1) A population of wild-flowers was scored for flower color. There were 205 blue

Question

1) A population of wild-flowers was scored for flower color. There were 205 blue (BB) plants, 1988 violet (BR) plants and 866 red (RR) plants. Please round your answer correctly to 4 decimal digits. What is the frequency of the blue (B) allele?

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

3)

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.

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

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

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

7).What is the number for the degrees of freedom?

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

Could really use help in this area of Genetics. Much appreciated

Explanation / Answer

Solution 1) Here, we have a population of wildflowers, out of which 205 blue (BB) plants, 1988 Violetiolet (BR) plants and 866 red (RR) plants.

Total number of plants will be 205+1988+866 = 3059

Also, we know that these are the diploid flowers, so the total number of alleles will be 3059 x 2 = 6118

and the alleles for each type of flowers will be BB = 410 B alleles, BR = 1988 B and 1988 R alleles, and RR = 1732 R alleles.

so, the frequency of B allele would be = 410+1988 / 6118 = 2398/6118 = 0.3919

Answer 2) Similarly, frequency of R allele = 1732+1988/6118 = 0.6080

Answer 3) According to the Hardy-Weinberg law,

expected Frequency of BB = (frequency of B allele)2 = 0.3919 x 0.3919 = 0.1536

expected Frequency of BR = 2xfrequency of B allelexfrequency of R allele = 2 x 0.3919x 0.6080 = 0.4765

expected Frequency of RR = frequency of R allele x frequency of R allele = 0.6080 x 0.6080 = 0.3696

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