4 Introduction to Mendelian Genetics

Question 1

intro
mendelian Genetics

Answer 1

Population genetics is concerned with the collection of genes belonging to all members
of a population, called the “gene pool” or “genetic material/information” of a population.
Central questions in population genetics are: What is the structure of the gene pool?
How does it change in time? What causes these changes? In this chapter we introduce Mendelian genetics and use difference equations (maps) to study how the gene pool of a population changes over time.

: introduction to Mendelian genetics => laws of inheritance
=> theoretical formulation of natural selection

Mendel’s inheritance laws + Darwin’s theory => evolution, Maths: evolution with nonlinear maps

Question 2

intro

Answer 2

in this chapter we assume that the trait under discussion
is produced by two alleles at a single locus. We are interested in whether genetic diversity
persists, or if one allele is lost and the other takes over the whole population. In other words,
for Mendel’s peas, we ask: in the long run, how likely is it that we will we end up with
only rounded peas? If both alleles persist the population is “polymorphic” at the locus. To
answer questions like this, we need to understand how the structure of the gene pool changes
over time at a given locus.
Mathematical population genetics was developed in the 1920s-1930s by Ronald Fisher,
John B. S. Haldane and Sewall Wright who laid out the mathematical formulation of Darwin’s
theory of evolution by natural selection. They built a theory based on Mendel’s observations
translated into his quantitative laws of inheritance (Mendelian genetics)

Question 3

4.1 Mendelian population genetics
Mathematical population genetics was developed by Fisher, Wright and Haldane in 1920-30
based on Mendel’s work. Mendel derived empirically the laws of inheritance by observing peas

Answer 3

Mendel observed that 75% of peas were rounded and 25% wrinkled
=> 2 gene variants called alleles, say R or r, coding for “rounded” (R) or “wrinkled” (r), at a “locus” (specific position of a gene on a chromosome), with R “dominant” over r (“recessive”)

=> 1 allele from each parent are paired, giving 3 possible pairings called “genotypes”: RR, Rr or rr. These give the “trait expression” or “phenotype”. RR and Rr give rounded peas (75%) and rr wrinkled peas (25%)

=> genotype and phenotype frequencies can vary over time
=> if both alleles persist the population is “polymorphic”;
otherwise one allele takes over the entire population

Question 4

How is genetic variability maintained?

Answer 4

The answer was provided by Mendel’s principles of inheritance (1866): organisms have discrete genotypes that get reshuffled and paired, not blended, during mating and this ensures genetic variability on which natural selection can operate, see Fig. 4.1. Nowadays, natural selection is broadly accepted as the biological explanation of adaptation.

Question 5

4.2 Mendel’s inheritance laws

Answer 5

Chromosomes made up of strands of DNA carrying hereditary information. Genes are DNA segments
that determine specific traits. Based on observations, Mendel formulated the laws of inheritance:

segregation
during the formation of gametes (sexual cells) alleles segregate from each other => gametes carry 1 allele for each gene

• Independent assortment:
  genes for different traits are sorted separately from one another=> inheritance of one trait independent of inheritance of another
• Dominance:
  in heterozygous diploid organisms carrying two different alleles, the allele that would determine the trait is dominant. The other allele, that is entirely masked by the expression of the dominant allele, is the recessive allele. However, an allele is not necessary either dominant or recessive

Mendel’s laws + Darwin’s principles => population genetics => foundations of evolution

Modern interpretation: a new organism, zygote (“egg”), is formed from the fusion of sexual cells
(gametes) of both parents . Then cells duplicate => diploid organisms composed of many cells

Question 6

Chromosomes

Answer 6

Chromosomes contain genes that can exist in different
variants (“alleles”). Same gene for both parents at same locus but possibly different alleles (A or a) => pairing and reshuffling lead to gentoypes AA or Aa or aA or aa
=> variability of traits, i.e. phenotypes
(e.g. color of wings in moths)

Question 7

diploid
Haploid
gametes
locus
alleles
heterozygotes

Answer 7

• diploid cells carry 2 instances of each gene, while haploid carry 1 such instance
• gametes are sexual cells and haploid; zygotes are diploid
• locus: position of a gene on a chromosome
• for 2 alleles, A and a at a locus, there can be 3 genotypes: AA, aa and Aa
• 2 same alleles (AA or aa) => homozygote; 2 different alleles (Aa or aA) => heterozygote
• In heterozygotes, expressed trait is (generally) associated with the dominant allele (when there is one

Question 8

assumptions

Answer 8

random mating, large population, diploid organisms,
discrete time t (generations) and non-overlapping generations

Question 9

N diploid individuals, each with 2 instances of gene (allele) per locus

Answer 9

of AA,Aa and aa individuals respectively

2N genes, A or a, at a given locus

Frequency of allele A is
p (= fraction of A in the population of size 2N) and
frequency of allele a is
q=1-p
3 genotypes: AA, aa and Aa
#AA Aa and aa individuals
N_{AA}, N_Aₐ, and Nₐₐ

Question 10

AA Aa and aa individuals

genotype frequencies
AA,Aa aa

N_{AA}, N_Aₐ, and Nₐₐ

Answer 10

D = N_AA/N (AA frequency),
H = N_Aa/N (Aa freq.),
R = N_aa/N (aa freq.).

