Huntington's disease is caused by the allele h, which is completely dominant to the normal recessive allele h. consider a couple
where the man has the genotype hh, and the woman has the genotype hh. what is the chance that their child will develop huntington's disease?
1 answer:
Thr genotypes of the parents are Hh heterozygote (the father) and hh recessive homozygote (the mother).
The father has a dominant allele (H) that causes him to develop Huntington disease.
As you can see in the punnet square presented below, he will pass this H allele to a half of his offspring, giving them Huntington's disease too.
So, the possibility of the offspring of this couple having Huntington's disease is 50%.
The father has a dominant allele (H) that causes him to develop Huntington disease.
As you can see in the punnet square presented below, he will pass this H allele to a half of his offspring, giving them Huntington's disease too.
So, the possibility of the offspring of this couple having Huntington's disease is 50%.
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Answer:
Because the offspring of a male and female are always going to be xx, xx, xy, or xy.
Explanation:
Remember that men have different reproductive chromesomes than women. Women have 2 x chromesomes(xx) while men have 1 x and 1 y(xy). These are given down to their offspring.
So we can write the this out in a punnet square as(link down below).
After you see the visual, you will find that the genotypes are xx, xx, xy, xy.
As I said above, women have sex chromesomes xx, while men have sex chromesomes xy. When you match the genotypes on the punnet square with the ones we know from men and women, we find that two of them match with men and two match with female.
So with this, we can conclude that 2/4 of offspring will always be male and 2 will be female. And as you well know 2/4 is 1/2 which is 50%.
