Question

# 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

Comparing two groups
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
$$\frac{2}{3}.$$

2020-11-09

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.
$$p + q r =1$$
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
$$P = P(p, q) = 2(1 - p - q)q + 2(1 - p - q) p + 2pq$$
$$P(p, q) = 2p - 2p^{2} + 2q - 2q^{2} - 2pq$$
As p, q, r is the proportion of species it ranges from 0 to 1.
Therefore, $$p \geq 0, q \geq 0, 1 - p - q \geq 0$$ this implies
$$p + q \leq 1$$
Therefore, domain of P is
$$D = {(p, q) 0 \leq p \leq 1, q \leq 1 - p}$$ which is closed set
bounded by lines $$p = 0, q = 0 and p + q = 1.$$
Now to find critical points, consider equation
$$P(p, q) = 2p - 2p^{2} + 2q - 2q^{2} - 2pq$$
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
$$0 \leq q \leq 1.$$
Therefore, $$P(0, q) = 2q - 2q^{2}$$
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 $$0 \leq p \leq 1.$$
Therefore, $$P(p, 0) = 2p - 2p^{2}$$
Which represents downward parabola with maximum value at vertex $$(\frac{1}{2}, 0)$$
$$P(\frac{1}{2}, 0) = 2(0) - 2(0)^{2} + 2(\frac{1}{2}) - 2(\frac{1}{2})^{2} - 2(0)(\frac{1}{2}) = \frac{1}{2}$$
Therefore,
$$P(\frac{1}{2}, 0) = \frac{1}{2}$$
Also along line $$p + q = 1$$
where $$0 \leq p \leq 1,$$
$$P(p, q) = P(p, 1 - p) = 2p - 2p^{2} + 2(1 - p) - 2(1 - p)^{2} - 2p(1 - p)$$
$$= 2p - 2p^{2} + 2 - 2p - 2 + 4p - 2p^{2} - 2p + 2p^{2}, 0 \leq p \leq 1$$
$$P(p, 1 - p) = 2p - 2p^{2}$$
Which is downward parabola whose vertex is at $$(\frac{1}{2}, \frac{1}{2})$$
Therefore, maximum value is
$$P(\frac{1}{2}, \frac{1}{2}) = 2(\frac{1}{2}) - 2(\frac{1}{2})^{2} = \frac{1}{2}$$
Therefore, comparing all values at boundary with value of P at the critical point
The absolute maximum value of P(p, q) on D is $$\frac{2}{3}.$$
Therefore, the maximum value of P is $$\frac{2}{3}.$$
Conclusion:
The maximum value of P is $$\frac{2}{3}$$.