The final step in the hybridization of radioactive DNA and normal DNA is to:
1 answer:
You might be interested in
Answer:
25% colorblind daughter: 25% colorblind son: 25% carrier daughters with normal vision: 25% normal son.
Explanation:
The genotype of a color-blind man is X^cY and the genotype of the heterozygous carrier female is X^cX. A cross between X^cY and X^cX would produce a progeny in following ratio=
25% colorblind daughter: 25% colorblind son: 25% carrier daughters with normal vision: 25% normal son.
Therefore, the couple is likely to have 50% normal son and 50% affected son. Likewise, the couple is likely to have 50% normal daughters and 50% colorblind daughters.
