1. 5 Agents of evolutionary change
Mutation
Gene Flow
Non-random mating
Genetic Drift
Natural Selection

2. C-Notes: Hardy-Weinberg Equilibrium
Stnd: 7e
C-Notes: Hardy-Weinberg Equilibrium
3/19/2014
Objective:
SWBAT interpret the Hardy-Weinberg equation in order to predict allele trends in populations.

3. What is the Hardy-Weinberg 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).
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4. Why is the Hardy-Weinberg 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 Hardy-Weinberg equilibrium, because allele frequencies do not change from generation to generation.
Why is the Hardy-Weinberg Theorem unlikely to appear in nature?
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5. How to find the value to help solve the Hardy-Weinberg Equation? Key:
Frequencies of Alleles:
Dominant : P
Recessive: q
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
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6. How to solve Hardy-Weinberg Problems(Steps)?
Examine the information to determine what pieces of information you have been given about the population
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.
Take the square root of q^2 to find q.
Determine p by subtracting q from 1 (ex: P=1-q)
Determine p^2 by multiplying p by itself (ex: p^2 = p x p)
Determine 2Pq by multiplying p times q times 2.
Check that your calculations are correct by adding up the values for:
p^2 + q^2 +2Pq = 1
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7. 2 types of Alleles in a population: p + q =1
How to use the Hardy-Weinberg Equation when there is 2 types of alleles in a population?
Equation Part 1:
2 types of Alleles in a population:
p + q =1     
Where:
p = the frequency of the dominant allele   
q = the frequency of the recessive allele
All the alleles in a population will total 100% (or 1 in this case)
7th block

8. 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

9. 3 types of Alleles in a population: p2 + 2pq + q2 = 1
How to use the Hardy-Weinberg Equation when there is 3 types of alleles in a population?
Equation Part 2:
3 types of Alleles in a population:
p2 + 2pq + q2 = 1     
p2 = % homozygous dominant  
2pq = % heterozygous   
q2 = % homozygous recessive
This determines the genotype frequencies in a population at Hardy-Weinberg equilibrium.
(Note that the square root of p2 = p  and the square root of q2 = q)
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10. 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

11. H-W formulas B b B BB Bb b Bb bb Alleles: p + q = 1 Individuals:
p2 + 2pq + q2 = 1 B b BB Bb bb BB Bb bb

12. Using Hardy-Weinberg equation
population: 100 cats
84 black, 16 white
How many of each genotype?
q2 (bb): 16/100 = .16
q (b): √.16 = 0.4
p (B): = 0.6
p2=.36
2pq=.48
q2=.16 BB Bb bb
Must assume population is in H-W equilibrium!
What are the genotype frequencies?

13. Using Hardy-Weinberg equation
p2=.36
2pq=.48
q2=.16
Assuming H-W equilibrium BB Bb bb
Null hypothesis
p2=.20
p2=.74
2pq=.10
2pq=.64
q2=.16
q2=.16
Sampled data 1:
Hybrids are in some way weaker.
Immigration in from an external population that is predomiantly homozygous B
Non-random mating... white cats tend to mate with white cats and black cats tend to mate with black cats.
Sampled data 2:
Heterozygote advantage.
What’s preventing this population from being in equilibrium. bb Bb BB
Sampled data
How do you explain the data?
How do you explain the data?

14. Using Hardy-Weinberg equation
population: 100 cats
84 black, 16 white
How many of each genotype?
q2 (bb): 16/100 = .16
q (b): √.16 = 0.4
p (B): = 0.6
p2=.36
2pq=.48
q2=.16 BB Bb bb
Must assume population is in H-W equilibrium!
What are the genotype frequencies?

15. Using Hardy-Weinberg equation
p2=.36
2pq=.48
q2=.16
Assuming H-W equilibrium BB Bb bb
Null hypothesis
p2=.20
p2=.74
2pq=.10
2pq=.64
q2=.16
q2=.16
Sampled data 1:
Hybrids are in some way weaker.
Immigration in from an external population that is predomiantly homozygous B
Non-random mating... white cats tend to mate with white cats and black cats tend to mate with black cats.
Sampled data 2:
Heterozygote advantage.
What’s preventing this population from being in equilibrium. bb Bb BB
Sampled data
How do you explain the data?
How do you explain the data?

16. Hardy-Weinberg Lab Data
Mutation
Gene Flow
Genetic Drift
Selection
Non-random mating

17. Hardy Weinberg Lab: Equilibrium
Original population
Case #1 F5
18 individuals
36 alleles
p (A): 0.5
q (a): 0.5
total alleles = 36
p (A): (4+4+7)/36 = .42
q (a): (7+7+7)/36 = .58 AA 4 Aa 7 aa 7 AA
.25 Aa
.50 aa
.25 AA
.22 Aa
.39 aa
.39
How do you explain these data?

