If you expected a 1:1 ratio of brown and yellow cockroaches, but the cross produced 22 brown and 18 yellow cockroaches, you probably wouldn’t be too surprised, even though it wasn’t a perfect 1:1 ratio. In this case, it seems reasonable to assume that chance produced the deviation between the expected and the observed results. But if you observed 25 brown and 15 yellow cockroaches, would you still assume that this result represents a 1:1 ratio? Something other than chance might have caused this deviation. Perhaps the inheritance of this characteristic is more complicated than was assumed, or perhaps some of the yellow progeny died before they were counted. Clearly, we need some means of evaluating how likely it is that chance is responsible for the deviation between the observed and the expected numbers.

To evaluate the role of chance in producing deviations between observed and expected values, a statistical test called the chi-square goodness-of-fit test is used. This test provides information about how well the observed values fit the expected values. Before we learn how to use this test, however, it is important to understand what it does and does not indicate about a genetic cross. The chi-square test cannot tell us whether a genetic cross has been correctly carried out, whether our results are correct, or whether we have chosen the correct genetic explanation for those results. What it does indicate is the probability that the difference between the observed and the expected values is due to chance. In other words, it indicates the likelihood that chance alone could produce the deviation between the expected and the observed values.

If we expected 20 brown and 20 yellow progeny from a genetic cross, the chi-square test gives the probability that we might observe 25 brown and 15 yellow progeny simply owing to chance deviations from the expected 20:20 ratio. The hypothesis that chance alone is responsible for any deviation between observed and expected values is sometimes called the null hypothesis. Statistical tests such as the chi-square test cannot prove that the null hypothesis is correct, but they can help us decide if we should reject it. When the probability calculated from the chi-square test is high, we assume that chance alone produced the deviation, and we do not reject the null hypothesis. When the probability is low, we assume that some factor other than chance—some significant factor—produced the deviation. For example, the mortality rate of the yellow cockroaches might be higher than that of the brown cockroaches. When the probability that chance produced the deviation is low, we reject the null hypothesis.

To use the chi-square goodness-of-fit test, we first determine the expected results. The chi-square test must always be applied to numbers of progeny, not to proportions or percentages. Let’s consider a locus for coat color in domestic cats, for which black color (B) is dominant over gray (b). If we crossed two heterozygous black cats (Bb × Bb), we would expect a 3:1 ratio of black and gray kittens. Imagine that a series of such crosses yields a total of 50 kittens—30 black and 20 gray. These numbers are our observed values. We can obtain the expected numbers by multiplying the expected proportions by the total number of observed progeny. In this case, the expected number of black kittens is ¾ × 50 = 37.5 and the expected number of gray kittens is ¼ × 50 = 12.5.

The chi-square (χ²) value is calculated by using the following formula:

where ∑ means the sum. We calculate the sum of all the squared differences between observed and expected values and divide by the expected values. To calculate the chi-square value for our black and gray kittens, we would first subtract the number of expected black kittens from the number of observed black kittens (30 − 37.5 = −7.5) and square this value: −7.5² = 56.25. We then divide this result by the expected number of black kittens, 56.25/37.5 = 1.5. We repeat the calculations on the number of expected gray kittens: (20 − 12.5)²/12.5 = 4.5. To obtain the overall chi-square value, we add the (observed − expected)²/expected values: χ² = 1.5 + 4.5 = 6.0.

The next step is to determine the probability associated with this chi-square value, which is the probability that the deviation between the observed and the expected results could be due to chance. This step requires us to compare the calculated chi-square value (6.0) with theoretical values that have the same degrees of freedom in a chi-square table. The degrees of freedom represent the number of ways in which the expected classes are free to vary. For a chi-square goodness-of-fit test, the degrees of freedom are equal to n − 1, in which n is the number of different expected phenotypes. Here, we lose one degree of freedom because the total number of expected progeny must equal the total number of observed progeny. In our example, there are two expected phenotypes (black and gray); so n = 2, and the degree of freedom equals 2 − 1 = 1.

Now that we have our calculated chi-square value and have figured out the associated degrees of freedom, we are ready to obtain the probability from a chi-square table (Table 3.4). The degrees of freedom are given in the left column of the table and the probabilities are given at the top; within the body of the table are chi-square values associated with these probabilities. First, we find the row for the appropriate degrees of freedom; for our example with 1 degree of freedom, it is the first row of the table. Then we find where our calculated chi-square value (6.0) lies among the theoretical values in this row. The theoretical chi-square values increase from left to right, and the probabilities decrease from left to right. Our chi-square value of 6.0 falls between the value of 5.024, associated with a probability of 0.025, and the value of 6.635, associated with a probability of 0.01.

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Thus, the probability associated with our chi-square value is less than 0.025 and greater than 0.01. So there is less than a 2.5% probability that the deviation that we observed between the expected and the observed numbers of black and gray kittens could be due to chance.

Most scientists use the 0.05 probability level as their cutoff value: if the probability of chance being responsible for the deviation between the observed and the expected values is greater than or equal to 0.05, they assume that chance is responsible for that deviation. When the probability is less than 0.05, scientists assume that chance is not responsible and that a significant difference from the expected values exists. The expression significant difference means that a factor other than chance is responsible for the difference between the observed values and the expected values. In regard to the kittens, perhaps one of the genotypes had a greater mortality rate before the progeny were counted, or perhaps other genetic factors skewed the observed ratios.

In choosing 0.05 as their cutoff value, scientists have agreed to assume that chance is responsible for deviations between observed and expected values unless there is strong evidence to the contrary. Bear in mind that even if we obtain a probability of, say, 0.01, there is still a 1% probability that the deviation between the observed and the expected numbers is due to nothing more than chance. Calculation of the chi-square value is illustrated in Figure 3.12. TRY PROBLEM 35