Notes/Biology/Paper 4/Inheritance
CAIEA Level9700§16

Inheritance

Haploid and diploid cells, meiosis and the eight named stages, crossing over and independent assortment, random fusion at fertilisation, every genetics term, monohybrid / codominant / multi-allele / sex-linked / dihybrid / dihybrid-test-cross / autosomal-linkage / epistasis crosses, the chi-squared test, the four named gene→protein→phenotype examples (TYR/albinism, HBB/sickle cell, F8/haemophilia, HTT/Huntington's), structural vs regulatory genes, repressible vs inducible enzymes, the lac operon, transcription factors in eukaryotes, and the gibberellin/DELLA-repressor mechanism.

240 min read 13 sub-topics
207
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2021–2025 · 37 papers
16 marks
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Every child is a unique combination of traits from two parents — this is inheritance. Sexual reproduction depends on meiosis, a special cell division that halves the chromosome number in the parent and shuffles the genes so that no two gametes (and no two offspring) are identical. The Cambridge 9700 §16 topic covers three intertwined questions: (1) how is the chromosome number halved and reshuffled — meiosis, crossing over, and the random orientation of homologs on the metaphase plate; (2) what rules describe the way traits pass from parents to offspring — Mendelian crosses, codominance, multiple alleles, sex linkage, autosomal linkage, epistasis, test crosses, and the chi-squared test that tells you whether your observed ratio matches the prediction; (3) how a gene becomes a phenotype — four named examples (TYR/tyrosinase/albinism, HBB/haemoglobin/sickle cell, F8/factor VIII/haemophilia, HTT/huntingtin/Huntington's) and the gene-control layer (structural vs regulatory genes, the lac operon, transcription factors, the gibberellin → DELLA breakdown that releases a transcription factor).

The route through the note is: §01 haploid/diploid and what a homologous pair is; §02 the eight named stages of meiosis and what the chromosomes, nuclear envelope, cell-surface membrane and spindle do at each stage; §03 identifying meiotic stages from photomicrographs; §04 crossing over and independent assortment — the two meiotic sources of genetic variation; §05 random fusion of gametes at fertilisation — the third source; §06 the genetics vocabulary; §07 monohybrid crosses (dominance, codominance, ABO multi-allele); §08 sex linkage and X-linked inheritance; §09 dihybrid crosses, autosomal linkage, epistasis; §10 the test cross and the chi-squared test; §11 the four named gene→protein→phenotype chains; §12 gene control — the lac operon, transcription factors, the gibberellin → DELLA mechanism.

Across 2021–2025 this is one of the heavier A2 topics on Paper 4 — 207 leaf parts, 603 marks, mean difficulty 2.45, across 37 papers. The mark schemes test it in many guises: monohybrid and dihybrid Punnett squares (the student draws them), pedigree interpretation, three-generation family trees with codominance and sex linkage in the same family, four-o'clock or blood-group crosses with the ABO multi-allele system, autosomal-linkage crosses that give a 3:1 instead of 9:3:3:1, chi-squared calculations with a critical-value table, lac-operon diagrams in the "repressed" and "induced" state, and a growing number of questions that link a named gene to its protein to its phenotype (TYR → tyrosinase → albinism, and the three other named chains). The same content shows up on Paper 1 as multiple-choice, so the same diagrams are reused. The diagrams in this note are hand-drawn SVGs (bio13-* keys) — no real photomicrographs are required for §16 (the diagnostic features can be taught with stylised drawings), and the four named gene-protein chains are linked to the protein's biology with a single chain diagram plus a prose summary.

