p 6 Self-test Mutation and selection- Requires Respondus LockDown B ohammed Kham

Question

p 6 Self-test Mutation and selection- Requires Respondus LockDown B ohammed Khamis: Attempt 2 Question 19 (1 point) d Assume there is a new population of individuals who are all of the Genotype AA So the allelic frequency of A' -1.0. Mutation results in a second allele a' appearing in the population. The mutation rate (H) of A' to a' o.00004 (4 x 109. There is also a known back mutation M) rate of 'a' to 'A' of 0.00001 (1 x 105). The population is in Hardy-Weinberg equilibrium with the exception of mutation. a) After one generation, the allelic frequency of the 'A' allele wilb? b) After one generation, the allelic frequency of the 'al allele will be? e) The equilibrium frequency for the for the 'A allele will be? allele will be?

Explanation / Answer

Let us assume the frequency of A = p

frequency of a (when it will appear) = q

As per the question,

p = 1.0

So, currently q = 0 (All the individuals of population have genotype AA)

Now, after one generation, allele 'a' would appear due to mutation

Rate of mutation from A to a = 0.00004

So, the frequency of A would slightly decrease i.e. pfinal = pactual - (pactual x mutation rate from A to a)

where, pfinal is frequency of A after one generation

pactual is frequency of A in current generation

So, pfinal = 1.0 - (1.0 x 0.00004)

pfinal = 0.99996

As we know p + q = 1, So, qfinal i.e. frequency of allele 'a' after one generation = 1- pfinal

qfinal = 0.00004

Answer to Q1 : Allelic frequency of A after one generation = 0.99996

Answer to Q2 : Allelic frequency of a after one generation = 0.00004

Now, equilibrium frequencies,

Equilibrium frequency of an allele depends on both forward and back mutations and are independent of initial frequencies of alleles

Forward mutation will decrease the allelic frequency of A while back mutation will increase its frequency

Similarly, Forward mutation will increase the allelic frequency of a while back mutation will decrease its frequency

Given, Rate of forward mutation (Mut A to a) = 0.00004

Rate of backward mutation (Mut a to A) = 0.00001

Now, equilibrium allelic frequency of A = Rate of backward mtation/ (rate of forward mutation + rate of backward mutation)

equilibrium allelic frequency of A = (Mut a to A)/ [(Mut a to A) + (Mut A to a)] = 0.00001/(0.00001+0.00004)

equilibrium allelic frequency of A = 0.2

So, Answer to Q3 equilibrium allelic frequency of A = 0.2

Answer to Q4  

equilibrium allelic frequency of a = 1- equilibrium allelic frequency of A

equilibrium allelic frequency of a = 1-0.2 = 0.8

Answer to Q5

This type of equilibrium (when both A and a are present in population) is stable equilibrium because the allele frequencies will not change over time.

The initial population consisting of only A allele was in unstable equilibrium because there were chances of mutation.

Answer to Q6

Stable equlibrium: It is that situation when the allelic frequencies remain constant over generations i.e the value of p and q will remain constant. This can be when (i) both alleles are in population and rate of mutations is fixed

or (ii) only one allele is present in population and rate of mutation is taken to be zero.

Unstable equilibrium: It is that situation when the allelic frequencies do not remain constant i.e if only allele A is present in population but the mutation can occur. This implies, there is always a chance of allele a to appear and it can change the allelic frequencies.

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