mjl1717
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Can anyone explain this law mathematically ,clinically , and lastly in laymens terms?
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| meg
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If,
1)mutation does not occur
2)natural selection does not occur
3)population is infinitely large
4)all members of teh population breed
5)mating is random
6) there is no migration in and out of population,
or in other words, if no mechanism that can cause evolution to occur, then gene pool frequencies will remain unchanged. This is the Hardy-Weinberg equilibrium.
for a trait controlled by a pair of alleles (A and a),
let p represent the frequency of dominant allele for a gene in the pool
and q represent the frequency of its recessive allele
in other words, p= all the alleles in individuals who are homozygous dominant (AA) and half of alleles in individuals who are heterozygous (Aa)
hence p= AA+1/2Aa
likewise, q= aa+1/2 Aa
Because there are only 2 alleles in this case, the frequency of one plus the freq of the other must be 100% which means p+q =1. So if we know p, we can calculate q and vice versa
Since p+q=1
(p+q)square =1
p2(here 2 is for square!) +2pq+q2 =1
p2 is the frequency of the homozygous dominant(AA) in the population
2pq the heterozy (Aa)
and q2 is the homozyg (aa)
From USMLE step 1 point of view, most of the genetics questions related to this principle seems like they give the frequency of the diseased people for an autosomal recessive (most commonly tay sachs). So they give q2. Then they could ask what are the chances of the disease in an offspring if mating occurs with someone else from that population. so in a sense it is the mating between a q2 and 2pq. so using q2 , get q, then get p which is 1-q, then calculate 2pq. From that we go on to calculate whatever the question is asking......
for eg, if the frequency of tay sachs is 1/100 in population. What is the chances of an offspring born of a mating between a diseased individual and a heterozygous?
q2= 1/100
q=1/10
p=9/10
2pq= 2*9/10 * 1/10 = 18/100 (actually most of the time we can omit the p) that is 2*1/10 which is 1/5
Now, the chances of the offspring having taysachs will be
q2 * 2pq * 1/2 (1/2 because only Aa and aa offsprings possible)
1/100 *1/5 *1/2 = 1/1000
Hope I am right and did not confuse! 
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| mjl1717
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Thank You very much :idea:
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