When generations
matter, we write D_t, H_t R_t, p_t and q_t
t denotes the generation

Question 11

Initial genotype frequencies

Allele frequency at t=0

Answer 11

D₀ = N_AA/N at t=0 (AA frequency),
H₀ = N_Aa/N at t=0 (Aa freq.),
R₀ = N_aa/N at t=0 (aa freq.).

p₀ = D₀ +(H₀/2) for A

q₀ = 1-p₀ = R₀ + (H₀/2) for a

Dt≥1, Ht≥1, Rt≥1 generally differ from D0, H0, R0

Question 12

ALLELE FREQUENCIES

Answer 12

Since each AA contributes two A alleles and each Aa contributes a single A and a single a allele
=> allele frequencies:
p = [2N_AA + N_Aa]/2N = D +(H/2) ,

q = [2N_aa + N_Aa]/2N = R + (H/2)

Question 13

4.3 Hardy-Weinberg principle (HWP)

Answer 13

When all genotypes have the same survival and reproduction potential (=fitness), HWP gives
simple expression for frequency of genotypes:

HWP: allele frequencies remain constant ⇒
p ≡ p₀ and q ≡ q₀ = 1 − p

Question 14

HWP applied

allele frequencies

Answer 14

HWP: After 1 generation, the genotype frequencies remain constant = random pairing of A and a =>product of allele frequencies:

D ≡ Dt≥1 = p²,
H ≡ Ht≥1 = 2pq ,
R ≡ Rt≥1 = q²

Allele frequencies remain constant:
p = p₀ (A freq. at t > 0 = A freq. at t = 0)
q = qv (a freq. at t > 0 = a freq. at t = 0)

Question 15

Example: black and white wings in moths. Allele A: black (dominant), allele a: white (recessive). At t=0: 45% of AA and 20% of Aa.
What fraction of population will have dark wings in the long run?

Answer 15

Assuming that HWP holds. Initial A frequency
p₀ = D₀ +(H₀/2) =0.45 +(0.2/2)=0.55

q₀ = 1-p₀ =0.45
initial frequency of allele a

According to HWP:
p = p₀ = 0.55 q = q₀ = 0.45 remain constant
so
frequencies are constant for t ≥ 1)
at t=1:
frequency of Aa is
H₁ = 2pq =2.0.55.0.45=0.495 =H_t for t ≥ 1)
frequency of AA is
D₁ = p²=0.55^2=0.3025=D_t for t ≥ 1)
frequency of aa is
R₁ = q²=0.452 t ≥ 1 = 0.2025 = R_t (for t ≥ 1)

DIAGRAMS
Since A (black wing) is dominant, the fraction of of moths with dark wings is D₁+H₁
⇒ Long-run fraction of dark months is 0.3025 + 0.495 = 0.7975, that is 79.75%

Question 16

DIAGRAMS

Answer 16

lecture W4L1

Question 17

HWP remark

Answer 17

pₜ = Dₜ + (Hₜ/2) = p²ₜ₋₁ + pₜ₋₁(1 − pₜ₋₁) = pₜ₋₁

for t ≥ 1 ⇒ pₜ = p₀ = p

=> allele frequencies remain constant:
pₜ = p₀ = p and
qₜ = q₀ = q = 1 − p

Question 18

complete

Answer 18

4-43 notes lecture

Question 19

Evolution with mutations but without selection pressure

Answer 19

Assumptions made in deriving the Hardy-Weinberg principle:
we assumed the absence of mutations. In reality, since duplication/reproduction is never perfect, mutations are always present. Here, we briefly discuss the influence of mutations on the allele frequencies in the absence of selection pressure

(We assumed no selection,, fitness etc)

Question 20

Mutations

Answer 20

Mutations can be caused by transcription errors and may for example change allele A into a. They are likely to be deleterious and arise without reference to adaptive needs. They occur with small probabilities, but are important since they help maintain genetic variability.
Here, we consider two alleles at a locus, in large random-mating diploid populations.

Question 21

4.4 Evolution with mutations but without selection pressure:

Since the allelic frequencies will now generally vary from one generation to another:

denote by p_t the frequency of allele A in generation t,
and by q_t the frequency of allele a in generation t (with qt = 1 − pt).

