Genetic Contributions to Disease - Pathophysiology

1. Two brothers are diagnosed with muscular dystrophy due to different mutations in the DMD gene. Patient A has a single base substitution creating a premature stop codon in exon 10. Patient B has a 3-base-pair deletion in exon 12 that removes one amino acid but maintains the reading frame. Which statement is the most likely prediction of their clinical phenotypes?
   Answer: Patient A will have a more severe phenotype due to protein truncation. When analyzing genetic disorders caused by different mutations in the same gene, you must consider how each type of mutation affects protein structure and function. The type of mutation—not just which gene is affected—determines the severity of the phenotype. Patient A's single base substitution creates a premature stop codon in exon 10, leading to nonsense-mediated decay or production of a severely truncated dystrophin protein. This truncated protein lacks most of its functional domains and cannot properly anchor the cytoskeleton to the cell membrane, resulting in severe membrane instability and muscle fiber degeneration typical of Duchenne muscular dystrophy. Patient B's 3-base-pair deletion removes one amino acid but maintains the reading frame, producing a nearly full-length dystrophin protein with only a single amino acid missing. While this protein may have reduced function, it retains most of its structural integrity and functional domains, likely resulting in a milder phenotype similar to Becker muscular dystrophy. Option A is incorrect because different mutations in the same gene can produce vastly different phenotypes depending on their effects on protein function. Option B is wrong because a single amino acid deletion typically causes less disruption than complete protein truncation. Option C is incorrect because the mutation types themselves are highly predictive of phenotype severity, regardless of family history. Remember this principle: nonsense mutations that truncate proteins generally cause more severe phenotypes than in-frame deletions that preserve protein structure. The reading frame matters enormously in determining disease severity.
2. A child is diagnosed with Prader-Willi syndrome. Karyotyping is normal (46,XY), and microarray analysis shows no deletion in the 15q11-13 region. Subsequent DNA methylation analysis reveals an exclusively maternal methylation pattern in this critical region. What is the underlying genetic mechanism?
   Answer: Maternal uniparental disomy of chromosome 15 Prader-Willi syndrome results from the absence of paternally expressed genes in the 15q11-13 region. An exclusively maternal methylation pattern indicates that the child inherited both copies of chromosome 15 from the mother and none from the father. This is known as maternal uniparental disomy (UPD) and is a known cause of Prader-Willi syndrome, as the necessary paternal contribution is missing.
3. A man and a woman, both of whom have congenital deafness due to an autosomal recessive condition, have a child together who has completely normal hearing. Assuming the deafness trait is fully penetrant in both parents, which genetic principle best explains this outcome?
   Answer: Locus heterogeneity Locus heterogeneity occurs when mutations in different genes can produce the same phenotype. Since the parents are both affected with a recessive disorder, they are homozygous for a pathogenic mutation. If the mutation were in the same gene (e.g., GJB2), all their children would also be homozygous and affected. The birth of an unaffected child indicates that the parents have mutations in different deafness-causing genes (e.g., Parent 1 is aaBB, Parent 2 is AAbb). The child's genotype would be AaBb, making them an unaffected double heterozygote.
4. A young couple is seeking counseling regarding their risk of having a child with a cleft lip and palate, a multifactorial disorder. The prospective father was born with a severe bilateral cleft lip and palate. There is no other family history. Which of the following additional factors would most significantly increase the recurrence risk for their child?
   Answer: The couple's first child is also born with a cleft lip and palate. For multifactorial traits, the empirical recurrence risk increases significantly with the number of affected first-degree relatives. While having one affected parent already increases the risk above the general population risk, having a previously affected sibling in addition to the affected parent confers the highest risk among the choices provided.
5. A man with hemophilia A, an X-linked recessive condition, and a woman with no family history of the disorder have a daughter. This daughter then has a son with a phenotypically normal man. What is the probability that this grandson will have hemophilia A?
   Answer: 50% The man with hemophilia A (XhY) will pass his X chromosome (Xh) to all of his daughters. Therefore, his daughter is an obligate carrier (XHXh), with a 100% probability. When this carrier daughter has children, she has a 50% chance of passing the Xh allele to each child. For a son, who receives a Y chromosome from his father, there is a 50% chance he will inherit the Xh from his mother and be affected. The overall probability is 1 (daughter is a carrier) * 1/2 (chance of passing Xh to a son) = 1/2 or 50%.
6. A deletion on paternal chromosome 15q11-13 results in Prader-Willi syndrome. What is the most likely clinical presentation if an identical microdeletion occurs on the maternally inherited chromosome 15?
   Answer: Angelman syndrome The 15q11-13 region is subject to genomic imprinting. Prader-Willi syndrome is caused by the loss of paternally expressed genes in this region. Angelman syndrome is caused by the loss of a maternally expressed gene, UBE3A, located in the same region. Therefore, a deletion on the maternal chromosome 15 results in Angelman syndrome, a distinct neurological disorder characterized by developmental delay, seizures, and a happy disposition.
