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Dihybrid Corn Lab Report

# Dihybrid Corn Lab Report - = 173(expected purple smooth 2...

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Sharron K. Smith January 30, 2008 Wednesday 10am Lab Biology Lab 1240-09 Special Assignment Dihybrid Corn Chi-Square Analysis 1. What is a chi-square? (O - E) X 2 = E It is a statistical test It helps to answer the question: “Is the observed data a satisfactory approximation of the expected data?” It helps one to decide whether to accept or reject a hypothesis 2. The Big Question: How can we be more certain that the observed phenotypes of the F2 dihybrid corn agree with the expected 9:3:3:1 phenotypic ratio? 3. A. Number of Kernels with Phenotypes Phenotypes Observed (O) 1. Purple Smooth (9/16) 163 2. Purple Shriveled (3/16) 73 3. Yellow Smooth (3/16) 53 4. Yellow Shriveled (1/16) 18 307(Total) 3. B. Determining the Expected Values 1. 307 x 9/16

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Unformatted text preview: = 173 (expected purple smooth) 2. 307 x 3/16 = 56 (expected purple shriveled) 3. 307 x 3/16 = 56 (expected yellow smooth) 4. 307 x 1/16 = 19 (expected yellow shriveled) 3. C. Chi-Square Table (O - E) 2 Phenotype Observed Expected (O - E) (O - E) 2 E Purple Smooth 163 173 163-173=-10 2 = 1/173=-10 100 0.00578 Purple Shriveled 73 56 73-56=17 17 2 = 289/56= 289 5.16 Yellow Smooth 53 56 53-56= -3-3 2 = 9/56= 9 0.161 Yellow Shriveled 18 19 18-19= -1-1 2= 1/19= 1 0.0526 3. D. Sigma Value 0.00578 + 5.16 + 0.161 + 0.0526 = 5.38 3. E. Degrees of Freedom Calculation d.f. = 4-1 = 3 3. F. Interpretation of Summed X 2 Value Probability of obtaining an X 2 value as large or larger: 7.82...
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Dihybrid Corn Lab Report - = 173(expected purple smooth 2...

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