Biology · Meiosis & Patterns of Inheritance
Meiosis, chromosome segregation and recombination
Label two homologues before they copy, then keep those identities on their daughter chromatids. The first meiotic division sends homologues to different sides; the second separates their sisters. Swapping those two instructions changes both the count and the inheritance route.
Begin with the cell-division lesson and Heredity, Variation & Evolution: recognise homologous chromosomes, sister chromatids, a locus and an allele. DNA, Replication & Repair explains how copying makes two sister DNA molecules. Gene Expression helps with the later distinction between transmitting a sequence and expressing a trait. Here, fractions and conditional probabilities are explained alongside the cases. All chromosome-count examples use a specified diploid model; the final lessons introduce extensions rather than requiring advanced genetics beforehand.
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Trace homologues and sisters through two divisions, explain one crossover model and compare two precisely stated segregation failures.
One copying step, two sorting steps
After replication, each homologue has two sister chromatids. A paired pair of homologues is a bivalent; its four chromatids can also be called a tetrad. Homologous chromosomes carry corresponding loci, whereas sister chromatids began as copies of the same chromosome. These relationships are different even when all four lines look alike.
In the standard meiosis I pathway, homologues orient toward opposite spindle poles while the sister pair stays together at its centromeric region. In meiosis II, sisters orient toward opposite poles and separate. Cohesion and its controlled release help make these successive operations possible. There is no ordinary second round of DNA replication between the two divisions.
Our two-pair model starts the divisions with four replicated chromosomes. After meiosis I, each of two nuclei contains two replicated chromosomes: one from each homologous pair. After meiosis II, each of four nuclei contains two one-chromatid chromosomes. Counts here describe completed stages; a whole dividing cell during anaphase needs a separately specified count.
Change combinations without losing the accounting
During the commonly taught prophase I pathway, homologues pair and reciprocal crossing over can exchange DNA between nonsister chromatids. In our selected bivalent, one homologue carries AB and the other ab at two marked loci. Copying gives chromatids AB, AB, ab and ab. One exchange between the loci, involving one chromatid from each homologue, gives AB, Ab, aB and ab.
The exchange leaves two unchanged parental combinations and creates two recombinant combinations in this particular tetrad. A still appears twice, a twice, B twice and b twice across its four products. We rearranged existing alleles at the marked loci; no allele was invented and no chromosome set was added.
Crossovers and chromosome connections also contribute to reliable segregation in many organisms. Their details vary: male Drosophila meiosis provides a well-studied case without the usual meiotic crossing over. Do not turn this one-exchange drawing into a claim that every pair must cross over in every organism. Molecular recombination can also yield outcomes beyond the reciprocal exchange drawn here.
Follow four chromatids through one selected exchange
Read the diagram in words
AB, AB, ab and ab become AB, Ab, aB and ab after one exchange between the two marked loci. Chromatids 1b and 2a exchange right-hand ancestry; 1a and 2b retain their parental combinations. Two products are recombinant between the markers.
| Locus | Relative position |
|---|---|
| A/a locus | 30 |
| B/b locus | 80 |
- Relative position range
- 0–100
- Centromere position
- 10
- Exchange position
- 55
Homologue 1, sister a Displayed chromatid Description A/a locus B/b locus 1a Homologue 1, sister a A B 1a AB: parental A B Relative drawing position Description Relative position range Segment ancestry Homologue 1, sister a 0–100 1 AB: parental 0–100 1 Homologue 1, sister b Displayed chromatid Description A/a locus B/b locus 1b Homologue 1, sister b A B 1b Ab: recombinant A b Relative drawing position Description Relative position range Segment ancestry Homologue 1, sister b 0–100 1 Ab: recombinant 0–55 1 Ab: recombinant 55–100 2 Homologue 2, sister a Displayed chromatid Description A/a locus B/b locus 2a Homologue 2, sister a a b 2a aB: recombinant a B Relative drawing position Description Relative position range Segment ancestry Homologue 2, sister a 0–100 2 aB: recombinant 0–55 2 aB: recombinant 55–100 1 Homologue 2, sister b Displayed chromatid Description A/a locus B/b locus 2b Homologue 2, sister b a b 2b ab: parental a b Relative drawing position Description Relative position range Segment ancestry Homologue 2, sister b 0–100 2 ab: parental 0–100 2
| Reciprocal chromatid product | Description |
|---|---|
| 1b ↔ 2a | Right of 55 |
| 2a ↔ 1b | Right of 55 |
- Sisters share the same homologue number
- Exchange involves 1b and 2a
- Four chromatid products of this selected tetrad
- A, a, B and b each occur twice
Specify which separation failed
Nondisjunction is a failure of the relevant chromosomes to separate normally. Suppose only the long homologous pair fails at meiosis I in our n = 2 model. Both long homologues go to one side, while the short pair separates normally. If meiosis II proceeds normally, two products receive three chromosomes each and two receive one each.
Instead, let meiosis I be normal and let only the sisters of the long chromosome fail to separate in one meiosis II division. The four products then contain 3, 1, 2 and 2 chromosomes. These two alternative failures produce different distributions because they occur at different points in the sorting tree.
An extra or missing individual chromosome is aneuploidy, rather than a gain or loss of a complete set. The diagrams assign neither survival nor the chance of an actual failure. The finished products could differ in viability and maturation; counting them is not predicting the proportions among living offspring.
