Genetics

You are called to examine a one hour old baby who was born with a facial defect. The patient's ears appear to be low-set. On auscultation, there is a holostolic murmur along the lower left sternal border. Basic labs return notable for hypocalcemia. What is the most likely underlying mechanism responsible for the abnormalities seen in this newborn? A. Trisomy 21 B. Trisomy 18 C. Trinucleotide repeat expansion D. Partial deletion on chromosome 22 E. Uniparental disomy

D. Partial deletion on chromosome 22 which is DiGeorge syndrome.

Assume there is a microRNA that participates in regulating the expression of a particular cyclin kinase inhibitor. How might an alteration in this microRNA lead to uncontrolled cell proliferation? (A) Overexpression of the microRNA, so it acts as an oncogene. (B) Reduced expression of the microRNA, so it acts as a tumor suppressor. (C) A total loss of activity of the microRNA, so it cannot bind to its target mRNA. (D) A loss of specificity of the microRNA for its target, so different mRNAs are not targeted. (E) No change in microRNA activity can lead to uncontrolled cell proliferation.

The answer is A. Cyclin kinase inhibitors act as brakes on the cell cycle. If the cyclin kinase inhibitor can be removed from the cell, then the cell cycle could proceed in an uncontrolled fashion. MicroRNAs reduce the amount of protein product formed from the target mRNA. In order to eliminate the production of the cyclin kinase inhibitor, the microRNA would need to be overexpressed, such that all target mRNAs are bound, and translation of the gene product is halted. Reducing the expression of the microRNA would lead to overexpression of the cyclin kinase inhibitor, and more control of the cell cycle. This is also the case if the microRNA lost all activity (overproduction of its target), or lost its specificity (again, overproduction of the target).

Li-Fraumeni syndrome results from which one of the following? Choose the one best answer. (A) Inability to recognize DNA damage (B) Inability to regulate CDKs (C) Inability to regulate a tyrosine kinase (D) Inability to regulate gene transcription (E) Inability to activate a heterotrimeric G-protein

The answer is A. Li-Fraumeni syndrome results from an inherited mutation in p53, the guardian of the genome. This protein monitors the DNA for damage, and if damage is found, acts as a transcription factor to arrest the cell cycle, allow the DNA damage to be repaired, and then to allow the cycle to proceed. If the DNA damage cannot be repaired, then apoptosis is induced so that the cell will not replicate the damaged DNA. P53 does not regulate the CDKs, tyrosine kinases, or G-proteins. Loss of p53 expression will alter gene transcription (repair enzymes will not be induced, nor will apoptosis be induced if the damage cannot be repaired), but does not hinder the normal regulation of gene transcription.

A tumor suppressor gene is best described by which of the following? A. A gain-of-function mutation leads to uncontrolled proliferation. B. A loss-of-function mutation leads to uncontrolled proliferation. C. When it is expressed, the gene suppresses viral genes from being expressed. D. When it is expressed, the gene specifically blocks the G1/S checkpoint. E. When it is expressed, the gene induces tumor formation

The answer is B. Tumor-suppressor genes balance cell growth and quiescence. When they are not expressed (via loss-of-function mutations), the balance shifts to cell proliferation and tumorigenesis (thus, A is incorrect). Answer C is incorrect because tumor-suppressor genes do not act on viral genes, answer D is incorrect because tumorsuppressor genes are not specifically targeted to just one aspect of the cell cycle, and answer E is incorrect because a loss of expression of tumor-suppressor genes leads to tumor formation, not expression of these genes.

The mechanism through which Ras becomes an oncogenic protein is which of the following? A. Ras remains bound to GAP. B. Ras can no longer bind cAMP. C. Ras has lost its GTPase activity. D. Ras can no longer bind GTP. E. Ras can no longer be phosphorylated by MAP kinase.

The answer is C. Ras, when it is oncogenic, has lost its GTPase activity and thus remains active for a longer time. Answer A is incorrect because GAP proteins activate the GTPase activity of Ras, and this mutation would make Ras less active. cAMP does not interact directly with Ras (thus, B is incorrect), and if Ras could no longer bind GTP, it would not be active (hence, D is also incorrect). Ras is not phosphorylated by the MAP kinase (thus, E is incorrect).

Chromosomal translocations can lead to uncontrolled cell growth due to which one of the following? (A) Interference with mitosis (B) Interference with DNA synthesis (C) Unequal crossing over during mitosis (D) Inappropriate expression of translocated genes (E) Loss of gene expression

The answer is D. For most translocations that lead to uncontrolled cell growth, a gene is inappropriately expressed because it has been moved adjacent to a constitutive promoter (such as the myc gene next to the immunoglobulin promoter in Burkitt lymphoma). The dysregulation of cell proliferation does not occur owing to problems with mitosis or DNA replication, nor with crossing over. In most cases, the problem is an increased or inappropriate expression of the translocated gene, and not a loss of gene expression.

When p53 increases in response to DNA damage, which of the following events occurs? A. p53 induces transcription of cdk4. B. p53 induces transcription of cyclin D. C. p53 binds E2F to activate transcription. D. p53 induces transcription of p21. E. p53 directly phosphorylates the transcription factor jun.

The answer is D. When p53 increases in response to DNA damage, it acts as a transcription factor and induces the transcription of p21, which blocks phosphorylation of Rb. Rb then stays bound to transcription factor E2F, and E2F cannot induce the genes required for the G1-to-S transition. p53 does not induce transcription of either cyclin D or cdk4 (thus, A and B are incorrect), nor does p53 bind to E2F (thus, C is incorrect), nor does p53 contain kinase activity (thus, E is incorrect).

A child with severe epilepsy, autistic behavior, and developmental delay has characteristics of a condition known as Angelman syndrome (105830). Because of the syndromic nature of the disorder and the developmental delay, a karyotype is performed that shows a missing band on one chromosome 15. Which of the following best describes this abnormality? A. Interstitial deletion of 15 B. Terminal deletion of 15 C. Pericentric inversion of 15 D. Paracentric inversion of 15 E. 15q−

The answer is a. (Lewis, pp 241-266. Scriver, pp 3-45. Murray, pp 396-414.) A missing band suggests an interstitial (internal) deletion rather than removal of the distal short or long arm (known as a terminal deletion). The shorthand notation 15q− implies a terminal deletion of the long arm of chromosome 15. Pericentric (surrounding the centromere) or paracentric (not including the centromere) inversions result from crossover of a chromosome with itself and then breakage and reunion to produce an internal inverted segment. Interstitial deletion 15q11q13 is seen in approximately 50% of patients with Prader-Willi and Angelman syndromes. Other patients with these syndromes inherit both chromosomes 15 from their mother (Prader-Willi) or both from their father (Angelman's), a situation known as uniparental disomy. Genomic imprinting of the 15q11q13 region is different on the chromosome inherited from the mother than on the chromosome inherited from the father. The normal balance of maternal and paternal imprints is thus disrupted by deletion or uniparental disomy, leading to reciprocal differences in gene expression that present as Angelman or Prader-Willi syndromes.

A newborn with ambiguous genitalia and a 46,XY karyotype develops vomiting, low serum sodium concentration, and high serum potassium. Which of the following proteins is most likely to be abnormal? A. 21-hydroxylase B. An ovarian enzyme C. 5β-reductase D. An androgen receptor E. A testicular enzyme

The answer is a. (Lewis, pp 377-496. Scriver, pp 4077-5016. Murray pp 434-455.) Sex steroids are synthesized from cholesterol by side-chain cleavage (employing a P450 enzyme) to produce pregnenolone. Pregnenolone is then converted to testosterone in the testis, to estrogen in the ovary, and to corticosterone and aldosterone in the adrenal gland. The enzymes 3β-hydroxysteroid dehydrogenase, 21-hydroxylase, 11βhydroxylase, and 18-hydroxylase modify pregnenolone to produce other sex and adrenal steroids. Deficiencies in adrenal 21-hydroxylase can thus lead to inadequate testosterone production in males and produce ambiguous external genitalia. Such children can also exhibit low sodium and high potassium due to deficiency of the more distal steroids corticol and aldosterone. 5β-reductase converts testosterone to dihydrotestosterone, and its deficiency produces milder degrees of hypogenitalism without salt wasting. Deficiency of the androgen receptor is called testicular feminization, producing normal looking females who may not seek medical attention until they present with infertility.

A child is born with spina bifida, a defect in the lower spinal cord and meninges that may cause bladder and lower limb dysfunction. The family history reveals that the father had a small spina bifida that was repaired by surgery. Which of the following is the most critical aspect of the medical evaluation as it pertains to genetic counseling? A. A search for additional anomalies to determine if the child has a syndrome B. A karyotype on the child C. A serum folic acid level on the child D. A spinal x-ray on the mother E. A spinal x-ray on the father

The answer is a. (Lewis, pp 397-416. Scriver, pp 3-45.) Spina bifida is a defect of neural tube development that can be partially prevented by encouraging preconceptional folic acid supplementation in women desiring to become pregnant. Examination for subtle evidence of dysmorphology in children with major birth defects is necessary to rule out a syndrome. Syndromes often exhibit Mendelian or chromosomal inheritance.

Females occasionally have symptoms of X-linked recessive diseases such as Duchenne muscular dystrophy, hemophilia, or color blindness. Which of the following is the most common explanation? A. Nonrandom lyonization B. X chromosome trisomy (47,XXX) C. X autosome-balanced translocation that disrupts the particular X chromosome locus D. Turner syndrome (45,X) E. 46,XY karyotype in a female

The answer is a. (Lewis, pp 75-94. Scriver, pp 3-45.) Females have two alleles for each locus on the X chromosome because of their 46,XX karyotype. One normal allele is by definition sufficient for normal function in X-linked recessive disorders, so that females with one abnormal allele are carriers instead of affected individuals. Only when the companion normal allele is disrupted or missing does the abnormal allele cause disease. The Lyon hypothesis predicts that X inactivation is early, irreversible, and random, but some females inactivate only the X chromosome carrying the normal allele. X autosome translocations may disrupt an X chromosome locus and cause disease because the translocated autosome must remain active to avert embryonic death; nonrandom inactivation of the normal X chromosome thus ablates expression of its normal allele. Females with Turner syndrome, like males with 46,XY karyotypes, have only one X chromosome and can be affected with X-linked recessive diseases. Conversely, females with triple X or trisomy X syndrome have three alleles at each X chromosome locus and are not affected with X-linked recessive disorders. Since choices c, d, and e each require two genetic changes, they are less common than choice a.

A 1-year-old Hispanic boy has a normal birth and infantile history except for delay in sitting up, crawling, and standing (delayed motor milestones). He begins the unusual habit of chewing on his fingers and lips and, in one instance, bites through the lip and leaves a large wound. His physician documents an elevated serum uric acid and suspects Lesch-Nyhan syndrome (MIM*300322). In considering potential therapy, the physician reads that purines are overproduced in gout and Lesch-Nyhan syndrome, causing hyperuricemia, yet the hypoxanthine analog allopurinol is only effective in gout. Allopurinol does not treat the neurologic symptoms of Lesch-Nyhan syndrome because it does not do which of the following? A. Decrease de novo purine synthesis B. Decrease de novo pyrimidine synthesis C. Diminish urate synthesis D. Increase phosphoribosylpyrophosphate (PRPP) levels E. Inhibit xanthine oxidase

The answer is a. (Murray, pp 331-342. Scriver, pp 2537-2570.) Some forms of gout derive from deficiency of phosphoribosyl pyrophosphate (PRPP) synthase, the first step of purine synthesis (eg, MIM*311850). Other patients may have a partial deficiency of hypoxanthine-guanine phosphoribosyl transferase (HGPRTase), which salvages hypoxanthine and guanine by transferring the purine ribonucleotide of PRPP to the bases and forming inosinate and guanylate, respectively. In all of these patients, the hypoxanthine analogue allopurinol has two actions: (1) it inhibits xanthine oxidase, which catalyzes the oxidation of hypoxanthine to xanthine and then to uric acid stones and tissue deposits; and (2) it forms an inactive allopurinol ribonucleotide from PRPP in a reaction catalyzed by HGPRTase, thereby decreasing the rate of purine synthesis. In contrast, because of the total loss of HGPRTase activity in Lesch-Nyhan patients, the allopurinol ribonucleotide cannot be formed. Thus, PRPP levels are not decreased and de novo purine synthesis continues unabated. The gouty arthritis caused by urate crystal formation is relieved in Lesch-Nyhan patients, but their neurological symptoms (mental deficiency, self-mutilation with compulsive chewing of fingers and lips) are not.

403. Ectrodactyly is an autosomal dominant trait that causes missing middle fingers (lobster claw malformation). A grandfather and grandson both have ectrodactyly, but the intervening father has normal hands by x-ray. Which of the following terms applies to this family? A. Incomplete penetrance B. New mutation C. Variable expressivity D. Germinal mosaicism E. Anticipation

The answer is a. (Murray, pp 812-828. Scriver, pp 3-45. Sack, pp 97-158. Wilson, pp 23-39.) Incomplete penetrance applies to a normal individual who is known from the pedigree to have an allele responsible for an autosomal dominant trait. Variable expressivity refers to family members who exhibit signs of the autosomal dominant disorder that vary in severity. When this severity seems to worsen with progressive generations, it is called anticipation. A new mutation in the grandson would be extremely unlikely given the affected grandfather. The father could be an example of somatic mosaicism if a back-mutation occurred to allow normal limb development, but there is no reason to suspect mosaicism of his germ cells (germinal mosaicism).

A family is seen for routine prenatal counseling because the mother of two normal children is age 35, the arbitrary "advanced maternal age" when the approximate 1 in 100 risk for fetal chromosome disorders is deemed significant. Family history reveals that the mother's parents and her husband all have had onset of high blood pressure (hypertension) at early ages, and two of mother's grandparents died of strokes that may be hypertension-related. The mother also had some hypertension in the third trimester of her last pregnancy. Recognizing that hypertension is a multifactorial trait, which of the following is the most appropriate explanation and counseling for the couple? A. Multifactorial determination indicates an interaction between the environment and a single gene, implying a 50% risk for eventual hypertension in mother and offspring B. Multifactorial determination indicates an interaction between the environment and multiple genes, implying a 5-10% risk for eventual hypertension in mother and offspring C. Multifactorial determination results from multiple postnatal environmental factors, implying a 75-100% risk for eventual hypertension in mother and a 5-10% risk to offspring D. Multifactorial determination results from multiple pre- and postnatal environmental factors, implying a 75-100% risk for eventual hypertension in mother and a 5-10% risk to offspring E. Multifactorial determination implies action of multiple genes independent of environmental factors, implying low risks for hypertension in mother's next pregnancy but a 5-10% risk for eventual hypertension in offspring

The answer is b. (Lewis, pp 135-154. Scriver, pp 193-202.) Many common disorders tend to run in families but are not single-gene or chromosomal disorders. These disorders exhibit multifactorial determination, a theoretical mechanism envisioned as the interaction of multiple genes (polygenic inheritance) and environmental factors. For quantitative traits like height, it is easy to visualize how the alleles at multiple loci plus environmental factors (e.g., nutrition) might make additive contributions toward a given stature. For qualitative traits such as cleft lip/palate and other congenital anomalies, a threshold is envisioned that divides normal from abnormal phenotypes. Individuals with more clefting alleles, in combination with an unfavorable intrauterine environment, can cross the threshold and manifest the anomaly. Environmental factors in multifactorial determination are thus prenatal as well as postnatal. For adult diseases like coronary artery disease, strokes, or hypertension, the quantitative trait can be viewed as degree of artery occlusion or blood pressure and the threshold as events like myocardial or cerebral infarction. The likelihood of inheriting hypertension-promoting alleles is increased if there are several affected relatives as in the family under discussion, translating the usual 3-5% risk for a multifactorial disorder in a first-degree relative to the 5-10% risk estimated in answer b. Recurrence risks for multifactorial disorders are modified by family history because the multiple predisposing alleles cannot be determined by testing. Current research is attempting to find single nucleotide polymorphisms (SNPs) that travel with these multiple predisposing alleles, allowing more definitive susceptibility testing and risk prediction for the occurrence and transmission of multifactorial traits.

