Population Genetics - Mutation Flashcards
Dominance in selection
Real cases often don’t follow strict dicotomy –> we can quatify the degree of dominance and define the coefficiant for an allele
- Most of the time its not purley dominant or recessive –> can quantofy the degree of dominannce
h
Dominance coefficient for an allele –> describes the degree of intermediate state
***Always a value between 0 and 1
Meaning of selection coefficants 0, 0.25, 0.5
0 –> no selection against that
0.25 – 25% cost to have that genotype
0.5 –> 50% decrease in fitness relative to other fitness
% Cost/decrease in fitness
Modifying for dominance
Quantify s for allle (rather than genotypes) and the modify with a dominance coeffciants
Meaning of h = 0 for AA
h = 0 –> there is no dominance in a –> A is complteley dominant to a
- There is no affact of a in Aa
Selection against Aa is 0
***Add inmage slide 19 (postclass)
Modifying s to account for degree of dominance
Use 1 selection coefficient
AA = h
Aa = hs (selection coefcinat X SC)
aa = s
Example
h = 0
s of AA = 0
s of Aa = hs
s of aa = s
HERE – A is completely dominant to a
Meaning of h = 1 for AA
a is completely dominant to A
- Affect of SC s – s of aa is domiant
Selection against Aa is 1 X s
Example of Quantofying the degree of dominance
WAA = 1
WAa = 0.75
Waa = 0.5
s AA = 0
s Aa = hs
s aa = s
SAA = 1-1 = 0 (1 - RF)
SAa = 1 - 0.75 = 0.25
Saa = 1 - 0.5 = 0.5 –> means s = 0.5
1 - WAa = hs
1 - WAa = h X 0.5
1 - 0.75 = h X 0.5
h = 0.5
Example – Quantify the degree of dominance
s AA = 0
s Aa = hs
s aa = s
WAA = 1
WAa = 0.55
Waa = 0.5
TO FIND S:
s aa = 1 - RF aa –> 1 - 0.5 = 0.5
S Aa = 1- RF Aa = 1 - 0.55 = 0.45
0.45 = hs
0.45 = 0.5h
h = 0.9
LOGIC:
Saa = 1-RF
S aa = 1 - 0.5
S aa = 0.5
hs = 1 - WAa
1 - 0.55 = hs
1 - 0.55 = 0.5 X h
h = 0.9
Meaning of h = 0.9
Dominance coefficient is closer to 1 = phenotypic affect of a is bigger than A
- Bigger affect if a; less affect of A
Meaning of all h values
h tells us about the degree of similarity between heterozygote and homozygote fitness
If h = 1 –> Aa is exactly like the deleterious homozygotes
If h = 0 –> Aa is exactly like the selectively favored homozygote
If h = 0.5 –> The fitness of Aa is exactly intermediate
Graphing the dominance
REMEMBER – we looking at h from the persecutive of a (is Aa like AA or aa) –> get straight line –> Aa falls on the line
at h = 0 –> A is dominant and a is recessive
(Because is at WAA)
- At h = 0 Aa has the same fitness as AA –> therefore a is completely recessive
h = 1 – Aa is like aa –> aa is dominant
(Because is at Waa)
- Aa has the same fitness as aa –> therefore aa is completely dominant
***Image on lide 31
Quantifying mutation rate
u (Mu)
Mutation (overall)
Process of evolutionary force – acts in background
Mutation rate for one BP at one locus
All of the rates are very low (often order of magnitudes less than 1) for rate at a single BP –> probability of a given BP chnaging from one generation to the next is low –> BUT when scale up over entire genome = get 30-40 mutations occuring
- Chance at any one mutation is small mu BUT the chance of Mu across genome is large
