Question

To assess Mendel's law of segregation using tomatoes, a true- breeding tall variety (SS) is crossed with a true-breeding short variety \((s s) .\) The heterozygous tall plants \((S s)\) were crossed to produce the two sets of \(\mathrm{F}_{2}\) data as follows: $$\begin{array}{cc} \text { Set I } & \text { Set II } \\ 30 \text { tall } & 300 \text { tall } \\ 5 \text { short } & 50 \text { short } \end{array}$$ (a) Using chi-square analysis, analyze the results for both datasets. Calculate \(\chi^{2}\) values, and estimate the \(p\) values in both cases. (b) From the analysis in part (a), what can you conclude about the importance of generating large datasets in experimental settings?

Short answer

Answer: Large datasets are important in experimental settings as they provide stronger statistical evidence. In the case of smaller dataset (Set I), we could not reject Mendel's law of segregation while with the larger dataset (Set II), we had strong evidence to reject it. Poor results from smaller datasets can lead researchers to incorrect conclusions, thus underlining the importance of generating large datasets in experimental settings.

Step-by-step solution

Writing Down Observed Frequencies

Firstly, we need to write down the observed frequencies (O) from the data provided for set I and set II. Set I: Tall plants observed frequency (O_tall Set I) = 30 Short plants observed frequency (O_short Set I) = 5 Set II: Tall plants observed frequency (O_tall Set II) = 300 Short plants observed frequency (O_short Set II) = 50

Calculate Expected Frequencies

Now, we need to calculate the expected frequencies (E) for both datasets. The expected ratio of tall to short plants according to Mendel's law of segregation is 3:1. Total plants in Set I = 30 + 5 = 35 Expected frequencies in Set I: E_tall Set I = (3/4) * 35 = 26.25 E_short Set I = (1/4) * 35 = 8.75 Total plants in Set II = 300 + 50 = 350 Expected frequencies in Set II: E_tall Set II = (3/4) * 350 = 262.50 E_short Set II = (1/4) * 350 = 87.50

Calculate Chi-square Values

Next, we calculate the Chi-square values for both datasets using the formula: \(\chi^2 = \Sigma\frac{(O-E)^2}{E}\) Set I: \(\chi^2_{Set I} = \frac{(30-26.25)^2}{26.25} + \frac{(5-8.75)^2}{8.75} \approx 0.535 + 1.607 = 2.142\) Set II: \(\chi^2_{Set II} = \frac{(300-262.50)^2}{262.50} + \frac{(50-87.50)^2}{87.50} \approx 5.350 + 20.286 = 25.636\)

Estimating P-values

Now, we estimate the p-values using the calculated Chi-square values and 1 degree of freedom (df): Set I: p-value for \(\chi^2_{Set I} = 2.142\) with 1 df is approximately 0.15. Set II: p-value for \(\chi^2_{Set II} = 25.636\) with 1 df is approximately <0.001.

Conclusions

(a) For Set I, the p-value is 0.15 which indicates that there is not enough evidence to reject Mendel's law of segregation. For Set II, the p-value is <0.001 which indicates that there is strong evidence to reject Mendel's law of segregation. (b) From the analysis in part (a), we can conclude that large datasets are important in experimental settings as they provide stronger statistical evidence. In the case of smaller dataset (Set I), we could not reject Mendel's law of segregation while with the larger dataset (Set II), we had strong evidence to reject it. Poor results from smaller datasets can lead researchers to incorrect conclusions, thus underlining the importance of generating large datasets in experimental settings.