Final answer:
In a cross between a heterozygous male bird for a Z-linked mutation and a wild type female, 50% of the male progeny will be mutant males, as the male has a 50% chance of passing on the mutant Z chromosome. Option b) 1/2 is the correct answer.
Step-by-step explanation:
If a male bird that is heterozygous for a recessive Z-linked mutation is crossed to a wild type female, the proportion of the progeny that will be mutant males can be determined by understanding how Z-linked genes are inherited.
Birds have a ZW sex-determination system, where males are ZZ (homozygous) and females are ZW (heterozygous). A male bird that is heterozygous for a Z-linked trait has one Z chromosome with the mutation and one Z chromosome without it, while a wild type female would have a normal Z and a W chromosome.
When these two mate, the male can pass on either the mutant Z or the normal Z, and the female can pass on her Z or W chromosome. The offspring possibilities are as follows:
- Z (mutant) from father and Z from mother = mutant male
- Z (normal) from father and Z from mother = normal male
- Z (mutant) from father and W from mother = normal female
- Z (normal) from father and W from mother = normal female
So, considering the only way to get a mutant male is when the mutant Z from the father is passed on, 50% of the male progeny will be mutant. Therefore, the correct answer to what proportion of the progeny will be mutant males is b) 1/2.