Question

Deep in the backwoods of Kentucky, there is a single town with a population of 100 individuals. 16 of these people have a rare condition that makes their skin blue. This blue skin is caused by a recessive allele in a single gene that is necessary for the proper synthesis of hemoglobin. The dominant allele is for normal pigmentation and these are the only two alleles that exist in this gene. Individuals must be homozygous recessive to actually have the blue skin phenotype. 1. What are the allelic frequencies for the two alleles of this gene in this population? [express as decimals]. 2. What percentage of this population is homozygous dominant? Heterozygous? Homozygous recessive? 3. What five conditions must be met for this population to be completely free from evolution?

Answer 1

Let us suppose that the given population is in Hardy-Weinberg equilibrium, in which the genetic variation remains constant from one generation to the next in the absence of disturbing factors.

For a population in Hardy-Weinberg equilibrium, the sum of frequency of dominant allele and recessive allele equals one.

i.e, p + q = 1

Where, p is the frequency of dominant allele

q is the frequency of recessive allele

Given that the total population is 100.

The dominant allele is for normal skin colour and the recessive allele makes the skin blue.

Also, given that the number of individuals with blue skin is 16.

Since, p + q = 1,

      p^2+2pq+q^2 = 1

Where, p^2 is the frequency of homozygous dominant individuals

2pq is the frequency of heterozygous individuals

q^2 is the frequency of homozygous recessive individuals

(1)Allelic frequencies of the alleles of the gene:

Given that q^2 = 16/100 = 0.16 [number of homozygous recessive individuals/total population]

Therefore, q. = 0.4

Since p + q = 1,

                     p = 1 - 0.4

                        = 0.6

Thus, the allele frequencies are p = 0.6 and q = 0.4

(2)To determine the percentage of individuals who are

(a) Homozygous dominant

The frequency of homozygous dominant individuals is given by p^2.

We found out that p = 0.6

Therefore, p^2 = 0.36

That is, 36 out of 100 individuals are homozygous dominant.

Thus, 36% of the total population is homozygous dominant.

(b) Heterozygous individuals

The frequency of heterozygous individuals is given by 2pq.

Since p=0.6 and q= 0.4,

2pq = 2 × 0.6 × 0.4

       = 0.48

Therefore 48 out of 100 individuals are heterozygous

That is 48% of individuals are heterozygous.

(c) Homozygous recessive

The frequency of homozygous recessive individuals is given by q^2

We know, q^2 = 0.16

Therefore, 16 out of 100 individuals are homozygous recessive.

Thus, 16% of individuals are homozygous recessive.

(3) For a population to be in Hardy-Weinberg equilibrium or non-evolving state, it must meet 5 conditions which are:

• No mutation

Mutation is a change that occurs in the DNA that results in formation of newer alleles in a population and thus contributes to evolution.

• Random mating

In random mating, the organisms mate randomly without any genotypic preferences. Non random mating results in evolution.

• No gene flow

Gene flow occurs when migration occurs. When no individuals or gametes enter or exit the population, then evolution does not occur.

• Very large population size

Genetic drift occurs in a population of finite size. Though it occurs in any population, the effect is stronger in smaller populations.

• No natural selection

Natural selection occurs when an allele or a combination of alleles of different genes makes an organism more or less fit, that is the ability to survive and reproduce in an environment. Natural selection also plays an important role in evolution.

These are the five assumptions to be met by a population not to evolve or to be in Hardy-Weinberg equilibrium.