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Question 1Genetics and Genomics Core Knowledge (molecular, cytogenetic, biochemical, population and statistical principles)
An autosomal recessive disorder affects 1 in 40,000 newborns in a population that meets Hardy-Weinberg assumptions. What is the approximate carrier frequency?
- 1 in 20
- 1 in 400
- 1 in 100 Correct answer.
- 1 in 200
Explanation
Disease incidence equals q² = 1/40,000, so the mutant allele frequency q is 1/200. Carrier frequency is 2pq, which approximates 2q = 1/100 because p is nearly 1.
Question 2Genetics and Genomics Core Knowledge (molecular, cytogenetic, biochemical, population and statistical principles)
Which change in a coding region directly produces a premature termination codon?
- A substitution that creates a UAA, UAG, or UGA codon Correct answer.
- An in-frame deletion of three coding bases
- A single-base insertion within an exon
- A substitution within the 5' untranslated region
Explanation
UAA, UAG, and UGA are the three stop codons, so a substitution that converts a sense codon into one of them is a nonsense mutation and truncates the protein. Insertions, untranslated-region changes, and in-frame deletions act through different mechanisms.
Question 3Genetics and Genomics Core Knowledge (molecular, cytogenetic, biochemical, population and statistical principles)
A cytogenetics report reads 46,XY,t(9;22)(q34;q11.2). How should this be interpreted?
- A Robertsonian fusion of the long arms of chromosomes 9 and 22
- An unbalanced translocation causing partial trisomy 9
- A balanced reciprocal translocation between chromosomes 9 and 22 Correct answer.
- A pericentric inversion with breakpoints in 9q34 and 22q11.2
Explanation
The designation t indicates a translocation, and the two breakpoints with exchange of distal segments without net loss of material indicate a balanced reciprocal translocation. Balanced carriers have 46 chromosomes and typically no phenotype.
Question 4Genetics and Genomics Core Knowledge (molecular, cytogenetic, biochemical, population and statistical principles)
Phenylketonuria most commonly results from deficiency of which enzyme?
- Homogentisate 1,2-dioxygenase
- Phenylalanine hydroxylase Correct answer.
- Branched-chain alpha-ketoacid dehydrogenase
- Cystathionine beta-synthase
Explanation
Classic phenylketonuria is caused by biallelic pathogenic variants in PAH, which encodes phenylalanine hydroxylase, the enzyme converting phenylalanine to tyrosine. Loss of activity causes phenylalanine accumulation and neurotoxicity if untreated.
Question 5Genetics and Genomics Core Knowledge (molecular, cytogenetic, biochemical, population and statistical principles)
A woman carries a heteroplasmic mitochondrial DNA variant. Which statement about transmission in her family is correct?
- Neither child can inherit the variant unless the father is also affected
- Only her sons can inherit the variant and pass it on
- Both sons and daughters transmit the variant in a mendelian ratio
- Her daughters can transmit the variant, and her sons can inherit it Correct answer.
Explanation
Mitochondrial DNA is transmitted through the maternal lineage, so both sons and daughters can inherit a mother's mtDNA variant, but only daughters can pass it to the next generation. Inheritance is not mendelian because of heteroplasmy and random segregation of organelles.
Question 6Genetics and Genomics Core Knowledge (molecular, cytogenetic, biochemical, population and statistical principles)
Which karyotype is expected in a healthy carrier of a Robertsonian translocation between chromosomes 13 and 14?
- 46,XY,inv(13)(q10q14)
- 45,XY,der(13;14)(q10;q10) Correct answer.
- 46,XY,t(13;14)(q10;q10)
- 46,XY,del(13)(q14)
Explanation
A Robertsonian translocation fuses two acrocentric chromosomes at the centromere, so the carrier has 45 chromosomes with one derived chromosome and no net loss of euchromatin. Carriers are typically phenotypically normal but are at risk for unbalanced offspring.
