GM Review Additional

Practice Test 1 (1-50 questions) Question 1 James, a 3-week-old infant with ectrodactyly (a rare malformation of the hands and feet) is brought to you by his parents, Steve and Sonya. James has one older sibling who has normal hands and feet. Although neither parent has hand or foot abnormalities, members of Steve’s family are affected with ectrodactyly. A family history reveals the following pedigree, where symbols representing individuals affected with ectrodactyly are shaded and symbols representing individuals with no clinical symptoms are not shaded. What mode of inheritance is most suggested by this pedigree? A. Autosomal recessive with variable expressivity B. X-linked dominant with reduced penetrance C. Y-linked with evidence of allelic heterogeneity D. Autosomal dominant with reduced penetrance E. X-linked recessive with a high degree of pleiotropy Question 2 The figure shown here displays the results of a DNA fragment analysis experiment involving marker D21S369. Genotypes were obtained for four individuals. The size (in base pairs) of each identified fragment is found in the box under each allele peak. Which statement best describes the DNA sequence of D21S369? A. D21S369 is a minisatellite B. D21S369 is a monoallelic marker C. D21S369 is a single nucleotide polymorphism (SNP) D. D21S369 is a dinucleotide microsatellite repeat This is an example of a dinucleotide repeat, with multiple alleles, each different from the other by a factor of two nucleotides. It is amplified using the polymerase chain reaction (PCR) with DNA primers that flank the repeat unit and the alleles are separated from one another by separation by capillary or gel electrophoresis (see Chapter 4). Question 3 Which karyotype represents an individual with a balanced chromosome complement? A. 69, XYY B. 45, XX, der(14;21)(q10;q10) C. 46, XX, del(4p) D. 46, XY, +der(1)t(1;3)(p31;p21) –3 E. 46, XY, dup(12)(q11.2;q21.2) This is the karyotype of an individual with a balanced Robertsonian translocation between the long arms of chromosomes 14 and 21. This individual lacks one normal 14 and one normal 21 and has instead a long chromosome composed of the q arms of both chromosomes (note that there are only 45 centromeres – translocation of two acrocentric chromosomes results in the loss of one centromere). As the short arms of acrocentric chromosomes are very small and contain no essential genetic material, their loss does not result in a clinical phenotype. However, the offspring of these balanced individuals are at risk of inheriting missing or extra parts of the involved chromosomes (see Chapter 18). Question 4 Robin is affected by an autosomal dominant disorder inherited from her mother. She is married to Chad, who is unaffected and has no history of the disorder in his family. Robin and Chad have two unaffected children. Studies suggest that for every 100 individuals who inherit mutations in the gene of interest, only 50 actually show clinical symptoms. The new mutation rate for this disorder is essentially zero. Based on this, what is the probability that their next child will present with the clinical signs of the disease? A. 3/4 B. 1/2 C. 1/4 D. 1/8 E. Virtually 0 We can assume that Robin carries one copy of the mutation. The probability that she passes the mutation to the next pregnancy is 1/2 and the probability that a fetus with the mutation will express clinical symptoms is also 1/2. Since these are independent events, they can be multiplied to obtain the answer of 1/4. Question 5 An adoption study was initiated in an attempt to determine whether depression has a genetic basis or is due strictly to environmental factors. Adults with depression who were adopted as infants were recruited into the study as cases. The frequency of depression was compared between the biological and non-biological adoptive relatives of the cases. If depression has a strong genetic component, what is the most likely finding of this study? A. Depression rates are similar between the biological and non-biological relatives. B. Depression rates are higher for the biological relatives. C. Depression rates are higher for the non-biological relatives. This addresses the question of whether a disease ‘runs in families’ because of shared genetic or environmental factors. Adopted individuals share the genetic factors of their biological family, but the environmental factors of their non-biological family. If depression is strongly influenced by genetic factors, there should be a higher incidence of depression among the biological family (see Chapter 15). Question 6 What is the inheritance pattern that best fits the disease shown in the pedigree shown here? A. Autosomal