HARDY-WEINBERG PRINCIPLE
Teacher Section
I. Advance Preparation -
The teacher provides 2 paper lunch bags, 60 red beans, and 40 white beans for each group of 4 students. Mark one bag male and the other bag female.
This may take up to 1 hour of preparation.
II. Introduction -
The Hardy-Weinberg Principle deals with the frequency, or percentage, of alleles in a population. It states that the frequency of an allele, or rather the number of times an allele is present in a population, is constant from one generation to the next. In order for this to occur certain conditions must be met. The conditions are:
1. The population must mate at random.
2. Alleles cannot be lost through migration.
3. Alleles are not changed by mutation.
4. Alleles are not lost by selective pressure or discrimination.
III. Student Objectives -
1. The student will calculate p + q = 1.
2. The student will calculate each of the following frequencies for an ideal Hardy- Weinberg situation, for a migrating population and a selective condition:
p, q, p2, q2, 2pq
3. The student will calculate p2 + 2pq + q2 = 1.
IV. Class Time Needed -
One class period for each activity.
V. Materials -
For each group of 4 students:
2 paper lunch bags
60 red beans
40 white beans
VI. Procedure -
1. Provide each group of students 60 red beans, 40 white beans, and 2 bags (one marked male and the other marked female.)
2. Assume that a red bean represents a dominant allele and that a white bean represents a recessive allele. Therefore, the homozygous dominant individual would be represented by 2 red beans, the homozygous recessive individual would be represented by 2 white beans, and the heterozygous individual would be represented by one red bean and one white bean.
ACTIVITY I:
PART I: Calculate the frequency of the dominant and the recessive alleles.
1. p = frequency of dominant allele = # of red beans = total # of beans
(Answer: 60 = 0.6)
100
2. q = frequency of recessive allele = # of white beans =
Total # of beans
(Answer: 40 = 0.4)
100
PART II: Calculate p + q
1. Take the values for p and q as determined in PART I and add them together:
p + q = (Answer: 0.6 + 0.4 = 1)
2. Extension: Multiply and calculate (p + q)2 =
Answer: (p + q)(p + q) = p2 + 2pq + q2 = 0.36 + 0.48 + 0.16 = 1
or (0.6 + 0.4)2 = (1)2 = 1
(Notice: the frequency of the heterozygote is represented as 2pq since the student will draw out approximately twice as many heterozygotes as homozygotes of either color.)
PART III: Determine the frequency of p2 (homozygous dominant), q2 (homozygous recessive), and 2pq (heterozygote).
1. Students place 30 red beans and 20 white beans in the bag marked male.
2. Students place 30 red beans and 20 white beans in the bag marked female.
3. One bean from each bag is drawn until all the beans are drawn. Tally the combinations in the table in the student section.
4. Calculate the frequencies of p2, q2, and 2pq:
p2 = # of beans in the double red bean drawing = Total # of beans
(Ideal answer: 0.36)
q2 = # of beans in the double white bean drawing =
Total # of beans
(Ideal answer: 0.16)
2pq = # of red and white beans drawn together = Total # of beans
(Ideal answer: 0.48)
ACTIVITY II: This activity will demonstrate the effect that migration of a small portion of a population will have on the frequencies of the alleles.
1. Reapeat the procedure of Part III in ACTIVITY I except that the student will draw only 10 pairs of beans from the respective bags. This bean population represents a migration. Now the Hardy-Weinberg Principle will no longer apply. The new frequencies should be different from those in ACTIVITY I.
2. Determine the frequencies of p, q, p2, q2, and 2pq of this new population.
p = # of red beans =
Total # of beans
q = # of white beans =
Total # of beans
p2 = # of beans in a double red bean drawing =
Total # of beans
q2 = # of beans in a double white bean drawing =
Total # of beans
2pq = # of red and white beans drawn together =
Total # of beans
ACTIVITY III: This activity will demonstrate the effect that selection or discrimination of one of the alleles will have on the frequencies of all of the alleles.
1. Repeat the procedure in Part III of ACTIVITY I. Use all of the beans in the bags for this activity.
2. Remove or discard all the homozygous recessive (white/white) bean combinations.
3. Recount the allele combinations to determine the new frequencies for p, q, p2, q2, and 2pq.
p = # of red beans =
Total # of beans
q = # of white beans =
Total # of beans
p2 = # of beans in a double red bean drawing =
Total # of beans
q2 = # of beans in a double white bean drawing = 0
Total # of beans
2pq = # of red and white beans drawn together =
Total # of beans
(Results for p2 and 2pq will be approximately the same as in activity I -
ie. p2 = 0.36 and 2pq = 0.48, however since all the white/white/ bean combinations have been removed q2 will be 0.)