The genesis of Prader-Willi syndrome by inheritance of two normal chromosomes from a single parent is an example of which of the following? A. Germinal mosaicism B. Genomic imprinting C. Chromosome deletion D. Chromosome rearrangement E. Anticipation

The answer is b. (Lewis, pp 241-266. Scriver, pp 3-45. Murray, pp 396-414.) In humans and other mammals, the source of genetic material may be as important as its content. Mice manipulated to receive two male pronuclei develop as abortive placentas, whereas those receiving two female pronuclei develop as abortive fetuses. The different impact of the same genetic material according to whether it is transmitted from mother or father is due to genomic imprinting. The term imprinting is borrowed from animal behavior and refers to parental marking during gametogenesis—the physical basis may be DNA methylation or chromatin phasing. Both maternally derived and paternally derived haploid chromosome sets are thus necessary for normal fetal development. This is why parthenogenesis does not occur in mammals. The imprint is erased in the fetal gonads and reestablished based on fetal sex. Certain cases of Prader-Willi syndrome are disorders of imprinting with the absence of the paternally imprinted chromosome 15.

A 6-year-old girl is referred to a physician for evaluation. She is known to have mild mental retardation and a ventricular septal defect (VSD). On physical examination, the patient is noted to have some facial dysmorphism, including a long face, a prominent nose, and flattening in the malar region. In addition, the patient's speech has an unusual quality. Which of the following descriptions best explains the patient's condition? A. Sequence B. Syndrome C. Disruption D. Deformation E. Single birth defect

The answer is b. (Lewis, pp 241-266. Scriver, pp 3-45.) The child described in the question has multiple independent anomalies that are characteristic of a syndrome. Although they are likely to be causally related, they do not appear to be sequential. These problems do not appear to be caused by the breakdown of an originally normal developmental process as in a disruption, nor do they appear to be related to a nondisruptive mechanical force as in a deformation.

Assume that frequencies for the different blood group alleles are as follows: A =0.3; B =0.1; O =0.6. What is the expected percentage of individuals with blood type B? A. 7% B. 13% C. 27% D. 36% E. 45%

The answer is b. (Lewis, pp 267-282. Scriver, pp 3-45.) It is important to remember that individuals with blood type A can have either genotype AA or AO, and individuals with blood type B can have either genotype BB or BO. Therefore, the frequency of blood type A is the frequency of homozygotes— that is, 0.3 × 0.3—plus the frequency of heterozygotes—that is, 2 (0.3) × 0.6—for a total of 0.45. The frequency of blood type B is 0.1 ×0.1+2 (0.1) × 0.6 for a total of 0.13. The frequency of individuals with blood type O is simply the frequency of homozygotes—that is, 0.6 ×0.6 =0.36.

A woman who married her first cousin wants to know the risk of having a child with cystic fibrosis (219700) because her grandmother, who is also her husband's grandmother, died of cystic fibrosis. Which of the following is her risk? A. 1 /8 B. 1 /16 C. 1 /60 D. 1 /120 E. 1 /256

The answer is b. (Lewis, pp 75-94, 267-282. Scriver, pp 3-45.) The grandmother has cystic fibrosis, so her children are obligate carriers. Each cousin therefore has a one-half chance of being a carrier. The woman's risk is 1 /2 × 1 /2 × 1 /4 = 1 /16 chance of having an affected child. This illustrates the effects of consanguinity.

The development of DNase therapy has dramatically improved survival and frequency/severity of lung infections in children with cystic fibrosis (219700). Over 30 common mutant alleles for Caucasions with cystic fibrosis have now been characterized at the DNA level, allowing screening of pregnant couples with high sensitivity. Which of the following techniques will be required for newborn screening of cystic fibrosis or other diseases using DNA-based technology? A. Fluorescent in situ hybridization (FISH) for deletions surrounding the disease locus B. Subtelomeric FISH analysis C. DNA array or chip analysis D. Southern analysis to display mutant alleles E. Comparative genomic hybridization (CGH)

The answer is b. (Lewis, pp 397-416. Scriver, pp 5121-5188. Murray, pp 396-414.) A method for rapid and inexpensive screening of multiple mutant alleles in large numbers of individuals will be required before genetic screening can be performed using DNA analysis. The likely method will involve DNA arrays or chips where thousands of DNA probes can be affixed to arrays on microbeads or glass slides, hybridized in replicate to individual DNAs, and passed through machines for reading of deficiencies.These techniques are undergoing explosive development, but cannot yet screen for thousands of potential mutant alleles in an individual (e.g., with neurofibromatosis—162200—or Marfan syndrome—154700), much less for multiple mutant alleles in each of the 3 million annual American newborns. Comparative genomic hybridization analyzes an individual DNA with probes across the genome, highlighting regions where there is extra or missing DNA. This technique is excellent for demonstrating extra or missing chromosome material but may not detect rearrangements. Fluorescent in situ hybrization may employ control and test labeled DNA probe to highlight their specific locus on a chromosome, showing two signals for individuals with paired autosomes. An absence of test signal on one chromosome when the control signal is present shows there is a submicroscopic deletion as in the DiGeorge or Williams syndromes. Subtelomere FISH analysis extends this to many DNA probes, highlighting all 44 autosome and the 2 X chromosome ends—subtle rearrangements will shift the position of the FISH signal, and deletions or duplications involving a target region will yield a missing or extra signal. Southern analysis is labor intensive and has limited use in this era of automated allele detection and DNA sequencing. The prime justification for newborn screening is that early diagnosis results in benefits through prevention or treatment. Although early identification of PKU carriers through diagnosis of their affected newborns is a benefit of newborn screening, its chief rationale is the prevention of mental retardation by early diagnosis and lowering of dietary protein intake. False-positive screens are the most frequent problem with screening, and repeat screens are sometimes needed for borderline results.

A girl seems normal at birth but begins flinching at loud noises (enhanced startle response) at age 6 months. Ophthalmologic examination reveals a central red area of the retina surrounded by white tissue (cherry red spot). The child initially can sit up, but then regresses so that she cannot roll over or recognize her parents. Her physician suspects a lipid storage disease (neurolipidosis). If the diagnosis is correct, what is the risk that the next child of these parents will be affected with the same disease? A. 1 /2 B. 1 /4 C. 3/4 D. 1 /12 E. 1 /24

The answer is b. (Lewis, pp 75-94. Scriver, pp 3827-3876.) Metabolic diseases usually exhibit autosomal or X-linked recessive inheritance. Autosomal recessive inheritance is most likely because the affected patient is female. In this case, the parents are obligate carriers and there is a onefourth chance (25% recurrence risk) that their next child will be affected. The symptoms suggest Tay-Sachs disease (272800), an autosomal recessive disorder involving severe neurodegeneration and early death.

The cause of Tay-Sachs disease (272800) is best described by which of the following? A. Excess of a lysosomal enzyme in blood due to defective uptake B. Deficiency of a lysosomal enzyme that digests proteoglycans C. Deficiency of a membrane receptor that takes up proteoglycans D. Deficiency of a mitochondrial enzyme that degrades glycogen E. Deficiency of a mitochondrial triglyceride lipase

The answer is b. (Lewis, pp 75-94. Scriver, pp 3827-3876. Murray, pp 197-204.)The lysosomal enzyme hexosaminidase A is deficient in Tay-Sachs disease. The enzyme cleaves aminohexose groups from gangliosides, complex lipids formed from ceramide (a derivative of sphingosine). Ceramide is synthesized in the endoplasmic reticulum from palmitoyl coenzyme A (16-carbon acyl CoA) and serine in a reaction catalyzed by pyridoxal phosphate. Uridine diphosphoglucose (UDP-glucose) or UDP-galactose moieties and sialic acid groups are then added in the Golgi apparatus and the gangliosides contribute to myelin in nerve cells. Neurolipidoses like TaySachs disease lack certain lysosomal enzymes necessary to degrade the gangliosides, causing severe effects on nerve cells (neuro-degeneration). A parallel group of disorders called mucopolysaccharidoses result from the absence of lysosomal enzymes that degrade complex carbohydrate chains and their associated proteins (called proteoglycans). Proteoglycans are more widely distributed than gangliosides, occurring in the ground substance of many tissues. Accumulation of the glycosaminoglycans from these proteoglycans thus causes a wide spectrum of symptoms including coarsening of the face and hair, cardiopulmonary problems, and bony deformities such as kyphosis (beaked spine). There is a specific lysosomal receptor that recognizes mannose-6-phosphate on certain lysosomal enzymes and targets them to lysosomes. Mutations in this receptor can cause increased blood levels and lysosomal deficiencies of several enzymes that are normally targeted to lysosomes. One such disease is I (inclusion) cell disease (252500). The slow accumulation of abnormal gangliosides in lipidoses and of abnormal proteoglycans in mucopolysaccharidoses causes a characteristic clinical course of normal early development that plateaus and then regresses. The age of regression and lifespan vary widely among the lysosomal storage diseases, with Tay-Sachs being one of the most severe.

Two parents are both affected with albinism (203100, 203200), but have a normal child. Which of the following terms best applies to this situation? A. Allelic heterogeneity B. Locus heterogeneity C. Variable expressivity D. Incomplete penetrance E. New mutation

The answer is b. (Lewis, pp 75-94. Scriver, pp 5587-5628.) Albinism is one of many genetic diseases that exhibit locus heterogeneity, which means that mutations at several different loci can produce identical phenotypes. The two McKusick numbers provide a clue that there is more than one locus for albinism, both causing autosomal recessive disease. Each parent must be homozygous for a mutant allele from one albinism locus but heterozygous or homozygous normal at the other locus. Their child would then be an obligate carrier for each type of albinism. A new mutation in the child is also possible, converting one of the parental mutant alleles to normal, but this would be very rare. Autosomal dominant disorders often vary in severity within families (variable expressivity) but occasionally are clinically silent in a person known to carry the abnormal allele (incomplete penetrance).

In another family with Charcot-Marie-Tooth disease (CMT), restriction analysis using sites flanking the CMT gene on 17 yields one large abnormal fragment and one smaller fragment that is seen in controls. What is the probable inheritance mechanism in this family? A. X-linked recessive B. Autosomal dominant C. Autosomal recessive D. Multifactorial E. X-linked dominant

The answer is b. (Lewis, pp 75-94. Scriver, pp 5759-5788.) The large fragment could derive from a mutation ablating one flanking restriction site or from extra DNA inserted between the restriction sites. The fact that there are two DNA fragment sizes in the affected individual but one in controls suggests alteration of only one of the two homologous CMT regions on chromosome 17. The production of disease by alteration of one homologous locus (one abnormal allele) causes autosomal dominant inheritance. This form of CMT is caused by a duplication of the PMP22 gene, a gene encoding a peripheral myelin protein. The extra copy of PMP22 increases protein abundance and interferes with nerve conduction. DNA duplication is one form of atypical inheritance discovered through DNA analysis.

Juvenile diabetes mellitus is a disorder of carbohydrate metabolism caused by insulin deficiency. The disease often follows a viral infection with inflammation of the pancreatic βcells, but also exhibits genetic predisposition with a 40 to 50% concordance rate in monozygous twins and clustering in families. Juvenile diabetes mellitus is best described as a A. Congenital disorder B. Multifactorial disorder C. Mendelian disorder D. Sporadic disorder E. Sex-limited disorder

The answer is b. (Murray, pp 812-828. Scriver, pp 3-45. Sack, pp 1-40. Wilson, pp 1-20.) Many common diseases are caused by a combination of environmental and genetic factors, and are described as multifactorial diseases. Examples include diabetes mellitus, schizophrenia, alcoholism, and many common birth defects such as cleft palate or congenital dislocation of the hip. The proportion of genetically identical monozygous twins who share a trait such as diabetes mellitus provides a measure of the genetic contribution to etiology (hereditability). Mendelian disorders are completely determined by the genotype of an individual, and exhibit 100% concordance in identical twins. Sporadic disorders have no genetic predisposition and do not cluster in families except by chance or through similar environmental exposure. Congenital disorders are present at birth, in contrast to juvenile diabetes mellitus, which usually presents during childhood. Sex-limited disorders occur predominantly in males or females, in contrast to the approximately equal sex distribution of juvenile diabetes mellitus.

A physician makes the diagnosis of 47,XYY in a 16-year old boy. Which of the following options is most appropriate for the physician during the counseling session that follows the chromosome result? A. Recommend karyotyping of the parents B. Explain that the recurrence risk for such chromosomal aberrations is about 1% C. Urge that the school receive a copy of the karyotype since these boys often have behavior problems D. Recommend testosterone supplementation when the boy reaches puberty E. Inform the parents that their child will be sterile

The answer is b. (Murray, pp 812-828. Scriver, pp 3-45. Sack, pp 57-84. Wilson, pp 123-148.) The recurrence risk for aneuploidies caused by meiotic nondisjunction is about 1% in addition to the maternal age-related risk. It is not known why the risk for aneuploidy increases slightly after an affected child is born, but parental karyotypes are almost always normal. Surveys of penal institutions have revealed an increased incidence of 47,XYY individuals, but other conditions with mental disability are increased as well. As with other chromosomal syndromes, the phenotype of 47,XYY is variable and can be found coincidentally in normal males. It would therefore be inappropriate to label a child as abnormal in school unless there have been previous concerns about a medical disorder. Males with Klinefelter's syndrome (47,XXY), rather than those with 47,XYY syndrome, are often sterile and may require supplementation with male hormones.

