Hardyweinberg
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Transcript of Hardyweinberg
AP Biology
5 Agents of evolutionary changeMutation Gene Flow
Genetic Drift Selection
Non-random mating
AP Biology
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
W. Weinbergphysician
G.H. Hardymathematician
AP Biology
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
bbBbBB
AP Biology
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
bbBbBB
AP Biology
H-W formulas Alleles:
p + q = 1
Individuals:
p2 + 2pq + q2 = 1
bbBbBB
BB
B b
Bb bb
B bB BB Bb
b Bb bb
AP BiologyWhat are the genotype frequencies?What are the genotype frequencies?
Using Hardy-Weinberg equation
q2 (bb): 16/100 = .16
q (b): √.16 = 0.40.4
p (B): 1 - 0.4 = 0.60.6
q2 (bb): 16/100 = .16
q (b): √.16 = 0.40.4
p (B): 1 - 0.4 = 0.60.6
population: 100 cats84 black, 16 whiteHow many of each genotype?
population: 100 cats84 black, 16 whiteHow many of each genotype?
bbBbBB
p2=.36p2=.36 2pq=.482pq=.48 q2=.16q2=.16
Must assume population is in H-W equilibrium!Must assume population is in H-W equilibrium!
AP Biology
Using Hardy-Weinberg equation
bbBbBB
p2=.36p2=.36 2pq=.482pq=.48 q2=.16q2=.16
Assuming H-W equilibriumAssuming H-W equilibrium
Sampled data Sampled data bbBbBB
p2=.74p2=.74 2pq=.102pq=.10 q2=.16q2=.16
How do you explain the data? How do you explain the data?
p2=.20p2=.20 2pq=.642pq=.64 q2=.16q2=.16
How do you explain the data? How do you explain the data?
Null hypothesis Null hypothesis
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
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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3
AP Biology
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.
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
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 can’t 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 = 0.5The frequency of the dominant (B) allele p = 0.5
http://www.phschool.com/science/biology_place/labbench/lab8/samprob1.html
WHAT ARE THE GENOTYPIC FREQUENCIES?
You know pp from problembb 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 bb25% of population are BB50% of population are Bb
http://www.phschool.com/science/biology_place/labbench/lab8/samprob1.html
Within a population of butterflies, the color brown (B) is dominant over the color white (b). And, 40% of all butterflies are white.
Image from: BIOLOGY by Miller and Levine; Prentice Hall Publishing ©2006
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%
PRACTICE HARDY WEINBERG1 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 populationCc = 4.68% of populationcc = .06% of population
Now you can answer questions about the population:
How many people in this population are heterozygous?
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?
0.0468 (1700) = 79.5 ~ 80 people are Cc
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 populationCc = 4.68% of populationcc = .06% of population
