How does mutation affect evolution?

 

Model

Start with chalcone synthetase locus

Wildtype allele A1 codes for blue flowers

Suppose that A2 is a mutant allele at this locus

Produces an ineffective enzyme that cannot produce pigment

A loss of function that codes for white flowers

Initial allele frequencies

p = freq(A) allele

q = freq(a) allele

Stipulate symbols for mutation rates

Let m = rate/generation at which A mutates to a

And n = rate/generation at which a mutates back to A

Assume that mutations occur while gametes are in gamete pool (after shedding, before fusion)

So, gametes enter gamete pool in frequencies of p and q

But then allele frequencies are altered by mutation

Frequency of forward mutations

Frequency of A gametes in population times rate at which they can mutate to a = pm

Frequency of backward mutations

Frequency of a gametes in population times rate at which they can mutate to A = qn

Hence,

Frequency of A gametes after mutation events

p* = p - mp + nq

Frequency of a gametes after mutation events

q* = q + mp - nq

Note

p* + q* = p - mp + nq + q + mp - nq = p + q = 1

Calculate genotype frequencies for next generation (assume H-W equilibrium)

Freq(AA) = (p*)2 = (p - mp + nq )2

Freq(Aa) = 2(p*q*) = 2(p - mp + nq )(q + mp - nq)

Freq(aa) = (q*)2 = (q + mp - nq)2

Calculate rate of change of frequencies

Dp = p* - p = (p - mp + nq ) p = nq - mp

Dq = q* - q = (q + mp - nq) q = mp - nq

Is mutation an important evolutionary force?

What would rate of evolution be if only evolutionary forces was mutation?

Suppose that new mutation arises

In a population fixed for A , new allele a arises by new mutation

Mutation rate (Aa) is 11.2 x 10-6 = m

Back-mutation rate (aA) is 2.5 x 10-6 = n

Then,

Dp = p* - p = (p - mp + nq ) p = nq - mp = 2.5 x 10-6 (0) 11.2 x 10-6 (1) = -11.2 x 10-6

Dq = q* - q = (q + mp - nq) q = mp - nq = 11.2 x 10-6 (1) 2.5 x 10-6 (0) = 11.2 x 10-6

Ignore back-mutation

Dp = p* - p = -mp

Dq = q* - q = mp


Extra points on mutation as evolutionary force:

Can derive more general equation if we assume that family size follows a Poisson distribution with mean family size = k

Im not prepared to derive this equation, and dont expect you to be able to do so
But, if you understand probability distributions, you could do so

Where x1 = probability mutation will persist to 2nd generation

k = average family size when family size follows a Poisson distribution

When average family size (k = 2), then

Also, dont expect you to be able to use this equation

Approximately 36.8% of time, new mutant will be lost in one generation

As family size increases, this probability decreases

So, more chance of mutant persisting in population

Selection is major force that prevents detrimental alleles from increasing in frequency




Modeling mutation-selection balance

 

First, assume that mutant is deleterious recessive

Selection against deleterious recessive

Increase in frequency due to mutation

Because selection and mutation are opposing forces, they balance each other to create an equilibrium

So, at some point,

Or, ,

If we assume that the mutant is rare, then q2 is very small and denominator of right side of the equation can be treated as unity

Then, cancel p from each side to get

And, equilibrium allele frequency is


Second, assume that mutant is deleterious dominant

Selection against deleterious dominant

Increase in frequency due to mutation

Because selection and mutation are opposing forces, they balance each other to create an equilibrium

Or, ,

If we assume that the mutant is rare, then q2 is very small and all term with q2 go to zero

Also, if mutant is rare, then qm is vanishingly small

And equilibrium allele frequency is

Think about the form of this equation

Again, when selection is strong and mutation rate is low, equilibrium frequency of mutant allele is low

Apply these equations to the delphiniums studied by Nick Waser and Mary Price

Observed average frequency of albino plants = 7.4 x 10-4

i.e. less than one albino plant per thousand individuals

Assumed that this is the equilibrium phenotype frequency

Also measured selection against albinos by pollinators

Found that albinos produced from 20% (in artificial populations) to 70% (in natural populations) the number of seed found on blue-flowered wildtypes

So, s = 0.3 to 0.8

Next calculated expected equilibrium phenotype frequency at mutation-selection equilibrium

First, assumed that albinism is due to a single gene with a recessive white mutant allele

So, assumed white-flowered plants are aa homozygous mutants

Calculated equilibrium phenotype frequency by assuming selection against recessive homozygous mutant (equations above)

Designated frequency of mutant allele (a) = q
Assumed H-W equilibrium
Frequency of mutant homozygote = q2
Hence, equilibrium phenotype frequency at mutation selection balance =

Plug in observed values

Low range of selection
High range of selection

Second, assumed that albinism is due to a single gene with a partly or completely dominant white mutant allele

So, assumed white-flowered plants are either homozygous (aa) or heterozygous (Aa)

Calculated equilibrium phenotype frequency by assuming selection against dominant or co-dominant mutant

Designated frequency of mutant allele (A) = q
Important assumptions
Assumed H-W equilibrium

Frequency of the mutant phenotype = 2pq

Mutant phenotype class probably consists of only heterozygotes

Because the rarity of the mutant ensures a low frequency of homozygous mutants

And because the dominant homozygotes is probably lethal

Further, assumed (recognized) that equilibrium frequency of wild type, recessive, allele is very nearly one
Thus, the equilibrium frequency of the mutant phenotype at mutation selection balance is
Because and

Plug in observed values

Low range of selection
High range of selection