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Genetics/Chi Square Question: Should I accept or reject the null hypothesis. Why?

Genetics/Chi Square Question: Should I accept or reject the null hypothesis. Why?

Biology 2e
Biology 2e
2nd Edition
ISBN:9781947172517
Author:Matthew Douglas, Jung Choi, Mary Ann Clark
Publisher:Matthew Douglas, Jung Choi, Mary Ann Clark
Chapter12: Mendel's Experiments And Heredity
Section: Chapter Questions
Problem 29CTQ: Why is it more efficient to perform a test cross with a homozygous recessive donor than a homozygous...
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Genetics/Chi Square Question: Should I accept or reject the null hypothesis. Why?
II. CHI-SQUARE Transpose: x- 0.01 1. The student tested the hypothesis that a dumpy wing is inherited as a recessive trait by performing X a. What ratio did the student hypothesize? b. Did the X' analysis support the hypothesis? c. What do the data suggest about the dumpy mutation? -0.009 (2.77) = -0.005949880668 4.19 1= 0.009 (2.77)+ 0.01 4. 19 The student hypothesized that the F2 data (792:208) fit Mendel's 3:1 monohybrid ratio for recessive = -0.005949880668 + 0.01 genes. = 0.004 d? = 0.004 < 0.005 Еxpected ratio Deviation (O- E) Deviation (d) = the null hypothesis Observed (O) Expected (E) (3/4) (1/4) 792-750 = 42 (42) = 1764 208-250 = -42(-42) = 1764 792 750 1764/750 = 2.35 250 |208 1764/250 = 7.056 or 7,06 X' = 9.41 Total = 1000 P= 0.004050119332 or 0.004 dt- 2-1 = 1 6.64 0.01 9.41 10.83 0.001 10.83 – 6.64 0.001 – 0.01 9.41 - 6.64 x- 0.01 4.19 -0.009 2.77 х- 0.01
Transcribed Image Text:II. CHI-SQUARE Transpose: x- 0.01 1. The student tested the hypothesis that a dumpy wing is inherited as a recessive trait by performing X a. What ratio did the student hypothesize? b. Did the X' analysis support the hypothesis? c. What do the data suggest about the dumpy mutation? -0.009 (2.77) = -0.005949880668 4.19 1= 0.009 (2.77)+ 0.01 4. 19 The student hypothesized that the F2 data (792:208) fit Mendel's 3:1 monohybrid ratio for recessive = -0.005949880668 + 0.01 genes. = 0.004 d? = 0.004 < 0.005 Еxpected ratio Deviation (O- E) Deviation (d) = the null hypothesis Observed (O) Expected (E) (3/4) (1/4) 792-750 = 42 (42) = 1764 208-250 = -42(-42) = 1764 792 750 1764/750 = 2.35 250 |208 1764/250 = 7.056 or 7,06 X' = 9.41 Total = 1000 P= 0.004050119332 or 0.004 dt- 2-1 = 1 6.64 0.01 9.41 10.83 0.001 10.83 – 6.64 0.001 – 0.01 9.41 - 6.64 x- 0.01 4.19 -0.009 2.77 х- 0.01
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