Genetics Quiz: Recurrence Risk From Pedigrees

What this quiz covers

This quiz focuses on Recurrence Risk From Pedigrees, 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.

Question 1

1. 0% (correct answer)
2. 25%
3. 50%
4. 100%

Explanation: The woman's father is affected with an autosomal dominant condition, so her prior risk of inheriting the mutation was 50%. However, the genetic test result supersedes the risk assessment based on the pedigree. The test has confirmed she does not carry the mutation. Therefore, she cannot pass it on to her children. Assuming the test is accurate and there are no other sources of the mutation (like a de novo mutation in the child, which is a negligible background risk), the risk for her child to inherit this specific familial condition from her is 0%. The partner's negative family history further supports that the risk is effectively zero.

Question 2

1. 0% (correct answer)
2. 25%
3. 50%
4. 100%

Explanation: Mitochondrial disorders are inherited exclusively through the maternal line. Mitochondria, which contain their own DNA, are present in the cytoplasm of the egg cell but are generally not found in the head of the sperm that fertilizes the egg. Therefore, a mother passes her mitochondrial DNA to all of her offspring (both male and female). A father does not pass his mitochondrial DNA to any of his offspring. Since individual II-3 is male, his children will not inherit his mitochondria, and thus will not inherit his mitochondrial disorder. The recurrence risk is effectively 0% (or equivalent to the background population risk).

Question 3

1. 1/4
2. 1/2 (correct answer)
3. 3/4
4. 1

Explanation: This problem involves locus heterogeneity, where mutations in different genes cause the same phenotype. Let Gene A be the autosomal dominant form and Gene B be the autosomal recessive form.

1. The man is affected with the AD form. His genotype is A/a; B/B. (He must be heterozygous for the dominant allele, and we assume he is homozygous normal for the recessive gene since it's a different condition).
2. The woman is affected with the AR form. Her genotype is a/a; b/b. (She must be homozygous for the recessive allele, and we assume she is homozygous normal for the dominant gene).
3. Let's analyze the inheritance of each gene separately for their child.
   • For Gene A, the cross is A/a x a/a. The probability of the child inheriting the 'A' allele and being affected is 1/2. The probability of inheriting 'a' and being unaffected (by this gene) is 1/2.
   • For Gene B, the cross is B/B x b/b. All children will have the genotype B/b. Since this is a recessive condition, all children will be unaffected carriers (by this gene).
4. The child will have retinitis pigmentosa if they are affected by either condition. Since no child can be affected by the AR form (genotype B/b), the only risk is from the AD form. The probability of being affected is therefore the probability of inheriting the 'A' allele, which is 1/2.

Question 4

1. 1/4
2. 1/6
3. 1/8
4. 1/12 (correct answer)

Explanation: This problem involves a two-step risk calculation with a Bayesian update.

1. The paternal grandfather was affected (Hh), so the woman's father had a 1/2 prior probability of inheriting the allele (H).
2. The father is 55 and unaffected. We must update his risk. Let F_H = Father has H allele; F_h = Father does not. Prior P(F_H) = 1/2. The chance of being unaffected at 55 given he has the allele is 1 - penetrance = 1 - 0.5 = 0.5. The chance of being unaffected if he doesn't have the allele is 1.
3. Using Bayes' theorem, P(F_H | unaffected at 55) = [(0.5)(1/2)] / [(0.5)(1/2) + (1)(1/2)] = 0.25 / (0.25 + 0.5) = 0.25 / 0.75 = 1/3. This is the father's current risk of carrying the allele.
4. The woman's probability of inheriting the allele from her father is his risk multiplied by the transmission probability: (1/3) * (1/2) = 1/6.
5. The probability that her child will inherit the allele from her is her risk multiplied by the transmission probability: (1/6) * (1/2) = 1/12.

Question 5

1. Essentially zero, as these were independent de novo mutations.
2. Approximately 5-15%, reflecting empirical data for this scenario. (correct answer)
3. Exactly 25%, consistent with autosomal recessive inheritance.
4. Exactly 50%, consistent with one parent being a carrier.

Explanation: While a single affected child with unaffected parents suggests a de novo (new) mutation with a very low recurrence risk, having two affected children makes two independent de novo events extremely improbable. The most likely biological explanation is germline mosaicism in one of the parents. This means a proportion of the parent's germ cells (sperm or eggs) carry the mutation, while their somatic cells do not, so they are phenotypically normal. The recurrence risk is therefore substantial, but dependent on the unknown fraction of mutant gametes. It is not 50% (which would imply one parent is heterozygous in all cells) or 25% (the risk for recessive disorders). Empirical data for such situations in AD disorders like achondroplasia suggest a recurrence risk in the range of 5-15%.

