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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 :

Oseni

Answer:

See the answer below

Explanation:

The Chi-square ([tex]X^2[/tex]) is calculated by the formula:

[tex]X^2[/tex] = [tex]\frac{(O - E)^2}{E}[/tex] 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                                  [tex]X^2[/tex]

Round yellow          104       9/16 x 160 = 90          [tex]\frac{(104 - 90)^2}{90}[/tex] = 2.18

Round green            27        3/16 x 160 = 30           [tex]\frac{(27 - 30)^2}{30}[/tex] = 0.30

wringled yellow         27        3/16 x 160 = 30           [tex]\frac{(27 - 30)^2}{30}[/tex] = 0.30

wrinkled green          2          1/16 x 160 = 10              [tex]\frac{(2 - 10)^2}{10}[/tex] = 6.40

Total  [tex]X^2[/tex] value to two decimal places = 9.18

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

Critical  [tex]X^2[/tex] value at 95% probabillity level = 7.815

Since the calculated  [tex]X^2[/tex] si more than the critical  [tex]X^2[/tex] value, the null hypothesis is rejected.

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