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  #1  
Old 11-19-2010
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Genetics Hardy-Weinberg Carrier of AR diseases?

If the question asks about the carrier of an autosomal recessive disease, do we use 2pq or pq?

e.g. if disease X (which we know is AR) has an incidence of 1/625, what is the carrier? And why?
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Originally Posted by zein1 View Post
If the question asks about the carrier of an autosomal recessive disease, do we use 2pq or pq?

e.g. if disease X (which we know is AR) has an incidence of 1/625, what is the carrier? And why?

If the disease is AR and homozygous gene then you should use q2 but heterozygous gene use 2pq. Chances of inheriting a trait: Disease 25%, Carrier 67%.

One child is born with two normal genes (normal)
Two children are born with one normal and one abnormal gene (carriers, without disease)
One child is born with two abnormal genes (at risk for the disease)
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Old 11-19-2010
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So what is the carrier? is pq or 2pq? Do you divide by 2 to get the carrier?
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Originally Posted by zein1 View Post
So what is the carrier? is pq or 2pq? Do you divide by 2 to get the carrier?
Autosomal Recessive Diseases

p2+2pq+q2=1

Frequency of normal allele = p
Frequency of mutant allele = q
Frequency of homozygous normal = p2
Frequency of heterozygous carriers = 2pq
Frequency of homozygous affected = q2

Lets do an example using above equation:

PKU

Prevalence of PKU is 1 in 10 000 live births.
What is the carrier frequency?

q2 = 1/10000
want to calculate 2pq

we know if we take the square of q
q = 1/100 or 0.01
P = 1-0.01 = 0.99 which is close to 1

Since p is approximately 1, then

Carrier frequency = 2q = 2(1/100) =1/50 = 0.02
Which means in this population 1 in every 50 people will carry this gene.

I hope this may help you clear your doubt.
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To simplify things for Step 1 exam;
  • Carriers of any disease: 2pq
  • Prevalence of affected persons:
    • XR: males (q), females (q^2)
    • AR: q^2
    • AD: 2q
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Many thanks for your help, guys, But that last point confused me:
Quote:
Carriers of any disease: 2pq

So for example, if incidence of an AR disease is 1/100. then:
Carrier = 2pq = 2 x 1 x 1/100 = 1/50
Is that right?
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Also, why in AD do we consider affected = q`2? isn't it q`2 + 2pq?
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Quote:
Originally Posted by zein1 View Post
Many thanks for your help, guys, But that last point confused me:

So for example, if incidence of an AR disease is 1/100. then:
Carrier = 2pq = 2 x 1 x 1/100 = 1/50
Is that right?
No, this is wrong.

Incidence (or prevalence) of an AR disease is q^2. Thus, q^2 here = 1/100. In order to get q, you will need to get the root of 1/100, which equals to 0.1 (1/10). Then, carriers = 2pq = 2 x 1 x 0.1 = 0.2 (1/5).
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Quote:
Originally Posted by zein1 View Post
Also, why in AD do we consider affected = q`2? isn't it q`2 + 2pq?
In AD diseases, frequency of affected persons is simply equal to carrier frequency, since all carriers are also affected. Thus, 2pq or 2q (since p is approx. = 1) is used for both frequencies of carriers and affected persons.
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Well that clears things, thanks a lot
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Quote:
Originally Posted by zein1 View Post
Well that clears things, thanks a lot
No problem. Please refer to my post #5 above, and just memorize it as-is! It helped me a lot in the exam and in Qbanks. Just don't forget to get the root when given an incidence/prevalence of an AR disease or an XR disease in females (since both are = q^2) if the question is asking about the carrier prevalence (2q).
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Many thanks, and if you have any idea about my other question:
Complete Penetrance
Please tell me what you know, genetics is a bit cruel to me..
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