1 A population of wildflowers was scored for flower color Th

1. A population of wild-flowers was scored for flower color. There were 387 blue (BB) plants, 1790 violet (BR) plants and 793 red (RR) plants. Please round your answer correctly to 4 decimal digits.

(a) What is the frequency of the blue (B) allele?

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(b) What is the frequency of the red (R) allele? Please round your answer correctly to 4 decimal digits.

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(c) A chi-square test was conducted to test the hypothesis that the population genotypic frequencies are consistent with Hardy-Weinberg equilibrium frequencies.

Fill in the spaces below with the expected numbers for each phenotypic (genotypic) class in the chi-square test.

What is the expected number of blue (BB) plants? Please round your answer correctly to 4 decimal digits.

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(d) What is the expected number of violet (BR) plants? Please round your answer correctly to 4 decimal digits.

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(e) What is the expected number of red(RR) plants? Please round your answer correctly to 4 decimal digits.

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(f) Keeping all decimal digits during your calculations and rounding your answer to 3 decimal digits, what is the calculated chi-square value?

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(g) What is the number for the degrees of freedom?

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(h) What is the critical value? Round to two decimal places.

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(I)

What is your conclusion from the chi-square test?

Select one: Fail to reject Ho or Reject Ho

Solution

Answer:

1. a. B (Blue) = 387 + (1/2 * 1790) / 2970 = 1282 / 2970 = 0.4316

b. R (Red) = 793 + (1/2 * 1790) / 2970 = 0.5684

c. Expected frequency of BB plants = p² = 0.4316 * 0.4316 = 0.1862

So, expected number of blue plants = 0.1862 * 2970 = 553.247 = 553 plants

d. Expected frequency of BR plants = 2pq = 2 * 0.4316 * 0.5683 = 0.4905

So, expected number of violet plants = 0.4905 * 2970 = 1456.9 = 1457 plants

e. Expected frequency of RR plants = q² = 0.5684 * 0.5684 = 0.3230

So, expected number of red plants = 0.323 * 2970 = 959.54 = 960 plants