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Genetics Quiz

Genetics Quiz: Penetrance And Expressivity

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

Question 1 / 20

0 of 20 answered

A specific mutation in the CFTR gene causes cystic fibrosis, leading to a spectrum of health issues including chronic lung infections, pancreatic insufficiency, and male infertility. While essentially all individuals with this genotype develop some health problems, the severity of the lung disease can range from mild to life-threatening. Which pair of genetic concepts correctly describes the multiple organ systems affected and the differing severity of lung disease, respectively?

Select an answer to continue

What this quiz covers

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

How to use this quiz

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

A specific mutation in the CFTR gene causes cystic fibrosis, leading to a spectrum of health issues including chronic lung infections, pancreatic insufficiency, and male infertility. While essentially all individuals with this genotype develop some health problems, the severity of the lung disease can range from mild to life-threatening. Which pair of genetic concepts correctly describes the multiple organ systems affected and the differing severity of lung disease, respectively?

  1. Incomplete penetrance and variable expressivity
  2. Pleiotropy and variable expressivity (correct answer)
  3. Polygenic inheritance and incomplete penetrance
  4. Pleiotropy and uniform expressivity

Explanation: The fact that a single gene mutation (in CFTR) affects multiple, distinct organ systems (lungs, pancreas, reproductive tract) is the definition of pleiotropy. The observation that the severity of the lung disease varies among individuals with the same genotype is the definition of variable expressivity. The stem says "essentially all individuals...develop some health problems," which points to complete or near-complete penetrance, making choices with incomplete penetrance less likely. Polygenic inheritance refers to a trait controlled by multiple genes, which is not the case here.

Question 2

Marfan syndrome results from mutations in the FBN1 gene and is inherited as an autosomal dominant trait. Affected individuals can have a wide range of clinical features, including being unusually tall, having long limbs and fingers (arachnodactyly), and developing severe cardiovascular problems like aortic aneurysms. The severity and combination of these features differ significantly among patients, even within the same family sharing the identical FBN1 mutation. Which genetic concept is best illustrated by this variation in the type and severity of symptoms among affected individuals?

  1. Variable expressivity (correct answer)
  2. Incomplete penetrance
  3. Allelic heterogeneity
  4. Locus heterogeneity

Explanation: The question focuses on the variation in the severity and combination of symptoms among individuals who share the same causative mutation. This phenomenon, where a single genotype produces a range of phenotypic expressions, is defined as variable expressivity. Incomplete penetrance refers to individuals with the genotype who show no phenotype, which is not the focus here. Allelic heterogeneity refers to different mutations in the same gene causing the condition, while locus heterogeneity refers to mutations in different genes causing a similar condition; neither describes variation among those with the identical mutation.

Question 3

A 30-year-old man tests positive for a pathogenic variant in the BRCA1 gene. This specific variant is inherited in an autosomal dominant manner and has an estimated lifetime penetrance of 60% for breast cancer in males. Which of the following is the most accurate counseling statement regarding his personal risk for developing breast cancer?

  1. He will eventually develop breast cancer, but its severity cannot be predicted.
  2. He has a 30% chance of developing breast cancer, calculated by multiplying his 50% inheritance risk by the 60% penetrance.
  3. Each of his children will have a 60% chance of developing breast cancer.
  4. He has a 60% chance of developing breast cancer in his lifetime due to this specific genetic variant. (correct answer)

Explanation: Penetrance is the probability that an individual with the causative genotype will express the phenotype. Since the man has already tested positive for the variant, his chance of having the genotype is 100%. Therefore, his lifetime risk of developing the associated cancer is equal to the penetrance, which is 60%. Choice A is incorrect because 40% of carriers will not develop the cancer (incomplete penetrance). Choice B is incorrect because he already has the variant; the 50% inheritance probability is no longer relevant for him. Choice C incorrectly applies the penetrance value to his children's risk of disease instead of their risk of inheritance (which is 50%).

Question 4

An autosomal dominant allele 'T' causes a particular trait. The frequency of this allele in a large, randomly mating population is 0.02. If the penetrance of the allele is 75%, what is the approximate expected frequency of individuals expressing the trait in this population?