Spina bifida exhibits female predilection and recurrence risks of 3% for first-degree relatives and 0.5% for second-degree relatives. A father and child have spina bifida, but the mother is normal. What is the risk that the couple's next child will have spina bifida? A. >1% B. <1% C. >6% D. <6% E. 10%

The answer is c. (Lewis, pp 135-154. Scriver, pp 193-202.) The father and child are affected with spina bifida. The next child will be related to them as a primary (first-degree) relative. The existence of two affected primary relatives predicts a recurrence risk of >6%. Had the child had a Mendelian syndrome, the risk could have been as high as 25% from autosomal recessive inheritance.

Most isolated congenital anomalies exhibit which of the following? A. Mendelian inheritance B. Chromosomal inheritance C. Multifactorial inheritance D. Maternal inheritance E. Atypical inheritance

The answer is c. (Lewis, pp 135-154. Scriver, pp 193-202.) When present as an isolated (rare) anomaly, spina bifida (meningomyelocele) exhibits multifactorial inheritance. Chromosomal inheritance usually causes a syndrome with predictable frequency rather than isolated anomalies. Atypical inheritance (genomic imprinting, trinucleotide repeat instability, mitochondrial inheritance) has not been implicated in neural tube defects.

A male infant is being considered for adoption, and his older half-sister is known to have developed hydrocephalus, an accumulation of cerebrospinal fluid in the brain ventricles. Hydrocephalus is a multifactorial disorder, and the prospective parents wish to know the chance the boy will develop hydrocephalus. In order to estimate this risk, the physician must determine what proportion of genes the brother and half-sister will have in common? A. One B. One-half C. One-fourth D. One-eighth E. One-sixteenth

The answer is c. (Lewis, pp 267-282. Scriver, pp 3-45.) Although all individuals, other than identical twins, are genetically unique, each individual will have some genes in common with their relatives. The more closely related, the more genes individuals have in common. First-degree relatives, such as siblings, parents, and children, share one-half of their genes. Second-degree relatives share one-fourth, and third-degree relatives share one-eighth. Full siblings will have half their genes in common, half-siblings (with only one common parent) one-fourth. Risks for multifactorial disorders like hydrocephalus fall off rapidly with decreasing degrees of relationship, with average figures of 3-5% for first degree relatives (sharing 1 /2 their genes), 0.5-0.7% for second-degree relatives (sharing one-fourth their genes as for this child and his sister), dropping to near baseline incidence (1-3 per thousand for congenital anomalies) for third-degree relatives like first cousins.

A newborn infant presents with poor feeding, vomiting, jaundice, and an enlarged liver. The urine tests positive for reducing substances, indicating the presence of sugars with aldehyde groups. Which of the following processes is most likely to be abnormal? A. Conversion of glucose to galactose B. Conversion of lactose to galactose C. Conversion of activated galactose to activated glucose D. Excretion of glucose by the kidney E. Excretion of galactose by the kidney

The answer is c. (Lewis, pp 377-396. Scriver, pp 1553-1588. Murray, pp 102-110.) This infant may have galactosemia (230400), a deficiency of galactose-1-phosphate uridyl transferase (GALT). Galactose from lactose in breast milk or infant formula is phosphorylated by galactokinase, activated to uridine diphosphogalactose (UDP-galactose) by GALT, and converted to UDP-glucose by UDP-galactose epimerase. The elevation of galactose metabolites is thought to cause liver toxicity, and their urinary excretion produces reducing substances. Infants with the signs and symptoms listed are placed on lactose-free formulas until enzyme testing is complete. Deficiencies of epimerase or kinase can cause mild forms of galactosemia.

Marfan syndrome is caused by which of the following mechanisms? A. Mutation that prevents addition of carbohydrate residues to the fibrillin glycoprotein B. Mutation in a carbohydrate portion of fibrillin that interferes with targeting C. Mutation that disrupts the secondary structure of fibrillin and blocks its assembly into microfibrils D. Mutation in a lysosomal enzyme that degrades fibrillin E. Mutation in a membrane receptor that targets fibrillin to lysosomes

The answer is c. (Lewis, pp 377-396. Scriver, pp 5287-5312. Murray, pp 539-540.) Mutations in structural proteins often exhibit autosomal dominant inheritance, while mutations in enzymes often exhibit autosomal recessive inheritance. Structural proteins such as collagen or fibrillin must interact to form scaffolds in the extracellular matrix of connective tissue. Mutation at one of the homologous autosomal loci can introduce an abnormal polypeptide throughout the scaffold much like a misshapen brick in a wall—the distorted polypeptide from the abnormal locus subverts that from the normal locus and weakens the connective tissue matrix, causing autosomal dominant disease. Sometimes the abnormal polypeptide complexes with normal polypeptides and causes them to be degraded, a mechanism called protein suicide. The suicidal effects of mutations at some loci are referred to generally as "dominant negative" mutations. Fibrillin is a glycoprotein used to form a scaffold in the connective tissue filaments called microfibrils. It is distributed in the suspensory ligament for the lens of the eye, the aorta, and the bones and joints, accounting for the symptoms of Marfan syndrome (154700). Similar pathogenetic mechanisms occur in the osteogenesis imperfectas (e.g., 166200) with multiple fractures and in the Ehlers-Danlos syndromes (e.g., 130060) with skin fragility (scarring) and vascular disease due to mutations in various collagens. The mutations disrupt the α-helix secondary structure of collagens, which is dependent on the glycine-X-Y triplet amino acid repeats; the distorted collagen polypeptides then disrupt the collagen fibrils with symptoms dependent on its tissue distribution (2 types of fibrillin and more than 15 types of collagen are known).

A newborn presents with ambiguous genitalia, having an enlarged clitoris or small phallus and labial fusion or hypoplastic scrotum. The newborn's sex can most reliably be established by which of the following? A. Buccal smear to determine if there are one or two Barr bodies B. Buccal smear to determine if there is one Barr body or none C. Peripheral blood karyotype D. Bone marrow karyotype E. Polymerase chain reaction (PCR) using primers specific for the long arm of the Y chromosome

The answer is c. (Lewis, pp 377-496. Scriver, pp 4077-5016. Murray pp 396-414.) A peripheral blood karyotype provides the most reliable examination of the sex chromosomes. A bone marrow karyotype is more rapid (it uses rapidly dividing bone marrow cells) but usually has less resolution for defining subtle X and Y chromosome rearrangements. A buccal smear would theoretically show one Barr body in females (representing inactivation of one X chromosome) and none in males. In practice, this test is not very reliable and is rarely used. Detection of material of the Y long arm by polymerase chain reaction (PCR) would be useful but does not examine the Y short arm that contains the sex-determining region.

Neonatal screening is mandated in all states, but examines different numbers of diseases. Most commonly tested are phenylketonuria (PKU— 261600), galactosemia (230400), congenital hypothyroidism, and sickle cell anemia (603903). Recently, a supplemental newborn screen using tandem mass spectrometry is being adopted by many states, allowing recognition of the more common organic acidemias and fatty acid oxidation disorders. Which of the following is the most important characteristic to qualify a disorder for newborn screening? A. A highly accurate diagnostic test B. A high frequency of disease C. An advantage for treatment from early diagnosis D. Use of microbial technology like the Guthrie method E. A minimal incidence of false positive tests

The answer is c. (Lewis, pp 397-416. Scriver, pp 175-192.) Genetic screening requires not only a highly accurate diagnostic test but also one that can be adapted to testing of large numbers of individuals. Key for neonatal screening is that the disease can be ameliorated because of early diagnosis—some countries have tried screening for Duchenne muscular dystrophy (310200) but most do not because there is no treatment advantage from early diagnosis. Neonatal screening is set up so there will be more false positives than false negatives; this requires some work for pediatricians in obtaining and interpreting repeat screens, but is considered far preferable to missing a child with preventable disease consequences. The Guthrie test was the first to allow screening of large populations, as exemplified by the test for phenylketonuria (PKU—261600): infant blood from a heel or finger stick is placed on filter paper discs and mailed to the central screening laboratory. Discs are arrayed on agar plates containing a competitive inhibitor of bacterial growth (thienylalanine), which must be overcome by sufficient amounts of phenylalanine for bacterial colonies to be visible. Rapid scanning of agar plates with hundreds of filter discs is thus possible by eye, and discs surrounded by bacterial growth constitute a positive result. A recent problem for newborn screening is the trend towards early infant discharge (24 hours or less). If the infant blood sample is obtained too early before adequate dietary intake, blood levels of phenylalanine or other metabolites may not be elevated and a false negative result will be obtained. Many hospitals request that parents return with their infant for proper screening. The supplemental newborn screen by tandem mass spectrometry can be justified because the aggregate incidence of its 30 detected disorders is 1 in 5-6000, well above that for currently screened metabolic disorders such as PKU (1 in 10-12,000) and galactosemia (1 in 40,000).

Tay-Sachs disease (272800) is a neurolipidosis that causes cherry red spots in the eye, "startle" responses in infancy, neurodegeneration, and death. The disorder is autosomal recessive, caused by deficiency of the lysosomal enzyme hexaminidase A with resultant accumulation of complex glycolipids in brain. What is the risk that the grandmother of an affected child is a carrier (heterozygote) for Tay-Sachs disease? A. 100% B. 67% C. 50% D. 25% E. Virtually 0

The answer is c. (Lewis, pp 75-94. Scriver, pp 3827-3876.) Parents of children with autosomal recessive disorders are obligate carriers if nonpaternity and rare examples of uniparental disomy (inheritance of both chromosomal homologues from the same parent) are excluded. Normal siblings have a 2/3 chance of being carriers because they cannot be homozygous for the abnormal allele. Grandparents have a one-half chance of being carriers because one or the other must have transmitted the abnormal allele to the obligate carrier parent. First cousins share a set of grandparents of whom one must be a carrier. There is a one-half chance for the aunt or uncle to be a carrier and a one-fourth chance for the first cousin. Halfsiblings share an obligate carrier parent and have a one-half chance of being carriers. These calculations assume a lack of mutations (Tay-Sachs is rare) and a lack of coincidental alleles (no consanguinity).

A woman with cystic fibrosis (219700) marries her first cousin. What is the risk that their first child will have cystic fibrosis? A. 1 /2 B. 1 /4 C. 1 /8 D. 1 /16 E. 1 /32

The answer is c. (Lewis, pp 75-94. Scriver, pp 5121-5188. Murray, pp 415-433.) The McKusick number for cystic fibrosis (219700) begins with 2, indicating an autosomal recessive disorder. The genotype of the affected woman with cystic fibrosis is therefore best represented as the two lowercase letters cc. Her parents are obligate carriers for the disorder (genotypes Cc), and one of her grandparents must also be a carrier (barring new mutations). Her first cousin then has a one-fourth chance of being a carrier, since one of their common grandparents is a carrier, one of his parents has a one-half chance of being a carrier, and he has a one-half chance of inheriting the c allele from his parent. The affected woman can only transmit c alleles to her fetus, while her cousin has one-half chance of transmitting his c allele if it is present. Thus, the probability that the first child will have cystic fibrosis is one-fourth (cousin is carrier)×one-half (cousin transmits c allele) = 1 /8 (fetus has cc genotype).

An infant with severe muscle weakness is born to a mother with mild muscle weakness and myotonia (sustained muscle contractions manifested clinically by the inability to release a handshake). The mother's father is even less affected, with some frontal baldness and cataracts. Worsening symptoms in affected individuals of successive generations suggest which of the following inheritance mechanisms? A. Genomic imprinting B. Heteroplasmy C. Unstable trinucleotide repeats D. Multifactorial inheritance E. Mitochondrial inheritance

The answer is c. (Lewis, pp 95-112. Scriver, pp 3-45.) Anticipation refers to the worsening of the symptoms of disease in succeeding generations. The famous geneticist L.S. Penrose dismissed anticipation as an artifact, but the phenomenon has been validated by the discovery of expanding trinucleotide repeats. Steinert myotonic dystrophy is caused by unstable trinucleotide repeats near a muscle protein kinase gene on chromosome 19; the repeats are particularly unstable during female meiosis and may cause a severe syndrome of fetal muscle weakness and joint contractures. Variable expressivity could also be used to describe the family in the question, but the concept implies random variation in severity rather than progression with succeeding generations. Diseases that involve triplet repeat instability exhibit a bias for exaggerated repeat amplification during meiosis (e.g., women with the fragile X syndrome or myotonic dystrophy and men with Huntington chorea). The explanation for this bias is unknown.

Which phrase best defines the genetic disease category of multifactorial inheritance? A. Mendelian inheritance B. Mitochondrial inheritance C. Most common type of human genetic disease D. Major cause of miscarriages E. Maternally derived

The answer is c. (Murray, pp 812-828. Scriver, pp 3-45. Sack, pp 205-222. Wilson, pp 45-57.)Genetic disorders may be classified into several major categories. Multifactorial disorders, the most common type of human genetic disease, represent the composite effects of multiple genes, each of which contributes a component to the disorder. Environmental factors also play a role in multifactorial disorders. Many common diseases, such as coronary artery disease or diabetes mellitus, are multifactorial disorders. Chromosomal disorders are caused by the deletion or duplication of either pieces of chromosomes or entire chromosomes and are a common cause of miscarriage. Single-gene disorders, also known as Mendelian disorders, are due to defects in single genes. Since mitochondria are cytoplasmic organelles inherited via the cytoplasm of the ovum, Mitochondrial disorders may be maternally inherited.

Chromosomal analysis reveals a 47,XYY karyotype. Which of the following descriptions best fits this abnormality? A. Autosomal trisomy B. A male with Klinefelter's syndrome C. Sex chromosome aneuploidy D. A female with Turner's syndrome E. Sex chromosome triploidy

The answer is c. (Murray, pp 812-828. Scriver, pp 3-45. Sack, pp 57-84. Wilson, pp 123-148.) The 47,XYY karyotype is an example of sex chromosome aneuploidy, as are Klinefelter's syndrome (47,XXY), Turner's syndrome (45,X), and triple X syndrome (47,XXX). Sex chromosome mixoploidy implies mosaicism, such as 45,X/46,XX with two cell lines in one individual. Autosomal trisomies include Down's syndrome [47,XX+21 (trisomy 21)], Patau's syndrome [47,XX+13 (trisomy 13)], and Edwards' syndrome [47,XY+18 (trisomy 18)].

The error in meiosis that produces a 47,XYY karyotype is best described by A. Meiosis division I of paternal spermatogenesis B. Meiosis division I of maternal oogenesis C. Meiosis division II of paternal spermatogenesis D. Meiosis division II of maternal oogenesis E. Meiosis division II in either parent

The answer is c. (Murray, pp 812-828. Scriver, pp 3-45. Sack, pp 57-84. Wilson, pp 123-148.) The sex chromosomes with differently named homologues allow easy visualization of chromosome sorting during meiosis. Female meiosis only involves X chromosomes; thus, Y chromosomal abnormalities must arise during paternal meiosis or occur spontaneously in offspring. Nondisjunction at paternal meiosis I produces XY secondary spermatocytes and a 24,XY gamete. Fertilization with a 23,X ovum yields a 47,XXY individual (Klinefelter's syndrome). Only nondisjunction at paternal meiosis II produces a 24,YY gamete that yields a 47,XYY individual after fertilization.