Question 7Genetics and Genomics Core Knowledge (molecular, cytogenetic, biochemical, population and statistical principles)
A child has severe developmental delay, absent speech, ataxia, and paroxysms of inappropriate laughter. Testing shows a deletion of 15q11-q13 on the maternal chromosome. Which diagnosis fits?
- Rett syndrome
- Prader-Willi syndrome
- Angelman syndrome Correct answer.
- Chromosome 15q duplication syndrome
Explanation
Angelman syndrome results from loss of maternal 15q11-q13 contribution, most often by deletion, and features severe speech impairment, ataxia, and a happy demeanor with inappropriate laughter. Prader-Willi syndrome arises from loss of the paternal contribution at the same region.
Question 8Genetics and Genomics Core Knowledge (molecular, cytogenetic, biochemical, population and statistical principles)
A screening test has fixed sensitivity and specificity. What happens to its positive predictive value as disease prevalence in the tested population rises?
- Positive predictive value increases Correct answer.
- Negative predictive value increases
- Sensitivity increases
- Specificity increases
Explanation
Sensitivity and specificity are properties of the test and do not change with prevalence. Positive predictive value rises with prevalence because a larger share of positive results are true positives, while negative predictive value falls.
Question 9Genetics and Genomics Core Knowledge (molecular, cytogenetic, biochemical, population and statistical principles)
A single-base substitution abolishes the invariant GT dinucleotide at the 5' end of an intron. What is the most likely consequence?
- Intron retention that alters the reading frame Correct answer.
- A silent change, since introns are not translated
- An in-frame insertion of three nucleotides
- A synonymous amino acid substitution
Explanation
The GT dinucleotide is part of the highly conserved splice donor signal recognized by the spliceosome. Its loss usually causes intron retention or activation of a cryptic site, either of which typically disrupts the reading frame and produces an abnormal protein.
Question 10Genetics and Genomics Core Knowledge (molecular, cytogenetic, biochemical, population and statistical principles)
A woman has one son with Duchenne muscular dystrophy and no other family history. Because about one third of cases arise from new mutations, her carrier probability is 2/3. She then has a second son who is unaffected. Using Bayesian analysis, what is her carrier probability now?
- 2/3
- 1/3
- 1/2 Correct answer.
- 1/4
Explanation
The prior probabilities are 2/3 carrier and 1/3 non-carrier. The likelihood of an unaffected son is 1/2 for a carrier and 1 for a non-carrier, giving a posterior of (2/3 x 1/2) / [(2/3 x 1/2) + (1/3 x 1)] = 1/2.
Question 11Genetics and Genomics Core Knowledge (molecular, cytogenetic, biochemical, population and statistical principles)
Congenital myotonic dystrophy, with severe neonatal hypotonia and respiratory failure, occurs almost exclusively when the expanded repeat is transmitted by which parent?
- Only a mother who is herself congenitally affected
- The father
- Either parent, with equal frequency
- The mother Correct answer.
Explanation
Congenital myotonic dystrophy is associated with very large CTG expansions in DMPK, and these expansions occur almost exclusively during maternal transmission. The mother is usually only mildly affected or identified after the birth of the affected child.
Question 12Genetics and Genomics Core Knowledge (molecular, cytogenetic, biochemical, population and statistical principles)
Two maternal relatives carry the same mitochondrial DNA variant, yet one has severe cardiomyopathy and the other is asymptomatic. Which explanation fits best?
- Tissue-specific differences in the fraction of mutant mitochondrial DNA Correct answer.
- Mitochondrial genes are subject to X-inactivation
- The variant is imprinted in a parent-of-origin manner
- Paternal and maternal mitochondrial DNA recombine in somatic cells
Explanation
Heteroplasmy varies among tissues and among individuals because mitochondria segregate randomly during development. A variant causes dysfunction only when the mutant load in a vulnerable tissue exceeds its threshold, which explains the variable severity.
Question 13Genetics and Genomics Core Knowledge (molecular, cytogenetic, biochemical, population and statistical principles)
A child with Prader-Willi features has no 15q11-q13 deletion on microarray. Methylation analysis shows a maternal-only pattern at that region. Which mechanism explains this?