Answer 21

We consider the Hardy-Weinberg scenario of no selection,
and assume that in generation t a fraction upt of the genes, i.e. a fraction u of the organisms
carrying the allele A, at this locus mutates into allele a, according to A → a, with a mutation
probability u. We also assume that the reverse mutation, a → A, occurs with a probability
v, and therefore from generation t to t + 1 a fraction vqt of the genes at this locus (fraction
v of the organisms carrying the allele allele a) mutates into A. In practice, one typically has
u, v ≈ 10−6 − 10−4
.

Question 22

Here: assume no selection, just mutations:
fraction u of A becomes a by error.
fraction v of a becomes A by error

pₜ₊₁ =

Answer 22

write the difference eq for the freq
pₜ₊₁ = (1 − u)pₜ + vqₜ
= (1 − u)pₜ + v(1 − pₜ)
= v + (1 − u − v)pₜ
fraction from A to A and a to A

A becomes a by small prob u etc

with initial cond given p_0

full sol
pₜ = p∗ + (p_0 − p∗)(1 − u − v)ᵗ
p∗ asymp stable
term (1 − u − v)ᵗ rapid decay in t as less than 1
approaches rapidly stationary state consisting
of finite fractions p* and q=1-p of both alleles

1st-order linear map solved:
p=v+(1-u-v)p
p=v/(u+v)
Thus Mutations: mechanism sustaining genetic variability as both variables coexist
(u and v typically small, p becomes finite number)

Question 23

4.5 Selection & fitness in population genetics

Answer 23

In population genetics, the fitness of an organism is
the expected number of offspring that an individual
contributes to the next generation
=>measures survival and reproduction potential
=>compute change in allele frequencies

(assumptions on HWP all genotypes have same repro and survival potential~ equally fit

natural selection: some traits render an organism more likely to survive and reproduce)

Pop genetics describes the influence of selection on evolution by introducing the concept of fitness

quantity reflects a propensity or probability to survive and reproduce in a given environment

To compute the change in gene frequencies, we need a measure of the survival and reproduction potential of the different genotypes. This is provided by the concept fitness

Question 24

absolute fitness

Answer 24

• The absolute fitness Wₓ of a genotype X is the average number of zygotes of type X in the next generation to which a type X in this generation contributes to.

Wₓ is the probability that a type X survives to breed, multiplied by its expected number of offspring

Question 25

relative fitness

Answer 25

The relative fitness wₓ ≡ Wₓ/Wᵧ of a genotype X relative to another genotype Y of absolute fitness Wᵧ

is the ratio of the absolute fitness Wₓ of X relative to the absolute fitness Wᵧ of a reference genotype Y
wₓ=Wₓ/Wᵧ with wᵧ= Wᵧ/Wᵧ

Question 26

Example: Case of a simple diploid pop
Geonotype A survival probability of 2/3 and 5 offspring=>
Genotype B survival probability of 1/2 and surviving type B has 4 offspring=>

relative fitness

Answer 26

diagram:
not all survive, of those that survive produce offspring dep on probability

W_A = (2/3) · 5 = 10/3
W_B = (1/2) · 4 = 2

Fitness of genotype B relative to genotype A: set (reference)
define w_A=W_A/W_A=1 and

w_b≡ W_B/W_A = 2/(10/3)=6/10=0.6

Hence
w_A = 1, w_B = 0.6 or
fitness of genotype B is 60% that of genotype A
wB : wA = 0.6

In practice, natural selection operates at every stage of an individual’s life, e.g. there is viability selection and variable fecundity

Question 27

4.6 Evolution in diploid populations under natural selection

Answer 27

Time-variation of the genetic material, driven by natural selection, is accompanied by changes in the allele frequencies. We are going to see that even “weak selection” can drastically change the frequency of alleles and genotypes through the generations.

How genetic material is passed on over generations

Question 28

4.6 Evolution in diploid populations under natural selection

simple model

Answer 28

simple model for the evolution in diploid populations
consider twoalleles, A and a at a specific locus,
allele A is dominant and a is recessive
resulting genotypes are AA, aa and Aa
Natural selection occurs because organisms with
different genotypes have generally different fitnesses
hence differ in their potential to survive (viability) and to reproduce (fertility).