Looking at change of one allele at one locus from one generation to the next = use very small numbers
***Those mutation rate values on the previous slide were
very small numbers: ~10-8 for humans
How does mutation affect H-W
Example
Start:
90% A – P = 0.9
10% a – q = 0.1
Mutation occurs in germline producing gametes – Parental genotype –> gamete – gamete now has mutation)
Mutation rate = 10^-4 (relatuvley high rate of mutation)
Convert 1 A –> a out of every 10,000 gametes in gamete pool
Change in allele frequncey –
p = 0.8991; q = 0.10009 –. get very small chnage in popultion with a relativley high mutation rate = small chnage in one gneration
1 out if every 10,000 gamete
Overall: Get very small chnage in popultion with a relativley high mutation rate = small change in one generation
Mutation rate of 10^-4
Mutation rate = 10^-4 (relatuvley high rate of mutation)
Convert 1 A –> a out of every 10,000 gametes
MEANS – chnages the allele frequncey directley by this proportion (convert copies of A –> a at a rate of 1 in 10,000 – means 1 out of 10,000 gametes chnages from A to a –> chnages the allle frequncey by this proportion)
Result: Barley any change in one generation
Mutation rate over time (scaling up)
Mutationrate can add up through time
Rate of chnage in one generation:
dP = -up
- Because p –> q = use negitive mutation rate
dP = chnage in one generation
We can scale this up over any number of generations:
Pn = P0e^-un
P0 = starting allele frequcey
n = number of generations
HERE = use exponetial (in one generation we just multiplied)
Pn = p after n generstions from a starting point of P0
Quantofying mutation rate
Quantofy this as the chnage in p based on the mutation rate of p allele converted to the q allele ( A to a)
P –> q alllele = A –> a
Example – scaling up mutation rate
u (mutation rate) = 10^-4
100 generations
Start – P = 0.6; q = 0.4
Pn = P0e^-un
Pn = 0.6e^-(10^-4 X 100)
P0 = 0.6
Pn = ?
u = 10^-4
n = 100
Pn = 0.6e^-(0.0001 X 100)
Pn = 0.594
Allele frequncey change is still very small (even after 100 generations)
NOTE – make sure you mutiply the exponents first then do to that multiplied number (Do 10^-4 X 100) then e to that number
Allele frequcney chnage from mutation after 100 generation
Allele frequncey change is still very small (even after 100 generations)
- Still not much chnage but adds up over very long amounts of time
Example 2 –
u (mutation rate) = 10^-4
1000 generations
Start – P = 0.6; q = 0.4
Pn = 0.6e^-(0.0001 X 1000)
Pn = 0.6e^-0.1
Pn = 0.6 X 0.904
Pn = 0.542
SHOWS – even after 1000 generations = still get small change
Pn + qn = 1.0
qn = 0.458
qn 0.4 –> 0.458
Pn + qn
STILL EQUALS 1.0
Example 3 – Humans have been around for ~250,000 years. Let’s assume that the average generation time has been constant at 20 years. If the population started being
fixed for one allele (p0 = 1) at a locus with a mutation rate of 10-6, what would the allele frequencies be at that
locus today?
n – 250,000/20 = 12500
(have 250,000 –> need in generations – have 20 years per generation)
^^^ 250,000 years X 1 gen/20 years = 250,000/20 years
P0 = 1
u = 10^-6
Pn = ?
