11. Distance between gene A and B is 10 cM.

If the F1 genotype  A     B    was test
a      b
crossed then what is probability of obtaining

A      B    genotype
a      b
(1) 90          (2) 45
(3) 10          (4) 5

Genetic Recombination Frequency for 10 cM Distance between Gene A and B

Introduction

In genetics problems, distances like 10 centimorgans (cM) between two linked genes directly indicate the percentage recombination between them, which is crucial for predicting progeny genotypes in test crosses. For CSIR NET, GATE and similar exams, understanding how to convert map distance into parental and recombinant gamete frequencies allows precise calculation of probabilities such as the frequency of AB/ab offspring when the F₁ is in coupling phase.

Concept: 10 cM and Recombination

One centimorgan corresponds to 1% recombination between two loci; therefore, 10 cM means a 10% recombination frequency between gene A and gene B.

With two linked genes, a 10% recombination frequency means 10% recombinant gametes in total and 90% parental gametes in total.

Phase of F₁ and Gamete Types

The F₁ genotype is A B / a b, where dominant alleles A and B are on the same homologous chromosome (coupling or cis phase).

• Parental gametes (no crossover): AB and ab
• Recombinant gametes (with crossover): Ab and aB

Since total recombinants = 10%, each recombinant class (Ab and aB) = 10% / 2 = 5%.
Similarly, total parentals = 90%, so each parental class (AB and ab) = 90% / 2 = 45%.

Test Cross and Required Progeny

In a test cross, the F₁ (A B / a b) is crossed with aabb.
The test-cross parent aabb can only produce gamete ab.
Therefore, each progeny genotype frequency equals the corresponding gamete frequency from the F₁.

Required genotype in the question: AB/ab.
This genotype arises when the F₁ produces an AB gamete (recombinant) and the tester contributes ab gamete.

Frequency of AB gamete from F₁ = 5%, so probability of AB/ab progeny = 5%.

Explanation of Options

Use the fact that “10 cM = 10% recombination; each recombinant class = 5%; each parental class = 45%” for discussion.

Option	Value (%)	Interpretation for AB/ab Progeny
(1)	90	Total parental gametes (AB + ab), not single AB class, so incorrect.
(2)	45	Frequency of each parental gamete type (AB or ab), valid only if AB were parental, but here AB is recombinant, so incorrect.
(3)	10	Total recombinants (Ab + aB) combined, not single AB/ab genotype, so incorrect.
(4)	5	Frequency of a single recombinant gamete (AB) and thus of AB/ab progeny in a test cross; correct.

Key Exam Points

• Distance in cM numerically equals recombination percentage for two loci: 10 cM → 10% recombination.
• In a dihybrid with linked genes, divide total recombination equally between the two recombinant gamete types and the remaining percentage equally between the two parental types.
• In a test cross with aabb, the percentage of a particular progeny genotype directly equals the percentage of the corresponding gamete from the F₁.