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These activities were developed by Dr. K. E. Lotterhos at Northeastern University with support from the National Science Foundation.

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Ecological and Conservation Genomics

Dr. K. E. Lotterhos

F-statistic challenges

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Instructions: Make a copy of this for yourself. You can move the icons around like paper strips to solve the problems.

Ecological and Conservation Genomics

Dr. K. E. Lotterhos

Paper Strip Challenges for understanding F-statistics

Your challenge: for each of the 8 scenarios presented below, see if you can figure out what patterns of genotypes within populations gives the values of F_IS and F_ST. With your proposed solution, work through the mathematical calculations presented above to prove that your solution is correct!

Note that two of the scenarios below are impossible, and some of them are biologically implausible but they may help you understand F-statistics.

In these activities, we represent:

• Each population with a circle
• The chromosomes for each individual within each population with paper strips
• The allele is represented by the color of the paper strip

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Ecological and Conservation Genomics

Dr. K. E. Lotterhos

Haploids vs. diploids

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Haploids have one copy of each chromosome. A population of haploids with 6 individuals, 1 that has the allele A (represented by blue) and the other 5 that have the allele a (represented by green) would be:

Diploids have two copies of each chromosome. A population of 4 diploid individuals, with the genotypes AA, Aa, Aa, and aa

• Circle = population
• Paper strip = chromosome
• Color = allele

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Ecological and Conservation Genomics

Dr. K. E. Lotterhos

Haploid Challenge #1

Create two subpopulations with F_ST = 0 for a single locus with two alleles, and haploid individuals.

Draw the case of two populations with six individuals within each population. (Note that for the haploid challenges, the inbreeding coefficient (F_IS) cannot be calculated because there are no heterozygotes.) Hint: There are many different solutions.

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Another solution

One solution

Ecological and Conservation Genomics

Dr. K. E. Lotterhos

Haploid Challenge #2

Create two subpopulations with F_ST = 1 for a single locus with two alleles, and haploid individuals.

Draw the case of two populations with six individuals within each population. (Note that for the haploid challenges, the inbreeding coefficient (F_IS) cannot be calculated because there are no heterozygotes.)

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Ecological and Conservation Genomics

Dr. K. E. Lotterhos

Diploid Challenges

Create two subpopulations with F_ST = ___ and F_IS = ___ for a single locus with two alleles, and diploid individuals.

Draw the case of two populations with four diploid individuals within each population.

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p₁ = allele frequency of blue A in population 1 (q₁ = 1-p₁)

p₂ = allele frequency of blue A in population 2 (q₂ = 1-p₂)

H_s = Expected heterozygosity averaged across subpopulations/demes

H₀ = Observed heterozygosity averaged across subpopulations/demes

p_tot = allele frequency of blue A when all subpopulations are combined together into a single population

H_T = Expected heterozygosity when all subpopulations are combined together into a single population

F_IS = 1 - H_o/H_S = inbreeding coefficient

F_ST = 1 - H_S/H_T = fixation index

Ecological and Conservation Genomics

Dr. K. E. Lotterhos

Diploid Challenge #1

Create two subpopulations with F_ST = 0 and F_IS = 0 for a single locus with two alleles, and diploid individuals.

Draw the case of two populations with four diploid individuals within each population. (Hint: there is more than 1 solution)

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p₁ = (2*AA₁+Aa₁)/(2N₁)=

p₂ = (2*AA₂+Aa₂)/(2N₂)=

H_s = (2p₁q₁ + 2p₂q₂)/2 =

H₀ = (%Hetero₁ + %Hetero₂)/2 =

p_tot = (2*AA_tot+Aa_tot)/(2N_tot) =

H_T = 2p_tot(1-p_tot) =

F_IS = 1 - H_o/H_S =

F_ST = 1 - H_S/H_T =

Ecological and Conservation Genomics

Dr. K. E. Lotterhos

Diploid Challenge #2

Create two subpopulations with F_ST = 0 and F_IS = -1 for a single locus with two alleles, and diploid individuals.

