Genetics Quiz: X Linked Inheritance

What this quiz covers

This quiz focuses on X Linked Inheritance, 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. He inherited the allele from his mother. (correct answer)
2. His mother is phenotypically affected by the disorder.
3. He inherited the allele from his father.
4. His maternal grandfather was affected by the disorder.

Explanation: A male's genotype is XY. He inherits the Y chromosome from his father and the X chromosome from his mother. Since the disorder is X-linked, the allele for it must be on his X chromosome. Therefore, he must have inherited the X chromosome carrying the recessive allele from his mother. His mother could be a phenotypically normal carrier (heterozygous) or be affected (homozygous recessive), so we cannot be certain of her phenotype. His father and maternal grandfather are irrelevant to the direct transmission of his specific X chromosome.

Question 2

1. Random inactivation of one X chromosome in somatic cells (correct answer)
2. Codominance of the normal and recessive alleles
3. Incomplete penetrance of the dominant allele
4. A high rate of somatic mutation in skin precursor cells

Explanation: The mosaic pattern of expression in females heterozygous for an X-linked trait is a classic example of lyonization, or random X-chromosome inactivation. Early in embryonic development, one of the two X chromosomes in each somatic cell is randomly and permanently inactivated. All descendant cells of that cell will have the same X chromosome inactivated, leading to clonal patches of tissue expressing either the maternally or paternally inherited allele.

Question 3

1. The DMD status of his mother
2. The DMD status of his maternal grandfather
3. His own phenotype regarding DMD (correct answer)
4. The number of male relatives he has with DMD

Explanation: The woman's genotype is X^D X^d. A daughter inherits one X from her mother and one from her father. To be a carrier ($X^D$ $X^d$), she could inherit X^d from her mother and X^D from her father, or X^D from her mother and X^d from her father. The father's contribution depends on his genotype. If he is unaffected ($X^D$ Y), he can only contribute X^D. If he is affected ($X^d$ Y), he can only contribute X^d. Therefore, knowing his own phenotype (affected or unaffected) directly determines which X chromosome he will pass to a daughter and is essential for the calculation.

Question 4

1. $1/16$
2. $1/8$ (correct answer)
3. $1/4$
4. $1/2$

Explanation: The woman's brother is affected ($X^hY$), which means their mother must be a carrier ($X^HX^h$). The woman, being a daughter of a carrier mother, has a 1/2 probability of being a carrier herself. For her to have an affected son, three independent events must occur: 1) she must be a carrier (P=1/2), 2) she must have a son (P=1/2), and 3) she must pass her X^h allele to that son (P=1/2). The overall probability is the product of these individual probabilities: $1/2 \times 1/2 \times 1/2 = 1/8$.

Question 5

1. The son will have the condition, but the daughter will not.
2. The daughter will have the condition, but the son will not. (correct answer)
3. Both the son and the daughter will have the condition.
4. There is a 50% chance for each child to have the condition, regardless of sex.

Explanation: Let the dominant allele be X^A and the recessive be X^a. The man's genotype is X^A Y, and his partner's is X^a X^a. A son inherits his father's Y chromosome and his mother's X^a, making his genotype X^a Y (unaffected). A daughter inherits her father's X^A chromosome and her mother's X^a, making her genotype X^A X^a (affected). Therefore, any daughter they have will be affected, and any son they have will be unaffected.

Question 6

1. 1 solid female : 1 solid male
2. 2 solid females : 1 solid male (correct answer)
3. 1 solid female : 2 solid males
4. 3 solid individuals : 1 striped individual

Explanation: The cross is X^S X^s (heterozygous female) × X^S Y (solid male). The potential offspring genotypes are X^S X^S (solid female), X^S X^s (solid female), X^S Y (solid male), and X^s Y (striped male). The X^s allele is lethal in hemizygous males, so the X^s Y offspring will not survive. The surviving offspring are X^S X^S, X^S X^s, and X^S Y. This gives a phenotypic ratio of 2 solid females to 1 solid male.

Question 7

1. X-linked recessive
2. X-linked dominant (correct answer)
3. Autosomal dominant
4. Sex-influenced, dominant in males

Explanation: The fact that affected fathers do not pass the trait to sons but pass it to all daughters is the hallmark of X-linked dominant inheritance. The father gives his Y chromosome to his sons and his only X (which carries the dominant allele) to his daughters. The observation that affected mothers pass it to half their children of either sex is also consistent with this mode.

