syndrome as a result of having an extra copy of chromosome 21 in every cell.

    Examiner tip: Non-disjunction can also occur during meiosis II or in the other parent's gametogenesis; you only need to explain one consistent pathway clearly, using meiosis I here as specified.

32. 32.

    Marking analysis: A learner attempts the following task: “Explain the role of non-disjunction during meiosis I in causing Down syndrome (trisomy 21).” Their response addresses only this point: “States that non-disjunction is the failure of homologous chromosomes (chromosome 21) to separate properly during anaphase I.” Evaluate the response against the complete 4-mark task. Identify what earns credit and state every additional requirement needed for full marks.

    [4 marks] · no calculator

    Marking points

    • Recognises credit for the stated point: States that non-disjunction is the failure of homologous chromosomes (chromosome 21) to separate properly during anaphase I.
    • Checks this criterion, crediting equivalent work already shown or identifying what is missing: States that this results in one secondary oocyte (or spermatocyte) receiving both copies of chromosome 21, and the other receiving none.
    • Checks this criterion, crediting equivalent work already shown or identifying what is missing: States that if a gamete with two copies of chromosome 21 is fertilised by a normal gamete, the resulting zygote has three copies of chromosome 21 (trisomy 21) instead of the normal two.
    • Checks this criterion, crediting equivalent work already shown or identifying what is missing: States that this individual develops with Down syndrome as a result of having an extra copy of chromosome 21 in every cell.

    Examiner tip: Check each criterion, including equivalent work already shown. Do not demand an extra step merely to rewrite an equivalent answer. Do not award the same idea twice or infer work the learner did not show.

33. 33.

    Explain how directional selection can change the allele frequencies of a population over time, using antibiotic resistance in bacteria as an example.

    [4 marks] · no calculator

    Worked explanation

    1. Heritable susceptibility varies in the bacterial population; useful resistance may pre-exist or arise through undirected mutation or gene transfer.
    2. Under antibiotic exposure, resistant variants leave more descendants than susceptible variants.
    3. Resistance alleles increase in frequency, shifting the population toward one end of the susceptibility distribution. This is directional selection under that environment, not purposeful mutation.

    Common mistake: The antibiotic's selection effect does not mean every surviving individual acquired resistance because it was exposed.

    Marking points

    • States that within a bacterial population there is already variation, with a small number of individuals carrying a resistance allele (e.g. due to a pre-existing mutation).
    • States that exposure to the antibiotic acts as a selection pressure, killing susceptible (non-resistant) bacteria.
    • States that resistant bacteria survive and reproduce, passing the resistance allele to their offspring.
    • States that over successive generations, the frequency of the resistance allele increases in the population, shifting the population's average phenotype towards resistance (directional selection).

    Examiner tip: Antibiotic exposure selects heritable resistance; it does not create mutations because they are needed. Resistance variants may pre-exist or arise during exposure through mutation or gene transfer, after which selection can increase their frequency.

34. 34.

    Marking analysis: A learner attempts the following task: “Explain how directional selection can change the allele frequencies of a population over time, using antibiotic resistance in bacteria as an example.” Their response addresses only this point: “States that within a bacterial population there is already variation, with a small number of individuals carrying a resistance allele (e.g. due to a pre-existing mutation).” Evaluate the response against the complete 4-mark task. Identify what earns credit and state every additional requirement needed for full marks.

    [4 marks] · no calculator

    Marking points

    • Recognises credit for the stated point: States that within a bacterial population there is already variation, with a small number of individuals carrying a resistance allele (e.g. due to a pre-existing mutation).
    • Checks this criterion, crediting equivalent work already shown or identifying what is missing: States that exposure to the antibiotic acts as a selection pressure, killing susceptible (non-resistant) bacteria.
    • Checks this criterion, crediting equivalent work already shown or identifying what is missing: States that resistant bacteria survive and reproduce, passing the resistance allele to their offspring.
    • Checks this criterion, crediting equivalent work already shown or identifying what is missing: States that over successive generations, the frequency of the resistance allele increases in the population, shifting the population's average phenotype towards resistance (directional selection).

