The evolution of populations & Hardy- Weinberg Equilibrium FRIDAY, SEPTEMBER 5, 2014.
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Transcript of The evolution of populations & Hardy- Weinberg Equilibrium FRIDAY, SEPTEMBER 5, 2014.
![Page 1: The evolution of populations & Hardy- Weinberg Equilibrium FRIDAY, SEPTEMBER 5, 2014.](https://reader033.fdocuments.net/reader033/viewer/2022051620/56649e685503460f94b64d25/html5/thumbnails/1.jpg)
The evolution of populations & Hardy-Weinberg EquilibriumFRIDAY, SEPTEMBER 5, 2014
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Warm-up What do you think is the most important component in order for evolution to occur? (Hint: think about the definition of evolution).
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Homework Be a leader assignment… due Monday, September 8!◦Directions: Solve your problem on a separate sheet of paper and staple it to your given worksheet slip. Hand it in on Monday.
◦Be prepared to show your group how you solved the problem!
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Lamarck vs. Darwin
Inheritance of acquired characteristics Natural selection
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Evolution is change in the genetic composition of a population from generation to generation
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But what is a more concise definition for evolution that will help us determine quantitatively if evolution is occurring??
Evolution is the change in allele frequencies over time
How can we measure allele frequencies?
How can we track changes in allele frequencies over time?
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Genetic variation exists in the population (this is key for evolution to occur!)
Certain alleles produce traits that are more adaptive, i.e., promote greater survival and reproduction
These alleles get passed on to the next generation
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Evolution is the change in allele frequencies in a population over generations Population: group of individuals of the same species that live in the same area and interbreed, producing fertile offspring. Gene pool: genetic makeup of a population
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Evolution is the change in allele frequencies in a population over generations
Mechanisms that cause allele frequency change:◦ Natural selection◦ Genetic drift◦ Gene flow◦ Mutation◦ Non-random mating
Only natural selection causes adaptive evolution
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Mini genetics review
Alleles are different versions of that geneExample:
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Combinations of alleles
Example for a population in which alleles for a gene are ‘R’ or ‘r’ Homozygous recessive: rr Homozygous dominant: RR Heterozygous: Rr
Skittles gene pool
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Now, how do we measure changes in allele frequencies in populations over time?
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First, we must know the starting point
What are the allele frequencies in the population right now? We cannot measure change unless we know the initial state.
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Use Hardy-Weinberg Principle to quantify evolution
Godfrey Harold "G. H." Hardy Wilhelm Weinberg
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The idea is to track allele frequencies
AA
AA
AA
AA AA
Aa
Aa
Aa
aa
aa
What is the frequency of allele A?What is the frequency of allele a?
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The idea is to track allele frequencies
AA
AA
AA
AA AA
Aa
Aa
Aa
aa
aa
The frequency for allele A = 13/20 or 0.65The frequency for allele a = 7/20 or 0.35
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AA
AA
AA
AA AA
Aa
Aa
Aa
aa
aa
Generation 1
?
Generation 2
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Hardy-Weinberg Principle Allele frequencies of alleles and genotypes in a population will remain constant from generation to generation if all assumptions are met
A gene pool that remains constant is said to be in Hardy-Weinberg equilibrium
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AA
AA
AA
AA AA
Aa
Aa
Aa
aa
aa
Generation 1
?
Generation 2
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AA
AA
AA
AA AA
Aa
Aa
Aa
aa
aa
Generation 1
A0.65
a0.35
A0.65
AA0.42
Aa0.23
a0.35
Aa0.23
aa0.12
Generation 2
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AA
AA
AA
AA AA
Aa
Aa
Aa
aa
aa
f(A) = 0.65f(a) = 0.35
A0.65
a0.35
A0.65
AA0.42
Aa0.23
a0.35
Aa0.23
aa0.12
f(A) = ?f(a) = ?