18. Hardy Weinberg Lab: Selection
Original population
Case #2 F5
15 individuals
30 alleles
p (A): 0.5
q (a): 0.5
total alleles = 30
p (A): (9+9+6)/30 = .80
q (a): (0+0+6)/30 = .20 AA 9 Aa 6 aa AA
.25 Aa
.50 aa
.25 AA
.60 Aa
.40 aa
How do you explain these data?

19. Hardy Weinberg Lab: Heterozygote Advantage Original population
Case #3 F5
15 individuals
30 alleles
p (A): 0.5
q (a): 0.5
total alleles = 30
p (A): (4+4+11)/30 = .63
q (a): (0+0+11)/30 = .37 AA 4 Aa 11 aa AA
.25 Aa
.50 aa
.25 AA
.27 Aa
.73 aa
How do you explain these data?

20. Hardy Weinberg Lab: Heterozygote Advantage Original population
Case #3 F10
15 individuals
30 alleles
p (A): 0.5
q (a): 0.5
total alleles = 30
p (A): (6+6+9)/30 = .70
q (a): (0+0+9)/30 = .30 AA 6 Aa 9 aa AA
.25 Aa
.50 aa
.25 AA .4 Aa .6 aa
How do you explain these data?

21. Hardy Weinberg Lab: Genetic Drift
Original population
Case #4 F5-1
6 individuals
12 alleles
p (A): 0.5
q (a): 0.5
total alleles = 12
p (A): (4+4+2)/12 = .83
q (a): (0+0+2)/12 = .17 AA 4 Aa 2 aa AA
.25 Aa
.50 aa
.25 AA
.67 Aa
.33 aa
How do you explain these data?

22. Hardy Weinberg Lab: Genetic Drift
Original population
Case #4 F5-2
5 individuals
10 alleles
p (A): 0.5
q (a): 0.5
total alleles = 10
p (A): (0+0+4)/10 = .4
q (a): (1+1+4)/10 = .6 AA Aa 4 aa 1 AA
.25 Aa
.50 aa
.25 AA Aa .8 aa .2
How do you explain these data?

23. Hardy Weinberg Lab: Genetic Drift
Original population
Case #4 F5-3
5 individuals
10 alleles
p (A): 0.5
q (a): 0.5
total alleles = 10
p (A): (2+2+2)/10 = .6
q (a): (1+1+2)/10 = .4 AA 2 Aa 2 aa 1 AA
.25 Aa
.50 aa
.25 AA .4 Aa .4 aa .2
How do you explain these data?

24. Hardy Weinberg Lab: Genetic Drift
Original population
Case #4 F5
5 individuals
10 alleles
p (A): 0.5
q (a): 0.5
AA Aa aa p q AA
.25 Aa
.50 aa
.25
How do you explain these data?

25. Any Questions??

26. HARDY-WEINBERG PRACTICE PROBLEMS p + q = 1 p2 + 2 pq + q2 = 1

27. 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 = = 0.5

28. Once you know q you can figure out p . . . p + q = 1 p + q = 1
Once you know q
you can figure out p
p + q = 1
p + q = 1
p = 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

29. 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

30. 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 = =
p = =
aa = 0.4 = 40%
Aa = 2 (0.632) (0.368) = =46.5%
AA = (0.3676) (0.3676) = .135 = 13.5%
Image from: BIOLOGY by Miller and Levine; Prentice Hall Publishing ©2006

31. 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

32. 1/1700 have cystic fibrosis
q2 = 1/1700
q =
q = 0.024
p = 1 – = 0.976
Frequency of C = 97.6%
Frequency of c = 2.4%
NOW FIND THE GENOTYPIC FREQUENCIES

33. CC or p2 = (0.976)2 = .953
Cc or 2pq = 2 (0.976) (0.024) =
cc = 1/1700 =
CC = 95.3% of population
Cc = 4.68% of population
cc = .06% of population

34. Now you can answer questions about the population:
How many people in this population are heterozygous?
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?
(1700) = 79.5 ~ 80 people are Cc
Cc more likely to survive than CC.
c will increase in population

35. 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 = q = 0.03
p = 1 – 0.03= 0.97
Frequency of C = 97%
Frequency of c = 3%

36. CC or q2 = (0.976)2 = .953
Cc or 2pq = 2 (0.976) (0.024) =
cc = 1/1700 =
CC = 95.3% of population
Cc = 4.68% of population
cc = .06% of population