Answer:
1/4 or 0.25
Explanation:
Given in the question,
Genotype for male chicken ZZ
Genotype for female chicken ZW
phenotype genotype
Barred trait B
Unbarred trait b
rose comb C
single comb c
male [tex]Z^{bc} Z^{bC}[/tex]
female [tex]Z^{Bc}W^{c}[/tex]
The proportion of the resulting progeny which is expected to be barred males with single combs is 1/4 or 0.25
Z^{Bc} W^{c}
Z^{bc} Z^{bc} Z^{Bc} Z^{bc}W^{c}
Z^{bC} Z^{bC} Z^{Bc} Z^{bC}W^{c}