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Old 04-22-2012
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Genetics Hardy-Weinberg Allele value?

Ok, so i want to get this hardy weinberg thing done once and for all!

What allele is equal to 1? and why?!
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p + q = 1
p2 + 2pq + q2 = 1

p = A (dominant allele)
q= a (recessive allele)
2pq = Aa
p2 = AA
q2 = aa
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but for simplicity they regard one of those alleles 1, right? Which one is it and why?!
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Quote:
Originally Posted by Renaissance View Post
but for simplicity they regard one of those alleles 1, right? Which one is it and why?!
i dont know about that...
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I think what you are referring to is the P allele in recessive diseases.

Since the frequency of the Dominant allele is very close to the value of 1 (may be 0.9 etc) in Recessive diseases, it can be taken as 1 to simply the calculations.


2pq will really become 2(1)(q) ...and so on.
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I think you might be referring to the fact that since q is almost equal to 0 (q equals 0.01), you can pretty much round it to zero to simplify equations.
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Quote:
Originally Posted by jjsanchezramirez View Post
I think you might be referring to the fact that since q is almost equal to 0 (q equals 0.01), you can pretty much round it to zero to simplify equations.
If you do that then 2pq = 0 ? ... that would mean that there are no heterozygous carriers
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What I mean is exemplified by the following example using PKU, an inherited metabolic disease.

Children with PKU can not process phenylalanine, an amino acid of the protein, so that phenylalanine accumulates in the blood causing brain damage and mental retardation. This disease is caused by a recessive gene when a situation of aa homozygotes occurs. Where p is the allele frequency of the healthy allele and q the "defective", calculate the incidence of aa combination carriers.

If we make a cross of two carriers Aa, where it remains hidden recessive gene, the genotypes obtained in the next generation will be:

AA (p ) Aa (pq)
Aa (pq) aa (q )

The three genotypes AA: Aa: aa appear in a relation p : 2pq: q . If we add, we obtain the unit:

p + 2pq + q = (p + q) = 1.

The frequency of patients with PKU genotyping is a 0.0001, a value corresponding to q . The frequency q gene to be the square root of 0.0001, ie, 0.01. The disease has an incidence of 1 in 10,000, but the gene frequency is 100 times, 1 100. Genes are found in a frequency pair Aa

2pq = 2q (1 - q) = 2 0.01 (1 - 0.01) = 0.0198.

About 2% of individuals carrying the European population, therefore, this gene in the pair Aa, which gives an idea of ​​how persistent it can be a recessive gene remained "underground" in heterozygosity.

This value, although not 0, is very close to 0.
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Yes I understand that but you cannot take Q to be 0 in order to simplify the equations. Q is the frequency of the disease in Autosomal Recessive diseases. It will always be a very low value .. like 1 in 1000 or 1 in 10,000 or 1 in 1, 000, 000 but you cannot substitute 0 in equations. Right? or Am I missing something here?

Thanks for the example though, it was a good read.
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That is my mistake. I did not mean to simplify equations. But it does make things easier. If you assume q is a value very close to 0, then you will know that

2pq is also very close to 0, and

p is very close to 1.

I hope this makes things easier!
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Quote:
Originally Posted by jjsanchezramirez View Post
That is my mistake. I did not mean to simplify equations. But it does make things easier. If you assume q is a value very close to 0, then you will know that

2pq is also very close to 0, and

p is very close to 1.

I hope this makes things easier!
Yeah, that makes more sense. I guess you can substitute it in p + q = 1 equation since Q is such a low value that P is almost close to 1.

Thanks
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