Answer:
C. "Because you have had one child with Tay-Sachs, you must each carry the allele. Any child you have has a 25% chance of having the disease
Explanation:
Tay-Sachs disease is an autosomal recessive disorder that means a child will acquire the phenotype or we can say disease only when both the alleles which cause disease are present in the child.
Here, in this condition neither parent has Tay-Sachs but they gave birth to a child who is suffering from the disease then it simply means that both the parents are carrier of this disease i.e. they both are heterozygous.
The cross and probability of healthy and diseased child is shown as under:
Mother Father
Parents Tt Tt
/ | | \
Off-springs TT Tt Tt tt
<u>Out of these 4 probable off-springs only 3 with genotype TT, Tt & Tt will be healthy while the 1 with genotype tt will be diseased.</u><u> </u><u>So, the probability of diseased child is 1/4 which is 25%. </u>
Also, the birth of children is an independent event so if one child has already been born with this disease then it doesn't mean that the next 3 children will be healthy for sure i.e. birth of first child cannot influence the next child. In short we can also say that in the next birth, there is still 25% chance of the child being born with this disease because previous birth has nothing to do with next birth.
<span>amount of nitrogen, oxygen, and other elements dissolved in the water. Hope this helps!</span>
Your answer is Dominant. Hope it helps
