need help with your account or subscription? click here to email us (or see the contact page)
join telegramNEW! discord
jump to exam page:
search for anything ⋅ score predictor (โ€œpredict me!โ€)

You must be logged in to vote!
Retired NBME 22 Answers

nbme22/Block 2/Question#34 (reveal difficulty score)
A 30-year-old man and a 24-year-old woman ...
1 in 600 ๐Ÿ” / ๐Ÿ“บ / ๐ŸŒณ / ๐Ÿ“–
tags:

 Login (or register) to see more


 +27  upvote downvote
submitted by โˆ—drdoom(1206)
get full access to all contentpick a username

Calculations for dad. The probability of the father being a carrier is 2/3 since it is known that he doesnโ€™t have the disease. Then the probability of him passing it on to his kid is 1/2, thus:

  • Probability of dad being carrier = 2/3
  • Probability of dad passing on disease allele = 1/2

Calculations for mom. With the Hardy-Weinberg Principle, you can figure out the probability of the mother being a carrier:

q = sqrt(1/40,000) = 1/200

So, 2pq = 2 * 1/200 * 199/200, which is approx 1/100.

For the child to get the allele from mom, two things need to happen: (1) mom must be a carrier [โ€œheterozygoteโ€] and (2) mom must pass the allele to child:

  • Probability of mom being carrier = 1/100
  • Probability of mom passing on disease allele = 1/2

Puting it all together. Now, combine all together:

= (probability of dad being carrier) * (probability of dad passing on disease allele) * (probability of mom being carrier) * (probability of mom passing on disease allele)

= 2/3 * 1/2 * 1/100 * 1/2
= 1 in 600

get full access to all contentpick a username
kernicterusthefrog  To quote Thorgy Thor, drag queen: "ew, Jesus, gross" +52
niboonsh  This question makes me want to vomit +14
drdoom  lol +
5thgencephalosporin  okay wow +
tekkenman101  You can make this a lot quicker by using simple rules for autosomal recessive diseases. 1) Unaffected parent with affected lineage will be a carrier 2/3 of the time (Aa) 2) frequency in population is super low so you can ignore P and just use 2(q) in order to calculate carrier frequency. So take the square root of the homozygous recessive frequency (1/40,000) and just plug it in: 2(.005) = .01 3) The odds of their child being affected with the two parents being assumed hypothetical carriers is 1/4 (aa) or .25. Dad's carrier chance (2/3) x Mom's carrier chance (.01) x child's chance of being recessive (.25) +1

Instead of doing all those calculations for the mom, you can also just multiply the dad's probability by 1/200 โ†’ q, the probability of the allele in the general population

+1/- hungrybox(1277)

You must be logged in to vote!

 +4  upvote downvote
submitted by โˆ—keycompany(351)
get full access to all contentpick a username

Mandelian Genetics:

Man has 2/3 chance of being a carrier. (He does not have the disease). Woman carrier risk must be calculated with p^2 + 2pq + q^2. q^2 = 1/40,000 q = 1/200, p is roughly = 1 2pq = 1/100 = Carrier frequency .

Risk of having a child thus equals 2/3 x 1/100 x 1/4 = 1/600 Because: 2/3 = Man Carrier risk 1/100 = Female Carrier Risk 1/4 = Chance they each pass on the recessive gene to their offspring.

get full access to all contentpick a username
hello  See my explanation if you need more words to explain this explanation +2


You must be logged in to vote!

 +3  upvote downvote
submitted by โˆ—hello(429)
get full access to all contentpick a username

This the same as @keycompany's explanation with more explanation those that need it...

The patient is unaffected by the particular autosomal recessive disease and his brother has the autosomal recessive disease --> this means each of their parents carries the AR allele. Now, for the patient to be unaffected given that both his parents have the AR allele, it means that the patient has a 2/3 chance that he is simultaneously unaffected and a carrier.

The patient's partner is unaffected and normal Hardy-Weinberg genetics (as stated in the problem)..when this is specifically mentioned, you are to assume that the partner has a carrier frequency for the AR allele, which equal to 2pq.

