Genes A and B are located on the same chromosome and are 20 % recombination apart. A plant heterozygous for both genes in coupling phase (genotype AB/ab) is crossed with a plant homozygous recessive (ab/ab). What is the probability that an offspring will inherit the recombinant genotype Ab?
Answer
10%
The correct answer is 10% — Thus, 10% of the progeny will have genotype Ab.
Step-by-step solution
- Recognize that the heterozygous parent has two chromosome types: AB (non‑recombinant) and ab (non‑recombinant).
- Recombination occurs in 20% of meioses, producing the two recombinant types: Ab and aB, each in half of the recombinant gametes.
- Calculate the frequency of the recombinant gamete Ab from the heterozygous parent:
- • 20% of gametes are recombinant.
- • Half of these recombinant gametes are Ab, so 0.20 × 0.5 = 0.10 (10%).
- The homozygous recessive parent can only provide the ab gamete.
- Therefore, the probability that an offspring receives Ab from the heterozygous parent and ab from the other parent is 10%.
- Thus, 10% of the progeny will have genotype Ab.
Common mistake: Applying independent assortment and giving 25% instead of using the 20% recombination rate.
Concept tested: Linked genes and recombination frequency
Published · How this solution was prepared and checked
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