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By the end of this lesson you should be able to explain and apply each part of this topic — Independent Assortment — and its Exceptions, How Crossing Over Works, Expected vs Observed Ratios and Calculating Cross-Over Value — and use these ideas accurately in exam-style questions.
Spec Mapping — OCR H420 Module 6.1.2 — Patterns of inheritance, content statements covering autosomal linkage, crossing over during prophase I of meiosis, and the use of recombination frequencies to detect linkage and to construct genetic maps (refer to the official OCR H420 specification document for exact wording). This lesson is the natural sequel to monohybrid and sex-linkage lessons and the prerequisite for understanding why observed dihybrid ratios often deviate from Mendel's expected 9:3:3:1 — the topic of the next two lessons.
The historical context bears repeating. Thomas Hunt Morgan (paraphrased) and his "fly room" group at Columbia accumulated Drosophila mutants whose inheritance violated Mendel's law of independent assortment. Alfred Sturtevant (1913, paraphrased) realised that the relative frequencies of recombination between pairs of linked genes could be used to construct a linear genetic map of the chromosome — the first such map (for the X chromosome of Drosophila) was drawn in his bedroom that night when he was 19. This insight set the stage for half a century of classical genetic mapping that culminated in the molecular-era Human Genome Project.
Mendel's laws predict that genes for two different traits assort independently during meiosis, producing the classic 9:3:3:1 ratio in a dihybrid cross. But Mendel got lucky — he chose seven pea traits each controlled by a gene on a different chromosome. Real life is not so tidy. Genes that happen to lie on the same chromosome are said to be linked, and they do not assort independently — they tend to be inherited together. Crossing over partly breaks this linkage, and the frequency of recombinant offspring can even be used to map gene order on a chromosome. OCR A-Level Biology A specification module 6.1.2(d) requires you to understand autosomal linkage, crossing over and how linkage modifies expected genetic ratios.
Key Definitions:
- Linked genes — genes that lie on the same chromosome and tend to be inherited together.
- Autosomal linkage — linkage of genes on any chromosome other than a sex chromosome.
- Recombination — the production of new allele combinations via crossing over.
- Crossing over — the exchange of corresponding segments between non-sister chromatids of homologous chromosomes during prophase I of meiosis.
- Chiasma (plural: chiasmata) — the visible point of crossing over.
- Parental (non-recombinant) offspring — carry the same allele combinations as the parents.
- Recombinant offspring — carry new combinations produced by crossing over.
- Cross-over value (COV) — the percentage of recombinant offspring among all offspring.
In meiosis I, homologous chromosome pairs line up at the metaphase plate and are distributed randomly to the two daughter cells. For genes on different chromosomes, this randomness means that one gene's inheritance has no effect on the other's — this is Mendel's law of independent assortment. A heterozygous dihybrid AaBb can make four equally frequent gametes: AB, Ab, aB, ab.
But if A and B are on the same chromosome, then the dominant alleles start out together on one chromosome and the recessive alleles together on the other. Meiosis simply separates the whole chromosomes, so the gametes will nearly all be AB or ab — the parental combinations. Only crossing over during prophase I can produce recombinant Ab or aB gametes.
During prophase I, homologous chromosomes pair up (synapsis) and form a structure called the bivalent (or tetrad — four chromatids). Non-sister chromatids exchange corresponding pieces at points called chiasmata. The exchange is reciprocal: the segment lost from one chromatid is replaced by an identical segment from the other.
flowchart LR
A[Parental: ABC and abc] --> B[Crossing over between B and C]
B --> C[Recombinants: ABc and abC]
The further apart two genes are on a chromosome, the more likely a chiasma will fall between them, and the higher the cross-over value. Genes very close together are rarely separated; genes very far apart may cross over so often that they effectively assort independently.
Consider a dihybrid test cross in fruit flies. Let:
Both genes are on chromosome 2 (linked). Cross a dihybrid BbVv heterozygote (from a pure-bred cross BBVV × bbvv, so B and V are on one chromosome and b and v on the other) with a homozygous recessive bbvv.
COV(%)=total number of offspringnumber of recombinant offspring×100
Using Morgan's data:
A cross-over value of 17% corresponds to a map distance of 17 centimorgans (cM). Morgan used this idea to build the first genetic maps — the cross-over values between pairs of linked genes revealed their relative positions on the chromosome.
If you know COVs for pairs of genes, you can deduce their order. Suppose:
The simplest consistent order is A — B — C, with B between A and C. This technique, called linkage mapping, underpinned classical genetics before DNA sequencing was possible.
Linkage does not last forever. Over many generations, repeated meioses break up linked combinations. In populations, the degree to which two alleles are still found together more than expected is called linkage disequilibrium. It is an important concept in medical genetics — it allows researchers to track disease genes using nearby marker alleles.
Without linkage (independent assortment), a dihybrid cross AaBb × AaBb gives the classic 9 : 3 : 3 : 1 phenotypic ratio. With linkage, this ratio is distorted in the direction of the parental combinations.
| Cross | Expected (unlinked) | Observed (if linked) |
|---|---|---|
| Test cross BbVv × bbvv | 1:1:1:1 | Mostly parentals, few recombinants |
| Dihybrid BbVv × BbVv | 9:3:3:1 | Excess of parental combinations, deficit of recombinants |
An exam question may give you data from a cross and ask you to show that the observed ratio deviates significantly from the expected 9:3:3:1 — evidence of linkage (see the chi-squared test lesson).
