Chapter 2, Problem 8EQ

For Mendel’s data for the experiment in Figure 2.8, conduct a chi square analysis to determine if the data agree with Mendel’s law of independent assortment.

Summary Introduction

To review:

Chi-square analysis for determining Mendel’s law of independent assortment.

Introduction:

Mendel’s law of independent assortment states that when crossing takes place between individuals, the alleles get sorted into different gametes independently. Chi-square test is the test which is used to determine the difference between the observed value and the expected value. It either accepts or rejects the hypothesis proposed from the test.

Explanation of Solution

In Mendel’s analysis of two-factor crosses, he crossed two true-breeding pea plants that possess two different characters. One of the strains is round, yellow seeds (RRYY) which is dominant and the other strain is wrinkled, green seeds (rryy) which is a recessive genotype. The data obtained in Mendel’s experiment is given below as:

Parental cross	F1 generation	F2 generation
Round, yellow X wrinkled green seeds	All round, yellow	315round, yellow seeds
				108 round, green seeds
				101 wrinkled, yellow seeds
				32 wrinkled, green seeds

In the data, a total of 556 (315+108+101+32) offspring were obtained in the F₂ generation. The observed data is consistent with 9:3:3:1 ratio. The cross between the F₂ offspring is shown below.

Parents: RrYy and RrYy

Gametes:

Parent 1: RY, Ry, Ry, and Ry

Parent 2: RY, Ry, Ry, and Ry

Cross using Punnett square:

Gametes	RY	Ry	rY	ry
RY	RRYY	RRYy	RrYY	RrYy
Ry	RRyY	RRyy	RrYy	Rryy
rY	RrYY	RrYy	rrYY	rrYy
Ry	RrYy	Rryy	rrYy	rryy

The number of offspring with different phenotypes is calculated below using the probability formula,

$\text{probability}=\frac{\text{Number of favourable events}}{\text{Number of total events}}$

Thus, by using this formula, the expected number of offspring with different phenotypes is calculated below:

For round pod and yellow seed,

$\begin{array}{c}\text{number}\text{of}\text{offspring}\text{=}\frac{\text{9}}{\text{16}}\times \text{556}\\ \text{=}\text{313 }\end{array}$

For yellow pod and wrinkled seed,

$\begin{array}{c}\text{Number of offspring =}\text{556}\times \frac{\text{3}}{\text{16}}\\ \text{ =104 }\end{array}$

For green pod and round seed,

$\begin{array}{c}\text{Number of offspring =}\text{556}\times \frac{3}{16}\\ \text{ =104}\end{array}$

For green pod and wrinkled seed,

$\begin{array}{c}\text{Number of offspring =}\text{556}\times \frac{1}{16}\\ =35\end{array}$

Hence, 313 (round, yellow), 104 (wrinkled yellow), 104(round, green), and 35 (wrinkled, green) offspring are expected in each phenotype.

By using the chi-square test, the observed value and the expected value are taken in the formula which is represented as:

$\begin{array}{l}{\text{γ}}^{\text{2}}\text{= }\frac{{\left({\text{O}}_{\text{1}}{\text{-E}}_{\text{1}}\right)}^{\text{2}}}{{\text{E}}_{\text{1}}}\text{+ }\frac{{\left({\text{O}}_{\text{2}}{\text{-E}}_{\text{2}}\right)}^{\text{2}}}{{\text{E}}_{\text{2}}}\text{+ }\frac{{\left({\text{O}}_{\text{3}}{\text{-E}}_{\text{3}}\right)}^{\text{2}}}{{\text{E}}_{\text{3}}}\text{+ }\frac{{\left({\text{O}}_{\text{4}}{\text{-E}}_{\text{4}}\right)}^{\text{2}}}{{\text{E}}_{\text{4}}}\hfill \\ {\text{γ}}^{\text{2}}\text{= }\frac{{\left(\text{315-313}\right)}^{\text{2}}}{\text{313}}\text{+ }{\frac{\left(\text{108-104}\right)}{\text{104}}}^{\text{2}}\text{+ }\frac{{\left(\text{101-104}\right)}^{\text{2}}}{\text{104}}\text{+ }\frac{{\left(\text{32-35}\right)}^{\text{2}}}{35}\hfill \\ {\text{γ}}^{\text{2}}\text{= 0}\text{.012 + 0}\text{.153 + 0}\text{.086 + 0}\text{.25}\hfill \\ {\text{γ}}^{\text{2}}\text{= 0}\text{.5}\hfill \end{array}$.

Hence, thechi-square value of the observed and the expected data is 0.5.

We get a value of 0.51. Degree of freedom is n-1, where n= 3 with four categories of observations.

$\begin{array}{c}\text{degree of freedom = n-1 }\\ \text{= 3-1 }\\ \text{= 2}\end{array}$

The P value is checked in the chi-square table for the value of 0.5.If the P-value lies within the range of 0.80 and 0.95, it indicates the deviation between expected results and observed results. The value is between 80% and 95%. Thus, the hypothesis that the data agrees with Mendel’s law of independent assortment is accepted.

Conclusion

Therefore, it can be concluded that chi-square analysis ranges between 0.8 and 0.9%, so there is only a slight deviation between the compared values. Thus, chi-square analysis confirms that the null hypothesis and the results are in favor of the law of independent assortment.