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).

• 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?

Based on the given information, let's analyze the data:

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.

i) Is the wingless gene autosomal or Z-linked?
- Since all the wingless chickens are female and the mutant wingless phenotype is due to a mutation at a single gene, it indicates that the wingless gene is 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)
- We know that the wingless gene is Z-linked, and the F1 male is phenotypically normal-winged. So, the genotype of the normal-winged F1 male parent would be ZAZA (homozygous dominant for the wingless gene on the Z chromosome).

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)
- The F1 female is also phenotypically normal-winged. Since the wingless gene is Z-linked, the genotype of the normal-winged F1 female parent could be either ZAZA (homozygous dominant for the wingless gene on the Z chromosome) or ZAZa (heterozygous for the wingless gene on the Z chromosome).

iv) What is the genotype of the wingless female offspring of the cross? (AA, Aa, aa, A_, ZAZA, ZAZa, ZaZa, ZAZ, ZAW, or ZaW).
- All the wingless offspring are female. Since the wingless gene is Z-linked, the only possible genotype for the wingless female offspring would be ZaW (homozygous for the recessive wingless allele on the Z chromosome and possessing W chromosome).

In conclusion:
- The wingless gene is Z-linked.
- The genotype of the normal-winged F1 male parent of the cross is ZAZA.
- The genotype of the normal-winged F1 female parent of the cross could either be ZAZA or ZAZa.
- The genotype of the wingless female offspring of the cross is ZaW.

To answer the questions, we need to analyze the provided information and make deductions based on the patterns observed in the offspring.

b) From the given data, we know that the wingless phenotype is observed only in female offspring. This suggests that the gene responsible for the wingless phenotype is sex-linked, specifically Z-linked, as the wingless gene is only expressed in females who have one Z chromosome.

i) The wingless gene is Z-linked.

ii) As mentioned, the F1 male parent is phenotypically normal-winged and produces a 3:1 ratio of normal to wingless offspring when crossed with the F1 female. This indicates that the F1 male must be heterozygous for the wingless gene. Based on the symbols provided, the genotype of the normal-winged F1 male parent can be represented as ZAZa, where "Za" represents the recessive wingless allele.

iii) Similarly, the normal-winged F1 female parent must also be heterozygous for the wingless gene to produce a 3:1 ratio of normal to wingless offspring. Thus, the genotype of the normal-winged F1 female parent can also be represented as ZAZa.

iv) Since the wingless phenotype is observed only in female offspring, it means that the female offspring receive the recessive allele from both the F1 male and F1 female parents. Therefore, the genotype of the wingless female offspring is ZaZa.

In summary:
i) The wingless gene is Z-linked.
ii) The genotype of the normal-winged F1 male parent is ZAZa.
iii) The genotype of the normal-winged F1 female parent is ZAZa.
iv) The genotype of the wingless female offspring is ZaZa.