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
23.
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Expert Answer

Daphne Broadhurst
Answered 2020-11-09 Author has 109 answers

Maximum value of P=23
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=1pq
Substitute this value of r in P=2pq+2pr+2rq, then it becomes
P=P(p,q)=2(1pq)q+2(1pq)p+2pq
P(p,q)=2p2p2+2q2q22pq
As p, q, r is the proportion of species it ranges from 0 to 1.
Therefore, p0,q0,1pq0 this implies
p+q1
Therefore, domain of P is
D=(p,q)0p1,q1p which is closed set
bounded by lines p=0,q=0andp+q=1.
Now to find critical points, consider equation
P(p,q)=2p2p2+2q2q22pq
Differentiating P partially with respect to p,
Pp(p,q)=24p2q
Differentiating P partially with respect to q,
Pq(p,q)=24q2p
Setting partial derivatives equal to 0, obtain the equations,
24p2q=0 and 24q2p=0
Therefore,
2p+q=1 and p+2q=1
Solving these system of equations,
p+2(12p)=1
p=13
Substitute p=13 in 2p+q=1
q=12(13)=13
Thus the critical point is (13,13)
Therefore,
p(13,13)=2(13)2(13)2+2(13)2(13)22(13)(13)
=4169=1269=69=23
p(13,13)=23
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
0q1.
Therefore, P(0,q)=2q2q2
Which represents downward parabola with maximum value at vertex (0,12)
P(12,0)=2(12)2(12)2+2(0)2(0)22(0)(12)=12
Therefore,
P(0,12)=12
Similarly, along the line q=0,
p ranges between 0 and 1 that is 0p1.
Therefore, P(p,0)=2p2p2
Which represents downward parabola with maximum value at vertex (12,0)

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