Frequency of allele A pₜ in generation t ≥ 0 is
Frequency of allele a in generation t ≥ 0 is qₜ = 1 − pₜ

Question 29

4.6 Evolution in diploid populations under natural selection

A first-order nonlinear map (Fisher-Haldane-Wright eq.) gives a relation between

Answer 29

A first-order nonlinear map (Fisher-Haldane-Wright eq.) gives a relation between p_{t+1} and p_t

Question 30

4.6 Evolution in diploid populations under natural selection

life cycle

Answer 30

Life cycle consists of different stages:
different phases for which selection acts on
Selection operates at each stage of life cycle: viability (survival) and fecundity (reproduction) selection

1) zygotic phase at time (generation) t when zygotes that have just been formed proceed to adulthood,
(viability selection operated between differentiating between survival probability/potential; must survive into adulthood to pass on)

2) then the breeding phase (generation t),
(how many offspring can be produced)

3) followed by the gametic phase (generation t), and then a new zygotic phase =>start of the new generation t+1
(fecundicity selection operates in this phase; organisms compete for mating and reproduction)

gen t Zygotes to adults to parents to Gametes to zygotes gen t+1

Question 31

4.6 Evolution in diploid populations under natural selection

selection in life cycle

Answer 31

By analysing how the selection operates in each of the zygotic, breeding and gametic phases of the life cycle => Fisher-Haldane-Wright equation (detailed derivation in Appendix C of lecture notes

Can produce equation relating to absolute fitness

Question 32

Fisher-Haldane-Wright equation

Answer 32

pₜ₊₁ = [pₜ(pₜW_{AA}+qₜW_{Aa})]/[pₜ²W+2pₜqₜW_{Aa} +qₜ²W_{aa}]

The Fisher-Haldane-Wright equation (FHWE) is a
nonlinear (relation) first-order map
shows that pₜ dep on fitness W

responsible for change of allele freq, genome freq, total pop size

No selection: fitness all the same
W_AA=W_aa=W_Aa means
pₜ₊₁= pₜ and pₜ=p
back to Hardy-Weinberg

(nonlinear scalar map giving a difference eq for allele A at the locus for a system of diploid individuals ,

Question 33

Selection also responsible for changes of total population size Nₜ

Answer 33

Nₜ₊₁=
(W_AApₜ² +2W_Aa pₜqₜ + W_aa qₜ²) Nₜ
nonlinear first-order map

nonlinear first-order map

Question 34

Different ways of writing the FHWE

Answer 34

Using relative fitness with w_aa=1 as reference
w_aa=W_AA/W_aa
w_Aa= W_Aa/W_aa
(dividing the absolute fitness)

FHWE can be written:
pₜ₊₁ =
pₜ+ (pₜ(1-pₜ)/w¯ ) [ (w_{AA}-w_Aa)pₜ+(w_{Aa}-1)(1-pₜ)]

where
w¯= w_AApₜ² + 2pₜqₜw_{Aa} + qₜ²
is the mean (average) relative fitness, in principle it depends on t as dep on p and q which dep on time

Question 35

These different expressions of are equivalent expressions of the FHWE

Answer 35

FHWE can also be written as
how freq changes from one gen to another

pₜ₊₁ -pₜ≡ δp
= pₜqₜ (w_A − w_a)/w¯
= pₜ(1 − pₜ)(w_A − w_a)/w¯

where the
w_A(pₜ) = [w_AApₜ² + w_Aapₜqₜ]/[pₜ² + pₜqₜ]
= w_AApₜ + w_Aaqₜ

w_a(pₜ) = [qₜ² + w_Aapₜqₜ]/[qₜ² + pₜqₜ]
= qₜ² + w_Aapₜ
are the mean fitness of allele A and a, resp

these quantities are the weighted average of the relative fitness, average RF by p

Question 36

FHWE can also be written as
how freq changes from one gen to another

pₜ₊₁ -pₜ≡ δp
= pₜqₜ (w_A − w_a)/w¯
= pₜ(1 − pₜ)(w_A − w_a)/w¯

Why is this convenient?

Answer 36

δp > 0:
i.e. frequency of A increases, if w_A> w_a
=> the frequency of A increases if A has a higher mean fitness than a

δp < 0:
i.e. frequency of A decreases if w_A< w_a
=> the frequency of A decreases if A has a lower mean fitness than a

(if the mean fitness of allele A is bigger than a’s then frequency over time)

delta p looks at this change in freq over one gen)

Question 37

FIRST ORDER NON LINEAR SCALAR EQ

FIXED POINTS OF FHWE

pₜ₊₁ -pₜ≡ δp
= pₜqₜ (w_A − w_a)/w¯
= pₜ(1 − pₜ)(w_A − w_a)/w¯

Answer 37

When δp = 0 ⇒ p_{t+1} = p_{t} = p∗

Clearly p=1, p=0 (or q=1) associated with A (for p=1) or a (for p=0) taking over the entire population (ENTIRE POP CONSIST OF ALLELE ONLY)

when p=0 or 1 delta p vanishes thus these are fixed points of the map

Question 38

Can there be a fixed point associated with the coexistence of alleles at respective frequencies p∗ and 1-p∗ ? When it is the case, the population is “polymorphic”

