Question

Suppose you are interested in following two independent traits in snap peas- seed texture \((\mathrm{S}=\mathrm{smooth}, \mathrm{s}=\) wrinkled \()\) and seed color \((\mathrm{Y}=\) yellow \(\mathrm{y}=\) green \()-\) in a second-generation cross of heterozygous parents. Mendelian theory states that the number of peas classified as smooth and yellow, wrinkled and yellow, smooth and green, and wrinkled and green should be in the ratio 9: 3: 3: 1 . Suppose that 100 randomly selected snap peas have \(56,19,17,\) and 8 in these respective categories. Do these data indicate that the 9: 3: 3: 1 model is correct? Test using \(\alpha=.01\).

Short answer

Answer: Yes, the observed distribution matches the expected Mendelian 9:3:3:1 ratio at a significance level of α=0.01, as the calculated Chi-square statistic (0.6765) is less than the critical value (6.635).

Step-by-step solution

Calculate the expected counts

According to Mendelian theory, the expected ratio of smooth yellow, wrinkled yellow, smooth green, and wrinkled green peas should be 9:3:3:1. Sum up the observed counts (56+19+17+8=100) and multiply the total count by the expected ratios to get the expected counts: Smooth-yellow: \(\frac{9}{16}\times100=56.25\) Wrinkled-yellow: \(\frac{3}{16}\times100=18.75\) Smooth-green: \(\frac{3}{16}\times100=18.75\) Wrinkled-green: \(\frac{1}{16}\times100=6.25\) The expected counts are 56.25, 18.75, 18.75, and 6.25, respectively for each category.

Calculate the Chi-square statistic

The Chi-square statistic is calculated using the formula: \(X^2 = \sum{\frac{(O_i - E_i)^2}{E_i}}\) Where \(O_i\) = Observed count for each class, and \(E_i\) = Expected count for each class. Now, calculate the Chi-square statistic for the given problem: \(X^2 = \frac{(56-56.25)^2}{56.25} + \frac{(19-18.75)^2}{18.75} + \frac{(17-18.75)^2}{18.75} + \frac{(8-6.25)^2}{6.25} = 0.0018 + 0.0033 + 0.1674 + 0.504 = 0.6765\) The calculated Chi-square statistic is 0.6765.

Determine the degrees of freedom

The degrees of freedom (df) for a Chi-square test with independent traits are calculated using the formula: \(df = (number\ of\ rows - 1) \times (number\ of\ columns - 1)\) In this case, we have 2 traits, each with 2 categories: \(df = (2-1) \times (2-1) = 1 \times 1 = 1\) The degrees of freedom for this problem is 1.

Compare the Chi-square statistic to the critical value

The critical value of the Chi-square distribution at a significance level of \(\alpha = 0.01\) and 1 degree of freedom (\(df=1\)) can be found from Chi-square table or using statistical software (in this case, it is \(\approx 6.635\)). Since our calculated Chi-square statistic of 0.6765 is less than the critical value of 6.635, we do not have enough evidence to reject the null hypothesis that the observed distribution matches the expected Mendelian 9:3:3:1 ratio.

Conclusion

At a significance level of \(\alpha = 0.01\), we do not have enough evidence to reject the null hypothesis that the observed counts of smooth-yellow, wrinkled-yellow, smooth-green, and wrinkled-green peas follow the expected Mendelian 9:3:3:1 ratio.

Key concepts

Chi-square test

The Chi-square test is a statistical tool used to determine whether there is a significant difference between observed and expected frequencies in one or more categories. In genetics, it's commonly employed to test the goodness of fit to Mendelian ratios in breeding experiments. To perform the test, you need to follow several steps.

First, calculate the expected frequencies based on the genetic ratio or model you're testing—like the 9:3:3:1 ratio in Mendelian genetics for dihybrid crosses. Then, calculate the Chi-square statistic using the formula: \[ X^2 = \sum{\frac{(O_i - E_i)^2}{E_i}} \] where \(O_i\) represents the observed frequency for each category, and \(E_i\) stands for the expected frequency.

Example Calculation

In the context of the snap peas problem, the observed data are compared to the expectations predicated on Mendelian inheritance. By plugging these values into the formula, you get a Chi-square statistic that quantitatively reflects the discrepancies between what you observed and what Mendelian genetics would predict. To make a conclusion, this statistic is then compared against a critical value from the Chi-square distribution, which changes based on the degree of freedom in your test and the chosen significance level.

Expected ratio

The expected ratio in a genetics problem is the theoretically calculated proportion of offspring that one expects to observe if the genetic cross follows Mendelian inheritance patterns. For example, a dihybrid cross involving two heterozygous pea plants (with genotypes SsYy for seed texture and color) would result in an expected ratio of 9 smooth-yellow (SSYY or SsYy), 3 wrinkled-yellow (ssYy), 3 smooth-green (Ssyy), and 1 wrinkled-green (ssyy) offspring based on the principle of independent assortment.

When conducting a Chi-square test, you use these expected ratios to calculate the expected frequencies by taking the total number of observed organisms and multiplying by the expected ratio for each category. For the problem given with 100 snap peas, the expected frequencies were calculated by multiplying the ratio for each category by the total number of peas. This approach assumes the sample size is large enough for the expected ratios to be representative of the actual probabilities. It's essential for the analyst to ensure that the calculated expected frequencies are sufficiently large to satisfy the Chi-square test's requirements, typically each expected frequency should be 5 or higher for the test to be valid.

Null hypothesis

The null hypothesis is a fundamental concept in statistical testing, serving as a default assumption that there is no effect or no difference between groups or treatments. In the context of genetics and the Chi-square test, the null hypothesis typically states that there is no significant difference between the observed frequencies of offspring categories and the expected frequencies, assuming Mendelian inheritance patterns hold true.

When you perform a Chi-square test, you're essentially testing the validity of this null hypothesis. If the Chi-square statistic is lower than the critical value from the Chi-square distribution table at a chosen significance level (like 0.01 or 1%), you don't have enough evidence to reject the null hypothesis. This suggests that your observed data fit well with the expected Mendelian ratio. Conversely, if the Chi-square statistic is higher than the critical value, this indicates that the observed frequencies significantly deviate from the expected ones, potentially leading to the null hypothesis being rejected.

Interpreting the Null Hypothesis

In our snap peas example, the null hypothesis states that the observed distribution of smooth-yellow, wrinkled-yellow, smooth-green, and wrinkled-green peas does not significantly differ from the 9:3:3:1 ratio expected under Mendelian genetics. The low Chi-square value obtained indicated that we cannot reject the null hypothesis, implying that the data are consistent with the expected patterns of inheritance which supports the presence of Mendelian genetics in the snap pea traits being observed.