Chapter 6, Problem 14EQ

A trait in garden peas involves the curling of leaves. A two-factor cross was made by crossing a plant with yellow pods and curling leaves to a wild-type plant with green pods and normal leaves. All ${\text{F}}_{\text{1}}$ offspring had green pods and normal leaves. The ${\text{F}}_{\text{1}}$ plants were then crossed to plants with yellow pods and curling leaves. The following results were obtained:  

117 green pods, normal leaves  

115 yellow pods, curling leaves   

78 green pods, curling leaves   

80 yellow pods, normal leaves

A. Conduct a chi square analysis to determine if these two genes are linked.

B. If they are linked, calculate the map distance between the two genes. How accurate do you think this calculated distance is?

Summary Introduction

To review:

A. A chi-square analysis to determine the linking of genes responsible for pod color and leaf style.

B. The map distance between the two genes if they are linked, and mention whether the map distance is accurate or not.

Introduction:

Chi-square analysis is a method extensively used for evaluating the goodness of fit among theexpected hypothesis and then experimental data observed. Likewise, this method is used for determining a two-factor cross outcome which is constant to either independent assortment of genes or the linkage.

Explanation of Solution

A. Chi-square analysis is calculated by the following steps:

Step 1: Proposing a hypothesis.

Mendel’s ratio, which is 1:1:1:1 ratio among the four phenotypes. This is not in accordance with the observed data, so it can be proposed that the 2 genes responsible for the pod color, leaf style obeys law of independent assortment given by Mendel. Hence, from the hypothesis, expected values can be calculated. Now, as the data conflicts along with this proposed hypothesis, it is anticipated that a chi-square analysis can rejectsucha hypothesis favoring the hypothesis that supports the linkage of genes. It can be assumed that the alleles show the segregation, and all the 4 phenotypes can be considered as equally viable.

Step 2: On the basis of hypothesis, one cancalculate the expected value of each 4 phenotypes.As each of the phenotypes have an equal probability of occurrence, the probability of each phenotype, then will be $\frac{\text{ 1 }}{\text{4}}$. The observed value of F₂ generation had a total of $390\left(117+115+78+80\right)$ individuals. Post this, the expected number of offspring with each phenotype is calculated:

$\begin{array}{c}{\text{P }}_{\text{expected phenotype}}\text{=}\frac{\text{ 1 }}{\text{4}}\text{x 390 }\\ \text{= 98}\end{array}$

Hence, the expected phenotype of each offspring is 98.

Step 3: Apply the chi-square formula.

Here, we take into consideration the previously calculated, observed (O) and expected values (E) (from Step 2). In this problem, the data contains 4 phenotypes.

$\begin{array}{l}{\text{γ}}^{\text{2}}\text{= }\frac{{\left({\text{O}}_{\text{1}}{\text{-E}}_{\text{1}}\right)}^{\text{2}}}{{\text{E}}_{\text{1}}}\text{+ }\frac{{\left({\text{O}}_{\text{2}}{\text{-E}}_{\text{2}}\right)}^{\text{2}}}{{\text{E}}_{\text{2}}}\text{+ }\frac{{\left({\text{O}}_{\text{3}}{\text{-E}}_{\text{3}}\right)}^{\text{2}}}{{\text{E}}_{\text{3}}}\text{+ }\frac{{\left({\text{O}}_{\text{4}}{\text{-E}}_{\text{4}}\right)}^{\text{2}}}{{\text{E}}_{\text{4}}}\hfill \\ {\text{γ}}^{\text{2}}\text{= }\frac{{\left(\text{117-98}\right)}^{\text{2}}}{98}\text{+ }{\frac{\left(\text{115-98}\right)}{98}}^{\text{2}}\text{+ }\frac{{\left(\text{78-98}\right)}^{\text{2}}}{98}\text{+ }\frac{{\left(\text{80-98}\right)}^{\text{2}}}{98}\hfill \\ {\text{γ}}^{\text{2}}\text{= 3}\text{.68 + 2}\text{.95 + 4}\text{.08 + 3}\text{.31}\hfill \\ {\text{γ}}^{\text{2}}\text{= 14}\text{.02}\hfill \end{array}$

Hence, the chi-square value is 14.02.

Step 4 (Interpreting the calculated chi-square value):

The interpretation is done with the help of a chi-square table, which is in turn dependent on the value of the degree of freedom.

According to the hypothesis, the law of segregation as well as the law of independent assortment determine the four phenotypes. Therefore, the two categories (n=2), which are recombinant, and non-recombinant can only be predicted by thelaw of independent assortment. Hence, the degree of freedom, based on this hypothesis of independent assortment, will be n-1.Therefore, the degree of freedom is calculated as:

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

The chi-square value is 14.02, which is a huge value (it was supposed to be is 7.81). This indicates a very large deviation between the observed values and the expected values. With the 1 degree of freedom, there is a large deviation that occurs by chance alone. Therefore, the hypothesis is rejectedas the two genes are hypothesized to assort independently and to acceptan alternative hypothesis which states that the genes are linked.

B. To calculate the map distance between the two genes, we need to use the formula of map distance shown below:

$\text{Map distance = }\frac{\text{Number of recombinant offspring}}{\text{ Total number of offspring}}\text{ x 100 }$

Recombinant offspring are the ones which have the least number of the offspring, 78 and 80 respectively. The non-recombinant/parental-offspring are the ones which have the highest number of offspring, 117 and 115 respectively. Map distance is now calculated as:

$\begin{array}{c}\text{map distance}\text{=}\frac{\text{78 + 80}}{\text{390 }}\text{X100}\\ \text{=}\frac{\text{158}}{\text{390}}\text{X100}\\ \text{=}\text{40}\text{.51mu}\end{array}$

As the distance between the genes is less than 50mu, this distance can be considered as the true distance between the genes.

Conclusion

Therefore, it can be concluded that chi-square analysis helps in identifying which of the two genes are assorting independently and which are linked to each other.The results are:

A. The hypothesisof independent assortment of the gene is rejectedas the chi-square’s value isover 7.815 whichmeans that 2 genes shows linkage pattern.

B. The distance between the loci of the two genes is 40.51 map unit, which is close to 50mu, so it is relativelyaccurate.