Measuring Evolution of Populations Image from: https://s-media-cache-ak0.pinimg.com/236x/0a/c4/6c/0ac46cac37e5afb70010f25beedec39e.jpg

CONDITIONS FOR HARDY WEINBERG EQUILIBRIUM NO Mutation NO Movement IN or OUT Random mating LARGE population NO Natural Selection http://pngimg.com/upload/hands_PNG905.png

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

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

The Hardy-Weinberg Equation p2 + 2pq + 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)

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

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

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

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

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

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 = ______

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 = 1- 0.5 = _____ 0.5   0.5 0.25 0.25 0.5 (0.5)2 0.25 0.5 If p = ____ then p2 = ______ = ______

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

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%

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

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

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 = 1- 0.6 = _____ 0.4   0.6 0.36 0.16 0.4 (0.4)2 0.16 0.4 If p = ____ then p2 = ______ = ______

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

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%

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

LET’S TRY ONE Dragon Hardy Weinberg