Before you start you should be able to
  • Mitosis and the cell cycle (Paper 1 §5) — the contrast with meiosis I and meiosis II hinges on knowing that mitosis separates sister chromatids while meiosis I separates homologous chromosomes

  • The central dogma and translation (Paper 1 §6) — that a gene codes for a polypeptide, and a polypeptide (often an enzyme) determines a phenotype

  • Enzymes and active sites (Paper 1 §3) — that loss-of-function mutations typically inactivate an enzyme and produce a recessive phenotype, while gain-of-function mutations are typically dominant (used in §11 for the four named examples)

  • The gibberellin mechanism from a2-04 §08 — the gibberellin → DELLA-repressor breakdown is shared between the plant-coordination topic and §16.3.4, so this note points back rather than re-teaching the whole gibberellin biology

By the end of this page you can
  • explain the meanings of the terms haploid (n) and diploid (2n)

  • explain what is meant by homologous pairs of chromosomes

  • explain the need for a reduction division during meiosis in the production of gametes

  • describe the behaviour of chromosomes in plant and animal cells during meiosis and the associated behaviour of the nuclear envelope, the cell surface membrane and the spindle (names of the main stages of meiosis, but not the sub-divisions of prophase I, are expected: prophase I, metaphase I, anaphase I, telophase I, prophase II, metaphase II, anaphase II and telophase II)

  • interpret photomicrographs and diagrams of cells in different stages of meiosis and identify the main stages of meiosis

  • explain that crossing over and random orientation (independent assortment) of pairs of homologous chromosomes and sister chromatids during meiosis produces genetically different gametes

  • explain that the random fusion of gametes at fertilisation produces genetically different individuals

  • explain the terms gene, locus, allele, dominant, recessive, codominant, linkage, test cross, F1, F2, phenotype, genotype, homozygous and heterozygous

  • interpret and construct genetic diagrams, including Punnett squares, to explain and predict the results of monohybrid crosses and dihybrid crosses that involve dominance, codominance, multiple alleles and sex linkage

  • interpret and construct genetic diagrams, including Punnett squares, to explain and predict the results of dihybrid crosses that involve autosomal linkage and epistasis (knowledge of the expected ratios for different types of epistasis is not expected)

  • interpret and construct genetic diagrams, including Punnett squares, to explain and predict the results of test crosses

  • use the chi-squared test to test the significance of differences between observed and expected results (the formula for the chi-squared test will be provided, as shown in the Mathematical requirements)

  • explain the relationship between genes, proteins and phenotype with respect to the: TYR gene, tyrosinase and albinism; HBB gene, haemoglobin and sickle cell anaemia; F8 gene, factor VIII and haemophilia; HTT gene, huntingtin and Huntington's disease

  • explain the role of gibberellin in stem elongation including the role of the dominant allele, Le, that codes for a functional enzyme in the gibberellin synthesis pathway, and the recessive allele, le, that codes for a non-functional enzyme

  • describe the differences between structural genes and regulatory genes and the differences between repressible enzymes and inducible enzymes

  • explain genetic control of protein production in a prokaryote using the lac operon (knowledge of the role of cAMP is not expected)

  • state that transcription factors are proteins that bind to DNA and are involved in the control of gene expression in eukaryotes by decreasing or increasing the rate of transcription

  • explain how gibberellin activates genes by causing the breakdown of DELLA protein repressors, which normally inhibit factors that promote transcription

01

Haploid, diploid and homologous chromosomes

Syllabus requirement · §16.1

explain the meanings of the terms haploid (n) and diploid (2n); explain what is meant by homologous pairs of chromosomes.

Every cell in your body (except the gametes) carries 46 chromosomes — half from your mother, half from your father. The two halves are not random; they pair up by shape, size and the genes they carry, and it is this pairing that makes inheritance predictable. The 46-chromosome number is a diploid number; the gametes (sperm and egg) carry a haploid number — 23. When a sperm and an egg fuse at fertilisation, the diploid number is restored in the zygote, and from there every cell of the new individual is built by mitosis from that 46-chromome cell.

