Population Genetics - Mutation

Question 1

Dominance in selection

Answer 1

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

Question 2

h

Answer 2

Dominance coefficient for an allele –> describes the degree of intermediate state

***Always a value between 0 and 1

Question 3

Meaning of selection coefficants 0, 0.25, 0.5

Answer 3

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

Question 4

Modifying for dominance

Answer 4

Quantify s for allle (rather than genotypes) and the modify with a dominance coeffciants

Question 5

Meaning of h = 0 for AA

Answer 5

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)

Question 6

Modifying s to account for degree of dominance

Answer 6

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

Question 7

Meaning of h = 1 for AA

Answer 7

a is completely dominant to A
- Affect of SC s – s of aa is domiant

Selection against Aa is 1 X s

Question 8

Example of Quantofying the degree of dominance

WAA = 1
WAa = 0.75
Waa = 0.5

s AA = 0
s Aa = hs
s aa = s

Answer 8

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

Question 9

Example – Quantify the degree of dominance

s AA = 0
s Aa = hs
s aa = s

WAA = 1
WAa = 0.55
Waa = 0.5

Answer 9

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

Question 10

Meaning of h = 0.9

Answer 10

Dominance coefficient is closer to 1 = phenotypic affect of a is bigger than A
- Bigger affect if a; less affect of A

Question 11

Meaning of all h values

Answer 11

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

Question 12

Graphing the dominance

Answer 12

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

Question 13

Quantifying mutation rate

Answer 13

u (Mu)

Question 14

Mutation (overall)

Answer 14

Process of evolutionary force – acts in background

Question 15

Mutation rate for one BP at one locus

Answer 15

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

Question 16

How does mutation affect H-W

Example

Answer 16

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

Question 17

Mutation rate of 10^-4

Answer 17

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

Question 18

Mutation rate over time (scaling up)

Answer 18

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

Question 19

Quantofying mutation rate

Answer 19

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

Question 20

Example – scaling up mutation rate

u (mutation rate) = 10^-4
100 generations
Start – P = 0.6; q = 0.4

Answer 20

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

Question 21

Allele frequcney chnage from mutation after 100 generation

Answer 21

Allele frequncey change is still very small (even after 100 generations)
- Still not much chnage but adds up over very long amounts of time

Question 22

Example 2 –

u (mutation rate) = 10^-4
1000 generations
Start – P = 0.6; q = 0.4

Answer 22

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

Question 23

Pn + qn

Answer 23

STILL EQUALS 1.0

Question 24

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?

Answer 24

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

Question 25

Mutation as a force of evolutionary change

Answer 25

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

Question 26

What does selection need

Answer 26

Selection requires genetic variation – some prvious mutation events are a prerequisite for evolution by NS

Question 27

What does mutation produce

Answer 27

In the case of favorable alleles –> mutation occasionally produces fodder for adaptations shaped by NS
- Have mutation –> THEN selection can act

Question 28

Example – Mutation + Selection working together

Answer 28

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

Question 29

Mutation working with selection

Answer 29

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

Question 30

Mutation Vs. Mutation + selection

Answer 30

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

Question 31

Selection weeding out deleterious alleles

Answer 31

Selection weeds out deleterious alleles as
mutation adds new copies
- Selection = wants to get rid of deleterious alleles

Question 32

Most mutations

Answer 32

Most mutations = deleterious or neutral – only a handful are good

Question 33

Selection vs. Deletrious alleles

Answer 33

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

Question 34

If u is weak and s is strong why doesn’t selection just push out deletrious alleles

Answer 34

ANSWER – dominance

Question 35

Selection vs. Deletrious alleles equillirbium

Answer 35

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

Question 36

Where is equillirboum beteween selection vs. deletrious alleles

Answer 36

Where that point occurs is affected by dominance vs. recessive

Question 37

Calculating equilbroum point between selection and mutation – ONLY if deleterious is entirley recessive

Answer 37

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

Question 38

Calculating mutation selection balance for anything more than purley recessive (if have effect in Aa)

Answer 38

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

Question 39

Mutation selection balance in purely recessive vs. with some dominance

Answer 39

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

Question 40

Effect of dominance on mutation-selection equillirbium

Answer 40

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

Question 41

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

Answer 41

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

Question 42

Why is CF maintained at 2%

Answer 42

The answer is likely an overdominant effect of heterozygote resistance to Typhoid (which interacts with the same proteins when it infects lung tissue)

Question 43

Constaints on NS’s ability to optimize popultions

Answer 43

1. 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
2. Mutation can interact with the weak effects of selection on deleterious mutations at low frequencies to maintain low fitness genotypes in populations

Question 44

Equilbroum point between selection and mutation

Answer 44

Equillirbium point in teh system – looks at change in frequncey based on u and negitive change in allele frequncey based on SC

Question 45

Recessive diseases in popultion

Answer 45

Have autosomal recesisve –> have many in popultions because selection is weak and mutation adds them = have lots of carreiers

Question 46

CF example (mine)

Answer 46

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

Question 47

How do we get mutation rates for CF

Answer 47

By looking at the kids with CF and parent who is not carrier

Question 48

Maintenance of CF in popultion

Answer 48

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