  1. 1.5%
  2. 2.0%
  3. 3.0% (correct answer)
  4. 4.0%

Explanation: First, calculate the frequencies of the genotypes that can cause the trait using the Hardy-Weinberg equation. Let p = frequency of T = 0.02, and q = frequency of t = 0.98. The causative genotypes are TT and Tt. Frequency(TT) = p² = (0.02)² = 0.0004. Frequency(Tt) = 2pq = 2(0.02)(0.98) = 0.0392. Second, calculate the frequency of affected individuals by multiplying the frequency of each causative genotype by the penetrance (0.75). Affected from TT = 0.0004 * 0.75 = 0.0003. Affected from Tt = 0.0392 * 0.75 = 0.0294. The total frequency of individuals expressing the trait is the sum of these two frequencies: 0.0003 + 0.0294 = 0.0297. This is approximately 0.030, or 3.0%.

Question 5

An individual heterozygous for an autosomal dominant allele that causes Polydactyly partners with an individual who is homozygous recessive. The condition has 90% penetrance. What is the probability that their first child will be phenotypically unaffected?

  1. 10%
  2. 45%
  3. 50%
  4. 55% (correct answer)

Explanation: A child can be phenotypically unaffected in two ways: 1) they do not inherit the dominant allele, or 2) they inherit the dominant allele but do not express the trait (non-penetrance). Let 'A' be the dominant allele and 'a' be the recessive allele. The cross is Aa × aa. The probability of the child inheriting the 'a' allele from the heterozygous parent is 1/2, resulting in genotype aa and an unaffected phenotype. P(aa) = 0.5. The probability of the child inheriting the 'A' allele is 1/2. If the child has the 'A' allele (genotype Aa), the penetrance is 90%, so the probability of being non-penetrant (unaffected) is 1 - 0.90 = 0.10. The probability of this combined event is P(Aa) × P(unaffected|Aa) = 0.5 × 0.10 = 0.05. The total probability of being unaffected is the sum of the probabilities of these two mutually exclusive events: P(unaffected) = P(aa) + P(unaffected and Aa) = 0.5 + 0.05 = 0.55, or 55%.

Question 6

Two different families present with members affected by hypercholesterolemia. In Family 1, all affected individuals have a deletion of exon 4 in the LDLR gene and exhibit moderate to severe symptoms. In Family 2, all affected members have a missense mutation in exon 10 of the same LDLR gene and consistently present with a very mild form of the condition. The difference in clinical presentation between these two families is best described as an example of:

  1. Variable expressivity
  2. Incomplete penetrance
  3. Locus heterogeneity
  4. Allelic heterogeneity (correct answer)

Explanation: The scenario describes two different mutations (alleles) within the same gene (LDLR) that lead to clinically distinct forms of the same disease (hypercholesterolemia). This phenomenon is known as allelic heterogeneity. Variable expressivity is a plausible distractor, but it refers to the variation in phenotype among individuals sharing the same allele, which is not what is being compared here. Locus heterogeneity would involve mutations in different genes causing the condition. Incomplete penetrance is not mentioned in the scenario.

Question 7

An individual heterozygous for an autosomal dominant allele that causes Polydactyly partners with an individual who is homozygous recessive. The condition has 90% penetrance. What is the probability that their first child will be phenotypically unaffected?

  1. 10%
  2. 45%
  3. 50%
  4. 55% (correct answer)

Explanation: A child can be phenotypically unaffected in two ways: 1) they do not inherit the dominant allele, or 2) they inherit the dominant allele but do not express the trait (non-penetrance). Let 'A' be the dominant allele and 'a' be the recessive allele. The cross is Aa × aa. The probability of the child inheriting the 'a' allele from the heterozygous parent is 1/2, resulting in genotype aa and an unaffected phenotype. P(aa) = 0.5. The probability of the child inheriting the 'A' allele is 1/2. If the child has the 'A' allele (genotype Aa), the penetrance is 90%, so the probability of being non-penetrant (unaffected) is 1 - 0.90 = 0.10. The probability of this combined event is P(Aa) × P(unaffected|Aa) = 0.5 × 0.10 = 0.05. The total probability of being unaffected is the sum of the probabilities of these two mutually exclusive events: P(unaffected) = P(aa) + P(unaffected and Aa) = 0.5 + 0.05 = 0.55, or 55%.

Question 8

A geneticist identifies a pathogenic variant responsible for a dominant disorder that is present in 1 out of every 200 individuals in a population. Clinical records show that the prevalence of the actual disease is 1 out of every 250 individuals. Which of the following statements is the most accurate conclusion?

  1. The penetrance of the variant is 125%.
  2. The expressivity of the disease is 80%.
  3. The penetrance of the variant is 80%. (correct answer)
  4. The prevalence of the variant is lower than the prevalence of the disease.

