Question

Mendelian Genetics Experiments are conducted with hybrids of two types of peas. If the offspring follow Mendel’s theory of inheritance, the seeds that are produced are yellow smooth, green smooth, yellow wrinkled, and green wrinkled, and they should occur in the ratio of 9:3:3:1, respectively. An experiment is designed to test Mendel’s theory, with the result that the offspring seeds consist of 307 that are yellow smooth, 77 that are green smooth, 98 that are yellow wrinkled, and 18 that are green wrinkled. Use a 0.05 significance level to test the claim that the results contradict Mendel’s theory.

Short answer

There is enough evidence to conclude that theobserved frequencies of the 4 types of seeds do not occur in the expected ratio as proposed by Mendel.

Thus, results contradict Mendel’s theory.

Step-by-step solution

Given information

The frequency of 4 different kinds of seeds are recorded. The expected frequencies should occur in the ratio 9:3:3:1.

Check the requirements

Assume that each experimental unit is selected randomly.

The test meets the requirements if the expected values are larger than 5.

Let O denote the observed frequencies of people of different races.

The following values are obtained:

\(\begin{aligned}{l}{O_1} = 307\\{O_2} = 77\\{O_3} = 98\\{O_4} = 18\end{aligned}\)

The sum of all observed frequencies is computed below:

\(\begin{aligned}{c}n = 307 + 77 + .... + 18\\ = 500\end{aligned}\)

Let E denote the expected frequencies.

Let the yellow smooth seed type be denoted by 1, green smooth seed type be denoted by 2, yellow wrinkled seed type be denoted by 3 and green wrinkled seed type be denoted by 4.

The expected frequencies will occur in the given ratio of 9:3:3:1.

Therefore, the expected frequencies are computed below:

\(\begin{aligned}{c}{E_1} = \frac{9}{{9 + 3 + 3 + 1}}\left( {500} \right)\\ = \frac{9}{{16}}\left( {500} \right)\\ = 281.25\end{aligned}\)

\(\begin{aligned}{c}{E_2} = \frac{3}{{9 + 3 + 3 + 1}}\left( {500} \right)\\ = \frac{3}{{16}}\left( {500} \right)\\ = 93.75\end{aligned}\)

\(\begin{aligned}{c}{E_3} = \frac{3}{{9 + 3 + 3 + 1}}\left( {500} \right)\\ = \frac{3}{{16}}\left( {500} \right)\\ = 93.75\end{aligned}\)

\[\begin{aligned}{c}{E_4} = \frac{1}{{9 + 3 + 3 + 1}}\left( {500} \right)\\ = \frac{1}{{16}}\left( {500} \right)\\ = 31.25\end{aligned}\]

Thus, the requirements are satisfied.

State the hypotheses

The hypotheses are,

\({H_0}:\)The observed frequencies of the 4 types of seeds are in the expected ratio as proposed by Mendel.

\({H_a}:\)The observed frequencies of the 4 types of seeds do not occur in the expected ratio as proposed by Mendel.

Compute the test statistic

The table below shows the necessary calculations:

Outcome	O	E	\(\left( {O - E} \right)\)	\({\left( {O - E} \right)^2}\)	\(\frac{{{{\left( {O - E} \right)}^2}}}{E}\)
Yellow Smooth	307	281.25	25.75	663.0625	2.357556
Green Smooth	77	93.75	-16.75	280.5625	2.992667
Yellow Wrinkled	98	93.75	4.25	18.0625	0.192667
Green Wrinkled	18	31.25	-13.25	175.5625	5.618

The value of the test statistic is equal to:

\(\begin{aligned}{c}{\chi ^2} = \sum {\frac{{{{\left( {O - E} \right)}^2}}}{E}} \\ = 2.357556 + 2.992667 + ...... + 5.618\\ = 11.161\end{aligned}\)

Thus,\({\chi ^2} = 11.161\).

Let k be the number of seed types, which is 4.

The degrees of freedom for\({\chi ^2}\)is computed below:

\(\begin{aligned}{c}df = k - 1\\ = 4 - 1\\ = 3\end{aligned}\)

The critical value of\({\chi ^2}\)at

\(\alpha = 0.05\)with 3 degrees of freedom is equal to 7.815.

The p-value is equal to 0.011.

Since the test statistic value is greater than the critical value and the p-value is less than 0.05, the null hypothesis is rejected.

State the conclusion

There is enough evidence to conclude that theobserved frequencies of the 4 types of seeds do not occur in the expected ratio as proposed by Mendel.

Thus, results contradict Mendel’s theory at 0.05 level of significance.