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Measuring Evolution of Populations Hardy Weinberg. 5 Agents of evolutionary change. Mutation. Gene Flow. Non-random mating. Chemical Changes to DNA. Migration. Sexual Selection. Genetic Drift. Selection. Natural Selection Differential Survival. Small population.
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MeasuringEvolution of Populations Hardy Weinberg
5 Agents of evolutionary change Mutation Gene Flow Non-random mating ChemicalChanges to DNA Migration Sexual Selection Genetic Drift Selection Natural SelectionDifferential Survival Small population
Hardy-Weinberg equilibrium • Hypothetical situation • serves as null hypothesis • non-evolving population REMOVE all agents of evolutionary change • no genetic drift(very large population size ) • no gene flow (no migration in or out) • no mutation (no chemical change to DNA) • random mating (no sexual selection) • no natural selection (equal survival)
Example of strong selection pressure • Tay Sachs • primarily in Ashkenazi Jews & Cajuns • recessive disease = aa • lysosomal storage disease • lack of one functional digestive enzyme in lysosome • build up undigested fat in brain cells • children die before they are 5 years old So where do new cases of Tay-Sachs come from?
Example of heterozygote advantage • Sickle cell anemia • inherit a mutation in gene coding for one of the subunits in hemoglobin • oxygen-carrying blood protein • normal allele = Hb • mutant allele = Hs • recessive trait = HsHs • low oxygen levels causes RBC to sickle • clogging small blood vessels • damage to organs • 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… HbHs
Malaria Single-celled eukaryote parasite (Plasmodium) spends part of its life cycle in red blood cells 1 liver 2 3
Heterozygote Advantage • In tropical Africa, where malaria is common: • homozygous dominant (normal) • reduced survival or reproduction from malaria: HbHb • homozygous recessive • reduced survival & reproduction from sickle cell anemia: HsHs • heterozygote carriers • survival & reproductive advantage: HbHs Hypothesis: In malaria-infected cells, the O2 level is lowered enough to cause sickling which kills the cell & destroys the parasite. Frequency of sickle cell allele & distribution of malaria
The Hardy-Weinberg Formula Determining Genetic Equilibrium… the BASELINE The Hardy-Weinberg formula can be used to determine if a population is in genetic equilibrium p2(AA) + 2pq (Aa) +q2(aa) = 1.0 The frequency of the dominant allele (A) plus the recessive allele (a) equals 1.0 p + q = 1.0
The Hardy-Weinberg Principle • The Hardy-Weinberg principle describes a population that is not evolving • If a population does not meet the criteria of the Hardy-Weinberg principle, it can be concluded that the population is evolving.
Finding out whether a population is evolving. The frequencies of wing-color alleles among all of the individuals in this hypothetical population of morpho butterflies are not changing; thus, the population is not evolving. Fig. 18-3a, p. 280
490 AA butterflies dark-blue wings 490 AA butterflies dark-blue wings 490 AA butterflies dark-blue wings 420 Aa butterflies medium-blue wings 420 Aa butterflies medium-blue wings 420 Aa butterflies medium-blue wings 90 aa butterflies white wings 90 aa butterflies white wings 90 aa butterflies white wings Starting Population Next Generation Next Generation Fig. 18-3b, p. 280
Hardy-Weinberg Sample • Pink is dominant over black. • Calculate q2: Count the individuals that are homozygous recessive in the illustration above. Calculate the percent of the total population they represent. This is q2. q2 = 25%
Find q. Take the square root of q2 to obtain q, the frequency of the recessive allele. q = .5
Find p. The sum of the frequencies of both alleles = 100%, p + q = l. You know q, so what is p, the frequency of the dominant allele? p = 1 – q p = 1 - .5 p = .5
Find 2pq. The frequency of the heterozygotes is represented by 2pq. This gives you the percent of the population that is heterozygous for white coat: q = .5 p = .5 2pq = 2(.5)(.5) 2pq = .5 This represents the carriers of the recessive alleles
This shows the allele frequency in generation 1. • One would study the allele frequencies over several generations to see if the population is evolving.
Concept Check • What is the usefulness of the H-W Law? • Homework • Complete the H-W Online Tutorial by your next period. • http://www.montereyinstitute.org/courses/AP%20Biology%20I/course%20files/multimedia/lesson23/lessonp.html?showTopic=2
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