Question

BIOL 1010

Hardy-Weinberg practice problems

The Hardy-Weinberg principle allows us to determine whether evolution has occurred. The principle essentially states that if no evolution is occurring, then an equilibrium of allele frequencies will remain in effect in each succeeding generation of sexually reproducing individuals.

In order for equilibrium to hold (no evolution is occurring), the following five conditions must be met:

1.No mutations must occur (so that new alleles do not enter the population).

2.No gene flow can occur (no migration of individuals into, or out of, the population).

3.Random mating must occur.

4.Thepopulation must be large so that no genetic drift (random chance) can cause the allele frequencies to change.

5.No selection can occur so that certain alleles are not selected for, or against.

Obviously, the Hardy-Weinberg equilibrium cannot exist in real life. Some or all of these forces all act on living populations at various times, and evolution at some level occurs in all living organisms. The Hardy-Weinberg equations are useful because they allow us to detect some allele frequencies that change from generation to generation, and so allow a simple method of determining that evolution is occurring.
There are two formulas:        p2 + 2pq + q2 = 1  and  p + q = 1

    p = frequency of the dominant allele in the population
    q = frequency of the recessive allele in the population

    p2 = frequency of homozygous dominant genotype
    q2 = frequency of homozygous recessive genotype
    2pq = frequency of heterozygous genotype

You are given enough information to plug in the numbers, and then just solve for whatever it is you are looking for.

Example:

If the frequency of the homozygous recessive genotype tt is 36%, then q2 = 0.36 and so q = 0.6.

And since q = 0.6, and p + q = 1, then p = 1– q =  0.4 and p2 = 0.16 (TT)

And from that, we can get the percentage of heterozygotes (Tt) by finding 2pq = 2(0.4)(0.6) = 0.48 = 48%

So in this population, 16% are TT, 36% are tt, and 48% are Tt.  And 16% + 36%+ 48% = 100%

Example:

If the frequency of the homozygous dominant genotype TT is 25%, then p2 = 0.25 and so p = 0.5

And since p = 0.5, and p + q = 1, then q = 1– p =  0.5 and q2 = 0.25 (tt)

And from that, we can get the percentage of heterozygotes (Tt) by finding 2pq = 2(0.5)(0.5) = 0.5 = 50%

So in this population, 25% are TT, 50% are tt, and 25% are Tt.  And 25% + 50%+ 25% = 100%

For both of these examples, these are the ratios of genotypes we would expect to find if no evolution is happening – if the populations are in Hardy-Weinberg equilibrium. If we then sample the actual population and find different ratios, then we know that evolution is occurring.

Using these concepts, answer the following questions:

1.    You have sampled a population in which you know that the frequency of the homozygous recessive genotype (aa) is 9%. Using that 9%, calculate the following:

   

    The frequency of the "a" allele.

    The frequency of the "A" allele.

   

The frequencies of the genotypes "AA" and "Aa."

    The frequencies of the two possible phenotypes if "A" is completely dominant over "a."

2.    Within a population of butterflies, the color brown (B) is dominant over the color     white (b). And, 40% of all butterflies are white. Given this information, calculate the following:

   

    The percentage of butterflies in the population that are heterozygous.

   

   

    The frequency of homozygous dominant individuals.

3.    A very large population of randomly-mating laboratory mice contains 35% white mice. White coloring is caused by the homozygous recessive genotype, "aa".

Calculate the     frequencies for all genotypes and alleles for this population.

4.    Cystic fibrosis is a recessive condition that affects about 1 in 2,500 babies in the Caucasian population of the United States. Please calculate the following.

    The frequency of the recessive allele in the population.

    The frequency of the dominant allele in the population.

    The percentage of heterozygous individuals (carriers) in the population.

Answer 1

1.   The frequency of the "a" allele.

The frequency of aa is 9%, which means that q² = 0.9, by definition. If q² = 0.9, then q = 0.3, again by definition. Since q equals the frequency of the a allele, then the frequency is 30%.

The frequency of the "A" allele.

?Since q = 0.3, and p + q = 1, then p = 0.7; the frequency of A is by definition equal to p, so the answer is 70%.

The frequencies of the genotypes "AA" and "Aa."

?The frequency of AA is equal to p², and the frequency of Aa is equal to 2pq. So, using the information above, the frequency of AA is 49% (i.e. p² is 0.7 x 0.7 = 0.49) and Aa is 42% (2pq = 2 x 0.7 x 0.3 = 0.42).

The frequencies of the two possible phenotypes if "A" is completely dominant over "a."

?As per the question the dominant phenotype will show if either the homozygous "AA" or heterozygous "Aa" genotypes occur. The homozygous aa genotype controls recessive phenotype. So, the frequency of the dominant phenotype and sum of the frequencies of AA and Aa are equal, and the recessive phenotype is denoted by frequency of aa. Therefore, the dominant frequency is (49% + 42%= 91%) 91% and, in the first part of this question above, as already given that the recessive frequency is 9%.

2. We have to find out p and q. So, since white is recessive (i.e. bb), and 40% of the butterflies are white, then bb = q² = 0.4. To find out q, which is the frequency of the recessive allele in the population, find the square root of q² which works out to be 0.632 (i.e. 0.632 x 0.632 = 0.4). So, q = 0.63. As p + q = 1, thus p= 1 - 0.63 = 0.37.

To find out the percentage of butterflies in the population that are heterozygous we have to find 2pq and answer is 2 (0.37) (0.63) = 0.47.

Then the frequency of homozygous dominant individuals would be p² or (0.37)² = 0.14.

3.  35% are white mice, which = 0.35. It shows the frequency of the aa genotype (or q²). The square root of 0.35 is 0.59, which is equal to q. Since p = 1 - q then 1 - 0.59 = 0.41. The frequency of each allele can be calculated from the frequency of the remaining genotypes (AA and Aa ). AA = p² = 0.41 x 0.41 = 0.17; Aa = 2pq = 2 (0.59) (0.41) = 0.48; and as before aa = q² = 0.59 x 0.59 = 0.35. When all the genotype frequencies are summed up it equals to 1.

4. The frequency of the recessive allele in the population.

1 / 2500 = .0004 <---- this is (q²) , then q = .02

The frequency of the dominant allele in the population.

p + .02 = 1, p = .98

The percentage of heterozygous individuals (carriers) in the population.  

2pq = 2(.98)(.02) = .04