II. Is Your Class In Hardy-Weinberg Equilibrium?

1. Differences between observed and expected genotypes may be due to chance or may indicate that the sample population is not in Hardy-Weinberg equilibrium. To help decide which of these explanations is more likely, one uses a statistic test called the Chi-Square. Use the Student Allele Server at the DNA Learning Center WWW site to test Hardy-Weinberg equilibrium for your class.

a. Go to Allele Server

b. In the upper right corner, click on the 'log in' link.

c. Click on 'REGISTER'

d. Choose a username and password, then click on 'register.'

e. After registering, you will automatically return to the Sequence Server (it's a bug, we're working on it). Type http://www.bioservers.org/sad/ to return to the Allele Server.

f. Click on 'MANAGE GROUPS.' When the window opens, use the pull-down menu to select 'Your Groups.' Then click on 'ADD GROUP' to create a database for your class.

g. Name your group and select the private or public option. BEFORE YOU CLICK 'OK,' scroll down to see the remaining two entry fields. You must enter a password and the number of students in your database if you wish to edit the group at a later date. We suggest you do this anyway in case you make a mistake in data entry.

h. Click on 'EDIT GROUP' to add your data. Click on the INDIVIDUALS tab at the top of the window. Use the pull-down menus to select the correct data for the first individual (i.e. haplotype, sex, label). When you are done, click 'Add.'

i. Repeat step h until you have finished entering your data. Click 'Done.'

j. Click on 'OK' in the 'MANAGE GROUPS' window.

k. Now you need to get your group into the workspace. Click on 'MANAGE GROUPS.' Use the pull-down menu to select 'Your Groups.' Click on the checkbox to the left of your group. Click 'OK.'

l. To see if your group is in Hardy-Weinberg equilibrium, choose 'Chi-Square' from the pull-down menu on the RIGHT SIDE of the screen. Click on the round button beneath this menu to select your group. Click 'ANALYZE.'

m. The Allele Server will perform a Chi-Square analysis using your observed genotype frequencies and the genotype frequencies predicted from a population in Hardy-Weinberg equilibrium. The predicted genotype frequencies are calculated using the same approach you used in Part I.

The bottom of the analysis page shows pie charts for the observed and expected genotype frequencies. Do these pie charts look substantially different? The Chi-Square test provides a statistical measure for the difference between the two sets of frequencies. In general, the higher the Chi-Square value, the greater the difference between the observed versus the expected frequencies.

The Chi-Square test for Hardy-Weinberg equilibrium assumes the "null hypothesis" - that is, the observed genotype frequencies are not significantly different from those predicted for a population in equilibrium. As with any theoretical value, the genotype frequencies predicted by the equilibrium equation almost always differ from the frequencies observed in a real population. The problem is to discern when the observed versus expected values differ due to chance and when they are truly different.

A probability value, or p-value, is used to evaluate the significance of a Chi-Square. Scientists give a wide margin for differences that may occur by chance by setting the cutoff for significance at p-value of 0.05 (5%) or less. This means that one may expect a Chi-Square of this value to occur by chance in 5% of genotype comparisons. Conversely, there is a 95% probability that the differences between observed versus expected genotype frequencies are "real." Social scientists expand the probability window by saying that p-values between 0.5 and 0.10 "approach significance."

For example, a p-value of 0.34 means that there is a 34% probability that the genotype differences are due to chance and 66% chance that they are not due to chance. This p-value is not significant, the null hypothesis is upheld, and we say that the population is in Hardy-Weinberg equilibrium. A p-value of 0.02 means that there is a 2% probability that the genotype differences are due to chance and 98% chance that they are not due to chance. This p-value is significant, the null hypothesis is rejected, and we say that the population is not in Hardy-Weinberg equilibrium.

What is the p-value for your Chi-Square? Is it less than .05? If so, can you suggest any factors that might account for why your observed population is not in Hardy-Weinberg equilibrium?

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DNA Learning Center, Cold Spring Harbor Laboratory
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