IB Biology HL Topic 4 — Genetics & Inheritance Paper 1 & 2 Core idea ~10 min read

Gene Linkage

Independent assortment only works if the two genes are on different chromosomes. Put them on the same chromosome and they travel together — and every ratio you learned for dihybrid crosses changes.

📚 What you need to know

Linked and unlinked, side by side

Unlinked genes sit on separate chromosomes, so meiosis shuffles them independently. Linked genes sit on the same piece of DNA, so whichever alleles started together tend to finish together.

The only thing that changes is which chromosome the genes are on UNLINKED A a B b two separate chromosome pairs four gamete types, equal numbers LINKED A a B b both genes on one chromosome pair mostly AB and ab gametes Linked alleles stay in the combination they arrived in only crossing over can break them apart
In the right-hand panel, A and B are stuck on the same chromosome. A gamete that receives that chromosome receives both of them.

Writing linked genotypes

Writing a linked genotype as TtGg hides the information you need. You must show which alleles are on which chromosome.

The standard notation puts the alleles from one chromosome together in brackets: (TG)(tg) means T and G are on one chromosome, t and g on its homologous partner. That individual mostly makes TG and tg gametes.

Compare it with (Tg)(tG). Exactly the same four alleles, but arranged differently — and this individual mostly makes Tg and tG gametes instead. Same genotype in the old notation, completely different offspring.

If a question gives you brackets, use brackets in your answer. Dropping them turns a linkage question back into an ordinary dihybrid, which is exactly the mistake the question is testing for.

The test cross that reveals linkage

Take a newt with two genes on the same chromosome: tail length (T normal, t short) and scale colour (G green, g white). Cross a double heterozygote with a double homozygous recessive.

If the genes were unlinked, the heterozygote would make four gamete types in equal numbers — TG, Tg, tG, tg — and the offspring would be 1 : 1 : 1 : 1 across four phenotypes.

If the genes are completely linked as (TG)(tg), the heterozygote makes only two gamete types: TG and tg. The other parent contributes only tg. So there are only two possible offspring, in a 1 : 1 ratio.

Gametes(tg) from parent 2
(TG) from parent 1(TG)(tg) — normal tail, green scales
(tg) from parent 1(tg)(tg) — short tail, white scales
The signature of linkage in a test cross unlinked → 1 : 1 : 1 : 1   |   linked → mostly 1 : 1, with a few recombinants
Complete linkage is a simplification. In reality crossing over during meiosis breaks the linkage in a small number of cells, so a few recombinant offspring do appear. That is why real data shows two large classes and two small ones rather than a clean 1 : 1.

How strong is the linkage?

Linkage is not all-or-nothing. It depends on how far apart the two loci are.

Crossing over happens at random points along the chromosome. Two genes far apart have a lot of DNA between them, so there are many places a crossover could fall and separate them. Two genes close together have very little DNA between them, so a crossover between them is rare and they nearly always stay joined.

The strength of linkage is measured in centimorgans. One centimorgan corresponds to a 1% chance that the two genes are separated by crossing over, and the unit is named after Thomas Hunt Morgan, whose fruit fly experiments provided the proof of linkage.

Worked examples

WORKED EXAMPLE 1

A newt with genotype (TG)(tg) is test crossed with a (tg)(tg) newt. Predict the offspring, assuming complete linkage.

Step 1: gametes from parent 1 T is joined to G, and t is joined to g, so only (TG) or (tg) Step 2: gametes from parent 2 Homozygous recessive for both, so only (tg) Step 3: combine (TG)(tg) and (tg)(tg) 1 normal tail, green scales : 1 short tail, white scales Two phenotypes instead of four. If the genes were unlinked you would expect four classes in equal numbers — so this result on its own is evidence of linkage.
WORKED EXAMPLE 2

A dihybrid cross expected to give 9 : 3 : 3 : 1 instead gave 240 : 25 : 22 : 33 out of 320 offspring. Explain what this suggests.

Step 1: compare with the expected numbers Expected: 180 : 60 : 60 : 20 Step 2: describe the pattern Two classes are much higher than expected and two are much lower Step 3: identify which The over-represented classes are the parental combinations; the under-represented ones are the recombinants Step 4: explain Alleles that arrived together stayed together, so they did not assort independently The two genes are linked — on the same chromosome A chi-squared test on this data gives 72.93 against a critical value of 7.82, so the difference is significant, not chance.

💡 Exam tip

⚠ Common mix-up

Up next: Identifying Recombinants — how crossing over produces new allele combinations, and how to calculate a recombination frequency.

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