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

Identifying Recombinants

If linked genes always stayed together, a test cross would give exactly two kinds of offspring. It never does — a few offspring turn up with combinations neither parent had. Those are recombinants, and counting them tells you how far apart the two genes sit.

📚 What you need to know

Crossing over makes new combinations

In prophase I of meiosis, homologous chromosomes pair up. Non-sister chromatids touch at points called chiasmata, break, and rejoin to the other chromatid. Whole blocks of alleles are swapped.

The result is chromatids carrying combinations that did not exist before. A chromatid that arrived carrying A and B might leave carrying A and b.

One crossover, two new combinations BEFORE A B a b chiasma forms here chromatids break and rejoin AFTER A B A b a B a b parental recombinant recombinant parental two of the four chromatids are changed A crossover only affects two of the four chromatids so recombinants are always the minority when genes are linked
The two untouched chromatids keep the parental combinations. Only the two that took part in the chiasma carry something new.
A crossover between two linked genes only counts if the chiasma forms between the two loci. A crossover somewhere else on the chromosome swaps DNA, but leaves those two alleles sitting together as before.

Spotting recombinants with a test cross

You cannot see genotypes, only phenotypes. A test cross fixes that: cross the individual with one that is homozygous recessive for both genes.

That partner can only donate recessive alleles, so it contributes nothing visible. Every offspring phenotype is a direct read-out of the gamete it received from the parent being tested.

🧩 Finding the recombinants in a data table

  1. Identify the parent’s allele combinations — for (AB)(ab), those are AB and ab.
  2. Match each offspring phenotype to the gamete it must have received.
  3. Any offspring showing AB or ab is parental. Anything else — Ab or aB — is recombinant.
  4. Confirm with the numbers. The two largest classes are parental; the two smallest are recombinant.
  5. Add the two recombinant classes together before calculating the frequency.
Recombination frequency RF = (number of recombinant offspring ÷ total offspring) × 100%

Distance and recombination frequency

Chiasmata form at random points along the chromosome. So the chance of one falling between two genes depends on how much DNA lies between them.

For linked genes RF is always below 50%, because the two parental classes stay in the majority. An RF of about 50% means the genes are behaving as if unlinked — either they are on different chromosomes, or they are so far apart on the same one that crossovers between them are almost certain.

This is how genetic maps were built. Morgan and his students crossed thousands of fruit flies, measured RF for pair after pair of genes, and used those percentages as distances. A 1% recombination frequency became one centimorgan — a map drawn entirely from counting offspring.

Worked examples

WORKED EXAMPLE 1

A plant with genotype (PL)(pl) is test crossed. Of 1000 offspring: 412 purple/long, 388 red/round, 105 purple/round, 95 red/long. Identify the recombinants and calculate the recombination frequency.

Step 1: what were the parental combinations? The tested parent is (PL)(pl), so PL and pl are parental Step 2: match phenotypes to gametes Purple/long = PL, red/round = pl → parental. Purple/round = Pl, red/long = pL → recombinant Step 3: count the recombinants 105 + 95 = 200 Step 4: apply the formula RF = (200 ÷ 1000) × 100 = 20% RF = 20%, so the loci are about 20 centimorgans apart The two big classes are parental and the two small ones are recombinant — the standard pattern for linked genes.
Offspring phenotypeGamete receivedTypeNumber
Purple flowers, long pollenPLParental412
Red flowers, round pollenplParental388
Purple flowers, round pollenPlRecombinant105
Red flowers, long pollenpLRecombinant95
WORKED EXAMPLE 2

Genes X and Y give a recombination frequency of 4%. Genes X and Z give 31%. Deduce which pair of loci lies closer together, and explain why recombinants appear at all.

Step 1: compare the frequencies 4% is much lower than 31% Step 2: link frequency to distance A low frequency means a chiasma rarely forms between the two loci, so there is little DNA between them X and Y are closer together Step 3: why any recombinants at all? Crossing over in prophase I exchanges sections between non-sister chromatids, breaking the linkage in some cells Both values are below 50%, so both pairs are linked — X and Z are just far enough apart to be separated often.

💡 Exam tip

⚠ Common mix-up

That completes Genetics & Inheritance. Up next: Gene Pools & Speciation — what happens to all these alleles at the scale of a whole population.

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