HARDY-WEINBERG PRINCIPLE
Answer Key
I. 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.
II. Procedure -
1. To investigate the Hardy-Weinberg Principle you will be working in a group with 3 other students. Each group should have 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 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 = 60 = 0.6 total # of beans 100
2. q = frequency of recessive allele = # of white beans = 40 = 0.4 total # of beans 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 = 0.6 + 0.4 = 1.0
2. Extension: Multiply and calculate (p + q)2 =
(p + q)(p + q) = p2 + 2pq + q2 = 0.36 + 0.48 + 0.16 = 1.0
or (0.6 + 0.4)2 = (1)2 = 1
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 beans have been drawn. Tally the combinations in the table below.
(Answer: The frequencies that the students obtain should be close to the ideal. The answers written here are the ideal.)
Red/Red Red/White White/White
Total:
4. Calculate the frequencies of p2, q2, and 2pq:
p2 = # of beans in the double red bean drawing = 0.36 Total # of beans
q2 = # of beans in the double white bean drawing = 0.16
Total # of beans
2pq = # of red beans and white beans drawn together = 0.48 Total # of beans
5. Gather class data in a table on the board and copy the data into the following table:
Class Data
Red/Red Red/White White/White
Total:
Calculate the class total frequencies of p2, q2, 2pq.
p2 = # of beans in the double red bean drawing = Answers vary per class.
total # of beans
q2 = # of beans in the double white bean drawing = Answers vary per
total # of beans class.
2pq = # of red beans and white beans drawn together = Answers vary
total # of beans per class.
Analysis Questions
1. What is the value of p + q? 1
2. What is the value of (p + q)2? 1
3. For this activity p = 0.6, what is the value of p2? 0.36
How close did your answer for p2 come to this value? (Students answers may not be exactly 0.36 but it should be close.)
4. For this activity q = 0.4, what is the value of q2? 0.16
How close did your answer for q2 come to this value? (Students answers may not be exactly 0.16 but it should be close.)
5. What is the value for 2pq? 2 x 0.6 x 0.4 = 0.48
How close did your answer for 2pq come to this value? (Students answers may not be exactly 0.48 but it should be close.)
6. Why must the value for p + q be equal to 1? The value for p + q must be equal to 1 because p and q represent the entire quantity of alleles for that gene. Since there is no more nor no less than the whole, the value must be 1.
ACTIVITY II: This activity will demonstrate that even a small migration of the
population will effect the frequencies of the alleles.
Procedure
Repeat the procedure of Part III in ACTIVITY I except this time 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.
(Answer: The values for p, q, p2, q2, and 2pq will vary for each student group. This is the importance of this activity as it shows that migration does change the frequencies of the alleles.)
Red/Red Red/White White/White
Total:
2. Determine the frequencies of p, q, p2, q2, and 2pq of this new population.
p = # of red beans = Answers will vary.
total # of beans
q = # of white beans = Answers will vary.
total # of beans
p2 = # of beans in a double red bean drawing = Answers will vary.
total # of beans
q2 = # of beans in a double white bean drawing = Answers will vary.
total # of beans
2pq = # of red and white beans drawn together = Answers will vary.
total # of beans
Analysis Questions
1. Compare the value you obtained for p in this activity with the value you obtained in ACTIVITY I.
(Values should vary for each student group.)
2. Compare the value you obtained for q in this activity with the value you obtained in ACTIVITY I.
(Values should vary for each student group.)
3. How have these values affected 2pq? 2pq should be different if both p and q have been altered.
Compare your values with the rest of the class. Class values should be similar.
4. Have the number of homozygous dominant, homozygous recessive, or heterozygous individuals been altered significantly by this migration of a few individuals?
(Answers will vary depending on the values of p and q each group obtained.)
5. What is the value of p + q? 1.0
6. What is the value of (p + q)2? 1.0
ACTIVITY III: This activity will demonstrate the effect that selection or discrimination upon a certain allele combination will have on the frequencies of all the alleles.
Procedure
Repeat the procedure in Part III of ACTIVITY I. Use all of the beans in the bags for this activity, but this time remove or discard all the homozygous recessive (white/white) bean combinations. Recount the allele combinations to determine the new frequencies for p, q, p2, q2, and 2pq.
Red/Red Red/White White/White
0
Total:
p = # of red beans = Students answers will vary. (Since there are no
total # of beans white/white combinations, ideally the frequency would be 60/84 = 0.71)
q = # of white beans = Students answers will vary. (Ideally the value total # of beans would be 24/84 = 0.29.)
p2 = # of beans in a double red bean drawing = Students answers will
total # of beans vary. (Ideally the value would be 36/84 = 0.43)
q2 = # of beans in a double white bean drawing = 0
total # of beans
2pq = # of red beans and white beans drawn together = Students answers
total # of beans will vary. (Ideally the value would be 48/84 = 0.57.
Analysis questions -
1. Compare the p value you obtained in this activity with the p value in ACTIVITY I.
The p value is greater because the total # of beans (the denominator) is less.
2. Compare the q value you obtained in this activity with the q value in ACTIVITY I.
The q value is less because the number of white beans (the numerator) is less.
3. The value for q should be less than the value you obtained for q in the first activity. Explain why that is.
Some of the white beans have been discarded, therefore the numerator is less causing the entire value to be less.
4. What happens to the p value? Explain. The p value increases as the q value decreases because the sum of p and q will always equal 1.
Procedure
Now refill the bags with equal amounts of the colored beans from the red/red drawing and red/white drawing. In other words, divide the red beans and the remaining white beans equally between the two bags. Again draw the beans from each bag and tally the pairs in the table. Again discard the white/white bean combinations. Calculate the new values of p, q, p2, q2, and 2pq.
Red/Red Red/White White/White
0
Total:
p = # of red beans = Students answers will vary.
total # of beans
q = # of white beans = Students answers will vary.
total # of beans
p2 = # of beans in a double red bean drawing = Students answers will
total # of beans vary.
q2 = # of beans in a double white bean drawing = Students answers will
total # of beans vary.
2pq = # of red beans and white beans drawn together = Students answers
total # of beans will vary.
4. What has happened to q? The value for q has decreased.
Why is q not 0? The value for q is not 0 because q appears in 2pq (the heterozygote), therefore there is a value for q.
Will q ever be 0? The value for q will never be 0 because there will always be heterozygote individuals, i.e. 2pq, thus q will perpetuate throughout generations.
5. Predict what would happen if the selection or discrimination occurred against the individuals expressing the dominant gene.
Expression of the dominant gene indicates any individual with the dominant allele will be eliminated. Therefore in one generation any individual with the p or dominant allele will be gone and p will be 0.
What would happen to the value of p? p would equal 0
What is the value of p + q? 1.0
6. Can you deduce why harmful human alleles are usually recessive rather than dominant?
If the harmful allele were dominant it would disappear in one generation, however by being recessive the allele is always present in the population.