Answer:
Let the "barred" allele be caller B and the "non-white" allele b. Since chickens use chromosomes [tex]Z[/tex] and [tex]W[/tex] to determine sex, hens would have chromosomes [tex]ZW[/tex], and roosters would have chromosomes [tex]ZZ[/tex]. A Z-linked gene is represented as a superscript on the [tex]Z[/tex] chromosome, [tex]Z^{B}[/tex] for the dominant allele and [tex]Z^{b}[/tex] for the ressesive allele.
A barred hen would have a copy of B on its Z chromosome, a non-barred rooster would have both copies of b on both Z chromosomes. Using Punnet squares to represent the crosses we get the following cases:
[tex]\begin{center}\begin{tabular}{ |c|c|c|c| }\ & Z^{B} & W \\ \ Z^{b} & Z^{B}Z^{b} & Z^{b}W \\ \ Z^{b} & Z^{B}Z^{b} & Z^{b}W \\ \end{tabular}\end{center}[/tex]
That is a ratio of two barred heterozygote roosters to two non-barred hens. Crossing them we get:
[tex]\begin{center}\begin{tabular}{ |c|c|c|c| }\ & Z^{b} & W \\ \ Z^{B} & Z^{B}Z^{b} & Z^{B}W \\ \ Z^{b} & Z^{b}Z^{b} & Z^{b}W \\ \end{tabular}\end{center}[/tex]
That is a ratio of one barred heterozygote rooster to one barred hen to one non-barred rooster to one non barred hen.