- Paternal uniparental disomy of chromosome 15
- A pathogenic variant in UBE3A
- Maternal uniparental disomy of chromosome 15 Correct answer.
- Trisomy 15 confined to the placenta
Explanation
A maternal-only methylation pattern at 15q11-q13 means the paternal contribution is absent, which causes Prader-Willi syndrome. Maternal uniparental disomy produces exactly this pattern when both copies of chromosome 15 come from the mother.
Question 14Genetics and Genomics Core Knowledge (molecular, cytogenetic, biochemical, population and statistical principles)
A phenotypically normal man carries a pericentric inversion of chromosome 2. Which statement about his reproductive risk is accurate?
- Recombinant chromosomes occur only in homozygotes for the inversion
- Offspring are not at risk because recombination is fully suppressed
- Gametes will contain acentric and dicentric chromosomes
- Crossing over in the inversion loop can yield recombinant chromosomes with duplications and deletions Correct answer.
Explanation
During meiosis the inverted and normal homologs pair as a loop. A crossover inside a pericentric loop produces chromosomes that duplicate one distal segment and delete the other, which can cause malformations or pregnancy loss in offspring.
Question 15Genetics and Genomics Core Knowledge (molecular, cytogenetic, biochemical, population and statistical principles)
First cousins have a coefficient of relationship of 1/8. What is the inbreeding coefficient of a child born to first cousins?
- 1/8
- 1/32
- 1/64
- 1/16 Correct answer.
Explanation
The inbreeding coefficient of a child equals half the coefficient of relationship between the parents, because the child is homozygous only if it inherits the same ancestral allele from both parents. Half of 1/8 is 1/16.
Question 16Genetics and Genomics Core Knowledge (molecular, cytogenetic, biochemical, population and statistical principles)
Direct preparation of chorionic villi at 11 weeks shows 47,XX,+21 in all cells, while amniocentesis at 16 weeks shows 46,XX. What is the most likely explanation?
- Maternal cell contamination of the amniotic fluid sample
- True fetal mosaicism that amniocentesis failed to detect
- Confined placental mosaicism with a euploid fetus Correct answer.
- A culture artifact confined to the amniocyte flask
Explanation
Trisomic cells limited to the cytotrophoblast are a well-recognized form of confined placental mosaicism and may not reflect the fetal karyotype. A normal amniocentesis, which samples fetal cells directly, indicates that the fetus is most likely euploid.
Question 17Genetics and Genomics Core Knowledge (molecular, cytogenetic, biochemical, population and statistical principles)
Which statement about the typical 22q11.2 deletion is correct?
- An inherited deletion in nearly all cases
- A recurrent ~3 Mb deletion mediated by low-copy repeats Correct answer.
- A rearrangement visible on routine G-banded karyotype
- A trinucleotide repeat expansion within TBX1
Explanation
Most 22q11.2 deletions span about 3 Mb and arise from nonallelic homologous recombination between low-copy repeat blocks flanking the region. Roughly 90% are de novo, and the deletion is below the resolution of G-banded karyotyping.
Question 18Genetics and Genomics Core Knowledge (molecular, cytogenetic, biochemical, population and statistical principles)
In a consanguineous population, which deviation from Hardy-Weinberg expectations is seen at autosomal loci?
- Unchanged genotype frequencies with altered allele frequencies
- An excess of heterozygotes and a deficit of homozygotes
- A deficit of affected homozygotes for rare recessive alleles
- An excess of homozygotes and a deficit of heterozygotes Correct answer.
Explanation
Inbreeding increases the chance that two alleles at a locus are identical by descent, which raises homozygosity at the expense of heterozygosity. Allele frequencies are unchanged, but the incidence of rare recessive disease rises.
Question 19Genetics and Genomics Core Knowledge (molecular, cytogenetic, biochemical, population and statistical principles)
Most cases of trisomy 21 result from which mechanism?
- Inheritance of a Robertsonian translocation
- Nondisjunction in maternal meiosis I Correct answer.