Answer 38

If both alleles persist the population is “polymorphic” at the locus.

is there a coexistent fixed point

Question 39

Fixed points of FHWE
pₜ₊₁ -pₜ≡ δp
= pₜqₜ (w_A − w_a)/w¯
= pₜ(1 − pₜ)(w_A − w_a)/w¯

Answer 39

by solving δp = 0
⇒ lim_{t→∞} pₜ = p∗
(stationary frequency, independent of t)

Stationary frequencies
q∗ = 1 − p∗

physical points
p∗ = 1,q∗ = 0 all carrying allele A
p∗ = 0,q∗ = 1 all allele a

physical coexistence point:
0< p∗<1 (fractions)
given by
w_A(p∗) = w_a(p∗)

relating to mean fitness of both alleles being the same

set wₐₐ=1
gives
p∗= (1- w_Aa)/( 1+ w_AA - 2w_Aa)
which is physical if
0<p∗<1
physical if w_Aa > max(w_AA,1)
or w_Aa < min(w_AA,1)
, i.e. if heterozygotes are fitter than all homozygotes,
or less fit then all homozygotes,

then coexistence of both alleles and polymorphic population is possible (if p∗is stable)

Question 40

Note when using FHWE

Answer 40

Idealized diploid non-overlapping populations evolve according to natural selection =>
we assume just two alleles at the locus, given p_0 and interested in evolution over many generations

2 alleles, A and a,
at a single locus, and
t=0,1,2,… (generations)

allele frequencies obey the Fisher-Haldane-Wright equation (FHWE), a nonlinear first-order map

Question 41

FHWE:
P_{t}
q_t

Answer 41

pₜ : frequency of allele A in generation t,
qₜ = 1 − pt : frequency of allele a in generation t.

p_0 = 1 − q_0 is known initial condition.
and
To be physical, for ∀t ≥ 0 we need
0 ≤ pₜ = 1 − qₜ ≤ 1

Question 42

The principles of Darwin’s evolution by natural selection

Answer 42

Darwin’s theory assumes that slightly advantageous modifications in the genetic material are preserved and passed on through generations. For this, it necessitates heredity and variability.
: (1) Not all produced offspring can survive;

(2) traits vary among individuals;
(3) rates of survival/reproduction differ;
(4) traitdifferences are heritable;
(5) offspring of parents better adapted replace deceased individuals.

Question 43

Reminder: mean fitness of allele A and a are

Answer 43

w_A(pₜ)
= w_AApₜ + w_Aaqₜ
=(w_AApₜ + w_Aa pₜ) + w_Aa

w_a(pₜ) = w_aa qₜ + w_Aapₜ
=(w_Aa - w_aa pₜ) +w_aa

Question 44

Which fixed points of the FHWE are stable?
Various scenarios to be considered (see Q3 of Example Sheet 2

Answer 44

Which fixed points of the FHWE are stable?
Various scenarios to be considered (see Q3 of Example Sheet 2

Question 45

Example: selection favours a dominant advantageous allele (Scenario 1)

Assume that A is dominant and advantageous, and a is recessive

Answer 45

so
w_aa = 1 and WLOG
w_AA = w_Aa = 1 + s > 1 (with s > 0)

AA and Aa have same fitness because A is dominant and a is recessive
1+s > waa = 1 because A is advantageous over a

By sub. into FHWE, we get:
pₜ₊₁ = f(pₜ),
where f(p) =
p + (sp(1-p)²)/(1 +sp(2-p)
looking at stability:
⇒ f′(p) =
(1+s)(1+sp²)/[(1+sp(2-p))²]
Fixed points are solutions of p=f(p)
only fp are 0,1 no coexistence as fitness cond prev not satisfied
Linear stability analysis
f′(p = 0) = 1 + s > 1 UNSTABLE
f′(p = 1) = 1 ⇒ no conclusion from linear stability analysis. Use another method

For instance, a cobweb diagram shows that p = 1 is here asymptotically stable
Other approach: note that in this case, the iterates of the FHWE form an increasing sequence (see Q3 of Example Sheet 2)
=> by monotone convergence theorem, p=1 is asymptotically stable

Question 46

REMARK SELECTION STRENGTH

Answer 46

Time to reach p=1 greatly depends on the selection strength s:

convergence to p=1 is ~100 faster when s=0.2 than when s=0.002 (weak selection)
DIAGRAM
closer to 1
S shapes but for less gens converge to fp for larger s, faster for larger s

more advantegouse A then quicker to take over

Question 47

Example: selection favours a recessive but advantageous allele (Scenario 2)