Draw the case of two populations with four diploid individuals within each population.

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p₁ = (2*AA₁+Aa₁)/(2N₁)=

p₂ = (2*AA₂+Aa₂)/(2N₂)=

H_s = (2p₁q₁ + 2p₂q₂)/2 =

H₀ = (%Hetero₁ + %Hetero₂)/2 =

p_tot = (2*AA_tot+Aa_tot)/(2N_tot) =

H_T = 2p_tot(1-p_tot) =

F_IS = 1 - H_o/H_S =

F_ST = 1 - H_S/H_T =

Ecological and Conservation Genomics

Dr. K. E. Lotterhos

Diploid Challenge #3

Create two subpopulations with F_ST = 0 and F_IS = 1 for a single locus with two alleles, and diploid individuals.

Draw the case of two populations with four diploid individuals within each population. (Hint: there is more than one solution.)

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p₁ = (2*AA₁+Aa₁)/(2N₁)=

p₂ = (2*AA₂+Aa₂)/(2N₂)=

H_s = (2p₁q₁ + 2p₂q₂)/2 =

H₀ = (%Hetero₁ + %Hetero₂)/2 =

p_tot = (2*AA_tot+Aa_tot)/(2N_tot) =

H_T = 2p_tot(1-p_tot) =

F_IS = 1 - H_o/H_S =

F_ST = 1 - H_S/H_T =

Ecological and Conservation Genomics

Dr. K. E. Lotterhos

Diploid Challenge #4

Create two subpopulations with F_ST = 1 and F_IS = -1 for a single locus with two alleles, and diploid individuals.

Draw the case of two populations with four diploid individuals within each population.

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p₁ = (2*AA₁+Aa₁)/(2N₁)=

p₂ = (2*AA₂+Aa₂)/(2N₂)=

H_s = (2p₁q₁ + 2p₂q₂)/2 =

H₀ = (%Hetero₁ + %Hetero₂)/2 =

p_tot = (2*AA_tot+Aa_tot)/(2N_tot) =

H_T = 2p_tot(1-p_tot) =

F_IS = 1 - H_o/H_S =

F_ST = 1 - H_S/H_T =

Ecological and Conservation Genomics

Dr. K. E. Lotterhos

Diploid Challenge #5

Create two subpopulations with F_ST = 1 and F_IS = 0 for a single locus with two alleles, and diploid individuals.

Draw the case of two populations with four diploid individuals within each population.

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p₁ = (2*AA₁+Aa₁)/(2N₁)=

p₂ = (2*AA₂+Aa₂)/(2N₂)=

H_s = (2p₁q₁ + 2p₂q₂)/2 =

H₀ = (%Hetero₁ + %Hetero₂)/2 =

p_tot = (2*AA_tot+Aa_tot)/(2N_tot) =

H_T = 2p_tot(1-p_tot) =

F_IS = 1 - H_o/H_S =

F_ST = 1 - H_S/H_T =

Ecological and Conservation Genomics

Dr. K. E. Lotterhos

Diploid Challenge #6

Create two subpopulations with F_ST = 1 and F_IS = 1 for a single locus with two alleles, and diploid individuals.

Draw the case of two populations with four diploid individuals within each population.

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p₁ = (2*AA₁+Aa₁)/(2N₁)=

p₂ = (2*AA₂+Aa₂)/(2N₂)=

H_s = (2p₁q₁ + 2p₂q₂)/2 =

H₀ = (%Hetero₁ + %Hetero₂)/2 =

p_tot = (2*AA_tot+Aa_tot)/(2N_tot) =

H_T = 2p_tot(1-p_tot) =

F_IS = 1 - H_o/H_S =

F_ST = 1 - H_S/H_T =

Ecological and Conservation Genomics

Dr. K. E. Lotterhos