Transcript HARDY-WEINBERG Practice - local.brookings.k12.sd.us

Measuring
Evolution of Populations
SLIDE SHOW MODIFIED FROM KIM [email protected]
5 Agents of evolutionary change
Mutation
Gene Flow
Genetic Drift
AP Biology
Non-random mating
Selection
Populations & gene pools
 Concepts
a population is a localized group of
interbreeding individuals
 gene pool is collection of alleles in the
population

 remember difference between alleles & genes!

allele frequency is how common is that
allele in the population
 how many A vs. a in whole population
AP Biology
Evolution of populations
 Evolution = change in allele frequencies
in a population


hypothetical: what conditions would
cause allele frequencies to not change?
non-evolving population
REMOVE all agents of evolutionary change
1. very large population size (no genetic drift)
2. no migration (no gene flow in or out)
3. no mutation (no genetic change)
4. random mating (no sexual selection)
5. no natural selection (everyone is equally fit)
AP Biology
Hardy-Weinberg equilibrium
 Hypothetical, non-evolving population

preserves allele frequencies
 Serves as a model (null hypothesis)


natural populations rarely in H-W equilibrium
useful model to measure if forces are acting on
a population
 measuring evolutionary change
G.H. Hardy
AP mathematician
Biology
W. Weinberg
physician
Hardy-Weinberg theorem
 Counting Alleles
assume 2 alleles = B, b
 frequency of dominant allele (B) = p
 frequency of recessive allele (b) = q

 frequencies must add to 1 (100%), so:
p+q=1
BB
AP Biology
Bb
bb
Hardy-Weinberg theorem
 Counting Individuals



frequency of homozygous dominant: p x p = p2
frequency of homozygous recessive: q x q = q2
frequency of heterozygotes: (p x q) + (q x p) = 2pq
 frequencies of all individuals must add to 1 (100%), so:
p2 + 2pq + q2 = 1
BB
AP Biology
Bb
bb
H-W formulas
 Alleles:
p+q=1
B
b
B
B
b
BB Bb
b
Bb
 Individuals:
p2
+ 2pq +
q2
=1
BB
AP Biology
BB
Bb
Bb
bb
bb
bb
Using Hardy-Weinberg equation
population:
100 cats
84 black, 16 white
How many of each
genotype?
p2=.36
BB
q2 (bb): 16/100 = .16
q (b): √.16 = 0.4
p (B): 1 - 0.4 = 0.6
2pq=.48
Bb
q2=.16
bb
are population
the genotype
frequencies?
MustWhat
assume
is in
H-W equilibrium!
AP Biology
Using Hardy-Weinberg equation
p2=.36
Assuming
H-W equilibrium
2pq=.48
q2=.16
BB
Bb
bb
p2=.20
=.74
BB
2pq=.64
2pq=.10
Bb
q2=.16
bb
Null hypothesis
Sampled data
How do you
explain
the data?
AP
Biology
Application of H-W principle
 Sickle cell anemia

inherit a mutation in gene coding for
hemoglobin
 oxygen-carrying blood protein
 recessive allele = HsHs
 normal allele = Hb

low oxygen levels causes
RBC to sickle
 breakdown of RBC
 clogging small blood vessels
 damage to organs

AP Biology
often lethal
Sickle cell frequency
 High frequency of heterozygotes
1 in 5 in Central Africans = HbHs
 unusual for allele with severe
detrimental effects in homozygotes