The vocabulary the syllabus uses:

  • Diploid (2n) — a cell that contains two complete sets of chromosomes, one set inherited from each parent. Human somatic cells are 2n = 46; the cells of a pea plant are 2n = 14; the cells of a fruit fly are 2n = 8. The number n (the haploid number) varies between species but is constant within a species.
  • Haploid (n) — a cell that contains one complete set of chromosomes. Human gametes are n = 23; pea gametes are n = 7; fruit-fly gametes are n = 4. In flowering plants the cells of the gametophyte (pollen grain, embryo sac) are haploid; in mammals the gametes are the only haploid cells in the life cycle.
  • Haploid number (n) = number of distinct chromosomes in a set. Humans have 23 distinct chromosomes in a set, so n = 23 and 2n = 46. A fruit fly has 4 distinct chromosomes per set, so n = 4 and 2n = 8.

The number n is not "half the diploid number" in a way that holds for every cell — the n is what's in a set, and the set is the unit. A common error is to write "n is the number of chromosome pairs". The pairs are counted in the diploid; the n counts the chromosomes in one of the two sets.

What "homologous" actually means

A homologous pair is a pair of chromosomes — one inherited from the mother, one from the father — that:

  • are the same size,
  • have the same shape (the centromere is in the same position),
  • carry the same genes at the same loci (loci are the fixed positions along the chromosome where a particular gene sits).

The two chromosomes of a homologous pair are not identical, however. They may carry different versions of the same gene (different alleles) at one or more loci. For example, both homologs of chromosome 7 carry a gene for eye colour at the same locus; the maternal homolog might carry the allele for brown eyes, the paternal the allele for blue eyes.

Autosomes vs sex chromosomes. Of the 23 pairs in humans, 22 are autosomes — they are true homologous pairs and they govern all the non-sex traits. The 23rd pair is the sex chromosomes — XX in females, XY in males. The X and Y are very different in size and in the genes they carry (the Y is much smaller and carries mostly male-determining genes), so they are not a true homologous pair in the same sense as the autosomes. They still pair up during meiosis I (the pseudoautosomal regions let the X and Y line up at the metaphase plate) so they can separate, but the gene content is largely different.

ABCchromosome 1AbChomologcentromeresame locus,different alleles

Fig 16.1 A pair of homologous chromosomes — same loci, possibly different alleles

Same locus, different alleles

A gene sits at a fixed position (a locus) on a chromosome. Different versions of the same gene at the same locus are called alleles. If the two homologs in a pair carry different alleles at a locus, the cell (or the person) is heterozygous at that locus. If the two homologs carry the same allele, the cell is homozygous at that locus.

The syllabus does not introduce the words "homozygous" and "heterozygous" until §16.2, but the concept belongs here: at every locus, the two homologous chromosomes either agree (homozygous) or disagree (heterozygous). The agreement-or-disagreement is the substrate of every inheritance pattern the rest of the note will explore — dominance, codominance, sex linkage, autosomal linkage, epistasis.

The sex chromosomes are not a true homologous pair

The X and Y chromosomes differ in size (X is about three times the length of Y) and in gene content. The Y carries mostly male-determining genes (e.g. SRY); the X carries many genes that have nothing to do with sex. The two still pair up during meiosis I so they can separate to different gametes, but the pairing is restricted to small pseudoautosomal regions and is not a full homologous pairing. A gene on the X has no corresponding allele on the Y, which is the structural basis of X-linked inheritance (see §08).

Counting chromosomes and identifying homologs

9700/41 M/J 2024 Q6(a)3 marks

In humans the TYR gene is located on chromosome 11. Fig. 6.1 shows the homologous pair for chromosome 11.

(a) State the diploid number of chromosomes in a human somatic cell.
(b) State the haploid number in a human gamete.
(c) With reference to Fig. 6.1, describe what is meant by a homologous pair of chromosomes.

Show full working
  1. 1

    Step 1 — read the question. The stem names the TYR gene on chromosome 11 and points to Fig. 6.1 — a picture of the chromosome-11 homologous pair. The first two parts are bookwork: humans have 46 chromosomes in a somatic cell, 23 in a gamete.

