The purpose of this Applet is to examine the ratio of additive genetic variance to dominance genetic variance at a single locus. To execute this Applet, enter the genetic value for the heterozygote Aa. This value can range between -1.0 and 1.0. A value of -1.0 implies that allele a is dominant, a value of 0.0 implies that the system is totally additive, and a value of 1.0 implies that allele A is dominant. Values between -1.0 and 0.0 (or between 0.0 and 1.0) imply partial dominance. The second value to enter is the frequency of allele a. Naturally, this value must lie between 0.0 and 1.0.

Do the following exercises:

• Set the value of the heterozygote to 0.0, so that gene action is completely additive. Now enter any value for the frequency of allele a and click on Submit. Notice the value next to the label "Percent Additive." This gives the percentage of the genetic variance that is additive, i.e., the additive genetic variance divided by the quantity (additive + dominance variance). Enter 5 different values for the frequency of allele a and record the Percent Additive variance. What does this tell you about how gene action contributes to the ratio of additive to dominance variance?
• Set the value of the heterozygote to 1.0. In this case allele A is completely dominant to allele a. Not let the frequency of allele a range from .10 to .90 in increments of .10, clicking on the Submit button each time. Record the frequency of allele a and the result of the "Percent Additive" column. What does this tell you about the other major factor that contributes to the ratio of additive to dominance variance?
• To reinforce the learning from the previous exercises, let us examine the variance components from a rare recessive disorder. Set the genetic value for the heterozygote to 1.0 and the frequency of allele a to .01. This is roughly the case for phenylketonuria. Click on Submit and record the Percent Additive variance?
• To further reinforce this, let's examine the situation of Huntington's Disease, a rare autosomal dominant. Again, set the value of the heterozygote to 1.0, but this time, let the frequency of allele a be .99. Again record the Percent Additive Variance.
• Compare and contrast the results of exercise (3) with exercise (4). In both cases, the gene action (i.e., the genetic value of the heterozygote) is the same. What accounts for the discrepancy between the figures for Percent Additive? What general lesson does this confirm about the relationship of "biological gene activity" to "statistical variance components?"