406. Mr. Smith is affected with Crouzon's syndrome (123500) and has craniosynostosis (i.e., premature closure of the skull sutures) along with unusual facies that includes proptosis secondary to shallow orbits, hypoplasia of the maxilla, and a prominent nose. His son and brother are also affected, although two daughters and his wife are not. Mr. and Mrs. Smith are considering having another child. Their physician counsels them that the risk that the child will be affected with Crouzon's syndrome is A. 100% B. 67% C. 50% D. 25% E. Virtually 0

The answer is c. (Murray, pp 812-828. Scriver, pp 3-45. Sack, pp 97-158. Wilson, pp 23-39.) In an autosomal dominant pedigree, there is a vertical pattern of inheritance. Assuming the disorder is not the result of a new mutation, every affected person has an affected parent. The same is true of X-linked dominant pedigrees. However, male-to-male transmission, as seen in this family, excludes the possibility of an X-linked disorder. A person with an autosomal dominant phenotype has one mutant allele and one normal allele. These people randomly pass one or the other of these alleles to their offspring, giving a child a 50% chance of inheriting the mutant allele and therefore being affected with the disorder. This risk is unaffected by the genotypes of the previous offspring.

The regulation of transferrin receptors is studied in tissue culture. There is increased synthesis of transferrin receptor protein with no changes in transferrin mRNA transcription. Which of the following is the most plausible explanation? A. Change in amounts or types of transcription factors B. Allosteric regulation of transferrin receptor function C. Activation of transferrin receptor function by a protein kinase D. Stabilization of transferrin mRNA E. Increased GTP levels to accelerate protein elongation

The answer is d. (Lewis, pp 377-396. Murray, pp 580-597. Scriver, pp 3127-3162.) The regulation of mammalian gene expression is selective: specific genes are up- or downregulated by controls at the gene dosage, mRNA transcription, mRNA splicing, mRNA stability, or protein function levels. Under conditions of iron deficiency, transferrin receptor mRNA is stabilized so that more protein is synthesized. Regulation thus occurs at the protein translation level, without changes in transferrin mRNA transcription through transcription factors or transferrin receptor activity through interaction with small molecules (allostery) or through phosphorylation by protein kinases. Overall increases in rates of RNA transcription or protein elongation are not employed for gene regulation by mammalian cells.

The major blood group locus in humans produces types A (genotypes AA or AO), B (genotypes BB or BO), AB (genotype AB), or O (genotype OO). For parents who are type AB and type O, what are the possible blood types of their offspring? A. Type AB child B. Type B child C. Type O child D. Type A or B child E. Type B or AB child

The answer is d. (Lewis, pp 75-94. Scriver, pp 3-45.) Diploid persons have two alleles per autosomal locus, with one being transmitted to each gamete (Mendel's law of segregation). The key to blood group problems is to recognize that a blood type is ambiguous regarding possible alleles—type A persons may have AA or AO genotypes. Once the possible genotypes are deduced from the blood types, potential offspring will represent all combinations of parental alleles. Parents with AB and OO genotypes can only have offspring with genotypes AO (type A) or BO (type B).

A couple is referred to the physician because their first three pregnancies have ended in spontaneous abortion. Chromosomal analysis reveals that the wife has two cell lines in her blood, one with a missing X chromosome (45,X) and the other normal (46,XX). Her chromosomal constitution can be described as A. Chimeric B. Monoploid C. Trisomic D. Mosaic E. Euploid

The answer is d. (Murray, pp 812-828. Scriver, pp 3-45. Sack, pp 57-84. Wilson, pp 123-148.) The case described represents one of the more common chromosomal causes of reproductive failure, Turner mosaicism. Turner's syndrome represents a pattern of anomalies including short stature, heart defects, and infertility. Turner's syndrome is often associated with a 45,X karyotype (monosomy X) in females, but mosaicism (i.e., two or more cell lines with different karyotypes in the same individual) is common. However, chimerism (i.e., two cell lines in an individual arising from different zygotes, such as fraternal twins who do not separate) is extremely rare. Trisomy refers to three copies of one chromosome, euploidy to a normal chromosome number, and monoploidy to one set of chromosomes (haploidy in humans).

The proper cytogenetic notation for a female with Down's syndrome mosaicism is A. 46,XX,+21/46,XY B. 47,XY,+21 C. 47,XXX/46,XX D. 47,XX,+21/46,XX E. 47,XX,+21(46,XX)

The answer is d. (Murray, pp 812-828. Scriver, pp 3-45. Sack, pp 57-84. Wilson, pp 123-148.)Mosaicism occurs when a chromosomal anomaly affects one of several precursor cells of an embryo or tissue. The two or more karyotypes that characterize the mosaic cells are separated by a slash in cytogenetic notation. The notation 47,XX,+21 denotes a cell line typical of a female with trisomy 21 (Down's syndrome), while 46,XX is the karyotype expected for a normal female.

405. A couple comes to the physician's office after having had two sons affected with a similar disease. The first-born son is tall and thin and has dislocated lenses and an IQ of 70. He has also experienced several episodes of deep vein thromboses. The chart mentions deficiency of the enzyme cystathionine-β-synthase, but a diagnosis is not given. The second son was treated from an early age with pyridoxine (vitamin B6) and is less severely affected. No other family members are affected. While taking a family history, the physician discovers that the parents are first cousins. The 38-year-old mother is pregnant, and amniocentesis has demonstrated that the fetus has a 46,XY karyotype. The risk that the fetus will be affected with the same disease is A. 100% B. 67% C. 50% D. 25% E. Virtually 0

The answer is d. (Murray, pp 812-828. Scriver, pp 3-45. Sack, pp 97-158. Wilson, pp 23-39.) The family history and the likelihood that the boys have a metabolic disease suggest autosomal recessive inheritance. Autosomal recessive conditions tend to have a horizontal pattern in the pedigree. Although there may be multiple affected individuals within a sibship, parents, offspring, and other relatives are generally not affected. Most autosomal recessive conditions are rare; however, consanguinity greatly increases the likelihood that two individuals will inherit the same mutant allele and pass it along to their offspring. The recurrence risk for the fetus will be that for an autosomal recessive condition with carrier parents—1/4 or 25%. This risk is not affected by the sex of the fetus. The disease caused by cystathionine-βsynthase (CS) deficiency is homocystinuria (236300). S-adenosylmethionine accepts methyl groups and is converted to S-adenosylhomocysteine, which yields homocysteine; homocysteine is converted to cystathionine by CS. Methionine and homocysteine (dimerized to homocystine) accumulate, and homocystine is excreted in urine. Pyridoxine is a cofactor for CS and is beneficial in some forms of homocystinuria. Other causes of homocystinuria include defective cobalamin (vitamin B12) metabolism.

A couple present for genetic counseling because three of their first four children have had pyloric stenosis, a narrowing of the valve between stomach and duodenum that presents with severe vomiting and loss of gastric hydrochloric acid (hypochloremic alkalosis). When parents with three affected children have a higher recurrence risk than parents with one affected child, the disease in question is likely to exhibit which of the following modes of inheritance? A. Autosomal dominant inheritance B. Autosomal recessive inheritance C. X-linked recessive inheritance D. X-linked dominant inheritance E. Multifactorial determination

The answer is e. (Lewis, pp 135-154. Scriver, pp 193-202.) An increasing recurrence risk according to the number of relatives affected is characteristic of polygenic inheritance. The more affected relatives there are, the more evidence there is that an individual's genetic background is shifted toward the threshold for a particular trait; for example, the expectation for tall parents with tall grandparents is to have tall children. Inheritance risks for Mendelian disorders are unaffected by outcomes in prior offspring because they reflect segregation ratios for a known (or deduced) pair of alleles (chance has no memory). Pyloric stenosis was once a lethal disease but now is easily repaired by surgery (recently, by less invasive laparoscopic surgery). The diagnosis can be difficult, often requiring palpation of the abdomen for 20-30 minutes when the baby has vomited and is relaxed. The characteristic "olive" mass can be felt in the mid-abdomen and can now be confirmed by ultrasound study.

Deletions of 11p13 may result in Wilm tumor, aniridia, genitourinary malformations, and mental retardation (WAGR syndrome). In some patients, however, not all features are seen. Additionally, individual features of this syndrome may be inherited separately in a Mendelian fashion. Limited features may also be seen in patients without visible chromosomal deletions. Which of the following is the most likely mechanism for this finding? A. Mitochondrial inheritance B. Imprinting C. Germ-line mosaicism D. Uniparental disomy E. Contiguous gene syndrome

The answer is e. (Lewis, pp 241-266. Scriver, pp 3-45.) Contiguous gene syndromes, also known as microdeletion syndromes, occur when deletions result in the loss of several different closely linked loci. Depending on the size of the deletion, different phenotypes may result. Mutations in the individual genes may result in isolated features that may be inherited in a Mendelian fashion.

If all SS individuals in the Minnesota population were sterilized, the SS genotype frequency in the next generation would be which of the following? A. Reduced by 2/3 B. Reduced by 1 /2 C. Reduced by 1 /3 D. Reduced to 0 E. Approximately the same

The answer is e. (Lewis, pp 267-282. Scriver, pp 4571-4636.) Even if SS individuals were prevented from contributing to the next generation by sterilization, breeding between AS individuals would replenish SS genotype frequencies. This stability of populations in accord with the HardyWeinberg law is often referred to as the Hardy-Weinberg equilibrium. During the decades of 1900 to 1920 in America, the eugenics movement succeeded in passing laws obligating sterilization of those with mental disabilities. These laws were based on two false premises—the idea that mental retardation is always due to Mendelian transmission (ignoring chromosomal and multifactorial disease) and the idea that elimination of affected people will always change gene frequencies.

An African American couple with a normal family history wants to know their chance of having a child with sickle cell anemia (603903). The incidence of sickle cell trait is 1 in 8 for African Americans. Which of the following is the risk in this case? A. 1 /8 B. 1 /16 C. 1 /60 D. 1 /120 E. 1 /256

The answer is e. (Lewis, pp 267-282. Scriver, pp 4571-4636.) The African American man and woman each have a one-eighth chance of having sickle trait. They have a 1 /64 × 1 /4 = 1 /256 chance of having a child with sickle cell anemia. There is also a 1 /64 × 1 /2 = 1 /128 chance that their child will have sickle trait.

The frequency of Tay-Sachs carriers among Ashkenazi Jews is 1 /30. The frequency of Tay-Sachs carriers among whites of Western European descent is approximately 1 /300. If a mother is an Ashkenazi Jew and a father is a white from Western Europe, what is the chance that a child of this union will have Tay-Sachs disease? A. 1 /120 B. 1 /240 C. 1 /3600 D. 1 /9000 E. 1 /36,000

The answer is e. (Lewis, pp 75-94, 267-282. Scriver, pp 3827-3876.) To determine the joint probability of two or more independent events, the product of their separate probabilities must be determined. If the parents were both Ashkenazi Jews, they would have a 1/30 chance of carrying an abnormal gene for Tay-Sachs disease; for each pregnancy, they would have a 1/2 chance of passing that gene along should they carry it. The probability that all of these four independent events would occur is 1/30 × 1/2 × 1/30 × 1/2 = 1/3600. The joint probability for a mother who is an Ashkenazi Jew and a father who is not is 1/30 × 1/2 × 1/300 × 1/2 = 1/36,000.

A sibling donor is found for a patient with Tay-Sachs disease, and the physician writes to the patient's insurance company explaining the diagnosis of Tay-Sachs disease and the reasons for the bone marrow transplant. Not only does the insurance company refuse payment for transplantation, it also discontinues coverage for the family based on anticipated medical expenses. From the ethical perspective, these events fall under which of the following categories? A. Patient confidentiality B. Nondisclosure C. Informed consent D. Failure to provide ongoing care E. Discrimination

The answer is e. (Lewis, pp 397-416. Scriver, pp 3827-3876.) The physician is obligated to describe a patient's disease accurately in the medical record and to share such records with legally entitled entities, such as health insurance companies. Although care should be exercised that records containing confidential information are not shared inappropriately, there was no such breach of confidentiality in this case. If the physician had declined further care without appropriate notice, then this would be a breach of ongoing care. However, insurance companies and managed care plans have excluded patients because of prior conditions or excessive expenses (i.e., capitation limits). This does constitute discrimination, but application of the Americans with Disabilities Act to patients with genetic diseases is not yet routine. These dilemmas will grow dramatically with the increasing ability to test for genetic diseases and predispositions. Although the administration of exogenous normal enzyme (enzyme therapy) or transplantation to provide a cellular source of normal enzyme has been successful in correcting lysosomal deficiencies, the enzymes fail to cross the blood-brain barrier in sufficient amounts to remit neurological symptoms in patients with lipidoses. This form of enzyme therapy has the advantage of targeting the defective organelle via the mannose-6-phosphate residues on the enzyme. It is very expensive but effective in lipidoses that have few neurological symptoms, such as Gaucher disease (230800).

Which of the following statements about hemophilia A (306700) is true? A. The extrinsic clotting pathway is impaired B. The cleavage of fibrinogen is impaired C. Tissue factor activation is impaired D. Activation of factor XII is impaired E. Activation of factor X is impaired

The answer is e. (Lewis, pp 75-94. Scriver, pp 4367-4392. Murray, pp 598-608.) Hemophilia A is caused by deficiency of factor VIII and hemophilia B by deficiency of factor IX. Both factors are involved in the intrinsic blood coagulation pathway that results in activation of factor X. Alternatively, factor X can be activated by tissue factors through the extrinsic blood coagulation pathway. Activated factors X and V produce thromin from prothrombin, which in turn cleaves fibrinogen to produce fibrin monomers. The fibrin monomers are polymerized and cross-linked to produce a fibrin polymer, which interacts with platelets and other factors to produce a blood clot. The genes for factor VIII and factor IX are on the X chromosome, making hemophilia A and B X-linked recessive diseases.

