Question

You are studying genetics in domestic chickens (Gallus gallus) and you find a mutant with no wings. Assume that the mutant wingless phenotype is due to a mutation at a single gene.
In contrast to mammals, chickens use ZZ/ZW sex-linkage. That is, male chickens have two copies of the sex chromosome, the Z chromosome, while females have one copy of Z and one copy of W. The W is like the Y in mammals; it carries very few genes.
You should use the following symbols in this problem:
• If the gene is autosomal, use:
• “A” for the allele with the dominant phenotype
• “a” for the allele with the recessive phenotype
• If the gene is sex-linked (Z-linked) use:
• “ZA” for the allele with the dominant phenotype
• “Za” for the allele with the recessive phenotype
• “W” for the W-chromosome:
• b) You cross an F1 male to an F1 female (both phenotypically normal-winged). In their 400 offspring, you see a 3:1 ratio of normal to wingless. Upon closer examination, you find that all the wingless chickens are female. Based on these data:
• i) Is the wingless gene autosomal or Z-linked?
• ii) What is the genotype of the normal-winged F1 male parent of the cross? (AA, Aa, aa, A_, ZAZA, ZAZa, ZaZa, ZAZ, ZAW, or ZaW)
• iii) What is the genotype of the normal-winged F1 female parent of the cross? (AA, Aa, aa, A_, ZAZA, ZAZa, ZaZa, ZAZ, ZAW, or ZaW)
• iv) What is the genotype of the wingless female offspring of the cross? (AA, Aa, aa, A_, ZAZA, ZAZa, ZaZa, ZAZ, ZAW, or ZaW).

Answer 1

• a) You cross a pure breeding male with normal wings to a pure breeding wingless female. All of the F1 offspring have normal wings. Based on this result, which of the following statements is true:

a-) Therefore Cross of Parent AA X aa or ZAZA X ZaW therefore
The normal wing phenotype is dominant; the wingless phenotype is recessive.
Please help me with part b?