chpt. 23 the evolution of populations-- population genetics

Chpt. 23 The Evolution of

Populations-- Population Genetics

Individuals are selected…

Populations evolve

Individuals are selected…

Populations Populations evolveevolve

PopulationsPopulations = unit of evolution

Natural selectionNatural selection =

mechanism of evolution

GradualismGradualism =

accumulation of small changes in

gene pool over LONG periods of

time

Hardy-WeinbergHardy-Weinberg TheoremTheorem


examines the gene structure of a NON-evolving population

Hardy-WeinbergHardy-Weinberg TheoremTheorem examines the gene

structure of a NON-evolving population.

Obviously, this is not common, however, gives a base-line / model NULL HYPOTHESIS for determining if and why populations evolve

Hardy-WeinbergHardy-Weinberg TheoremTheorem even though alleles are shuffled and recombined during meiosis and random

fertilization.This has no effect on This has no effect on the the overalloverall gene pool gene pool

percentages.percentages.

Not SWIMMING pool….

A

a

A

aA

a

A

a

a

a

A

A

A

a GENE POOL!!!


Gene pool frequenciesGene pool frequencies

(percentages) will (percentages) will

remainremain unchangedunchanged if if nono

mechanismmechanism thatthat cancan

causecause evolutionevolution toto

occuroccur actsacts onon aa

populationpopulation..

Hardy-WeinbergHardy-Weinberg TheoremTheoremGene pool frequencies Gene pool frequencies

will remain unchanged will remain unchanged if:if:

Mutations are not occurring



Natural selection is not occurring



Population is LARGE



EVERYONE breeds…



EVERYONE randomly mates…



EVERYONE produces the same number of offspring



NOONE migrates in or out of the population… everyone stays

5 Agents of evolutionary change:Mutation Gene Flow (migration)

Genetic Drift (same # of offspring) Selection

Non-random mating


remember:remember: H.W. H.W. explains: explains: the the

frequencyfrequency of allelesof alleles remainsremains constantconstant in a in a population… population… unless unless

acted upon by agents acted upon by agents OTHER THAN sexual OTHER THAN sexual recombination.recombination.

Hardy-Weinberg Hardy-Weinberg Principle Principle

p2 + 2pq + q2 = 1p2 + 2pq + q2 = 1

Mathematical statement Mathematical statement about the relative about the relative frequency of alleles frequency of alleles (genotypes) in a (genotypes) in a population.population.p +q = 1p +q = 1


Frequency of alleles Frequency of alleles remainsremains constantconstant in a in a population, population, unless unless acted upon by agents acted upon by agents OTHER THAN sexual OTHER THAN sexual recombination.recombination.InheritanceInheritance doesdoes notnot causecause changeschanges inin alleleallele frequencyfrequency..


p + q = 1p + q = 1

p = frequency of p = frequency of dominant dominant allele allele


p + q = 1p + q = 1

q = frequency of q = frequency of recessive recessive allele allele


in most cases, we in most cases, we only know the only know the phenotypic phenotypic frequenciesfrequencies

Mathematical statement Mathematical statement about the relative about the relative frequency of alleles frequency of alleles (genotypes) in a (genotypes) in a population.population.


qq2 2 = # of aa = # of aa individualsindividuals

qq22 = frequency of = frequency of homozygotehomozygote recessiverecessive individualsindividuals


pp2 2 = # of AA = # of AA individualsindividuals

pp22 = frequency of = frequency of homozygotehomozygote dominantdominant individualsindividuals


p = # of (AA) + 2 p = # of (AA) + 2 (# Aa)(# Aa)

p = frequency of p = frequency of dominantdominant alleleallele


q = # of (aa) + 2 q = # of (aa) + 2 (# Aa)(# Aa)

q = frequency of q = frequency of recessiverecessive alleleallele


2pq2pq = # of Aa = # of Aa individualsindividuals

2pq = frequency of 2pq = frequency of heterozygoteheterozygote individualsindividuals

Hardy-Weinberg Hardy-Weinberg Principle Principle by comparing genotypic by comparing genotypic frequencies from one frequencies from one generation to the next, generation to the next, you can learn you can learn whetherwhether oror notnot evolution has evolution has occurred…occurred…

Hardy-Weinberg Hardy-Weinberg Principle Principle if genotypic if genotypic frequencies have frequencies have changed from your changed from your original count… original count…

evolution has evolution has occurred!occurred!