    Examiner tip: Check each criterion, including equivalent work already shown. Do not demand an extra step merely to rewrite an equivalent answer. Do not award the same idea twice or infer work the learner did not show.

35. 35.

    Distinguish between a gene pool and a gene, and explain how genetic drift can reduce genetic diversity in a small population.

    [4 marks] · no calculator

    Worked explanation

    1. A gene occupies a locus and specifies a functional product; the gene pool comprises all alleles across all loci in a population.
    2. Random differences in which individuals survive or reproduce sample that pool unevenly. In a small population, each lost lineage represents a larger fraction of the pool.
    3. An allele can therefore be lost or fixed by chance, reducing within-population variation without any fitness advantage. Drift can operate alongside selection rather than replacing it.

    Common mistake: Genetic drift is not movement of individuals between populations; that is migration/gene flow.

    Marking points

    • States that a gene is a heritable factor at a specific locus that controls a characteristic, while a gene pool is the complete set of all alleles of all genes present in a population.
    • States that genetic drift is random change in allele frequencies from one generation to the next, due to chance events in which individuals survive and reproduce.
    • States that in a small population, chance events have a proportionally larger effect on allele frequencies than in a large population, so an allele can be lost (or fixed) purely by chance.
    • States that this loss of alleles over time reduces the overall genetic diversity of the gene pool.

    Examiner tip: Genetic drift is unrelated to fitness or adaptation — unlike natural selection, it changes allele frequencies by chance alone, which is why its effect is strongest in small populations.

36. 36.

    Marking analysis: A learner attempts the following task: “Distinguish between a gene pool and a gene, and explain how genetic drift can reduce genetic diversity in a small population.” Their response addresses only this point: “States that a gene is a heritable factor at a specific locus that controls a characteristic, while a gene pool is the complete set of all alleles of all genes present in a population.” Evaluate the response against the complete 4-mark task. Identify what earns credit and state every additional requirement needed for full marks.

    [4 marks] · no calculator

    Marking points

    • Recognises credit for the stated point: States that a gene is a heritable factor at a specific locus that controls a characteristic, while a gene pool is the complete set of all alleles of all genes present in a population.
    • Checks this criterion, crediting equivalent work already shown or identifying what is missing: States that genetic drift is random change in allele frequencies from one generation to the next, due to chance events in which individuals survive and reproduce.
    • Checks this criterion, crediting equivalent work already shown or identifying what is missing: States that in a small population, chance events have a proportionally larger effect on allele frequencies than in a large population, so an allele can be lost (or fixed) purely by chance.
    • Checks this criterion, crediting equivalent work already shown or identifying what is missing: States that this loss of alleles over time reduces the overall genetic diversity of the gene pool.

    Examiner tip: Check each criterion, including equivalent work already shown. Do not demand an extra step merely to rewrite an equivalent answer. Do not award the same idea twice or infer work the learner did not show.

37. 37.

    Explain how the loop of Henle establishes a concentration gradient in the medulla that enables the kidney to produce concentrated urine.

    [5 marks] · no calculator

    Worked explanation

    1. Salt leaves the ascending limb, actively in the thick segment and passively in a thin segment where present, while little water follows across its water-impermeable wall.
    2. The descending limb is water-permeable, so the hyperosmotic medulla draws water out and concentrates the tubular fluid.
    3. Opposite flow and repeated local transport effects multiply the medullary gradient; urea recycling adds to inner-medullary osmolality.
    4. The collecting duct uses, rather than creates alone, that gradient. With ADH-enhanced permeability, water exits osmotically, leaving a smaller volume of concentrated urine.

    Common mistake: Describing the entire ascending limb as an active salt pump misses its thin segment and obscures different transport mechanisms.

    Marking points

    • States that the ascending limb is relatively impermeable to water; its thick portion reabsorbs NaCl using energy-dependent transport, whereas the thin ascending limb transports salt passively.
    • States that salt reabsorption and countercurrent multiplication help establish the medullary osmotic gradient; urea recycling also contributes, particularly in the inner medulla.
    • States that the descending limb is relatively water-permeable, so water leaves by osmosis into the more concentrated interstitium, concentrating tubular fluid.
    • States that this countercurrent arrangement (fluid flowing in opposite directions in the two limbs) maintains and amplifies the concentration gradient along the length of the loop.
    • States that this steep medullary concentration gradient is what allows water to be reabsorbed osmotically from the collecting duct later, concentrating the final urine.