Question 6

1. 0% (correct answer)
2. 25%
3. 50%
4. 100%

Explanation: Mitochondrial disorders are inherited exclusively through the maternal line. Mitochondria, which contain their own DNA, are present in the cytoplasm of the egg cell but are generally not found in the head of the sperm that fertilizes the egg. Therefore, a mother passes her mitochondrial DNA to all of her offspring (both male and female). A father does not pass his mitochondrial DNA to any of his offspring. Since individual II-3 is male, his children will not inherit his mitochondria, and thus will not inherit his mitochondrial disorder. The recurrence risk is effectively 0% (or equivalent to the background population risk).

Question 7

1. 1/4
2. 1/2 (correct answer)
3. 3/4
4. 1

Explanation: This problem involves locus heterogeneity, where mutations in different genes cause the same phenotype. Let Gene A be the autosomal dominant form and Gene B be the autosomal recessive form.

1. The man is affected with the AD form. His genotype is A/a; B/B. (He must be heterozygous for the dominant allele, and we assume he is homozygous normal for the recessive gene since it's a different condition).
2. The woman is affected with the AR form. Her genotype is a/a; b/b. (She must be homozygous for the recessive allele, and we assume she is homozygous normal for the dominant gene).
3. Let's analyze the inheritance of each gene separately for their child.
   • For Gene A, the cross is A/a x a/a. The probability of the child inheriting the 'A' allele and being affected is 1/2. The probability of inheriting 'a' and being unaffected (by this gene) is 1/2.
   • For Gene B, the cross is B/B x b/b. All children will have the genotype B/b. Since this is a recessive condition, all children will be unaffected carriers (by this gene).
4. The child will have retinitis pigmentosa if they are affected by either condition. Since no child can be affected by the AR form (genotype B/b), the only risk is from the AD form. The probability of being affected is therefore the probability of inheriting the 'A' allele, which is 1/2.

Question 8

1. 1/9
2. 1/18
3. 1/36 (correct answer)
4. 1/8

Explanation: When tackling X-linked inheritance problems, you need to work through the genetics systematically, considering both the woman's possible genotype and the probability of each pregnancy outcome. Since the woman's brother has an X-linked recessive disorder, their mother must be at least a carrier (she passed the recessive allele to her son). This means the woman has a 50% chance of being a carrier herself. Let's call the normal allele X and the recessive allele x. If the woman is a carrier (Xx), each pregnancy has these outcomes: 25% normal female (XX), 25% carrier female (Xx), 25% normal male (XY), and 25% affected male (xY). So there's a 25% chance of an affected child per pregnancy. The probability that this fourth child will be affected equals: P(woman is carrier) × P(affected child | woman is carrier) = $\frac{1}{2} \times \frac{1}{4} = \frac{1}{8}$ Wait - but we have additional information! The woman already had three healthy sons. If she were a carrier, each son had a 50% chance of being affected. Having three healthy sons makes it less likely she's a carrier. Using Bayesian reasoning: P(carrier | three healthy sons) = $\frac{(1/2)(1/2)^3}{(1/2)(1/2)^3 + (1/2)(1)^3} = \frac{1/16}{1/16 + 1/2} = \frac{1}{9}$ Therefore: $\frac{1}{9} \times \frac{1}{4} = \frac{1}{36}$ Answer choice A (1/9) represents the updated probability she's a carrier but ignores the 1/4 pregnancy risk. Answer choice B (1/18) and D (1/8) reflect calculation errors or incomplete reasoning. Remember: previous offspring outcomes provide valuable information that updates genetic probabilities - don't ignore this data in inheritance calculations.

