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-Weinberg Law states that the proportion of individuals in a population who carry two different
alleles is P=2pq+2pr+2rq where p, q, and r are the proportions of A, B, and O in the
population. Use the fact that to show that p+q+r=1 is at most 2/3.

Answer 1

Answer:

The fact that shows that p+q+r=1 is at most 2/3 is well explained from what we have below.

Step-by-step explanation:

Given that:

The maximum of function P = 2pq+2pr+2rq is at most 2/3.

where ;

p+q+r=1

From  p+q+r=1

Let make r the subject ; then r = 1 - p - q

If we replace the value for r into the given function; we have:

P = 2pq+2pr+2rq

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

P = 2pq + 2p - 2p² - 2pq + 2q -2pq -2q²

P = 2p - 2p² + 2q -2pq -2q²

By applying the standard method for determining critical points we obtain:

$P_p$ = -4p - 2q + 2 = 0

$P_q$ = -4q - 2p + 2 = 0

Divide all through by 2

$P_p$ = -2p - q + 1 = 0

$P_q$ = -2q - p + 1 = 0

$P_p$ = -2p - q  = - 1

$P_q$ = -2q - p  = - 1

Multiplying both sides by - ; we have:

$P_p$ = 2p + q  =  1   ----- (1)

$P_q$ = 2q + p  =  1    ------(2)

From equation (1) ; let make q the subject ; then

2p + q = 1

q = 1 - 2p

Now; let us replace the value of q = 1-2p into equation (2) , we have:

2q + p  =  1  

2(1-2p)+ p = 1

2 - 4p+p = 1

2 - 3p = 1

3p = 2-1

p = 1/3

Replacing the value of p = 1/3 into equation (1) ; we have :

2p + q  =  1  

2(1/3) + q = 1

2/3 + q = 1

q = 1 -2/3

q = 1/3

Taking the second derivative D; we have

$D = P_{pp}P_{qq}-p^2_{pq}$

D = -4 × -4 - (2²)

D = 16 - 4

D = 12

This implies that the given function has a minimum or maximum at its critical point for which  12 > 0

Therefore , the value for function p = q = r  since  p, q, and r are the proportions of A, B, and O in the  population.

Hence r = 1/3

P=2pq+2pr+2rq

P = 2(1/3)(1/3) + 2(1/3)(1/3) + 2(1/3)(1/3)

P = 6/9

P = 2/3