recessive B. Autosomal dominant C. Sex-linked recessive D. Sex-linked dominant Notice individuals are affected in each generation – a hallmark of dominant disorders. In addition, there is no male-to-male transmission, there are approximately twice as many affected females as males, and affected fathers transmit the disease to all of their daughters – all of which suggest a sex-linked dominant disorder (see Chapter 7). Question 7 Type 1 albinism is an autosomal recessive disorder that results from a mutation in tyrosinase, resulting in the lack of pigmentation. John and Susan, who are married, are both healthy and not affected with type 1 albinism, although John’s father was affected as was Susan’s maternal grandmother. Susan is pregnant. What is the probability that this fetus will be affected with type 1 albinism? A. 1/4 B. 1/8 C. 1/16 D. 1/24 E. 1/64 Individuals with autosomal recessive disorders have mutations in each copy of the causative gene. For a child of John and Susan to have type 1 albinism, each parent must inherit one copy of the mutation and then pass it to the child. John’s father had type 1 albinism so John inherited one copy of the tyrosinase mutation from him. There is a 1/2 chance he will pass this along to a child. We know Susan’s mother is an obligate carrier for type 1 albinism (she had to inherit one of the two mutant copies from Susan’s grandmother, who was affected.) Subsequently, there is a 1/4 chance Susan will pass along a tyrosinase mutation to a child – a 1/2 chance she received the mutation her mother inherited from Susan’s grandmother multiplied by a 1/2 chance that she would pass along that mutation to a child. So there is a 1/8 total chance the fetus will inherit two copies of the tryrosinase mutation (see Chapter 7). Question 8 Suppose a research study shows that people who suffer from test anxiety are homozygous for a mutation in the EXAM gene. Individuals without test anxiety have the following sequence at the very beginning of the translated region of their EXAM genes: 5′-ATG ACG TTT GAA ATT CAG TCT AGA-3′ Met Thr Phe Glu Ile Gln Ser Arg Affected individuals have the following sequence: 5′-ATG ACG TTT GAA ATT TAG TCT AGA-3′ Met Thr Phe Glu Ile STOP The mutation identified among affected individuals is most likely an example of which of the following types? A. Missense B. Gain of function C. Nonsense D. Frameshif E. Dominant negative A nonsense mutation is a substitution that leads to the generation of one of the stop codons, resulting in premature termination of the translation machinery (see Chapter 2). Question 9 Ned and Stacey are the parents of Mark, a child affected with a fully penetrant, autosomal recessive disorder that is easily diagnosed at birth, and occurs in the population with an incidence of 1/3600. Neither Ned nor Stacey has this disorder themselves. Their next child, Tony, is born without any apparent signs of the disease. Tony grows up and marries Maria, a woman with no known family history of the disorder. The chance that Tony and Maria’s first child will be affected with the same disorder that affects Mark is closest to which of the following numbers? A. 1/45 B. 1/120 C. 1/180 D. 1/240 E. 1/360 Three things must occur for Tony and Maria to have an affected child. 1. Tony must be a carrier for the disorder. As Tony doesn’t show any symptoms of the disorder (which is fully penetrant), he does not carry both copies of the mutation. Subsequently, there are three equally likely remaining genotype options for Tony – homozygous normal, heterozygous with the mutation received from his mother, or heterozygous with the mutation received from his father. The overall probability that Tony is a carrier is therefore 2/3 (see Chapter 7). 2. Maria must also be a carrier. Because Maria has no family history of disorder, we assume her risk of being a carrier is equivalent to the carrier frequency in the population. This can be deduced using Hardy-Weinberg equilibrium (see Chapter 8). The incidence of this disorder in the population is 1/3600. As mentioned in the text (p. 126), a rough estimate of the carrier frequency can be obtained by doubling the square root of the population frequency – ≈ 2 × 1/60 = 1/30. This is the chance that Maria is a carrier. 