13. The distance between gene A and B is 10 cM. If a genotype    A    b     is selfed, the
a    B
percentage of progeny with genotype aabb will be
(1) 10%            (2) 25%
(3) 0.25%         (4) 0.01%

Question and concept

The question states: “The distance between gene A and B is 10 cM. If a genotype A b / a B is selfed, the percentage of progeny with genotype aabb will be (1) 10% (2) 25% (3) 0.25% (4) 0.01%.” Genes are linked in trans (repulsion phase) because each chromosome carries one dominant and one recessive allele (Ab / aB).

Genetic distance 10 cM means recombination frequency r = 10% = 0.10 between A and B. Thus, 10% of gametes are recombinant and 90% are parental in any meiosis for this heterozygote.

Step 1: Gamete frequencies from Ab/aB

For an Ab/aB individual:

• Parental gametes: Ab and aB together = 1 − r = 0.90, so each parental type = 0.90 ÷ 2 = 0.45.

• Recombinant gametes: AB and ab together = r = 0.10, so each recombinant type = 0.10 ÷ 2 = 0.05.

So gamete types and frequencies are:

• Ab = 0.45

• aB = 0.45

• AB = 0.05

• ab = 0.05

Since the plant is selfed, both parents produce the same four gametes with these frequencies.

Step 2: Getting aabb in the progeny

Genotype aabb can arise only when:

• One gamete is ab (a and b) from parent 1, and

• The other gamete is ab from parent 2.

Thus:

• P(aabb) = P(ab from parent 1) × P(ab from parent 2)

• = 0.05 × 0.05

• = 0.0025 = 0.25%.

Therefore, the correct option is 0.25% (option 3).

Option-wise explanation

• Option (1) 10%: This equals the total recombination frequency (r = 10%), which represents all recombinant gametes (AB and ab together), not the final double-recessive genotype in the selfed progeny. Only a small subset of these gametes combine as ab × ab, so 10% is far too high.

• Option (2) 25%: A value of 25% might occur for aabb if the two genes assorted independently in a standard dihybrid self (AaBb × AaBb) with no linkage; here genes are linked at 10 cM, and recombination is low, so this Mendelian 1/4 proportion does not apply.

• Option (3) 0.25%: Correct. It comes from 5% ab gametes from each parent combining (0.05 × 0.05 = 0.0025 = 0.25%). This matches the expected proportion of aabb progeny when Ab/aB individuals (genes 10 cM apart, trans configuration) are selfed.

• Option (4) 0.01%: This would correspond to 0.0001 frequency, as if each ab gamete occurred at 1% and then multiplied (0.01 × 0.01). However, with a recombination frequency of 10%, each recombinant class appears at 5%, not 1%, so 0.01% is too small.

Thus, the percentage of progeny with genotype aabb when Ab/aB (distance 10 cM) is selfed is 0.25%, corresponding to option (3).