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Hardy-Weinberg Principle?
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Asclepius.
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Hi! can someone help me out, I struggle with the application and maths aspect of the Hardy-Weinberg principle,
could someone explain to me in simple terms how to do it, thanks in advance
could someone explain to me in simple terms how to do it, thanks in advance
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theomarch
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Hi,
The Hardy-Weinberg equation is simply a mathematical model of proabitility of alleles in the gene pool.
p2 + 2pq + q2 = 1 where; p2 = homozygous dominant individuals
q2 = homozygous recessive individuals
2pq = heterozygous individuals
The equation is actually derived from very simple probability laws.
In GCSE maths tree diagrams you would multipy along the lines to find the probability of various combinations of objects.
What this translates to in an equation: P(A) x P(B) = P(AB)
So instead of putting numbers into this equation lets put P for being homozygous dominant. Thus you need PP as the genotype.
P(P) x P(P) = P(PP) = P(P2 )
Similarly for pp (homozygous recessive)
P(p) x P(p) = P(p2 ) -- As we cannot use p2 twice we just use q2 instead.
It gets a bit more complicated for heterozygous (Pp)
P(P) x P(p) = P(Pp) -- however relating back to GCSE maths you can have Pp and then pP also
P(p) x P(P) = P(pP)
We now replace the lowercase p with q to avoid confusion as mentioned earlier.
P(Pq) + P(qP) = 2Pq = 2pq
Therefore p2 + 2pq + q2 = 1, the equation equals one because all of the possible results are included in the equation i.e. one of them must happen.
I hope this helps! and feel free to ask for any further clarification.
-Theo
The Hardy-Weinberg equation is simply a mathematical model of proabitility of alleles in the gene pool.
p2 + 2pq + q2 = 1 where; p2 = homozygous dominant individuals
q2 = homozygous recessive individuals
2pq = heterozygous individuals
The equation is actually derived from very simple probability laws.
In GCSE maths tree diagrams you would multipy along the lines to find the probability of various combinations of objects.
What this translates to in an equation: P(A) x P(B) = P(AB)
So instead of putting numbers into this equation lets put P for being homozygous dominant. Thus you need PP as the genotype.
P(P) x P(P) = P(PP) = P(P2 )
Similarly for pp (homozygous recessive)
P(p) x P(p) = P(p2 ) -- As we cannot use p2 twice we just use q2 instead.
It gets a bit more complicated for heterozygous (Pp)
P(P) x P(p) = P(Pp) -- however relating back to GCSE maths you can have Pp and then pP also
P(p) x P(P) = P(pP)
We now replace the lowercase p with q to avoid confusion as mentioned earlier.
P(Pq) + P(qP) = 2Pq = 2pq
Therefore p2 + 2pq + q2 = 1, the equation equals one because all of the possible results are included in the equation i.e. one of them must happen.
I hope this helps! and feel free to ask for any further clarification.
-Theo
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tocka90
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https://youtu.be/2bpAONAHlLE
A level biology with miss jercha!!!
A level biology with miss jercha!!!
(Original post by Asclepius.)
Hi! can someone help me out, I struggle with the application and maths aspect of the Hardy-Weinberg principle,
could someone explain to me in simple terms how to do it, thanks in advance
Hi! can someone help me out, I struggle with the application and maths aspect of the Hardy-Weinberg principle,
could someone explain to me in simple terms how to do it, thanks in advance
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