Answer 47

We consider that A is recessive but advantageous, and a dominant => fitnesses relative to aa:

wAa = waa = 1
Aa and aa have same fitness, set to 1,because a is dominant and A is recessive
and wAA = 1 + s with s > 0
> 1 because A is
advantageous

By sub. into FHWE, we get the map
pₜ₊₁ = pₜ + spₜ (1-pₜ )²)[pₜ /[1 +spₜ²]] =f(pₜ)
=pₜ+ spₜ²(1-pₜ)/[1+spₜ²]

f′(p) = [1 + s(2 − p)p]/[(1 + sp²)²]

for fixed points solve p=f(p)

Only fixed points are p=1 (all carry A) and
p=0 (all carry a), both physical.
linear stability analysis

f ‘(p=1)=1/(1+s)<1 => p=1 is asymptotically stable
f ‘(p=0) =1 => no conclusion from linear analysis, but is clearly unstable.
This can be checked using another method, like the cobweb diagram

Alternatively, we can note that the iterates of the FHWE form an increasing sequence {p_0,p_1,p_2…} and bounded above by 1
=> This is the unique asymptotically stable fixed point

In this scenario, the advantageous allele A takes over, as in Scenario 1, even if it is recessive.
However, for same value of s, this now takes longer than under Scenario 1. This is because
heterozygotes now don’t benefit from selection advantage (since A is here recessive)

Question 48

difference in scenario 1 and 2

Answer 48

A takes over in scenario 2,
stable even if recessive
now we see since the heterozygote same fitness as small aa
A recessive and for it to take over takes longer for similar value of s than in scenario 1

in 1: we have heterozygote have same fitness as A: wAA=w Aa allele A spreads when genotype AA is also carried by heterozygote

not the case in scenario 2: only homozygotes carry 2 alleles and these are the ones that spread

thus in both cases we get p=1 but how it converges is faster in scenario one because both heterozygotes and homozygotes carry A spread
so convergence is faster for same s

Question 49

Example: selection favours a semi-dominant & advantageous allele (Scenario 3

Answer 49

We now consider that A and a are neither completely dominant or recessive => fitness of heterozygotes Aa is between that of of homozygotes, say
w_AA> w_Aa > w_aa

For concreteness, we consider that A is advantageous and semi-dominant, and specify:
w_AA = 1 + 2s,
w_Aa = 1 + s and
w_aa = 1 with s > 0

By sub. into FHWE, we get the map
p_{t+1}= p_t + [sp_t(1-p_t)]/[1+sp_t]

fixed points:
p=f(p)
sp(1-p)/(1+sp)=0 p=1,1 only fps both physical
(no coexistence fp as the fitness is between doesnt meet prev condition)

p=1 is asymptotically stable
p=0 is unstable

Question 50

Extra MATH5567M topic: evolution with selection and mutations

Answer 50

We discussed the influence of mutations in the absence of selection, and in we focused on the role of selection and ignored mutations. Here, we consider the joint influence
of selection and mutations on the allele frequency changes over the generations. ( Fisher-Haldane-Wright equation but now modify it to take mutations into account).

Question 51

Allele frequencies vary due to the joint effect of selection and mutations.
Evolution can lead to different forms of social behaviour => how to make sense of these?

Answer 51

Due to mutations, prior to starting a new zygotic phase and a new generation t + 1, a
fraction 1 − u of the A alleles (a fraction (1 − u)pt of the population) consists of unmutated
(genuine) A gametes and the remaining fraction upt consists mutated a gametes (mutated
from A parent). Similarly, due to mutations a fraction (1 − v)qt of the population consists of
unmutated a gametes (from a parent), while the remaining fraction vqt consists of mutated
A gametes (mutated from a parent). Proceeding as in the absence of selection

Question 52

modified Fisher-Haldane-Wright eq. (FHWe) to account for mutations.

Answer 52

Diploid population, with allele A or a at a single locus

assume that there are mutations from one generation to the next, with small probability u an A (from A parent) can become an alelle a, and an a (from a parent can become an
allele A with a small prob. v

small probability that
A → a prob u
a → A, prob v
A → A prob 1−u
a → a prob 1−v
with 0 < u ≪ 1, 0 < v ≪ 1

in the next generation
At end of gen. t:
Fractions (1 − u)pₜ of unmutated A and vqₜ of mutated A

Question 53

FHWE without mutations:
pₜ₊₁

Uses fitnesses

Answer 53

pₜ₊₁ = (w_A/w~)pₜ
and
qₜ₊₁=(w_a/w~)qₜ

pₜ:
A frequency in generation t = 0, 1, . . .
qₜ=1- pₜ: a frequency in generation t = 0, 1, .

w_A : mean fitness of allele A
w_a : mean fitness of allele a
w~ : average population fitness

Question 54

Usually u,v «s«1 (weak selection).

