Question

Please provide details for this problem. I got 3.24; therefore accept the null hypothesis.

The trait for round peas (R) is dominant to wrinkled peas (r), and yellow peas (Y) is dominant to green peas (y). A dyhbrid cross between two heterozygous pea plants is performed (RrYy x RrYy). The following results are observed:
104 round yellow
27 round green
27 wrinkled yellow
2 wrinkled green
What is the chi-square value for this experiment to two decimal places?
For the chi-square calculated, do you accept the null hypothesis? Yes or No

Answer 1

See the answer below

Step-by-step explanation:

The Chi-square (
`X^2`) is calculated by the formula:

`X^2` =
`((O - E)^2)/(E)` where O = observed frequency and E = expected frequency

A standard dihybrid cross is expected to give 9:3:3:1 phenotypic ratio.

Phenotype O E
`X^2`

Round yellow 104 9/16 x 160 = 90
`((104 - 90)^2)/(90)` = 2.18

Round green 27 3/16 x 160 = 30
`((27 - 30)^2)/(30)` = 0.30

wringled yellow 27 3/16 x 160 = 30
`((27 - 30)^2)/(30)` = 0.30

wrinkled green 2 1/16 x 160 = 10
`((2 - 10)^2)/(10)` = 6.40

Total
`X^2` value to two decimal places = 9.18

Degree of freedom = (n-1) = 4 -1 = 3

Critical
`X^2` value at 95% probabillity level = 7.815

Since the calculated
`X^2` si more than the critical
`X^2` value, the null hypothesis is rejected.