Genetics of Organisms and χ² Analysis

            Imagine that you toss a coin 20 times and the results are 12H : 8T. Is this a fluke or is something wrong with your coin? Statistics can be used to determine whether the differences between expected and observed results are significant. To determine whether data fit expected values, we use the chi square (χ²) test. This test allows us to say whether deviations from expected values are due to chance alone or if some other factor is at play.

It’s a good idea to familiarize yourself with some terminology before going on.

•          A null hypothesis states that there is no statistically significant difference between the expected and observed results. If a statistical test shows that there is a difference, the alternative hypothesis is accepted.

•          Is the difference between the expected and observed results significant?

            χ² = Σ (o-e)²    where o = observed number of individuals

                           e         e = expected number of individuals

Σ = the sum of the values

•          Degrees of freedom is the number of possible outcomes minus one. For example, if there are two possible phenotypes in a cross, (2 - 1) = 1 degree of freedom.

•          p value - If the χ² calculated is greater than the critical value (read from a table) then the null hypothesis is rejected. The minimum probability for rejecting a null hypothesis in science is generally 5%. The p value of 0.05 means that you would expect to see similar data only 5% of the time if the null hypothesis were correct. That means you are 95% certain that the data you gathered do not fit your predicted results. In a research study, you would now consider reasons why the data do not fit your prediction. Reasons could include a small sample size, error in collection, or a problem in your experimental design.

Example 1 

For the cross Gg x Gg we would predict the F1 to be 3 green : 1 albino.

The null hypothesis here predicts that the data from the experimental cross will fit the expected 3:1 ratio.

You should get χ² = 5.14

Find the p value. In the table, the critical value for p = 0.05 is 3.84. This means, if the calculated χ² is greater than or equal to the value in the table, the null hypothesis is rejected. In this example, 5.14 > 3.84 so we reject the null hypothesis that there is no statistically significant difference between the observed and expected results. In other words, something more than chance must be responsible for the observed difference.

Example 2

In a study of incomplete dominance in tobacco seedlings, the following data were collected:

P Gg x Gg

F₁ 22 GG : 50 Gg : 12 gg (green : yellow green : albino)

Do the data support or contradict the null hypothesis? (² = 5.43; critical value = 5.99; null hypothesis isaccepted)

Protocol

Imagine a cross between two fruit flies where P X^rX^r x X^Ry

1. Describe and name the mutation. (eye color, sex distribution implies sex-linkage)

2. Write a null hypothesis that describes the mode of inheritance of the trait. (the observed data will fit theexpected ratio for the expression of a sex-linked trait 1:1:1:1)

3. Predict the expected genotypic and phenotypic ratios for both F₁ and F₂.

4. Do the actual results differ from the expected results? (yes. expect a 1:1:1:1 ratio)

5. Based on the results, is the trait

a) sex-linked or autosomal? (sex-linked)

b) dominant or recessive? (dominant)

6. Are the deviations for the phenotypic ratio of the F₂ generation within the limits expected by chance? To answer this question, statistically analyze the data using the chi-square analysis. Calculate the chi-square statistic for the F₂ generation in the chart below. Refer to the chi square distribution table to determine the p value that is associated with your chi-square statistic.

a) How many degrees of freedom are there? (3)

b) Referring to the critical value chart, what is the 5% probability (p) value for these date? (7.815)

7. According to the probability value, can you accept or reject the null hypothesis? (accept)

8. How confident are you that the variations you observed are due to chance alone? (50-80%)