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Old 06-24-2012
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Default HY genetic question....

A recent study of achondroplastic dwarfism documented 7 new cases out of 250,000 in X population. What is the allele frequency in that population?
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Old 06-24-2012
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isn't achondroplasia autosomal dominant? can we figure out the allele in autosomal dominant diseases?
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Quote:
Originally Posted by Renaissance View Post
isn't achondroplasia autosomal dominant? can we figure out the allele in autosomal dominant diseases?
Yes... we can.
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Old 06-24-2012
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Originally Posted by Dr. Mexito View Post
Yes... we can.
Isn't there some info missing?
Bc those 7 cases can be homozygots dominant and/or heterozygots, so how can we distinguish that?
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Old 06-24-2012
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7/500,000 ???
feels soo wrong... :sorry:
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Old 06-25-2012
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if there's an option which says 7 in 250000, pick it. Autosomal dominant diseases generally tend to be heterozygous. They also have zero carrier rate(if we assume near 100% penetrance, which I think is a feature of achondroplasia)

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Old 06-25-2012
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Quote:
Originally Posted by Casandra View Post
Isn't there some info missing?
Bc those 7 cases can be homozygots dominant and/or heterozygots, so how can we distinguish that?

NO, nothing is missing here.

1. Seven new cases out of 250,000 means that achondroplastic dwarfism occurs a disease frequency of 0.00008 (7/250,000)

2. All of the genotypes containing the achondropasia disease gene include only the heterozygotes because homozygosity is fatal (all AA die intra-utero). This is different in Huntington disease (HD) because in HD, AA and Aa have same lethality.

3. Having said this, 2pq = 0.00008 = disease frequency.

4. p = A. "p" is usually really small in AD diseases and q = a ~ 1 (q=1-p), thus... 2pq = 2p(1) = 2p = 0.00008 = disease frequency.

5. If 2p = 0.00008, then you all will agree with me that p = 0.00014, right?

6. If that is true, then the allele frequency is p = 0.00014
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Old 06-25-2012
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Thank you..just one small correction. 7/250,000 is 0.000028.

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Quote:
Originally Posted by Dr. Mexito View Post
NO, nothing is missing here.

1. Seven new cases out of 250,000 means that achondroplastic dwarfism occurs a disease frequency of 0.00008 (7/250,000)

2. All of the genotypes containing the achondropasia disease gene include only the heterozygotes because homozygosity is fatal (all AA die intra-utero). This is different in Huntington disease (HD) because in HD, AA and Aa have same lethality.

3. Having said this, 2pq = 0.00008 = disease frequency.

4. p = A. "p" is usually really small in AD diseases and q = a ~ 1 (q=1-p), thus... 2pq = 2p(1) = 2p = 0.00008 = disease frequency.

5. If 2p = 0.00008, then you all will agree with me that p = 0.00014, right?

6. If that is true, then the allele frequency is p = 0.00014
I didn't know that homozygous is fatal - that's crucial info here thanks for great explanation!
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Great concept, thnx!

Can we always replace the normal allele (q or a in this case) with 1?
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Quote:
Originally Posted by smanthrav View Post
Thank you..just one small correction. 7/250,000 is 0.000028.

Sent from my Desire HD
shouldnt the denominator be 500,000? i.e the total no. of alleles...
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Old 06-26-2012
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Idea!

thanks for the Q but we dont really need to calculate the disease frequency.... its more simple...
as the q asks us allele frequency... we can get that simply by
no.of the allele/total no. of alleles
no. of alleles = 7 (as the disease is AD with fatal homozygosity)
total no. of alleles = 500,000 (each person has 2 alleles)
ergo 7/500,000 = 0.000014 ....

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