Final answer:
The correct answer is 50%, meaning half of the male offspring of a heterozygous male bird and a wild type female would be mutant, since male birds inherit one Z chromosome from each parent and have a 50% chance of inheriting the Z-linked mutation.
Step-by-step explanation:
The correct direct answer to the question is: 50% of the progeny will be mutant males. When a male bird is heterozygous for a recessive Z-linked mutation, it means that one of his Z chromosomes carries the mutation, while the other does not. Since birds have a ZW sex determination system, where males are ZZ (homogametic) and females are ZW (heterogametic), the offspring's sex and phenotype depend on which chromosome they inherit from their father.
When the heterozygous male (ZMZw) mates with a wild type female (ZW), their offspring will be as follows: Male offspring will inherit the Z chromosome from their mother and one of the father's Z chromosomes. Therefore, half the male progeny will receive the mutant Z chromosome (ZMZ) and be mutant, while the other half will receive the wild type Z chromosome (ZwZ) and be of wild type phenotype. This results in a 50% chance of mutant males. Female offspring will inherit the father's Z chromosome and the mother's W chromosome, giving them only one copy of the Z-linked gene. Since the mutation is recessive and the females' only Z chromosome is wild type, none of the female offspring will be mutant. Thus, the explanation in 200 words for the expected proportion of mutant males is a 50% likelihood.