ap biology measuring evolution of populations ap biology there are 5 agents of evolutionary change...
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AP BiologyAP Biology
MeasuringEvolution of Populations
AP Biology
There are 5 Agents of evolutionary changeMutation Gene Flow
Genetic Drift Selection
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. Weinbergphysician
G.H. Hardymathematician
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 = 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
bbBbBB
AP Biology
H-W formulas Alleles: p + q = 1
Genotypes: p2 + 2pq + q2 = 1
bbBbBB
BB
B b
Bb bb
AP BiologyWhat are the genotype frequencies?What are the genotype frequencies?
Using Hardy-Weinberg equationpopulation: 100 cats84 black, 16 whiteHow many of each genotype?
population: 100 cats84 black, 16 whiteHow 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 BiologyWhat equation is used for genotype frequency calculation?What equation is used for genotype frequency calculation?
Using Hardy-Weinberg equation
q2 (bb)= 16/100 = .16
q (b): √.16 = 0.40.4
Now work out pNow work out p
p (B)= 1 - 0.4 = 0.60.6
q2 (bb)= 16/100 = .16
q (b): √.16 = 0.40.4
Now work out pNow work out p
p (B)= 1 - 0.4 = 0.60.6
population: 100 cats84 black, 16 whiteHow many of each allele?
population: 100 cats84 black, 16 whiteHow many of each allele?
bbBbBB
AP Biology
Using Hardy-Weinberg equationq2 (bb): 16/100
= .16
q (b): √.16 = 0.40.4
p (B): 1 - 0.4 = 0.60.6
Now work out Now work out genotype genotype frequenciesfrequencies
q2 (bb): 16/100 = .16
q (b): √.16 = 0.40.4
p (B): 1 - 0.4 = 0.60.6
Now work out Now work out genotype genotype frequenciesfrequencies
population: 100 cats84 black, 16 whiteHow many of each genotype? p2 + 2pq + q2 = 1
population: 100 cats84 black, 16 whiteHow many of each genotype? p2 + 2pq + q2 = 1
bbBbBB
p2=0.36p2=0.36 2pq=0.482pq=0.48 q2=0.16q2=0.16
AP Biology
Hardy Weinberg problemsNow 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 = HsHs
normal allele = Hb
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 = 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?Why is the Hs allele maintained at such high levels in African populations?
Suggests some selective advantage of being heterozygous…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
Single-celled eukaryote parasite (Plasmodium) spends part of its life cycle in red blood cells
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AP Biology
Heterozygote Advantage In tropical Africa, where malaria is common:
homozygous dominant (normal) die or reduced reproduction from malaria: HbHb
homozygous recessive die or reduced reproduction from sickle cell anemia: HsHs
heterozygote carriers are relatively free of both: HbHs
survive & reproduce more, more common in population
Frequency of sickle cell allele & distribution of malaria
AP BiologyAP Biology
Any Questions??Any Questions??Any Questions??