Hardy-Weinberg conditions?

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What is the generalized equation which can be used to calculate genotype frequencies in a population which meets Hardy-Weinberg conditions?

Jul 7th, 2015

Thank you for the opportunity to help you with your question! the solution is as such:

Hardy-Weinberg equilibrium equation

In this equation (p² + 2pq + q² = 1), p is defined as the frequency of the dominant allele and q as the frequency of the recessive allele for a trait controlled by a pair of alleles (A and a).   In other words, p equals all of the alleles in individuals who are homozygous dominant (AA) and half of the alleles in people who are heterozygous (Aa) for this trait in a population.  In mathematical terms, this is

p = AA + ½Aa

Likewise, q equals all of the alleles in individuals who are homozygous recessive (aa) and the other half of the alleles in people who are heterozygous (Aa).

q = aa + ½Aa

Because there are only two alleles in this case, the frequency of one plus the frequency of the other must equal 100%, which is to say

p + q = 1

Since this is logically true, then the following must also be correct:

p = 1 - q

There were only a few short steps from this knowledge for Hardy and Weinberg to realize that the chances of all possible combinations of alleles occurring randomly is

(p + q)² = 1

or more simply

p² + 2pq + q² = 1

In this equation, p² is the predicted frequency of homozygous dominant (AA) people in a population, 2pq is the predicted frequency of heterozygous (Aa) people, and q² is the predicted frequency of homozygous recessive (aa) ones. 


References: http://anthro.palomar.edu/synthetic/synth_2.htm

Please let me know if you need any clarification. I'm always happy to answer your questions.awaiting a positive feedback. It was lovely helping you. Please do contact or text me directly for more work assistance. Looking forward to help you again.
Jul 7th, 2015

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