Genetics Quiz: Allele And Genotype Frequencies

20 questions · exam conditions

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Allele And Genotype FrequenciesQuestion 1 of 20

A population's genetic structure is described by the following haplotype counts for two linked loci: AB = 400, Ab = 200, aB = 100, ab = 300. Based on these data, what is the frequency of the 'A' allele?

0.30

0.50

0.60

0.70

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Genetics Quiz

Genetics Quiz: Allele And Genotype Frequencies

Practice Allele And Genotype Frequencies in Genetics with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

What this quiz covers

This quiz focuses on Allele And Genotype Frequencies, giving you a quick way to practice the rules, question types, and explanations that matter most for Genetics.

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Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.

All questions

Question 1

0.30
0.50
0.60 (correct answer)
0.70

Explanation: The frequency of an allele can be calculated by summing the frequencies of all haplotypes that contain it. The information about linkage is extraneous to this specific question.

Calculate total number of haplotypes: Total = 400 + 200 + 100 + 300 = 1000.
Calculate haplotype frequencies:\n - $f(AB) = 400 / 1000 = 0.4$ \n - $f(Ab) = 200 / 1000 = 0.2$ \n - $f(aB) = 100 / 1000 = 0.1$ \n - $f(ab) = 300 / 1000 = 0.3$ \n3. Calculate the frequency of the 'A' allele: The 'A' allele is present in the AB and Ab haplotypes.\n $p_A = f(AB) + f(Ab) = 0.4 + 0.2 = 0.6$ .\n\nDistractor A (0.30) is the frequency of the 'ab' haplotype. Distractor B (0.50) is the frequency of the 'B' allele ( $f(B) = f(AB) + f(aB) = 0.4 + 0.1 = 0.5$ ). Distractor D (0.70) is a miscalculation.

Question 2

A large mainland population of butterflies with an allele frequency of $p=0.8$ for a wing-spot gene colonizes a new island. The founding group consists of 20 individuals from the mainland. Concurrently, 30 butterflies from a different island population, where the same allele has a frequency of $p=0.3$ , also arrive. Assuming these 50 butterflies form a single, randomly mating new population, what is the initial allele frequency for $p$ on the island?

0.50 (correct answer)
0.55
0.60
0.80

Explanation: The allele frequency in an admixed population is the weighted average of the frequencies from the source populations, weighted by their proportional contribution to the new population.\n1. Source population 1:

N_1 = 20

p_1 = 0.8

.\n2. Source population 2:

N_2 = 30

p_2 = 0.3

.\n3. Total size of the new population:

N_{total} = N_1 + N_2 = 20 + 30 = 50

.\n4. The new allele frequency

p_{new}

is calculated as:

p_{new} = \frac{(N_1 \times p_1) + (N_2 \times p_2)}{N_{total}} = \frac{(20 \times 0.8) + (30 \times 0.3)}{50} = \frac{16 + 9}{50} = \frac{25}{50} = 0.50

.\n\nDistractor B is the unweighted, simple average of the two frequencies

(0.8 + 0.3)/2 = 0.55

. Distractor C is an incorrect calculation. Distractor D is the allele frequency of the larger source population, ignoring the contribution of the smaller one.

Question 3

In a human population at Hardy-Weinberg equilibrium for the ABO blood group locus, the frequency of the $I^B$ allele is 0.1 and the frequency of the O phenotype is 49%. What is the frequency of the $I^A$ allele?

0.20 (correct answer)
0.30
0.40
0.70

Explanation: For the ABO system with alleles

I^A

(p),

I^B

(q), and

i

(r), we have

p+q+r=1

.\n1. The frequency of the O phenotype corresponds to the genotype

ii

, so its frequency is

r^2

.\n2. Given that

f(O) = 0.49

, we can find

r

r = \sqrt{0.49} = 0.7

.\n3. We are given the frequency of the

I^B

allele,

q = 0.1

.\n4. Using the equation

p+q+r=1

, we can solve for

p

(the frequency of

I^A

p = 1 - q - r = 1 - 0.1 - 0.7 = 0.2

.\n\nDistractor B (0.30) would be the answer if

r

was mistaken for 0.6. Distractor C (0.40) might result from miscalculation. Distractor D (0.70) is the value of

r

, the frequency of the

i

allele.

Question 4

The incidence of an autosomal recessive condition is 1 in 6,400 individuals in a population assumed to be in Hardy-Weinberg equilibrium. What is the approximate frequency of heterozygous carriers for this condition?

0.000156
0.0125
0.0247 (correct answer)
0.9875

Explanation: This is a multi-step calculation based on the Hardy-Weinberg equilibrium principle.\n1. The incidence of the condition corresponds to the frequency of the homozygous recessive genotype (aa), which is

q^2

. So,

q^2 = 1/6400 = 0.00015625

.\n2. The frequency of the recessive allele (a) is

q = \sqrt{q^2} = \sqrt{1/6400} = 1/80 = 0.0125

.\n3. The frequency of the dominant allele (A) is

p = 1 - q = 1 - 0.0125 = 0.9875

.\n4. The frequency of heterozygous carriers (Aa) is

2pq = 2 \times 0.9875 \times 0.0125 \approx 0.0247

.\n\nDistractor A is

q^2

, the incidence of the disease. Distractor B is

q

, the frequency of the recessive allele. Distractor D is

p

, the frequency of the dominant allele.