    Examiner tip: Distinguish the water-permeable descending limb from the water-impermeable ascending limb, and the transport properties of thin versus thick ascending segments. The collecting duct uses the medullary gradient, with ADH controlling its water permeability.

38. 38.

    Marking analysis: A learner attempts the following task: “Explain how the loop of Henle establishes a concentration gradient in the medulla that enables the kidney to produce concentrated urine.” Their response addresses only this point: “States that the ascending limb is relatively impermeable to water; its thick portion reabsorbs NaCl using energy-dependent transport, whereas the thin ascending limb transports salt passively.” Evaluate the response against the complete 5-mark task. Identify what earns credit and state every additional requirement needed for full marks.

    [5 marks] · no calculator

    Marking points

    • Recognises credit for the stated point: States that the ascending limb is relatively impermeable to water; its thick portion reabsorbs NaCl using energy-dependent transport, whereas the thin ascending limb transports salt passively.
    • Checks this criterion, crediting equivalent work already shown or identifying what is missing: States that salt reabsorption and countercurrent multiplication help establish the medullary osmotic gradient; urea recycling also contributes, particularly in the inner medulla.
    • Checks this criterion, crediting equivalent work already shown or identifying what is missing: States that the descending limb is relatively water-permeable, so water leaves by osmosis into the more concentrated interstitium, concentrating tubular fluid.
    • Checks this criterion, crediting equivalent work already shown or identifying what is missing: States that this countercurrent arrangement (fluid flowing in opposite directions in the two limbs) maintains and amplifies the concentration gradient along the length of the loop.
    • Checks this criterion, crediting equivalent work already shown or identifying what is missing: States that this steep medullary concentration gradient is what allows water to be reabsorbed osmotically from the collecting duct later, concentrating the final urine.

    Examiner tip: Check each criterion, including equivalent work already shown. Do not demand an extra step merely to rewrite an equivalent answer. Do not award the same idea twice or infer work the learner did not show.

39. 39.

    Explain the role of antidiuretic hormone (ADH) in regulating the water permeability of the collecting duct, and predict the effect of dehydration on ADH secretion.

    [4 marks] · no calculator

    Worked explanation

    1. Dehydration increases blood solute concentration and lowers water potential, stimulating hypothalamic osmoreceptors and increased posterior-pituitary ADH release.
    2. ADH promotes aquaporin insertion into collecting-duct membranes, increasing water permeability.
    3. Water moves into the concentrated medullary surroundings and back to circulation by osmosis, reducing urine volume and conserving water. As the original deviation falls, ADH stimulation decreases.

    Common mistake: ADH raises permeability; the loop/medullary gradient supplies the driving force, not an ADH-powered water pump.

    Marking points

    • States that ADH is released from the posterior pituitary gland and travels in the blood to the kidney.
    • States that ADH increases the permeability of the collecting duct wall to water, by causing aquaporin channels to be inserted into the cell membranes.
    • States that this allows more water to be reabsorbed from the filtrate into the blood by osmosis (down the medullary concentration gradient), producing a smaller volume of more concentrated urine.
    • Predicts that dehydration will increase ADH secretion, because osmoreceptors in the hypothalamus detect the resulting higher (more concentrated) blood solute concentration.

    Examiner tip: ADH controls permeability, not active transport, at the collecting duct — the actual osmotic gradient driving water reabsorption was already established by the loop of Henle.

40. 40.

    Marking analysis: A learner attempts the following task: “Explain the role of antidiuretic hormone (ADH) in regulating the water permeability of the collecting duct, and predict the effect of dehydration on ADH secretion.” Their response addresses only this point: “States that ADH is released from the posterior pituitary gland and travels in the blood to the kidney.” Evaluate the response against the complete 4-mark task. Identify what earns credit and state every additional requirement needed for full marks.

