2. In a dihybrid testcross, the following results were obtained: A-B-, 121: A-bb

Question

2. In a dihybrid testcross, the following results were obtained: A-B-, 121: A-bb, 107, aaB-,115 and aabb, 93. Do these data approximate a dihybrid testcross ratio with independent assonment? Complete the following table and answer the related questions. 34 The Chi-Square Test Phenotype A-B A-bb O -E st- 121 107 115 93 aaB- aahb Totals Instructor's calculation of x*40367 a In interpreting this z' value, you have b. In this case, do you accept/reject the hypothesis that these data approximate a dihybrid test- degrees of freedom. cross ratio with independent assortmen C. What is the probability that the deviations are due to chance alone?

Explanation / Answer

Given below is is the phenotypic fequencies of the observed inside the table.

The expected ratio will be 9:3:3:1 theoritical for a dihybrid cross , The fractional ratios for these four phenotypes are 9/16.3/16,3/16 and 1/16.

To calculate the expected number , multiply the number of each grain phenotype by the expected fractional ratio of the grain phenotype.

expected (E) for

A-B- = 436*9/16 = 245.25

A-bb= 436*3/16 = 81.75

aaB- = 436*3/16 = 81.75

aabb = 436*1/16 = 27.25

A) In Interpreting this X2 = 117.019, You have 4-1 = 3 degree of freedom .( Degree of freedom = number of phenotypes -1) so Df is 3

B ) In this case , do uou accept/reject the hypothesis that these data approximate a dihybrid test cross with independent Assortment. Rejected

C) What is the proablity that the deviations are due to chance alone 0.05

Phenotypes Observed (O) Expected(E) (O-E) (O-E)2 (O-E)2 /E A-B- 121 245.25 -124.25 15438.0625 62.948 A-bb 107 81.75 25.25 637.5625 7.79892 aaB- 115 81.75 33.25 1105.5626 13.5237 aabb 93 27.25 65.75 4323.0625 158.6444 Total 436 436 117.019

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