H-Weezy

In a population of 1000, 7% show the recessive phenotype. Solve using what is given to solve the rest of the Hardy-Weinberg equation.

GIVEN = 7% * 100 = q^2 = 0.07

Step 1) use the q^2, (Homozygous recessive individual) to determine the frequency of the recessive allele by unsquaring it (square root it).

q^2 = 0.07

When unsquared, the frequency of the recessive allele equals:

q = 0.26

Step 2) After determining what the q is subtract it from 1 to determine the value of p:

1- 0.26 = 0.74

 So, p = 0.74

Step 3) After solving for p, square it to find the value of p^2, the Homozygous dominant individual

0.74 * 0.74 = 0.55

So,  p^2 = 0.55

Step 4) Plug in p and q into the equation 2pq to verify that the values calculated are precise

2pq = 2(0.74)(0.26) = 0.38

So, 2pq = 0.38

Step 5) Double check results into the Hardy-Weinberg equation:

p^2 + 2pq + q^2 = 1

0.55 + 0.38 + 0.07 = 1

If the equation results in 1 as the answer, move onto determining the number of individuals present in the population.

PART 2

Step 6) In order to determine about how many individuals are Heterozygous, Homozygous recessive, and Homozygous dominant, simply multiply each value times the total population, which in this case, is 1000.

q^2 : 0.07 * 1000 = 70 Homozygous recessive individuals

p^2: 0.55 * 1000 = 550 Homozygous dominant individuals

2pq: .38 * 1000 = 380 Heterozygous individuals

 Step 7) (optional) Double check each count of individuals. Just add them all together, if they equal 1000, then your work is officially done. If not, double check your math in steps 1-5 and seek help from an instructor.

p^2 + 2pq + q^2 = 1

550 + 380 + 70 = 1000