1 Given genotype frequencies, calculate allele frequencies in a gene pool ! Alleles = A, a Genotypes = AA, Aa, aa Frequency of allele A: f (A) = f (AA) + 1/2 f (Aa) Frequency of allele a: f (a) = f (aa) + 1/2 f (Aa) f(A) + f(a) = 1.0 p + q = 1.0 (allele frequencies) p = 1 - q or, q = 1 - p AA Aa Aa aa A a A a 2pq + q 2 = 1 f [AA] + 2 [f(Aa)] + f [aa] = 26a
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Given genotype frequencies, calculate allele frequencies in a gene pool !
Given genotype frequencies, calculate allele frequencies in a gene pool !. Alleles = A, a Genotypes = AA, Aa, aa Frequency of allele A: f (A) = f (AA) + 1/2 f (Aa) Frequency of allele a: f (a) = f (aa) + 1/2 f (Aa). f(A) + f(a) = 1.0 p + q = 1.0 (allele frequencies) - PowerPoint PPT Presentation
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1
Given genotype frequencies, calculate allele frequencies in a gene pool !
Alleles = A, a
Genotypes = AA, Aa, aa
Frequency of allele A: f (A) = f (AA) + 1/2 f (Aa)
Frequency of allele a: f (a) = f (aa) + 1/2 f (Aa)f(A) + f(a) = 1.0 p + q = 1.0 (allele frequencies) p = 1 - q or, q = 1 - p
AA Aa
Aa aa
A
a
A a
p2 + 2pq + q2 = 1 f [AA] + 2 [f(Aa)] + f [aa] = 1
26a
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Hardy-Weinberg Equilibrium
Parental generation: 2 alleles, r and R
f (R) + f (r) = 1.0
p + q = 1.0 p = 0.1, q = 0.9
In the next generation (F1):
p2 + 2pq + q2 = 1 predicts allele freqs.
F1 genotype Genotype Allele freq.
p2 (.01) RR p = 0.01+.18/2= .1
q2 (.81) rr q = 0.81+.18/2 = .9 2pq (.18) Rr
27a-1
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Hardy-Weinberg Equilibrium
Parental allele frequencies p and q predict F1 generation genotype frequencies, by the formula p2 + 2pq + q2 = 1
27a-2
Note: parental generation genotype frequencies do NOTpredict F1 generation genotype frequencies!!
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Hardy-Weinberg Equilibrium
Conclusions:
1)Allele frequencies are conserved (i.e., the same) from one generation to the next.
2) genotype frequency reaches Hardy-Weinberg equilibrium in one generation
27a-2
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Hardy-Weinberg Law: allele frequencies in a populationremain constant from generation to generation …..
•IF random mating•IF all genotypes are equally viable•IF not disturbed by mutation, selection or whatever
But
Only
Bottom line: Only in an IDEAL population is genetic diversity conserved forever.
Hardy-Weinberg’s caveats:
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Sex = Random sampling of a gene poolPopulation of 10 individuals (N = 10)
Phenotypes Red WhiteWhite
Genotypes RR,Rr, rr
Allele frequencies R = 0.6 r = 0.4
Parental gene pool
10 genotypes
20 alleles
Parental gametes
Probability of
F1 r = .4; R =.6
R r r R R R R r r R rr R R R r R r R R
25A -1
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F1 genotypes and phenotypes
Genotype Frequency Phenotype Frequency
Rr or rR .48 R_ .48+.36=.84
rr .16 rr .16
RR .36
25A -2
= 1
= 1-------------
---------------
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If allele frequencies are P and Q in the parental generation, howdo we calculate what they will be in the F1 generation?
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If allele frequencies are P and Q in the parental generation, howdo we calculate what they will be in the F1 generation?
Genotype frequencies in F1 are calculated by:p2 + 2pq + q2 = 1
From which we can calculate p and q for F1
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If the fraction of the population with allele “A” at a given locus is .7, and the fraction of the population with “a” at the locus is .3, what will be the expected genotype frequencies in F1, , the nextgeneration?
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If the fraction of the population with allele “A” at a given locus is .7, and the fraction of the population with “a” at the locus is .3, what will be the expected genotype frequencies in the F1?
Allele frequencies in P (parental generation):A = .7 = pa = .3 = q
Expected genotypes and their frequencies in F1:AA = p2 = .49aa = q2 = .09Aa = 2pq = .42
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What will be the expected phenotypes and their ratios in thisexample?
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What will be the expected phenotypes and their ratios in thisexample?
Allele frequencies in P (parental generation):A = .7 = pa = .3 = q
H = % heterozygous genotypes for a particular locus
= % heterozygous individuals for a particular locus
= probability that a given individual randomly selected from the population will be heterozygous at a given locus
29f
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Aa aa aa AA aa
aA aa AA aA AA
aa aa AA aa aA
H = 4/15
H = ?
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Heterozygosity defined
H (“H-bar”) = average heterozygosity for all loci in a population.
H estimated = % heterozygous loci
those examined
29f
H = 2pq
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Calculating H (assuming simple dominance and Hardy-Weinberg Eq.)
calculate H if q2 = 0.09
f (a) = 0.3 = q q2 = 0.09f (A) = 0.7 = p p2 = 0.49
2 pq = 0.42H = 2pq = 0.42 (for only 2 alleles)
29a - 1
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Calculating H (but if……)Codominance Genotype Phenotype N
AA Red 50 Aa Pink 22 aa White 10
total = 82H = 22/82 (don’t need Hardy - Weinberg)
29a - 1
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Calculating H for 3 alleles: p, q, r
p = 0.5
q = 0.4
r = 0.1
H = 2pq + 2qr + 2pr
= .40 + .08 + .10
= .58
pp pq pr
r
r
q
p
p q
pq qq qr
pr qr rr
29a
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1% of golden lion tamarins have diaphragmatic hernias, a condition expressed only in the homozygous recessive genotype. Calculate the number of heterozygous individuals in the wild population (N = 508). Assume Hardy-Weinberg equilibrium and simple dominance.
Genotypes: AA Aa aa
F1 generation p2 = ? = f (AA) 2pq = ? = f (Aa) = H q2 = .01 = f (aa)
29ez - 1
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q = 01. = 0.1
p + q = 1 so p = 1 - qp = 1 - .1 = .9
H = 2pq = 2 x .9 x .1 = 0.18
Nheterozygous = .18 x 508 = 91
29ez - 2
q2 = .01
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H and P
H = heterozygosity = the percent of heterozygous genotypes inthe population for that locus