Question

3. A trait in garden peas involves the curling of leaves. A dihydrid cross was made involving a plant with yellow pods (yy) and curling leaves (cc) to a wild-type plant with green pods (YY) and normal leaves (CC). All the F, offspring had green pods and normal leaves. The F, plants were then self-fertilized, the phenotypes and number of offspring of each phenotype are indicated in the following, 220 green pods, normal leaves 19 yellow pods, curling leaves 61 green pods, curling leaves 60 yellow.pods normal leaves. 3a. Conduct a Chi-square analysis to see if these two genes are linked, X (p 0.05, df 3)-7.82 (Hint: if these two genes are not linked, it will follow the law of independent assortment)

Answer 1

As per the question,

Observed number of offspring with green pods and normal leaves(O1) = 220

Observed number of offspring with green pods and curling leaves (O2) = 61

Observed number of offspring with yellow pods and normal leaves (O3) = 60

Observed number of offspring with yellow pods and curling leaves (O4) = 19

Total number of offspring= 220+61+60+19

Total number of offspring= 360

In order to conduct a Chi square analysis to see if two genes are linked, we propose a hypothesis that two genes are not linked and they follow independent assortment.

On the basis of the hypothesis that two genes are not linked and they follow independent assortment, we can calculate expected number of offspring of each phenotype as given below,

$Expected number of offsprings with green pods and normal leaves(E_{1}) = Expected ratio from independent assortment of two genes \times Total offsprings$

$Expected number of offsprings with green pods and normal leaves(E_{1}) = \frac{9}{16} \times 360$

Expected number of offspring with green pods and normal leaves(E1) = 202.5

Similarly, we can calculate expected number of offspring with green pods and curling leaves (E2), expected number of offspring with yellow pods and normal leaves (E3), expected number of offspring with yellow pods and curling leaves (E4). While calculating number of offspring with green pods and curling leaves (E2), we will use 3/16 as expected ratio from independent assortment of two genes. Expected ratio from independent assortment of two genes for calculating  expected number of offspring with yellow pods and normal leaves (E3) will also be 3/16. While calculating number of offspring with yellow pods and curling leaves (E4), we will use 1/16 as expected ratio from independent assortment of two genes.

Expected number of offspring with green pods and normal leaves(E1) = 202.5

Expected number of offspring with green pods and curling leaves (E2) = 67.5

Expected number of offspring with yellow pods and normal leaves (E3) = 67.5

Expected number of offspring with yellow pods and curling leaves (E4) = 22.5

Now, we will apply Chi-square formula.

$\chi^{2} = \sum \frac{(observed number of offsprings- expected number of offsprings)}{expected number of offsprings}^{2}$

$\chi^{2} = \frac{(O_{1}- E_{1})}{E_{1}}^{2} + \frac{(O_{2}- E_{2})}{E_{2}}^{2} + \frac{(O_{3}- E_{3})}{E_{3}}^{2} + \frac{(O_{4}- E_{4})}{E_{4}}^{2}$

$\chi^{2} = \frac{(220 - 202.5)}{202.5}^{2} + \frac{(61- 67.5)}{67.5}^{2} + \frac{(61- 67.5)}{67.5}^{2} + \frac{(19- 22.5)}{22.5}^{2}$

$\chi^{2} = 1.5 + 0.6+ 0.6 + 0.5$

$\chi^{2} = 3.2$

Degree of freedom is 3 and for chi square value of 3.2, probability (p) value is 0.20. This is less than p value of 0.05. So, chi square analysis shows that our hypothesis that two genes are not linked is rejected.This shows that two genes are linked and do not follow the law of independent assortment.