From a Mendelian Ratio to a Statistical Model
A genetic model may predict a particular ratio of offspring phenotypes, such as 3:1 or 9:3:3:1. Observed offspring counts will rarely match those proportions exactly: even if the model is correct, random variation can make some categories more common and others less common in a particular sample. A goodness-of-fit test asks whether the observed differences are larger than would be reasonably expected from that variation.
The key statistical step is to turn the ratio into probabilities. The terms in a ratio are relative weights, not probabilities by themselves. Divide each term by the sum of all terms to obtain the probability for its corresponding category. Those probabilities specify the null model used to calculate expected offspring counts.
For example, a 3:1 ratio has four total ratio units. Its category probabilities are \(3/4\) and \(1/4\), not 3 and 1. With 80 offspring, the expected counts are \(80(3/4)=60\) and \(80(1/4)=20\). Expected counts can be fractional in general; they are model-based averages, not counts that must occur in an actual sample.
Hypotheses and the Goodness-of-Fit Statistic
Suppose there are \(k\) mutually exclusive phenotype categories, and each offspring is recorded in exactly one category. The null hypothesis states that the category probabilities follow the probabilities implied by the specified ratio. The alternative is that the offspring distribution does not follow that ratio: at least one category probability differs.
Let \(O_i\) be the observed count and \(E_i=np_i\) the expected count in category \(i\). Pearson’s chi-square statistic adds the squared discrepancy in each category, scaled by its expected count. Large values mean the observed counts are far from what the null model predicts.
When the ratio probabilities are fixed in advance and the chi-square approximation is appropriate, the reference distribution has \(df=k-1\). The degrees of freedom reflect that the category counts sum to the fixed total \(n\): once \(k-1\) counts are known, the last count is determined. The p-value is the probability, assuming \(H_0\), of obtaining a chi-square statistic at least as large as the one calculated.
Conditions for Using the Test
The arithmetic alone does not establish that a chi-square p-value is valid. Before using it, check that the genetic model and data collection support the stated probabilities and that the approximation is adequate.
- The ratio is specified before examining the counts. The probabilities must come from the genetic prediction being tested, rather than being chosen after seeing which ratio best matches the data.
- Categories are mutually exclusive and exhaustive. Each offspring belongs to one, and only one, category included in the ratio.
- Offspring outcomes are independent under the sampling model. One offspring's recorded category should not determine another's. For example, the data should not count the same offspring more than once.
- The genetic assumptions are credible for the question. The cross, phenotype classification, and biological setting should reasonably support the stated Mendelian prediction. Factors such as misclassification or differential survival could produce a distribution other than the predicted ratio.
- Expected counts are adequate for the chi-square approximation. A common guideline is that every expected count is at least 5. If this condition is not met, the usual chi-square calibration may be unreliable; the small-count alternatives are a separate issue.
A test can assess whether the data are consistent with the stated ratio; it cannot establish that the biological assumptions are true. A small p-value may signal a mismatch, but it does not by itself identify why the mismatch occurred.
Worked Example: A 3:1 Ratio
Suppose, hypothetically, that 100 offspring from a cross are classified into two phenotype categories. A prespecified genetic model predicts a 3:1 ratio. The observed counts are 78 in the first category and 22 in the second. Test the model at \(\alpha=0.05\).
Convert the ratio to probabilities. The ratio sum is \(3+1=4\), so the null probabilities are \(p_1=3/4=0.75\) and \(p_2=1/4=0.25\). The expected counts are \(E_1=100(0.75)=75\) and \(E_2=100(0.25)=25\).
Check the conditions. Assume the ratio and categories were chosen before counting, each offspring was independently classified once, and the cross and classification reasonably satisfy the model. The expected counts, 75 and 25, are both at least 5, so the usual chi-square approximation is appropriate.
Calculate the statistic.
There are \(k=2\) categories, so \(df=2-1=1\). The upper-tail p-value for \(\chi^2=0.480\) with 1 degree of freedom is approximately \(0.488\). Since \(0.488>0.05\), do not reject \(H_0\). These data do not provide strong evidence against the 3:1 model. This is not proof that the ratio is correct; it means the observed difference is not unusual under the model.
Worked Mendelian Examples
Worked Example: A 9:3:3:1 Ratio
In a hypothetical cross, a genetic model specified in advance predicts four phenotype categories in a 9:3:3:1 ratio. Among 160 offspring, the observed counts are 110, 26, 18, and 6 in the corresponding categories. Test whether the data fit the predicted ratio.
