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Lab Exercise: Population Genetics/Hardy-Weinberg
When a population is at genetic equilibrium the frequency of gene alleles does not change. Evolution is a
process resulting in changes in the genetic makeup of populations through time. Several factors can work to change
allele frequencies resulting in evolution. These factors disrupt genetic equilibrium and are called evolutionary agents.
This lab exercise demonstrates the effect of these agents on the genetic makeup of a model population of birds.
There is a population of birds that has a gene coding for the size of their beak. There are three phenotypes for this trait:
long-pointy beaks (AA), medium beaks (Aa), and short-rounded beaks (aa). A scientist counted 605 long-pointy beaks,
310 medium beaks, and 85 short-rounded beaks in a population of 1000 birds. Use the Hardy-Weinberg equation to
determine the allele and genotype frequencies. (Use the daisy example in the PowerPoint as a guide. The on-line lesson
videos are also helpful.)
1000 birds total
605 long beaks (AA)
310 medium beaks (Aa)
85 short beaks (aa)
How many copies of the bird beak alleles are there in this population? (Hint: each bird has two alleles for each gene)
How many of these copies are A alleles?
How many of these copies are a alleles?
What are the allele frequencies (or solve for p and q)?
p = frequency of A = (number of A alleles/total number of alleles) =
q = frequency of a =
What are the genotype frequencies (or solve for p2, 2pq, and q2)
Frequency of AA (solve for p2)
Frequency of Aa (solve for 2pq)
Frequency of aa (solve for q2)
Double check the work by substituting in the values
p2 + 2pq + q2 = 1
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The birds in this population only have one type of food to eat. They grab insects in deep crevices of tree bark. The
longer the beak, the better the bird is at foraging for food. The short beaks just can’t reach into the bark far enough.
The scientist returns to count the population after a month. Of the original population of 1000 birds 571 long beak
birds, 133 medium beak birds, and 23 short beak birds now make up the population. These are the only birds that were
able to get enough food to survive. Thus they are the only birds that will reproduce. Solve for the allele and genotype
frequencies for the survivors.
727 birds total
571 long beaks (AA)
133 medium beaks (Aa)
23 short beaks (aa)
How many copies of the bird beak alleles are there in this population?
How many of these copies are A alleles?
How many of these copies are a alleles?
What are the allele frequencies (or solve for p and q)?
p = frequency of A =
q = frequency of a =
What are the genotype frequencies (or solve for p2, 2pq, and q2)
Frequency of AA (solve for p2)
Frequency of Aa (solve for 2pq)
Frequency of aa (solve for q2)
Double check the work by substituting in the values
p2 + 2pq + q2 = 1
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Compare the original population with the population under natural selection. Record each value for the two population
counts in the table.
p
Dominant allele
frequency
q
Recessive allele
frequency
p2
2pq
q2
AA genotype
Aa genotype
aa genotype
Original
Population
Population 1
month later
(under pressure
of natural
selection)
Questions
1. What happened to the dominant allele frequency?
2. What happened to the recessive allele frequency?
3. If the original population had 1000 offspring how many would have short beaks? (Multiply 1000 X q2)
4. If the population under natural selection had 1000 offspring how many would have short beaks? (Multiply 1000 X q2)
5. How has natural selection affected this population?
6. What would happen to the allele frequencies if a group of these birds migrated to a new habitat that had trees with
thinner bark?
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Hardy-Weinberg problems
p2 + 2pq + q2 = 1
p+q=1
p = one allele frequency (A)
q = the other allele frequency (a)
p2 = homozygous dominant genotype frequency (AA)
2pq = heterozygous genotype frequency (Aa)
q2 = homozygous recessive genotype frequency (aa)
1. If p is 0.57 in a given population, what is q?
2. If q is 0.56 in a given population, what is p?
3. In a population of 1000 shrews, 750 had straight fur (AA and Aa) while 250 had curly fur. Solve for q2.
Hint: q2 is the frequency of the homozygous recessive genotype (in other words the “percentage” of aa
individuals).
4. In a population of 300 sloths the frequency of the homozygous long furred individuals was 0.46. The frequency of the
homozygous short furred individuals was 0.09. What is the frequency of the heterozygotes? (Hint: p2 + 2pq + q2 = 1)
5. In a population of 100 snakes, 37 had yellow spots (AA), 28 had green spots (Aa), and 35 had blue spots (aa)
(Incomplete dominance). What is the q value? (Hint: divide the total number of “a”s by the total number of alleles in
the population)
6. The p2 value for a population of butterflies is 0.25.
What is the p value?
What is the q value?
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