1

BSCI 363: read the rest of chapter 9

CONS 670: read the rest of chapter 7, and chapter 9

2

stabilizing directional disruptive

As natural selection begins

After selection has occurred

26

P decreasesH depends on genotype favored by selection

3

stabilizing directional disruptive

As natural selection begins

After selection has occurred

26

P

H

P

H

P

H? ? ?

4

Dynamic Effects:

Natural Selection

maintains allele frequencies in equilibrium

with environmental demands

vs.

Genetic Drift

pulls allele frequencies away from environmental

equilibrium

31a-2

5

5 causes of microevolution

1) genetic drift - stochastic variation in inheritance

2) Assortative mating

3) Mutation

4) Natural selection

5) Migration (gene flow)

31a-2

6

Emigration / Immigration

Donor populationRecipient population

Pollen grains

emigration from one population and immigrationinto the other; breeding = Gene flow

Migration (m) of breeding individuals results in increased H and increased P

41f4

7

Models of gene flow based on population structure

metapopulation

subpopulation

1. Continent to island model e.g., Madagascar (source - sink model)

2. Equivalent island modele.g., Philippines

30-1

8

3. Stepping-stone model e.g., Hawaiian Islands

4. Isolation by distance model (continuous habitat)e.g., Amazon forest

Genetic neighborhood

30-2

9

Gene flow results in homogenization of allele frequencies on“islands” of equivalent size. *

* assume thorough gene flow between populations

30e

Beforegene flow:

After:

A = .7

A = .6

A = .5A = .4

A = .55

A = .55 A = .55

A = .55.7.6.5.4

X = .55

m

10

Changes in allele frequency due to migrationmij = gene flow = # breeding immigrants from donor population j

size of recipient population i

migrants (m) moving from donor (j) to recipient (i)

Change in allele frequency (q) in population i:

Before AfterRecipient i qi qi’ = (1-mij) qi + mij qj

Donor j qj qj

j i“jump” “into”

41A

11

Gene flow example

Before Ne = 200 Ne = 300qj = 0.9 qi = 0.5

After qj’ = 0.9 qi’ = 0.51

mij = 5 = 0.0167 300qi’ = (1 - mij) qi + mij(qj) = (1 - 0.0167) (.5) + (0.0167) (0.9) = (0.5067) = 0.51

(If number of immigrants = 50, then qi’ = 0.57)

Donor population (j) Recipient (i)

41e

5 individuals

12

Gene flow: major points

1) High mij homogenizes allele frequencies in two populations

2) Rate of gene flow influences Ne of recipient population and metapopulation

3) A small amount of gene flow may counteract genetic drift and conserve genetic diversity in small populations

4) Allele frequency in the donor population is assumed to be unchanged after gene flow to recipient population

5) Size of donor population does not influence allele frequencies in recipient populations

6) Applications: calculate number of individuals needed to introduce into recipient population of known size to maintain its genetic diversity.

41f1

13

Directional selection in peppered moths (Biston betularia) in England

2 phenotypes: black moth, mottle white moth

Prior to 1600 (Industrial revolution)black form approximately 1%white form approximately 99%

After 1600 (widespread industrial pollution, smoke and soot)black form approximately 90%white form approximately 10%

Now (local pollution from smokestacks)Near pollution source Away

black form 50% 10%white form 50% 90%

Outbreeding depression?

41

14

Selection and gene flow:colonization along an environmental gradient

42

Cold-adapted favored

Warm-adapted favored

m

m

m

m

m

15

Effect of inbreeding on HSelfing: In a population with f (a) = f (A) = 0.5 At Hardy-Weinberg equilibrium, genotypic frequencies are

p2 + 2pq + q2 = 1AA Aa aa

Parental genotypic frequencies: .25 .50 .25

F1 homozygotes .25 .25

F1 heterozygotes .125 .25 .125

F1 genotypes .375 .25 .375

Conclusion: Frequency of heterozygotes is reduced by 50% witheach generation of selfing. But there is no loss of allelic diversity: f (a) = f (A) = 0.5 43

16

(1-s)HS = --------------- 2pq

(1- s/2)

HS = equilibrium heterozygote frequency (random + selfing) s = proportion of selfing

The case of selfing with some random mating too

The frequency of heterozygotes will always fall between 2pq and 0

44?

17

.5

0

.1

.2

.3

.4

Ht

Time in generations0 20

Brother-sister (sibs)

Selfing

45-1

The loss of heterozygosity through time caused by inbreeding

18Generations

521 43

Los

s of

Ht

0.75

0.50

0.25

Full-sibs

Half-sibs

Doublefirstcousins

Firstcousins

45-2

19

Genetic consequences of inbreeding1) decrease in heterozygosity, no change in P (allelic diversity)

(the more related the individuals, the faster the loss of H)2) increases the probability of a zygote receiving identical alleles (homologous alleles), which will result in increased expression of recessive alleles.3) increased phenotypic expression of deleterious alleles (strongly selected against) - often results in decreased size, reproduction, vigor, etc., which decrease fitness (i.e., inbreeding depression)4) increase in phenotypic variability resulting from a deviation from the mean genotypes in non-inbred individuals

43e-1

Genetic load

20

Inbreeding coefficient

Sewall Wright (1923)

F = the probability that an individual will receive two equal alleles, at a specific locus, that are from the same ancestor.

Autozygous = identical by descent

allozygous = not identical by descent

F = probability that an individual will be autozygous at a given locus1 - F = probability that an individual will be allozygous at a given locus

43e-2

21

Calculate Junior’s inbreeding coefficients from this pedigree:

AB CDMom Dad

AC

CC

C = .5

C = .5

C = .5Sis

Junior (or couldbe DD from Dad)

Probability of C from Dad to Sis to Junior = .25Probability of C from Dad to Junior = .50 Probability of Jr. inheriting CC from Dad = .25 X .50 = .125Probability of Junior inheriting DD from Dad = .125

F = .125 + .125 = .25 = probability of Jr. being autozygous

31