19. A hypothetical biochemical pathway for the formation of eye color in insect is given below.

Two autosomal recessive mutants ‘a’ and ‘b’ are identified which block the pathway as shown above. Considering that the mutants are not linked, what will be the phenotype of the F2 progeny if crosses were made between parents of the genotype aaBB x AAbb, and the F1 progeny are intercrossed?
(1) 9 orange-brown: 3 orange; 3: brown: 1 colorless
(2) 9 orange-brown: 7 colorless
(3) 1 orange: 2 colorless
(4) 15 orange-brown: 1 colorless

Introduction
The CSIR NET December 2014 Life Sciences paper included a classic dihybrid genetics problem on eye colour in an insect, based on a biochemical pathway and two autosomal recessive mutants a and b that block pigment formation. Understanding how these mutants affect brown and orange pigment synthesis allows prediction of F2 phenotypic ratios when parents aaBB and AAbb are crossed and the F1 are intercrossed. This article explains the pathway logic, derives the ratio step‑by‑step, and evaluates every option from the exam perspective.

Pathway and genotype–phenotype logic

The pathway shows that substrate X becomes a colourless intermediate and then brown pigment via gene A, while substrate Y becomes a colourless intermediate and then orange pigment via gene B. Mutant allele a blocks conversion to brown pigment, and mutant allele b blocks conversion to orange pigment; both mutants are recessive, so only aa or bb genotypes stop their respective branches. When both functional pigments are produced together (A_B_), the eye appears orange‑brown (wild type), whereas loss of one pigment gives either brown (aaB_) or orange (A_bb), and loss of both gives colourless (aabb).

Cross design and F1 genotype

Parents in the question are aaBB (no brown, only brown gene mutant) and AAbb (no orange, only orange gene mutant). Each parent can produce only one gamete type, so their cross yields F1 genotype AaBb, which is heterozygous at both loci and phenotypically orange‑brown because both A and B are present in dominant form. Thus, intercrossing the F1 effectively means a standard dihybrid cross AaBb × AaBb with independent assortment, as the mutants are not linked.

Deriving the F2 phenotypic ratio

From AaBb × AaBb, the usual Mendelian genotypic ratio 9:3:3:1 is produced for A_B_, A_bb, aaB_, and aabb respectively. Mapping these genotype classes onto the eye‑colour phenotypes defined by the pathway gives:

• A_B_ → both brown and orange pigments → orange‑brown eyes (9/16)

• A_bb → brown only blocked, orange only present? No, here B is recessive mutant, so orange blocked; only brown pigment via A → brown eyes (3/16)

• aaB_ → A recessive mutant, brown blocked; only orange pigment via B → orange eyes (3/16)

• aabb → both enzymes defective, no pigment → colourless eyes (1/16)
  Arranging these as required yields a 9 orange‑brown : 3 orange : 3 brown : 1 colourless phenotypic ratio in the F2 generation.

Evaluating option (1)

Option (1): 9 orange‑brown : 3 orange : 3 brown : 1 colourless.
This option exactly matches the mapping of the 9:3:3:1 dihybrid genotypic classes onto the four phenotypes produced by the pathway. Because the genes are unlinked and each mutant affects only its own pigment branch, this is the correct phenotypic ratio for the F2 progeny from AaBb × AaBb, so option (1) is correct.

Evaluating option (2)

Option (2): 9 orange‑brown : 7 colourless.
A 9:7 ratio is typical of complementary gene interaction, where at least one dominant allele at both loci is required for a single phenotype and any homozygous recessive combination gives the alternative phenotype. In this pathway, however, single‑gene mutants (aaB_ or A_bb) still produce one pigment and therefore yield orange or brown eyes, not colourless, so 7/16 of the progeny do not become colourless; hence this option does not fit the described biochemistry and is incorrect.

Evaluating option (3)

Option (3): 1 orange : 2 colourless.
This ratio would imply that most F2 progeny lack one or both pigments, which contradicts the expected dihybrid segregation from AaBb × AaBb where 15/16 individuals carry at least one dominant allele at one or both loci. Moreover, the dihybrid cross clearly predicts substantial numbers of orange‑brown and brown phenotypes that this option omits, so option (3) is inconsistent with both Mendelian expectations and the pathway model.

Evaluating option (4)

Option (4): 15 orange‑brown : 1 colourless.
A 15:1 ratio usually indicates duplicate gene action, where either of two dominant alleles can independently produce the same phenotype, making only the double recessive different. Here A and B do not produce the same pigment; they make brown and orange pigments respectively, giving distinct phenotypes when only one gene is functional, so A_bb and aaB_ cannot both show the same orange‑brown colour, making this 15:1 pattern impossible and option (4) incorrect.

Therefore, the correct answer is option (1): 9 orange‑brown : 3 orange : 3 brown : 1 colourless.