Question

In fruit flies, purple eyes and ebony body are traits that display autosomal recessive patterns of inheritance. In a

Answer 1

Answer:

d. The calculated chi-square value is 10.48, and the critical value is 11.35.

Explanation:

The correct option would be option d.

Using the formula for calculating the Chi square ($X^2$):

$X^2 = \frac{(observed frequency - expected frequency)^2}{expected frequency}$

For normal eye, normal body;

$X^2 = \frac{(187 - 171)^2}{171}$ = 1.50

For normal eyes, ebony body;

$X^2 = \frac{(49 - 57)^2}{57}$ = 1.12

For purple eyes, normal body;

$X^2 = \frac{(41 - 57)^2}{57}$ = 4.50

For purple eyes, ebony body;

$X^2 = \frac{(27 - 19)^2}{19}$ = 3.37

Total $X^2$ = 1.50 + 1.12 + 4.50 + 3.37 = 10.48

Degree of freedom = 4 - 1 = 3

Critical value of $X^2$ = 11.35

Hence, the calculated chi-square value is 10.48, and the critical value is 11.35.

The correct option is d.