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
ANSWERED
asked 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 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
\(\frac{2}{3}.\)

Answers (1)

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 \(\frac{2}{3}\).

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