Osteogenesis imperfecta (166200) is an autosomal dominant disorder that causes thin, bluish sclerae (whites of the eyes), deafness, and multiple bone fractures. Parents have two children with osteogenesis imperfecta, but themselves exhibit no signs of the disease. Which of the following genetic mechanisms is the most likely explanation for two offspring of normal parents to have an autosomal dominant disease? A. Variable expressivity B. Uniparental disomy C. New mutations D. Germinal mosaicism in one parent E. Incomplete penetrance

The answer is e. (Lewis, pp 75-94. Scriver, pp 5241-5287.) If two individuals in a sibship are affected with an autosomal dominant disease, then the usual implication is that one of the parents has the abnormal allele. Parents with one normal and one abnormal allele have a 50% chance of transmitting the abnormal allele with each pregnancy. Complicating the recognition of autosomal dominant inheritance are incomplete penetrance, where there are no signs of the disease phenotype after all relevant medical evaluations, and variable expressivity, where a parent may have more subtle disease than the offspring. Incomplete penetrance applies to this family because the parents have no signs or symptoms of disease. If a mutation occurs in the primordial germ cells, then these cells may have abnormal alleles despite the lack of these alleles in the rest of the body tissues (germinal mosaicism). Germinal mosaicism was thought to be very rare until testing for type I collagen gene mutations in osteogenesis imperfecta allowed verification of germinal mosaicism in this condition. Germinal mosaicism explained why autosomal recessive inheritance had been incorrectly postulated for families with normal parents and multiple affected children. Once a child has received the abnormal allele through the gamete of the mosaic parent, the child has the abnormal allele in all cells, with the usual 50% risk of transmission. The characterization of mutations in the α1 or α2 chains of type I collagen in the osteogenesis imperfectas allowed proof of germ-line mosaicism through paternal sperm studies and corrected the false impression that such families were examples of autosomal recessive inheritance.

A woman with no family history of color blindness (304000) marries a color-blind man. What are the risks for this couple of having a son or daughter who is color-blind? A. 100% B. 75% C. 50% D. 25% E. Virtually 0

The answer is e. (Lewis, pp 75-94. Scriver, pp 5955-5976.)The common forms of color blindness are X-linked recessive, as indicated by the initial 3 of the McKusick number (304000). The couple's daughters will be obligate carriers—that is, carriers implied by the pedigree. Using a lowercase c to represent the recessive color blindness allele, the woman is most likely XCXC, while her husband is XcY. The Punnett square below indicates that all daughters will be carriers (XcXC), while sons will be normal (XCY). Note again that loci on the X chromosome cannot be transmitted from father to son, since the son receives the father's Y chromosome.

Assuming that all alleles derive from a single locus, match the mating of an Aa father with an aa mother and their probabilities for genotypes in offspring. A. 1 AA B. 1⁄2 AA, 1⁄2 aa C. 1⁄4 AA, 1⁄2 Aa, 1⁄4 aa D. 1⁄2 AA, 1⁄2 Aa E. 1⁄2 Aa, 1⁄2 aa

The answer is e. (Murray, pp 812-828. Scriver, pp 3-45. Sack, pp 1-40. Wilson, pp 1-20.) During meiotic segregation, each parental gamete receives one allele from every genetic locus. The probability of a parental allele being transmitted to offspring is thus 1/2, and the probability of a given genotype appearing in offspring is thus a joint probability. For a maternal Aa versus paternal aa mating, the probability of maternal alleles A or a being transmitted is 1/2, and the probability of transmission of the paternal allele a is 1. The joint probability for an Aa genotype in offspring is thus 1/2 ×1 =1/2, the same as for an aa genotype. For a maternal Aa versus paternal Aa mating, the probabilities for AA, Aa, or aa genotypes in offspring are all 1/2 × 1/2 = 1/4, but the Aa genotype can occur in two ways (A from mother, a from father, or vice versa).

Which of the following karyotypes is an example of aneuploidy? A. 46,XX B. 23,X C. 69,XXX D. 92,XXXX E. 90,XX

The answer is e. (Murray, pp 812-828. Scriver, pp 3-45. Sack, pp 57-84. Wilson, pp 123-148.) Aneuploidy involves extra or missing chromosomes that do not arise as increments of the haploid chromosome number n. Polyploidy involves multiples of n, such as triploidy (3n = 69,XXX) or tetraploidy (4n = 92,XXXX). Diploidy (46,XX) and haploidy (23,X) are normal karyotypes in gametes and somatic cells, respectively. A 90,XX karyotype represents tetraploidy with two missing X chromosomes, which has been seen in one patient who had features that resembled those of Turner's syndrome.

A child with cleft palate, a heart defect, and extra fifth fingers is found to have 46 chromosomes with extra material on one homologue of the chromosome 5 pair. This chromosomal abnormality is best described by which of the following terms? A. Polyploidy B. Balanced rearrangement C. Ring formation D. Mosaicism E. Unbalanced rearrangement

The answer is e. (Murray, pp 812-828. Scriver, pp 3-45. Sack, pp 57-84. Wilson, pp 123-148.) Chromosomal abnormalities may involve changes in number (i.e., polyploidy and aneuploidy) or changes in structure (i.e., rearrangements such as translocations, rings, and inversions). Extra material (i.e., extra chromatin) seen on chromosome 5 implies recombination of chromosome 5 DNA with that of another chromosome to produce a rearranged chromosome. Since this rearranged chromosome 5 takes the place of a normal chromosome 5, there is no change in number of the autosomes (nonsex chromosomes) or sex chromosomes (X and Y chromosomes). The question implies that all cells karyotyped from the patient (usually 11 to 25 cells) have the same chromosomal constitution, ruling out mosaicism. The patient's clinical findings are similar to those occurring in trisomy 13, suggesting that the extra material on chromosome 5 is derived from chromosome 13, producing an unbalanced karyotype called dup(13) or partial trisomy 13.

402. Gardner's syndrome is an autosomal dominant condition characterized by multiple polyps of the intestines, bony tumors, skin cysts, and a high risk of intestinal cancer. A family is encountered in which a greatgrandfather, grandmother, and father are affected with Gardner's syndrome and develop intestinal cancer in their thirties. The father brags that none of his four children have inherited Gardner's syndrome because they lack skin cysts and have not had cancer. The chance that at least one child has inherited the Gardner's syndrome allele, and the reason the children have not manifested cancer, are A. 1/4, ascertainment bias B. 1/2, variable cancer predisposition C. 3/4, early-onset disease manifestation D. 13/16, incomplete medical evaluation E. 15/16, later-onset disease manifestation

The answer is e. (Murray, pp 812-828. Scriver, pp 521-524. Sack, pp 85-96. Wilson, pp 23-39.) The father is affected with Gardner's syndrome (175100), an autosomal dominant disease. Therefore, each of his four children has a 1/2 chance of receiving the allele that causes Gardner's syndrome and a 1/2 chance of receiving the normal allele. The probability that none of his four children received the allele for Gardner's syndrome is thus the joint probability of four independent events, computed by the product 1/2 × 1/2 × 1/2 × 1/2 = 1/16. The probability that at least one child has received the abnormal Gardner's syndrome allele is thus 1 − 1/16 = 15/16. Gardner's syndrome is one of many genetic disorders that may not be obvious in early childhood. Intestinal cancer in particular has a later onset, with 50% of patients being affected by age 30 to 35. More extensive evaluation of the children for internal signs of disease (e.g., the bony tumors) is required before the father can conclude that he has not transmitted the gene. Lateonset disorders are an important category of adult genetic disease, and presymptomatic testing for these diseases is a novel application of DNA diagnosis.

A man with early-onset emphysema undergoes protein electrophoresis for analysis of α1-antitrypsin (AAT) deficiency (107400). The result shows two electrophoretic bands that react with AAT, one at the normal position and one at an abnormal position. Which of the following best describes this result? A. The man is homozygous and has normal AAT activity B. The man is heterozygous and has normal AAT activity C. The man is homozygous and has deficient AAT activity D. The man is homozygous and has an altered AAT protein E. The man is heterozygous and has an altered AAT protein

The answer is e. (Scriver, pp 5559-5586. Lewis, pp 377-396. Murray, pp 21-29.) Serum protein electrophoresis separates proteins according to their structure and charge. Two bands for AAT in this man imply that two types of AAT protein with different structures or charges are present. The electrophoresis does not reveal whether the abnormal AAT protein has normal or abnormal activity. The McKusick number indicates that AAT deficiency is autosomal dominant, implying that two homologous loci encode AAT proteins. The man is thus heterozygous, one locus encoding a normal and one an abnormal protein. The AAT locus is located on chromosome 14 within a family of protease inhibitors called serpins. Altered AAT proteins termed M, S, or Z variants have normal inhibitory activity but are defective in their rates of secretion across the liver membrane into the blood. Lower levels of AAT protein apparently expose lung proteins to damage, causing emphysema. Heterozygotes are usually not affected, so the man may have emphysema because of cigarette smoking or other factors. Homozygous ZZ individuals may have liver disease in addition to lung disease because the abnormally secreted AAT accumulates in liver cells.

Which of the following statements best describes a characteristic of oncogenes? A. All retroviruses contain at least one oncogene. B. Retroviral oncogenes were originally obtained from a cellular host chromosome. C. Proto-oncogenes are genes, found in retroviruses, which have the potential to transform normal cells when expressed inappropriately. D. The oncogenes that lead to human disease are different from those that lead to tumors in animals. E. Oncogenes are mutated versions of normal viral gene products.

The answer is B. Retroviral oncogenes were originally obtained from genes on a host's chromosome. All retroviruses do not contain oncogenes (thus, A is incorrect); proto-oncogenes are found in host cells, not viruses (thus, C is incorrect); the oncogenes that lead to human disease are very similar to those that are mutated in animals (thus, D is incorrect); and oncogenes are a misexpressed or mutated version of normal cellular genes, not viral genes (thus, E is incorrect).

Individuals with achondroplastic dwarfism have about 80% fewer viable offspring than do normal persons, but the incidence of achondroplasia seems to have remained constant for generations. These observations imply which of the following? A. Decreased fitness, negative selection, and relatively high mutation rates B. Increased fitness, negative selection, and relatively high mutation rates C. Decreased fitness, positive selection, and relatively low mutation rates D. Increased fitness, positive selection, and relatively low mutation rates E. Decreased fitness, positive selection, and relatively high mutation rates

The answer is a. (Lewis, pp 267-282. Scriver, pp 3-45.)If an abnormal allele is as likely to be transmitted to the next generation as its corresponding normal allele, it is said to have a fitness of 1. Loss of fitness (decrease in allele frequency after one generation) is also referred to as negative selection. The decreased fitness of achondroplast alleles that are eliminated by negative selection must be balanced by new mutations if the disorder has not disappeared or declined in incidence. Thus, the mutation rate of achondroplasia would be expected to be high relative to those of more benign dominant diseases.

In Burkitt lymphoma, there is increased expression of a hybrid protein with an amino-terminus similar to immunoglobulin (Ig) heavy chain and an unknown carboxyterminus. Which of the following best explains this phenomenon? A. Chromosome translocation that brings together an Ig heavy chain with an oncogene B. Chromosome duplication involving a segment with an oncogene C. Chromosome translocation involving a segment with Ig heavy chains D. Chromosome deletion removing an oncogene E. Chromosome deletion removing a tumor suppressor gene

The answer is a. (Lewis, pp 355-376. Scriver, pp 521-552. Murray, pp 396-414.) Chromosome translocations may often promote tumors in somatic cells by placing regulatory genes next to promoters that aberrantly increase their expression. Burkitt lymphoma, a B cell lymphoma that usually occurs in childhood, often involves reciprocal translocation of chromosomes 8 and 14. The result of this is to place the c-myc protooncogene from 8q24 into the immunoglobulin heavy chain locus at 14q32. Because immunoglobulin genes are actively transcribed, this move alters the normal regulatory control of c-myc. Another example is the Philadelphia chromosome, a shortened chromosome 22 caused by translocation t(9:22)(q34:q11). This translocation is seen in almost all patients with chronic myelogenous leukemia (CML) and in a percentage of patients with acute lymphoblastic leukemia (ALL). The Philadelphia chromosome is seen with increased frequency in individuals with Down syndrome.

A couple decide to have prenatal diagnosis because their previous child has Tay-Sachs disease. Which of the following prenatal diagnostic techniques is optimal for fetal diagnosis? A. Chorionic villi sampling (CVS) B. Percutaneous umbilical blood sampling C. Amniotic fluid α-fetoprotein levels D. Maternal serum α-fetoprotein (MSAFP) E. Fetal x-rays

The answer is a. (Lewis, pp 397-416. Scriver, pp 3827-3876.) Most enzymes are expressed in chorionic villi or amniocytes and allow prenatal diagnosis of metabolic disorders through cell culture and enzyme assay. Percutaneous umbilical blood sampling (PUBS), or cordocentesis, offers another strategy if the enzyme is normally present in leukocytes. However, transabdominal aspiration of the umbilical cord is difficult and must be performed later in pregnancy (18+weeks) than CVS (8-10 weeks). α-fetoprotein (AFP) is not known to be involved in any metabolic disorders, but it is used as an index of fetal tissue differentiation and integrity. Amniotic or maternal serum α-fetoprotein (MSAFP) is most often used to detect, respectively, neural tube defects or chromosomal disorders and would not be useful in a case of normal fetal development with hexosaminidase A deficiency.

A male child presents with delayed development and scarring of his lips and hands. His parents have restrained him because he obsessively chews on his lips and fingers. Which of the following is likely to occur in this child? A. Increased levels of 5-phosphoribosyl-1-pyrophosphate (PRPP) B. Decreased purine synthesis C. Decreased levels of uric acid D. Increased levels of hypoxanthine-guanosine phosphoribosyl transferase (HGPRT) E. Glycogen storage

The answer is a. (Lewis, pp 75-94. Scriver, pp 2537-2570. Murray, pp 293-302.) The child has Lesch-Nyhan syndrome (308000), an X-linked recessive disorder that is caused by HGPRT enzyme deficiency. HGPRT is responsible for the salvage of purines from nucleotide degradation, and its deficiency elevates levels of PRPP, purine synthesis, and uric acid. PRPP is also elevated in glycogen storage diseases due to increased amounts of carbohydrate precursors.

401. Autosomal recessive conditions are correctly characterized by which of the following statements? A. They are often associated with deficient enzyme activity B. Both alleles contain the same mutation C. They are more variable than autosomal dominant conditions D. Most persons do not carry any abnormal recessive genes E. Affected individuals are likely to have affected offspring

The answer is a. (Murray, pp 812-828. Scriver, pp 3-45. Sack, pp 97-158. Wilson, pp 23-39.) Autosomal recessive conditions tend to have a horizontal pattern in the pedigree. Men and women are affected with equal frequency and severity. It is the pattern of inheritance most often seen in cases of deficient enzyme activity (inborn errors of metabolism). Autosomal recessive conditions tend to be more severe than dominant conditions and are less variable than dominant phenotypes. Both alleles are defective but do not necessarily contain the exact same mutation. All individuals carry 6 to 12 mutant recessive alleles. Fortunately, most matings involve persons who have mutations at different loci. Since related persons are more likely to inherit the same mutant gene, consanguinity increases the possibility of homozygous affected offspring.

