A man with a genetic disease marries a woman who does not carry the disease. It is not possible for their sons to have the disea
se, but their daughters will inherit it. The disease must be
1 answer:
<h2>X linked dominant </h2>
Explanation:
In X linked dominant if the father carries the abnormal X gene,all of his daughters will have the disease because daughters always inherit their father's X chromosome and if the mother carries the abnormal X gene, half of all their children (daughters and sons) will inherit the disease tendency
- In this case man is the one who is diseased and female is normal so daughters in this case will surely inherit the disease
- When a cross is made,all the sons are normal whereas daughters are diseased
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Answer:
Probability of having a child with genotype Ww is 50%.
Explanation:
This question involves a single gene coding for hairline shape in humans. The allele for widow's peak hairline (W) is dominant over the allele for straight hairline (w). This means that allele W will mask the phenotypic expression of allele w in a heterozygous state.
In a cross between a mother with genotype, Ww and father with genotype, ww, the following gametes will be produced by each parent:
Mother (Ww)- W and w
Father (ww)- w and w
Using these gametes in a punnet square (see attached image), four possible offsprings will be produced with genotypes: Ww, Ww, ww and ww.
Ww(2) : ww(2) or Ww(1) : ww(1)
Hence, 1 out of 2 children produced by the parents will have a genotype Ww, i.e. 1/2. Therefore, the percentage probability is 1/2 × 100= 50%.
