20. The following is a schematic representation of a hypothetical pathway involved in formation of eye color in an insect species.

Genes A and B are linked and have a map distance of 10cM. Females With genotypes a⁺ab⁺b are test crossed. Further. in these females. the two genes are linked in cis. a⁺ and b⁺ represent wild type alleles, while a and b are null alleles. The progeny of the test cross have individuals with four different eye colours.
What is the expected ratio of individuals with eye color Red: Vermillion: Brown: White in the progeny?
(1) 9:3:3:1           (2) 1:1:1:1
(3) 9:1:1:9           (4) 1:9:9:1

Genes A and B controlling insect eye color are linked with 10 cM distance, so the test cross of a female a⁺a b⁺b (cis) with recessive male produces four eye colours in a specific non-Mendelian ratio: the correct expected phenotypic ratio of Red:Vermilion:Brown:White is 9:1:1:9.​

Introduction

This CSIR NET June 2017 genetics question describes a hypothetical insect eye colour pathway where gene A converts white pigment X to vermilion and gene B converts white pigment Y to brown, and together these pigments yield red eyes. Genes A and B are linked at 10 cM, and a cis heterozygous female a⁺a b⁺b is test crossed with aabb, producing four eye colours whose frequencies depend on recombination between linked genes.​

Genetic setup and phenotypes

• a⁺ = functional A allele, converts pigment X so vermilion pigment is produced.​

• a = null A allele, no vermilion pigment from X (pigment X remains white).​

• b⁺ = functional B allele, converts pigment Y so brown pigment is produced.​

• b = null B allele, no brown pigment from Y (pigment Y remains white).​

Phenotypes in terms of pigments:

• Red: both vermilion and brown present (A⁺ and B⁺ present).​

• Vermilion: only vermilion present (A⁺ present, B null).​

• Brown: only brown present (B⁺ present, A null).​

• White: neither pigment present (both A and B null).​

The test-cross female genotype is a⁺b⁺/ab (cis), male is ab/ab.​

Gamete frequencies with 10 cM linkage

Map distance 10 cM means 10% recombinant gametes and 90% parental gametes are produced by the heterozygous female.​

• Parental gametes: a⁺b⁺ and ab → together 90%, so each is 45%.​

• Recombinant gametes: a⁺b and ab⁺ → together 10%, so each is 5%.​

When these gametes fertilise ab sperm, zygotes are:

• a⁺b⁺/ab → genotype A⁺B⁺ (red).

• ab/ab → genotype aabb (white).

• a⁺b/ab → A⁺bb (vermillion).

• ab⁺/ab → aaB⁺ (brown).​

Thus expected proportions:

• Red: 45%

• White: 45%

• Vermilion: 5%

• Brown: 5%​

Dividing by 5 gives the ratio Red:Vermilion:Brown:White = 9:1:1:9.​

Phenotypic ratio table

Option-by-option explanation

The options are:

1. 9:3:3:1

2. 1:1:1:1

3. 9:1:1:9

4. 1:9:9:1​

• Option 1: 9:3:3:1

  • This is the classical Mendelian dihybrid F2 ratio for two independently assorting genes in simple dominance without linkage or test cross.​

  • In this question, the cross is a test cross (heterozygote × double recessive) and the genes are linked at 10 cM, so the parental types must be more frequent than recombinant types; therefore 9:3:3:1 is not applicable.​

• Option 2: 1:1:1:1

  • A 1:1:1:1 ratio arises in a test cross when the two genes are unlinked and assort independently, producing equal frequencies of the four genotype combinations.​

  • Because genes A and B are only 10 cM apart, parental phenotypes (red and white) must greatly outnumber recombinants (vermillion and brown), so equal 1:1:1:1 classes contradict the given map distance.​

• Option 3: 9:1:1:9

  • This option reflects 90% parental and 10% recombinant gametes from a heterozygote with linked genes 10 cM apart, matching the calculated red 45%, vermilion 5%, brown 5% and white 45%.​

  • It correctly makes red and white (parental) much more common than vermilion and brown (recombinants), so option 3 is the correct answer.​

• Option 4: 1:9:9:1

  • Here recombinants (vermillion and brown) would be nine times more frequent than parentals (red and white), implying a very high recombination rate contrary to the small map distance of 10 cM.​

  • This pattern is not biologically consistent with linkage; therefore option 4 is incorrect.