- Nondisjunction in paternal meiosis II
- Postzygotic nondisjunction causing mosaicism
Explanation
About 95% of Down syndrome cases result from free trisomy 21, and in most of these the extra chromosome is maternal in origin, usually from meiosis I nondisjunction. Robertsonian translocations account for only a small percentage of cases.
Question 20Genetics and Genomics Core Knowledge (molecular, cytogenetic, biochemical, population and statistical principles)
A carrier of a balanced reciprocal translocation has a normal SNP chromosomal microarray. Which property of microarray testing explains this result?
- Balanced rearrangements do not change copy number Correct answer.
- SNP arrays cannot detect loss of heterozygosity
- Translocations always involve pericentromeric heterochromatin
- Microarrays cannot detect copy-number changes larger than 10 Mb
Explanation
Microarrays measure copy number and, for SNP platforms, allelic imbalance; a balanced translocation changes chromosome structure without gain or loss of material. Such rearrangements therefore require karyotyping or FISH for detection.
Question 21Genetics and Genomics Core Knowledge (molecular, cytogenetic, biochemical, population and statistical principles)
For an X-linked recessive lethal disorder with mutation rate μ per generation, what is the equilibrium frequency of the mutant allele?
- μ/3
- The square root of μ
- 3μ Correct answer.
- μ
Explanation
One third of X chromosomes are carried by males, where a recessive lethal allele is fully exposed to selection and lost. Equilibrium requires new mutation to replace that loss, giving q = 3μ/s; for a lethal allele s = 1 and q = 3μ.
Question 22Genetics and Genomics Core Knowledge (molecular, cytogenetic, biochemical, population and statistical principles)
A woman whose brother has cystic fibrosis has a prior carrier probability of 2/3. A mutation panel that detects 90% of carrier mutations in her ethnic group is negative. What is her residual carrier risk?
- 1/30
- 1/10
- 1/3
- 1/6 Correct answer.
Explanation
The prior odds are 2/3 carrier versus 1/3 non-carrier. A negative panel is 10 times more likely in a non-carrier than in a carrier, giving posterior odds of (2/3 x 0.1) : (1/3 x 1), which reduces to a probability of 1/6, about 17%.
Question 23Genetics and Genomics Core Knowledge (molecular, cytogenetic, biochemical, population and statistical principles)
Unaffected parents have two sons with Duchenne muscular dystrophy caused by an identical pathogenic variant. Neither parent carries the variant in blood. What is the best explanation for the recurrence?
- Two independent de novo mutations at the same nucleotide
- Non-paternity in the second pregnancy
- Autosomal dominant inheritance with incomplete penetrance
- Gonadal mosaicism in the mother Correct answer.
Explanation
Recurrence of the same DMD variant in two sons of a non-carrier mother indicates that a fraction of her oocytes carries the mutation, that is, gonadal mosaicism. This scenario is a classic reason that recurrence risk after a single de novo X-linked case is not negligible.
Question 24Genetics and Genomics Core Knowledge (molecular, cytogenetic, biochemical, population and statistical principles)
A linkage study reports a LOD score of 3.0 between a marker and a disease locus. How is this value interpreted?
- A 3% probability that the two loci are linked
- A map distance of 3 cM between the loci
- Odds of 1000 to 1 favoring linkage over no linkage Correct answer.
- Exclusion of linkage at a recombination fraction of 0.3
Explanation
The LOD score is the base-10 logarithm of the odds ratio comparing linkage at a given recombination fraction with no linkage. A value of 3.0 corresponds to odds of 10^3, or 1000:1, which is the conventional threshold for accepting linkage.
Question 25Genetics and Genomics Core Knowledge (molecular, cytogenetic, biochemical, population and statistical principles)
A researcher pools samples from two genetically distinct subpopulations and analyzes them as a single group at an autosomal locus. Compared with the average of the separate subpopulations, what will the combined sample show?
- Excess heterozygotes in the smaller subpopulation only
- More heterozygotes than expected from the pooled allele frequencies
- Fewer heterozygotes than expected from the pooled allele frequencies Correct answer.