    Mark 1. The student who writes '46' without the symbol 2n loses the symbol mark.

  2. 2

    Step 2 — state the diploid number. 2n = 46 (somatic cells are diploid; humans have 46 chromosomes in total, organised as 23 homologous pairs plus the sex chromosomes).

    Mark 2. The MS credits '46' with the 2n label.

  3. 3

    Step 3 — state the haploid number. n = 23 (gametes are haploid and contain one of each pair).

    Mark 3. The MS credits '23' with the n label.

  4. 4

    Step 4 — describe a homologous pair. A homologous pair is a pair of chromosomes — one maternal, one paternal — that are the same size, have the centromere in the same position, and carry the same genes at the same loci (here, the TYR locus on each chromosome 11). The two chromosomes may carry different alleles at the TYR locus.

    Mark 4. The MS credits the three matching features plus the alleles-may-differ note.

Answer

(a) 2n = 46. (b) n = 23. (c) A homologous pair is a pair of chromosomes — one from each parent — that are the same size, have the centromere in the same position, and carry the same genes at the same loci (the TYR locus on each chromosome 11); the two chromosomes may carry different alleles at that locus.

When a question points to a Fig., your description must mention the same gene (or same locus) the figure labels — that is what the MS is checking the student noticed.

Your turn — haploid, diploid and homologous

  1. 13 marks

    A pea-plant somatic cell contains 14 chromosomes.

    (a) State the diploid and haploid numbers for Pisum sativum.
    (b) A pollen grain from a pea plant contains 7 chromosomes. Explain why.

    Stuck? Show hint

    Somatic = body cell (diploid). Pollen grain = gametophyte cell (haploid). The diploid number is twice the haploid.

    Show solution
    1. 1

      (a) Diploid number 2n = 14; haploid number n = 7. A somatic (body) cell is diploid, so 2n = 14. The haploid number n is half of 2n, so n = 7.

      Marks 1 and 2. One mark for each number with the correct symbol.

    2. 2

      (b) A pollen grain is a gametophyte cell — it develops from a haploid spore produced by meiosis in the anther, so all its cells are haploid (n = 7). A pollen grain therefore contains one set of 7 chromosomes, not two.

      Mark 3. The MS credits the link to meiosis in the anther.

    Answer

    (a) 2n = 14, n = 7. (b) A pollen grain is the male gametophyte; it develops from a haploid spore (produced by meiosis in the anther), so all its cells are haploid and contain 7 chromosomes.

  2. 22 marks

    A student writes: "Humans have 46 chromosomes arranged in 23 homologous pairs, including the X and the Y which are a homologous pair like any other."

    Identify two errors in this statement.

    Stuck? Show hint

    Look at the second clause carefully — "like any other" is the part to challenge.

    Show solution
    1. 1

      Error 1: the X and Y are NOT a true homologous pair "like any other". The X and Y differ in size and gene content — the Y is much smaller and carries mostly male-determining genes. They pair up during meiosis I only across small pseudoautosomal regions, not along their full length.

      Mark 1. The MS credits the recognition that X/Y are not a true homologous pair.

    2. 2

      Error 2: a male has X and Y (NOT a homologous pair); a female has two X chromosomes (which ARE a true homologous pair). So in half the population the 23rd pair is a true homologous pair, and in half it is not. The statement treats all humans as if they were male.

      Mark 2. The MS credits the recognition that X/Y sex determination makes the 23rd pair asymmetric in males.

    Answer

    Error 1: the X and Y are not a true homologous pair "like any other" — they differ in size and gene content, and only pair up across small pseudoautosomal regions during meiosis I. Error 2: a male has X and Y (not a true homologous pair), while a female has two X chromosomes (a true homologous pair) — the statement does not distinguish the two sexes.

Practise haploid, diploid and homologousReal past-paper questions · Haploid and diploid; homologous chromosomes

The rest of this note

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Can you do all of these?

