PV92 PCR/Informatics KitPV92 PCR/Informatics Kit

Population Genetics and Population Genetics and InformaticsInformatics

To estimate frequency of To estimate frequency of Alu Alu

within a population:within a population:

Amplify Amplify Alu Alu insert from insert from representative sample populationrepresentative sample population

Calculate the expected allelic and Calculate the expected allelic and genotypic frequenciesgenotypic frequencies

Perform Chi-squared TestPerform Chi-squared Test

Calculating Allelic and Genotypic Calculating Allelic and Genotypic FrequenciesFrequencies

Within your class, how unique is your particular combination of Alu alleles? By calculating an allele frequency, you can begin to answer this question. An allele frequency is the percentage of a particular allele within a population of alleles. It is expressed as a decimal. You can calculate an allele frequency for the Alu PV92 insertion in your class by combining all your data. For example, imagine that there are 38 students in your class and the genotype distribution within the class is as follows: GenotypeGenotype +/+ +/- -/- TotalTotal

(N)(N)

# of people# of people 25 5 825 5 8 3838

Calculating Allelic Frequencies: + Calculating Allelic Frequencies: + alleleallele

Total number of alleles = 2N = Total number of alleles = 2N = 2(38)2(38) = 76 = 76

Number of + alleles =Number of + alleles =

+/+=+/+= 25 with two 25 with two + + alleles alleles = 50 + = 50 + allelesalleles

+/- = +/- = 5 with one 5 with one + + allelesalleles = 5 + = 5 + allelesalleles

Total= Total= 55 + 55 + allelesalleles

Frequency of + Frequency of + = = number of + allelesnumber of + alleles = = 55 55 = = 0.720.72 total alleles total alleles 7676

Calculation for the – alleles would be similar, and Calculation for the – alleles would be similar, and the result would be .28the result would be .28

Notice: (+ allele) + (- allele) = 1.0Notice: (+ allele) + (- allele) = 1.0

Calculating Genotypic Calculating Genotypic FrequenciesFrequencies

How does the distribution of Alu genotypes in your class compare with the distribution in other populations? For this analysis, you need to calculate a genotype frequency, the percentage of individuals within a population having a particular genotype. Remember that the term allele refers to one of several different forms of a particular genetic site whereas the term genotype refers to the specific alleles that an organism carries. You can calculate the frequency of each genotype in your class by counting how many students have a particular genotype and dividing that number by the total number of students.

Given the ethnic makeup of your class, might you expect something different?

How can you estimate what the expected frequency should be?

Calculating Calculating ObservedObserved Genotypic FrequenciesGenotypic Frequencies

GenotypeGenotype +/+ (p2)(p2) +/- (2pq)(2pq) -/- (q2)(q2) TotalTotal (N) (N)

# of people# of people 25 5 825 5 8 38 38

Observed Observed 0.66 0.13 0.21 0.66 0.13 0.21 1.001.00

frequencyfrequency

Calculation:Calculation:+/+ genotypic frequency == # with genotype

total number of people (N)total number of people (N)

== 25/3825/38 = .66= .66

Alu and Population GeneticsAlu and Population Genetics If within an infinitely large population no mutations

are acquired, no genotypes are lost or gained, mating is random, and all genotypes are equally viable, then that population is said to be in Hardy-Weinberg equilibrium. In such populations, the allele frequencies will remain constant generation after generation. Genotype frequencies within this population can then be calculated from allele frequencies by using the equation:

p2 + 2pq + q2 = 1.0 p and q are the allele frequencies for two

alternate forms of a genetic site. The genotype frequency of the homozygous condition is either p2 or q2 (depending on which allele you assign to p and which to q). The heterozygous genotype frequency is 2pq.

Alu and Population GeneticsAlu and Population GeneticsHardy-Weinberg EquilibriumHardy-Weinberg Equilibrium

pp22 + 2pq + q + 2pq + q22 = 1 = 1

+/+ = p+/+ = p22

+/- = 2pq+/- = 2pq-/- = q-/- = q22

pppp pqpq

qqqqpqpq

p p

pp

qq

qq

Using Hardy-WeinbergUsing Hardy-Weinberg

Determine Determine pp22,, 2pq, and q2pq, and q22 values= values= Expected genotypic frequenciesExpected genotypic frequencies

p = 0.72 , so q = 0.28p = 0.72 , so q = 0.28 since since p + q = 1p + q = 1 pp22 ++ 2pq2pq + q+ q22 = 1= 1

(0.72)(0.72)22 + 2 (0.72)(0.28)+ 2 (0.72)(0.28) + (0.28)+ (0.28)22 = 1= 1 0.520.52 ++ 0.400.40 + + 0.080.08 = 1= 1

p2 = 0.52p2 = 0.52 2pq = 0.402pq = 0.40 q2 = 0.08q2 = 0.08

Calculate Calculate Expected Expected Number of Number of GenotypesGenotypes

Expected number of genotype =Expected number of genotype =

Genotypic frequency x population Genotypic frequency x population number (N)number (N)

GenotypeGenotype Expected Expected numbernumber+/+ +/+ 0.52 x 38 = 0.52 x 38 = 2020+/+/- - 0.40 x 38 = 0.40 x 38 = 1515

--//- - 0.08 x 38 = 0.08 x 38 = 33

Chi Squared TestChi Squared Test Chi-square, a statistical test used for comparing

observed frequencies with expected frequencies. The larger the chi-square value, the greater is the difference between the observed and the expected values.