3. Both Tony and Maria must pass their mutation on to a child. If both Tony and Maria are carriers, the chance they would have an affected child is 1/4 (see Chapter 7). To obtain the final probability for this problem, multiply the probabilities from all three events – 2/3 × 1/30 × 1/4 = 2/360 = 1/180. Question 10 Anne is 16 weeks pregnant and undergoes maternal serum screening to test her maternal serum alpha fetoprotein levels (MSAFP). Her MSAFP level is 4.5 multiples of the median (MoM). Assuming Anne is carrying a single fetus and her pregnancy is correctly dated for gestational age, for which of the following disorders should Anne be offered additional testing? A. Trisomy 13 B. Pulmonary hypoplasia C. Open spina bifida D. Cystic fibrosis E. Trisomy 21 Increased levels of MSAFP are associated with the presence of open neural tube defects, such as anencephaly. The generally accepted cut-off level above which further investigation is offered is 2.5 MoM. This value detects approximately 75% of screened open neural tube defects. Question 11 Karen is a 39-year-old woman who gives birth to an infant girl with flexed and overlapping fingers, rocker-bottom feet, low-set ears and a recessed jaw. Prenatal ultrasound had revealed the presence of large choroid plexus cysts in the cerebral ventricles of the fetus, which did not spontaneously disappear. The infant dies afer 3 weeks. Chromosome analysis of the infant is most likely to reveal which of the following karyotypes? A. 45, XX, der(13;21)(q10:q10) B. 47, XX, +18 C. 45, XO D. 47, XX, +13 E. 46, XX, –4p The above physical description is consistent with a diagnosis of trisomy 18. In addition, the presence of choroid plexus cysts on ultrasound is a so-called ‘sof marker’ sometimes observed among trisomy 18 fetuses (see Chapter 18). Question 12 For what purpose are prenatal screening tests best used? A. To diagnose a suspected disease in a fetus B. To diagnose a suspected disease in a parent C. To obtain fetal cells for chromosome karyotyping D. To identify individuals who should undergo diagnostic testing E. To examine the DNA of a fetus for suspected mutations in a specific gene Screening tests are not, in and of themselves, diagnostic. They are designed to separate most of the individuals who are at an increased risk of being affected with a disease from those who are less likely to have the disease. Confirmation of disease status requires a follow-up diagnostic test. Question 13 Which of the following prenatal diagnostic techniques routinely allows fetal chromosomes to be analyzed during the first trimester of pregnancy? A. Cordocentesis Question 14 Mark is affected with Duchenne muscular dystrophy. Submetacentric and acrocentric chromosomes undergo recombination C. Amniocentesis C. an X-linked recessive disorder. Sister chromatids of the homologous chromosomes separate D. Homologous chromosomes segregate to opposite poles B. Variable expressivity C. mRNA transcripts containing premature stop codons are actually degraded by a process known as nonsense-mediated decay. Some. These represent an example of which of the following? A. allowing independent movement of homologous chromosomes During the second meiotic stage. may present with café au lait spots and Lisch nodules.B. Question 16 Which of the following processes occurs during meiosis II? A. an X-linked disorder. Digenic inheritance D. align on the metaphase spindle and are separated (see Chapter 3). particularly if the mutation occurs near the 5′ end of the gene. Her father is also color blind. while others have life-threatening tumors surrounding the spinal cord. Length of DMD-encoded protein A premature stop codon will ofen lead to the expression of a truncated protein. The most likely consequence of this mutation is a reduction in which of the following? A. Length of DMD mRNA D. Amount of DMD DNA B. preventing the formation of protein altogether (see Chapter 2). joined by cohesion proteins at their centromere. XY sex reversal B. What is the likely cause of Elizabeth’s color blindness? A. Reduced penetrance B. introducing a premature stop codon. Chiasmata are resolved. Question 15 Nearly every individual affected with neurofibromatosis type 1 (NF1) exhibits clinical symptoms. leading to the formation of the synaptonemal complex E. Aside from being color blind. There is no history of color blindness on her mother’s side of the family. Generally. Note that in some cases. Radiography E. Locus heterogeneity E. Chorionic villus sampling D. her medical history is unremarkable. It is routinely performed at 11–12 weeks’ gestation. Maternal triple testing Chorionic villus sampling is a first trimester diagnostic method for obtaining DNA for direct testing or chromosome analysis. DNA pairing is initiated. Question 17 Elizabeth is a 25-year-old woman with color blindness. sister chromatids. Mutation analysis in his family identifies a single base pair substitution in exon 1 of the DMD gene. Length of DMD DNA C. Skewed X inactivation . Sex-influenced expression Variable expressivity refers to the variation observed in clinical symptoms among individuals with an identical genetic disorder (see Chapter 7). however. the shortened amino acid chain is unable to retain normal function. the inactivation pattern is highly unbalanced in favor of one chromosome. X inactivation occurs when the female embryo consists of around 5000 cells. At this point. At times however. Klinefelter syndrome C. 126). who presents with primary infertility. In Elizabeth’s case. either of the two X chromosomes can be inactivated in any particular cell and the choice is generally random. Turner syndrome Elizabeth is likely a manifesting heterozygote with skewed X inactivation. 