Answer:
25%
Explanation:
The female with brown teeth had a father with normal teeth. Since normal teeth is a recessive trait, the father would have been homozygous recessive for the trait and would have transmitted one recessive allele to the female. Therefore, the female is heterozygous dominant for brown teeth.
Let's assume that the allele for brown teeth is X^B while the one for normal teeth is X^b. The genotype of heterozygous dominant female would be X^BX^b and that of the male with normal teeth would be X^bY.
A cross between X^BX^b and X^bY gives progeny in following phenotype ratio= 1/4 daughter with brown teeth: 1/4 daughter with normal teeth: 1/4 son with brown teeth: 1/4 son with normal teeth. Therefore, the probability of having a daughter with brown teeth is = 25%
