A 0%

B 12.5%

C 25%

D 50%

E 100%

please educate me with rationale. thanx

1 - 8 of 8 Posts

A 0%

B 12.5%

C 25%

D 50%

E 100%

please educate me with rationale. thanx

A 0%

B 12.5%

C 25%

D 50%

E 100%

please educate me with rationale. thanx

because the child may inherete just one mutated allele of each gene so child will be carrier of each/both dis or will be normal

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45 Posts

because each parent is a carrier of a different autosomal recessive disease gene subunit mutation, the child born has a 0% chance of acquiring either of the disease mentioned

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5 Posts

Both disease are autosomal recessive --- so they need 2 chromosomes to MANIFEST .... (carriers have only one defective chromosome)

If carrier mother has chromosomal structure A-x (A = normal and x = gene for tay sachs)

And carrier father has chromosomal structure A-z (A = normal and z = gene for sandoff's)

their possible offsprings will have the following structure

- A-A

- A-z

- A-x

- x-z

But to MANIFEST the disease, the chromosomal structure needs to be x-x(tay sach's) or z-z(sandoff)...

So 0% of the offspring will MANIFEST either disease.

--- How many percent of their offspring willl be CARRIERS of either disease?

Ans - - 75%

---- How many percent of their offspring will be CARRIERS of both diseases?

Ans - - 25%

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590 Posts

Devareddy has a point - HEXA and HEXB are different genes on different chromosomes (15 and 5 respectively). So, if the parents are each carriers for only one of the diseases, the probability that the child will be a carrier for either is 50%, and the probability that the child will be a carrier for both is 25%. The probability that the child will have either disease is 0% if each parent for sure only carries one disease (if the wording of the question leaves open the possibility that a parent could be a carrier unknowingly for the other, then the probability of the child having either disease is 50% of the prevalence of that allele in the general population. Both are rare diseases, so the answer is closer to 0 than to 12 in any case).why have u taken A as the normal(dominant) allele for both the diseases why should nt we consider them as separate genes pls xplain......?

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19 Posts

both dz's are AR, so if both parents are carriers they only carry one of two recessive genes. Thereby making it impossible for any of their offspring to have have the disease (be homozygous) in either of the parent's carried dz's.Devareddy has a point - HEXA and HEXB are different genes on different chromosomes (15 and 5 respectively). So, if the parents are each carriers for only one of the diseases, the probability that the child will be a carrier for either is 50%, and the probability that the child will be a carrier for both is 25%. The probability that the child will have either disease is 0% if each parent for sure only carries one disease (if the wording of the question leaves open the possibility that a parent could be a carrier unknowingly for the other, then the probability of the child having either disease is 50% of the prevalence of that allele in the general population. Both are rare diseases, so the answer is closer to 0 than to 12 in any case).

0%.

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