(A) 1/4

(B) 1/9

(C) 1/12

(D) 1/36

(E) 1/44

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(A) 1/4

(B) 1/9

(C) 1/12

(D) 1/36

(E) 1/44

Would you explain for us how you came to this conclusion. I really would like to understand this tricky questionoption D. 1/36

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120 Posts

Mom (Aa) x Dad (Aa)

so Bob has a 2/3 chance of being a carrier

Sarah since she has the same parents also has a 2/3 chance of being a carrier

We say 2/3 since their brother Tom was the aa

3rd generation:

The child has (1/2) chance of being a boy and (1/2) chance of being a girl.

SO: (2/3)(2/3)(1/2)(1/2)(1/2)= 1/36

Hope that helps.

1. parents of Bob and Sarah are heterozygous Aa (that is why one of their sons died in past - he was with aa genotype);

2. Bob has possibility to be a heterozygous Aa 2/3, his sister Sarah has the same possibility 2/3, both of them hav 2/3*2/3=4/9 possibility to be Aa together,

3. We suppose that Bob and Sarah married people with AA genotype, so their chldren have possibility to be a geterozygous Aa 1/2 for Adam and 1/2 for Jennifer, their possibility together 1/2*1/2=1/4

4. Adam's and Jennifer's child has a possibility to have aa genotype and be sick 1/4 (because we suppose that parents Aa),

5. 4/9*1/4*1/4=1/36

Hi,

It doesn't seem to be correct!

The problem should gave us the frequencies of the allels in population and with that we can calculate the probability of being carrier for Bob's wife and Sarah's husband.

I think for these types of question which we don't have the allel's frequencies, we should consider Bob's wife and Sarah's husband to be phenotipically and genotypically healthy 'cause this disease is considered to be UNCOMMON.

if anyone is Consentaneous me, plz say!

isnt the calculation wrong here [??] it looks like ur answer should be 1/18:notsure:

When you grid the parents Bob's and Sarah's parents were:1st generation:

Mom (Aa) x Dad (Aa)

so Bob has a 2/3 chance of being a carrier

Sarah since she has the same parents also has a 2/3 chance of being a carrier

We say 2/3 since their brother Tom was the aa

: We're not given any info about Bob's wife or Sarah's husband so they have a 1/2 (50%) chance of being carriers as well.2nd generationBob's son

3rd generation:

Adam(??) X Sarah's daughter Jennifer(??)

The child has (1/2) chance of being a boy and (1/2) chance of being a girl.

SO: (2/3)(2/3)(1/2)(1/2)(1/2)= 1/36

Hope that helps.

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