If the frequency of an allele is 0.75, how many copies of this allele should be in a population of 300 diploid individuals?

Hardy-Weinberg law states, the allelic frequencies of a gene that is, q and p follow the relationship p^2+q^2+2pq = 1, if the population is in equilibrium.

Therefore p^2+q^2+2pq = 1, can be used to calculate the percentage of the allelic frequencies i.e. p^2 and q^2

Allelic frequency = 0.75.

Therefore, the percentage of the allele in the population would be given by

0.75^2 = 0.5625 = 56%.

And the total number of individuals with this allele in the population:

number of individuals = 56×300÷100

= 168 or

150 nearest figure.

marianaegarciaperedo marianaegarciaperedo · Answer 2 · 2021-11-25T16:38:22+01:00

Assuming that each h0m0zyg0us individual carries two copies of the allele, and each heter0zyg0us individual carries one copy of the allele, the total number of copies of the allele in this population (N=300) is 450.

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Available data:

The frequency of an allele = 0.75
Total number of individuals, N = 300
Diploid individuals

We need to calculate the number of copies of this allele present in the population. We will get to it in three main steps.

⇒⇒ FIRST STEP:

Let us calculate all the allelic and genotypic frequencies in the population.

Following the Hardy-Weinberg equilibrium theory, an h0m0zyg0us genotypic frequency equals the square potential of the allelic frequency. So,

Let us assume a diallelic gene for which X is the dominant allele and x is the recessive one.

We can use the already known allelic frequency to get its genotypic frequency.

f(X) = p = allelic frequency = 0.75
F(XX) = p² = genotypic frequency = 0.75 ² = 0.5625

Now, we can get the other allele frequency by clearing the following equation.

p + q = 1

0.75 + q = 1

q = 1 - 0.75

q = 0.25

Knowing the other allele frequency, we can use it to calculate the other genotypic frequency

f(x) = q = allelic frequency = 0.25
F(xx) = q² = genotypic frequency = 0.25 ² = 0.0625

Finally, we can use both allelic frequencies to get the heter0zyg0us genotypic frequency.

2pq = heter0zyg0us genotypic frequency = 2 x 0.75 x 0.25 = 0.375

So, up to here, we know that:

Allelic frequencies

f(X) = p = 0.75

f(x) = q = 0.25

Genotypic frequencies

F(XX) = p² = 0.5625

F(Xx) = 2pq = 0.375

F(xx) = q² = 0.0625

⇒⇒ SECOND STEP:

We will get the number of individuals with each genotype. To do it, we need to multiply each genotypic frequency by N.

H0m0zyg0us dominant individuals, XX = 0.5625 x 300 = 168.75

Heter0zyg0us individuals, Xx = 0.375 x 300 = 112.5

H0m0zyg0us recessive individuals, xx = 0.0625 x 300 = 18.75

⇒⇒ THIRD STEP:

Finally, we need to get the copies of the X allele in the population.

We know that,

H0m0zyg0us individuals carry two copies of the allele

Heter0zyg0us individuals carry one copy of the allele

If,

1 h0m0zyg0us individual --------------------- 2 copies of the X allele

168.75 individuals ---------------------------------X = 337.5 copies of the X allele

1 heter0zyg0us individual --------------------- 1 copy of the X allele

112.5 individuals ------------------------------------X = 112.5 copies of the X allele

Now, we need to add these numbers:

337.5 (h0m0zyg0us) + 112.5 (heter0zyg0us) = 450 copies of the X allele.

In a population of 300 diploid individuals, there are 450 copies of an X allele.

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