Question

Suppose a population of flour beetles has 10,000 individuals. There are two alleles possible for the gene that determines body color: red (B), and black (b). BB and Bb beetles are red, while bb individuals are black. Assume the population is at Hardy–Weinberg equilibrium, with equal frequencies of the two alleles. What would be the expected frequency of red beetles?

Answer 1

Answer:

$0.75$

Explanation:

Given frequency of the two alleles i.e "B" and "b" are equal .

As per Hardy–Weinberg equilibrium, sum of frequency of the two alleles is equal to one.

i.e $p+ q= 1$

Since $p=q$

$p$ and $q$ will be equal to 0.5

Frequency of genotype "BB"

$= p^{2} \\= 0.5^{2} \\= 0.25$

Frequency of genotype "bb"

$= q^{2} \\= 0.5^{2} \\= 0.25$

As per second equation of Hardy–Weinberg equilibrium,

$p^{2} + q^{2} + 2pq=1$

Substituting the values of p and q in above equation we get -

$0.25+0.25+2pq=1\\2pq = 1-0.25-0.25\\2pq = 0.5$

Red beetles have genotype BB and Bb

number of red beetles with genotype (BB)

$= 0.25 * 10000\\= 2500$

number of red beetles with genotype (BB)

$= 0.5 * 10000\\= 5000$

Total red beetles

$2500+ 5000= 7500$

Frequency of red beetles

$= \frac{7500}{10000} \\= 0.75$