Hardy-Weinberg Equilibrium Practice Problems/Quiz Questions

What is the difference between micro- and macroevolution?

Microevolution describes the evolution of organisms in populations, while macroevolution describes the evolution of species over long periods of time.

Population genetics is the study of:

How selective forces change the allele frequencies in a population over time

Which of the following populations is not in Hardy-Weinberg equilibrium:

A. A population with 12 homozygous recessive individuals (yy), 8 homozygous dominant individuals (YY), and 4 heterozygous individuals (Yy)

B. A population in which the allele frequencies do not change over time

C. p² + 2pq + q² = 1

D. A population undergoing natural selection

D. A population undergoing natural selection

One of the original Amish colonies rose from a ship of colonists that came from Europe. The ship's captain, who had polydactyly, a rare dominant trait, was one of the original colonists. Today, we see a much higher frequency of polydactyly in the Amish population. This is an example of:

A. Natural selection

B. Genetic drift

C. Founder effect

D. b and c

D. b and c

T/F: A dominant trait is more likely to become frequent in a population because it is dominant to other traits

False

T/F: A population that is in Hardy-Weinberg equilibrium is not evolving

True

Which of the following is NOT an assumption of the Hardy-Weinberg Model:

A. Assortative mating

B. An infinitely sized population

C. No migration

D. No mutation or selection

A. Assortative mating

Give the formula for the following frequencies:

1. Frequency of p allele

2. Frequency of qq genotype

3. Frequency of pp genotype

4. Frequency of of pq genotype

1. 1-q

2. q²

3. p²

4. 2pq

What happens to the allele frequency of the recessive allele when Mendelian genetics rules are followed by a randomly mating population?

The population can be observed in Hardy-Weinberg equilibrium

Which of the following best describes the Modern Synthesis of Biology:

A. A theory to explain how biodiversity developed

B. A model for how allele frequencies will stay the same when there are no mechanisms for evolution

C. The combining of Mendelian genetics with evolutionary theory

D. A mathematical framework to explain genetic diversity that does not create phenotypic diversity

C. The combining of Mendelian genetics with evolutionary theory

When can a real population be assumed to be in Hardy-Weinberg Equilibrium (HWE)?

When the allele frequencies are observed to be unchanging over time

If 2/1000 individuals have the Sickle Cell phenotype, what is the frequency of heterozygotes? Assume the population is in HWE

1. What information is given to us?

• Allele frequency or genotype frequency?

• “Sickle Cell phenotype” = genotype frequency of recessive allele

2. Find one allele frequency.

• p? q?

• “Sickle Cell” recessive genotype = q2 = 2/1000

• q = √(2/1000) = 0.045

3. What is the problem asking for?

• Allele frequency or genotype frequency?

• “heterozygotes” = genotype frequency of pq and qp

• If q = 0.045, and p = 1 – q, then p = 0.955

• Genotype frequency of pq and qp = 2pq = 2*0.045*0.955 = 0.086

The frequency of recessive homozygotes at a particular locus is 0.36. Assuming no evolutionary forces are acting, what do you expect the frequency of the homozygous dominant genotype to be?

1. What information is given to us?

• Allele frequency or genotype frequency?

• “recessive homozygotes” = genotype frequency of recessive allele

2. Find one allele frequency.

• p? q?

• recessive genotype = q2 = 0.36

• q = √(0.36) = 0.6

3. What is the problem asking for?

• Allele frequency or genotype frequency?

• “homozygous dominant” = genotype frequency of p2

• If q = 0.6, and p = 1 – q, then p = .4

• Genotype frequency of p2 = (.4)2 = 0.16

What are the observed frequencies of the AA, AS, and SS genotypes?

f(AA) = 9365/12387 = 0.756

f(AS) = 2993/12387 = 0.242

F(SS) = 29/12387 = 0.002

Compare the observed frequency of heterozygotes to the expectation we calculated earlier from the observed frequency of the Sickle Cell phenotype

The observed frequency of heterozygotes (0.242) is significantly higher than the expected frequency calculated under HWE (0.086). There is a distinct excess of heterozygotes (AS) in the population. Conclusion: No, the population is NOT in Hardy-Weinberg Equilibrium. The observed genotype frequencies differ substantially from the expected Hardy-Weinberg frequencies, indicating that evolutionary forces (specifically heterozygote advantage / balancing selection, which is classic for the sickle cell allele in malaria-endemic regions) are actively acting on this population.

What are the observed frequencies of the A and S alleles?

Count method: f(p) = 2(N_pp) + 1(N_pq)+0(N_qq) / 2(N_Tot)

f(A) = (2*9365 + 2993)/(2*12387) =0.877

F(S) = (2*29 + 2993)/(2*12387) =0.123

Assume a locus in a diploid population of 100 individuals has the genotype frequencies: AA = 64; Aa = 20; and aa = 16. What is the observed allele frequency (p) of the A allele? Is it in HWE?

First, calculate observed frequencies.

f(A) = (2*64 + 20)/(2*100) f(a) = (2*16 + 20)/(2*100)

=0.74 =0.26

Then, calculate expected frequencies.

f(AA) = (0.74)² = 0.5476 f(Aa) = 2(0.74)(0.26) = 0.3848 f(a) = (0.26)² = 0.0676

But is it significantly different?

Χ² = (64 − 54.76)²/54.76 + (20 − 38.48)²/38.48 + (16 − 6.76)²/6.76 = 23.07

23.07 > 3.841 so the difference is significant

Chi-Square analysis can ONLY be used on population counts!

Polydactyly (F) is dominant phenotype in which individuals born with a single copy of the allele grow more than 10 fingers. Embryos that inherit two copies of the polydactyly allele (FF) are inviable and die early in pregnancy. Please keep four (4) decimal places.

1. About 1 in 750 births produce the polydactyly phenotype (Ff). What is the observed frequency of the polydactyly phenotype?

1/750 = 0.0013

What is the observed frequency of the non-polydactyly (f) allele?

1 heterozygote in 750 births means 1 F allele in 1500 alleles

f(f) = 1499/1500 = 0.9993

If polydactyly were in Hardy-Weinberg equilibrium (HWE), what would the predicted frequency of the polydactyly phenotype be?

1 – 0.9993 = f(f) = 0.0007

2(0.9993)(0.0007) = 0.0014

Is polydactyly in HWE?

(749 − 748.9504)²/748.9504 + (1 − 1.05)²/1.05 + (0 − 0.00000049)²/0.00000049 = 0.002383

0.002383 < 3.841

Not significant, therefore the gene is in HWE