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%
I am sure that the answer is D
Not sure but I think the answer is D because D for DIS N ya that’s what I think
This would depend on if the other parents has it or not. If they do not, they have a 50/50 chance that the trait will become recessive. That said, it could still be passed on to the child's kids in later generations.
<span>Here are the choices: </span>
(a) deletion(b) inversion (c) duplication(d) insertion
Answer: (d) insertion
This mutation occurs when (an) additional nucleotide base pair/s attach themselves into an existing DNA strand. This can possibly result to diseases such as cystic fibrosis and myotonic dystrophy.