Explanation: Penetrance is the ratio of individuals with the phenotype (disease) to individuals with the genotype (variant). The frequency of the genotype is 1/200. The frequency of the phenotype is 1/250. Penetrance = (Frequency of Phenotype) / (Frequency of Genotype) = (1/250) / (1/200) = 200/250 = 0.80, or 80%. Choice A is a nonsensical value for penetrance. Choice B misuses the term expressivity, which refers to the range of symptoms, not a calculated percentage in this context. Choice D is contradicted by the data (1/200 > 1/250).

Question 9

A new autosomal dominant syndrome is characterized by two main features: vision loss and hearing impairment. A study of 500 individuals, all heterozygous for the same causative mutation, reveals the following: 100 individuals remain completely asymptomatic; 250 develop only vision loss; 50 develop only hearing impairment; and 100 develop both vision and hearing loss. What is the penetrance of this syndrome?

  1. 20%
  2. 60%
  3. 80% (correct answer)
  4. 100%

Explanation: Penetrance is the proportion of individuals with a given genotype who exhibit any part of the associated phenotype. The total number of individuals with the mutation is 500. The number of asymptomatic individuals is 100. Therefore, the number of symptomatic (affected) individuals is 500 - 100 = 400. The penetrance is the number of affected individuals divided by the total number of carriers: 400 / 500 = 0.80, or 80%. The different combinations of symptoms (vision only, hearing only, both) are an example of variable expressivity but do not change the overall calculation of penetrance.

Question 10

A rare autosomal dominant neurological disorder is studied. In a large population at Hardy-Weinberg equilibrium, the frequency of the disease-causing allele (A) is 0.001. A survey finds the prevalence of the disorder (the proportion of affected individuals) to be 1 in 1,250. Assuming all affected individuals are heterozygotes due to the rarity of the allele, what is the penetrance of the A allele?

  1. 20%
  2. 40% (correct answer)
  3. 80%
  4. 100%

Explanation: First, calculate the expected frequency of the heterozygous genotype (Aa) in the population using the Hardy-Weinberg equation. The frequency of the allele A (p) is 0.001, so the frequency of the allele a (q) is 1 - 0.001 = 0.999. The frequency of heterozygotes is 2pq = 2 * 0.001 * 0.999 ≈ 0.002. This is equivalent to 1 in 500 individuals (1/0.002). The observed prevalence of the phenotype is 1 in 1,250. Penetrance is the ratio of the observed phenotype frequency to the causative genotype frequency. Penetrance = (Phenotype Frequency) / (Genotype Frequency) = (1/1,250) / (1/500) = 500 / 1,250 = 0.40, or 40%.

Question 11

Which of the following scenarios best illustrates the principle of variable expressivity without clear evidence of incomplete penetrance?

  1. In a family with an autosomal dominant cancer syndrome, 2 of 5 members with the mutation develop cancer, while 3 remain cancer-free.
  2. All individuals with a specific mutation for osteogenesis imperfecta have brittle bones, but some experience hundreds of fractures while others have only a few. (correct answer)
  3. A mutation for a neurological disorder is present in 1% of the population, but the disorder is only seen in 0.5% of the population.
  4. An individual with an autosomal dominant mutation is asymptomatic but passes the allele to a child who is severely affected by the disorder.

Explanation: Variable expressivity refers to the range of phenotypic severity among individuals with the same genotype. Choice B is the best example: all individuals are affected (complete penetrance), but the degree to which they are affected (number of fractures) varies greatly. Choice A and C are clear examples of incomplete penetrance. Choice D demonstrates incomplete penetrance in the parent. While the child is affected, the key distinction is that in scenario B, there is no mention of non-penetrant individuals.

Question 12

An individual who is heterozygous (Aa) for a dominant allele causing a disorder with 80% penetrance has a child with an individual who is homozygous recessive (aa). What is the probability that their child will be a phenotypically normal carrier of the disease allele?

  1. 10% (correct answer)
  2. 20%
  3. 40%
  4. 50%

Explanation: This question asks for the probability of a specific combined outcome: the child must have the causative genotype AND be phenotypically normal. First, the probability that the child inherits the disease allele 'A' from the heterozygous parent is 1/2 or 50%. This means the child's genotype is Aa. Second, given the child has the Aa genotype, the probability of expressing the phenotype is the penetrance, 80%. The probability of not expressing the phenotype (being a non-penetrant carrier) is 1 - penetrance = 1 - 0.80 = 0.20 or 20%. The total probability of being a phenotypically normal carrier is the product of these two probabilities: P(inheriting 'A') × P(non-penetrance | has 'A') = 0.50 × 0.20 = 0.10, or 10%.