    [4 marks] · no calculator

    Marking points

    • Recognises credit for the stated point: States that ADH is released from the posterior pituitary gland and travels in the blood to the kidney.
    • Checks this criterion, crediting equivalent work already shown or identifying what is missing: States that ADH increases the permeability of the collecting duct wall to water, by causing aquaporin channels to be inserted into the cell membranes.
    • Checks this criterion, crediting equivalent work already shown or identifying what is missing: States that this allows more water to be reabsorbed from the filtrate into the blood by osmosis (down the medullary concentration gradient), producing a smaller volume of more concentrated urine.
    • Checks this criterion, crediting equivalent work already shown or identifying what is missing: Predicts that dehydration will increase ADH secretion, because osmoreceptors in the hypothalamus detect the resulting higher (more concentrated) blood solute concentration.

    Examiner tip: Check each criterion, including equivalent work already shown. Do not demand an extra step merely to rewrite an equivalent answer. Do not award the same idea twice or infer work the learner did not show.

41. 41.

    Outline the roles of follicle-stimulating hormone (FSH) and luteinising hormone (LH) in controlling the menstrual cycle.

    [4 marks] · no calculator

    Worked explanation

    1. Anterior-pituitary FSH promotes follicle development, and the developing follicle produces oestrogen that rebuilds the uterine lining.
    2. Sustained high oestrogen promotes positive feedback and an LH surge, triggering ovulation rather than a gradual LH rise being sufficient by itself.
    3. LH also supports corpus-luteum formation; its progesterone maintains the endometrium after ovulation. Thus pituitary signals coordinate ovarian events and uterine preparation indirectly through ovarian hormones.

    Common mistake: FSH does not directly maintain the post-ovulation uterine lining; progesterone from the corpus luteum has that role.

    Marking points

    • States that FSH, secreted by the pituitary gland, stimulates the growth and development of ovarian follicles.
    • States that developing follicles secrete oestrogen, which stimulates the repair and thickening of the uterine lining (endometrium).
    • States that a surge in LH, triggered by a peak in oestrogen, causes ovulation — the release of the secondary oocyte from the mature follicle.
    • States that LH also stimulates the remains of the follicle to develop into the corpus luteum, which secretes progesterone to maintain the uterine lining.

    Examiner tip: The LH surge, not a gradual rise, is what directly triggers ovulation; it is caused by positive feedback from a high, sustained oestrogen level, not by LH acting alone.

42. 42.

    Marking analysis: A learner attempts the following task: “Outline the roles of follicle-stimulating hormone (FSH) and luteinising hormone (LH) in controlling the menstrual cycle.” Their response addresses only this point: “States that FSH, secreted by the pituitary gland, stimulates the growth and development of ovarian follicles.” Evaluate the response against the complete 4-mark task. Identify what earns credit and state every additional requirement needed for full marks.

    [4 marks] · no calculator

    Marking points

    • Recognises credit for the stated point: States that FSH, secreted by the pituitary gland, stimulates the growth and development of ovarian follicles.
    • Checks this criterion, crediting equivalent work already shown or identifying what is missing: States that developing follicles secrete oestrogen, which stimulates the repair and thickening of the uterine lining (endometrium).
    • Checks this criterion, crediting equivalent work already shown or identifying what is missing: States that a surge in LH, triggered by a peak in oestrogen, causes ovulation — the release of the secondary oocyte from the mature follicle.
    • Checks this criterion, crediting equivalent work already shown or identifying what is missing: States that LH also stimulates the remains of the follicle to develop into the corpus luteum, which secretes progesterone to maintain the uterine lining.

    Examiner tip: Check each criterion, including equivalent work already shown. Do not demand an extra step merely to rewrite an equivalent answer. Do not award the same idea twice or infer work the learner did not show.

43. 43.

    Explain the hormonal changes that occur if fertilisation does not take place, leading to menstruation.

    [3 marks] · no calculator

    Worked explanation

    1. Without an established pregnancy supplying rescue signals, the corpus luteum regresses near the cycle's end.
    2. Progesterone and oestrogen levels fall, removing hormonal support for the secretory endometrium.
    3. The lining breaks down and is shed as menstruation. Withdrawal of support is the causal change, not a surge of progesterone actively shedding the lining.

    Common mistake: Progesterone maintains the lining; its fall, not its rise, precedes menstruation in this scenario.

