For these problems, remember that the frequency of alleles, p + q = 1, where p represents the dominant allele and q
represents the recessive allele. When populations are in equilibrium, p2 + 2pq + q2 = 1, where p2 = homozygous
dominant genotype, 2pq = heterozygous genotype, and q2 = homozygous recessive genotype.
Estimating Allele Frequencies & Hardy-Weinberg
PART I - Hardy-Weinberg Problems (SHOW ALL WORK) – Due Thursday 2/21
1. In Drosophila (fruit fly), the allele for normal wing length is dominant over the allele for short wings. In
a population of 1000 individuals, 360 show the recessive phenotype. How many individuals would you
expect to be homozygous dominant for the trait?
We know q2 (homozygous recessive). We are looking for p2 (homozygous dominant).
q2 = 360/1000 = .36
q = .6
p = 1-q
p = .4
p2 = .16
.16 x 1000 = 160 individuals homozygous dominant
2. The allele for a widow's peak (hairline) is dominant over the allele for a straight hairline. In a
population of 500 individuals, 25% show the recessive phenotype. How many individuals would you
expect to be homozygous dominant and heterozygous for the trait?
q2 = .25
q = .5
p = 1-q
p = .5
p2 = .25
2pq = 2(.5)(.5) = .50
.25 x 500 = 125 individuals homozygous dominant (p2)
.50 x 500 = 250 individuals heterozygous (2pq)
3. The allele for a hitchhiker's thumb is dominant over a straight thumb. In a population of 1000
individuals, 510 show the dominant phenotype. How many individuals would you expect for each of the
three possible genotypes for this trait.
p2 + 2pq = 510/1000
q2 = (1000-510)/1000 = 490/1000 = .49
q = .7
p = 1-q
p = .3
p2 = .09
2pq = 2(.3)(.7) = .42
Homozygous dominant = .09 x 1000 = 90 individuals
Heterozygous = .42 x 1000 = 420 individuals
Homozygous recessive = .49 x 1000 = 490 individuals
4. This is a classic data set on wing coloration in the scarlet tiger moth (Panaxia dominula). Coloration in
this species had been previously shown to behave as a single-locus, two-allele system with incomplete
dominance. Data for 1612 individuals are given below:
White-spotted (AA) = 1469
Intermediate (Aa) = 138
Little spotting (aa) =5
Calculate the allele frequencies (p and q)
Total population = 1612
q2 = 5/1612 = .003
q = .055
p = 1-.055
p = .945
5. In a tropical forest there is a species of bird that has a variable tail length. Long is incompletely
dominant over short. In one population of 2000 birds, 614 have long tails, 973 have medium length tails,
and 413 birds have short tails.
a) What is the frequency of each allele in the population?
Total population = 2000
q2 = 413/2000 = .21
q = .46
p = 1-.46
p = .54
b) Is the population in Hardy Weinberg equilibrium? Explain!
If q = .46 and p = .54, then we would expect that:
q2 = (.46)2 = .21
p2 = (.54)2 = .29
2pq = 2(.54)(.46) = .50
So, we would expect that
.21 x 2000 = 420 short tails
.29 x 2000 = 580 long tails
.50 x 2000 = 1000 medium tails
We observed
413 short tails
614 long tails
973 medium tails
Using our chi-square analysis, we calculate:
(observed – expected)2 / expected for each group
= (413-420)2/420 + (614-580)2/580 + (973-1000)2/1000
= .12 + 1.99 + .72
= 2.83
This value is less than 5.99 (our critical value for 2 degrees of freedom) so the difference is
insignificant, therefore, this population IS in Hardy-Weinberg equilibrium.
6. A botanist is investigating a population of plants whose petal color is controlled by a single gene
whose two alleles are codominant. She finds 170 plants that are homozygous brown, 340 plants that are
homozygous purple, and 21 plants whose petals are brown-purple.
Is this population in Hardy- Weinberg equilibrium? Explain!
This example is a little trickier because the alleles are co-dominant. Remember that means that neither
allele is dominant over the other one. If we represent B as brown and P as purple, we find that
BB = 170/531 = .32
BP = 21/531 = .04
PP = 340/531 = .63
How do we determine expected frequencies?
We have to count alleles.
B alleles (170 x 2) + 21 = 361
P alleles (340 x 2) + 21 = 701
Total # of alleles = 701 + 361 = 1062, so
B = 361/1062 = .34 = q
P = 701/1062 = .66 = p
q2 = (.34)2 = .12
p2 = (.66)2 = .44
2pq = 2(.34)(.66) = .11
So, we would expect that
.12 x 531 = 63 BB
.44 x 531 = 233 PP
.11 x 531 = 25 BP
We observed
170 BB
340 PP
21 BP
Using our chi-square analysis, we calculate:
(observed – expected)2 / expected for each group
= (170-63)2/63 + (340-233)2/233 + (21-25)2/25
= 181.7 + 49.1 + .64
= 231.44
This value is greater than 5.99 (our critical value for 2 degrees of freedom) so the difference is
significant, therefore, this population is NOT in Hardy-Weinberg equilibrium.