The disease has a frequency in the population of 1/40000. This is q^2 --> q^2 = 1/40000. Solving for q, you get q = 1/200.

Carrier frequency is 2pq. However, for a rare disease, 2pq โ‰ˆ 2q. So, the carrier frequency for the partner = 2q = 2 * 1/200 = 1/100.

Now, the Q asks about an offspring of the patient and his partner being affected, so to have an affected child, there is a 1/4 chance of having an affected child (becasuse the patient and the partner both have a 1/2 chance of passing the allele --> you multiple these together 1/2 * 1/2).

So, multiply 2/3 * 1/100 * 1/4 = 1/600. This is P(patient being a carrier) * P(partner being a carrier) * P(having an affected offspring together).

get full access to all contentpick a username


You must be logged in to vote!

 +0  upvote downvote
submitted by beastfromtheastx(0)
get full access to all contentpick a username

The probability of the father being a carrier is 2/3 since it is known that he doesnโ€™t have the disease. Then the probability of him passing it on to his kid is:

1/2 * 2/3 = 1/3

With the Hardy-Weinberg Principle, you can figure out the probability of the mother being a carrier:

q = sqrt(1/40,000) = 1/200

So, 2pq = 2 * 1/200 * 199/200, which approx is 1/100, and the probability of the child getting this allele is 1/100 * 1/2 = 1/200

Thus:

1/200 * 1/3 = 1/600

get full access to all contentpick a username
brill45  You did this right but I think made it a bit more complicated than needed to be. To figure out the likelihood of a random person in the population passing on a recessive allele, you just need to do squareroot(q2) to get to q. And just use that q to multiply with the 1/3 you got earlier. The q value tells you the allele frequency in the population, whether the person is a carrier or homozygous recessive both. You still got it, but just letting you know the easier route in case you see it again! +2


You must be logged in to vote!

 +0  upvote downvote
submitted by kdckjn(0)
get full access to all contentpick a username

why there is probability of father being a carrier is 2/3

get full access to all contentpick a username
drdoom  We assume this is a recessive disease. In other words, you manifest the disease if you inherit a disease allele, d, from your mother and a disease allele, d, from your father, giving you an allele pairing denoted by dd. dd = you have the disease. The father does not have the disease; this means his genotype can only be 1 of three things: DD, Dd, or dD. That is, DD = D from his dad, D from his mom; Dd = D from his dad, d from his mom; or dD = d from his dad, D from his mom. Notice that in only two of these 3 possibilities does the father (potentially) have a disease allele (d) that he might pass on to his progeny. +

We assume this is a recessive disease. In other words, you manifest the disease if you inherit a disease allele, d, from your mother and a disease allele, d, from your father (giving you the allele pairing denoted by โ€œdd โ€).

dd = you have the disease.

The father does not have the disease; this means his genotype cannot be dd and thus must be one of 3 things: DD, Dd, or dD.

That is,

  • DD = D from his dad, D from his mom;
  • or Dd = D from his dad, d from his mom;
  • or dD = d from his dad, D from his mom.

Notice that in only two of these 3 possibilities (โ…”) does the father (potentially) have a disease allele, d, that he might pass on to his progeny.

+1/- drdoom(1206)

You must be logged in to vote!

Must-See Comments from nbme22

sacredazn on Unrearranged immunoglobulin gene
seagull on Decreased binding of RNA polymerase
seagull on Anticholinergic
mcl on Area labeled โ€˜Dโ€™
liverdietrying on Release of stored thyroid hormone from a ...
kernicterusthefrog on Displacement
keycompany on Negative nitrogen balance
joha961 on Displacement
imgdoc on Area labeled โ€˜Cโ€™ (Abducens nucleus, right)
alwaysanonymous on 25 mL/cm H2O
drdoom on 1 in 600
seagull on Glutamine
yotsubato on Phase variation
bubbles on Acute retroviral infection

search for anything NEW!