As exemplified by HLA-DQβ haplotypes in type I diabetes mellitus, an individual's HLA status may be relevant to genetic counseling for certain multifactorial diseases. The relation of HLA haplotypes to disease and the use of this information in genetic counseling are referred to as which of the following? A. Genetic linkage and the frequency of recombination B. Allele association and risk modification C. Positional cloning and gene isolation D. Gene mapping and gene segregation E. Genotyping and phenotypic correlation

The answer is b. (Lewis, pp 135-154. Scriver, pp 193-202.) Individuals affected with autoimmune disorders such as juvenile diabetes mellitus, ankylosing spondylitis, or rheumatoid arthritis often have increased frequencies of particular HLA alleles, termed allele associations. Genetic linkage differs from allele association in that the linking of allele and phenotype depends on the family context; one family may exhibit segregation of the nail-patella phenotype with allele A of the ABO blood group, whereas another family exhibits segregation with allele O. Allele association or linkage disequilibrium implies that the same allele is always seen at higher frequency in affected individuals from different families (e.g., HLA-B27 in ankylosing spondylitis). Allele association implies neither a genotypephenotype relation between allele and disease nor a common chromosomal location for allele and disease. It may indicate a role for the allele in facilitating disease pathogenesis. In contrast, genetic linkage places a disease gene on the chromosome map, facilitating its isolation by positional cloning. Gene mutations in various individuals can then be characterized, allowing genotype-phenotype correlations. HLA testing for autoimmune disorders, like cholesterol testing for heart disease, exemplifies the use of risk factors to modify risks for multifactorial diseases.

The frequency of galactosemia is approximately 1 in 40,000 live births. The frequency of the carrier state can be calculated as which of the following? A. 1 in 50 live births B. 1 in 100 live births C. 1 in 200 live births D. 1 in 500 live births E. 1 in 1000 live births

The answer is b. (Lewis, pp 267-282. Scriver, pp 1553-1588. Murray, pp 102-110.) The Hardy-Weinberg expansion, p2 + 2pq + q2, describes the frequency of genotypes for allele frequencies p and q. In the case of rare disorders (q2 < 1/10,000), p approaches 1. The heterozygote frequency 2pq is thus approximately 2q. In this case, q2 < 1/40,000, q = 1/200 and 2q = 1/100. Since carriers are still quite rare compared with normal individuals, the matching of rare recessive alleles is greatly enhanced when there is common descent through consanguinity.

Increased resistance to malaria is seen in persons with hemoglobin AS, where A is the normal allele and S is the allele for sickle hemoglobin. Which of the following terms applies to this situation? A. Founder effect B. Heterozygote advantage C. Genetic lethal D. Fitness E. Natural selection

The answer is b. (Lewis, pp 267-282. Scriver, pp 3-45.) Sickle cell anemia (603903) is the classic example of a disorder with a high frequency in a specific population because of heterozygote advantage. Persons who are heterozygous for this mutant allele (hemoglobin AS) have increased resistance to malaria and are therefore at an advantage in areas where malaria is endemic. Founder effect is a special type of genetic drift. In these cases, the founder or original ancestor of a population has a certain mutant allele. Because of genetic isolation and inbreeding in populations such as the Pennsylvania Amish, certain disorders such as maple syrup urine disease (248600) are maintained at a relatively high frequency. Fitness is a measure of the ability to reproduce. A genetic lethal implies that affected individuals cannot reproduce and, therefore, cannot pass on their mutant alleles. Natural selection is a theory introduced by Charles Darwin, which postulates that the fittest individuals have a selective advantage for survival.

Studies of the eye tumor retinoblastoma have revealed an Rb locus on the long arm of chromosome 13 that influences retinoblastoma occurrence. Patients with 13q−deletions often develop bilateral tumors (both sides), in contrast to more common forms of retinoblastoma that occur at one site. Which of the following phrases best explains this phenomenon? A. Rb is an oncogene B. Rb is a tumor suppressor gene C. Rb mutations ablate a promoter sequence D. Rb mutations ablate an enhancer sequence E. Rb mutations must always involve chromosome abnormalities

The answer is b. (Lewis, pp 355-376. Scriver, pp 521-524. Murray, pp 314-340.) The two-hit hypothesis was developed by Knudsen to explain why patients with hereditary retinoblastoma [germ-line mutations (180200)] have multiple, bilateral tumors while those with sporadic tumors (no family history) have single tumors. A germ-line mutation (first hit) alters one Rb allele and confers enhanced susceptibility to retinoblastoma. A somatic mutation (second hit) inactivating the other homologous Rb allele can then occur in any tissue. If it occurs in the retina, a tumor is born. The multiple tumors thus represent the sites at which somatic mutations have occurred in the retina. In sporadic cases, two somatic mutational events must take place. Since these somatic mutations are relatively rare events, it is extremely uncommon for more than one tumor to develop. It is curious that, although retinoblastoma susceptibility is inherited in a dominant fashion, tumor development is a recessive event, requiring the inactivation of both alleles. Genes such as Rb are called tumor suppressor genes, in contrast to oncogenes, in which only one of the two homologous alleles must be altered to initiate malignant transformation. Alteration of an enhancer or promoter site on one Rb allele would thus not be sufficient to cause cancer, since the other Rb allele would not be affected. Obvious chromosome changes such as 13q−are rare compared to other mutations that alter Rb function.

A couple request genetic counseling because the wife has contracted early-onset breast cancer at age 23. The husband has a benign family history, but the wife has several relatives who developed cancers at relatively early ages. Affected relatives include a sister (colon cancer, age 42), a brother (colon cancer, age 46), mother (breast cancer, age 56), maternal aunt (leukemia, age 45), maternal uncle (muscle sarcoma, age 49), and a nephew through the brother with colon cancer (leukemia, age 8). Which of the following is the correct conclusion from the family history? A. No genetic predisposition to cancer since most individuals have different types of cancer B. Possible autosomal dominant inheritance or multifactorial inheritance of cancer predisposition C. Germ-line mutations in an oncogene, with somatic mutations that suppress the oncogene D. Germ-line mutations in a tumor suppressor gene, with neoplasia from chemical exposure E. Mitochondrial inheritance of tumor predisposition evidenced by the affected maternal relatives

The answer is b. (Lewis, pp 355-376. Scriver, pp 521-552. Murray, pp 396-414.) Genetic predisposition to cancer is best understood by the Knudsen hypothesis, where two independent mutations or "hits" are required to produce neoplasia of a somatic tissue. In many hereditary cancers, the "first hit" is a germ-line mutation that is transmitted in families. Individuals who inherit this mutation are much more likely to develop cancer through a "second hit" in their somatic cells. The second hit can be any mutation that removes the homologous allele (loss of heterozygosity); mechanisms include missense mutation, chromosome deletion, and chromosome nondisjunction. For tumor suppressor genes like those responsible for neurofibromatosis 1 (162200) or the Li-Fraumeni syndrome (114480), the first hit removes one suppressor allele and the second hit removes the homologous suppressor allele. The family in the question is an example of a "cancer family" that exhibits the bone, breast, colon, and blood cancers that are typical of Li-Fraumeni syndrome. The mechanism involves mutations in the src tumor suppressor gene.

In the operating room, a child receives succinylcholine as a muscle relaxant to facilitate intubation and anesthesia. The operation proceeds until it is time for recovery, when the child does not begin breathing. A hurried discussion with the father discloses no additional problems in the family, but he does say that he and his wife are first cousins. Which of the following is the most likely possibility? A. An autosomal dominant disorder that interferes with succinylcholine metabolism B. An autosomal recessive disorder that interferes with succinylcholine metabolism C. An X-linked disorder that interferes with succinylcholine metabolism D. A lethal gene transmitted through consanguinity that affects the respiratory system E. Mismanagement of halothane anesthesia during the operation

The answer is b. (Lewis, pp 361-391. Murray, pp 396-414. Scriver, pp 233-238.) Succinylcholine is metabolized by a plasma enzyme formerly called pseudocholinesterase [now called butyrylcholinesterase (BChE) to designate its favored substrate]. Approximately 1 in 100 individuals are homozygous for a variant of BChE that has 60% activity, whereas 1 in 150,000 individuals are homozygous for a variant with 33% activity. The latter group exhibits prolonged recovery from succinylcholine-induced anesthesia, a phenotype known as succinylcholine apnea (177400). As with most enzyme defects, succinylcholine apnea exhibits autosomal recessive inheritance. The parents will be heterozygous for a BChE variant but have not undergone anesthesia to display the phenotype.

The diagnosis of osteogenesis imperfecta (166200) is most accurately performed by which of the following? A. PCR amplification and DNA sequencing of type I collagen gene segments to look for point mutations B. Gel electrophoresis of labeled type I collagen chains synthesized in fibroblasts C. PCR amplification and ASO hybridization to detect particular mutant alleles D. Northern blotting to evaluate type I collagen mRNAs E. Purification and trypsin digestion of type I collagen chains to visualize altered peptides after two-dimensional gel electrophoresis

The answer is b. (Lewis, pp 377-396. Scriver, pp 5241-5286. Murray, pp 396-414.) The spectrum of mutations in collagen (and fibrillin) disorders is very broad, making it more efficient to evaluate electrophoretic mobility of their polypeptide chains as a clue to structural abnormality. Almost every patient with osteogenesis imperfecta (and other collagen disorders) has a different type of mutation. PCR amplification followed by allele-specific oligonucleotide (ASO) hybridation to detect specific alleles is thus impractical—hundreds of PCR/ASO reactions would be required to screen for all of the possible mutant alleles. Similarly, DNA sequencing would be extraordinarily time-consuming and give many false positives due to nucleotide polymorphisms or silent mutations that do not cause structural abnormalities in the polypeptide. Northern blotting would detect mutations that affect RNA processing and generate RNAs of altered size, but these are a small fraction of possible collagen mutations. As DNA chip technology becomes practical, screening for thousands of mutations at a locus may be possible. DNA chips contain thousands of different oligonucleotides embedded on a solid matrix. Hybridization of colored or labeled gene fragments with randomized sequences from that gene on a chip gives signals corresponding to the gene sequences that are present. The chip can then be washed and used again. Hybridization of the chip with suitably digested DNAs from patients and controls can thus detect any variant gene fragment (complementary oligonucleotide). Use of numerous control DNAs would separate true mutant alleles (sequence variants associated with disease) from polymorphisms or silent mutations.

A woman who is at risk to be a carrier of hemophilia A desires prenatal diagnosis. She does not want her extended family to know about her pregnancy if the fetus is affected. Which of the following prenatal diagnostic techniques should be advised? A. Amniocentesis with western blot analysis of factor VIII B. Chorionic villus sampling with DNA analysis for factor VIII mutations C. Percutaneous umbilical blood sampling with testing of factor VIII levels D. Amniocentesis with DNA analysis for factor VIII mutations E. Chorionic villus sampling with assay of factor VIII activity

The answer is b. (Lewis, pp 75-94. Scriver, pp 4367-4392. Murray, pp 598-608.) Chorionic villus sampling (CVS) is performed at 8-10 weeks' gestation, before a woman is obviously pregnant. This technique preserves the confidentiality of prenatal decisions because diagnostic results are available by 11-12 weeks' gestation rather than the 18-20 weeks for standard amniocentesis. DNA analysis must be employed because factor VIII is not expressed in chorion or amniotic cells. Percutaneous umbilical blood sampling (PUBS) must be performed later in gestation (18+ weeks). Because some factor VIII gene mutations may give normal amounts of structurally abnormal factor VIII, activity rather than amounts of factor VIII protein must be measured for diagnosis.

Little People of America (LPA) is a support group for individuals with short stature that conducts many workshops and social activities. Two individuals with achondroplasia (100800), a common form of dwarfism, meet at an LPA convention and decide to marry and have children. What is their risk of having a child with dwarfism? A. 100% B. 75% C. 50% D. 25% E. Virtually 0

The answer is b. (Lewis, pp 75-94. Scriver, pp 5379-5398.) The genotype of each dwarf can be represented as Aa, with the uppercase A representing the achondroplasia allele. The Punnett square below demonstrates that 3/4 of the possible gamete combinations yield individuals with at least one A allele. Homozygous AA achondroplasia is a severe disease that is usually lethal in the newborn period. The increased likelihood of individuals with achondroplasia marrying each other because of their similar phenotypes is an example of assortative mating.

395. A female with Turner's syndrome is denoted by which of the following cytogenetic notations? A. 47,XX,+21 B. 45,X C. 47,XXX D. 46,XX,t(14;21) E. 45,XX,−21

The answer is b. (Murray, pp 812-828. Scriver, pp 3-45. Sack, pp 57-84. Wilson, pp 123-148.) Cytogenetic notation provides the chromosome number (e.g., 46), the sex chromosomes, and a shorthand description of anomalies. Examples include the following: 45,X indicates a female with monosomy X or Turner's syndrome; 47,XX+21 indicates a female with trisomy 21 or Down's syndrome; 46,XX,t(14;21) indicates a female with translocation Down's syndrome; 45,XX−21 indicates a female with monosomy 21. Note the absence of spaces between symbols, and the use of 47,XXX for sex chromosomal aneuploidy ("triple X" syndrome) rather than the more awkward 47,XX+X. (Note also that 45,X is sufficient for X chromosome monosomy, since absence of an X is indicated by the convention of listing sex chromosomes). Translocations that join two chromosomes with minuscule short arms (acrocentric chromosomes—13, 14, 15, 21, and 22) are called Robertsonian translocations. The joined acrocentric chromosomes in a Robertsonian translocation have a single centromere between them and are counted as one chromosome. A normal person who "carries" a Robertsonian translocation therefore has a chromosome number of 45, as in 45,XX,t(14;21). This female has a 5 to 20% risk of transmitting the Robertsonian 14:21 translocation to her offspring and having a child with Down's syndrome—e.g., 46,XX,t(14;21).

Neural tube defects, such as spina bifida and anencephaly, are best diagnosed by which of the following laboratory tests? A. Chorionic villus biopsy and karyotype at 10 weeks after the last menstrual period (LMP) B. Maternal serum α-fetoprotein (MSAFP) levels and ultrasound at 16 weeks after conception C. Amniotic fluid α-fetoprotein (AFP) levels and ultrasound at 16 weeks after the LMP D. Amniotic fluid acetylcholinesterase levels at 16 weeks after conception E. Amniotic fluid karyotype and ultrasound at 16 weeks after the LMP

The answer is c. (Lewis, pp 135-154, 397-416. Scriver, pp 193-202.) Any defect of the fetal skin may elevate the amniotic α-fetoprotein (AFP) level, causing a parallel rise of this substance in the maternal blood. Neural tube defects such as anencephaly or spina bifida elevate the AFP in amniotic fluid or maternal serum; other causes of increased AFP include fetal kidney disease with leakage of fetal proteins into amniotic fluid. Mild forms of spina bifida or meningomyelocele may be covered by the skin, so that the AFP is not elevated, and maternal serum AFP is less sensitive than amniotic fluid AFP for such cases. Ultrasound is required to detect covered neural tube defects that do not leak fetal AFP into the amniotic fluid and maternal blood. Acetylcholinesterase is an enzyme produced at high levels in neural tissue that is somewhat more specific than AFP for neural tube defects; it is used for confirmation rather than as a primary prenatal test. Chorionic villus biopsy is performed at about 10 weeks after the last menstrual period (LMP) and amniocentesis at 14-16 postmenstrual weeks. Because conception often occurs 2 weeks prior to the LMP, distinction between postconceptional and postmenstrual timing is important for early stages of pregnancy. Neural tube defects are usually localized, multifactorial anomalies rather than part of a malformation syndrome that can result from chromosomal aberrations. For this reason, documentation of the fetal karyotype by chorionic villus biopsy or amniocentesis does not influence the risk for neural tube defects.

