Measuring Evolution of Populations SLIDE SHOW MODIFIED FROM

Measuring Evolution of Populations SLIDE SHOW MODIFIED FROM

Measuring Evolution of Populations SLIDE SHOW MODIFIED FROM KIM [email protected] 5 Agents of evolutionary change Mutation Gene Flow Genetic Drift AP Biology Non-random mating Selection

Populations & gene pools Concepts a population is a localized 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 AP Biology

Evolution of populations Evolution = change in allele frequencies in a population hypothetical: what conditions would cause allele frequencies to not change? non-evolving population REMOVE all agents of evolutionary change 1. very large population size (no genetic drift) 2. no migration (no gene flow in or out) 3. no mutation (no genetic change) 4. random mating (no sexual selection) 5. no natural selection (everyone is equally fit)

AP Biology Hardy-Weinberg equilibrium Hypothetical, non-evolving population preserves allele frequencies Serves as a model (null hypothesis) natural populations rarely in H-W equilibrium useful model to measure if forces are acting on a population measuring evolutionary change

G.H. Hardy APmathematician Biology W. Weinberg physician 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 AP Biology Bb bb 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 AP Biology Bb bb H-W formulas Alleles: p+q=1 B

Individuals: p2 + 2pq + q2 = 1 BB AP Biology BB b Bb Bb B B b

BB Bb b Bb bb bb bb Using Hardy-Weinberg equation population: 100 cats 84 black, 16 white How many of each genotype?

p2=.36 BB AP Biology What q2 (bb): 16/100 = .16 q (b): .16 = 0.4 p (B): 1 - 0.4 = 0.6 2pq=.48 Bb q2=.16 bb are the

genotype is frequencies? Must assume population 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? AP Biology Application of H-W principle Sickle cell anemia inherit a mutation in gene coding for hemoglobin oxygen-carrying blood protein recessive allele = HsHs normal allele = Hb

low oxygen levels causes RBC to sickle breakdown of RBC clogging small blood vessels damage to organs AP Biology often lethal Sickle cell frequency High frequency of heterozygotes 1 in 5 in Central Africans = HbHs

unusual for allele with severe detrimental effects in homozygotes 1 in 100 = HsHs usually die before reproductive age Why is the Hs allele maintained at such high levels in African populations? Suggests some selective advantage of being heterozygous AP Biology Single-celled eukaryote parasite (Plasmodium) spends part of its

life cycle in red blood cells Malaria 1 2 AP Biology 3 Heterozygote Advantage In tropical Africa, where malaria is common:

homozygous dominant (normal) die of malaria: HbHb homozygous recessive die of sickle cell anemia: HsHs heterozygote carriers are relatively free of both: HbHs survive more, more common in population Hypothesis: In malaria-infected cells, the O2 level is lowered enough to cause sickling which kills the cell & destroys the parasite. AP Biology Frequency of sickle cell allele & distribution of malaria

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 cant 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 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 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

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