Step 5 - Interpreting The Results

We are now ready for the final step, interpreting the results of our chi-square calculation. For this we will need to consult a Chi-Square Distribution Table. This is a probability table of selected values of X2 (Table 3). 

Table 3: Chi-Square Distribution Table.

Statisticians calculate certain possibilities of occurrence (P values) for a X2 value depending on degrees of freedom. Degrees of freedom is simply the number of classes that can vary independently minus one, (n-1). In this case the degrees of freedom = 1 because we have 2 phenotype classes: resistant and susceptible.

The calculated value of X2 from our results can be compared to the values in the table aligned with the specific degrees of freedom we have. This will tell us the probability that the deviations (between what we expected to see and what we actually saw) are due to chance alone and our hypothesis or model can be supported.

In our example, the X2 value of 1.2335 and degrees of freedom of 1 are associated with a P value of less than 0.50, but greater than 0.25 (follow blue dotted line and arrows in Fig. 5). This means that a chi-square value this large or larger (or differences between expected and observed numbers this great or greater) would occur simply by chance between 25% and 50% of the time. By convention biologists often use the 5.0% value (p<0.05) to determine if observed deviations are significant. Any deviations greater than this level would cause us to reject our hypothesis and assume something other than chance was at play (see red circle on Fig. 5). If your chi-square calculated value is greater than the chi-square critical value, then you reject your null hypothesis. If your chi-square calculated value is less than the chi-square critical value, then you "fail to reject" your null hypothesis. 

Figure 5. Finding the probability value for a chi-square of 1.2335 with 1 degree of freedom. First read down column 1 to find the 1 degree of freedom row and then go to the right to where 1.2335 would occur. This corresponds to a probability of less than 0.5 but greater than 0.25, as indicated by the blue arrows.

Therefore in our tomato breeding example, we failed to reject our hypothesis that resistance to bacterial spot in this set of crosses is due to a single dominantly inherited gene (Rx-4). We can assume that the deviations we saw between what we expected and actually observed in terms of the number of resistant and susceptible plants could be due to mere chance. We can continue working with our current hypothesis. Remember, we have not “proven” our hypothesis at this point. Further testing in other crosses and populations will be used to provide additional evidence that our hypothesis accurately explains the mode of inheritance for Rx-4.

Launch the video tutorial below showing how to use a chi-square distribution chart, using this tomato breeding example.

Next we will go through a genotyping example, or you skip ahead in this lesson to a discussion about computer programs available when you have extensive data, as well as the strengths and weaknesses of the chi-square test.



In a BC2S4 IBC population (inbred backcross) with 197 tomato lines, you observed the following phenotypic data with regards to bacterial spot disease. Calculate a chi-square value for the hypothesis that resistance is dominantly inherited. Would you reject or fail to reject this hypothesis?

169 susceptible to bacterial spot

6 resistant to bacterial spot

Looks Good! ANSWER: For an IBC at the BC2S4 generation, we would expect 7 susceptible to 1 resistant, or 153 susceptible lines and 22 resistant. This gives a chi-square of 1.6732 for the susceptible class and 11.6364 for the resistant class, with an overall chi-square of 13.3096. . We have 1 degree of freedom (2 classes minus one). Using this information and the chi-square probability chart, we find a p-value of far less than 0.001. Therefore, we must reject our hypothesis of the phenotypic data supporting a dominant inheritance of bacterial spot in this IBC population.