Question

Difficulty: HardSex Determination and Sex-Linked Traits

In chickens, sex is determined by the ZZ-ZW mechanism, where males are ZZ and females are ZW. Barred plumage is controlled by a Z-linked dominant allele (ZBZ^B), whereas non-barred plumage is controlled by the recessive allele (ZbZ^b). A non-barred rooster (ZbZbZ^b Z^b) is mated with a barred hen (ZBWZ^B W). If an F1F_1 male offspring is subsequently crossed with an F1F_1 female offspring, what is the probability that a female in the F2F_2 generation will have barred feathers?

  1. A
    25% (1/4)
  2. 50% (1/2)Answer
  3. C
    75% (3/4)
  4. D
    100% (1)

Answer

50% (1/2)
In the F1×F1F_1 \times F_1 cross (ZBZb×ZbWZ^B Z^b \times Z^b W), the male parent produces gametes carrying ZBZ^B and ZbZ^b in equal frequency (50%50\% each). Female offspring inherit the WW chromosome from their mother and one ZZ chromosome from their father. Therefore, 50%50\% of the female offspring receive ZBZ^B and display barred feathers (ZBWZ^B W), while 50%50\% receive ZbZ^b and display non-barred feathers (ZbWZ^b W).

Step-by-Step Solution

1
Determine the genotypes of the P1P_1 parents and F1F_1 offspring
Parental genotypes: Male = ZbZbZ^b Z^b, Female = ZBWZ^B W. F1F_1 male genotype = ZBZbZ^B Z^b (barred male), F1F_1 female genotype = ZbWZ^b W (non-barred female).
The male parent passes a ZbZ^b chromosome to all offspring; female offspring inherit the WW chromosome from the mother.
2
Perform the Punnett square cross for the F1F_1 interbreed (ZBZb×ZbWZ^B Z^b \times Z^b W)
Gametes from F1F_1 male: ZBZ^B, ZbZ^b. Gametes from F1F_1 female: ZbZ^b, WW. F2F_2 genotypes: ZBZbZ^B Z^b (barred male, 25%), ZbZbZ^b Z^b (non-barred male, 25%), ZBWZ^B W (barred female, 25%), ZbWZ^b W (non-barred female, 25%).
Constructing the cross yields all possible F2F_2 genotypic combinations.
3
Calculate the specific probability among female offspring
Female genotypes in F2F_2 are ZBWZ^B W (barred) and ZbWZ^b W (non-barred) in equal proportions. Probability of barred among females = 12=50%\frac{1}{2} = 50\%.
The question asks specifically for the probability within the female subset of offspring, not the total offspring.

Key Concept

ZZ-ZW Sex Determination and Sex-Linked Inheritance Ratios
Estimated Time:1m 30s
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