In: Biology
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.
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