 1 in 100 = HsHs
 usually die before reproductive age
Why is the Hs allele maintained at such high
levels in African populations?
Suggests some selective advantage of
being heterozygous…
AP Biology
Single-celled eukaryote parasite
(Plasmodium) spends part of its
life cycle in red blood cells
Malaria
1
2
AP Biology
3
Heterozygote Advantage
 In tropical Africa, where malaria is common:



homozygous dominant (normal) die of malaria: HbHb
homozygous recessive die of sickle cell anemia: HsHs
heterozygote carriers are relatively free of both: HbHs
 survive more, more common in population
Hypothesis:
In malaria-infected
cells, the O2 level is
lowered enough to
cause sickling which
kills the cell &
destroys the parasite.
AP Biology
Frequency of sickle cell allele &
distribution of malaria
HARDY-WEINBERG
PRACTICE PROBLEMS
p+q=1
p2 + 2 pq + q2 = 1
Black (b) is recessive
to white (B)
Bb and BB pigs “look alike”
so can’t tell their alleles by observing their phenotype.
ALWAYS START WITH RECESSIVE alleles.
p= dominant allele
q = recessive allele
4/16 are black.
So bb or q2 = 4/16 or 0.25
q=
0.25 = 0.5
http://www.phschool.com/science/biology_place/labbench/lab8/samprob1.html
Once you know q
you can figure out p
... p+q=1
p+q=1
p + 0.5 = 1
p = 0.5
Now you know the allele frequencies.
The frequency of the recessive (b) allele q = 0.5
The frequency of the dominant (B) allele p = 0.5
WHAT ARE THE
GENOTYPIC FREQUENCIES?
You know pp from problem
bb or q2 = 4/16 = 0.25
BB or p2 = (0.5)2 = 0.25
Bb = 2pq = 2 (0.5) (0.5) = 0.5
25% of population are bb
25% of population are BB
50% of population are Bb
http://www.phschool.com/science/biology_place/labbench/lab8/samprob1.html
Within a population of butterflies, the color brown (B)
is dominant over the color white (b). And, 40% of all
butterflies are white.
q2 = 0.4
q=
0.4
= 0.6324
p = 1 - 0.6324 = 0.3676
aa = 0.4 = 40%
Aa = 2 (0.632) (0.368) = 0.465 =46.5%
AA = (0.3676) (0.3676) = .135 = 13.5%
Image from: BIOLOGY by Miller and Levine; Prentice Hall Publishing ©2006
PRACTICE HARDY WEINBERG
1 in 1700 US Caucasian newborns have
cystic fibrosis. C for normal is dominant
over c for cystic fibrosis.
Calculate the allele frequencies for
C and c in the population
Image from: BIOLOGY by Miller and Levine; Prentice Hall Publishing ©2006
1/1700 have cystic fibrosis
q2 = 1/1700
q=
0.00059
q = 0.024
p = 1 – 0.024 = 0.976
Frequency of C = 97.6%
Frequency of c = 2.4%
NOW FIND THE GENOTYPIC FREQUENCIES
CC or p2 = (0.976)2
=
.953
Cc or 2pq = 2 (0.976) (0.024) = 0.0468
cc = 1/1700 = 0.00059
CC = 95.3% of population
Cc = 4.68% of population
cc = .06% of population
Now you can answer questions about the population:
How many people in this population are heterozygous?
0.0468 (1700) = 79.5 ~ 80 people are Cc
It has been found that a carrier is better able to survive
diseases with severe diarrhea. What would happen to the
frequency of the "c" if there was a epidemic of cholera or
other type of diarrhea producing disease?
Cc more likely to survive than CC.
c will increase in population
The gene for albinism is known to be a recessive allele.
In Michigan, 9 people in a sample of 10,000 were found to
have albino phenotypes. The other 9,991 had skin
pigmentation normal for their ethnic group.
Assuming hardy-Weinberg equilibrium, what is the allele
frequency for the dominant pigmentation allele in this
population?
q2 = 9/10000
q=
0.0009
q = 0.03
p = 1 – 0.03= 0.97
Frequency of C = 97%
Frequency of c = 3%
CC or q2 = (0.976)2
=
.953
Cc or 2pq = 2 (0.976) (0.024) = 0.0468
cc = 1/1700 = 0.00059
CC = 95.3% of population
Cc = 4.68% of population
cc = .06% of population