- Unchanged heterozygosity with a shift in allele frequencies
Explanation
When subpopulations with different allele frequencies are combined, the pooled allele frequencies predict more heterozygotes than are actually observed, a phenomenon known as the Wahlund effect. Treating the pooled data as one population therefore causes spurious homozygote excess.
Question 26Genetics and Genomics Core Knowledge (molecular, cytogenetic, biochemical, population and statistical principles)
Which change describes a nonsense variant?
- A substitution creates a premature termination codon. Correct answer.
- An insertion of two nucleotides alters the reading frame.
- A substitution changes one amino acid to another.
- A substitution leaves the amino acid sequence unchanged.
Explanation
A nonsense variant converts an amino acid-coding codon into a stop codon, predictably truncating the protein. A missense variant substitutes one amino acid, a synonymous variant does not change the protein sequence, and a two-nucleotide insertion causes a frameshift.
Question 27Genetics and Genomics Core Knowledge (molecular, cytogenetic, biochemical, population and statistical principles)
Deficiency of which enzyme causes classic galactosemia?
- HEXA
- GALT Correct answer.
- PAH
- G6PD
Explanation
Classic galactosemia results from deficiency of galactose-1-phosphate uridylyltransferase, encoded by GALT. PAH deficiency causes phenylketonuria, HEXA deficiency causes Tay-Sachs disease, and G6PD deficiency causes oxidative hemolysis.
Question 28Genetics and Genomics Core Knowledge (molecular, cytogenetic, biochemical, population and statistical principles)
The Philadelphia chromosome in chronic myeloid leukemia most commonly produces which fusion gene?
- BCR-ABL1 Correct answer.
- RUNX1-RUNX1T1
- ETV6-RUNX1
- PML-RARA
Explanation
The Philadelphia chromosome arises from t(9;22)(q34;q11.2), joining BCR on chromosome 22 to ABL1 on chromosome 9. The resulting BCR-ABL1 fusion encodes a constitutively active tyrosine kinase.
Question 29Genetics and Genomics Core Knowledge (molecular, cytogenetic, biochemical, population and statistical principles)
At an autosomal locus under Hardy-Weinberg equilibrium, the recessive disease allele frequency is 0.2. What is the expected heterozygote frequency?
- 0.64
- 0.32 Correct answer.
- 0.16
- 0.04
Explanation
For a recessive allele frequency q = 0.2, p = 0.8, and the heterozygote frequency is 2pq. Thus 2(0.8)(0.2) = 0.32.
Question 30Genetics and Genomics Core Knowledge (molecular, cytogenetic, biochemical, population and statistical principles)
An autosomal recessive disorder has an incidence of 1 in 10,000 in a population at equilibrium. What is the approximate carrier frequency?
- 1/10,000
- 1/100
- 1/50 Correct answer.
- 1/25
Explanation
The disease incidence equals q squared, so q = 1/100 and p is approximately 99/100. The carrier frequency is 2pq, approximately 2/100, or 1/50.
Question 31Genetics and Genomics Core Knowledge (molecular, cytogenetic, biochemical, population and statistical principles)
Which abnormality is quantitative fluorescent PCR designed to detect rapidly in prenatal samples?
- A single-gene point mutation
- Trisomy 21 Correct answer.
- Imprinting methylation status
- A balanced reciprocal translocation
Explanation
Quantitative fluorescent PCR amplifies short tandem repeats to estimate copy number for chromosomes 13, 18, 21, X, and Y. It is used for rapid detection of common aneuploidies such as trisomy 21. It does not assess balanced rearrangements, point mutations, or methylation.
Question 32Genetics and Genomics Core Knowledge (molecular, cytogenetic, biochemical, population and statistical principles)
A female is heterozygous for an X-linked recessive pathogenic variant but has symptoms of the disorder. Which mechanism best explains this observation?
- Mitochondrial heteroplasmy
- Paternal imprinting of the gene
- Preferential inactivation of the normal X Correct answer.
