Three alleles (alternative versions of a gene) A, B, and O determine the four hu

Question

Three alleles (alternative versions of a gene) A, B, and O determine the four human blood types by occurring in pairs: type A (caused by AO or AA), type B (caused by be or BB), type O (caused by OO), and type AB (caused by 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 alleles in the population. Use the fact that p + q + r = 1 to show that P is at most 2/3.

Explanation / Answer

Good old lagrange multiplier problem here...

P= 2pq + 2pr + 2rq --> to be minimized

p + q + r = 1

LEts find the partial derivatives...
fp = 2q + 2r
fq = 2p + 2r
fr = 2p + 2q
gp = 1
gq = 1
gr = 1

Now, link em together using fp = mgp , fq = mgq etc...

2q + 2r = m(1)
2p + 2r = m(1)
2p + 2q = m(1)

Equating two at a time :

2q + 2r = 2p + 2r
2q= 2p
p = q

2p + 2r = 2p + 2q
2r = 2q
q = r

So, we have
p = q = r

And plug this into the constraint p + q+ r = 1
p + p + p = 1
3p = 1
p = 1/3

So, p =q = r = 1/3

Now using this, P = 2pq + 2pr + 2rq becomes....

2(1/3)(1/3) + 2(1/3)(1/3) + 2(1/3)(1/3)

2/9 + 2/9 + 2/9

2/3

Hence proved that P is atmost 2/3

Related Questions

Navigate

• Browse (All)
• Browse T
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.