Sixty flowering plants are planted in a flowerbed. forty of the plants are red-flowering homozygous dominant. twenty of the plants are white-flowering homozygous recessive. the plants naturally pollinate and reseed themselves for several years. in a subsequent year, 178 red-flowered plants, 190 pink-flowered plants, and 52 white-flowered plants are found in the flowerbed. use a chi-square analysis to determine if the population is in hardy-weinberg equilibrium

Considering that the critical value is lower than the chi-square value, with a significance level of 0.05 and 2 degrees of freedom, we can conclude that this population is not in Hardy-Weinberg equilibrium.

Available data:

after many generations

Hypothesis: The population is in H-W equilibrium

Let us firts get the genotypic frequencies in the original population.

F(RR) = p² = 40/60 = 0.667

F(rr) = q² = 20/60 = 0.333

Now, let us take the allelic frequencies, following H-W equilibrium.

To do it, we can use the recessive genotypic frequency, F(rr).

q² = 0.333

q = √0.333 = 0.577

Now, by clearing the following equation, we can get the dominant allelic frequency, p.

p + q = 1

p + 0.577 = 1

p = 1 - 0.577

p = 0.423

If this population is in H-W equilibrium, after many generations the genotypic frequencies should be as follow,

Now, assuming that this population is in H-W equilibrium, the allelic and genotypic frequencies will remain equal generation after generation.

So, the expected numbers in a population of 420 individuals are,

Now, to get to know if the population is in hardy-weinberg equilibrium, we will use chi square. And to calculate it, we will use the following equation

                                     X² = Σ(Obs-Exp)²/Exp  

                                  RR                      Rr                      rr                      

Observed                 178                     190                   52

Expected                  75.18                 204.96             139.86

(Obs-Exp)²/Exp        140.62              1.092                55.193                

X² = Σ(Obs-Exp)²/Exp = 140.62 + 1.092 + 55.193 ≅ 197

Now, we need to get the degrees of freedom.

Freedom degrees = n - 1 = 3 - 1 = 2

Now, using a Significance level of 5% = 0.05, and the degrees of freedom, we will look for the critical value.

Critical value = 5.991

Chi-square = 197

P₀.₀₅  < X²

5.991 < 197

The difference between the observed individuals and the expected individuals is statistically significant.

Not probably to have differe by random chances.

There is enough evidence to reject the null hypothesis.

The population is not in equilibrium H-W.

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The flowering population is in Hardy Weinberg equilibrium.

Step by Step Explanation:

Total initial population size = 60

As both the populations are homozygous so the frequency of alleles will be,

P = 80/120= 0.66

q = 40/120= 0.33

The predicted frequencies for genotypes once the population has reached Hardy-Weinberg

p2 = 0.4356

2pq= 0.4356

q2 = 0.1089

The number of plants with each type of flower in a papulation of 420 is,

Homozygous dominant= 185

heterozygous= 185

Homozygous recessive= 50

Chi square analysis:

The observed values for red-flowered plants, pink-flowered plants, and white-flowered plants are not significantly different from the expected values predicted by Hardy Weinberg equilibrium.

Phenotype      observed(o)    expected (e)          (o-e)2/e

Red                        178                 185                       0.26

Pink                       190                 185                       0.135

White                     52                   50                     0.08

                                       chi-square = 0.474

With 2 degree of freedom this chi-square gives a p value of 0.7 - 0.8, which is not significant.  

doctors i believe if i'm wrong sorry