7q11 B. small gonads and slight gynecomastia. 17p11. part of the 22q11 syndrome (DiGeorge/Sedlackova/velocardiofacial syndrome) caused by a three megabase microdeletion at chromosome 22q11 (see Chapter 18).2 The symptoms described are consistent with velocardiofacial syndrome. Turner syndrome Klinefelter syndrome is due to the presence of an additional X chromosome in males. 1/400 C. he is a tall and lanky individual with sparse body hair. 4p C. in part be determined by the use of the Hardy-Weinberg equilibrium equation. Question 18 Steve is a 16-year-old male with clef palate. At 6′4″. Question 20 John is a 32-year-old man. leaving the mutated copy active in most of her cells and resulting in the clinical phenotype (see Chapter 6). Steve’s mother shows a similar constellation of features. Marfan syndrome B. microcephaly. A testicular biopsy reveals an absence of germ cells in the seminiferous tubules.2 E. it is likely that the X chromosome carrying the normal gene for color vision is inactivated at a higher frequency.C. 1/6400 This answer can. As mentioned in the text (p. Based upon this description. Cellular analysis reveals the presence of a Barr body. Question 19 If an autosomal recessive disorder occurs in the US Ashkenasic Jewish population with an incidence of approximately 1/6400. randomly drawn from this population will both be carriers for the mutation is closest to which of the following? A. the chance that two people. 1/3200 E. 1/80 B. Androgen insensitivity D. 1/1600 D. Congenital adrenal hyperplasia E. The incidence in the Ashkenasic Jewish population is 1/6400. which is the most likely diagnosis for John? A. conotruncal defects. 15q11-12 D. a rough estimate of the carrier frequency can be obtained by doubling the square root of the population frequency – ≈ 2 × 1/80 = 1/40. midface hypoplasia and moderate developmental delay. The likelihood that two individuals drawn from the population will both be carriers is then 1/40 × 1/40 = 1/1600 (see Chapter 8). 22q11. Adults with Klinefelter . FISH analysis of Steve’s chromosomes would likely identify an abnormality in which of the following chromosomal regions? A. Biochemical analysis of serum from the infant shows reduced levels of hexosaminidase A. Mullerian inhibiting substance is produced.syndrome are slightly taller than average and all are infertile with small testes. the woman is noted to have blue-gray sclerae. but that male and female genital ducts (vas deferens. Question 22 As a general rule. blocking the virilizing effects of testosterone. There are several subtypes of this disorder. uterus etc. she had three fractures. reflecting the presence of the additional X chromosome (see Chapter 18). leading to a truncated or absent procollagen protein. Which of the following conditions best explains this constellation of characteristics? A. Development was normal until 6 months of age. Which of the following is the most likely diagnosis? . In early adulthood E. epididymis. External genitalia appear as female. Maternal androgen ingestion during pregnancy B. Collagen I pro-alpha1 (proα1(I)) B. Barr body analysis shows the presence of one Barr body. Galactose-1-phosphate uridyl transferase (GALT) D. Magnetic resonance imaging reveals that testes are present. During uterine development B. During puberty D. Cystathione beta synthase (CBS) C. Question 24 A 9-month-old infant presents with poor feeding and lethargy. ranging in severity. all in different bones. Which of the following genes is likely to be mutated in this woman? A. when do patients with metabolic disorders first present with clinical abnormalities? A. From birth to early childhood C. During childhood. fallopian tubes. Afer the age of 40 Most metabolic disorders are identified during infancy or early childhood when the child’s body is first responsible for metabolizing the proteins obtained from formula or food (see Chapter 11). however. but the vagina ends in a blind pouch (see Chapter 18). caused by the absence of functioning androgen receptors. Mild forms are usually due to nonsense mutations in the pro-alpha1 gene. Lack of estrogen receptors E. Mutations in 5-alpha