Find the expected counts. The ratio sum is \(9+3+3+1=16\). The probabilities are \(9/16\), \(3/16\), \(3/16\), and \(1/16\). Thus the expected counts are \(160(9/16)=90\), \(160(3/16)=30\), \(160(3/16)=30\), and \(160(1/16)=10\). All are at least 5.
Assume independent offspring outcomes, complete and nonoverlapping phenotype categories, and a cross for which the specified genetic model is reasonable. With the expected-count condition also satisfied, calculate:
Here \(df=4-1=3\). The upper-tail p-value is approximately \(0.0098\). Because this is below \(0.05\), reject \(H_0\): the counts provide evidence of a distributional mismatch with the specified 9:3:3:1 ratio.
The category contributions are approximately 4.444, 0.533, 4.800, and 1.600. The third category contributes most to the overall statistic: it has 12 fewer offspring than its expected count of 30. The first category has 20 more than expected and also contributes substantially. These comparisons help describe the mismatch, but the overall test is the test of the distribution as a whole; a large contribution alone is not a separate significance test for that category.
Worked Example: A 1:1 Ratio
Consider a hypothetical cross for which a prespecified model predicts a 1:1 ratio. Of 120 independently classified offspring, 68 are in the first category and 52 are in the second. Use \(\alpha=0.05\).
The ratio sum is 2, so each null probability is \(1/2\). The expected counts are \(120(1/2)=60\) in each category; both meet the expected-count guideline. Suppose the ratio was specified in advance, categories are exhaustive and nonoverlapping, and the cross and classification support the model.
The statistic is:
With \(df=2-1=1\), the p-value is approximately \(0.144\). Since \(0.144>0.05\), do not reject \(H_0\). The sample has eight more offspring than expected in the first category and eight fewer in the second, but that difference is not sufficiently unusual under the 1:1 model to provide evidence against it at the 5% level.
What Category-Level Differences Can Tell You
The overall chi-square test answers an omnibus question: do the counts, taken together, depart from the specified ratio more than chance variation would typically produce? The statistic’s summands show which categories contribute most to the discrepancy. A positive difference \(O_i-E_i\) means a category has more observed offspring than expected; a negative difference means it has fewer.
A useful category-level description reports both direction and scale. For instance, “the third category had 12 fewer offspring than the 30 expected” is clearer than saying only that it “deviated.” Pearson residuals, defined as \((O_i-E_i)/\sqrt{E_i}\), can also express the signed discrepancy relative to its expected-count scale. In the 9:3:3:1 example, the third category’s residual is \((18-30)/\sqrt{30}\), approximately \(-2.191\). This is a diagnostic description of its contribution, not by itself a conclusion that this category has a statistically significant difference.
A rejection does not identify the cause of a mismatch. Possible explanations include random variation that was unlikely under the model, a violated biological assumption, classification error, or another feature of how offspring were sampled. Conversely, a failure to reject does not prove the Mendelian mechanism; it indicates only that these data do not provide strong evidence of lack of fit.
Common Mistakes
- Treating ratio terms as probabilities. A 9:3:3:1 ratio does not mean probabilities 9, 3, 3, and 1. Divide each term by the sum, 16, to obtain probabilities that total 1.
- Using observed counts to define the expected ratio. The null probabilities must come from the prespecified genetic model, not from adjusting the ratio until it resembles the data.
- Claiming that a nonsignificant result proves the model. A p-value above \(\alpha\) means the data do not provide sufficient evidence against the model; it does not establish that the model is true.
- Interpreting the p-value as the probability that the ratio is correct. It is a probability of data at least this discrepant, calculated under \(H_0\), not a probability that \(H_0\) is true.
- Calling the largest category contribution a separate test result. Contributions locate discrepancies in the overall statistic. They do not automatically provide category-specific significance conclusions.
Check Your Understanding
Answer each question using the ideas and calculations in this tutorial.
- A model predicts a 3:1 ratio among 200 offspring. What are the two null probabilities and expected counts?
- For a 1:2:1 ratio, explain why the probabilities are not 1, 2, and 1.
- What does a p-value of 0.03 mean when testing a prespecified Mendelian ratio?
- If a goodness-of-fit test does not reject a ratio model, can you conclude that the genetic model is true? Explain.
- In a four-category test, what can the category contributions tell you, and what can they not establish on their own?