- A second pathogenic variant in the same gene
Explanation
In X-linked recessive disorders, heterozygous females are usually unaffected because roughly half of cells express the normal allele. Skewed X-inactivation favoring the chromosome carrying the normal allele can leave most relevant cells expressing the pathogenic allele and produce symptoms.
Question 33Genetics and Genomics Core Knowledge (molecular, cytogenetic, biochemical, population and statistical principles)
Hereditary retinoblastoma typically presents with bilateral, multifocal tumors. Which genetic mechanism best explains this pattern?
- One germline hit plus a second somatic hit Correct answer.
- Gain of an extra copy of chromosome 13
- A single somatic RB1 mutation
- Two independent germline mutations in RB1
Explanation
Hereditary retinoblastoma follows the two-hit model: an RB1 pathogenic variant is inherited in the germline, and a second somatic hit occurs in retinal cells. Because many retinal cells already carry the first hit, independent second hits can produce multiple tumors in both eyes.
Question 34Genetics and Genomics Core Knowledge (molecular, cytogenetic, biochemical, population and statistical principles)
A patient with suspected fragile X syndrome has no PCR product from the repeat region. What is the most appropriate next test?
- Repeating the same PCR with more cycles
- Chromosomal microarray
- G-banded karyotype
- Southern blot analysis Correct answer.
Explanation
Very large CGG expansions may fail to amplify by conventional PCR, producing no product. Southern blot analysis can detect large expanded alleles and is the standard reflex test in this setting.
Question 35Genetics and Genomics Core Knowledge (molecular, cytogenetic, biochemical, population and statistical principles)
Which karyotype is associated with Turner syndrome and includes an isochromosome Xq?
- 47,XX,+21
- 46,XX,del(5p)
- 46,XY,t(9;22)
- 46,X,i(X)(q10) Correct answer.
Explanation
An isochromosome Xq, written i(X)(q10), has two copies of the long arm and loss of the short arm. In a 46,X,i(X)(q10) karyotype, this abnormal X replaces one normal X and is a recognized cause of Turner syndrome.
Question 36Genetics and Genomics Core Knowledge (molecular, cytogenetic, biochemical, population and statistical principles)
Which condition must hold for autosomal genotype frequencies to remain at Hardy-Weinberg equilibrium across generations?
- Recurrent mutation
- Assortative mating
- Genetic drift
- No selection Correct answer.
Explanation
Hardy-Weinberg equilibrium assumes no selection, no mutation, no migration, no genetic drift, and random mating in a large population. Selection changes allele frequencies and therefore violates the equilibrium assumption.
Question 37Genetics and Genomics Core Knowledge (molecular, cytogenetic, biochemical, population and statistical principles)
A test has a positive likelihood ratio of 10. What does this value indicate?
- Risk rises by exactly 10 percentage points.
- Post-test odds are 10-fold higher. Correct answer.
- Sensitivity is 10 times specificity.
- Post-test probability is 10%.
Explanation
The positive likelihood ratio is the factor by which a positive result multiplies the pre-test odds. A value of 10 means the post-test odds are 10 times the pre-test odds, not that the post-test probability is 10%.
Question 38Genetics and Genomics Core Knowledge (molecular, cytogenetic, biochemical, population and statistical principles)
A premature termination codon is located in the final exon of a gene, far from the last exon-exon junction. What is the most likely consequence?
- The mRNA will always be degraded by NMD.
- A truncated protein may be produced. Correct answer.
- The variant will be synonymous.
- Ribosomal read-through will restore full-length protein.
Explanation
Nonsense-mediated decay is generally triggered when a premature termination codon is more than about 50 to 55 nucleotides upstream of the last exon-exon junction. A termination codon in the final exon usually escapes NMD, so a truncated protein may be produced.
Question 39Genetics and Genomics Core Knowledge (molecular, cytogenetic, biochemical, population and statistical principles)
An autosomal recessive disorder in an isolated population is caused almost exclusively by one pathogenic variant, whereas many different variants cause the same disorder elsewhere. Which population-genetic process best explains this observation?
