The data from actual experiments will never exactly equal the expected probabilities. One would not really expect that the number of individuals showing the dominant phenotype in a monohybrid cross Aa x Aa will be exactly 3/4 of the total number of offspring. We expect the results to be close to the theoretical expectation, but how close is close enough? To answer this question we need to know the probability of getting a given result in an experiment. We then compare our results to a predetermined probability. Most often we use a probability of 0.05 for this comparison. Any outcome with a probability greater than 0.05 is acceptable. We consider deviations that occur more than 5% of the time to be due to chance. To understand this concept, consider a weather report that says there is a 25% chance of rain during the day. Most people would not carry an umbrella on such a day, but the statistician would. Suppose there is only a 10% chance of rain. The statistician still carries an umbrella. If the probability of rain is 5% or less, the statistician says that the chance of rain is so improbable that someone seeded the clouds to make it rain on that day.
In statistical terms we need to test whether an outcome of an experiment is due to chance or whether some factor influenced the results. We begin with a null hypothesis. In genetics the null hypothesis is usually the assumption that the data in the experiment are in accord with the expected theoretical distribution of events. For example, in the cross Aa x Aa, the null hypothesis is that 3/4 of the offspring will show the A trait and 1/4 of the offspring will show the a trait. If our results are so improbable that they could occur by chance only 5% of the time or less, we reject the null hypothesis and search for other explanations of our results. The way we determine the probability of getting the results we found in our experiment is to use a test called the chi-square test. the formula to calculate chi-square is:
Σ ((O-E)^{2} / E)
where O stands for the observed value of an event and E is the expected value for that same event. Let's calculate chi-square for a particular data set. In swine normal parents produced 136 normal offspring and 42 hydrocephalics. If hydrocephaly in swine is due to a recessive gene we would expect 2 heterozygous parents to produce offspring in a 3:1 ratio of normal to hydrocephalic. The null hypothesis in this experiment is that 3/4 of the swine were normal and 1/4 were hydrocephalic. There were a total of 178 pigs. The expected numbers of normal pigs is 3/4(178)= 133.5. The expected numbers of hydrocephalic pigs is 1/4(178)= 44.5.
O | E | (O-E) | (O-E)^{2}/E | |
Normal | 136 | 133.5 | 2.5 | 6.25/133.5= 0.047 |
Hydrocephalic | 42 | 44.5 | -2.5 | 6.25/44.5= 0.140 |
Chi-square = 0.047 + 0.140 = 0.187. OK so now we have chi-square, but how does that help us with the probability of getting these results. we need to know the probability of getting a chi-square of 0.187. to do this we must first determine number called the degrees of freedom (df) for our experiment. The degrees of freedom in a chi-square analysis will be equal to the number of classes in the experiment minus 1. In this case there are 2 classes; hydrocephaly and normal so df = 2-1=1. There is one degree of freedom in this experiment. Now we have to resort to a table of chi square probabilities. Page 63 in your text has such a table. Examine the row in the table for 1 df. Notice that the probability of a result decreases as chi-square gets larger. Our value of 0.187 has a probability between 0.50 and 0.90. It is very likely that our results differ from chance very little and we do not reject the null hypothesis that the data are in accord with a 3:1 ratio.
Example 2. In mice (also a mammal), 2474 mice were recovered and 2069 were normal and 405 were hydrocephalic.
O | E | (O-E) | (O-E)^{2}/E | |
Normal | 2069 | 1855.5 | 213.5 | 45582.25/1855.5= 24.566 |
Hydrocephalic | 405 | 618.5 | -213.5 | 45582.25/618.5= 73.698 |
chi square = 98.264 df = 1 |
Examining this chi-square value in our table shows that the p =<<0.001. We must reject the null hypothesis of a 3:1 ratio and search for an explanation for these data. Closer examination revealed that mother mice eat defective offspring. So the experimenters watched and found that of 389 mice, 280 were normal and 109 were hydrocephalic. Calculate the chi-square for these data. Do you now reject the null hypothesis of a 3:1 ratio of normal to hydrocephalic mice? You should find that hydrocephaly was due to a recessive allele after all. The message here is that one should not accept aberrant results blindly. Check for a factor that could be influencing the data.