The division at which separation fails changes the product pattern
Read the diagram in words
Normal segregation gives 2, 2, 2, 2 chromosomes. A selected long-homologue meiosis I failure gives 3, 3, 1, 1. A selected long-sister failure in one meiosis II division gives 3, 1, 2, 2. Each lane preserves eight copied DNA molecules in total.
| Chromosome identity | Chromosomal DNA copy identities |
|---|---|
| 1 | L1a, L1b |
| 2 | L2a, L2b |
| 3 | S1a, S1b |
| 4 | S2a, S2b |
Both divisions separate normally
| Nuclear product | Chromosomal DNA copy identities |
|---|---|
| 1 | L1a, L1b, S1a, S1b |
| 2 | L2a, L2b, S2a, S2b |
| Nuclear product | Chromosomal DNA copy identities | Chromosomes per indicated nucleus | Outcome |
|---|---|---|---|
| 1 | L1a, S1a | 2 | Balanced n = 2 product |
| 2 | L1b, S1b | 2 | Balanced n = 2 product |
| 3 | L2a, S2a | 2 | Balanced n = 2 product |
| 4 | L2b, S2b | 2 | Balanced n = 2 product |
Long homologues fail to separate at meiosis I
| Nuclear product | Chromosomal DNA copy identities |
|---|---|
| 1 | L1a, L1b, L2a, L2b, S1a, S1b |
| 2 | S2a, S2b |
| Nuclear product | Chromosomal DNA copy identities | Chromosomes per indicated nucleus | Outcome |
|---|---|---|---|
| 1 | L1a, L2a, S1a | 3 | One chromosome extra or missing |
| 2 | L1b, L2b, S1b | 3 | One chromosome extra or missing |
| 3 | S2a | 1 | One chromosome extra or missing |
| 4 | S2b | 1 | One chromosome extra or missing |
Long sisters fail in one meiosis II division
| Nuclear product | Chromosomal DNA copy identities |
|---|---|
| 1 | L1a, L1b, S1a, S1b |
| 2 | L2a, L2b, S2a, S2b |
| Nuclear product | Chromosomal DNA copy identities | Chromosomes per indicated nucleus | Outcome |
|---|---|---|---|
| 1 | L1a, L1b, S1a | 3 | One chromosome extra or missing |
| 2 | S1b | 1 | One chromosome extra or missing |
| 3 | L2a, S2a | 2 | Balanced n = 2 product |
| 4 | L2b, S2b | 2 | Balanced n = 2 product |
- L: long type
- S: short type
- 1 and 2 identify homologues
- a and b identify sisters
- Balanced n = 2 product
- One chromosome extra or missing
- Alternative models, not consecutive events
Worked example
Conserve eight DNA molecules across the products
After replication, four chromosomes contain eight chromatid DNA molecules. Normal meiosis I produces two nuclei with four molecules each: 2 × 4 = 8. Normal meiosis II produces four nuclei with two each: 4 × 2 = 8. No new replication is needed to supply the second division.
For the selected meiosis I failure, the final chromosomes contain one DNA molecule each. Their counts sum to 3 + 3 + 1 + 1 = 8. For the selected meiosis II failure, 3 + 1 + 2 + 2 = 8 too. Conservation of the total does not guarantee a balanced complement in every product. The alternative failure routes should not be added together.
Check your understanding
Why are the products after meiosis I haploid even though their chromosomes still have two chromatids?
Show reasoning
Each nucleus has one homologue of each chromosome type: one complete set. Sister copies add DNA amount, not another homologous set.
Our single crossover changes AB, AB, ab, ab into AB, Ab, aB, ab. Did half the meioses recombine?
Show reasoning
The drawing shows one selected meiosis, not a sample of meioses. Two of its four chromatids are recombinant between the marked loci; that is a product count within this tetrad.
Why does the chosen meiosis II failure leave two balanced products?
Show reasoning
Only one of the two first-division products suffers the sister-separation failure. The other divides normally, supplying two n = 2 products. This conclusion depends on the specifically limited failure model.
What to carry forward
Keep homologue identity, sister identity and the stage of separation visible throughout the model.
References
Resources supporting this lesson’s explanations and examples.
- Biology 2e — Meiosis and sexual life cycles (2018)Mary Ann Clark, Matthew Douglas and Jung Choi · OpenStaxOpen textbook sectionsBiology 2e — Meiosis and sexual life cycles (2018)Sexual reproduction and life cyclesNondisjunction and chromosome number
- Introduction to Genetics — Chromosomes, crosses and mapping (2023 adaptation of Open Genetics Lectures)Natasha Ramroop Singh · Thompson Rivers UniversityOpen university textbookIntroduction to Genetics — Chromosomes, crosses and mapping (2023 adaptation of Open Genetics Lectures)DNA amount and chromosome numberAssumptions of a dihybrid ratioConditional pedigree probabilitiesAllele phaseLinkage and recombinant fractionsGenetic maps and their limitsTwo-point and three-point crossesTypes of dominanceBiochemical basis of dominanceGene interactions and phenotype groupingPseudoautosomal regions
- Meiosis in male Drosophila (Spermatogenesis 2:167–184, 2012)Bruce D. McKee, Rihui Yan and Jui-He TsaiResearch abstract