- Heterozygote advantage at the disease locus
- Meiotic drive favoring the variant
- Founder effect and genetic drift Correct answer.
- Assortative mating increasing mutation rates
Explanation
A founder effect occurs when a population descends from a small number of founders carrying a limited set of alleles. Genetic drift can then increase the frequency of one pathogenic variant while other variants remain rare, producing reduced allelic heterogeneity.
Question 40Genetics and Genomics Core Knowledge (molecular, cytogenetic, biochemical, population and statistical principles)
A child has dysmorphism and pigmentary anomalies suggesting mosaic trisomy 8, but a blood karyotype is normal. Which test is most appropriate to detect low-level mosaicism?
- Chromosomal microarray of blood
- Interphase FISH for chromosome 8 Correct answer.
- Metaphase karyotype of five cells
- Sanger sequencing of a candidate gene
Explanation
Interphase FISH examines individual nuclei and can detect low-level mosaic aneuploidy when a sufficient number of cells is scored. Chromosomal microarray is less sensitive for low-level mosaicism, and five metaphases are far too few to exclude it.
Question 41Clinical Assessment, Diagnosis and Natural History of Genetic Conditions Across the Lifespan
A term newborn has a newborn screening result showing elevated TSH with low free T4. The infant is asymptomatic and feeding well. What is the most appropriate management?
- Obtain a thyroid ultrasound before any treatment is started
- Repeat the filter-paper screening test at 2 weeks of age
- Begin levothyroxine now while confirming with serum thyroid function tests Correct answer.
- Defer treatment until serum tests confirm the diagnosis at 1 month
Explanation
Congenital hypothyroidism is a screening emergency because thyroid hormone is required for normal brain development in the first weeks of life. Treatment with levothyroxine should begin as soon as confirmatory serum testing is drawn, without waiting for the results. Imaging may characterize anatomy but must never delay therapy.
Question 42Clinical Assessment, Diagnosis and Natural History of Genetic Conditions Across the Lifespan
A man with achondroplasia and his unaffected partner seek reproductive counseling. What is the recurrence risk for each of their children?
- Less than 1%
- 25%
- 50% Correct answer.
- 100%
Explanation
Achondroplasia is an autosomal dominant condition caused by activating FGFR3 variants, and an affected heterozygote transmits the variant to half of offspring. Most cases arise from new mutations, which explains why unaffected parents of an affected child have a low recurrence risk. Homozygous achondroplasia is lethal and would require both parents to be affected.
Question 43Clinical Assessment, Diagnosis and Natural History of Genetic Conditions Across the Lifespan
A 10-year-old with Down syndrome is seen for routine health supervision. Which screening is recommended at this age?
- Annual bone marrow biopsy
- Routine colonoscopy beginning at age 10
- Routine echocardiography every year
- Annual thyroid function testing Correct answer.
Explanation
Children with Down syndrome have a substantially increased risk of autoimmune hypothyroidism, so thyroid function should be checked at least annually. Echocardiography is performed in infancy and then as clinically indicated rather than yearly in every child. Colorectal cancer screening begins in adulthood, and marrow biopsy is not a screening test.
Question 44Clinical Assessment, Diagnosis and Natural History of Genetic Conditions Across the Lifespan
A 14-year-old girl presents with short stature, delayed puberty, and a bicuspid aortic valve. Karyotype shows 45,X. Which additional finding is most characteristic of this condition?
- Intellectual disability in most affected individuals
- Primary ovarian insufficiency Correct answer.
- Macroorchidism
- Tall stature relative to the family
Explanation
In Turner syndrome, accelerated follicular atresia leads to gonadal failure and primary ovarian insufficiency, so most adolescents require estrogen replacement for puberty and bone health. Short stature is typical rather than tall stature, and intellect is usually normal. Macroorchidism is not applicable to a phenotypic female.
Question 45Clinical Assessment, Diagnosis and Natural History of Genetic Conditions Across the Lifespan
A 40-year-old man develops chorea, irritability, and progressive cognitive decline; his father died of a similar illness at age 55. Which inheritance pattern best explains this family history?