Question 9

1. 1/9
2. 1/18
3. 1/36 (correct answer)
4. 1/8

Explanation: When tackling X-linked inheritance problems, you need to work through the genetics systematically, considering both the woman's possible genotype and the probability of each pregnancy outcome. Since the woman's brother has an X-linked recessive disorder, their mother must be at least a carrier (she passed the recessive allele to her son). This means the woman has a 50% chance of being a carrier herself. Let's call the normal allele X and the recessive allele x. If the woman is a carrier (Xx), each pregnancy has these outcomes: 25% normal female (XX), 25% carrier female (Xx), 25% normal male (XY), and 25% affected male (xY). So there's a 25% chance of an affected child per pregnancy. The probability that this fourth child will be affected equals: P(woman is carrier) × P(affected child | woman is carrier) = $\frac{1}{2} \times \frac{1}{4} = \frac{1}{8}$ Wait - but we have additional information! The woman already had three healthy sons. If she were a carrier, each son had a 50% chance of being affected. Having three healthy sons makes it less likely she's a carrier. Using Bayesian reasoning: P(carrier | three healthy sons) = $\frac{(1/2)(1/2)^3}{(1/2)(1/2)^3 + (1/2)(1)^3} = \frac{1/16}{1/16 + 1/2} = \frac{1}{9}$ Therefore: $\frac{1}{9} \times \frac{1}{4} = \frac{1}{36}$ Answer choice A (1/9) represents the updated probability she's a carrier but ignores the 1/4 pregnancy risk. Answer choice B (1/18) and D (1/8) reflect calculation errors or incomplete reasoning. Remember: previous offspring outcomes provide valuable information that updates genetic probabilities - don't ignore this data in inheritance calculations.

Question 10

1. 0% (correct answer)
2. 25%
3. 50%
4. 100%

Explanation: The woman's father is affected with an autosomal dominant condition, so her prior risk of inheriting the mutation was 50%. However, the genetic test result supersedes the risk assessment based on the pedigree. The test has confirmed she does not carry the mutation. Therefore, she cannot pass it on to her children. Assuming the test is accurate and there are no other sources of the mutation (like a de novo mutation in the child, which is a negligible background risk), the risk for her child to inherit this specific familial condition from her is 0%. The partner's negative family history further supports that the risk is effectively zero.

Question 11

1. 1/4
2. 1/6
3. 1/8
4. 1/12 (correct answer)

Explanation: This problem involves a two-step risk calculation with a Bayesian update.

1. The paternal grandfather was affected (Hh), so the woman's father had a 1/2 prior probability of inheriting the allele (H).
2. The father is 55 and unaffected. We must update his risk. Let F_H = Father has H allele; F_h = Father does not. Prior P(F_H) = 1/2. The chance of being unaffected at 55 given he has the allele is 1 - penetrance = 1 - 0.5 = 0.5. The chance of being unaffected if he doesn't have the allele is 1.
3. Using Bayes' theorem, P(F_H | unaffected at 55) = [(0.5)(1/2)] / [(0.5)(1/2) + (1)(1/2)] = 0.25 / (0.25 + 0.5) = 0.25 / 0.75 = 1/3. This is the father's current risk of carrying the allele.
4. The woman's probability of inheriting the allele from her father is his risk multiplied by the transmission probability: (1/3) * (1/2) = 1/6.
5. The probability that her child will inherit the allele from her is her risk multiplied by the transmission probability: (1/6) * (1/2) = 1/12.

Question 12

1. Essentially zero, as these were independent de novo mutations.
2. Approximately 5-15%, reflecting empirical data for this scenario. (correct answer)
3. Exactly 25%, consistent with autosomal recessive inheritance.
4. Exactly 50%, consistent with one parent being a carrier.

Explanation: While a single affected child with unaffected parents suggests a de novo (new) mutation with a very low recurrence risk, having two affected children makes two independent de novo events extremely improbable. The most likely biological explanation is germline mosaicism in one of the parents. This means a proportion of the parent's germ cells (sperm or eggs) carry the mutation, while their somatic cells do not, so they are phenotypically normal. The recurrence risk is therefore substantial, but dependent on the unknown fraction of mutant gametes. It is not 50% (which would imply one parent is heterozygous in all cells) or 25% (the risk for recessive disorders). Empirical data for such situations in AD disorders like achondroplasia suggest a recurrence risk in the range of 5-15%.

Question 13

1. 0 (correct answer)
2. 1/4
3. 1/2
4. 1

Explanation: For an X-linked dominant disorder, an affected male (genotype X^DY) will pass his Y chromosome to all of his sons and his X chromosome (X^D) to all of his daughters. His partner is unaffected (genotype X^dX^d).

1. Any son they have will receive the Y from the father and an X^d from the mother, resulting in genotype X^dY. All sons will be unaffected.
2. Any daughter they have will receive the X^D from the father and an X^d from the mother, resulting in genotype X^DX^d. All daughters will be affected.
3. The question asks for the probability of having an unaffected daughter. Since all daughters will be affected, this probability is 0.

Question 14

1. 1/15 (correct answer)
2. 1/12
3. 1/10
4. 2/5

Explanation: This problem requires using Bayesian analysis to update the probability that the unaffected father (III-2) carries the disease allele.

