Question

Name Once the data for experiment # 3 has been generated, analyze it. At this point, you already know which of the aldehyde oxidase alleles is dominant and you have six predicted outcomes of crosses. Using this information and the spot assay results, hypothesize as to the actual genotypes of the P generation of these progeny based on YOUR results. (Hints: Look at the ratios of aldehyde oxidase positive test results to negative test results. Compare them to the ratios you predicted) Based on all observations, hypothesize the genotypes of the parent flies, making sure to name the alleles correctly Name Name Exercise # 3 Determining Parental Genotypes Using the Evidence from Progeny: For this experiment, you will be predicting the genotypes of the Parental generation, based upon the data generated from a spot assay test of the progeny. Perform the spot assay as directed. Complete table 3 below with the results Table 3: Results of Assay Tests on Progeny of Unknown Parents AO Activity (+/-) AO Activity (+/-) Fly No. Fly No. 13 + 14 15 posikve 1 Ngatire 3 15 4 16 + 5 17 6 18 + 7 19 + 8 20 9 + 21 10 22 + 11 23 12 24 Positive control results (fly 1 a) wide composit J 4

Answer 1

Null hypothesis: Half of the parents has Ao Ao genotype which produce aldehyde oxidase and half of the parents has ao ao genotype.

Now this is tested by chi square test:

(0-EYIE (0-E Deviation Expected (E) Observed Category/ Event (0-E) (o) 0.75 0.75 1.5 AO 12 15 Present -3 AO Absent 12 24 24 Totals Chi Square Analysis Deviation is due to some other variable significant deviation; hypothesis is rejected Deviation is dueto chance, nonsignificant deviation evidence supports hypothesis of the y Cal va Table 1.5 Thee umbers pr am A sa sar a rn d re ri ve Ust UIP avistion () ts mess

Now, chi square from table at degree of freedom 1 and at 0.05 is 3.841.

Thats why chi square experimental is less than chi square from table. That's why our null hypothesis is accepted. There is deviation occurs due to chance only