An individual with genotype a⁺ b⁺ / a b shows a patch of yellow body cells with short and thick hairs due to somatic recombination involving the linked genes a and b during mitosis (option 2).

Introduction

A classic genetics question describes an organism where allele a⁺ gives grey body and dominant allele b⁺ gives long thin hairs, while recessive alleles a and b cause yellow body and short thick hairs, respectively. An individual with genotype a⁺ b⁺ / a b unexpectedly shows a localized patch of yellow cells with short and thick hairs, and the task is to identify the most likely chromosomal event producing this somatic mosaicism.​

Question restatement and genotype logic

• Allele a⁺ (wild type) = grey body; recessive a = yellow body.​

• Allele b⁺ (wild type) = long thin hairs; recessive b = short thick hairs.​

• a⁺ and b⁺ are dominant over a and b, so germline genotype a⁺ b⁺ / a b normally gives overall grey body with long thin hairs.​

The appearance of a patch (clone) of cells that are yellow with short thick hairs means that in those cells the genotype has become a a / b b, i.e. homozygous for both recessive alleles. Because only a localized patch is affected, this change must have occurred in a single somatic cell followed by mitotic divisions, creating a genetic mosaic.​

Correct option: (2) Somatic recombination involving a and b

Somatic (mitotic) recombination between homologous chromosomes can occur during mitosis, leading to reciprocal exchange of chromosome segments and loss of heterozygosity distal to the crossover point. In an a⁺ b⁺ / a b individual where the genes are linked in trans (a⁺ with b⁺ on one chromosome, a with b on the homolog), a crossover between a and b or between the centromere and the genes can generate daughter cells that are homozygous for a b (mutant) and a⁺ b⁺ (wild type).​

If such a crossover happens in an early somatic precursor cell, one of the daughter lineages becomes a a / b b and expresses the double mutant phenotype—yellow body with short thick hairs—giving rise to a visible patch on an otherwise wild‑type background. This phenomenon, known as mitotic recombination–induced mosaicism or twin‑spot formation, is well documented in Drosophila for traits like yellow body and altered bristle morphology, matching the description in the question.​

Therefore, somatic recombination involving genes a and b during mitosis is the most likely cause of the yellow, short‑hair patch, making option (2) correct.​

Why option (1) Non‑disjunction of homologous chromosomes during mitosis is incorrect

Non‑disjunction is the failure of homologous chromosomes or sister chromatids to separate properly, producing aneuploid daughter cells with abnormal chromosome numbers. In somatic tissues this usually leads to cell death, reduced viability, or severe developmental defects, not a clean, viable clone showing only a specific recessive phenotype while retaining the rest of the genome.​

In this question, the patch shows normal growth but only changes for the a and b traits, which is more consistent with a localized loss of heterozygosity at those loci through mitotic recombination rather than whole‑chromosome gain or loss through non‑disjunction. Hence option (1) does not best explain the observed mosaic patch.​

Why option (3) Translocation occurring in a few somatic cells is incorrect

Translocation involves the exchange or movement of chromosome segments between non‑homologous chromosomes, often disrupting genes or altering their regulation but not specifically converting a heterozygous region into a homozygous recessive state. A somatic translocation might cause loss-of-function if a break interrupts a gene or places it in an unfavorable context, but it typically affects only one allele and does not neatly generate a a a / b b genotype from a⁺ b⁺ / a b.​

Moreover, a balanced translocation would usually leave gene copy number and dominance relationships intact, whereas an unbalanced one may cause broad cellular dysfunction, not just a discrete yellow, short‑hair patch with normal viability. Thus, translocation is an unlikely explanation, so option (3) is rejected.​

Why option (4) Mutation of both alleles in the somatic cell is unlikely

For the patch to be a a / b b, independent somatic mutations would have to convert both dominant alleles a⁺ and b⁺ to their recessive forms in the same cell lineage. The probability of two specific loss‑of‑function mutations happening together in one somatic cell is extremely low compared with a single recombination event between existing mutant and wild‑type alleles on homologous chromosomes.​

Additionally, the recessive alleles a and b already exist on the homologous chromosome in this individual, so mitotic recombination is a much more parsimonious mechanism to uncover them by generating homozygosity, rather than invoking new mutations. Therefore, option (4) is not the most likely event.​

Key exam takeaways

• A localized patch of recessive phenotype in a heterozygous individual indicates somatic mosaicism and suggests a mitotic event rather than a germline mutation.​

• When both linked recessive traits appear together in a patch, somatic recombination between homologous chromosomes causing loss of heterozygosity (a⁺ b⁺ / a b → a a / b b and a⁺ a⁺ / b⁺ b⁺) is the most likely mechanism.​

• Non‑disjunction, translocations, and multiple independent somatic mutations are less efficient explanations because they either affect chromosome number, disturb many loci, or are statistically rare events compared with a single crossover.​