A standard karyotypic analysis is ordered for a girl with heart defects, developmental delay, and an unusual appearance. The results are normal, but a colleague recommends performing fluorescent in situ hybridization (FISH) analysis on the patient's chromosomes, using probes for chromosome 22. Only one signal is seen for each chromosomal spread. Which of the following statements regarding these analyses is true? A. The initial karyotype results are inconsistent with the FISH results B. This is a normal result C. A small deletion is present on one of the patient's number 22 chromosomes D. FISH is only helpful when the initial karyotype results are abnormal E. The chromosome with the positive signal is paternal in origin

The answer is c. (Lewis, pp 241-266. Scriver, pp 3-45.) Fluorescent in situ hybridization (FISH) analysis is a technique in which molecular probes that are specific for individual chromosomes or chromosomal regions are used to identify these regions. FISH probes frequently identify chromosomal regions that are submicroscopic and therefore may be useful when standard karyotypic analysis is normal. In this case, the fact that only one signal is present, despite the fact that there are two number 22 chromosomes, indicates that a submicroscopic deletion has occurred. The parental chromosome of origin cannot be determined using this technique unless that parent also carries a similar deletion and his or her chromosomes are evaluated. Submicroscopic deletion at band 22q11 causes a spectrum of disorders ranging from DiGeorge anomaly to Shprintzen syndrome (192430).

Screening of an African American population in Minnesota yields allele frequencies of 7/8 for the A globin allele and 1 /8 for the sickle globin allele. A companion survey of 6400 of these people's ancestors in central Africa reveals 4600 individuals with genotype AA, 1600 with genotype AS (sickle trait), and 200 with genotype SS (sickle cell disease—603903). Compared to their descendants in Minnesota, the African population has which of the following? A. A lower frequency of AS genotypes consistent with inbreeding B. A lower frequency of AS genotypes consistent with malarial exposure C. A higher frequency of AS genotypes consistent with heterozygote advantage D. A higher frequency of AS genotypes consistent with selection against the S allele E. Identical A and S allele frequencies as predicted by the Hardy-Weinberg law

The answer is c. (Lewis, pp 267-282. Scriver, pp 4571-4636.) Under certain conditions, the Hardy-Weinberg law allows one to interconvert genotype and allele frequencies in a population by using the formula (p + q)2 = p2 + 2pq + q2. For a locus with two alleles, p represents the frequency of the more common allele, q of the less common allele, and p + q = 1. The Minnesota population therefore has p2 = 7/8 × 7/8 = 49/64 (4900 individuals) with the AA genotype, 2pq = 2 × 7/8 × 1/8 = 14/64 (1400 individuals) with sickle trait (AS genotype), and q2 = 1/8 × 1/8 = 1/64 (100 individuals) with sickle cell disease (SS genotype). The African population has a higher frequency of AS and SS genotypes caused by heterozygote advantage for the AS genotype that confers resistance to malaria.

Many family studies employing DNA have the potential to demonstrate nonpaternity. If the physician ordering these analyses does not discuss this possibility with the couples involved, he or she is in violation of which of the following? A. Patient confidentiality B. Patient rights C. Informed consent D. Standards of care E. Malpractice guidelines

The answer is c. (Lewis, pp 377-396. Scriver, pp 3-45. Murray, pp 396-414.) Informed consent requires that the patient be informed of all adverse effects that might result from a procedure. Evidence for nonpaternity may result from various types of DNA analysis and should be discussed with the concerned parties at the time of blood collection. Some physicians speak to the mother and father separately about this issue to maximize the opportunity for independent decision making.

A patient with the Marfan syndrome (154700) is evaluated at a clinic. He is noted to have a tall, thin body habitus, loose joints, and arachnodactyly (spider fingers). Ophthalmologic examination reveals lens dislocation. Echocardiogram reveals dilation of the aortic root. A family history reveals that the patient's parents are medically normal, but that his paternal grandfather and great-grandfather died in their forties with lens dislocation and dissecting aortic aneurysms. A sister is found to have a similar body habitus, dilation of the aortic root, and normal lenses. The different findings in these different family members with the same disease are best described by which of the following terms? A. Pleiotropy B. Founder effect C. Variable expressivity D. Incomplete penetrance E. Genetic heterogeneity

The answer is c. (Lewis, pp 377-396. Scriver, pp 5287-5312.) Although a mutation at a single locus generally alters a single gene, the result being the abnormal synthesis or lack of production of a single RNA molecule or polypeptide chain, the results of this mutation may be farreaching. When there are multiple phenotypic effects involving multiple systems, the result is referred to as pleiotropy. Penetrance is the all-or-none expression of an abnormal genotype, whereas expressivity is the degree of expression of that genotype. Incomplete or reduced penetrance implies that some individuals have a mutant allele with absolutely no phenotypic expression of that allele. Variable expressivity implies that all individuals with a mutant allele have some phenotypic effects, although the severity and range of effects differ in different people. Marfan syndrome (154700) exhibits pleiotropy of its single-gene mutation by causing lens dislocation, loose connective tissue (joint laxity, tall stature, sternal and vertebral deformities), and fragile aortic tissue that can lead to aortic valve insufficiency or aortic dissection. Individuals with the disease exhibit variable combinations and severity of these symptoms due to variable expressivity of this single gene.

Which of the following is most likely in an untreated child with PKU? A. Elevated tyrosine B. Increased skin pigmentation C. Decreased skin pigmentation D. Normal phenylalanine hydroxylase levels E. Elevated alanine

The answer is c. (Lewis, pp 397-416. Scriver, pp 1667-1724. Murray, pp 249-263.) Decreased melanin can occur in PKU because melanin is produced from phenylalanine and tyrosine. The defect in most children with PKU is deficiency of phenylalanine hydroxylase. Rare children have deficiency of biopterin cofactor due to a defect in its synthetic enzyme that is also autosomal recessive. Phenylalanine is converted to tyrosine by phenylalanine hydroxylase, so deficient tyrosine can occur in children on restrictive diets.

A patient presents to the physician's office to ask questions about color blindness. The patient is color-blind, as is one of his brothers. His maternal grandfather was color-blind, but his mother, father, daughter, and another brother are not. His daughter is now pregnant. What is the risk that her child will be color-blind? A. 100% B. 50% C. 25% D. 12.5% E. Virtually 0

The answer is c. (Lewis, pp 75-94. Scriver, pp 3-45.) Males always transmit their single X chromosome to their daughters. Therefore, a daughter of a male affected with an X-linked disorder is an obligate carrier for that disorder. When the condition is X-linked recessive, as with most forms of color-blindness, the daughter is unlikely to show any phenotypic evidence that she is carrying this abnormal gene. Offspring of female carriers are of four types: (1) female carrier with one normal and one mutant allele, (2) normal female with two normal alleles, (3) affected male with a single mutant allele, and (4) normal male with a single normal allele. The chance of having an affected child is thus one-fourth or 25%. If the obligate carrier female gives birth to a son, the chance of the son being color-blind is 50%.

Polycystic kidney disease (173900) is a significant cause of renal failure that presents from early infancy to adulthood. Early-onset cases tend to affect one family member or siblings, whereas adult-onset cases often show a vertical pattern in the pedigree. Which of the following offers the best explanation of these facts? A. Pleiotropy B. Allelic heterogeneity C. Locus heterogeneity D. Multifactorial inheritance E. Variable expressivity

The answer is c. (Lewis, pp 95-112. Scriver, pp 5467-5492.) Polycystic kidney disease (173900) occurs in two distinctive genetic forms—adult-onset and infantile. Infantile disease is autosomal recessive, whereas adult-onset disease is autosomal dominant. Confusion between these types can occur due to variable expressivity in the adult, dominant form. Occasional onset in young children may occur in adult-type disease. Consistency of early onset, the presence of consanguinity, and the lack of vertical transmission distinguish the infantile, recessive form. Polycystic kidney disease is an example of genetic heterogeneity, in which different mutations may cause similar phenotypes. This may be further divided into allelic and nonallelic (locus) heterogeneity. Allelic heterogeneity implies that there are different mutations at the same locus that both result in similar disease [i.e., the many fibrillin mutations in Marfan syndrome (154700)]. In locus heterogeneity, mutations occur at different loci, yet the phenotype is similar. Locus heterogeneity also explains why certain disorders, such as polycystic kidney disease, Charcot-Marie-Tooth disease, sensorineural hearing loss, and retinitis pigmentosa may be inherited in several different fashions. A general rule predicts that the autosomal recessive forms of these diseases will be more severe, the autosomal dominant forms less so. It is especially important to recognize the possibility of genetic heterogeneity when counseling patients in regard to recurrence risks.

Hyperuricemia in Lesch-Nyhan syndrome is due to a defect in which of the following pathways? A. Purine biosynthesis B. Pyrimidine biosynthesis C. Purine salvage D. Pyrimidine salvage E. Urea cycle

The answer is c. (Murray, pp 331-342. Scriver, pp 2537-2570.) Uric acid is a purine derivative, increased by purine salvage reactions that convert purines, purine ribonucleosides, and purine deoxyribonucleoside to mononucleotides (incorrect answer d). Such salvage reactions require much less energy than de novo synthesis (incorrect answers a, b). The liver is the major site of purine nucleotide biosynthesis and provides excess purines for other tissues that cannot synthesize purines. A defect in hypoxanthine-guanine phosphoribosyl transferase, one of the enzymes of purine salvage, is responsible for purine overproduction and subsequent hyperuricemia observed in Lesch-Nyhan syndrome. Carbamoyl phosphate involved in pyrimidine synthesis is also an intermediate of the urea cycle, but different enzymes convert this substrate in the two pathways (incorrect answer e).

The standard karyotype is performed by photomicroscopy of cells at which mitotic stage? A. Interphase B. Prophase C. Metaphase D. Anaphase E. Telophase

The answer is c. (Murray, pp 812-828. Scriver, pp 3-45. Sack, pp 57-84. Wilson, pp 123-148.) The standard karyotype is an arrangement of chromosomes from one cell that is undergoing division at metaphase. At other mitotic stages, the chromosomes are not sufficiently condensed or are too dispersed to allow counting and comparison of pairs under the microscope. After growth, metaphase arrest, separation, hypotonic treatment, and fixing of white blood cells, smearing on a slide yields only about 3% cells that can be analyzed (metaphase spreads). In high-resolution chromosome analysis, less condensed chromosomes in late prophase may be analyzed (prometaphase analysis); however, this process is extremely timeconsuming and usually requires focus on a particular chromosome region (e.g., chromosome 15 in a patient suspected of Prader-Willi syndrome, a condition marked by obesity and mental retardation).

A 45-year-old male is hospitalized for treatment of myocardial infarction. His father and a paternal uncle also had heart attacks at an early age. His cholesterol is elevated, and lipoprotein electrophoresis demonstrates an abnormally high ratio of low- to high-density lipoproteins (LDL to HDL). Which of the following is the most likely explanation for this problem? A. Mutant HDL is not responding to high cholesterol levels B. Mutant LDL is not responding to high cholesterol levels C. Mutant caveolae proteins are not responding to high cholesterol levels D. Mutant LDL receptors are deficient in cholesterol uptake E. Intracellular cholesterol is increasing the number of LDL receptors

The answer is d. (Lewis, pp 135-154. Scriver, pp 2863-2914. Murray, pp 219-230.) This man has familial hypercholesterolemia (143890), an autosomal dominant phenotype defined by studying men who experienced heart attacks at young ages. Mutations in the LDL receptor lead to decreased cellular cholesterol uptake and increased serum cholesterol. Since LDL has a high cholesterol content, the LDL fraction is elevated compared to the HDL fraction on lipoprotein electrophoresis. In normal individuals, the LDL is taken up by its specific receptor and imported via caveolae to the cell interior. Cholesterol then produces feedback inhibition on the rate-limiting enzyme of cholesterol synthesis (hydroxymethylglutaryl CoA reductase) and also leads to a decrease in the number of LDL receptors. In rare cases, two individuals with familial hypercholesterolemia marry and produce a child with homozygous familial hypercholesterolemia. These children develop severe atherosclerosis and xanthomas (fatty tumors) at an early age.

A child is referred for evaluation because of low muscle tone and developmental delay. Shortly after delivery the child was a poor feeder and had to be fed by tube. In the second year, the child began to eat voraciously and became obese. He has a slightly unusual face with almond-shaped eyes and downturned corners of the mouth. The hands, feet, and penis are small, and the scrotum is poorly formed. The diagnostic category and laboratory test to be considered for this child are which of the following? A. Sequence, serum testosterone B. Single birth defect, serum testosterone C. Deformation, karyotype D. Syndrome, karyotype E. Disruption, karyotype

The answer is d. (Lewis, pp 241-266. Scriver, pp 3-45.) This child has several minor anomalies, a major anomaly that affects the genitalia, and developmental delay. These multiply affected and embryologically unrelated body regions suggest a syndrome rather than a sequence. Because of the multiple anomalies and developmental delay, the first diagnostic test to be considered is a karyotype rather than a test for specific organ function, such as serum testosterone.

Availability of DNA testing for many single disease traits has allowed routine prenatal screening of couples for disorders prevalent in their ethnic group. Which of the following genetic disorders has a similar incidence in different ethnic groups and would not be subject to different criteria for screening? A. Cystic fibrosis B. Thalassemias C. Tay-Sachs disease D. Down syndrome E. Sickle cell anemia

The answer is d. (Lewis, pp 267-282. Scriver, pp 3-45.) Down syndrome, a chromosomal disorder, has virtually the same frequency of 1 in 600 births in all ethnic groups. Screening for Down syndrome is carried out during pregnancy with triple test/quad screens offered to all individuals (with very poor sensitivity or specificity) and chorionic villus sampling/ amniocentesis (with high accuracy but some miscarriage risk) to woman above age 35. Allele frequencies may differ among populations when there has been geographic isolation, founder effect, or selection for certain alleles based on different environments. Although African Americans have intermixed with whites in the United States for over 400 years, they retain a higher frequency of sickle cell alleles (603903), which are thought to protect individuals from malarial infection. Each ethnic group has frequencies of polymorphic alleles that reflect its origin; for example, Ashkenazi Jews have a higher frequency of Tay-Sachs alleles (272800); Greeks and other Mediterranean peoples of β-thalassemia alleles (603902); Asian peoples of α-thalassemia alleles (141800); and whites of cystic fibrosis alleles (219700). Specific genetic differences (polymorphisms) in the mitochondrial and Y chromosomes have allowed reconstruction of migrations that show correlations between genetic homogeneity and language.

