 # Three alleles (alternative versions of a gene) A, B, and O determine the four blood types A (AA or AO), B (BB or BO), O (OO), and AB. The Hardy-Weinbe Kaycee Roche 2020-11-08 Answered
Three alleles (alternative versions of a gene) A, B, and O determine the four blood types A (AA or AO), B (BB or BO), O (OO), and AB. The Hardy-Weinberg Law states that the proportion of Boas in a population who carry two different alleles is
$P=2pq+2pr+2rq$
where p, g, and rare the proportions of A, B, and O in the population.
Use the fact that $p+q+r=1$ to show that P is at most
$\frac{2}{3}.$
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Maximum value of $P=\frac{2}{3}$
1) Concept:
Use the concept of absolute maximum and minimum.
To find the absolute maximum or minimum value of a continuous function f on the closed bonded set D:
1. Find the values of f at the critical points of f in D.
2. Find the extreme values of f on the boundary of D
3. The largest value of step 1 and step 2 is absolute maximum value and smallest value being the absolute minimum value.
2) Given:
Hardy-Weinberg Law:
The proportion of individuals in a population who carry two different allers is
$P=2pq+2pr+2rq$
Where p, q and r be A, B and O blood group populations.
$p+qr=1$
3) Calculation:
Consider Hardy-Weinberg Law,
The proportion of individuals in a population who carry two different alleles is
$P=2pq+2pr+2rq$
Where p, q and r be the A, B and O blood group populations.
By using fact, $p+q+r=1$
write r in terms of p and q
$r=1-p-q$
Substitute this value of r in $P=2pq+2pr+2rq$, then it becomes
$P=P\left(p,q\right)=2\left(1-p-q\right)q+2\left(1-p-q\right)p+2pq$
$P\left(p,q\right)=2p-2{p}^{2}+2q-2{q}^{2}-2pq$
As p, q, r is the proportion of species it ranges from 0 to 1.
Therefore, $p\ge 0,q\ge 0,1-p-q\ge 0$ this implies
$p+q\le 1$
Therefore, domain of P is
$D=\left(p,q\right)0\le p\le 1,q\le 1-p$ which is closed set
bounded by lines $p=0,q=0andp+q=1.$
Now to find critical points, consider equation
$P\left(p,q\right)=2p-2{p}^{2}+2q-2{q}^{2}-2pq$
Differentiating P partially with respect to p,
${P}_{p}\left(p,q\right)=2-4p-2q$
Differentiating P partially with respect to q,
${P}_{q}\left(p,q\right)=2-4q-2p$
Setting partial derivatives equal to 0, obtain the equations,

Therefore,

Solving these system of equations,
$p+2\left(1-2p\right)=1$
$p=\frac{1}{3}$
Substitute
$q=1-2\left(\frac{1}{3}\right)=\frac{1}{3}$
Thus the critical point is $\left(\frac{1}{3},\frac{1}{3}\right)$
Therefore,
$p\left(\frac{1}{3},\frac{1}{3}\right)=2\left(\frac{1}{3}\right)-2\left(\frac{1}{3}{\right)}^{2}+2\left(\frac{1}{3}\right)-2\left(\frac{1}{3}{\right)}^{2}-2\left(\frac{1}{3}\right)\left(\frac{1}{3}\right)$
$=\frac{4}{1}-\frac{6}{9}=\frac{12-6}{9}=\frac{6}{9}=\frac{2}{3}$
$p\left(\frac{1}{3},\frac{1}{3}\right)=\frac{2}{3}$
Now find values of P on the boundary of D consisting three lines.
Along the line $p=0$, q ranges between 0 and 1 that is
$0\le q\le 1.$
Therefore, $P\left(0,q\right)=2q-2{q}^{2}$
Which represents downward parabola with maximum value at vertex $\left(0,\frac{1}{2}\right)$
$P\left(\frac{1}{2},0\right)=2\left(\frac{1}{2}\right)-2\left(\frac{1}{2}{\right)}^{2}+2\left(0\right)-2\left(0{\right)}^{2}-2\left(0\right)\left(\frac{1}{2}\right)=\frac{1}{2}$
Therefore,
$P\left(0,\frac{1}{2}\right)=\frac{1}{2}$
Similarly, along the line $q=0$,
p ranges between 0 and 1 that is $0\le p\le 1.$
Therefore, $P\left(p,0\right)=2p-2{p}^{2}$
Which represents downward parabola with maximum value at vertex $\left(\frac{1}{2},0\right)$

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