Question

Prairie lupine, Lupinus lepidus, grows throughout Oregon's High Desert country. Most produce blue flowers, but occasionally, plants with white flowers may be found. Color is controlled by a single gene that shows incomplete dominance such that one can distinguish three distinct phenotypes: dark blue (BB), white (bb) and pale blue (Bb). Genotypes BB Bb bb Observed genotypes 570 260 170 If this population were in Hardy-Weinberg Equilibrium, what proportion of the population would you expect to be dark blue [1]? Pale blue [2]? White [3]? Compare your values for observed and expected genotype numbers across all genotypes. Does this population appear to be in Hardy-Weinberg equilibrium? Explain the implication of your answer for full credit. [4]

Answer 1

Observed Frequencies:

Dark Blue (BB) Flowers = 570
Pale Blue (Bb) Flowers = 260
White (bb) Flowers = 170

Observed Frequency of BB individuals = 570/1000 = 0.57
Observed Frequency of Bb individuals = 260/1000 = 0.26
Observed Frequency of bb individuals = 170/1000 = 0.17

Total Number of Individuals = 1000

The allele frequency for the Dark Blue (B) allele = Number of B alleles in population/Total Number of alleles
    = (570 * 2 + 260)/(1000 * 2) = 1400/2000 = 0.7

Allele frequency for the White (b) allele = 1 - 0.7 = 0.3

Expected frequency of BB individuals = p² = 0.7 * 0.7 = 0.49
Expected number of BB individuals = Expected Frequency * Total Number of Indiviuals = 0.49 * 1000 = 490

Expected frequency of Bb individuals = 2pq = 2 * 0.7 * 0.3 = 0.42
Expected number of Bb individuals = Expected Frequency * Total Number of Indiviuals = 0.42 * 1000 = 420

Expected frequency of bb individuals = q² = 0.3 * 0.3 = 0.09
Expected number of bb individuals = Expected Frequency * Total Number of Indiviuals = 0.09 * 1000 = 90

Null Hypothesis: The population is in Hardy-Weinberg proportions.

Individuals	Observed (O)	Expected (E)	O - E	(O - E)2	(O - E)2 / E
Dark Blue	570	490	80	6400	13.06
Pale Blue	260	420	-160	25,600	60.95
White	170	90	80	6400	71.11
					$\chi^{2}$ = 145.12

df = 1 (As there are 3 genotypes and 2 alleles, the degree of freedom is a difference between the two)

The $\chi^{2}$ value for this data is 145.12. The critical value at 95% confidence level for 1 degrees of freedom is 3.84. Since the value of $\chi^{2}$ we obtained is larger than the critical value, we reject the Null Hypothesis. The population is not in Hardy Weinberg proportions.

From the data, we see that the heterozygotes (Bb genotype) are much less frequent than expected. Therefore, the Dark Blue and the White flowering plants are preferred and the population is undergoing Disruptive selection or Diversifying Selection, where the extreme phenotypes are better at surviving than the average, either due to reproductive or survival benefits.