Pn = 1e^-(10^-6 X 12500)
Pn = 0.988
HAVE 1 –> 0.988
SHOWS – mutation is necessary BUT it is not suffecunct
- Mutation by itself does almsot nothing – mutation is not what causes evolutionary chnage -
- Mutation is needed (need varaition) – necessary for evolution but not a good force by itself – works with other forceses
- Other forces act on mutation and dirve chnage in allele frequncey
Mutation as a force of evolutionary change
Mutation is necessary for change BUT it is not sufficient
- Mutation by itself does almost nothing – mutation is not what causes evolutionary change
- Mutation is needed (need variation) – necessary for evolution but not a good force by itself – works with other forces
- Other forces act on mutation and drive change in allele frequency
Mutation is necessary for long-term evolutionary change, but it is no where near sufficient to explain genetic and phenotypic change in nature
What does selection need
Selection requires genetic variation – some prvious mutation events are a prerequisite for evolution by NS
What does mutation produce
In the case of favorable alleles –> mutation occasionally produces fodder for adaptations shaped by NS
- Have mutation –> THEN selection can act
Example – Mutation + Selection working together
Start with low varaition (low diversty) –> have salt stress
IF just start with high inbred fly line with low genetic doveristy –> ADD salt stress –> No flies survive
- No varaition that allows the flies to survive
RESULT – ALL die
Controls (just mutation) – no salt stress BUT have muttaion accumalation
- Not imposing perging of varaition – have varaition but no salt added –> allow mutation to act but no selection because no salt
RESULT – some survive – just mutation - get varaition in ability to to surive salt
- Because of the little amouny of varaition get some ability to handle salt
- The control populations evolved degree of improve salt tolerance through mutation alone
Salt seelction (mutation + selection) – heterogeneous salt stress envirnmnet
- Have mutations + exposing to salt stress
- Have patches of salt and salty food –> If they can eat the salt = less competition = selection
RESULT – salt tolerance improved dramatically
- The selection lines show that selection acting on that new variation leads to much larger changes
OVERALL – tells us that mutations does make benefical mutations + might have benefical effect
- IF impose stress = new varaition metters – when you have selection = you can see much more rapdi chnage
- New varaition = NS can act on it = evolve more rapdily with weak varaition + Strong seleection
Mutation working with selection
Experiment = tells us that mutations does make beneficial mutations + might have beneficial effect
- IF impose stress = new variation matters – when you have selection = you can see much more rapid change
- New variation = NS can act on it = evolve more rapidly with weak variation + Strong selection
Mutation Vs. Mutation + selection
Mutation + Selection – The selection lines show that selection acting on that new variation leads to much larger changes
Mutation – The control populations evolved
degree of improve salt tolerance through mutation alone
Selection weeding out deleterious alleles
Selection weeds out deleterious alleles as
mutation adds new copies
- Selection = wants to get rid of deleterious alleles
Most mutations
Most mutations = deleterious or neutral – only a handful are good
Selection vs. Deletrious alleles
Evolutionary force of selection and force of mutations work against each other
Selection = wants to get rid of deletrious alleles
Have two forces
1. Mutation adding in new mutants (u)
2. Selection taking out mutants from popultion (s)
u = weak –> se;ection pushes together – push against each other
S = strong
QUESTION – if u is weak and s is strong why doesn’t selection just push out deletrious alleles
If u is weak and s is strong why doesn’t selection just push out deletrious alleles
ANSWER – dominance
Selection vs. Deletrious alleles equillirbium
You can have a stable equillirbirum point where forceses are balanced – we can define an equillirbium point where these forces are balanced
- Can calculate teh expected allele frequncey of deletrious mutations at this point
Where equillirbium point occurs = affected by dominance vs. recessive
Where is equillirboum beteween selection vs. deletrious alleles
Where that point occurs is affected by dominance vs. recessive
Calculating equilbroum point between selection and mutation – ONLY if deleterious is entirley recessive
Can calculate an expected allele frequncey of deletrious mutation mainatined by the equillirbium – ONLY if deleterious is entirley recessive
q> = Square root u/s
Example
u = 10^-4 (mutation rate of 10^-4)
s = 0.98 (means that it is very deletrious – selection acts against it)
q> = square root 0.0001/0.98
q> = 0.01
MEANS – the frequncey of the trait is q^2 = 0.0001 –> 1 in 10,000 people have this debilitating disease and 1 in 100 are carriers
