21. The following is a hypothetical pathway for the development of wild type (red) eye colour in an insect:

Enzymes A and B are encoded by the genes a⁺ and b⁺, respectively. The following statements are made regarding inheritance of the genes involved in the development of eye colour:
A) When two heterozygous individuals of the genotype a⁺ab⁺b are mated, progenies with red, orange, brown and white eye colour will be observed irrespective of whether the genes are independently assorting or showing incomplete linkage.
B) When two heterozygous individuals of the genotype are mated, progenies with red, orange, brown and white eye colour will be observed in a ratio of 9:3:3:1, when the genes are independently assorting.
C) When an heterozygous individual of the genotype a⁺b/ a b⁺ is test crossed, progenies with red and white eye colour will be more in number if genes are linked.
D) When an heterozygous individual of the genotype a⁺b/ ab⁺ is test crossed, progenies with orange and brown eye colour will be more in number if genes are linked.
Which of the above statements is TRUE?
(1) A and C        (2) B and C
(3) A,B, and C    (4) A, B and D

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Question overview and pathway

The figure shows a hypothetical pathway for wild‑type red eye colour in an insect.
A colourless precursor W is converted by enzyme A into orange pigment X, and a colourless precursor Y is converted by enzyme B into brown pigment Z. Both X and Z together produce wild‑type red eye colour. Enzyme A is encoded by gene a+a+ and enzyme B by gene b+b+. Recessive mutants $a$ and $b$ are loss‑of‑function alleles that block their respective steps.

Thus:

• a+b+a+b+: both enzymes active → red eyes (orange + brown).

• a+bba+bb: only enzyme A active → orange eyes.

• aab+aab+: only enzyme B active → brown eyes.

• $aabb$: neither enzyme active → white (colourless) eyes.

This is a classic duplicate gene action with complementary pigments giving a 9:3:3:1 ratio in a dihybrid cross when the genes assort independently, and epistatic deviations when they are linked or tested.

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Step 1: Genotypes and phenotypes

For any cross, first list functional vs non‑functional alleles:

• Gene A:

  • $A$ (or a+a+) = functional enzyme A → orange pigment X.

  • $a$ = non‑functional → no orange pigment.

• Gene B:

  • $B$ (or b+b+) = functional enzyme B → brown pigment Z.

  • $b$ = non‑functional → no brown pigment.

Phenotypes:

• $A-B-$ (at least one dominant allele at both loci) → red.

• $A-bb$ → orange.

• $aaB-$ → brown.

• $aabb$ → white.

These phenotype classes are crucial for evaluating each option.

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Option A explained

Statement A: “When two heterozygous individuals of genotype a+a b+ba+ab+b (i.e., $AaBb\times AaBb$) are mated, progenies with red, orange, brown and white eye colour will be observed irrespective of whether the genes are independently assorting or show incomplete linkage.”

1. If genes assort independently (no linkage):
   The classic dihybrid cross $AaBb\times AaBb$ gives 16 genotypes in a 9:3:3:1 ratio.

   • 9 $A-B-$ → red.

   • 3 $A-bb$ → orange.

   • 3 $aaB-$ → brown.

   • 1 $aabb$ → white.
     All four phenotypes appear.

2. If genes are incompletely linked:
   With incomplete linkage, recombinant classes are reduced but not absent, so all four genotype classes still occur in some proportion.
   Therefore, red, orange, brown and white phenotypes still appear, though in altered ratios.

Hence, the key part “all four eye colours will be observed whether the genes are independently assorting or show incomplete linkage” is true.

Verdict for A: Correct.

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Option B explained

Statement B: “When two heterozygous individuals of the genotype a+a b+ba+ab+b are mated, progenies with red, orange, brown and white eye colour will be observed in a ratio of 9:3:3:1, when the genes are independently assorting.”

• As shown above, a standard dihybrid cross with independent assortment produces the Mendelian ratio 9:3:3:1 across four phenotypes.

• Here, the four phenotypes correspond exactly to the four genotype classes: red (9), orange (3), brown (3), white (1).

Thus, if the genes assort independently, the 9:3:3:1 ratio specifically is expected and matches the pathway explanation.

Verdict for B: Correct.

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Option C explained

Statement C: “When a heterozygous individual of genotype a+b/a b+a+b/ab+ is test‑crossed, progenies with red and white eye colour will be more in number if genes are linked.”

Genotype notation in the option is showing repulsion (trans) phase: one chromosome carries a+ba+b (functional A, mutant B), and the homolog carries ab+ab+ (mutant A, functional B). The test cross mate is $ab/ab$ (double recessive):

• If genes are unlinked:
  Gametes from a+b/ab+a+b/ab+ will be a+ba+b, ab+ab+, a+b+a+b+, and $ab$ in equal proportion (25% each). This gives four progeny phenotypes: orange, brown, red, and white, each at 1:1:1:1 ratio.

• If genes are tightly linked in repulsion:
  Recombination between a+ba+b and ab+ab+ is rare, so parental gametes a+ba+b and ab+ab+ dominate.

  • Parental gametes produce mostly orange (a+b/aba+b/ab) and brown (ab+/abab+/ab) phenotypes.

  • Recombinant gametes a+b+a+b+ and $ab$ are fewer, so red and white progenies are less frequent, not “more in number”.

Therefore, the statement claims the opposite of what linkage actually produces in a trans‑heterozygote test cross.

Verdict for C: Incorrect.

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Option D explained

Statement D: “When a heterozygous individual of genotype a+b/ab+a+b/ab+ is test‑crossed, progenies with orange and brown eye colour will be more in number if genes are linked.”

Using the same trans‑heterozygous arrangement and test cross as in option C:

• With linkage, parental chromosome combinations a+ba+b and ab+ab+ are transmitted more often.

• These parental gametes produce:

  • Orange eyes from a+ba+b paired with $ab$ (test parent) → a+a bba+abb.

  • Brown eyes from ab+ab+ paired with $ab$ → aa b+baab+b.

• Recombinants giving red and white appear in lower numbers.

Therefore, if the genes are linked in repulsion, orange and brown phenotypes predominate, exactly as stated.

Verdict for D: Correct.

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Final correct combination of statements

From the detailed analysis:

• A: True

• B: True

• C: False

• D: True

The option that includes A, B and D together is therefore the correct answer:
Answer: (4) A, B and D.

This approach—first mapping genotypes to phenotypes from the biochemical pathway, then applying Mendelian and linkage principles—provides a systematic method to solve similar CSIR NET, GATE, and MSc entrance questions on epistasis in insect eye colour inheritance.