Question

genetics problem

In Drosophila, the allele gl+ determines wild type eyes and gl determines glassy eyes. At a separate locus c+ determines wild type body color and c determines coal body. Both loci are autosomal. The original parents were purebred with antagonistic traits. The dihybrid F1 was test crossed with the following results: Wild type eyes, coal body Glassy eyes, wild type body Glassy eyes, coal body Wild type 496 504 315 285 1600 Total You want to determine if the data support whether the genes are linked or sort independently by using the chi squared test. a. State the null hypothesis for this experiment and calculate the chi squared value. YOU MUST SHOW ALL CALCULATIONS. b. What is your conclusion? And what was the genotype of the original parents?

Answer 1

Chi-square test x = 5 O = of served value E: Expected value Take-I Degree of freedom 5% Probality 8.84 5.99 7.81 9.49 Dihybrid cross so phenotypic ratio - 9:3: 3:1 06served value wild type 285 (gl+c+) wild type eyes, coal body - 496 (getes Glassy eyes, willd type body - 504 (gl ct) Glassy eyes, cocel body - 315 (gle) total – 1600 Expected value wild type – 160089/16 wild type eyes, coul Cody - 1600 X 3/16 Glassy eyes, wild type body - 1600x3/16 Glassy eyes, coal body – 1600x416 goo = 300 = 300 = 100

(E) Expected (0) observed 285 496 504 315 Phenotype W.T. W.T.E.C.B. G.E., W.T.B. G.E., C.B. (o-E) - 615 196 goo 300 300 204 (0-E) ² 378225 38416 41616 46995 ECO-E) 2 = 504482 215 Loo {E=1600 x = 5(01-E)² СЕ = < 504487 Z 1600 - 315.30 studied characters = 4 So Degree of freedom = 401 = 3 3= 7.81 (from table 1) well Hypothesis says that there is no difference 6/" the two variables in a hypothesis The chi-square value is greater than the 5% probality value of degree of freedom Conclusion - 315.30 77.81 These data fit in genetic data ne. follows mendetan now of genetics Genotype of original parents - gltglt eact & gegloc Caorlo (o) or (a)