EB7

Question 1

What is needed to maintain adaptation in the face of recurrent mutation

Answer 1

selection against deleterious mutations (purifying selection)

Question 2

Model reversible mutation no selection:

Final frequency of A

Answer 2

initial frequency of A x rate of mutation away from A + frequency of B x the rate of mutation of B to A
= p(1-u) + qv / = q(1-v) +pu

Question 3

Model reversible mutation no selection:

Change in q (deltaq) due to mutation

Answer 3

the final frequency of B minus the initial frequency of B

change in frequency of q = (q(1-v)+pu) -q = pu-qv

Question 4

what is the change in frequency of q at equalibrium

Answer 4

=0 the frequency of q does not change

Question 5

Model reversible mutation no selection:

what is the frequency of q relative to p if the change in q due to mutation is 0

Answer 5

frequency of q relative to p is equal to the rate of mutation of u relative to v
phat’u -qhatv =0
qhat/phat =u/v

Question 6

Model reversible mutation no selection:

what if the rate of mutation of u is 100x the rate of mutation of q

Answer 6

u = v/100
q'=p'/100
q'+p' = 1 therefore 1-p1=q'
therefore 1-p' =p'/100
100-100p' =p'
p' =100/101
p' =0.99 q' =0.01 (1%)

Question 7

Model: mutation/selection balance: haploids

what is the final frequency after selection

Answer 7

the initial frequency over the total fitness
p/(p+qw)
qw/(p +qw)

Question 8

Model: mutation/selection balance: haploids

what is the final frequency after selection and mutation

Answer 8

initial frequency over the total fitness X the rate of mutation away from allele or towards
final freq A = (p/p(+qw)(1-u)
final freq B = qw(p +qw)(p/(p+qw))*u

Question 9

what does

p/(p+qw))*u show

Answer 9

the newly created B mutants

p/(p+qw)*(1-u) shows the A alelle lost to B mutants

Question 10

Model: mutation/selection balance: haploids

what is the change in frequency of B equal to

Answer 10

the final frequency of B - the initial frequency
 ((qw/ (p + qw) X (p /(p + qw))u) - q
or
Or ((qw+pu)/p+qw)) -q

Question 11

what is q’ if u

Answer 11

q’ =u/hs

Question 12

what is q’ if u>hs

Answer 12

q’=1

Question 13

when is h relevant

Answer 13

only in diploid heterozygote as stands for dominance

Question 14

what if the rate of mutation is lower than the selection coefficient

Answer 14

then selection frequency of deleterious allele is given by mutation rate/selection ceofficient. therefore the more deleterious the alelle the freater the s and the lower the frequency as q’ =u/hs

Question 15

what if there is no selection against the mutation

Answer 15

s~0
all A will eventyually convert to B as in this model mutation is assumed to unidirectional.
q’ could be equal to 1 if the rate of mutation is much lower than the selection coefficient.

Question 16

Model mutation – selection balance: diploids.

how does the mutation selection balance equation change for diploids

Answer 16

same but 1 = w-hs
use the fitness of the heterozygote since the utation will first arise in a heterozygous background, this assumes selection against heterozygotes

Question 17

what does it mean if h~ 0

Answer 17

mutation is larfelt recessive
so selection is acting mostly against homozygote mutants and the equilibrium frequency will be higher as the frequency of e.g. B will need to get up to a higher frequency before appreicable numbers of homozygotes are formed.

Question 18

what does q’ = if we need to consider double homozygotess

Answer 18

SQRT(u/s)

Question 19

Hypothesis:

what must the fitness differences be for selection to be effective in removing delterious alelles

Answer 19

fitnness must be greater than mutation rate.

Question 20

what do the mutation rates need to be for selection to be effective at removing delterious alleles

Answer 20

mutation rates need to be low otherwise mutations will keep occuring.
therefore mutation rates have evolved to be low to allow selection tto be effective
*selection acts on proof reading mech to act on mutation rate

Question 21

what can selection mutation balance explain

Answer 21

why many chromosomes of drosoph and humans carry rate mutations that slihgly reduce fitness inn heterozyote state and are strongly deleterious when homozygous

Question 22

what is SMA

Answer 22

spinal musclar atropgy
a neurodegenerative disease: weakness and wasting of muscles controlling voluntary movement
*

Question 23

what is SMA caused by

Answer 23

loss of function/deletion in the telSMn locus *teleomeric survival motor neuron gene) Ch5.
it is a lethal autosomal recessive

Question 24

how common is SMA

Answer 24

second most common disease in caucasians after CF (100, 000 new borns)
q’ =0.01
s~= 0.9

Question 25

why is SMA maintained despite the strong selection against it

Answer 25

disease is recessive so use q' = SQRT(u/s) | u = q2s, u = (o.o1)^2 x -0.9 = 0.9*10-4

Question 26

what can u be estimated for (SMA)

Answer 26

no. of new mutations e.e 7/340 affected invididuals were due to new mutations obtain similar u.

Question 27

why is SMA still maintained

Answer 27

due to mutation selection blaance: slightly reduced fitness in hetero state and strongly reduced in homozgyous

Question 28

when a mutation appears how does the expected no. of individuals carrying the mutation (Nc) change with generations

Answer 28

``` g0 Nc = 1 g1 Nc = 1x selection against it = 1(1-s) with each generation background changes so (1-s) is squared G3 = 1(1-s)^3 Gn = 1(1-s)n ```

Question 29

what is the no. of carriers if the mutation is very bad

Answer 29

selection cef = 1 | Nc = 1-1 =0 carriers

Question 30

how can you calculate how many generations before a mutation goes extinct

Answer 30

``` sum up the Nc across generations sum (1-s)n = 1/s Nc (total =1/s) therefore expected no. of individuals carrying the mutation decreases the higher s (e.g. the more delteriois) ```

Question 31

what is the Nc of a mutation before is goes extinct proporttion to

Answer 31

inversely proportional to the magnitude of its deleterious effect (i.e. is selection coeff)

Question 32

more delterious the mutation

Answer 32

the lower the persistence through generationa and the fewer the no. of carriers. *strongly deleterious mutations disapear quickly whereas mildly persist for a long time.

Question 33

if a genetic disease has arisen from a new mutation

Answer 33

likely very bad as disease has killed off its previous carriers.

Question 34

how can we calculate s for a genetic disease

Answer 34

from the proportion of cases where neither parent has the disease

Question 35

give three examples of calculating s for diseases

Answer 35

Aperts syndrom: proportion due to new mutation = 95% s =0.95 Achrondroplasia: proportion due to new mutation = 80% s =80 Huntingtons: proportion dye to new mutation = 1% s=0.01