nCa onesmeo pexnali leropryp how ew kwon ttha hsit ttira fllswoo lnMdaenie gtsienec and is soatoluma ssierveec dna teefhorrmru hwo the ntserap erew rotzegsho?euy
I esdeugs a lot on iths enqsuito nda ogt lkuyc (:
enzyme deficiencys = AR
homozygous presence of CYP..."
QED: homozygous + AR = 25%
moxfniTae sha ot eb aodtizmlbee avi rsfit asps stbemmoali to na vciate ilotbmeeta x(neoei.d)fn ehT penitat ash creadsede tocecoinsnantr of eht izametdloeb ocprdut itianndcig ahtt hte ittnpse’a arpi of crhtooceym 50P4 lllseae ner’at beiilzgatonm nfmieaoxt rtcloec.yr heT tqnsoeui si sgkani atwh het escnhac aer eht sresit has eht asme engpty,eo ichhw dwluo eb 52% ;t&g-- 12/ * /21 = 4/1
If meso oen cna hlep me esdnrutdna cb mi a itb sn.udceof I dsarentdnu eht utothgh cposesr nad I ielazerd htta shit was an RA ssiaeed nad I aslo gto eth /41 fce,eftda 21/ rrceair adn 14/ udenaceff.t tuB I hoecs %0 bc I gdureif fi ti asw na RA sideesa het 1 licdh dlayaer diesades was hmosgoozuy aftcdefe (14/ ec.aef)tfd Whchi aeld em ot ntikh ttha teh toreh iertss swa theier a errarci ro nto tdafeecf at lal. Am I sutj rveo hgtnniik sith ro am I ton ulfyl nunsitrnddeag shwat igogn n?o
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
submitted by b1ackcoffee(50), 2020-04-13T07:09:55Z
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.