Answer:
The correct answer is - XDXd and XD Y
Explanation:
Sex-linked disorders are the disorders that are expressed only when there is two copy of the mutated allele or have not masked by the dominant allele of the gene. In DMD, the dominant allele is XD and the mutated allele is Xd. To express it there are the following genotype possible-
in women XdXd, and in men XdY, so these two genotypes are not possible for the parents in this case. If a mutated allele is present with the dominant allele it will be considered as carrier women but diseased will not be found in them.
So the correct genotype would be
mother: XDXd ( a carrier woman)
father : XDY (normal man)
<span>Primitive, single-celled life forms without organized nuclei are termed PROKARYOTES.</span>
I think the correct answer from the choices listed above is the first option. The circumstances that this condition would manifest would be if the child is a male and its mother has the recessive allele. X-linked recessive inheritance<span> is a mode of </span>inheritance<span> in which a </span>mutation<span> in a </span>gene<span> on the </span>X chromosome<span> causes the phenotype to be expressed in males </span><span>and in females who are homozygous for the gene mutation</span>