- have 1 in 10,000 people having negitive fitness effect – NS can’t do things against it = low mutation raye maintains allele frequcney because in recessive –> NS can’yt do anything
- u = mianatin 1 in 100 people being carreiers for deleterious mutation
Allele is maintained in popultion at rate of 0.01 even though selection is working against it
Calculating mutation selection balance for anything more than purley recessive (if have effect in Aa)
q> = u/hs
***Used if h is anything other than 0
Example
h = 0.05 (have a small amount of dominance – get some deleterious affect in Aa –> 5% deleterious affect)
u = 10^-4
2 = 0.98 (very deletrious)
NOW q> = 0.002
Used for anything other than pure recessive (if h is anytjing other than 0)
***h = 0.05 = have 5% deleterious affect
Mutation selection balance in purely recessive vs. with some dominance
Purley recessive (h=0) –> q> = 0.01
Not recessive (h=0.05 – has a little dominance) –> q> = 0.002
- Any part of deleterious in Aa (if h is anything other than 0) = get much lower frequncey of deleterious allele
- NS is more effective at purging out of popultion if have some effect in Aa –> now have stronger strength of selection
Have a 5X change in equilibrium frequency
SMALL values in h (small amounts of dominance) can have a big effect – If any of the negative trait peaks through in the
heterozygotes, the equilibrium frequency
becomes very, very low
- 0.002 is MUCH lower than the 0.01 seen in purley recessive –> any deletrious affect in Aa mattera a lot
BIG DIFFERNCES IN PURLEY RECESSIVE AND MOSTLY RECESSIVE
Effect of dominance on mutation-selection equillirbium
Small values of h (Small amounts of dominance) can have a big effect –> If any of the negative trait peaks through in the
heterozygotes, the equilibrium frequency
becomes very, very low
(Because if have any dominance = see some of the deleterious in Aa = selection acts against = frequency becomes very very low)
BIG DIFFERNCES IN PURLEY RECESSIVE AND MOSTLY RECESSIVE
Mutation Selection balance example – Cystic fibroceous
What we know:
q = 0.02
s = 1 (lethal)
u = 6.7 X 10^-7
Question – is the high allele frequncey maintained by selection mutation balance – Our question here is whether our real 𝑞 0.02 is anywhere
close to the expected value for q> under selection-mutation balance of those known parameters
HERE q> would be square root of 6.7 X 10^-7/1
q> = 0.0008 –> this is the value that we would expect to be maintained based on selection-mutations balance
0.0008 is NOT close to 0.02 –> the real value of q found in populations (0.02) is 25X higher than our expected value (q> = 0.0008)
Conlcude – we can conclude here that mutation-selection balance by itself DOES NOT explain why CF is carried by 2% of the popultion
Why is CF maintained at 2%
The answer is likely an overdominant effect of heterozygote resistance to Typhoid (which interacts with the same proteins when it infects lung tissue)
Constaints on NS’s ability to optimize popultions
- The outcome of natural selection doesn’t just depend on the overall fitness of an allele, it depends on the population context that the
allele exists in - Mutation can interact with the weak effects of selection on deleterious mutations at low frequencies to maintain low fitness genotypes in populations
Equilbroum point between selection and mutation
Equillirbium point in teh system – looks at change in frequncey based on u and negitive change in allele frequncey based on SC
Recessive diseases in popultion
Have autosomal recesisve –> have many in popultions because selection is weak and mutation adds them = have lots of carreiers
CF example (mine)
S = very close to 1 – almost always kills you before you can reproduce
Is CF maintained by selection-mutation balance?
Mutation rate = 6.7 X 10^-7 (includes any nucleotide that we can change that results in CF phenotype)
Does q match what we expect based on the numbers we have? – is CF allles maintained by mutation-selection balance
q> = 0.0008 –> THIS IS NOT 0.02 (wgicg is what we find in the popultion)
SHOWS – we can’r explain this as mutation selection balance for purley reseccive –> what is happening?
- Have higher than expected frequncey –> Rather than there being a negitive affect in Aa – it is likley that there is a positive affect in Aa – could be overdominace – Aa is the best outcome
How do we get mutation rates for CF
By looking at the kids with CF and parent who is not carrier
Maintenance of CF in popultion
Have higher than expected frequncey –> Rather than there being a negitive affect in Aa – it is likley that there is a positive affect in Aa – could be overdominace – Aa is the best outcome
CF = more common in northern Europe = interctaions with human diseases
Ex. Typhoid – intercats with the same protein that CF allele breaks –> though that having one ‘broken’ CF alllee (broken protein) = prevents typhoid from affecting lungs –> benefit = prevented from dying during typhoid fever = might explain rate if CF that we have
Have elevated mutation selection balance because of some small effect –> overide deleterious because of benefit in Aa
