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#1
01-27-2011
 USMLE Forums Guru Steps History: 1+CK+CS Posts: 378 Threads: 161 Thanked 102 Times in 51 Posts Reputation: 116
Applying Hardy Weinberg Equation to carrier state?

A carrier with cystic fibrosis asks about the probability of having a child with the disease.

disease prevalence 1/2500??

A- 1/25

B- 1/50

C- 1/100

D- 1/1000
 The above post was thanked by: aktorque (03-02-2011)

#2
01-27-2011
 USMLE Forums Master Steps History: 1+CK+CS Posts: 590 Threads: 31 Thanked 1,227 Times in 410 Posts Reputation: 1251
C. 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
 The above post was thanked by: aktorque (01-27-2011), Amenah (01-31-2011), Dr. Mexito (03-24-2012), kemoo (01-27-2011), Sabio (01-27-2011), Sparco (03-02-2011), venky2600 (08-18-2012)
#3
01-27-2011
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WOW Mondoshawan, excellent explanation.
#4
01-27-2011
 USMLE Forums Guru Steps History: 1+CK+CS Posts: 378 Threads: 161 Thanked 102 Times in 51 Posts Reputation: 116

Quote:
 Originally Posted by Mondoshawan 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

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
#5
01-27-2011
 USMLE Forums Master Steps History: 1+CK+CS Posts: 590 Threads: 31 Thanked 1,227 Times in 410 Posts Reputation: 1251

Quote:
 Originally Posted by kemoo1985 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
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)
 The above post was thanked by: aktorque (01-28-2011), axax (01-28-2011), Doc4Step1 (12-13-2012), dr z (01-27-2011), dr_ram (01-31-2011), Dr. Mexito (03-24-2012), drortho (11-24-2011), DrSK (10-07-2012), GGossling (02-01-2011), Hokolesqua (01-27-2011), kemoo (01-28-2011), kkat (11-27-2011), koolkiller88 (08-18-2012), mahwish (01-27-2011), mle2resident (02-02-2011), raheel_memon (03-02-2011), Sabio (01-27-2011), slowpoke (04-12-2012), venky2600 (08-18-2012)
#6
01-27-2011
 USMLE Forums Newbie Steps History: Not yet Posts: 7 Threads: 2 Thanked 0 Times in 0 Posts Reputation: 10

Quote:
 Originally Posted by Mondoshawan 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)
LOL....A.W.E.S.O.M.E
#7
01-31-2011
 USMLE Forums Scout Steps History: Not yet Posts: 55 Threads: 2 Thanked 93 Times in 27 Posts Reputation: 103

Quote:
 Originally Posted by Mondoshawan 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...
very funny looking hardy and weinberg...
keep it up. good luck.
#8
01-31-2011
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is it possible that the carrier in this question gets married to affected homozygous(aa)???
#9
01-31-2011
 USMLE Forums Master Steps History: 1+CK+CS Posts: 590 Threads: 31 Thanked 1,227 Times in 410 Posts Reputation: 1251

Quote:
 Originally Posted by rsin21 is it possible that the carrier in this question gets married to affected homozygous(aa)???
Not unknowingly....
#10
02-01-2011
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I think it's possible.

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.
 The above post was thanked by: Mondoshawan (02-02-2011)
#11
03-02-2011
 USMLE Forums Scout Steps History: 1 + CK Posts: 54 Threads: 15 Thanked 46 Times in 21 Posts Reputation: 56

Quote:
 Originally Posted by Mondoshawan 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)
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
 The above post was thanked by: Mondoshawan (03-02-2011)
#12
03-02-2011
 USMLE Forums Master Steps History: 1+CK+CS Posts: 590 Threads: 31 Thanked 1,227 Times in 410 Posts Reputation: 1251

Quote:
 Originally Posted by adorisz 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
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...
 The above post was thanked by: adorisz (03-02-2011)
#13
04-12-2012
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Quote:
 Originally Posted by Mondoshawan 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
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?
#14
04-12-2012
 USMLE Forums Addict Steps History: Not yet Posts: 135 Threads: 3 Thanked 73 Times in 48 Posts Reputation: 83

Quote:
 Originally Posted by rapiddo 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 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).

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 above post was thanked by: rapiddo (04-13-2012)
#15
08-18-2012
 USMLE Forums Veteran Steps History: 1+CK+CS Posts: 207 Threads: 38 Thanked 115 Times in 45 Posts Reputation: 125

I have a different way of looking at this questions, if anyone still checking this post , I would appreciate if they correct me if I was wrong.

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??
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#16
10-07-2012
 USMLE Forums Newbie Steps History: Not yet Posts: 4 Threads: 0 Thanked 0 Times in 0 Posts Reputation: 10

Quote:
 Originally Posted by rsin21 is it possible that the carrier in this question gets married to affected homozygous(aa)???
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 guess

 Tags Biostatistics-Epidemiology, Genetics-, Step-1-Questions

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