Question 13

Split-hand/foot malformation (SHFM) is a condition that can be inherited as an autosomal dominant trait. In an extensive study of a family with a single causative mutation, 100 individuals are determined to be carriers of the mutation. Of these, 70 show some form of SHFM, while 30 have completely normal hands and feet. Of the 70 affected individuals, 20 have malformations of all four limbs, while 50 have malformations affecting only one or two limbs. Which statement accurately quantifies a genetic principle in this family?

  1. The expressivity is 70% and the penetrance is variable.
  2. The penetrance is 30% and the expressivity is variable.
  3. The penetrance is 70% and the condition shows variable expressivity. (correct answer)
  4. The expressivity is 100% in penetrant individuals.

Explanation: Penetrance is the percentage of carriers who show the phenotype. Here, 70 out of 100 carriers are affected, so the penetrance is 70/100 = 70%. Expressivity describes the range of symptoms in affected individuals. Since affected individuals can have one, two, or all four limbs affected, the condition shows variable expressivity. Choice A confuses the terms. Choice B has the incorrect penetrance value (30% is the rate of non-penetrance). Choice D is incorrect because expressivity is variable, not uniform, among those who are penetrant.

Question 14

Which of the following scenarios best illustrates the principle of variable expressivity without clear evidence of incomplete penetrance?

  1. In a family with an autosomal dominant cancer syndrome, 2 of 5 members with the mutation develop cancer, while 3 remain cancer-free.
  2. All individuals with a specific mutation for osteogenesis imperfecta have brittle bones, but some experience hundreds of fractures while others have only a few. (correct answer)
  3. A mutation for a neurological disorder is present in 1% of the population, but the disorder is only seen in 0.5% of the population.
  4. An individual with an autosomal dominant mutation is asymptomatic but passes the allele to a child who is severely affected by the disorder.

Explanation: Variable expressivity refers to the range of phenotypic severity among individuals with the same genotype. Choice B is the best example: all individuals are affected (complete penetrance), but the degree to which they are affected (number of fractures) varies greatly. Choice A and C are clear examples of incomplete penetrance. Choice D demonstrates incomplete penetrance in the parent. While the child is affected, the key distinction is that in scenario B, there is no mention of non-penetrant individuals.

Question 15

Neurofibromatosis type 1 (NF1) is an autosomal dominant condition caused by mutations in the NF1 gene. Some individuals with a pathogenic allele develop hundreds of neurofibromas, café-au-lait spots, and skeletal abnormalities. Others with the same allele may only have a few café-au-lait spots and no other symptoms. Furthermore, a small percentage of individuals who carry a pathogenic allele show no clinical symptoms of the disorder throughout their lives.

Based on the passage, which concepts best explain the clinical observations associated with NF1?

  1. Incomplete penetrance only
  2. Variable expressivity only
  3. Incomplete penetrance and variable expressivity (correct answer)
  4. Pleiotropy and complete penetrance

Explanation: The passage describes two key phenomena. The fact that individuals with the same pathogenic allele can show a wide range of symptoms, from very mild (a few spots) to severe (hundreds of tumors), is a classic example of variable expressivity. The fact that a small percentage of individuals with the allele show no clinical symptoms at all is the definition of incomplete penetrance. Therefore, both concepts are required to fully explain the observations.

Question 16

A dominant allele 'G' is associated with a specific phenotype. In a survey of 1,000 individuals, genetic testing reveals 150 individuals with genotype Gg and 10 with genotype GG. Phenotypic assessment finds that 96 individuals in the total sample exhibit the trait. Assuming the penetrance is the same for both Gg and GG genotypes, what is the penetrance of the 'G' allele in this population?

  1. 60.0% (correct answer)
  2. 64.0%
  3. 85.0%
  4. 96.0%

Explanation: Penetrance is the proportion of individuals with a causative genotype who exhibit the corresponding phenotype. First, determine the total number of individuals with the causative genotype. This is the sum of individuals with Gg and GG genotypes: 150 + 10 = 160 individuals. The problem states that 96 of these individuals exhibit the phenotype. Penetrance is calculated as (Number of affected individuals) / (Total number of individuals with the causative genotype). Therefore, Penetrance = 96 / 160 = 0.60, or 60.0%.

Question 17

A 30-year-old man tests positive for a pathogenic variant in the BRCA1 gene. This specific variant is inherited in an autosomal dominant manner and has an estimated lifetime penetrance of 60% for breast cancer in males. Which of the following is the most accurate counseling statement regarding his personal risk for developing breast cancer?

