NBME 22 25% (NBME Answers)

NBME 22 Answers

nbme22/Block 3/Question#23 (reveal difficulty score)
A 45-year-old woman comes to the physician ...
25% 🔍 / 📺 / 🌳 / 📖
tags: biostats

+8 upvote downvote
submitted by ∗b1ackcoffee(115)

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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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cjdinurdreamz i used the same exact reasoning +

skonys Suffering from success over here because I just read it as there one locus "6*" and shes homozygous aa. Thus her sisters has a 25% chance of being aa too because punnet square. Most probably the wrong logic but hey, we out here. +

+4 upvote downvote
submitted by ∗bubbles(79)

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Can someone explain properly how we know that this trait follows Mendelian genetics and is autosomal recessive and furthermore how the parents were heterozygous?

I guessed a lot on this question and got lucky :(

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niboonsh Autosomal Dominant disorders usually present as defects in structural genes, where as Autosomal Recessive disorders usually present as enzyme deficiencies. P450 is an enzyme, so we are probably dealing with an autosomal recessive disorder. furthermore, the question states there was a "homozygous presence of p450.....". In autosomal recessive problemos, parents are usually heterozygous, meaning that 1/4 of their kiddos will be affected (aka homozygous), 1/2 of the kids will be carriers, and 1/4 of their kids will be unaffected. +38

nwinkelmann Is this how we should attack this probelm?: First clue stating endoxifen is active metabolite of Tamoxifen should make us recognize this undering first pass hepatic CYP450 metabolism? Once we know that, the fact that the metabolite is decrease suggests an enzyme defect, which is supported by patient's homozygous enzyme alleles. Then use the general rule that enzyme defects are AR whereas structural protein defects are AD inheritance patters. Once we know the pattern, think that most common transmission of AR comes from two carrier parents. So offspring alleles = 25% homozygous normal, 50% heterozygous carrier, and 25% homozygous affected, thus sister has a 25% of having the same alleles as patient (i.e. homozygous CYP450 2D6*4)? +6

impostersyndromel1000 we had the exact same thought process, so i too am hoping this is the correct way to approach it get reasoning friend +

ajss thanks for this explanation, I totally forgot about AR patterns are most likely enzymes deficiencies, this kind of make the question easier if you approach it that way, thanks +

+3 upvote downvote
submitted by ∗dentist(94)

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enzyme deficiencys = AR
homozygous presence of CYP..."

QED: homozygous + AR = 25%

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+1 upvote downvote
submitted by ∗its_raining_jimbos(29)

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Tamoxifen has to be metabolized via first pass metabolism to an active metabolite (endoxifen). The patient has decreased concentrations of the metabolized product indicating that the patient’s pair of cytochrome P450 alleles aren’t metabolizing tamoxifen correctly. The question is asking what the chances are the sister has the same genotype, which would be 25% --> 1/2 * 1/2 = 1/4

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medschul How do we know the parents are not homozygous +8

yotsubato Chances are they are not unless they had or are incestuous +

+0 upvote downvote
submitted by 07chess(3)

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Overthinking screwed me. The question stem asks for a chance for a sister to get the same alleles, not that she is homozygous with them. So I picked 75% instead of 25%. At the top of it, they put 2D6*4(I thought that means that pt. has 4 copies of alleles(two from each parent like HLA). They have to make questions impossible to understand.

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+0 upvote downvote
submitted by g8427(0)

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If some one can help me understand bc im a bit confused. I understand the thought process and I realized that this was an AR disease and I also got the 1/4 affected, 1/2 carrier and 1/4 unaffected. But I chose 0% bc I figured if it was an AR disease the 1 child already diseased was homozygous affected (1/4 affected). Which lead me to think that the other sister was either a carrier or not affected at all. Am I just over thinking this or am I not fully understanding whats going on?

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rush you have to think about each child individually, doesn't matter what the siblings have. The question states what are the odds of the child getting the disease. So regardless of the other siblings it still is Mom (1/2) dad (1/2) which makes it 1/4 AR +

titanesxvi But how do we know that the parents are heterozygous for the mutation +

need_answers we know that the parents have to be heterozygous Aa X Aa because on a 2x2 table, the only way the daughter could be homozygous for an AR is by having both parents be carriers (Aa) so the question was asking what are the chances the sister has the same alleles (aa) and there is only a 25% of having the same alleles. +

+0 upvote downvote
submitted by timekeeper36(0)

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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

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+0 upvote downvote
submitted by fukprometric(3)

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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

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