16. Two genes a and b are located at a distance of 10 cM. Individuals of the genotype AaBb are sib-mated. The two genes are linked in trans. What percentage of the progeny is expected to have the genotype aabb?
(1) 0.25         (2) 0.01
(3) 6.25         (4) 25

Two genes a and b are linked in trans configuration at a distance of 10 centiMorgans (cM), with heterozygous individuals of genotype AaBb sib-mated. The expected percentage of progeny with genotype aabb is 0.25% (option 2).

Explanation of the Problem

• The two genes a and b are linked, meaning they are located close together on the same chromosome.

• Distance of 10 cM corresponds to a 10% recombination frequency (r = 0.1).

• The genes are in trans configuration: One homologous chromosome carries the dominant A and recessive b (Ab), and the other carries recessive a and dominant B (aB).

• An AaBb individual in trans configuration produces four types of gametes:

  • Parental chromatids: Ab and aB each at frequency (1 – r)/2 = 0.45

  • Recombinant chromatids: AB and ab each at frequency r/2 = 0.05

• In sib-mating of AaBb x AaBb (both in trans), the genotype aabb progeny arises only from fusion of two ab recombinant gametes, so frequency:

  $$(0.05)\times (0.05)=0.0025=0.25\%$$

Explanation of Each Option

1. 0.25% (Correct): Matches the calculation based on recombination frequency and trans configuration. This is the percentage of aabb progeny expected in the sib-mated offspring.

2. 0.01%: This is much lower than expected. Likely a distractor or misinterpretation of recombination percentage.

3. 6.25%: This is 116161, the Mendelian expected frequency if genes were unlinked and assorting independently for aabb genotype in a dihybrid cross. But here, linkage and trans phase reduce recombinant frequency, so this does not apply.

4. 25%: Would represent expected frequency of aabb if the parent was homozygous recessive for these genes (aabb) or in classic independent assortment with homozygotes. Incorrect for linked genes in trans.

Introduction

Understanding the expected genotypic ratios in offspring from linked genes is fundamental in genetics. When two genes, a and b, are linked with a known distance of 10 centiMorgans and arranged in trans configuration in AaBb individuals, calculating the proportion of progeny with genotype aabb becomes an interesting problem in linkage genetics.

Detailed Explanation

Two genes a and b on the same chromosome at 10 cM distance have 10% recombination frequency, meaning 10% of gametes show crossover (recombinant gametes). In trans configuration for heterozygous parents AaBb, parental gametes are Ab and aB (45% each), and recombinant gametes are AB and ab (5% each).

When these individuals are sib-mated, progeny genotype aabb appears only when both parents contribute the recombinant ab gamete. The expected frequency is the product of their probabilities: 0.05 (from one parent) × 0.05 (from the other) = 0.0025 or 0.25%.

Other choices such as 6.25% or 25% are typical of independent assortment or homozygous parents and do not apply to linked genes in trans.

This calculation underscores how gene linkage and their cis or trans arrangement impact inheritance patterns in progenies.

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This content provides educators and students with a precise understanding to solve linkage genetics problems involving trans configuration, recombination frequency, and sib mating crosses for expected genotype percentages.