Take care not to confuse these:
An X-linked gene is both sex-linked and linked to other X-linked genes (because they are all on the same chromosome).
When you see a dihybrid cross that gives unusual numbers — especially an excess of the two "parental" classes over the "recombinant" classes — suspect linkage. Always calculate the cross-over value as the total number of recombinant offspring divided by the total number of offspring, multiplied by 100. Make sure you can identify which classes are parental and which are recombinant by looking at the parents' original genotypes.
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<text x="220" y="35" font-family="Helvetica, Arial, sans-serif" font-size="11" fill="#444">→ abCD recombinant gamete</text>
<text x="0" y="100" font-family="Helvetica, Arial, sans-serif" font-size="11" fill="#444">Plus parental ABCD and abcd from non-recombinant chromatids</text>
Crossing over between non-sister chromatids of paired homologues at chiasmata in prophase I produces recombinant chromatids. After meiosis is complete, the four haploid products include both parental (ABCD, abcd) and recombinant (ABcd, abCD) combinations. The further apart two loci are on the chromosome, the more likely a chiasma falls between them, and the higher the recombination frequency.
The cross-over value, or recombination frequency, is:
COV=total offspringnumber of recombinant offspring×100%
For Morgan's classic 1911 Drosophila cross (BV/bv × bv/bv):
COV=965+944+206+185206+185×100=2300391×100≈17%
This corresponds to a map distance of 17 centiMorgans (cM): 1 cM ≡ 1% recombination frequency for short distances. The cM is named after Morgan in recognition of the discovery.
To test whether the observed deviation from 1:1:1:1 is statistically significant we apply the chi-squared statistic:
χ2=∑E(O−E)2
with degrees of freedom = number of phenotypic classes − 1 = 4 − 1 = 3, and compare to the critical value 7.82 at p = 0.05. For Morgan's data the χ² is enormous (>1000), so we firmly reject the null hypothesis of independent assortment — strong evidence of linkage.
| Feature | Genes on different chromosomes | Linked genes on same chromosome |
|---|---|---|
| Behaviour in meiosis I | Independent assortment | Travel together unless crossing over occurs |
| Gamete frequencies (dihybrid) | AB : Ab : aB : ab = 1 : 1 : 1 : 1 | Parental classes much higher than recombinants |
| Test cross (AaBb × aabb) | 1:1:1:1 | Excess parentals, deficit recombinants |
| F2 from dihybrid intercross | 9:3:3:1 | Skewed toward parental phenotypes |
| Modified by | Cannot be linked | Crossing over (the further apart, the more recombinants) |
| Map distance | N/A | Recombination frequency × 100 = cM |
Synoptic Links — Connects to:
ocr-alevel-biology-membranes-cell-division / meiosis(crossing over occurs in prophase I; chiasmata are visible cytologically and are the source of genetic recombination).ocr-alevel-biology-genetics-inheritance / phenotypic-variation-monogenic-inheritance(Mendel's law of independent assortment is the unlinked case — the simpler scenario this lesson extends).ocr-alevel-biology-genetics-inheritance / dihybrid-crosses(the next lesson — explores 9:3:3:1 ratios and how linkage modifies them).ocr-alevel-biology-genetics-inheritance / chi-squared-test(the formal statistical tool for testing whether observed deviation from independent-assortment ratios is significant).ocr-alevel-biology-genetics-inheritance / sex-linkage(sex-linked genes are linked to each other on the X — this is a special case of autosomal linkage on a sex chromosome).ocr-alevel-biology-biodiversity-evolution / evolution-natural-selection(recombination maintains genetic variation in populations and combines beneficial mutations from different lineages — central to the long-term action of natural selection).
Practical Activity Group anchor: PAG 1 — Microscopy (chiasmata can be visualised in stained meiotic preparations of locust testis or lily anther) and PAG 11 — Research skills (planning) + PAG 12 — Research skills (reporting) (chi-squared analysis of test-cross data to detect linkage and compute recombination frequencies).
Question (9 marks): A geneticist crosses pure-breeding grey-bodied, long-winged Drosophila (BBVV) with pure-breeding black-bodied, vestigial-winged Drosophila (bbvv). All F1 are grey, long. The F1 are test-crossed with bbvv flies. Out of 2300 offspring, the geneticist counts 965 grey long, 944 black vestigial, 206 grey vestigial, and 185 black long.
(a) Use a chi-squared test to determine whether the data fit the 1:1:1:1 ratio expected from independent assortment. [4] (b) Suggest an explanation for the observed deviation. [2] (c) Calculate the recombination frequency and state what it represents. [3]
| Mark | AO | Awarded for |
|---|---|---|
| 1 | AO1 | Stating null hypothesis (data fit 1:1:1:1) |
| 2 | AO2 | Calculating expected values (each class: 2300/4 = 575) |
| 3 | AO2 | Computing χ² ≈ very large (≫7.82) |
| 4 | AO3 | Concluding "reject H₀ — deviation significant" using df = 3 and critical value 7.82 |
| 5 | AO2 | Identifying B and V loci as linked on same chromosome |
| 6 | AO3 | Explaining the excess of parental classes as the consequence of linkage |
| 7 | AO2 | Calculating recombinants = 206 + 185 = 391 |
| 8 | AO2 | Calculating COV = 391/2300 × 100 ≈ 17% |
| 9 | AO3 | Stating that 17% represents a map distance of 17 cM between B and V loci |
AO split: AO1 = 1, AO2 = 5, AO3 = 3.
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