Question

A population of wild-flowers was scored for flower color. There were 302 blue (BB) plants, 1857 violet (BR) plants and 811 red (RR) plants. Please round your answer correctly to 4 decimal digits. What is the frequency of the blue (B) allele? What is the frequency of the red (R) allele? Please round your answer correctly to 4 decimal digits. 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.

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

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

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

What is the number for the degrees of freedom?

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

What is your conclusion from the chi-square test?

Answer 1

Answer:

Explanation:

Hardy and Weinberg described all the possible genotypes for a gene with two alleles. The binomial expansion representing this is, p2 + 2pq + q2 = 1.0

Where,

p2 = proportion of homozygous dominant individuals (BB) = 313

q2 = proportion of homozygous recessive individuals (RR) = 857

2pq = proportion of heterozygotes (BR) = 1820

The proportion of BB individuals in the population is = 313/2990 = 0.1046

The proportion of BR individuals in the population is = 1820/2990 = 0.6086

The proportion of RR individuals in the population is = 2/82 = 0.2866

I).

a). The frequency of p allele in the population is, (B) = p2 + 1/2(2pq) = 0.1046 + ½ (0.6086) = 0.4089

b). The frequency of q allele in the population is, (R) = q2 + 1/2(2pq) = 0.2866 + ½ (0.6086) = 0.5909.

II).

The expected number of individuals with the BB genotype = 0.4089* 0.4089 = 0.1671*2990 = 499.6 = 500

The expected number of individuals with the BR genotype = 2*0.4089* 0.5909 = 0.4832*2990 = 1444.768 = 1445

The expected number of individuals with the RR genotype = 0.5909*0.5909 = 0.3491*2990 = 1043.809 = 1044

CHI - SQUARE (X2):

X2 = Σ(O - E)2 / E

Where O = Observed frequency

E = Expected frequency

Phenotype    O              E          (O-E)         (O-E)^2             (O-E)^2/E

BB                313           500      -187            34969             69.938

BR              1820         1445      375           140625           97.31834

RR              857           1044       1.5            2.25                0.155172

                2990       2989      189.5                                 167.4115

The calculated Chi-square value is = 167.4115

Degrees of freedom is = n-1 = 3-1 = 2

The P-value is < 0.00001, which is significant at p < 0.05. So, we reject the null hypothesis.

Conclusion: There is a significant difference between the observed and expected values.