Haploid, diploid and homologous chromosomes
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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.
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
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
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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.
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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.
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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.
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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.
(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
- 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
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(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.
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(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.
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- 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
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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.
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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.
AnswerError 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.
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The rest of this note
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: and codominant; both dominant to ; 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 and Y (no allele on Y); mother's gametes are and ; 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