The incidence of a genetic form of diabetes insipidus (304800) in North Americans was hypothesized to be related to immigration of affected individuals on the ship Hopewell that arrived in Halifax Nova Scotia several hundred years ago. If the disease allele were known as A, and residents near Halifax had 10 times the frequency of this allele as did those on mainland Canada, which of the following terms best describes this phenomenon? A. Selection for allele A B. Linkage disequilibrium with allele A C. Linkage to allele A D. Founder effect for allele A E. Assortative mating for allele A

The answer is d. (Lewis, pp 267-282. Scriver, pp 3-45.) Founder effects represent a special case of genetic drift in which rare alleles are introduced into a small population by the migration of ancestors. These founder mutant alleles can overcome selective disadvantage because they begin with high frequency in a small gene pool. Linkage disequilibrium describes an association between a particular polymorphic allele and a trait. Many autoimmune diseases exhibit association with particular human leukocyte antigen (HLA) alleles (i.e., HLA-B27 and ankylosing spondylitis). The association is not necessarily cause and effect (e.g., when viral infections that trigger a disease preferentially infect certain HLA genotypes). Genetic linkage implies physical proximity of the allele locus to the gene causing the disease. Linkage differs from allele association in that either allele A or a may be linked in a given family, depending on which allele is present together with the offending gene. Neither assortative mating (preferential mating by genotype) nor selection (advantageous alleles) applies to the situation under discussion. Later research using DNA polymorphisms has refuted the Hopewell hypothesis, showing that families with this form of diabetes insipidus (304800) are not related to ancestors from Halifax.

A man whose brother has cystic fibrosis (219700) wants to know his risk of having an affected child. The prevalence of cystic fibrosis is 1 in 1600 individuals. Which of the following is the risk in this case? A. 1 /8 B. 1 /16 C. 1 /60 D. 1 /120 E. 1 /256

The answer is d. (Lewis, pp 267-282. Scriver, pp 3-45.)According to the Hardy-Weinberg equilibrium, the frequency of heterozygotes (2pq) is twice the square root of the rare homozygote frequency (q2). The man in the question has a two-thirds chance of being a carrier. His wife has a onetwentieth chance of being a carrier. The rare homozygote frequency is 1 /600, so the square root is 1 /40 and the frequency of heterozygotes is thus 2 × 1 /40 = 1 /20. His risk for an affected child is 2/3 × 1 /20 × 1 /4 = 1 /120.

A normal 6-year-old girl has a strong family history of cancer, including several relatives with Li-Fraumeni syndrome, an autosomal dominant condition that predisposes to breast and colon cancer. Her parents request that she have genetic testing for a possible cancer gene. Which of the following is the major ethical concern about such testing? A. Nonmaleficence B. Beneficence C. Autonomy D. Informed consent E. Confidentiality

The answer is d. (Lewis, pp 355-376. Scriver, pp 521-552. Murray, pp 396-414.) Presymptomatic DNA testing of individuals in cancer families is increasingly available. However, testing of minors is controversial because they may not be old or mature enough to understand the personal, medical, and financial implications. They therefore cannot give truly informed consent. Beneficence is the ethical imperative to do good for patients, while nonmaleficence is the imperative to do no harm. Autonomy refers to a patient's right to make decisions regarding his or her health care, and confidentiality to the privilege of doctor-patient communication.

A 3-year-old girl is scheduled for a tonsillectomy. As she is prepared for the operating room, her father becomes agitated and insists on accompanying her. He says that he lost a son several years ago when the child did not wake up after an operation. Which of the following options is the best response to the father's anxiety? A. Postpone the operation until the psychiatric state of the father can be evaluated B. Proceed after explaining that problems in the father's siblings are unlikely to be transmitted to his daughter C. Proceed after reassuring the father that drug reactions are environmental and unlikely to have a genetic basis D. Postpone the operation until a more detailed family history is obtained E. Proceed after explaining that modern anesthetic procedures are much safer than in the past

The answer is d. (Lewis, pp 397-416. Scriver, pp 225-258. Murray, pp 396-414.) A family history is an important precedent for anesthesia, and awareness of individual differences is important when administering any drug. Pharmacogenetics is the area of study that examines genetic influences on drug metabolism. The extensive human genetic variation revealed by DNA analysis has important implications for pharmacology, since drug effects often vary according to each patient's unique genome.

Which of the following would make hyperuricemia very unlikely in a patient? A. Lesch-Nyhan syndrome B. Gout C. Xanthine oxidase hyperactivity D. Carbamoyl phosphate synthase deficiency E. Purine overproduction secondary to von Gierke disease

The answer is d. (Murray, pp 271-280, 331-342. Scriver, pp 2513-2570.) Carbamoyl phosphate (CAP) synthase I is found in mitochondrial matrix and is the first step in urea synthesis, condensing CO2 and NH4 +. Hyperammonemia occurs when CAP is deficient. CAP synthase II forms CAP as the first step in pyrimidine synthesis. Its complete deficiency would probably be a lethal mutation. When its activity is decreased, purine catabolism to uric acid is decreased, decreasing the possibility of hyperuricemia. In contrast, gout, Lesch-Nyhan syndrome, high xanthine oxidase activity, and von Gierke disease (glycogen storage disease type la [MIM*232200]) all lead to increased urate production and excretion.

A 10-year-old boy is referred to the physician because of learning problems and a lack of motivation in school. His family history is unremarkable. Physical examination is normal except for single palmar creases of the hands and curved fifth fingers (clinodactyly). The physician decides to order a karyotype. Which of the following indications for obtaining a karyotype would best explain the physician's decision in this case? A. A couple with multiple miscarriages, or a person who is at risk for an inherited chromosome rearrangement B. A child with ambiguous genitalia who needs genetic sex assignment C. A child with an appearance suggestive of Down's syndrome or other chromosomal disorder D. A child with mental retardation and/or multiple congenital anomalies E. A child who is at risk for cancer

The answer is d. (Murray, pp 812-828. Scriver, pp 3-45. Sack, pp 57-84. Wilson, pp 123-148.) The hallmarks of children with chromosomal anomalies are mental retardation and multiple congenital anomalies. In this case, the individual has learning problems that have not yet been recognized as mental retardation, and he has minor anomalies rather than major birth defects that cause cosmetic or surgical problems. The physician was astute to suspect a chromosomal anomaly even when the developmental disability and alterations in physical appearance were subtle. Other indications for a karyotype include a couple with multiple miscarriages, an individual at risk for inheriting or transmitting a chromosomal rearrangement, a child with ambiguous external genitalia, or an individual with characteristics of a chromosomal syndrome such as Down's, Turner's, or Klinefelter's syndrome. Chromosome translocations are characteristic of many types of cancer, but these occur in somatic cancer cells rather than in the patient's germ line.

404. A 4-year-old boy presents to the physician's office with coarse facies, short stature, stiffening of the joints, and mental retardation. Both parents, a 10-year-old sister, and an 8-year-old brother all appear unaffected. The patient's mother is pregnant. She had a brother who died at 15 years of age with similar findings that seemed to worsen with age. She also has a nephew (her sister's son) who exhibits similar features. Based on the probable mode of inheritance, the risk that her fetus is affected is A. 100% B. 67% C. 50% D. 25% E. Virtually 0

The answer is d. (Murray, pp 812-828. Scriver, pp 3-45. Sack, pp 97-158. Wilson, pp 23-39.) The fact that the mother of the affected child has an affected brother and an affected nephew through her sister suggests X-linked recessive inheritance. This is made more likely because the symptoms suggest a mucopolysaccharidosis (storage of glycosaminoglycans) and because one type exhibits X-linked recessive inheritance [Hunter's syndrome or MPS type II (309900)]. When evaluating the possibility of an X-linked disorder, it is important to remember the pattern of inheritance of the X chromosome. Females have two X chromosomes, which are passed along in a random fashion. They pass any given X chromosome to 50% of their sons and 50% of their daughters. For an X-linked recessive condition, those daughters who inherit the affected allele are heterozygous carriers of the disorder but are not affected. Since males have only one X chromosome, those who inherit the affected allele are affected with the disorder. Given X-linked recessive inheritance, the mother must have the abnormal allele on one of her X chromosomes (she is an obligate carrier) in order for her son and brother to be affected. The fetus thus has a 1/2 chance of being a boy and a 1/2 chance of being affected given male sex, resulting in a 1/4 (25%) overall risk of being affected.

Isolated cleft lip and palate is a multifactorial trait. The recurrence risk of isolated cleft lip and palate is which of the following? A. The same in all families B. Not dependent on the number of affected family members C. The same in all ethnic groups D. The same in males and females E. Affected by the severity of the cleft

The answer is e. (Lewis, pp 135-154. Scriver, pp 193-202.) Cleft lip with or without cleft palate [CL(P)] is one of the most common congenital malformations. Because of the genetic component of this trait, it tends to be more common in certain families. The more family members affected and the more severe the cleft, the higher the recurrence risk. In addition, CL(P) is more common in males and in certain ethnic groups (i.e., Asians > whites > African Americans).

Prader-Willi syndrome involves a voracious appetite, obesity, short stature, hypogonadism, and mental disability. At least 50% of Prader-Willi patients have a small deletion on the proximal long arm of chromosome 15. In detecting the Prader-Willi deletion, which of the following techniques would be most accurate? A. Standard karyotyping of peripheral blood leukocytes B. Northern blotting of mRNAs transcribed from the deletion region C. Restriction analysis to detect DNA fragments from the deletion region D. Rapid karyotyping of bone marrow E. Fluorescent in situ hybridization (FISH) analysis of peripheral blood lymphocytes using fluorescent DNA probes from the deleted region

The answer is e. (Lewis, pp 241-266. Scriver, pp 3-45.) The PraderWilli deletion is quite small and is not usually detected by standard metaphase karyotyping. Fluorescent in situ hybridization (FISH) is the most efficient and accurate method for detecting the deletion in Prader-Willi syndrome. Fluorescent DNA probes from the deletion region (chromosome band 15q11) give two signals in normal subjects and one signal in patients with a deletion. Detection of RNA or DNA fragments from this region would require quantitation to reveal one-half normal amounts, since genes on the homologous 15 chromosome would be normal. It is much easier to visualize one versus two fluorescent signals. Standard karyotypes typically display about 300 bands over the 23 chromosomes or about 10 bands on chromosome 10. This is adequate for detecting aneuploidy but inadequate for small deletions seen in conditions like Prader-Willi syndrome. Rapid karyotyping of bone marrow samples is possible because marrow contains actively dividing cells. Results are available in 2 to 3 h rather than the 2 to 3 days for standard karyotyping because peripheral blood T leukocytes must be stimulated to divide using lectins like phytohemagglutinin. Resolution of bone marrow karyotypes is usually even less than for standard karyotypes from blood, necessitating the use of FISH probes for accurate diagnosis. Newborns suspected of one of the common trisomies can have bone marrow karyotypes with FISH using probes from chromosomes 13, 18, and 21. Diagnosis is thus available in several hours, allowing guidance of management decisions.

Every prenatal evaluation should include which of the following diagnostic procedures? A. Level I ultrasound B. Chorionic villus sampling (CVS) C. Doppler analysis D. Amniocentesis E. Genetic counseling

The answer is e. (Lewis, pp 397-416. Scriver, pp 3-45.) Genetic counseling is an essential component of every prenatal diagnostic test. Couples must understand their risks and options before selecting a prenatal diagnostic procedure. There must also be adequate provisions for explaining the results. Because additional obstetric procedures, such as pregnancy termination, may follow prenatal diagnosis, obstetricians need to be comprehensive and thorough with the genetic counseling process.

Incontinentia pigmenti (308300) is an X-linked disorder that is lethal in utero for affected males. The findings vary in females and include pigmented skin lesions, dental abnormalities, patchy areas of alopecia, and mental retardation. Approximately 45% of cases are the result of new mutations. Which of the following descriptions of incontinentia pigmenti is most accurate? A. X-linked recessive inheritance with spontaneous abortions and few isolated cases B. X-linked dominant inheritance; 3:1 ratio of females to males in affected sibships C. X-linked recessive inheritance with spontaneous abortions and many isolated cases D. X-linked dominant inheritance, 1.5:1 ratio of females to males in affected sibships E. X-linked dominant inheritance with spontaneous abortions and many isolated cases

The answer is e. (Lewis, pp 75-94. Scriver, pp 3-45.) Women who are heterozygous at an X-chromosome locus will have a one-half chance to transmit their abnormal allele to sons or daughters with each pregnancy. When heterozygous women show disease symptoms, they are considered affected and the disorder is considered to be X-linked dominant. When an abnormal X chromosome allele is sufficiently severe to affect heterozygous women, the hemizygous males often do not survive the embryonic period and present as unrecognized early losses or later miscarriages. If the disorder does not affect the in utero viability of females, twice as many females as males are born. The decreased survival of affected males and reduced reproductive fitness of affected females also implies a high rate of new mutations and means that many isolated cases will occur. In disorders such as incontinentia pigmenti (308300), where affected individuals have variable manifestations, careful examination of the mother is required before assuming that an affected daughter is a new mutation. A pregnancy history will also be helpful, in that a history of early pregnancy losses would suggest the mother was a mildly affected heterozygote. The distinction between X-linked dominant and X-linked recessive diseases is somewhat arbitrary, since heterozygous women may show mild symptoms in disorders such as Duchenne muscular dystrophy (310200) or hemophilia A (306700). Severe symptoms in some female heterozygotes, together with male lethality, are good criteria for classification as X-linked dominant (as for Goltz syndrome—305600, Aicardi syndrome—304050, and incontinentia pigmenti).

The age of onset of a degenerative neurologic disease is 35. Epidemiologic study of affected persons indicates that most cases occur in the spring, are isolated (i.e., no neighbors or relatives are affected), and occur equally in men and women. However, a subset of cases consists of two affected siblings in a family. The best description of this disease is A. Inherited B. Genetic C. Sporadic D. Congenital E. Familial

The answer is e. (Murray, pp 812-828. Scriver, pp 3-45. Sack, pp 1-40. Wilson, pp 1-20.) The term familial indicates that a trait or disorder tends to cluster in families. A genetic disorder is one in which there is evidence that a gene or chromosome is involved in the susceptibility to the disease. Evidence for vertical transmission (e.g., father to daughter) is necessary for a disorder to be labeled inherited. Sporadic indicates that evidence for vertical transmission or familial clustering is lacking. Congenital simply means present at birth. Note that many congenital diseases (e.g., congenital AIDS) are not genetic, that adult-onset diseases may be genetic without being congenital, and that diseases may be familial (e.g., chickenpox) without being inherited or genetic. The eugenics movement was based on a fallacy about genetics, as it proposed breeding restrictions based on the assumption that all genetic traits (e.g., Down's syndrome) have a high risk of transmission.