Hardy Weinberg Equilibrium Practice Problems

Question 1

The Hardy Weinberg Principle states that …

Answer 1

Heredity alone cannot cause changes in the frequency of alleles in a population

Question 2

The frequency of alleles that make up a population will remain…

Answer 2

Constant, generation after generation, so that the population remains in equilibrium, in other world, it is not evolving, and allele frequencies do not change over time

Question 3

Thus evolution may be defined as…

Answer 3

A change in allele frequencies in a population over time

Question 4

For the genetic makeup of a population to change over time (evolve), forces (sometimes referred to as evolutionary agents)…

Answer 4

Must act to disrupt the Hardy-Weinberg equilibrium

Question 5

Terms

Answer 5

1) p + q = 1
2) p^2 + 2pq + q^2 = 1
3) p = frequency of the dominant allele in the population
4) q = frequency of the recessive allele in the population
5) p^2 = percentage of homozygous dominant individuals
6) q^2 = percentage of homozygous recessive individuals
7) 2pq = percentage of heterozygous in dividuals

Question 6

1. The allele for PTC tasting is inherited as a dominant. Use T for tasters and t for nontasters and answer the following questions
   A. What genotypes are able to taste PTC?
   B. What genotype are unable to taste PTC?
   C. If the T allele is found in 60% of the population, what is the frequency of the t allele?
   D. Using the Hardy-Weinberg formula, calculate the frequency of the homozygote dominant, heterozygote, and homozygote recessive individuals in the population

Answer 6

A. TT, Tt
B. tt
C. 40%
D. p^2 + 2pq + q^2 = 1, (0.6)^2 + 2(0.6)(0.4) + (0.4)^2 = 1, 0.36 + 0.48 + 0.16 = 1, 1 = 1, TT = 36%, Tt = 48%, tt = 16%

Question 7

2. You are interested in the frequency of people who are PTC tasters on campus. Fill in the chart

Answer 7

Phenotype/Genotype, # of students, # T Alleles, # t Alleles, Total # Alleles
Tasters / TT 10,20,0,20
Tasters / Tt 43,43,43,86
Non-tasters / tt 47,0,94,94
Totals 100,63,137,200

Question 8

2. Observed allele frequencies (how often the allele occurs in the population) Take total number of the allele in the population and divide by total number of alleles in that locus

Answer 8

T alleles = (63/200) = 0.315

# t alleles = (137/200) = 0.685

Question 9

2. Observed genotype frequencies for college students. Total number of people with that genotype divided by total number of people in the population

Answer 9

Tasters TT = 10/100 = 0.1
Tasters Tt = 43/100 = 0.43
Nontasters tt = 47/100 = 0.47

Question 10

2. Is the population in equilibrium? Calculate the expected genotype frequencies using the Hardy-Weinberg principle to determine it. p^2 + 2pq + q^2 = 1. Decimal points on exam to 4 digits

Answer 10

(0.315)^2 + 2(0.315)(0.685) + (0.685)^2 = 1
0.099225 + 0.43155 + 0.469225 = 1
0.0992 + 0.4316 + 0.4692 = 1

Question 11

2. Compare the expected genotype frequencies you got from the Hardy-Weinberg principle to the observed genotype frequencies you got from the college students. Are they similar? If so, the population is in equilibrium; if not, the population is evolving

Answer 11

Yes the population is in equilibrium
Expected, Actual
0.1, 0.099
0.43, 0.43
0.47, 0.47

Question 12

3. You are interested in the frequency of tongue rolling in a Central American village. Tongue rolling is inherited as a dominant. R for rollers, r for nonrollers. Fill out chart

Answer 12

Phenotype/Genotype, # of people, # R alleles, # r alleles, Total # alleles
Rollers RR 80,160,0,160
Rollers Rr 20,20,20,40
Nonrollers rr 100,0,200,200
Totals 200,180,220,400