- Autosomal recessive
- Autosomal dominant Correct answer.
- X-linked recessive
- Mitochondrial
Explanation
Huntington disease is caused by a CAG repeat expansion in HTT and is inherited in an autosomal dominant pattern, so an affected parent transmits a 50% risk to each child. Transmission can occur through either sex, and repeat instability produces anticipation. The vertical transmission across two generations with adult onset is classic.
Question 46Clinical Assessment, Diagnosis and Natural History of Genetic Conditions Across the Lifespan
A 16-year-old boy with gynecomastia, small testes, and tall stature has a 47,XXY karyotype. Which finding is the most likely explanation for his infertility?
- Obstruction of the epididymis
- Absence of the vas deferens
- Pituitary adenoma
- Progressive seminiferous tubule hyalinization Correct answer.
Explanation
The supernumerary X chromosome in Klinefelter syndrome causes progressive hyalinization and fibrosis of the seminiferous tubules, leading to azoospermia and infertility. Testosterone production may also decline over time. The defect is testicular rather than ductal.
Question 47Clinical Assessment, Diagnosis and Natural History of Genetic Conditions Across the Lifespan
A 6-week-old born with meconium ileus has a positive newborn screen for cystic fibrosis. Which test establishes the diagnosis?
- Stool elastase
- Chest radiograph
- Serum immunoreactive trypsinogen
- Sweat chloride testing Correct answer.
Explanation
Quantitative pilocarpine iontophoresis with sweat chloride measurement is the confirmatory test for cystic fibrosis; a value of 60 mmol/L or higher is diagnostic in this context. Immunoreactive trypsinogen is only a screening marker, and supportive tests such as stool elastase assess pancreatic function rather than diagnose the disease.
Question 48Clinical Assessment, Diagnosis and Natural History of Genetic Conditions Across the Lifespan
A 30-year-old woman of Mediterranean ancestry has microcytic anemia, a normal ferritin, and an elevated hemoglobin A2 level. Which diagnosis best fits these findings?
- Alpha-thalassemia trait
- Iron deficiency anemia
- Sideroblastic anemia
- Beta-thalassemia trait Correct answer.
Explanation
Beta-thalassemia trait produces microcytosis with a compensatory elevation of hemoglobin A2, and iron stores are normal because this is not an iron disorder. Iron deficiency would lower ferritin and typically reduce rather than raise hemoglobin A2. Alpha-thalassemia trait causes microcytosis with a normal hemoglobin A2.
Question 49Clinical Assessment, Diagnosis and Natural History of Genetic Conditions Across the Lifespan
A 20-year-old with Marfan syndrome has an aortic root diameter of 4.5 cm on echocardiography. Which management is most appropriate?
- Start beta-blockade with serial imaging Correct answer.
- Reassess with imaging in five years
- Avoid all competitive athletics indefinitely
- Proceed directly to aortic root replacement
Explanation
At 4.5 cm the aortic root is dilated but below the typical surgical threshold of about 5.0 cm for adults with Marfan syndrome, so beta-blockade or ARB therapy plus serial imaging is indicated to slow progression. Restriction of strenuous competitive sports is advisable but does not replace pharmacologic therapy and monitoring. Elective repair is reserved for larger diameters, rapid growth, or concerning family history.
Question 50Clinical Assessment, Diagnosis and Natural History of Genetic Conditions Across the Lifespan
A 62-year-old man develops progressive gait ataxia and intention tremor; his daughter has a son with fragile X syndrome. Which diagnosis best explains his neurologic findings?
- Huntington disease
- Multiple system atrophy
- FXTAS Correct answer.
- Spinocerebellar ataxia type 1
Explanation
Fragile X-associated tremor/ataxia syndrome occurs in older carriers of FMR1 premutations of roughly 55 to 200 CGG repeats, and such carriers can transmit an expanded full mutation to grandchildren. The combination of late-onset ataxia, intention tremor, and a family history of fragile X syndrome is characteristic. Other ataxias do not carry this reproductive link.