Phenot (O-E) (O-E(O-E/E 1.32 AB/ 121 144 109 107 | 101 aaB 115 109 12. - A b 0.0

Question

Phenot (O-E) (O-E(O-E/E 1.32 AB/ 121 144 109 107 | 101 aaB 115 109 12. - A b 0.037 30% (04 101 C. aab 93 2.35 Totals 430 436 Instructor's Calculation of X2-4.0367 a. In interpreting this X2 value, you have 3 degrees of freedom b. In this case do you accept/rejeet the hypothesis that these data approximate a dihybrid test cross ratio with independent assortment? Accept c. What is the probability that the deviations are due to chance alone? With this background in the meaning, calculation, and interpretation of X2, you can apply the test to some data you have collected yourself. B. Assigned Problems (5 pts)-Application of Chi-Square to New Data 1. Record your observed and expected data from your fly monohybrid cross or Dihybrid cross. Complete and calculate X2 (O-E (O-EVE Phenotyp Wild-type Mutant Totals (O-E) (classes) Observed ExpectedDeviations

Explanation / Answer

Monohybrid cross

A- Degree of freedom = number of alleles - 1

hence We have one degree of freedom

B- 1.64

C- p-20

D- Calculated value of chi-square shows that our cross follow the Mendelian inheritance in case of monohybrid cross

Dihybrid cross

A- We have three degrees of freedom

B- 2.37

C- p - 0.50

D- Calculated value of chi-square shows that our cross follow the Mendelian independent inheritance in case of Dihybrid cross

Observed Expected O-E (O-E)2 (O-E)2/E wild type 160 165 -5 25 0.151515 Mutant 60 55 5 25 0.454545 220 0.606061

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