9. A wet-mount slide preparation of a specimen is stained in order to
1 answer:
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Answer:
Normal daughters and Hemophiliac son
Explanation:
Females exhibit hemophilia and colorblindness in homozygous genotype only.
The genotype of hemophiliac woman= X^hX^h
The genotype of colorblind man= X^cY
A cross between X^hX^h and X^cY would produce progeny in the following phenotype ratio= 1 Normal but carrier daughters:1 Hemophiliac son
