Chapter 2, Problem 16EQ

E16. A cross was made between a plant that has blue flowers and purpleseeds to a plant with white flowers and green seeds. The following data were obtained:

${\text{F}}_1$ generation: All offspring have blue flowers with purple seeds

${\text{F}}_2$ generation: $103$ blue flowers, purple seeds

$49$ blue flowers, green seeds

$44$ white flowers, purple seeds

$\underset{\_}{\text{104}}$ white flowers, green seeds

Total: $300$

Start with the hypothesis that blue flowers and purple seeds aredominant traits and that the two genes assort independently. Calculate a chi square value. What does this value mean with regardto your hypothesis? If you decide to reject your hypothesis, whichaspect of the hypothesis do you think is incorrect (i.e., the ideathat blue flowers and purple seeds are dominant traits, or the ideathat the two genes assort independently)?

Summary Introduction

To analyze:

A cross between a plant with blue flower and purple seeds with a plant with white flower and green seeds resulted in the following gametes-

$\begin{array}{cc}\text{F1}& \text{100\%}\text{offsprings with}\text{blue flower and purple seeds}\\ \text{F2}& \begin{array}{c}\text{103}\text{blue flower and }\text{purple seeds}\\ \text{49}\text{blue flower and }\text{green seeds}\\ \text{44}\text{white flower and }\text{purple seeds}\\ \text{104}\text{white flower}\text{and }\text{green seeds}\end{array}\\ \text{Total}& \text{300}\end{array}$

Consider the hypothesis that blue flowers and purple seeds are dominant traits, and these two genes assort independently.

From the given information, calculate the chi-square value.

What does the value obtain depict regarding the hypothesis?

Question asked- If you decide to reject your hypothesis, which aspect of the hypothesis do you think is incorrect (i.e., the idea that blue flowers and purple seeds are dominant traits, or the idea that two genes assort independently)?

Introduction:

In genetic crosses, to know whether the data is significant or whether it fits in any of the Mendelian ratios, the Chi-square test $\text{2}\left(\text{χ}\right)$ is performed. When a genetic cross is performed, the genotypes are given, and the phenotypic ratio is predicted by a hypothesis. Chi-square test $\text{2}\left(\text{χ}\right)$ allows determining the accuracy of the prediction or hypothesis on observed and expected values or ratios. For Mendelian genetic crosses, Chi-square $\text{2}\left(\text{χ}\right)$ Goodness of Fit (one sample) test is performed.

The test is performed in a series of steps-

$\text{1}$. Hypotheses (Null and Alternative), $\text{2}$. Calculating chi-square test statistic, $\text{3}$. Determine the Degree of Freedom df and p-value, $4$. The decision about the null hypothesis, $5$. Conclusion.

Explanation of Solution

From the given information, the cross is made between two plants that resulted in the following data:

$\begin{array}{cc}\text{F1}& \text{100\%}\text{offsprings with}\text{blue flower and purple seeds}\\ \text{F2}& \begin{array}{c}\text{103}\text{blue flower and }\text{purple seeds}\\ \text{49}\text{blue flower and }\text{green seeds}\\ \text{44}\text{white flower and }\text{purple seeds}\\ \text{104}\text{white flower}\text{and }\text{green seeds}\end{array}\\ \text{Total}& \text{300}\end{array}$

The hypothesis is that purple seeds and blue flowers are dominant traits, and they assort independently.

In a dihybrid cross, if two heterozygous genes are crossed, the expected phenotypic ratio generally results with the $\text{9: 3: 3: 1}$ ratio.

$\begin{array}{cc}\text{9}& \text{blue flower and }\text{purple seeds}\\ \text{3}& \text{blue flower and }\text{green seeds}\\ \text{3}& \text{white flower and }\text{purple seeds}\\ \text{1}& \text{white flower}\text{and }\text{green seeds}\end{array}$

${\text{H}}_{\text{0}}\text{:}$= The two genes are independently assorted.

${\text{H}}_a\text{:}$= The two genes do not show independent assortment.

$\begin{array}{ccccc}\text{Offsprings}& \text{Observed (O)}& \text{Expected (E)}& {\text{(O-E)}}^{\text{2}}& \frac{{\text{(O-E)}}^{\text{2}}}{\text{E}}\\ \text{blue flower and }\text{purple seeds}& \text{103}& \left(\frac{9}{16}\times 300\right)=168.75& \text{4323}& \text{25}\text{.61}\\ \text{blue flower and }\text{green seeds}& \text{49}& \left(\frac{3}{16}\times 300\right)=56.25& \text{52}\text{.56}& \text{0}\text{.934}\\ \text{white flower and }\text{purple seeds}& \text{44}& \left(\frac{3}{16}\times 300\right)=56.25& \text{150}& \text{2}\text{.66}\\ \text{white flower}\text{and }\text{green seeds}& \text{104}& \left(\frac{1}{16}\times 300\right)=18.75& \text{7267}& \text{387}\text{.6}\\ Total& 300& & & 416.81\end{array}$

So, the test statistics ${\chi }^2=416.81$.

The degree of freedom is the number of phenotypes minus $\text{1}$.

$\text{4}\text{phenotypes-}\text{1}\text{=}\text{3}$.

For $\text{α=0}\text{.05}$ the critical value is $\text{7}\text{.82}$.

The estimated value is much greater than expected, hence, the hypothesis is rejected.

Hence, ${\text{H}}_{\text{0}}$ is rejected as the hypothesis does not explain it.

These genes may be found on the same chromosome and do not assort independently.

Conclusion

All the aspects are verified.