  • Define diploid (2n) as a cell with two complete sets of chromosomes (one from each parent) and haploid (n) as a cell with one set. n is the number of distinct chromosomes in a set, not 'half the diploid number'

  • Define a homologous pair as two chromosomes (one maternal, one paternal) that are the same size, have the centromere in the same position, and carry the same genes at the same loci (they may carry different alleles)

  • State that the autosomes are the 22 true homologous pairs in humans; the X and Y are not a true homologous pair (different size, different gene content, pair only across pseudoautosomal regions)

  • Explain why a reduction division is needed: gametes are haploid so that fertilisation restores the diploid number in the zygote

  • Describe the eight named stages of meiosis: prophase I, metaphase I, anaphase I, telophase I, prophase II, metaphase II, anaphase II, telophase II — and what the chromosomes, nuclear envelope, cell-surface membrane and spindle are doing at each

  • State that meiosis I separates homologous chromosomes; meiosis II separates sister chromatids (like mitosis)

  • Identify meiotic stages from a photomicrograph: prophase I = condensed chromosomes inside an intact nuclear envelope; metaphase I = paired homologs (bivalents) on the metaphase plate; anaphase I = homologs moving to opposite poles; telophase I = two cells with one chromosome from each pair

  • State the difference between anaphase I and anaphase II: anaphase I = whole chromosomes (each still with two chromatids) move apart; anaphase II = sister chromatids separate (as in mitosis)

  • Explain crossing over in prophase I: non-sister chromatids of a bivalent exchange segments at a chiasma, producing recombinant chromatids

  • Explain random orientation (independent assortment) at metaphase I: each pair of homologs lines up at the metaphase plate independently of every other pair, so the gametes receive all 2ⁿ possible combinations of maternal and paternal chromosomes (n = haploid number)

  • State that a single diploid parent makes 2ⁿ genetically different gametes (n = haploid number); in humans, n = 23, so 2²³ ≈ 8.4 million different gametes

  • Explain that the random fusion of gametes at fertilisation is a third source of variation: a parent that makes n gamete types and another that makes m gamete types can produce n × m different zygotes

  • Define the 14 genetic terms: gene, locus, allele, dominant, recessive, codominant, linkage, sex linkage, test cross, F1, F2, phenotype, genotype, homozygous, heterozygous

  • State that dominant and recessive describe the phenotype of the heterozygote; they do not describe the frequency of the allele in the population

  • Construct a Punnett square for a monohybrid cross: 2 × 2 with one row per paternal gamete and one column per maternal gamete; each box is the union of the row and column gametes

  • State the F2 ratio for a monohybrid cross with simple dominance: 3 dominant phenotype : 1 recessive phenotype (3 : 1)

  • State the F2 ratio for a monohybrid cross with codominance: 1 : 2 : 1 (three distinguishable phenotypes including the heterozygote)

  • State the F2 ratio for a dihybrid cross with independent assortment: 9 : 3 : 3 : 1

  • Use the ABO blood-group system as the multi-allele example: IAI^{A} and IBI^{B} codominant; both dominant to ii; one person has two of the three alleles, never more

  • State that an X-linked gene is on the X chromosome; a male (XY) has only one copy, so a single recessive allele is expressed (the male is hemizygous, not heterozygous)

  • Construct an X-linked Punnett square: father's gametes are XHX^{H} and Y (no allele on Y); mother's gametes are XHX^{H} and XhX^{h}; the sex split in the offspring is the giveaway of X-linkage

  • State that X-linked recessive conditions (haemophilia A from F8, haemophilia B from F9, red–green colour blindness, DMD) appear more often in males than in females

  • Define autosomal linkage as two or more genes on the same autosome, close enough that crossing over between them is rare; the F2 collapses from 9 : 3 : 3 : 1 toward 3 : 1 (in the limit of complete linkage)