Question 13

3. Calculate the observed allele frequencies of the R and r alleles using the formula p + q = 1

Answer 13

R alleles = 180/400 = 0.45

# r alleles = 220/400 = 0.55

Question 14

3. Calculate the observed genotype frequencies for each of the three genotypes

Answer 14

RR = 80/200 = 0.4
Rr = 20/200 = 0.1
rr = 100/200 = 0.5

Question 15

3. Calculate the expected genotype according to the HW formula

Answer 15

(0.45)^2 + 2(0.45)(0.55) + (0.55)^2
0.2025 + 0.4950 + 0.3025 =1

Question 16

3. Comparing your observed genotype frequencies to your expected genotype frequencies, is the population in equilibrium

Answer 16

No, population not in equilibrium because frequencies not what we expected
.4 —> .2
.1 —> .5
.5 —> .3

Question 17

4. You have sampled a population in which you know that the percentage of the homozygous recessive genotype (aa) is 36%. Using that 36% calculate the following
   A. The frequency of the “aa” genotype
   B. The frequency of the a allele
   C. The frequency of the A allele
   D. The frequencies of the genotypes “AA” and “Aa”
   E. The frequencies of the two possible phenotypes if A is completely dominant over a

Answer 17

A. 0.36
B. Sqrt 36 is 6 so 0.6
C. 0.4
D. 0.16 (p^2 = 0.4^2) 0.48 (2(0.4)(0.6))
E. (Add D results) 0.64

Question 18

5. Cystic fibrosis is a recessive condition that affects about 1 in 2,500 babies in the Caucasian population of the United States. Please calculate the following.
   A. The frequency of the recessive ALLELE in the population
   B. the frequency of the dominant allele in the population
   C. The percentage of heterozygous individuals (carriers) in the population

Answer 18

A. Sqrt 1/2500 = 0.02 (q)
B. 0.98 (p)
C. 2pq = 2 (0.02)(0.98) = 0.0392

Question 19

6. PKU is a rare autosomal recessive allele that affects 1 in 10,000 newborns. Please calculate the following
   A. The frequency of the recessive allele
   B. The frequency of the dominant allele
   C. The frequency of the heterozygous

Answer 19

A. Sqrt 1/10000 = 0.01 (q)
B. 0.99 (p)
C. 2(0.01)(0.99) = 0.0198

Question 20

7. Albinism is an autosomal recessive trait. In Europeans, the frequency of albinism varies from 1 in 10,000 to 1 in 20,000. In West Africans, the frequency is 5 in 10,000 and in some Native American tribes the frequency is 1/200.
   A. What are the allele frequencies for albinism in these populations?

Answer 20

European 1/10000 q = Sqrt 1/10000 = 0.01 p = 0.99
European 1/20000 q = Sqrt 1/20000 = 0.0071 p = 0.9929
West Africans q = Sqrt 5/10000 = 0.0224 p = 0.9776
Native Americans q= Sqrt 1/200 = 0.0707 or 0.071 p = 0.9293 or 0.929

Question 21

8. The following table gives GENOTYPE FREQUENCIES for four populations. Which are in HWE? Show calculations. For those not suggest hypothesis to explain why not at equilibrium
   Population, AA, Aa, aa
9. 25,50,25
10. 10,80,10
11. 40,20,40
12. 0,150,100
    *** = dont quite understand why

Answer 21

1. q=0.5 (cause 25 from aa +25 from Aa = 50/100 =0.5), p=0.5, (0.5)^2 +2(0.5)(0.5) + (0.5)^2 =1, 0.25 + 0.5 + 0.25 = 1 In equilibrium
2. q=0.5 (cause 10 from aa + 40 from Aa = 50/100 = 0.5), p=0.5, (0.5)^2 +2(0.5)(0.5) + (0.5)^2 =1, 0.25 + 0.5 + 0.25 = 1, Not in equilibrium because heterozygous advantage/ stabilizing selection
   ***3. q=0.5 (cause 40 from aa + 10 from Aa = 50/100 = 0.5), p=0.5, (0.5)^2 +2(0.5)(0.5) + (0.5)^2 =1, 0.25 + 0.5 + 0.25 = 1, Not in equilibrium because assortative mating/ disruptive selection
3. q= 100 from aa + 75 from Aa = 175/200 (total alleles) = 0.7, p=0.3, (0.3)^2 + 2(0.3)(0.7) + (0.7)^2 = 1, 0.09 + 0.42 + 0.49 = 1, Not in equilibrium because dominant homozygote is lethal (no AA)/ directional selection