When using the Chi-square analysis, we test the null hypothesis that there is no difference between samples (observed and expected) and we assume that if there is any difference, then it arose simply by chance and is not real.

Our null hypothesis is that your class is in Hardy-Weinberg equilibrium. Whether or not we can accept the null hypothesis is given by a p-value.

If the calculated p-value is less than 0.05, the null hypothesis is disproved; the population is not in Hardy-Weinberg equilibrium.

If the p-value is greater than 0.05, the population may be in Hardy-Weinberg equilibrium; we can not prove that it is not in Hardy-Weinberg equilibrium.

Chi Squared TestChi Squared Test

As an example, let’s say that Chi-square analysis of some data gives a p-value of 0.17. This means that there is a 17% probability that the difference between the observed and the expected values is due to chance. It also means that there is an 83% (100% - 17% = 83%) probability that the difference is not due to chance; the difference is real.

Chi SquaredChi Squared Test Test

E

EOX

22

Observed Observed Expected Expected (O-E)2(O-E)2

EE+/+ +/+ 2525 20 20 1.251.25 +/- +/- 55 15 15 6.666.66-/- -/- 88 3 3 8.338.33

Total Total 17.1217.12

X2 Critical Value (from statistics table) = 5.9X2 Critical Value (from statistics table) = 5.9

17.12 is above 5.9 so the ratio is not accepted.17.12 is above 5.9 so the ratio is not accepted.

Allele ServerAllele Server

Cold Spring Harbor LabCold Spring Harbor Lab

DNA Learning CenterDNA Learning Center

http:// vector.cshl.orghttp:// vector.cshl.org

Click on Resources

Click on BioserversClick on Bioservers

Click on Bioservers

Enter the Allele ServerEnter the Allele Server

Allele ServerAllele Server

Click on Manage Groups

Select Type of DataSelect Type of Data

Select Group

Your GroupYour Group

Scroll down to “Your Group”

Click on Add GroupClick on Add Group

Click on Add Group

Fill Out FormFill Out Form

Click on Edit GroupClick on Edit Group

Click on Edit Group

Fill out CompletelyFill out Completely

Click on Individuals TabClick on Individuals Tab

Click on Individuals Tab

Add Each Student With Add Each Student With InformationInformation

Add as much info as you canAdd as much info as you can Genotype ( +/+, +/-, -/- )Genotype ( +/+, +/-, -/- ) GenderGender Personal InfoPersonal Info

Click on DoneClick on Done

Done

Select Your GroupSelect Your Group

Select and then press OK

Click to AnalyzeClick to Analyze

Click Here

Then Click Here

Then Look at the Terse and Then Look at the Terse and Verbose TabsVerbose Tabs

ExtensionsExtensions Is your class in Hardy-Weinberg Is your class in Hardy-Weinberg

Equilibrium?Equilibrium? Compare your group to other existing Compare your group to other existing

groups.groups. Form an explanation for the origination Form an explanation for the origination

of Alu and how it spread throughout of Alu and how it spread throughout different populations.different populations.

Have students do manual calculations Have students do manual calculations first and then compare to the first and then compare to the computer generated version.computer generated version.

Plotting Alu PV92 on a World Plotting Alu PV92 on a World MapMap The Alu element first appeared tens of millions of years ago and since

that time, it has been increasing within our genome at the rate of about one copy every 100 years.

It is difficult to tell how Alu arose. It shows a striking similarity to a gene (called 7SL RNA) that performs a vital function in our metabolism. But Alu , it seems, has no function. It is self-serving and, like a parasite, takes advantage of us for its own replication without providing us any advantage to our own survival.

Most Alu elements are “fixed”; they are found at the same chromosomal site in every person on the planet. Fixed Alu elements must have arose very early in our evolution, well before Homo sapiens appeared. When modern humans did arise some 200,000 years ago, the vast majority of our Alu insertions came to us already intact in our DNA.

The Alu PV92 insertion, however, is not fixed. This insertion may or may not be present on one or both of a person’s number 16 chromosomes. Since not everyone has the Alu PV92 element, it must have arisen after the initial human population began growing.

It is a widely held belief that modern humans originated in Africa and then disseminated across the planet. Did the Alu PV92 insert arise in Africa or on some other continent during our spread across the globe?

In the following exercise, you will plot the “+” allele frequencies for various populations on a world map and make some determination as to where this Alu arose and how it might have spread across continents.

.18 .20

.96

.26

.09

.35

.30

.18

.52

.86

.15

.53

.12

Classroom How To…Classroom How To…

We have worksheets for calculating We have worksheets for calculating allelic and genotypic frequencies for allelic and genotypic frequencies for your class your class

BABEC has the materials for the BABEC has the materials for the world population dataworld population data