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nbme22/Block 2/Question#34
A 30-year-old man and a 24-year-old woman ...

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+27  upvote downvote
submitted by โdrdoom(1203)

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heterozygoteโ] 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`

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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)

+4  upvote downvote
submitted by โkeycompany(351)

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.

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hello  See my explanation if you need more words to explain this explanation +2

+3  upvote downvote
submitted by โhello(429)

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 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).

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+0  upvote downvote
submitted by beastfromtheastx(0)

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`

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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

+0  upvote downvote
submitted by kdckjn(0)

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

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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dd 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(1203)

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