X-Linked Inheritance - Genetics

When analyzing a pedigree for a rare genetic condition, which of the following findings would provide the strongest evidence for X-linked recessive inheritance as opposed to autosomal recessive inheritance?
Answer: An affected mother passes the condition to all of her sons. 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.
A woman's brother has hemophilia A, a rare X-linked recessive disorder. The woman is unaffected, and her partner is also unaffected. They are expecting their first child. What is the overall probability that they will have a son with hemophilia A?
Answer: $1/8$ 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: $1/2 \times 1/2 \times 1/2 = 1/8$ .
A man has a rare X-linked dominant condition. He and his partner, who does not have the condition, have two children: a son and a daughter. Which statement accurately predicts the phenotypes of their children?
Answer: The daughter will have the condition, but the son will not. 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.
A male is diagnosed with an X-linked recessive disorder. Which statement about the transmission of the causal allele to this individual can be made with absolute certainty, based only on the mode of inheritance?
Answer: He inherited the allele from his mother. 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.
A heterozygous female is crossed with a solid-colored male. What is the expected phenotypic ratio among the living offspring?
Answer: 2 solid females : 1 solid male 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.
A purple female is crossed with a blue male. What is the probability that an individual offspring will be phenotypically identical to its mother?
Answer: $1/4$ 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 $1/4$ .
A geneticist observes a pedigree for a trait where affected fathers never have affected sons, but their daughters are always affected. Additionally, affected mothers pass the trait to half of their sons and half of their daughters. Which mode of inheritance is most consistent with all these observations?
Answer: X-linked dominant 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.
A woman is a carrier for two different X-linked recessive traits: hemophilia A and color blindness. Her genotype is X^HC / X^hc, meaning the dominant alleles are on one X chromosome and the recessive alleles are on the other. The genes are far enough apart to assume a 50% recombination frequency. What is the probability she will have a son who is color-blind but does not have hemophilia?
Answer: $1/8$ 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 $1/4 \times 1/2 = 1/8$ .
If this F1 red-eyed female is crossed with a red-eyed male from a different, true-breeding stock, what proportion of their offspring is expected to have white eyes?
Answer: $1/4$ 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 $1/4$ .
Given that spontaneous mutations are significantly more common during spermatogenesis than oogenesis, in which of the following did the de novo mutation most likely arise?
Answer: A germline cell of the affected female's father 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.
A woman is a known carrier for Duchenne muscular dystrophy (DMD), an X-linked recessive disorder. Her partner wishes to determine the risk for their future children, but his family history is unknown. To calculate the probability that their first daughter will be a carrier of DMD, what is the most direct and essential piece of information needed about the partner?
Answer: His own phenotype regarding DMD 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.
Based on this allele frequency, the incidence of the disorder in females is expected to be what fraction of the incidence in males?
Answer: 1% of the incidence in males 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.
Assuming a single non-disjunction event was responsible, in which parent and during which meiotic division did it occur?
Answer: Meiosis II in the mother 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.
This mosaic phenotype is best explained by which of the following genetic principles?
Answer: Random inactivation of one X chromosome in somatic cells 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.

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