PV92 PCR/Informatics Kit Population Genetics and Informatics

To estimate frequency of Alu within a population: Amplify Alu insert from representative sample population Calculate the expected allelic and genotypic frequencies Perform Chi-squared Test

Calculating Allelic and Genotypic Frequencies 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: Genotype +/+ +/- -/- Total (N) # of people 25 5 8 38

Calculating Allelic Frequencies: + allele Total number of alleles = 2N = 2(38) = 76 Number of + alleles = +/+= 25 with two + alleles = 50 + alleles +/- = 5 with one + alleles = 5 + alleles Total= 55 + alleles Frequency of + = number of + alleles = 55 = 0.72 total alleles 76 Calculation for the – alleles would be similar, and the result would be .28 Notice: (+ allele) + (- allele) = 1.0

Calculating Genotypic Frequencies 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 Observed Genotypic Frequencies Genotype +/+ (p2) +/- (2pq) -/- (q2) Total (N) # of people 25 5 8 38 Observed 0.66 0.13 0.21 1.00 frequency Calculation: +/+ genotypic frequency = # with genotype total number of people (N) = 25/38 = .66

Alu 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 Genetics Hardy-Weinberg Equilibrium p2 + 2pq + q2 = 1 p q pp pq p +/+ = p2 +/- = 2pq -/- = q2 q pq qq

Using Hardy-Weinberg Determine p2, 2pq, and q2 values= Expected genotypic frequencies p = 0.72 , so q = 0.28 since p + q = 1 p2 + 2pq + q2 = 1 (0.72)2 + 2 (0.72)(0.28) + (0.28)2 = 1 0.52 + 0.40 + 0.08 = 1 p2 = 0.52 2pq = 0.40 q2 = 0.08

Calculate Expected Number of Genotypes Genotypic frequency x population number (N) Genotype Expected number +/+ 0.52 x 38 = 20 +/- 0.40 x 38 = 15 -/- 0.08 x 38 = 3

Chi 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 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 Squared Test Observed Expected (O-E)2 E +/+ 25 20 1.25 +/+ 25 20 1.25 +/- 5 15 6.66 -/- 8 3 8.33 Total 17.12 X2 Critical Value (from statistics table) = 5.9 17.12 is above 5.9 so the ratio is not accepted.

Cold Spring Harbor Lab DNA Learning Center Allele Server Cold Spring Harbor Lab DNA Learning Center

http:// vector.cshl.org Click on Resources

Click on Bioservers Click on Bioservers

Enter the Allele Server

Allele Server Click on Manage Groups

Select Type of Data Select Group

Your Group Scroll down to “Your Group”

Click on Add Group Click on Add Group

Fill Out Form

Click on Edit Group Click on Edit Group

Fill out Completely

Click on Individuals Tab

Add Each Student With Information Add as much info as you can Genotype ( +/+, +/-, -/- ) Gender Personal Info

Click on Done Done

Select Your Group Select and then press OK

Click to Analyze Then Click Here Click Here

Then Look at the Terse and Verbose Tabs

Extensions Is your class in Hardy-Weinberg Equilibrium? Compare your group to other existing groups. Form an explanation for the origination of Alu and how it spread throughout different populations. Have students do manual calculations first and then compare to the computer generated version.

Plotting Alu PV92 on a World Map 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 .12 .18 .86 .53 .30 .52 .09 .26 .35 .96 .15

Classroom How To… We have worksheets for calculating allelic and genotypic frequencies for your class BABEC has the materials for the world population data