Answer 54

2 Usually u,v «s«1 (weak selection).
Typically, u, v ∼ 10^−6 − 10^−4 and s ∼ 10^−3 − 10^−2

all w_a, w_A, w_AA, w_aa, w_Aa, w~ = 1 + O(s) ≈ 1 to leading order

Question 55

Modified FHWE with mutations
pₜ₊₁

Answer 55

pₜ₊₁=
(1-u)(w_Apₜ/w~) +v(w_aqₜw~)

fraction (1-u)pₜ of unmutated A: A-¹⁻ᵘ→ A

Fraction vqt
of mutated A :
a -ᵛ→ A

=pₜ + [(w_A-w~)pₜ]/w~ -uw_Apₜ/w~ +vw_aqₜ/w~
terms
A-ᵘ→ a
a -ᵛ→ A

Question 56

uw_A/w¯ ≈ u,
vw_a/w¯ ≈ v and all w’s = 1 + O(s) ⇒ FHWE with mutations:

δp

Answer 56

δp ≡ pₜ₊₁− pₜ
= [w_A(pₜ) − w~(pₜ)]pₜ− (u + v)pₜ + v

[First-order scalar nonlinear map for p]
[We assume 0 < s ≪ 1,
s is not as small as u and v

Question 57

FHWE with mutations:
FIXED POINTS
δp ≡ pₜ₊₁− pₜ
= [w_A(pₜ) − w~(pₜ)]pₜ− (u + v)pₜ + v

[First-order scalar nonlinear map for p]

Answer 57

Fixed point obtained by setting pₜ → p∗ and δp = 0 ⇒
[w_A(p∗) − w~(p∗)]p∗ − (u + v) p∗ + v = 0

FIXED POINT NO LONGER p=1, p=0 always mutation, never have A that takes over, always a chance mutated A into a etc

fp depends on w’s

Question 58

FHWE with mutations:Mutation-selection balance: say that is deleterious and recessive and is dominant, such that

Answer 58

A is not wiped out even if it is opposed by selection
w_AA = 1 − s,
with 0 < s ≪ 1, and w_aa = w_Aa = 1 (SAME FITNESS)

w~ = w_AApₜ² + 2w_Aapₜqₜ + qₜ²
= (1 − s)pₜ² + 2pₜqₜ + qₜ²
= 1 − spₜ²

Fitness of A
w_A = w_AApₜ +w_Aaqₜ
= (1 − s)pₜt + 1 − pₜ
= 1 − spₜ,

Fitness of a
w_a = w_Aapₜ + w_aaqₜ = pₜ + 1 − pₜ = 1

by subbing eqs 4.9-4.12

Question 59

=> FHWE with mutations and weak selection:
δp

Answer 59

δp=
pₜ [w_A − w~]/[w~]
− upₜ (w_A/w~)+ v[w_A/w~] qₜ
≈ −spₜ²qₜ − upₜ(1 − spₜ) + vqₜ
due to selection, frequency
of decreases, negative contribution, mutations + due to mutation , a-ᵛ→ A
frequency of increases

≈ −spₜ²qₜ − upₜ + vqₜ

this is simplified by using
[w_A-w~]/w~ ≈ −spₜ(1 − pₜ) = −spₜqₜ and
w_A/w~ ≈ 1 − spₜ
w_a/w~ ≈ 1

Question 60

FHWE with mutations and weak selection:

Coexistence fixed point given by

Answer 60

Selection is balanced by mutations:
coexistence of and at small freq p* of A

δp∗ = 0
⇒ 0 =
−s(p∗)²(1− p∗) − up∗ + v(1 − p∗)
cubic

Solve for p* (messy). Smarter: expect p* to be “small”
1 − p∗ ≈ 1, up∗ ≈ 0, vp∗ ≈ 0
p∗³≈ 0

(Since is deleterious, and thus opposed by selection, we expect p* to be small

⇒ Eq. for p∗ simplifies to 0 ≈
−s(p∗)² + v
⇒ p∗ ≈ sqrt(v/s) when u, v ≪ s ≪ 1
In fact, p∗ ≈ sqrt(v/s)+ O(v/s) which is small, as assumed, but generally not vanishingly small

When u, v ≈ 10−4 and s ≈ 0.01: v/s ≈ 0.01 and p∗ ≈ 0.1 (small but not negligible)

Question 61

mutation-selection balance

Answer 61

This example shows how mutations help maintain genetic variability: deleterious allele is eliminated by selection but this is balanced by mutations (heterozygotes act as a reservoir of A alleles). This is called mutation-selection balance

We have as expected mutations can have genetic variablity, if we have mutations we always end up with a non-negligible fraction from possible mutations
Indeed A is deletirious but heterozygotes have the same fitness as those that have the advantageous allele, since it is deletirous and recessive.
So they (heterozygotes) spread just like a , passing on both a and A that is why A remains in the pop and end up with deleterious allele in pop.

