Question

In a population of wild sweet pea plants that you assume are in Hardy-Weinberg equilibrium, 1% of plants express a recessive white flower color. A) What is the expected frequency of heterozygotes in the population? Number You have noticed a shift in pollinator frequency and you want to test whether this has resulted in selection on flower color. You sample 1000 plants and genotype them at this locus. B) How many of each genotype do you expect based on the expected frequencies calculated using the Hardy-Weinberg equilibrium? 250 W/W, 500 W/W, 250 w/w 810 W/W, 180 W/W, 10 w/w 100 W/W, 100 W/W, 800 w/w 604 W/W, 332 W/W, 64 w/w 410 W/W, 180 W/W, 410 w/w

Answer 1

We will use the Hardy Weinberg assumption were:

p = Dominant allele

q = Recessive allele

p + q = 1

p2 = frequency of homozygous dominant

q2 = frequency of homozygous recessive

2pq = frequency of heterozygous

Recessive frequency of recessive genotype = 1% = 0.01

Calculate the frequency of the recessive allele (q) = √0.010= 0.1

Calculate the frequency of the dominant allele

p = 1-q = 0.9

Calculate the frequency of homozygous dominant (p2)

p2 = 0.92 = 0.81

Calculate the frequency of heterozygous

2pq = 2 x 0.01 x 0.81 = 0.162

Now, to determine the number of each genotype in the population, multiply each frequency by the total number of the population (1000)

Number of recessive individuals = 1000x0.01 = 10

Number of homozygous dominant individuals = 1000 x 0.81= 810

Number of heterozygous individuals = 810 – 10 =180

The correct answer is 810 W/W, 180 W/w, 10 w/w

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