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AA
AA
AA
AA AA
Aa
Aa
Aa
aa
aa
f(A) = 0.65f(a) = 0.35
A0.65
a0.35
A0.65
AA0.42
Aa0.23
a0.35
Aa0.23
aa0.12
f(A) = 0.65f(a) = 0.35
Allele frequencies did not change, thus no evolution.
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The Hardy-Weinberg Principle
Allele frequencies in a population will remain constant if ALL of the following conditions are met:
1. The population is infinitely large2. Individuals mate randomly3. No gene flow4. No natural selection5. No mutations
If all conditions are met, then NO evolution. Allele frequencies will remain constant. This is the null hypothesis.
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The Hardy-Weinberg equation: p2 + 2pq + q2 = 1A
0.65a
0.35
A0.65
AA0.42
Aa0.23
a0.35
Aa0.23
aa0.12
![Page 26: The evolution of populations & Hardy- Weinberg Equilibrium FRIDAY, SEPTEMBER 5, 2014.](https://reader033.fdocuments.net/reader033/viewer/2022051620/56649e685503460f94b64d25/html5/thumbnails/26.jpg)
Understanding the equation
AA
AA
AA
AA AA
Aa
Aa
Aa
aa
aa
p2 + 2pq + q2 = 1p is the frequency of the dominant allele (A) = 0.65q is the frequency of the recessive allele (a) = 0.35
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p2 + 2pq + q2 = 1p is the frequency of the dominant allele (A) = 0.65q is the frequency of the recessive allele (a) = 0.35
p = f(AA) + ½ f(Aa)p = 0.5 + ½ 0.3 = 0.65
q = f(aa) + ½ f(Aa)q = 0.2 + ½ 0.3 = 0.35
Understanding the equation
AA
AA
AA
AA AA
Aa
Aa
Aa
aa
aa
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Using p and q from generation 1, solve for frequencies of predicted genotypes in generation 2 using HW
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Understanding the equation
A0.65
a0.35
A0.65
AA0.42
Aa0.23
a0.35
Aa0.23
aa0.12
p2 + 2pq + q2 = 1p is the frequency of the dominant allele (A) = 0.65q is the frequency of the recessive allele (a) = 0.35
p = f(AA) + ½ f(Aa)p = 0.5 + ½ 0.3 = 0.65
q = f(aa) + ½ f(Aa)q = 0.2 + ½ 0.3 = 0.35
Generation 2p = f(A) = 0.65q = f(a) = 0.35
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Understanding the equation
A0.65
a0.35
A0.65
AA0.42
Aa0.23
a0.35
Aa0.23
aa0.12
Generation 2p = f(A) = 0.65q = f(a) = 0.35
p2 + 2pq + q2 = 1p is the frequency of the dominant allele (A) = 0.65q is the frequency of the recessive allele (a) = 0.35
p = f(AA) + ½ f(Aa)p = 0.5 + ½ 0.3 = 0.65
q = f(aa) + ½ f(Aa)q = 0.2 + ½ 0.3 = 0.35
p2 = the predicted frequency of genotype AA = 0.422pq = the predicted frequency of genotype Aa = 0.46q2 = the predicted frequency of genotype aa = 0.12
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The Hardy-Weinberg equation: p2 + 2pq + q2 = 1
What is the frequency of alleles B and b?
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Allele Frequencies
a) red = 0.36, white = 0.16
b) red = 0.6, white = 0.4
c) red = 0.5, white = 0.5
d) Allele frequencies cannot be determined unless the population is in equilibrium.
Red short-horned cattle are homozygous for the red allele, white cattle are homozygous for the white allele, and roan cattle are heterozygotes. Population A consists of 36% red, 16% white, and 48% roan cattle. What are the allele frequencies?
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Allele Frequencies
a) red = 0.36, white = 0.16
b) red = 0.6, white = 0.4
c) red = 0.5, white = 0.5
d) Allele frequencies cannot be determined unless the population is in equilibrium.
Red short-horned cattle are homozygous for the red allele, white cattle are homozygous for the white allele, and roan cattle are heterozygotes. Population A consists of 36% red, 16% white, and 48% roan cattle. What are the allele frequencies?
