POPULATION GENETICS: HOW TO SOLVE HARDY WEINBERG PROBLEMS.
The Hardy-Weinberg model is a hypothetical model of a population at equilibrium
– in that the population is NOT evolving, or changing over time. In order to be at HW
equilibrium, a population must fulfill ALL of the following requirements:
 No genetic change (i.e. No mutations.)
 No gene flow (i.e. No immigration or emigration)
 No genetic drift (i.e. Large population size)
 No sexual selection (i.e. Random mating)
 No natural selection (i.e. All phenotypes equally likely to survive and
reproduce.)
Are these conditions realistic for actual populations? No! HW represents a null
hypothesis. The Hardy-Weinberg formulas allow us to detect frequencies that
change from generation to generation, thus allowing a simplified method of
determining that evolution is occurring.
So, if you want to argue that the population is evolving, you have to reject the null
hypothesis. To do that, you must be able to compare the frequencies in a
population that isn’t evolving to one that could be.
Before being able to solve HW problems, you must be able to answer the following:
1. What is the difference between a gene and an allele?
2. What is the difference between an allele and a genotype?
3. What is the difference between a phenotype and a genotype?
4. What is the difference between dominant and recessive?
5. What is the difference between homozygous and heterozygous?
6. What is the difference between a number, a frequency, and a percentage?
To solve HW problems, you need to memorize one equation, which specifies allelic
frequency in a population:
p+q=1
Where: p is the frequency of dominant alleles; and
q is the frequency of recessive alleles.
The number 1 represents the total alleles in a population. (This is based on a
two-allele system – alleles are either dominant or recessive. This won’t work
with multiple alleles systems!)
From the above equation, we can determine genotype frequencies. If p+q=1, then:
p2+2pq+q2=1
2
Where: p is the frequency of homozygous dominant individuals.
2pq is the frequency of heterozygous individuals.
q2 is the frequency of homozygous recessive individuals.
The number 1 represents the total genotypes in a population. (This is based
on a three-genotype system – genotypes are either homozygous dominant,
heterozygous or homozygous recessive. This won’t work with incomplete
dominance, sex-linkage, etc!)
Understanding HW problems is a matter of understanding both the concepts and the
math. We’ll use an example, and then complete several practice problems.
Example 1: In a certain population of 1000 fruit flies, 640 have red eyes while the
remaining flies have sepia eyes. The sepia eye trait is recessive to red eyes. How
many individuals would you expect to be homozygous for red eye color?
1. Understand what you have been given in the above problem.
a. The population is 1000. Of the 1000, 640 have red eyes and 360 have
sepia eyes. Calculate the frequency of each.
i. 640/1000 = 64 percent red eyes
ii. 360/1000 = 36 percent sepia eyes.
2. Red eyes are dominant. Sepia eyes are recessive. Therefore, red-eyed flies can be
homozygous dominant or heterozygous. Sepia-eyed flies are homozygous
recessive ONLY. So, the number of sepia-eyed flies will tell us the figure for q2.
a. If q2 = 36 percent, or .36, the q = .6, since the square root of .36 is .6.
3. List your knowns, and then solve for your remaining unknowns.
a. If q = .6 and p+q=1, then p=.4.
b. If q2=.36, and p=.4 and q=.6, then you can calculate p2 and 2pq.
i. p2 = .4*.4 = .16
ii. 2pq = 2*.4*.6 = .48
c. Check your math by making sure p2+2pq+q2=1
i. .16+.48+.36=1
4. Answer the question being asked – this is the most important step!
a. If .16 or 16 percent of the population is homozygous dominant or redeyed, homozygous flies, then .16*1000 is 160.
Directions: Complete the assigned practice problem(s) and be prepared to present
your problem. ALL groups will be tasked with practice problem 5!
1. In the United States, one out of approximately 10,000 babies is born with PKU, a
genetic disorder that interferes with their ability to digest proteins.
Approximately what percent of the population are heterozygous carriers of the
recessive PKU allele?
2. The allele for the hair pattern called "widows peak" is dominant over the allele
for no "widows peak." In a population of 1000 individuals, 510 show the
dominant phenotype. How many individuals would you expect of each of the
possible three genotypes for this trait?
3. In the U.S. about 16 percent of the population is Rh negative. The allele for Rh
negative is recessive to the allele for Rh positive. If the student population of a
high school in the U.S. is 2000, how many students would you expect for each of
the three possible genotypes?
4. In a certain population, the dominant phenotype of a certain trait occurs 91
percent of the time. What is the frequency of each of the three genotypes?
5. Sickle-cell anemia is a genetic disease. Normal homozygous individials (SS) have
normal blood cells that are easily infected with the malarial parasite. Thus, many
of these individuals become very ill from the parasite and many die. Individuals
homozygous for the sickle-cell trait (ss) have red blood cells that readily collapse
when deoxygenated. Although malaria cannot grow in these red blood cells,
individuals often die because of the genetic defect.
However, individuals with the heterozygous condition (Ss) have some sickling of
red blood cells, but generally not enough to cause mortality. In addition, malaria
cannot survive well within these "partially defective" red blood cells. Thus,
heterozygotes tend to survive better than either of the homozygous conditions.
If 9 percent of an African population is born with a severe form of sickle-cell
anemia (ss), what percentage of the population will be more resistant to malaria
because they are heterozygous (Ss) for the sickle-cell gene?