7. A genetic mutation is identified that alters the universally conserved GU sequence at the 5' boundary of an intron. Which of the following is the most likely consequence of this mutation on the mRNA transcript and resulting protein?
   Answer: The entire intron is retained in the mature mRNA, leading to a frameshift. The GU sequence at the 5' end of an intron is the splice donor site, which is critical for recognition by the spliceosome. A mutation here will disrupt normal splicing. This can lead to either the retention of the intron in the mature mRNA or the skipping of the preceding exon. Both outcomes typically result in a frameshift and the creation of a premature stop codon, leading to a severely altered or truncated protein.
8. The incidence of an autosomal recessive metabolic disorder in a population adhering to Hardy-Weinberg equilibrium is 1 in 40,000 births. What is the approximate frequency of heterozygous carriers for this disorder in the population?
   Answer: 1 in 100 In Hardy-Weinberg equilibrium, the frequency of affected individuals (genotype aa) is represented by $q^2$. Given $q^2 = 1/40,000$, the allele frequency $q$ is $\sqrt{1/40,000} = 1/200$. The frequency of the dominant allele $p$ is $1 - q$, which is $1 - 1/200 = 199/200$. The carrier frequency (genotype Aa) is calculated as $2pq$. So, $2 * (199/200) * (1/200)$. For rare diseases, $p$ is approximately 1. Therefore, the carrier frequency is approximately $2q = 2 * (1/200) = 1/100$.
9. A neurological disorder is observed to manifest at a progressively earlier age and with increasing severity in successive generations of a family. This phenomenon, known as anticipation, is most commonly associated with which underlying molecular mechanism?
   Answer: Expansion of an unstable trinucleotide repeat sequence during meiosis Anticipation is the characteristic feature of diseases caused by unstable trinucleotide repeat expansions, such as Huntington's disease and myotonic dystrophy. During meiosis, particularly maternal or paternal depending on the disease, the repeat sequence can expand. A larger number of repeats typically correlates with an earlier age of onset and more severe symptoms.
10. A man is affected with a rare genetic disorder. He has four children with an unaffected woman: two affected daughters and two unaffected sons. This pattern is observed in all known cases of affected fathers. What is the most likely mode of inheritance?
    Answer: X-linked dominant The inheritance pattern where an affected father passes the trait to all of his daughters and none of his sons is pathognomonic for X-linked dominant inheritance. The father (XAY) gives his only X chromosome (XA) to all his daughters, making them affected. He gives his Y chromosome to all his sons, who are therefore unaffected (as they inherit a normal X from their mother).
11. A patient is diagnosed with Marfan syndrome due to a mutation in the FBN1 gene. The patient exhibits a tall stature with long limbs, lens dislocation in the eyes, and is at high risk for aortic aneurysm. Which genetic principle best describes this constellation of diverse phenotypic effects from a single gene mutation?
    Answer: Pleiotropy Pleiotropy is the phenomenon where a single gene influences multiple, seemingly unrelated phenotypic traits. In this case, the mutation in the fibrillin-1 gene (FBN1) affects connective tissue throughout the body, leading to distinct pathologies in the skeletal, ocular, and cardiovascular systems.
12. A phenotypically normal couple has a child with achondroplasia, a fully penetrant autosomal dominant disorder. Genetic analysis confirms a de novo mutation in the child. They subsequently have a second child, who is also diagnosed with achondroplasia. What is the most likely genetic explanation for the recurrence of this disorder?
    Answer: Germline mosaicism in one of the parents While the first child's condition could be explained by a single de novo mutation, the recurrence of the same 'de novo' dominant disorder in a second child strongly suggests germline mosaicism. This means a subset of one parent's germ cells (sperm or eggs) carries the mutation, while their somatic cells are normal, explaining their unaffected phenotype. This confers a significant recurrence risk for their offspring.
13. A phenotypically normal woman experiences several first-trimester miscarriages. A cytogenetic analysis reveals a karyotype of 46,XX,t(2;8)(q21;p23). What is the most probable cause of her recurrent pregnancy losses?