![Page 35: The evolution of populations & Hardy- Weinberg Equilibrium FRIDAY, SEPTEMBER 5, 2014.](https://reader033.fdocuments.net/reader033/viewer/2022051620/56649e685503460f94b64d25/html5/thumbnails/35.jpg)
Let’s see another example
We sampled 200 individuals from a population:
128 individuals have the AA genotype
53 individuals have the Aa genotype
19 individuals have the aa genotype
What are the genotype frequencies?
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Let’s see another example
We sampled 200 individuals from a population:
128 individuals have the AA genotype (0.64)
53 individuals have the Aa genotype (0.26)
19 individuals have the aa genotype (0.10)
What are the allele frequencies?
Genotype frequencies
![Page 37: The evolution of populations & Hardy- Weinberg Equilibrium FRIDAY, SEPTEMBER 5, 2014.](https://reader033.fdocuments.net/reader033/viewer/2022051620/56649e685503460f94b64d25/html5/thumbnails/37.jpg)
Let’s see another example
We sampled 200 individuals from a population:
128 individuals have the AA genotype (0.64)
53 individuals have the Aa genotype (0.26)
19 individuals have the aa genotype (0.10)
p = AA + ½(Aa) = 0.64 + ½(0.26) = 0.77
q = aa + ½(Aa) = 0.10 + ½(0.26) = 0.23
Genotype frequencies
Allele frequencies
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Let’s see another example
We sampled 200 individuals from a population:
128 individuals have the AA genotype (0.64)
53 individuals have the Aa genotype (0.26)
19 individuals have the aa genotype (0.10)
p = AA + ½(Aa) = 0.64 + ½(0.26) = 0.77
q = aa + ½(Aa) = 0.10 + ½(0.26) = 0.23p + q = 0.77 + 0.23 = 1.0p2 + 2pq + q2 = (0.77)2 + 2(0.77)(0.23) + (0.23)2 = 1.0
Genotype frequencies
Allele frequencies
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Let’s work through another example
Albinism (aa) occurs on average 1 in 20,000 individuals in North America.
What is the frequency of the A and a allele in this population?
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Let’s work through another exampleAlbinism (aa) occurs on average 1 in 20,000 individuals in North America. What is the frequency of the A and a allele in this population?
The Hardy-Weinberg equation: p2 + 2pq + q2 = 1
q2 = f(aa) = 1/20,000 = 0.00005√ q2 = √ 0.00005q = 0.007 (frequency of a in the population)
p = 1 – qp = 1 – 0.007 p = 0.993 (frequency A in the population)
p2 + 2pq + q2 (0.993)2 + 2(0.993)(0.007) + (0.007)2 = 1.0
p2 = 98.6% (AA)2pq = 1.4% (Aa)q2 = 0.005% (aa)
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The Hardy-Weinberg Principle Allele frequencies in a population will remain constant if ALL of
the following conditions are met:1. The population is infinitely large2. Individuals mate randomly3. No genetic migration and mutation4. No natural selection5. No mutation
If ALL conditions are met, then there’s NO evolution. Allele frequencies will remain constant.
The Hardy-Weinberg equation: p2 + 2pq + q2 = 1
![Page 42: The evolution of populations & Hardy- Weinberg Equilibrium FRIDAY, SEPTEMBER 5, 2014.](https://reader033.fdocuments.net/reader033/viewer/2022051620/56649e685503460f94b64d25/html5/thumbnails/42.jpg)
AA
AA
AA
AA AA
Aa
Aa
Aa
aa
aa
f(A) = 0.65f(a) = 0.35
A0.65
a0.35
A0.65
AA0.42
Aa0.23
a0.35
Aa0.23
aa0.12
What will happen if these assumptions are not met?1. The population is infinitely large2. Individuals mate randomly3. No genetic migration and mutation4. No natural selection
f(A) = 0.65f(a) = 0.35
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Exit TicketWhat is the Hardy-Weinberg formula, and what does each component represent?
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Work on Clover Study