    Marking points

    • States that if fertilisation does not occur, the corpus luteum degenerates towards the end of the cycle.
    • States that this causes a sharp fall in progesterone (and oestrogen) levels.
    • States that since progesterone was maintaining the thickened uterine lining, its withdrawal causes the endometrium to break down and be shed as menstruation.

    Examiner tip: Menstruation is caused by hormone withdrawal (falling progesterone), not by a hormone actively 'triggering' shedding — the uterine lining is only maintained as long as progesterone remains high.

44. 44.

    Marking analysis: A learner attempts the following task: “Explain the hormonal changes that occur if fertilisation does not take place, leading to menstruation.” Their response addresses only this point: “States that if fertilisation does not occur, the corpus luteum degenerates towards the end of the cycle.” Evaluate the response against the complete 3-mark task. Identify what earns credit and state every additional requirement needed for full marks.

    [3 marks] · no calculator

    Marking points

    • Recognises credit for the stated point: States that if fertilisation does not occur, the corpus luteum degenerates towards the end of the cycle.
    • Checks this criterion, crediting equivalent work already shown or identifying what is missing: States that this causes a sharp fall in progesterone (and oestrogen) levels.
    • Checks this criterion, crediting equivalent work already shown or identifying what is missing: States that since progesterone was maintaining the thickened uterine lining, its withdrawal causes the endometrium to break down and be shed as menstruation.

    Examiner tip: Check each criterion, including equivalent work already shown. Do not demand an extra step merely to rewrite an equivalent answer. Do not award the same idea twice or infer work the learner did not show.

45. 45.

    A woman takes a combined oral contraceptive containing an oestrogen and a progestogen as prescribed. Explain how suppression of ovulation reduces the chance of pregnancy.

    [3 marks] · no calculator

    Worked explanation

    1. The supplied oestrogen and progestogen exert negative feedback on the hypothalamic-pituitary-ovarian axis.
    2. Reduced FSH/LH signalling suppresses follicular development and the ovulatory LH surge.
    3. Without ovulation there is no released oocyte for fertilisation, reducing pregnancy probability. Other effects include cervical-mucus changes, but the mechanism neither recreates identical pregnancy hormone levels nor guarantees complete protection.

    Common mistake: The combined pill primarily suppresses ovulation through feedback; it does not principally destroy sperm or reverse an established pregnancy.

    Marking points

    • States that the supplied oestrogen and progestogen exert negative feedback on the hypothalamic-pituitary-ovarian axis.
    • States that this negative feedback suppresses the secretion of FSH and LH.
    • Explains that suppression of follicular development and the LH surge usually prevents ovulation, so no oocyte is released for fertilisation when the method works as intended.

    Examiner tip: The principal mechanism is suppression of ovulation; progestogen also affects cervical mucus. Do not describe pill hormone levels as identical to pregnancy or promise complete contraceptive protection.

46. 46.

    Marking analysis: A learner attempts the following task: “A woman takes a combined oral contraceptive containing an oestrogen and a progestogen as prescribed. Explain how suppression of ovulation reduces the chance of pregnancy.” Their response addresses only this point: “States that the supplied oestrogen and progestogen exert negative feedback on the hypothalamic-pituitary-ovarian axis.” Evaluate the response against the complete 3-mark task. Identify what earns credit and state every additional requirement needed for full marks.

    [3 marks] · no calculator

    Marking points

    • Recognises credit for the stated point: States that the supplied oestrogen and progestogen exert negative feedback on the hypothalamic-pituitary-ovarian axis.
    • Checks this criterion, crediting equivalent work already shown or identifying what is missing: States that this negative feedback suppresses the secretion of FSH and LH.
    • Checks this criterion, crediting equivalent work already shown or identifying what is missing: Explains that suppression of follicular development and the LH surge usually prevents ovulation, so no oocyte is released for fertilisation when the method works as intended.

    Examiner tip: Check each criterion, including equivalent work already shown. Do not demand an extra step merely to rewrite an equivalent answer. Do not award the same idea twice or infer work the learner did not show.

Original, unofficial practice questions from SubjectScout — not official past-paper questions or complete syllabus coverage. Added English teaching explanations are AI-authored; independent teacher review is pending.