Calculate the Chi square for cross 1 and cross 2 from the data in the attached image.Cross1:PhenotypesRatioObservedExpected(O-E)2/E Totals X2 = Cross 2:PhenotypesRatioObservedExpected(O-E)2/E Totals X2 = F2 from Cross 1:Date counted:FemalesMalesRed eyesWhite eyesNormalVestigialVestigialwingsNormalwingswingswings581312F2 from Cross 2:Date counted:Females + MalesRed eyesWhite eyesVestigialwingsNormalVestigialwingsNormalwingswings3065310

Name: Calculate the Chi square for cross 1 and cross 2 from the data in the
Uploaded: 2024-03-20T06:43:04.000Z
Channel: Bartleby
Description: Calculate the Chi square for cross 1 and cross 2 from the data in the attached image.Cross1:PhenotypesRatioObservedExpected(O-E)2/E Totals X2 = Cross 2:PhenotypesRatioObservedExpected(O-E)2/E Totals X2 = F2

The Chi-square test is a test of significance used for the comparison of experimentally obtained results with those expected theoretically and based on the hypothesis. It was proposed by Karl Pearson in 1900. It is denoted by the symbol . The formula used to calculate chi-square ()is =∑(O-E)2Ewhere O= observed frequencyE= expected frequencyCross 1 Null Hypothesis: The phenotypic ratio for the given dihybrid cross is 9:3:3:1. Total number of offspring = 58+13+12+1= 84 Expected frequencies: Red eyes wiith normal wings = 916 × 84= 47.25Red eyes with vestigial wings = 316 × 84= 15.75White eyes with normal wings = 316 × 84= 15.75White eyes with vestigial wings = 116 × 84= 5.25 Red eyes with normal wings Red eyes with vestigial wings White eyes with normal wings White eyes with vestigial wings Observed frequency (O) 58 13 12 1 Expected frequency (E) 47.25 15.75 15.75 5.25 Deviation (O-E) 10.75 -2.75 -3.75 -4.25 Thus, will be = (10.75)247.25 + (-2.75)215.75 + (-3.75)215.75 + (-4.25)25.25= 115.562547.25 + 7.562515.75 + 14.062515.75 + 18.06255.25= 2.446 + 0.48 + 0.893 + 3.44 = 7.259 Hence, chi-square () is 7.259. Since there are 4 possible phenotypes, the degree of freedom (d.f) will be 3. Conclusion: The calculated chi-square value (7.259) is less than the tabulated chi-square value (7.82) at the 5% level of probability for 3 d.f. The hypothesis is, therefore, in agreement with the recorded facts. Cross 2 Null Hypothesis: The phenotypic ratio for the given dihybrid cross is 9:3:3:1. Total number of offspring = 50+6+53+10= 119 Expected frequencies: Red eyes with normal wings = 916 × 119= 66.9375Red eyes with vestigial wings = 316 × 119= 22.3125White eyes with normal wings = 316 × 119= 22.3125White eyes with vestigial wings = 116 × 119= 7.4375 Red eyes with normal wings Red eyes with vestigial wings White eyes with normal wings White eyes with vestigial wings Observed frequency (O) 50 6 53 10 Expected frequency (E) 66.9375 22.3125 22.3125 7.4375 Deviation (O-E) -16.9375 -16.3125 30.6875 2.5625 Thus, will be =(-16.9375)266.9375 + (-16.3125)222.3125 + (30.6875)222.3125 + (2.5625)27.4375= 286.87966.9375 + 266.09822.3125 + 941.72322.3125 + 6.5667.4375= 4.286 + 11.926 + 42.206 + 0.883= 59.301 Hence, chi-square () is 59.301. Since there are 4 possible phenotypes, the degree of freedom (d.f) will be 3. Conclusion: The calculated chi-square value (59.301) is greater than the tabulated chi-square value (7.82) at the 5% level of probability for 3 d.f. The hypothesis is, therefore, not in agreement with the recorded facts.