# AP Biology Measuring Evolution of Populations AP Biology There are 5 Agents of evolutionary change MutationGene Flow Genetic DriftSelection Non-random.

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AP Biology Measuring Evolution of Populations

AP Biology There are 5 Agents of evolutionary change MutationGene Flow Genetic DriftSelection Non-random mating

AP Biology Populations & gene pools  a population is a localized group of interbreeding individuals  gene pool is a collection of alleles in a particular population (remember difference between alleles & genes!)  allele frequency is how common that allele is in the population (how many of A or a in whole population)

AP Biology Evolution of populations  Evolution is a change in allele frequencies in a population  What conditions would cause allele frequencies not to change?  In a non-evolving population you would need to remove all agents of evolutionary change  Eg: very large population size (no genetic drift) no migration (no gene flow in or out) no mutation (no genetic change) random mating (no sexual selection) no natural selection (everyone is equally fit)

AP Biology Hardy-Weinberg equilibrium  In a non-evolving population allele frequencies are preserved (they are said to be in Hardy-Weinberg equilibrium), but:  natural populations are rarely in Hardy-Weinberg equilibrium  However it provides a useful model to measure if evolutionary forces are acting on a population W. Weinberg physician G.H. Hardy mathematician

AP Biology Hardy-Weinberg theory  Counting Alleles  assume 2 alleles = B, b  frequency of dominant allele (B) = p  frequency of recessive allele (b) = q  frequencies must add up to 1 (100%), so: p + q = 1 bbBbBB

AP Biology Hardy-Weinberg theory  Counting Genotypes  frequency of homozygous dominant: p x p = p 2  frequency of homozygous recessive: q x q = q 2  frequency of heterozygotes: (p x q) + (q x p) = 2pq  frequencies of all individuals must add to 1 (100%), so: p 2 + 2pq + q 2 = 1 bbBbBB

AP Biology H-W formulas  Alleles:p + q = 1  Genotypes:p 2 + 2pq + q 2 = 1 bbBbBB BbBbbb

AP Biology What are the genotype frequencies? Using Hardy-Weinberg equation population: 100 cats 84 black, 16 white How many of each genotype? population: 100 cats 84 black, 16 white How many of each genotype? bbBbBB Must assume population is in H-W equilibrium What does this mean? Must assume population is in H-W equilibrium What does this mean? First calculate frequency of b from the known number of bb genotypes.

AP Biology What equation is used for genotype frequency calculation? Using Hardy-Weinberg equation q 2 (bb)= 16/100 =.16 0.4 q (b): √.16 = 0.4 Now work out p 0.6 p (B)= 1 - 0.4 = 0.6 q 2 (bb)= 16/100 =.16 0.4 q (b): √.16 = 0.4 Now work out p 0.6 p (B)= 1 - 0.4 = 0.6 population: 100 cats 84 black, 16 white How many of each allele? population: 100 cats 84 black, 16 white How many of each allele? bbBbBB

AP Biology Using Hardy-Weinberg equation q 2 (bb): 16/100 =.16 0.4 q (b): √.16 = 0.4 0.6 p (B): 1 - 0.4 = 0.6 Now work out genotype frequencies q 2 (bb): 16/100 =.16 0.4 q (b): √.16 = 0.4 0.6 p (B): 1 - 0.4 = 0.6 Now work out genotype frequencies population: 100 cats 84 black, 16 white How many of each genotype? p 2 + 2pq + q 2 = 1 population: 100 cats 84 black, 16 white How many of each genotype? p 2 + 2pq + q 2 = 1 bb Bb BB p 2 =0.36 2pq=0.48 q 2 =0.16

AP Biology Hardy Weinberg problems Now have a go at sorting out some:  Allele frequencies  Genotype frequencies  Phenotype frequencies And use the frequencies to make deductions about data

AP Biology Application of H-W principle  Sickle cell anemia Caused by inheriting a mutation in the gene coding for haemoglobin  oxygen-carrying blood protein  recessive allele = H s H s  normal allele = H b  low oxygen level causes RBC to sickle  clogging small blood vessels  depriving tissues of oxygen  damage to organs  The condition is often lethal

AP Biology Sickle cell frequency  High frequency of heterozygotes  1 in 5 in Central Africans = H b H s  unusual for allele with severe detrimental effects in homozygotes  1 in 100 = H s H s  usually die before reproductive age Why is the H s allele maintained at such high levels in African populations? Suggests some selective advantage of being heterozygous…

AP Biology Malaria Single-celled eukaryote parasite (Plasmodium) spends part of its life cycle in red blood cells 1 2 3

AP Biology Heterozygote Advantage  In tropical Africa, where malaria is common:  homozygous dominant (normal)  die or reduced reproduction from malaria: H b H b  homozygous recessive  die or reduced reproduction from sickle cell anemia: H s H s  heterozygote carriers are relatively free of both: H b H s  survive & reproduce more, more common in population Frequency of sickle cell allele & distribution of malaria

AP Biology Any Questions??

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