1. The affected grandfather (II-1) has the disease genotype (let's use 'Aa'). The prior probability that his son (III-2) inherited the disease allele ('A') is 1/2.
2. We have new information: III-2 is unaffected. We use this to update his risk. Let P(Aa) be the probability III-2 has the genotype and P(aa) be the probability he doesn't. P(Aa)=1/2, P(aa)=1/2. The probability of being unaffected given the genotype is P(unaffected|Aa) = 1 - penetrance = 1 - 0.8 = 0.2. The probability of being unaffected with genotype aa is P(unaffected|aa) = 1.
3. Using Bayes' theorem, the posterior probability that III-2 has the 'Aa' genotype given he is unaffected is: P(Aa | unaffected) = [P(unaffected|Aa) * P(Aa)] / P(unaffected). The total probability of being unaffected is P(unaffected) = [P(unaffected|Aa) * P(Aa)] + [P(unaffected|aa) * P(aa)] = (0.2 * 1/2) + (1 * 1/2) = 0.1 + 0.5 = 0.6. Therefore, P(Aa | unaffected) = (0.2 * 1/2) / 0.6 = 0.1 / 0.6 = 1/6.
4. This (1/6) is the updated probability that III-2 is a non-penetrant carrier. The probability he passes this allele to his child (IV-1) is (1/6) * (1/2) = 1/12.
5. The final risk for the child to be affected is the probability of inheriting the allele multiplied by the penetrance: (1/12) * 0.8 = 0.8/12 = 8/120 = 1/15.

Question 15

1. 1/4
2. 1/3
3. 1/2 (correct answer)
4. 2/3

Explanation: This question tests whether a student can correctly apply basic Mendelian principles without getting confused by extraneous information.

1. The mother, II-2, is a known carrier of an X-linked recessive disorder. Her genotype is X^AX^a.
2. Her daughter, III-3, will inherit one X chromosome from her mother and one from her father.
3. The father, II-1, is unaffected, so his genotype is X^AY. He will pass his X^A chromosome to his daughter.
4. The mother (X^AX^a) has a 1/2 chance of passing on the X^A chromosome and a 1/2 chance of passing on the X^a chromosome.
5. For the daughter to be a carrier, she must inherit the X^a from her mother. The probability of this is 1/2.
6. The information about her brothers (one affected, one unaffected) confirms the mother's carrier status but does not change the independent probability for the daughter's inheritance. It is distractor information.

Question 16

1. Approximately 1% (correct answer)
2. Approximately 5%
3. Approximately 10%
4. Approximately 45%

Explanation: This question combines concepts of de novo mutation, germline mosaicism, and incomplete penetrance.

1. Individual III-1 is affected, but his parents are not. This points to a de novo mutation. In this case, the recurrence risk for a sibling (like III-2) is typically very low, attributable only to the possibility of germline mosaicism in a parent.
2. The empirical risk for recurrence in siblings due to germline mosaicism is often quoted as being in the range of 1-2%. Let's use 2% as a high-end estimate of the probability that the sister, III-2, inherited the mutation due to parental mosaicism.
3. Even if III-2 inherits the allele, she might not show symptoms due to incomplete penetrance. The question states she is unaffected. So we need to calculate P(III-2 has allele | III-2 is unaffected). However, the risk of her inheriting the allele from a mosaic parent is already very low, and this complex Bayesian update would only slightly modify a very small number.
4. The most direct approach is to estimate the risk to her child. The risk for her child to be affected is P(she has the allele) * P(she passes it on) * penetrance. The dominant probability is P(she has the allele), which is low due to germline mosaicism (let's say ~2%).
5. Risk to child ≈ P(III-2 has allele from mosaic parent) * (1/2) * (penetrance) ≈ (0.02) * (0.5) * (0.9) ≈ 0.009, or approximately 1%. The other choices are distractors: 45% would be (1/2)*0.9, the risk for an affected person's child. 5% and 10% are plausible but less accurate estimations.

Question 17

1. 0
2. 1/4
3. 1/2 (correct answer)
4. 1

Explanation: Y-linked inheritance means the gene for the trait is on the Y chromosome.

1. Only males can be affected, as only they have a Y chromosome.
2. An affected father passes his Y chromosome to all of his sons. Therefore, if a child is male, he will be affected (100% risk for sons).
3. An affected father passes an X chromosome to his daughters, so they cannot inherit a Y-linked trait.
4. The question asks for the probability of the pregnancy resulting in an affected male. This requires considering two independent probabilities: the probability of the child being male (1/2) and the probability of being affected if male (1).
5. The total probability is P(male) * P(affected | male) = (1/2) * 1 = 1/2.