Question 5

In haplodiploid honey bees, females are diploid and males (drones) are haploid. A queen bee has the genotype Bb for a particular gene. She mates with drones from a population where the frequency of the B allele is 0.4. What is the expected frequency of the B allele among her female (worker) offspring?

0.40
0.45 (correct answer)
0.50
0.70

Explanation: Female workers inherit one set of chromosomes from their mother (the queen) and one from their father (a drone). To find the allele frequency in the workers, we must average the allele frequencies from the egg and sperm pools.\n1. Allele frequency from the queen's eggs: The queen is genotype Bb. Due to meiosis, her gametes (eggs) will carry the B allele with a frequency of 0.5 and the b allele with a frequency of 0.5. So,

p_{egg} = 0.5

.\n2. Allele frequency from the drone's sperm: Drones are haploid. The allele frequency in the drone population's sperm is equal to the allele frequency in the drones themselves. So,

p_{sperm} = 0.4

.\n3. Allele frequency in the diploid female offspring: The frequency in the offspring is the average of the frequencies from the two gamete pools.

p_{worker} = (p_{egg} + p_{sperm}) / 2 = (0.5 + 0.4) / 2 = 0.9 / 2 = 0.45

.\n\nDistractor A (0.40) is the allele frequency from the drones. Distractor C (0.50) is the allele frequency from the queen. Distractor D is a miscalculation.

Question 6

A researcher genotyping a wild barley population finds the following counts for three alleles at one locus: Allele 1 = 150 copies, Allele 2 = 250 copies, Allele 3 = 100 copies. Assuming the population is in Hardy-Weinberg equilibrium, what is the expected frequency of individuals heterozygous for Allele 1 and Allele 3?

0.06
0.12 (correct answer)
0.30
0.50

Explanation: This question requires calculating allele frequencies from allele counts and then using the HWE principle for multiple alleles to find a specific heterozygote frequency.

Calculate allele frequencies:\n - Total alleles = 150 + 250 + 100 = 500.\n - Frequency of Allele 1 (p1) = 150 / 500 = 0.30.\n - Frequency of Allele 2 (p2) = 250 / 500 = 0.50.\n - Frequency of Allele 3 (p3) = 100 / 500 = 0.20.
Calculate expected heterozygote frequency:\n - The genotype for individuals heterozygous for Allele 1 and Allele 3 is A1A3.\n - In HWE, the frequency of a specific heterozygote is given by $2 \times p_i \times p_j$ .\n - $f(A1A3) = 2 \times p1 \times p3 = 2 \times 0.30 \times 0.20 = 0.12$ .\n\nDistractor A (0.06) is the product of the allele frequencies (p1 * p3), a common error where the '2' is forgotten. Distractor C (0.30) is the frequency of Allele 1. Distractor D (0.50) is the frequency of Allele 2.

Question 7

In a population assumed to be in Hardy-Weinberg equilibrium, a form of X-linked color blindness affects 9% of males. What is the expected frequency of females who are heterozygous carriers of the color blindness allele?

0.0081
0.090
0.164 (correct answer)
0.810

Explanation: For an X-linked trait in HWE, the frequency of the trait in males directly reflects the allele frequency in the population.\n1. The frequency of affected males is equal to the frequency of the recessive allele,

q

. Given that 9% of males are affected,

q = 0.09

.\n2. The frequency of the dominant allele is

p = 1 - q = 1 - 0.09 = 0.91

.\n3. In females, genotype frequencies follow the standard HWE formula:

p^2

2pq

q^2

.\n4. The frequency of heterozygous female carriers is

2pq = 2 \times 0.91 \times 0.09 = 0.1638 \approx 0.164

.\n\nDistractor A (0.0081) is

q^2

, the frequency of affected females. Distractor B (0.090) is

q

, the frequency of the allele and of affected males. Distractor D (0.810) is

p^2

, the frequency of homozygous dominant females.

Question 8

In a population at equilibrium, a completely recessive lethal allele has a frequency $q_0 = 0.05$ . After one generation of selection removes all homozygous recessive individuals before they can reproduce, what is the frequency of this lethal allele ( $q_1$ ) among the surviving adults?

