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The answer should be 1/100

The prevalence of the AR disease is 1/2500

That means that under Hardy Weinberg:

p^2 = 1/2500

so p = root(1/2500) = 1/50

q = 1-p = 49/50 (effectively 1)

2pq, which is the probablity that a random mate will be a heterozygous carrier of CF, is 2(1*1/50) = 1/25

A carrier has a 1/2 probability of passing on the bad allele, so the probability that a random partner will pass on a CF allele is 1/2*1/25 = 1/50.

Your patient is a carrier, so the probability that your patient will pass on the CF allele is 1/2.

The probability that both will pass on the CF allele is 1/50 * 1/2 = 1/100

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WOW Mondoshawan, excellent explanation.

the question not passing one allele but passing the diseaseThe answer should be 1/100

The prevalence of the AR disease is 1/2500

That means that under Hardy Weinberg:

p^2 = 1/2500

so p = root(1/2500) = 1/50

q = 1-p = 49/50 (effectively 1)

2pq, which is the probablity that a random mate will be a heterozygous carrier of CF, is 2(1*1/50) = 1/25

A carrier has a 1/2 probability of passing on the bad allele, so the probability that a random partner will pass on a CF allele is 1/2*1/25 = 1/50.

Your patient is a carrier, so the probability that your patient will pass on the CF allele is 1/2.

The probability that both will pass on the CF allele is 1/50 * 1/2 = 1/100

mean they have affected boy ( aa ) so two carrier have probability having a disease child is 1/4

child with carrier allele 1/2

if you have simple explanation

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590 Posts

It's hard to type it, so I drew it out - I hope this helps some!the question not passing one allele but passing the disease

mean they have affected boy ( aa ) so two carrier have probability having a disease child is 1/4

child with carrier allele 1/2

if you have simple explanation

(if it's too small, click on the image to go to the large version)

LOL....A.W.E.S.O.M.E :happy:It's hard to type it, so I drew it out - I hope this helps some!

(if it's too small, click on the image to go to the large version)

good job buddy...It's hard to type it, so I drew it out - I hope this helps some!

(if it's too small, click on the image to go to the large version)

very funny looking hardy and weinberg...)

keep it up. good luck.

is it possible that the carrier in this question gets married to affected homozygous(aa)???

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590 Posts

Not unknowingly....is it possible that the carrier in this question gets married to affected homozygous(aa)???

Some time ago I visited a patient who had undiagnosed cystic fibrosis until she was 23. Actually, she had comparatively few symptoms, and may still not have her diagnosis, because her sweat test was an "unconclusive borderline". In fact, she is probably waiting for DNA results.

Well, she is probably double positive for the mutation, and would be one of these cases.

Anyway, nobody is interested in this degree of detail. A question answer would never involve these very-very-very rare possibilities.

But that was a great opportunity to realize that patients don't read the books.

And that life is not a picnic.

I dont want to undermine your great drawing, but I thought that Psquared is the normal allele frequency and Qsquared is the prevalence of CF??

Can u please tell me if im wrong

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590 Posts

Ha ha! You're probably right - it's been a while since biology class! The calculation is the same in either case, though, so I guess whichever way we remember it is okay...

Can u please tell me if im wrong

Hi, thanks for the explanation. Just a clarification...If the patient had the disease, what is the probability that the patient will pass on the CF allele.The answer should be 1/100

The prevalence of the AR disease is 1/2500

That means that under Hardy Weinberg:

p^2 = 1/2500

so p = root(1/2500) = 1/50

q = 1-p = 49/50 (effectively 1)

2pq, which is the probablity that a random mate will be a heterozygous carrier of CF, is 2(1*1/50) = 1/25

A carrier has a 1/2 probability of passing on the bad allele, so the probability that a random partner will pass on a CF allele is 1/2*1/25 = 1/50.

Your patient is a carrier, so the probability that your patient will pass on the CF allele is 1/2.

The probability that both will pass on the CF allele is 1/50 * 1/2 = 1/100

Is is 1?

If the patient already had the disease, and also married a heterozygote (probability of which is 2pq = 1/25), then half of all their childred would be affected (homozygous reccessive) and the other half would be heterozygous. So the probability of having a child with CF is 1/50 (ie. 1/25*1/2).Hi, thanks for the explanation. Just a clarification...If the patient had the disease, what is the probability that the patient will pass on the CF allele.

Is is 1?

If you were asking what the probability of passing on a CF allele if one of the parents were affected, then yes it's 1. All of their children will have at least one allele, because one of the parents was homozygous recessive and will pass on one of the chromosomes.

The person in the question is carrier, chance of marrying carrier/normal partner is 1/2, we got q from above as 1/50, probability of their affected offspring is 1/25, and multiplying all probabilities will give us 1/100. Isn't it right that way too?? Happy-2

Well, CF patients have an increased risk of infertility, in women mucus builds up in the fallopian tubes and prevents conception (or causes ectopic pregnancy), in men there is usually agenesis of the vas deferens. So even if the carrier marries an homozygous the chance of having children is very low i guessis it possible that the carrier in this question gets married to affected homozygous(aa)???

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