12.6 : Chi-square Analysis

Mendel hypothesized that the phenotypic ratio of a monohybrid cross between individuals with a dominant trait, like purple flowers, and a recessive trait, like white flowers, would be approximately three to one provided the alleles are independently assorted.

When there is an assumption that data will fit a certain ratio, like Mendel’s 3:1 prediction, a null hypothesis is established. The null hypothesis assumes that there is no real difference between the expected and observed values and that any apparent differences are due to chance.

Consider a monohybrid cross between twenty tall plants and twenty short plants. The F2 generation of this cross yields 33 tall plants and 7 short plants. This is the observed value.

To test the null hypothesis, a chi-square analysis can be performed. First, create a table with three columns: phenotype, observed value, and expected value. The phenotype and observed values are already known, but the expected values must be calculated.

Calculate the expected values by dividing the total number of plants observed with the total number of expected plants - that is three plus one equals four. So, forty divided by four is ten.

Ten represents the “one” of the expected three to one frequency. Hence, there are ten plants with the recessive “short” gene.

Now, multiply ten by three to obtain the “three” of the expected three to one frequency. Thus, there are thirty plants with the dominant “tall” gene.

Calculate the difference between the observed and expected value in each row and square the results.

Then, divide this value by the expected value in each row and add them all together to obtain the chi-squared value - 1.2.

The degree of freedom, which is defined as the number of values that can vary in the final calculation of a statistic, must also be calculated. Obtain the degree of freedom using this formula, where n is the number of classes, or in this case phenotypes. Here, there is one degree of freedom.

Using the chi-squared value and degree of freedom, obtain the result on the probability chart. Here, 1.2 falls between 1.07 and 1.64. The p-values for these is 0.30 and 0.20, which is greater than 0.05. Thus, the null hypothesis is not rejected, indicating that the alleles for plant height are independently assorted.