0.0250
0.0476 (correct answer)
0.0500
0.0526

Explanation: When selection acts against a homozygous recessive genotype, the allele frequency in the next generation can be calculated by considering the composition of the surviving population. The formula for the new recessive allele frequency (

q_1

) after one generation of complete selection against the recessive homozygote is

q_1 = q_0 / (1 + q_0)

.\nGiven

q_0 = 0.05

:\n

q_1 = 0.05 / (1 + 0.05) = 0.05 / 1.05 \approx 0.0476

.\n\nDerivation: Initial genotype frequencies:

p_0^2

2p_0q_0

q_0^2

. After selection, the 'aa' (frequency

q_0^2

) are removed. The new total population size is proportional to

1 - q_0^2

. The frequency of the 'a' allele is half the frequency of the surviving heterozygotes, rescaled to the new population size:

q_1 = (p_0q_0) / (1 - q_0^2) = p_0q_0 / ((1-q_0)(1+q_0)) = p_0q_0 / (p_0(1+q_0)) = q_0 / (1+q_0)

.\n\nDistractor A is a miscalculation. Distractor C is the initial allele frequency,

q_0

, representing no change. Distractor D corresponds to the formula

q_0 / (1 - q_0)

, which represents a sign error in the denominator.

Question 9

Achondroplasia is a rare autosomal dominant disorder. In a particular population, its prevalence is approximately 1 in 25,000. Assuming most affected individuals are heterozygous, what is the most reasonable estimate for the frequency of the dominant allele that causes this disorder?

0.00002 (correct answer)
0.00004
0.00316
0.00632

Explanation: For a dominant disorder, the prevalence represents the frequency of affected individuals, which is

f(AA) + f(Aa) = p^2 + 2pq

.\n1. The prevalence is

1/25000 = 0.00004

.\n2. For a rare dominant allele,

p

is very small, and

q = 1-p

is very close to 1.\n3. Consequently, the

p^2

term (homozygous dominant) is negligible compared to the

2pq

term (heterozygous). Most affected individuals are heterozygotes.\n4. Therefore, we can approximate the prevalence as

2pq \approx 2p(1) = 2p

.\n5. So,

2p \approx 0.00004

.\n6. Solving for

p

gives

p \approx 0.00004 / 2 = 0.00002

.\n\nDistractor B is the prevalence of the disorder itself. Distractor C (0.00632) is the result of incorrectly treating it as a recessive disorder (

p = \sqrt{1/25000}

). Distractor D is double the value from distractor C.

Question 10

A geneticist uses pooled sequencing to estimate allele frequencies in a population of 500 diploid insects. The DNA from all individuals is combined and sequenced. At a target SNP, a total of 20,000 sequence reads are obtained. Of these, 4,000 reads correspond to allele 'G' and 16,000 reads correspond to allele 'A'. Based on this data, what is the estimated frequency of the genotype 'GG' in the population, assuming it is in Hardy-Weinberg equilibrium?

0.04 (correct answer)
0.20
0.32
0.64

Explanation: This question combines a modern data scenario (pooled sequencing) with classic HWE calculations.

Estimate allele frequencies from read counts: The proportion of reads for an allele estimates its frequency in the population's gene pool.
- Frequency of G (q) = (Reads for G) / (Total Reads) = 4,000 / 20,000 = 0.20.
- Frequency of A (p) = (Reads for A) / (Total Reads) = 16,000 / 20,000 = 0.80.
Calculate expected genotype frequency using HWE: The question asks for the frequency of the 'GG' genotype, which is q².
- f(GG) = q² = (0.20)² = 0.04.

Distractor B (0.20) is the frequency of the G allele (q). Distractor C (0.32) is the frequency of the heterozygote AG (2pq = 2 × 0.8 × 0.2 = 0.32). Distractor D (0.64) is the frequency of the AA genotype (p²).

Question 11

A gene controlling mammal coat color has three alleles with a dominance hierarchy: $C$ (full color) > $c^{ch}$ (chinchilla) > $c^h$ (Himalayan). In a population at Hardy-Weinberg equilibrium, the allele frequencies are $f(C) = 0.2$ and $f(c^h) = 0.5$ . What is the expected frequency of the chinchilla phenotype?

0.09
0.30
0.39 (correct answer)
0.51

Explanation: This is a multi-step problem involving a three-allele system.\n1. First, determine the frequency of the third allele,

c^{ch}

. Let

p = f(C)

q = f(c^{ch})

, and

r = f(c^h)

. Since

p+q+r=1

q = 1 - p - r = 1 - 0.2 - 0.5 = 0.3

.\n2. Identify the genotypes that produce the chinchilla phenotype. Due to the dominance hierarchy, chinchilla is produced by the genotypes

c^{ch}c^{ch}

and

c^{ch}c^h

.\n3. Calculate the frequencies of these genotypes assuming HWE:\n -

f(c^{ch}c^{ch}) = q^2 = (0.3)^2 = 0.09

\n -

f(c^{ch}c^h) = 2qr = 2 \times 0.3 \times 0.5 = 0.30

\n4. The total frequency of the chinchilla phenotype is the sum of these genotype frequencies:

0.09 + 0.30 = 0.39

.\n\nDistractor A (0.09) is the frequency of only the homozygous chinchilla genotype. Distractor B (0.30) is the frequency of the

c^{ch}

allele. Distractor D (0.51) is the frequency of the full color phenotype (

p^2 + 2pq + 2pr

Question 12

0.000156
0.0125
0.0247 (correct answer)
0.9875

q^2