  • State that a dihybrid test cross (AaBb × aabb) reveals linkage: independent assortment gives 1 : 1 : 1 : 1; complete linkage gives 1 : 1 (parentals only); the rare classes are the recombinants

  • Define epistasis as one gene masking the expression of another gene at a different locus; the commonest pattern is 9 : 3 : 4 (one allele of one gene is epistatic to the other gene)

  • Define a test cross as a cross of an unknown genotype with a homozygous recessive; the offspring phenotypes reveal the unknown parent's gametes

  • State the chi-squared test: χ² = Σ ( (O − E)² / E ) summed over every class; df = number of classes − 1; if χ² < critical value (at p = 0.05) accept H₀; if χ² ≥ critical value reject H₀

  • State the chi-squared decision rule in words: a small χ² means the data fit the expected ratio; a large χ² means they do not; accepting H₀ does not prove H₀ true, only that the data are not inconsistent with it

  • Apply the χ² test to a 3 : 1 monohybrid and a 9 : 3 : 3 : 1 dihybrid; identify which class is most-deviated from expected and use that to suggest linkage or another modifier

  • State the TYR → tyrosinase → melanin → pigmentation chain: TYR codes for tyrosinase, tyrosinase makes melanin from tyrosine, melanin gives skin/hair/eye colour; tyr/tyr is albino; albinism is autosomal recessive

  • State the HBB → β-globin → haemoglobin → sickle cell chain: HBB codes for β-globin, β-globin is part of HbA; the HbS variant (Glu→Val at position 6) polymerises at low O₂ and distorts the RBC; sickle cell is autosomal codominant (HbA/HbA, HbA/HbS, HbS/HbS all distinguishable)

  • State the F8 → factor VIII → clotting → haemophilia chain: F8 is on the X chromosome; loss-of-function mutation gives no factor VIII, uncontrolled bleeding; haemophilia A is X-linked recessive

  • State the HTT → huntingtin → neuronal toxicity → Huntington's chain: HTT CAG expansion gives a toxic gain-of-function huntingtin protein that damages striatal neurones; HD is autosomal dominant (one copy is enough)

  • State the general principle: loss-of-function mutations are usually recessive (TYR, HBB, F8); gain-of-function mutations are usually dominant (HTT). The biology of the protein sets the inheritance pattern

  • Use the Le/le allele: Le = functional gibberellin-synthesis enzyme; le = non-functional; le/le plants are dwarf because they cannot make gibberellin to break down DELLA

  • Define structural gene as a gene that codes for a functional polypeptide; define regulatory gene as a gene that codes for a product (often a protein) that controls the expression of other genes

  • Define repressible enzyme as one made continuously until something stops it; inducible enzyme as one made only when the substrate is present

  • Describe the lac operon: lacI (regulatory) codes for the LacI repressor; the operator sits between the promoter and the structural genes lacZ (β-galactosidase), lacY (permease), lacA (transacetylase)

  • State the two states of the lac operon: repressed (no lactose — repressor bound to operator — no transcription) and induced (lactose present — converted to allolactose — allolactose binds repressor — repressor released — RNA polymerase transcribes the structural genes)

  • Do NOT mention cAMP, CAP or catabolite activator protein — these are off-syllabus for 2025–2027

  • Define transcription factor as a protein that binds DNA and controls gene expression in eukaryotes, either by recruiting RNA polymerase (increase) or by blocking it (decrease)

  • Describe the gibberellin → DELLA chain: GA binds GID1 receptor → complex binds DELLA repressor → DELLA is ubiquitinated → DELLA is broken down by the 26S proteasome → TFs are released → TFs enter nucleus → α-amylase gene is switched ON → α-amylase hydrolyses starch in the endosperm

  • State the two-step wording: DELLA represses the TFs (not 'DELLA represses the gene directly') — the MS credits the two-step wording

Now do the questions
207 real Paper 4 parts from 2021–2025, sorted by difficulty, with mark schemes