Question 62

4.9 Extra MATH5567M topic: kin selection & Hamilton’s rule

What is the role of social behaviour on evolution?

In fitness terms, there are four possible outcomes to interactions

Answer 62

There are different forms of social behaviour: cooperation when actions result in fitness gains for both participants;
altruism is when the actor suffers a fitness cost and the receiver gets a fitness benefit;
selfishness is when the actor gets a fitness gain and the receiver suffers a fitness loss;
spite is when the
action results in fitness losses for both participants.

with some overlap between them.

Question 63

Kin selection:

Answer 63

Altruism can be explained if selection favours traits resulting in decrease of
personal fitness to benefit kins => “evolutionary altruistic strategy”

“Would I give my life to save a drowning brother?
No, but I would lay down my life for two brothers or eight cousins”. (J. B. S. Haldane)

Question 64

“Would I give my life to save a drowning brother?
No, but I would lay down my life for two brothers or eight cousins”. (J. B. S. Haldane)

Answer 64

This is because in a diploid population (no inbreeding) siblings share half of their genetic material (they have same parents),
two full siblings share the same genes
with a probability 1/2,

while cousins are 1/8 identical
while on average an uncle and his niece and first cousins are respectively 1/4 and 1/8 genetically identical

=> 2 full siblings or 8 cousins would carry the same genetic
material as Haldane.

Question 65

The degree of relatedness

Answer 65

Coefficient of relatedness r (between 0 and 1) gives
the probability that 2 individuals share an allele at a
given locus by descending from a common ancestor.

between two individuals

Formally, the degree r between any two individuals is the probability that they share an allele at a given locus inherited from a common ancestor.

In diploid organisms, the coefficient of relatedness r between a parent and child is 1/2, since half of the genetic material of the child is descended from that parent.

Question 66

pedigree chart

Answer 66

in which all minimal paths of descent starting from the
actors and ending to the receiver are traced

Since each parent contributes to 50% of the alleles of each offspring (one says that the probability that alleles are identical by descent is 50%), each step in the pedigree chart carries a probability 1/2. Hence, the coefficient of relatedness of two full siblings is the sum of the probabilities of being equal by descent through the mother (1/2 × 1/2) and through the father (1/2 × 1/2), resulting in r = 1/2

Question 67

Coefficient of relatedness r
between
parents and offspring
siblings
uncle/neice

grandparent

Answer 67

Hence r=1/2 between parents and offspring,
r=1/2 between full siblings,
r=1/4 between uncle/niece and grandparent/ grandchild
r=1/8 between cousins
To obtain r: “pedigree chart” in which we draw all
minimal paths of descent from actor to receiver, each path
weighted 1/2.

Question 68

Haldane and his cousin have a coefficient r=1/8

Answer 68

Haldane and his cousin have a coefficient r=1/8 because they are related
(i.e. received their common genes) via the following paths:
Haldane -> Haldane’s father -> Haldane’s uncle -> Haldane’s
0.5 x 0.5 x 0.5 = 1/8

Question 69

In 1964, W. D. Hamilton popularised the concept of kin selection and formulated what is known as Hamilton’s rule.

Answer 69

Hamilton’s rule says that kin selection causes genes to increase in frequency when the genetic relatedness of a recipient to an actor multiplied by the benefit to the recipient is greater than the reproductive cost to the actor.

Hamilton’s rule says that the recipient’s
genes should increase in frequency when
rB > C.

the frequency of the receiver’s genes increases, where
r: coefficient of relatedness, B: reproductive benefit to receiver, C: cost to actor (altruist)

Question 70

r for siblings

Answer 70

1/2 x 1/2 + 1/2 x 1/2 = 1/4

M F
BOTH RELATES 1/2 x 1/2 two paths
B S

Question 71

rB>C

Answer 71

If r is the coefficient of genetic relatedness of the recipient to
the actor of the altruistic behaviour (individual dispensing the behaviour), B is the additional reproductive benefit gained by the recipient (individual receiving the behaviour), and C is the
reproductive cost to the individual performing the act,

Question 72

hamiltons rule remark

Answer 72

In practice, it is hard to estimate B and C. Some recent observations (groups of red squirrels) seem to be in line with this “rule”. Hamilton rule can be derived mathematically, under
suitable assumptions, from the so-called “Price equation”.