I thought regarding this question as HLA gene distribution (en block).
Even if we discard everything and just think about the last sentence (question), for any gene, chance of sibling having exact same allele is 25%
suppose there are 4 alleles for a gene in parents, A,a, B, and b (what I mean to say is all 4 are different in both parents).
Now suppose one daughter has aB genotype. To have exact same alleles, other daughter must have
50% chance of a (among a and A from mother) multiplied by 50% chance of B (among b and B from father)
combined probability for any gene would be 25%.
Please, point out any fault in the reasoning.
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I esdeugs a lot on iths enqsuito nda ogt lkuyc (:
enzyme deficiencys = AR
homozygous presence of CYP..."
QED: homozygous + AR = 25%
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So I hate this question because you can't tell what the parent's genotypes are. This enzyme defect isn't something that you can tell photogenically so we can't just assume the parents are heterogeneous.
It honestly does't even matter if it is autosomal recessive or not since we only have to worry about homozygous off springs. Here is the math.
Parents possible genotypes:
A - Normal; a - abnormal.
Mom: Aa Dad: Aa (what the question is assuming)
Mom: Aa Dad: aa
Mom: aa Dad: Aa
Mom: aa Dad: aa (who is to say this isn't possible?)
(Mom AA or Dad AA is impossible since the offspring can't be aa)
From those genotype possibilities we can calculate better probabilities.
Since it's possible for parents to have any of these 4 genotypes we just assume that each possibility is 1/4.
Mom: Aa + Dad: Aa -> offspring aa = 1/4 (multiply this by 1/4 since we're assuming equal possibility for other genotypes) -> 1/16
Mom: aa + Dad: Aa -> offspring aa = 1/2 -> 1/2*1/4 = 1/8
Mom: Aa + Dad: aa -> offspring aa = 1/2 -> 1/2*1/4 = 1/8
Mom: aa + Dad: aa -> offspring aa = 1 -> 1*1/4 = 1/4
Then we add all the probabilities together 1/16 + 1/8 + 1/8 + 1/4 = 9/16
I know the question's answer is 25%... so I guess put 25% on the test. BUT if you want to actually understand, this is a better approximation. The only problem is we don't know the gene distribution of the alleles or I can give better math :P
Sorry it's long and confusing
There was a uworld question on this. HLA are passed down to offspring in haplotypes. Parent X has A and B haplotype, parent Y has C and D haplotype. These have Mendelian inheritance, i.e. AR. If the patient is homozygous for Cyp2D6, that means that Parent X and Parent Y both have a haplotype with cYP2D6. i.e: A haplotype has cyp2D6, C haplotype has cyp2D6
For her sister, there is a 50% chance mom passes down her A haplotype, and a 50% chance the dad passes down his C haplotype. 1/2 x 1/2 = 1/4