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.
Mitochondria, plastids and chloroplast all are surrounded by a double membrane
Answer:
A. 1/16
Explanation:
If we break the dihybrid cross of AaBb X AaBb into individual monohybrid crosses:
Aa X Aa :
A a
A AA Aa
a Aa aa
1/4 of the progeny will be homozygous for dominant allele (AA)
Similarly, in cross Bb X Bb 1/4 of the progeny will be homozygous for dominant allele (BB)
Hence, in cross AaBb X AaBb:
1/4 * 1/4 = 1/16 will be homozygous for both the dominant alleles.