Suppose there are 1,000 individuals in a population

Genotype Number Genotypic Frequency AA 490 0.49

Aa 420 0.42

aa 90 0.09

total 1000 1.00


Genotypic Frequency 0.49

0.42

0.09

total 1.00

Genotypic frequency = the proportion of a particular genotype found in a

population

AA

Aa

aa


Phenotype Number Phenotypic Frequency dominant 9100.91

recessive 900.09

total 10001.00


Phenotypic Frequency 0.91

0.09

total 1.00

Phenotypic frequency = the proportion of a particular phenotype found in a population


Allele Number Allele Frequency A 1400 0.7

a 4200.3

total 20001.00

480

Allele frequency

Pssst…(There are 1,000 copies of the flower color gene in this population of 500 total flowers…)

q =

q = frequen

cy of recessi

ve allele

However, we do

not know how

many a’s

there are

just by looking

at phenoty

pe

480

Genotypic frequency

q2 = 20/500

q2 = frequen

cy of recessi

ve genotyp

e

480

Genotypic frequency

q2 = .04

q2 = frequen

cy of recessi

ve genotyp

e

480

Allele frequency

q = .04

q = frequen

cy of recessi

ve allele

480

Allele frequency

q = .2

q = frequen

cy of recessi

ve allele

480

Allele frequency

q = .2

q = frequen

cy of recessi

ve allele

p = frequen

cy of dominan

t allele

p + q = 1p + .2 = 1p = 1 - .2p = .8

480

Some of the pink Some of the pink flowers will be AA and flowers will be AA and some will be Aasome will be Aa

p2 + 2pq + q2 = p2 + 2pq + q2 = 11

p = .8 q = .2

480

Some of the pink Some of the pink flowers will be AA and flowers will be AA and some will be Aasome will be Aa

.64 + .32 + .04 .64 + .32 + .04 = 1= 1

480

How many of the pink How many of the pink flowers will be AA and flowers will be AA and how many will be Aahow many will be Aa

.64 X 500 .64 X 500 individualsindividuals320 individuals are 320 individuals are AAAA

480

How many of the pink How many of the pink flowers will be AA and flowers will be AA and how many will be Aahow many will be Aa

.32 X 500 .32 X 500 individualsindividuals160 individuals are 160 individuals are AaAa

480

320 320 are AAare AA

160 160 are Aaare Aa

480 total480 total

Genetic structure of next generation

.8 x .8 = .64


.2 x .2 = .04


.2 x .8 = .16

.8 x .2 = .16 .32

aA

Aa


Under ideal conditions, the relative allele frequencies are the same in the offspring generation as in the parent generation.

pp22 + 2pq + + 2pq + qq22 = 1 = 1

IFIF: frequencies in a population deviate from Hardy-Weinberg (these are set numbers)…

THENTHEN: we know the population is evolving.

Oh Hardy, why did you quit your job at

ABERCROMBIE?

Weinburg, I keep telling you, I got

sick of changing genes!!!

changechange overover timetime is a result of is a result of changes in a changes in a population’s population’s frequency of frequency of genotypes genotypes / /

geneticgenetic

Lets Hardy!!What percentage of the human population are carriers for the allele for PKU?

1 out of 10,000 babies are born with this1 out of 10,000 babies are born with this recessiverecessive disease disease

Most of the time you will Most of the time you will begin by determining the begin by determining the

frequency of the frequency of the homozygoushomozygous recessiverecessive genotypegenotype

Find q2 Find q2 (frequency of homo (frequency of homo recessive)recessive)

Lets Hardy!!What percentage of the human population are carriers for the allele for PKU?

1 out of 10,000 babies are born with this1 out of 10,000 babies are born with this recessiverecessive disease disease

Hint: q2 = 1 / 10,000 Hint: q2 = 1 / 10,000 (frequency of aa)(frequency of aa)

Find qFind q

Find p Find p Find Find pqpq

AP Problems Using AP Problems Using Hardy-WeinbergHardy-Weinberg

• Solve for q2 (% of total)

• Solve for q (equation)

• Solve for p (1- q)

• H-W is always on the national AP Bio exam (but no calculators are allowed).


• Solve for q2 (% of total)

• Solve for q (equation)

• Solve for p (1- q)

• H-W is always on the national AP Bio exam (but no calculators are allowed)


population: 100 cats84 black, 16 whiteHow many of each genotype?

population: 100 cats84 black, 16 whiteHow many of each genotype?

q2 (bb): 16/100 = .16

q (b): √.16 = 0.40.4

p (B): 1 - 0.4 = 0.60.6

q2 (bb): 16/100 = .16

q (b): √.16 = 0.40.4

p (B): 1 - 0.4 = 0.60.6

bbBbBB

p2=.36p2=.36 2pq=.482pq=.48 q2=.16q2=.16


bbBbBB

p2=.36p2=.36 2pq=.482pq=.48 q2=.16q2=.16

Assuming H-W equilibriumAssuming H-W equilibrium

Sampled data Sampled data bbBbBB

p2=.20p2=.20 2pq=.642pq=.64 q2=.16q2=.16

How do you explain the data? How do you explain the data?

Null hypothesis Null hypothesis

chpt. 23 the evolution of populations-- population genetics

Documents

hardyweinberg theorem

hardyweinberg principle

large slide

hardyweinberg principle

hardyweinberg principle

gene structure

overall gene pool percentages

allele frequency