reductase C.) are absent. Blood analysis reveals normal circulating levels of testosterone and dihydrotestosterone. On examination. Testes are present. Sonic hedgehog (SHH) This patient likely has a mild form of osteogenesis imperfecta. Disabled androgen receptors D. but without the action of testosterone. Individuals with these mutations ofen experience a small number of fractures in childhood and in later life (due to postmenopausal bone loss). an autosomal dominant collagen I disorder. at which time the infant began to miss milestones. leading to the absence of all female internal structures as well. Fibrillin 1 (FBN1) E. XY). male internal structures do not form. Question 21 Karyotype analysis of an apparent female infant reveals that the baby is chromosomally male (46. Question 23 A 52-year-old woman comes into the clinic with her second broken bone in 15 months. Mutation of the SRYgene This is a case of androgen insensitivity. Blue sclerae are ofen a hallmark of this syndrome (see Chapter 19). liver and spleen. Glucosylceramide beta glucosidase C. who also has hemophilia A. Tay-Sachs disease Reduced or absent levels of hexosaminidase A is a hallmark of Tay-Sachs disease. Question 25 Mutations in which of the following genes inhibit chondrocyte proliferation within the long bones leading to rhizomelia and achondroplasia? A. A modified retrovirus A modified adeno-associated virus can accommodate inserts of up to 5 kb in size. Wilson’s disease B. Which of the following delivery vehicles would be the most likely candidate? A. Susan. Fibroblast growth factor receptor 3 (FGFR3) E. caused by mutations in the gene for clotting factor VIII. which is a disorder of lipid storage. Question 27 John is a 42-year-old man with hemophilia A. Sandhoff disease C. Fred passes his Y chromosome to the fetus. is married to Fred. 1 B. long-term integration and lasting stable gene expression are the goals of the protocol and the investigators wish to avoid eliciting any immune response. leading to progressive mental impairment and ultimately death (see Chapter 11). There is a 50% chance Susan will pass the normal X chromosome to her son and a 50% chance she will pass the X chromosome that carries the mutation in . 3/4 C. Lesch-Nyhan syndrome E. The failure to degrade sphingolipid results in its gradual accumulation in the brain. Susan and Fred are expecting their first child. integrates stably at a specific site on chromosome 19. 1/4 Hemophilia A is an X-linked recessive disorder. John carries this mutation on his X chromosome and passes it to Susan. and is able to infect nondividing cells without stimulating host immune responses. so his hemophilia status is actually irrelevant. Paired box gene 3 (PAX3) Mutations in the FGFR3 gene is responsible for achondroplasia. and ultrasound demonstrates the fetus is a boy. megalencephaly and spinal cord compression (see Chapter 6). 1/2 E. Hurler syndrome D.9 kb) neurotrophic gene is packaged into a vehicle and inserted directly into the nondividing cells of the hippocampus. 21-hydroxylase B. who is a carrier for hemophilia.A. Collagen III pro-alpha2 D. A modified adenovirus D. John’s daughter. characterized by rhizomelia. DNA packaged in a liposome E. 2/3 D. A modified adeno-associated virus B. A small (4. High gene transfer rates. A modified herpes virus C. The probability that Fred and Susan’s new son will have hemophilia A is closest to which of the following? A. Question 26 A patient undergoes gene therapy as part of a clinical trial for Alzheimer’s disease. leading to hemophilia (see Chapter 19). Question 28 A 13-year-old boy is brought for examination. At the age of 3. she exhibited hypotonia. The second mutation may occur in several of the retinoblasts. Amplification of both copies of Rb in an otherwise genetically normal cell D. and a learning disability. At birth.the factor VIII gene. Bilateral retinoblastoma generally occurs as an inherited disorder. the . Disomy of chromosome 15 in the sperm. small hands and feet. poor suck reflex and a failure to thrive. Question 29 A girl is identified with short stature. obesity. An out-of-frame deletion in the dystrophin gene D. resulting in multiple tumors in both eyes. which alters the translational reading frame (see Chapter 19). Many genes in this region are imprinted. Somatic occurrence of a single mutation in one Rb allele in an otherwise genetically normal cell B. a developmental disorder caused by loss of gene expression from the paternally-derived copy of chromosome 15q11-q13. A deletion of the 15q11-q13 region on the paternally-inherited chromosome E. leaving him wheelchair bound. while absence of a gene in this region from the maternal copy leads to Angelman