Question

In com, colored aleurone (R) is dominato colorless and green plant color is dan to yellow ( Two plants, each heterozygous for both characteristics are bestowed to honor gous recessives, and their progey are combined to produce the following boats colored green 100 colored yellow 97 colorless green 103 colorless yellow 100 Urechquare analysis to test these data for independent m ent of the two chances Note: The for Chi-squa values is on the last page! b. When the progeny of each of the two heterozygous plants are scored separately, the follow- ing results are obtained: PHENOTYPES PLANT1 PLANT 2 colored green colored yellow colorless green colorless yellow 92 Use chi-square analysis to test each data set for independent assortment. c. Explain why the X2 results in part A and part B differ. HINT: I have not told you ANYTHING about these two plants except that each is heterozygous. Try re-reading the textbook section "Coupling and Repulsion' in Pierce Chapter 7.

it should be something like this. please i need a detailed explaintion on the answers to understand whats going on.thanks

Answer 1

The total number of plants of each phenotype are given:

Colored green =100

Colored yellow =97

Colorless green = 103

Colorless yellow =100

Total = 400 plants

If the plants were showing independent assortment for each trait, they would segregate in the ratio 9:3:3:1

The expected number of plants in each category would be:

400/ 16 X 9 = 225

400/16 X 3 = 75

400/16 X 3 = 75

400/16 X 1 = 25

Chi square is calculated as, $\chi$ ² = (observed - expected)² / expected

When we calculate the chi square for each category, we get the following values,

Colored green = 69.44

Colored yellow = 6.45

Colorless green = 10.4

Colorless yellow = 225

The total chi square = 69.44 + 6.45 + 10.4 + 225 = 311.29

We need to compare this value with the value of $\chi$ ² given in the table, for degree of freedom = 2 (number of phenotype classes - 1 ) and confidence level = 0.05 (usually taken as standard).

The value from table = 7.815

Since the value of chi square which we obtained (311.29) is much higher than the value from the table (7.815), we reject the null hypothesis and conclude that these genes are not independently assorting.