1. Measuring Evolution of Populations
Image from:

2. CONDITIONS FOR HARDY WEINBERG EQUILIBRIUM
NO Mutation
NO Movement IN or OUT
Random mating
LARGE population
NO Natural Selection

3. Populations & gene pools
population = 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

4. Hardy-Weinberg equilibrium
POPULATION IN HW EQUILIBRIUM = IDEAL IT MEANS NO EVOLUTION IS HAPPENING Allele frequencies stay the same Rare in real populations
Way to tell if evolution is happening in a population
G.H. Hardy
mathematician
W. Weinberg
physician

5. The Hardy-Weinberg Equation
p pq + q2 = 1
p2 = the frequency of homozygous dominant genotype (T T)
2pq = the frequency of heterozygous genotype (T t)
q2 = the frequency of homozygous recessive genotype (t t)

6. The Hardy-Weinberg Equation
p + q = 1
p = the frequency of dominant ALLELE in population (T)
q = the frequency of recessive ALLELE in population (t)
BOZEMAN BIOLOGY Hardy Weinberg equation

7. T t T p + q = 1 t T t T T T t t t p q ALLELES in population T = _____

8. T t T t T t T T T t t t TT tt Tt Homozygous dominant = ______
Homozygous recessive = ______ Heterozygous = _____ tt Tt

9. T t T t pq pp qq p2 + 2pq + q2 = 1 p X p = p2 p X q = pq 2 q X q = q2
GENOTYPES in population
Homozygous dominant = ________
Homozygous recessive = ________
Heterozygous = __________
p X p
= p2
q X q
= q2
p X q
= pq 2

10. In a population of pigs color is determined by one gene
In a population of pigs color is determined by one gene. If the black allele (b) is recessive and the white allele (B) is dominant, what is the frequency of the black allele in this population?
p + q = 1
p2 + 2pq + q2 = 1
ALWAYS START WITH HOMOZYGOUS RECESSIVE (bb) = q2
GENOTYPE OF Black pigs = ______ White pigs = ______ or ______
b b
B B
B b
BOZEMAN BIOLOGY SOLVING HARDY WEINBERG PROBLEMS

11. In a population of pigs color is determined by one gene
In a population of pigs color is determined by one gene. If the black allele (b) is recessive and the white allele (B) is dominant, what is the frequency of the black allele in this population?
p + q = 1
p2 + 2pq + q2 = 1
4/16
25%
0.25
Black pigs = ______ = _____ = _____
0.25
SO bb q2 = ______

12. p + q = 1
p2 + 2pq + q2 = 1
b = ______ (q) B = ______ (p)
bb = ________ (q2) Bb = _______ (2pq) BB = _______ (p2)
0.5
If q2 = 0.25 q = _____ = _______
p + q = 1 OR 1 – q = p
p = = _____
0.5
0.5
0.25
0.25
0.5
(0.5)2
0.25
0.5
If p = ____ then p2 = ______ = ______

13. p + q = 1
p2 + 2pq + q2 = 1
b = ______ (q) B = ______ (p)
bb = ________ (q2) Bb = _______ (2pq) BB = _______ (p2)
0.5
If you know this : can you figure out 2pq?
p2 + 2pq + q2 = 1
____ + 2pq + ____ = 1
2pq = 1 – 0.5 = _____
0.5
0.25
0.5
0.25
0.25
0.25
0.5

14. p + q = 1
p2 + 2pq + q2 = 1
0.5
b= ______ (q) B = ______ (p)
bb = ________ (q2) Bb = _______ (2pq) BB = _______ (p2)
You can answer ?’s now What is the frequency of the white allele (B) in population?
B = ___ = _____ = ____
What percent of the population is HOMOZYGOUS DOMINANT?
BB = ____ = _____ = _____
0.5
0.25
0.5
0.25
50% p
0.5 p2
0.25
25%

15. In a population of fruit flies, 64% have red eyes and the remainder have sepia eyes. The sepia (r) eye trait is recessive to red (R) eyes. Calculate the frequencies in this population.
p + q = 1
p2 + 2pq + q2 = 1
ALWAYS START WITH HOMOZYGOUS RECESSIVE (rr) = q2
GENOTYPE OF Red eyes = ______ or _____ sepia eyes = ______
R r RR
r r

16. In a population of fruit flies, 64 % have red eyes and the remainder have sepia eyes. The sepia (r) eye trait is recessive to red (R) eyes. Calculate the frequencies in this population.
p + q = 1
p2 + 2pq + q2 = 1
If red eyes = ______= _____ ? % have sepia eyes = _____ = r r = q2
64%
0.64
0.36

17. p + q = 1
p2 + 2pq + q2 = 1
r = ______ (q) R = ______ (p)
rr = ________ (q2) Rr = _______ (2pq) RR = _______ (p2)
0.6
If q2 = 0.36 q = _____ = _______
p + q = 1 OR 1 – q = p
p = = _____
0.4
0.6
0.36
0.16
0.4
(0.4)2
0.16
0.4
If p = ____ then p2 = ______ = ______

18. p + q = 1
p2 + 2pq + q2 = 1
r = ______ (q) R = ______ (p)
rr = ________ (q2) Rr = ______ (2pq) RR = _______ (p2)
0.6
If you know this : can you figure out 2pq?
p2 + 2pq + q2 = 1
____ + 2pq + ____ = 1
2pq = 1 – 0.52 = _____
0.4
0.36
0.48
0.16
0.36
0.16
0.48

19. p + q = 1
p2 + 2pq + q2 = 1
0.6
r= ______ (q) R= ______ (p)
rr = ________ (q2) Rr = _______ (2pq) RR = _______ (p2)
You can answer ?’s now What is the frequency of the sepia allele (r) in population?
r = ___ = _____ = ____
What percent of the population is HETEROYGOUS?
Rr = ____ = _____ = _____
0.4
0.36
0.48
0.16
60% q
0.6
2pq
0.48
48%

20. Using Hardy-Weinberg equation
p2=.36
2pq=.48
q2=.16
Assuming H-W equilibrium BB Bb bb
p2=.20
p2=.74
2pq=.64
2pq=.10
q2=.16
q2=.16
Sampled data
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
Change in allele frequency = EVOLUTION happened

21. LET’S TRY ONE Dragon Hardy Weinberg