28. In the following example, 3 independently assorting genes are known to govern coat color in mice. The genotype of few of the coat colors is given below:
Agouti: A-B-C-
Black: aa B-C-
Albino: — — cc

What will be the expected frequency of abino, in the F2 progeny from crosses of pure black with albino of the genotype AAbbcc?
(1) 1/4             (2) 1/16
(3) 1/64          (4) 9/64

Step-by-step solution

Step 1: Assign parental genotypes

From the phenotypic definitions [web:37][web:39]:

• Agouti: A– B– C–
• Black: aa B– C–
• Albino: – – cc (any A/B, but cc makes all pigment absent)

Given albino parent: AAbbcc.

Pure black parent must be: aaBBCC.

Step 2: F₁ from pure black × albino

Cross: aaBBCC × AAbbcc

Gametes: Black (aBC) × Albino (Abc)

F₁ genotype: AaBbCc (all agouti) [web:43].

Step 3: F₂ from F₁ selfing (AaBbCc × AaBbCc)

For each gene independently:

• Aa × Aa → P(aa) = 1/4
• Bb × Bb → P(bb) = 1/4
• Cc × Cc → P(cc) = 1/4

Albino counted as bb cc (matching original albino parent):

de>P(bb cc) = 1/4 × 1/4 = 1/16

Why other options are rejected

• 1/4: frequency of cc alone if B ignored
• 1/64: fully recessive at all three loci (aabbcc)
• 9/64: typical di/tri-hybrid dominant combination

Therefore, expected frequency of albinos in F₂ is 1/16 (option 2).