1
Population Genetics:
The Hardy-Weinberg Rule
Background
This tasty activity for biology students demonstrates the
effects of selective predation on a small population.
Predators (students) selectively prey upon (eat) fish
(Goldfish® crackers) in a small population. Prey are
replaced with fish randomly selected from an ocean (a
mixing bowl full of Goldfish® crackers) to create the next generation. Students calculate
genotype frequencies for each generation using the Hardy-Weinberg rule. They then
examine their data to see if the genotype frequencies have shifted as a result of selective
predation.
Students work in lab groups of 3 to 4. As an option, you can have the lab groups pool their
results and calculate genotype frequencies for the aggregate data.
To review, the Hardy-Weinberg rule for a population at genetic equilibrium assumes the
following:
1. No genetic mutation is occurring within the population.
2. The breeding population is large.
3. The population is isolated from other populations of the species. No differential migration
occurs.
4. All members of the population survive and reproduce. There is no natural selection.
5. Mating is random within the population.
Genotype frequencies for a population in Hardy-Weinberg equilibrium
p2 + 2 pq + q2 = 1, where p2 is the Homozygous dominant frequency in the population,
and q2 is the homozygous recessive frequency in the population
p+q=1
example: if one out of every 100 people are homozygous recessive
for beak shaped nose then q2 = 1/100 = .01
Then q = square root of (.01) = .1 and p = 1-.1 = .9
Then p2 = .9 * .9 = .81 and x people will be homozygous
dominant for not having a beak shaped nose. That means
that 100 – 81 -1 = 18 people out of every 100 are carriers
for the beak shaped nose.
Note: This activity involves eating. Do not conduct the activity in a laboratory
where chemicals or other agents can contaminate the crackers. Make sure
students wash their hands before eating any crackers.
Materials

A printed copy of the activity

A large mixing bowl

A package of cheese-flavored
Pepperidge Farm Goldfish® crackers

A package of pretzel-flavored
Pepperidge Farm Goldfish® crackers
2
Procedure
1.
Empty both packages of Goldfish® crackers into the bowl and mix thoroughly.
2.
Take a random population of 10 fish (crackers) from the ocean (mixing bowl).
3.
Record on your chart under "Generation 1" the number of gold fish and brown fish in
the random population. You can calculate frequencies later. Gold fish (cheese-flavored
Goldfish®) express the recessive allele (f) and brown fish (pretzel-flavored Goldfish®)
express the dominant allele (F). Gold are ff brown are either Ff or FF
4.
Choose 3 gold fish from the first generation and eat them; if you do not have 3 gold
fish, fill in the missing number by eating brown fish.
5.
Choose at random 3 fish from the ocean—one fish for each one that died (was
consumed by you) and add them to your population. Do not use artificial selection.
6.
Record on your chart under "Generation 2" the number of gold fish and brown fish.
7.
Choose 3 gold fish from the second generation and eat them; if you do not have 3 gold
fish, fill in the missing number by eating brown fish.
8.
Choose at random 3 fish from the ocean and add them to your population. Do not use
artificial selection.
9.
Record on your chart under "Generation 3" the number of gold fish and brown fish.
10. Choose 3 gold fish from the third generation and eat them; if you do not have 3 gold
fish, fill in the missing number by eating brown fish.
11. Choose at random 3 fish from the ocean and add them to your population. Do not use
artificial selection.
12. Record on your chart under "Generation 4" the number of gold fish and brown fish.
13. Choose 3 gold fish from the fourth generation and eat them; if you do not have 3 gold
fish, fill in the missing number by eating brown fish.
14. Choose at random 3 fish from the ocean and add them to your population. Do not use
artificial selection.
15. Record on your chart under "Generation 5" the number of gold fish and brown fish.
16. Calculate the allele and genotype frequencies for each generation using the data in
Chart 1.
17 (Optional) Calculate the sum of the gold and brown fish counts of each generation for
all the lab groups, record them in Chart 2, and then calculate the allele and genotype
frequencies of the aggregate data for each generation.
Hints for calculating genotype frequencies (example 2)
If you know from counting that 16% of the fish in a population express the homozygous
recessive allele (ff), then the q2 value is 0.16, and q equals the square root of 0.16 =(0.4).
Thus, the frequency of the f allele is 0.4 or 40%. Since the sum of the F and f alleles must
be 1, the frequency of the F allele must be 0.6 or 60%. Using Hardy-Weinberg, you can
assume that in the population there are 0.36 FF (0.6 x 0.6), 0.48 Ff (2 x 0.4 x 0.6), and
0.16 ff.
Remember:
p2 + 2 pq + q2 = 1
and
p+q=1
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Conclusion
The Hardy-Weinberg rule is an important tool for detecting changes in the genotype
frequencies of a population. Applying the Hardy-Weinberg rule to the data shows that
selective predation of the gold fish results in the reduction of the ff allele’s frequency in
subsequent generations. It is important to recognize that gold fish and the ff allele will not
totally disappear from the population.
Although individuals expressing the ff allele may die, because of selective predation, before
they reproduce, surviving individuals that do not express the ff allele still carry it. If the
pressure of selective predation is removed, individuals expressing the ff allele survive,
reproduce, and the population expressing the ff allele increases, as does the frequency of
the ff allele.
Charts for Population Genetics:
The Hardy-Weinberg Rule
Generation
Gold
(ff)
Brown
(Fx)
Chart 1 (Lab group data)
q2=
q= freq. of
homozygous
gold gene
recessive
in
(gold)
population
p2= freq of
homozygous
Dominant
FF
2pq= freq of
heterozygous
Ff
1
2
3
4
5
Gold fish represent homozygous (ff) recessive fish:
q2 = number of ff fish / total number of fish (In our lab total number =10)
p2 = percent of fish who are homozygous FF dominant brown. Since they look the same as
heterozygous Ff brown fish we have to calculate a predicted number who are FF.
First calculate q = √q2
Then calculate p from p+q = 1
p=1–q
Now you can calculate the frequency or percent of FF fish by:
p2 = p * p
The percent of fish that are heterozygous i.e. have both genes Ff, is:
2pq = 2 * p * q
Chart 2 (Aggregate data for all lab groups) Total N=
Gold
Brown
Generation
q2
q
p
P2
(f)
(F)
1
2
3
4
5
AVERAGE
No. of Fish
Gold=
FF Fish
2pq
How many of our aggregate population are heterozyogous?_________________