1. Measuring Evolution of Populations
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2. CONDITIONS FOR HARDY WEINBERG EQUILIBRIUM

3. Populations & gene pools
_________ = group of interbreeding individuals
___________is collection of alleles in the population
remember difference between alleles & genes!
allele ___________ 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 (the English mathematician) and W. Weinberg (the German physician) independently worked out the mathematical basis of population genetics in Their formula predicts the expected genotype frequencies using the allele frequencies in a diploid Mendelian population. They were concerned with questions like "what happens to the frequencies of alleles in a population over time?" and "would you expect to see alleles disappear or become more frequent over time?"
G.H. Hardy
mathematician
W. Weinberg
physician

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

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

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

8. T t T t T t T T T t t t Homozygous dominant = ______
Homozygous recessive = ______ Heterozygous = _____

9. T t T t p2 + 2pq + q2 = 1 = _____ = _________ = ______
GENOTYPES in population
Homozygous dominant = ________
Homozygous recessive = ________
Heterozygous = __________
= _____
= ______
= _________

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 ______
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
Black pigs = ______ = _____ = _____
SO bb q2 = ______

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

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

14. p + q = 1 p2 + 2pq + q2 = 1 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 = ____ = _____ = _____

15. In a population of fruit flies, 36% 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 = ______

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

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

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

19. p + q = 1 p2 + 2pq + q2 = 1 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 HETEROZYGOUS?
Rr = ____ = _____ = _____

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