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Summary: Flies Assortment


Flies Assortment [phpbay], 5, "", "", "", "", "", "", "", "", "", "SEKW", "1", "", "4"[/phpbay]
[phpbay]Flies Assortment, 20, "", "", "", "", "", "", "", "", "", "irfac", "2", "", "4"[/phpbay]
Flies Assortment Flies Assortment 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]

Flies Assortment


Flies Assortment [phpbay], 5, "", "", "", "", "", "", "", "", "", "SEKW", "1", "", "4"[/phpbay] [phpbay]Flies Assortment, 20, "", "", "", "", "", "", "", "", "", "irfac", "2", "", "4"[/phpbay] Flies Assortment FliesAssortment 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]
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Date Added: 12/15/2010
Date Approved: 12/15/2010
By: Anonymous
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