    Answer: Formation of gametes with unbalanced chromosomal content during meiosis. When you encounter questions about chromosomal translocations and recurrent pregnancy loss, focus on how balanced translocations in parents can produce unbalanced gametes that lead to nonviable pregnancies. This woman has a balanced reciprocal translocation between chromosomes 2 and 8. While she appears phenotypically normal (indicating the translocation doesn't disrupt essential genes), her chromosomes form an unusual configuration during meiosis. The translocated chromosomes must pair as a quadrivalent structure to ensure proper segregation. However, this complex pairing often results in unbalanced segregation, producing gametes with duplications and deletions of chromosomal segments. These chromosomal imbalances in the developing embryo typically cause early pregnancy loss. Option A is incorrect because if an essential gene were disrupted at the breakpoints, the woman would likely show phenotypic abnormalities herself. Option B misses the point—the issue isn't general nondisjunction of other chromosomes, but specifically the segregation problems caused by the translocation itself. Option C incorrectly suggests somatic mosaicism affecting the uterus; however, the karyotype shows a constitutional translocation present in all cells, and the problem stems from gamete formation, not uterine function. The correct answer is D because balanced translocations in parents commonly produce unbalanced gametes through abnormal meiotic segregation, leading to embryonic chromosomal imbalances incompatible with survival. Remember: balanced translocations in phenotypically normal parents are a classic cause of recurrent pregnancy loss due to unbalanced gamete production, not gene disruption or other chromosomal abnormalities.
14. A man is diagnosed with Huntington's disease, a fully penetrant autosomal dominant condition. A review of his family history reveals that both of his parents are phenotypically normal and lived into their 80s. The man and his unaffected partner want to know the risk of passing the condition to their children. What is the probability that their first child will be affected?
    Answer: 50%, following standard dominant inheritance. The man's condition is due to a de novo mutation, as his parents were unaffected. However, once he has the mutation in his germline, the inheritance pattern for his offspring follows standard Mendelian genetics for a dominant trait. As an affected individual (heterozygous, Hh), he has a 50% chance of passing the pathogenic allele (H) to each child. The information about his parents explains the origin of the disease in him but does not alter his transmission risk.
15. A 50-year-old woman is diagnosed with colon cancer. Sequencing of the tumor tissue reveals a pathogenic mutation in the APC gene, which is commonly associated with the hereditary condition Familial Adenomatous Polyposis (FAP). However, sequencing of DNA from her peripheral blood leukocytes does not show this mutation. What is the most accurate counsel regarding the risk for her children to inherit this specific APC mutation?
    Answer: Her children are not at risk of inheriting this particular mutation as it is somatic in origin. The finding of the mutation in the tumor tissue but not in the patient's blood cells (leukocytes) indicates that this is a somatic mutation. Somatic mutations are acquired in a specific cell line during an individual's lifetime and are not present in the germ cells (eggs or sperm). Therefore, this mutation cannot be passed on to her offspring. The children's risk of FAP would be based on the rest of the family history, but they will not inherit this specific, somatically acquired mutation.
16. A woman is diagnosed with Leber hereditary optic neuropathy (LHON), a disease caused by a point mutation in the mitochondrial genome. She has two sons and one daughter. Her partner is unaffected. Which statement accurately describes the inheritance risk for her children?
    Answer: All of her children will inherit her mitochondria and are therefore at risk for the disease. Mitochondrial DNA (mtDNA) is inherited exclusively from the mother. All offspring, regardless of sex, inherit their mother's mitochondria. Therefore, all of her children will inherit the mtDNA mutation. The clinical expression can vary due to heteroplasmy (the proportion of mutant vs. wild-type mtDNA), but all are at risk.
17. Which two genetic concepts best describe the clinical findings in this family?
    Answer: Incomplete penetrance and variable expressivity Incomplete penetrance refers to the fact that some individuals with a pathogenic genotype do not exhibit the phenotype; this is seen in the family members who carry the mutation but show no clinical signs. Variable expressivity refers to the variation in phenotypic severity among individuals with the same genotype; this is seen in the range of symptoms from only café-au-lait spots to severe tumors.
18. A couple has a child with cystic fibrosis (CF), an autosomal recessive disorder. The mother's phenotypically normal sister is planning a family with an unaffected partner from the general population, where the carrier frequency for CF is 1/25. What is the approximate probability that this sister's first child will be affected with CF?
    Answer: 1/150 Since the sister has a sibling with CF (genotype aa), her parents must both be carriers (Aa). Among the phenotypically normal offspring of two carriers, the probability of being a carrier (Aa) is 2/3 (genotypes are AA, Aa, Aa; the aa genotype is excluded because she is unaffected). The probability her partner is a carrier is given as 1/25. The probability that two carriers have an affected child is 1/4. Therefore, the total risk is the product of these three probabilities: (2/3) (1/25) (1/4) = 2/300 = 1/150.
19. A deletion of two consecutive base pairs within the coding sequence of an exon is identified in a gene. Compared to a single base-pair substitution at the same location that results in a missense mutation, what is the most likely consequence of this two-base-pair deletion?
    Answer: An altered reading frame leading to a premature stop codon and a truncated protein A deletion of one or two base pairs (or any number not divisible by three) causes a frameshift mutation. This alters the triplet reading frame for all subsequent codons, leading to a completely different amino acid sequence downstream of the mutation. A frameshift almost invariably introduces a premature stop codon, resulting in a truncated and nonfunctional protein, which is generally a more severe outcome than a missense mutation.

Start this quiz