Test the cross for independent assortment? PLEASE HELP!?
A cross was made to produce D. melanogaster flies heterozygous for two pairs of alleles; dp+ and dp, which determines long versus short wings, and e+ and e, which determine gray vs. ebony body color. The following F2 data were obtained: Long wing, gray body 462 Long wing, ebony body167 Short wing, gray body127 Short wing, ebony body44 Test these data for agreement with the 9:3:3:1 ratio expected if the two pairs of alleles segregate independently.
Ho: The data agrees Ha: The data doesn't agree Significance level = α = 0.05 ______________________________________ First Convert the ratios in fractional proportions. 9 + 3 + 3 + 1 = 16 So you are testing for the following agreement. 1) (9/16) 2) (3/16) 3) (3/16) 4) (1/16) ================ Observed: 1)462 2)167 3)127 4)44 Total: 800 Expected: (9/16)*800 = 450 (3/16)*800 = 150 (3/16)*800 = 150 (1/16)*800 = 50 Test Statistic Chi-Square Goodness-of-Fit Test X² = ∑(observed - expected)² / expected X² = [(462-450)² / 450] + [(167-150)² / 150] + [(127-150)² / 150] + [(44-50)² / 50] X² = 6.493 degrees of freedom = 4 - 1 = 3 P-Value = 0.0898 We fail to reject Ho. It seems the data does agree with the given ratios.
May Bass Fly Assortments [affmage source="ebay" results="20"]Flies Assortment[/affmage]
Test the cross for
independent assortment? PLEASE HELP!?
A cross was made to produce D. melanogaster flies heterozygous for two pairs of alleles; dp+ and dp, which determines long versus short wings, and e+ and e, which determine gray vs. ebony body color. The following F2 data were obtained: Long wing, gray body 462 Long wing, ebony body167 Short wing, gray body127 Short wing, ebony body44 Test these data for agreement with the 9:3:3:1 ratio expected if the two pairs of alleles segregate independently.
Ho: The data agrees Ha: The data doesn't agree Significance level = α = 0.05 ______________________________________ First Convert the ratios in fractional proportions. 9 + 3 + 3 + 1 = 16 So you are testing for the following agreement. 1) (9/16) 2) (3/16) 3) (3/16) 4) (1/16) ================ Observed: 1)462 2)167 3)127 4)44 Total: 800 Expected: (9/16)*800 = 450 (3/16)*800 = 150 (3/16)*800 = 150 (1/16)*800 = 50 Test Statistic Chi-Square Goodness-of-Fit Test X² = ∑(observed - expected)² / expected X² = [(462-450)² / 450] + [(167-150)² / 150] + [(127-150)² / 150] + [(44-50)² / 50] X² = 6.493 degrees of freedom = 4 - 1 = 3 P-Value = 0.0898 We fail to reject Ho. It seems the data does agree with the given ratios.
May Bass Fly Assortments [affmage source="ebay" results="20"]Flies Assortment[/affmage]