Question

A Texan population of wild turkey (Meleagris gallopavo) shows an insertion/deletion polymorphism locus with alleles + and –. Eggs of different genotypes have the following probabilities of surviving to adulthood:

Genotype Viability

+ / + 60%

+ / – 70%

– / – 50%

Q.4) What will be the frequency of +/– heterozygotes in the adult survivors of the eggs described in the previous question?

	a.	0.41
	b.	0.43
	c.	0.45
	d.	0.47
	e.	0.49
	f.	0.51
	g.	0.53
	h.	0.55
	i.	0.57
	j.	0.59

The Previous Question was: The + allele is at a frequency of p = 0.4 in eggs. Assuming that the genotype frequencies at the egg stage are those expected under Hardy-Weinberg equilibrium, what is the mean viability of the eggs? (Use the absolute viabilities in question #1.)

I want help with Q4.

Answer 1

According to the Hardy-Weinberg equilibrium

(p+q)² = p2 +q² + 2pq

Therefore, for heterozygous population the frequency would be 2pq

Now we know that one allele + has frequency of 0.4 so the other allele – has frequency

Again according the Hardy-Weinberg equilibrium-

In a population for two alleles, there cumulative addition is always one.

Therefore, p+q=1

Now if p is 0,4 then q will be 0.6 (1-04 =0.6)

Now the population have Genotype Viability therefore, we use following formula-

$p=\frac{p\times G_{p}}{p\times G_{p}+q\times G_{q}}$

Here p is the frequency of allele p and Gp is the genotype viability for allele p and Gq is the genotype viability for allele q.

$p=\frac{0.4\times 0.5}{0.4\times 0.5+0.6\times 0.7}$

$p=\frac{0.2}{0.2+0.42}$

p = 0.32

now for q= 1-p

q= 0.68

Now the heterozygous frequency 2pq= 2× 0.32 ×0.68

2pq= 0.4352 or 0.43

Therefore the correct option is b.