Kaycee Roche

2020-11-08

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 represent the population proportions of A, B, and O.

Use the fact that $p+q+r=1$ to show that P is at most

$\frac{2}{3}.$

Daphne Broadhurst

Skilled2020-11-09Added 109 answers

Maximum value of $P=\frac{2}{3}$

1) Concept:

Use the concept of absolute maximum and minimum.

To determine a continuous function's exact maximum or minimum value 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 absolute maximum and lowest values of steps 1 and 2 are represented by their largest and smallest values, respectively.

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

$P=P(p,q)=2(1-p-q)q+2(1-p-q)p+2pq$

$P(p,q)=2p-2{p}^{2}+2q-2{q}^{2}-2pq$

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

$D=(p,q)0\le p\le 1,q\le 1-p$ which is closed set

bounded by lines $p=0,q=0andp+q=1.$

Now to find critical points, consider equation

$P(p,q)=2p-2{p}^{2}+2q-2{q}^{2}-2pq$

Differentiating P partially with respect to p,

${P}_{p}(p,q)=2-4p-2q$

Differentiating P partially with respect to q,

${P}_{q}(p,q)=2-4q-2p$

Setting partial derivatives equal to 0, obtain the equations,

$2-4p-2q=0\text{and}2-4q-2p=0$

Therefore,

$2p+q=1\text{and}p+2q=1$

Solving these system of equations,

$p+2(1-2p)=1$

$p=\frac{1}{3}$

Substitute $p=\frac{1}{3}\text{in}2p+q=1$

$q=1-2(\frac{1}{3})=\frac{1}{3}$

Thus the critical point is $(\frac{1}{3},\frac{1}{3})$

Therefore,

$p(\frac{1}{3},\frac{1}{3})=2(\frac{1}{3})-2(\frac{1}{3}{)}^{2}+2(\frac{1}{3})-2(\frac{1}{3}{)}^{2}-2(\frac{1}{3})(\frac{1}{3})$

$=\frac{4}{1}-\frac{6}{9}=\frac{12-6}{9}=\frac{6}{9}=\frac{2}{3}$

$p(\frac{1}{3},\frac{1}{3})=\frac{2}{3}$

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, $P(0,q)=2q-2{q}^{2}$

Which represents downward parabola with maximum value at vertex $(0,\frac{1}{2})$

$P(\frac{1}{2},0)=2(\frac{1}{2})-2(\frac{1}{2}{)}^{2}+2(0)-2(0{)}^{2}-2(0)(\frac{1}{2})=\frac{1}{2}$

Therefore,

$P(0,\frac{1}{2})=\frac{1}{2}$

Similarly, along the line $q=0$,

p ranges between 0 and 1 that is $0\le p\le 1.$

Therefore, $P(p,0)=2p-2{p}^{2}$

Which represents downward parabola with maximum value at vertex $(\frac{1}{2},0)$

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