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NBME 22 Answers

nbme22/Block 3/Question#23 (47.2 difficulty score)
A 45-year-old woman comes to the physician ...

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submitted by b1ackcoffee(44),

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.

cjdinurdreamz  i used the same exact reasoning +  

submitted by bubbles(69),
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naC msoeone neiplxa ypprerlo woh ew kwno tath sthi iartt osofwll ndaelMeni ientcgse nda is oaostluma isecveser dna rmoreehtufr woh het rpstean weer ztehyo?esguro

I eegduss a lot no hsti ueiotnsq nda ogt lkcuy (:

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. +36  
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 +  

submitted by dentist(53),
  • enzyme deficiencys = AR

  • homozygous presence of CYP..."

QED: homozygous + AR = 25%

submitted by its_raining_jimbos(22),
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medschul  How do we know the parents are not homozygous +6  
yotsubato  Chances are they are not unless they had or are incestuous +  

submitted by g8427(0),
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If eosm one anc hlpe em nedrtsdnau cb mi a tbi eusdnco.f I uaetrsddnn eht hotguht spceosr nad I rzlaedie ttah sthi was an AR siadees nad I osal ogt teh 41/ a,dfcefet 21/ rireacr nda /41 deu.ntfecfa tBu I hoesc %0 cb I feurgid if ti was an AR sisdeae the 1 idlch realayd dsseiade asw uogoshmozy tadeceff 1(/4 ciWhh eadl me ot thnik htat teh ehrto tisers aws ertieh a rrrieac or ont efedtcfa at la.l Am I sjtu eovr nkhgtini hits or ma I not ulfly grnnuienadtsd hstwa nggio ?no

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. +  

submitted by timekeeper36(0),

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

submitted by fukprometric(2),

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