Question 8

1. A somatic cell of the affected female early in her development
2. A germline cell of the affected female's mother
3. A germline cell of the affected female's father (correct answer)
4. A germline cell of the affected female's maternal grandmother

Explanation: The affected female ($X^A$ $X^a$) has unaffected parents (father $X^a$ Y, mother $X^a$ $X^a$). She must have received an X^a from her mother. Therefore, she must have received an X^A from her father. Since the father is phenotypically normal ($X^a$ Y), the mutation from X^a to X^A must have occurred in his germline during the formation of the sperm that conceived his daughter. While a mutation in the mother's germline is possible, the higher rate of mutation during spermatogenesis makes the paternal germline the most probable origin.

Question 9

1. Two unaffected parents have an affected offspring.
2. The condition is found in males much more frequently than in females.
3. An affected mother passes the condition to all of her sons. (correct answer)
4. An affected father has an unaffected daughter.

Explanation: An affected mother with an X-linked recessive trait has the genotype X^rX^r. She passes one of her X^r chromosomes to all of her sons, so all sons will have the genotype X^rY and be affected. This is a definitive feature of X-linked recessive inheritance that is not seen in autosomal recessive patterns. Distractor A is true for any recessive trait. Distractor B is suggestive but not definitive proof. Distractor D is possible for both autosomal recessive and X-linked recessive inheritance.

Question 10

1. None
2. $1/4$ (correct answer)
3. $1/3$
4. $1/2$

Explanation: The original cross was a female ($X^R$ X^?) with a white-eyed male ($X^r$ Y). Because she produced white-eyed male offspring ($X^r$ Y), she must have passed on an X^r allele. Therefore, her genotype is heterozygous ($X^R$ $X^r$). The new cross is this female ($X^R$ $X^r$) with a red-eyed male ($X^R$ Y). The possible offspring are X^R X^R (red female), X^R X^r (red female), X^R Y (red male), and X^r Y (white male). Only one of the four possible genotypes, X^r Y, results in a white-eyed phenotype. Thus, the expected proportion is $1/4$.

Question 11

1. $1/2$
2. $1/4$
3. $1/8$ (correct answer)
4. $0$

Explanation: The son needs to inherit an X chromosome with the genotype X^Hc (dominant allele for clotting, recessive for color vision). The mother's chromosomes are X^HC and X^hc. An X^Hc chromosome is a recombinant type. With a 50% recombination frequency, all four possible gamete types (parental $X^HC$, $X^hc$; recombinant $X^Hc$, $X^hC$) are produced in equal proportions (1/4 each). The probability of producing an egg with the X^Hc chromosome is 1/4. The probability of having a son is 1/2. The combined probability is $1/4 \times 1/2 = 1/8$.

Question 12

1. Equal to the incidence in males
2. Half of the incidence in males
3. The square root of the incidence in males
4. 1% of the incidence in males (correct answer)

Explanation: Let q be the frequency of the recessive allele, so q = 0.01. The incidence in males (genotype $X^rY$) equals the allele frequency q, since males are hemizygous. The incidence in females (genotype $X^rX$^r) is q^2 under Hardy-Weinberg equilibrium. The ratio of female incidence to male incidence is q^2/q = q = 0.01. Therefore, the incidence in females is 1% of the incidence in males.

Question 13

1. Meiosis I in the father
2. Meiosis II in the father
3. Meiosis I in the mother
4. Meiosis II in the mother (correct answer)

Explanation: The child's genotype is X^h X^h Y. The Y chromosome must come from the father. Therefore, the father contributed a normal Y sperm. The X^h X^h gamete must have come from the mother. The mother's genotype is X^H X^h. A non-disjunction in Meiosis I would separate the X^H and X^h homologs incorrectly, producing X^H X^h gametes. A non-disjunction in Meiosis II occurs after a normal Meiosis I. A secondary oocyte would contain two X^h sister chromatids, and if they fail to separate, an X^h X^h gamete is formed. This is the only way to produce the required gamete.

Question 14

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

Explanation: The purple female's genotype is X^R X^B. The blue male's genotype is X^B Y. The cross produces four possible offspring genotypes in equal proportions: X^R X^B (purple female), X^B X^B (blue female), X^R Y (red male), and X^B Y (blue male). The mother's phenotype is purple. Only the X^R X^B genotype results in a purple phenotype. This genotype occurs in 1 out of the 4 possible offspring, so the probability is $1/4$.