  1. He will eventually develop breast cancer, but its severity cannot be predicted.
  2. He has a 30% chance of developing breast cancer, calculated by multiplying his 50% inheritance risk by the 60% penetrance.
  3. Each of his children will have a 60% chance of developing breast cancer.
  4. He has a 60% chance of developing breast cancer in his lifetime due to this specific genetic variant. (correct answer)

Explanation: Penetrance is the probability that an individual with the causative genotype will express the phenotype. Since the man has already tested positive for the variant, his chance of having the genotype is 100%. Therefore, his lifetime risk of developing the associated cancer is equal to the penetrance, which is 60%. Choice A is incorrect because 40% of carriers will not develop the cancer (incomplete penetrance). Choice B is incorrect because he already has the variant; the 50% inheritance probability is no longer relevant for him. Choice C incorrectly applies the penetrance value to his children's risk of disease instead of their risk of inheritance (which is 50%).

Question 18

A rare autosomal dominant neurological disorder is studied. In a large population at Hardy-Weinberg equilibrium, the frequency of the disease-causing allele (A) is 0.001. A survey finds the prevalence of the disorder (the proportion of affected individuals) to be 1 in 1,250. Assuming all affected individuals are heterozygotes due to the rarity of the allele, what is the penetrance of the A allele?

  1. 20%
  2. 40% (correct answer)
  3. 80%
  4. 100%

Explanation: First, calculate the expected frequency of the heterozygous genotype (Aa) in the population using the Hardy-Weinberg equation. The frequency of the allele A (p) is 0.001, so the frequency of the allele a (q) is 1 - 0.001 = 0.999. The frequency of heterozygotes is 2pq = 2 * 0.001 * 0.999 ≈ 0.002. This is equivalent to 1 in 500 individuals (1/0.002). The observed prevalence of the phenotype is 1 in 1,250. Penetrance is the ratio of the observed phenotype frequency to the causative genotype frequency. Penetrance = (Phenotype Frequency) / (Genotype Frequency) = (1/1,250) / (1/500) = 500 / 1,250 = 0.40, or 40%.

Question 19

A 25-year-old woman's father was recently diagnosed with Huntington's disease, an autosomal dominant disorder with age-dependent penetrance. The penetrance is near 0% at age 25, rises to 50% by age 40, and approaches 100% by age 65. The woman is currently asymptomatic. Which statement is the most accurate assessment of her situation?

  1. Since she is asymptomatic at 25, her risk of having inherited the allele is now significantly less than 50%.
  2. Her overall probability of developing the disease by age 40 is 50%.
  3. Her risk of having inherited the allele is 50%, and her lack of symptoms at this age provides very little information about her carrier status. (correct answer)
  4. If she has a child, that child has a 50% chance of being a carrier, regardless of her future symptom development.

Explanation: As the child of an affected individual, her prior risk of inheriting the autosomal dominant allele is 50%. Because the penetrance of Huntington's disease is near zero at her current age of 25, being asymptomatic is expected whether she carries the allele or not. Therefore, her current health status does little to change the initial 50% probability that she is a carrier. Choice A is incorrect because being symptom-free at an age where symptoms are not expected does not significantly reduce her genetic risk. Choice B is incorrect; her overall risk of being symptomatic by 40 is 0.5 (chance to inherit) × 0.5 (penetrance at 40) = 25%. Choice D is a common error; the child's risk is 50% only if she is a carrier. Her overall risk of passing it on is 25% (0.5 chance she has it * 0.5 chance she passes it on).

Question 20

An autosomal dominant allele 'T' causes a particular trait. The frequency of this allele in a large, randomly mating population is 0.02. If the penetrance of the allele is 75%, what is the approximate expected frequency of individuals expressing the trait in this population?

  1. 1.5%
  2. 2.0%
  3. 3.0% (correct answer)
  4. 4.0%

Explanation: First, calculate the frequencies of the genotypes that can cause the trait using the Hardy-Weinberg equation. Let p = frequency of T = 0.02, and q = frequency of t = 0.98. The causative genotypes are TT and Tt. Frequency(TT) = p² = (0.02)² = 0.0004. Frequency(Tt) = 2pq = 2(0.02)(0.98) = 0.0392. Second, calculate the frequency of affected individuals by multiplying the frequency of each causative genotype by the penetrance (0.75). Affected from TT = 0.0004 * 0.75 = 0.0003. Affected from Tt = 0.0392 * 0.75 = 0.0294. The total frequency of individuals expressing the trait is the sum of these two frequencies: 0.0003 + 0.0294 = 0.0297. This is approximately 0.030, or 3.0%.