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?

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.

1: most common garden plants prefer loam — soils with a balance of different-sized mineral particles (approximately 40% sand, 40% silt, and 20% clay) with ample organic matter and pore space. however, some plants grow better in sandy soils, while others are well-adapted to clay soils.

2: available water capacity is the maximum amount of plant available water a soil can provide. it is an indicator of a soil's ability to retain water and make it sufficiently available for plant use. available water capacity is the water held in soil between its field capacity and permanent wilting point.

3: it is the same for all living things (plants and animals). it also includes very tiny organisms that our eyes cannot see. they are known as microorganisms, and include viruses, bacteria and fungi.

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