5 Agents of evolutionary change
Mutation
Gene Flow
Genetic Drift
Non-random mating
Natural Selection
Stnd: 7e
Objective:
C-Notes: Hardy-Weinberg
Equilibrium
3/19/2014
SWBAT interpret the HardyWeinberg equation in order
to predict allele trends in
populations.
What is the
HardyWeinberg
Equilibrium
Model?
• The Hardy-Weinberg law examines
changes in gene pools and explains
how evolution can select for certain
genes
– If a population is in Hardy-Weinberg
equilibrium and its members continue to
mate randomly generation after
generation, allele and genotype
frequencies would remain inherently
stable (constant), if certain assumptions
are met (5 Conditions).
Why is the
HardyWeinberg
Theorem
unlikely to
appear in nature?
• Hardy-Weinberg model was an important step to
defining the mechanism for evolution. Through
recognizing that no population can ever meet these 5
conditions, it identified situations that would lead to
allele frequency.
5 Necessary Conditions for Hardy-Weinberg equilibrium:
– Large Population >10,000 individuals
– No Mutations
– No Migration or Emigration
• (no transferring alleles between populations)
– No natural selection
– Random mating
**we do NOT see Hardy-Weinberg in NATURE
because not many population follows (meet) his
5 rules**
– useful model to measure if forces are acting on a population
•
measuring evolutionary change
A population that meets these conditions is said to be in HardyWeinberg equilibrium, because allele frequencies do not change
from generation to generation.
How to find the Key:
value to help
• Frequencies of Alleles:
solve the Hardy– Dominant : P
Weinberg
– Recessive: q
Equation?
• Frequencies of allele combination
(genotypes)
– Homozygous Diploid individuals:
• Homozygous Dominant(AA) : P^2
• Homozygous recessive(aa): q^2
– Heterozygous Diploid individuals:
• Heterozygous Dominant(Aa): 2Pq
How to solve
Hardy-Weinberg
Problems(Steps)?
1. Examine the information to determine
what pieces of information you have
been given about the population
2. Find out the value of P or q.
–
If this is achieved, then every other value in
the equation can be determined by simple
calculation.
3. Take the square root of q^2 to find q.
4. Determine p by subtracting q from 1
(ex: P=1-q)
5. Determine p^2 by multiplying p by itself
(ex: p^2 = p x p)
6. Determine 2Pq by multiplying p times q
times 2.
7. Check that your calculations are correct
by adding up the values for:
–
p^2 + q^2 +2Pq = 1
How to use the
Equation Part 1:
Hardy2 types of Alleles in a population:
Weinberg
Equation when • p + q =1
– Where:
there is 2 types of
– p = the frequency of the dominant allele
alleles in a
– q = the frequency of the recessive allele
population?
• All the alleles in a population will total
100% (or 1 in this case)
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
Bb
bb
How to use the
Equation Part 2:
Hardy3 types of Alleles in a population:
Weinberg
2 + 2pq + q2 = 1
•
p
Equation when
2 = % homozygous dominant
–
p
there is 3 types of
– 2pq = % heterozygous
alleles in a
2 = % homozygous recessive
–
q
population?
• This determines the genotype
frequencies in a population at HardyWeinberg equilibrium.
• (Note that the square root of p2 = p and
the square root of q2 = q)
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
Bb
bb
H-W formulas
• Alleles:
p+q=1
B
b
B b
B BB Bb
b Bb
• Individuals:
p2
+ 2pq +
q2
=1
BB
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!
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?
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
What assume
Must
are the genotype
populationfrequencies?
is in H-W equilibrium!
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?
Hardy-Weinberg
Lab Data
Mutation
2006-2007
Gene Flow
Genetic Drift
Selection
Non-random mating
Hardy Weinberg Lab: Equilibrium
Original population
18 individuals
36 alleles
p (A):
0.5
q (a):
0.5
AA
.25
Aa
.50
aa
.25
Case #1 F5
AA
Aa
aa
4
7
7
total alleles = 36
p (A): (4+4+7)/36 = .42
q (a): (7+7+7)/36 = .58
AA
.22
Aa
.39
How do you explain these data?
aa
.39
Hardy Weinberg Lab: Selection
Original population
15 individuals
30 alleles
p (A):
0.5
q (a):
0.5
AA
.25
Aa
.50
aa
.25
Case #2 F5
AA
Aa
aa
9
6
0
total alleles = 30
p (A): (9+9+6)/30 = .80
q (a): (0+0+6)/30 = .20
AA
.60
Aa
.40
How do you explain these data?
aa
0
Hardy
Original population
15 individuals
30 alleles
p (A):
0.5
q (a):
0.5
AA
.25
Aa
.50
aa
.25
Heterozygote
Weinberg
Lab:
Advantage
Case #3 F5
AA
Aa
aa
4
11
0
total alleles = 30
p (A): (4+4+11)/30 = .63
q (a): (0+0+11)/30 = .37
AA
.27
Aa
.73
How do you explain these data?
aa
0
Hardy
Original population
15 individuals
30 alleles
p (A):
0.5
q (a):
0.5
Heterozygote
Weinberg
Lab:
Advantage
Case #3 F10
AA
Aa
6
9
total alleles = 30
aa
0
p (A): (6+6+9)/30 = .70
q (a): (0+0+9)/30 = .30
AA
.25
Aa
.50
aa
.25
AA
.4
Aa
.6
How do you explain these data?
aa
0
Hardy Weinberg Lab: Genetic Drift
Original population
6 individuals
12 alleles
p (A): 0.5
q (a): 0.5
AA
.25
Aa
.50
aa
.25
Case #4 F5-1
AA
Aa
aa
4
2
0
total alleles = 12
p (A): (4+4+2)/12 = .83
q (a): (0+0+2)/12 = .17
AA
.67
Aa
.33
How do you explain these data?
aa
0
Hardy Weinberg Lab: Genetic Drift
Original population
5 individuals
10 alleles
p (A): 0.5
q (a): 0.5
AA
.25
Aa
.50
aa
.25
Case #4 F5-2
AA
Aa
aa
0
4
1
total alleles = 10
p (A): (0+0+4)/10 = .4
q (a): (1+1+4)/10 = .6
AA
0
Aa
.8
How do you explain these data?
aa
.2
Hardy Weinberg Lab: Genetic Drift
Original population
5 individuals
10 alleles
p (A): 0.5
q (a): 0.5
AA
.25
Aa
.50
aa
.25
Case #4 F5-3
AA
Aa
aa
2
2
1
total alleles = 10
p (A): (2+2+2)/10 = .6
q (a): (1+1+2)/10 = .4
AA
.4
Aa
.4
How do you explain these data?
aa
.2
Hardy Weinberg Lab: Genetic Drift
Original population
5 individuals
10 alleles
p (A): 0.5
q (a): 0.5
AA
.25
Aa
.50
Case #4 F5
AA Aa aa
1 .67 .33 0
p q
.83 .17
2
0
.8
.2
.4 .6
3
4
.4
.2
.6 .4
aa
.25
How do you explain these data?
Any Questions??
2007-2008
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