syndrome. A deletion of the maternal copy of 15q11-q13 C. If this occurs through two somatic mutations in the same cell. in which one mutated copy. The most common cause of Prader-Willi syndrome is a deletion of 15q11-q13 on the chromosome inherited from the father (see Chapter 7). leading to the marked obesity. A CTG expansion in the 3′ untranslated region of the dystrophia myotonica protein kinase (DMPK) gene C. she developed extreme hyperphagia and food seeking behavior. an X-linked recessive disorder. Somatic occurrence of mutations in both Rb alleles in an otherwise genetically normal cell C. Which of the following options represents the most likely genetic cause of these physical features? A. passed from generation to generation. first noted at age 4 years in the lower limbs has progressed with increasing age. A duplication of the peripheral myelin protein-22 gene (PMP22) B. Absence of the paternal copy of these genes leads to Prader-Willi syndrome. Which of the following best describes the most likely etiology of this case? A. leading to the tumor. followed by loss of the maternal copy of chromosome 15 in the early embryo B. meaning there is differential expression depending on whether the region is inherited from the father or the mother. In families in whom this form of retinoblastoma occurs. Slight muscle weakness. A missense mutation in the neurofibromin gene E. Both copies of the gene must be mutated for tumor development to occur. An in-frame deletion in the neuronal apoptosis inhibitory protein (NAIP) gene This boy likely has Duchenne muscular dystrophy (DMD). Two-thirds of individuals affected with DMD have a deletion in the dystrophin gene. A mutation in the imprinting center of 15q11-q13. leading to a fixed maternal imprint pattern This girl has Prader-Willi syndrome. sporadic retinoblastoma (a single tumor in one eye) develops. Paternal uniparental disomy for chromosome 15 D. Loss of function of both copies of the Rb gene due to inherited and somatic mutations Retinoblastoma is the classic example of a tumor suppressor. The second copy of the gene is somatically mutated. Loss of function of one copy of the Rb gene due to an inherited mutation E. Question 30 What change(s) in the retinoblastoma (Rb) gene are required for the formation of bilateral retinoblastomas in a young child? A. is present in all retinoblasts of the individual. Examination reveals calf pseudohypertrophy. All affected individuals were identified with cancer in their middle forties. expansion of this repeat region can exceed 2000 repeats. Li-Fraumeni syndrome E. A hallmark of HNPCC mutations is the presence of microsatellite instability in tumor DNA (see Chapter 14). 160 . In the congenital form of myotonic dystrophy. The recurrence risk for disease X is highest for a couple with an affected daughter C. A point mutation in the promoter region of the DMPK gene C. Which of the following syndromes best describes this family? A. MYH polyposis B. Cowden disease D. then the risk will be higher for relatives of an individual of the less frequently affected sex. Her mother and one of her five siblings have also been diagnosed with these types of tumors and another sibling has developed gastric cancer. Question 33 Assume a multifactorial disorder (disease X) is found around the world at a higher frequency among males than females. two mutations are required for tumor formation (see Chapter 14). a loss of function of the second copy leads to the inability to repair DNA mismatches. Which of the following is true? A.disorder seems to segregate in an autosomal dominant manner. The recurrence risk for disease X is independent of gender According to the threshold liability model of disease. When individuals inherit a mutated copy of one of these mismatch repair genes. A deletion that encompasses the entire DMPK coding region Myotonic dystrophy is caused by the expansion of a CTG sequence found in the 3′UTR of the corresponding DMPK gene. The recurrence risk for disease X is highest for a couple with an affected son B. Unlike FAP. the presence of numerous additional alleles not present in noncancerous cells. leading to an increased risk of malignancy due to unrepaired errors in oncogenes or tumor suppressors. Because the disorder described above is more common in males. What is the most likely molecular mechanism for congenital myotonic dystrophy? A. the recurrence risk will be higher for the couple with an affected daughter (see Chapter 9). only a small number of polyps are present in the colon. An expansion of a CTG triplet repeat in the 3′ region of the DMPK gene E. although at the cellular level. Question 34 An allele of the a2c-adrenergic receptor (del322-325) has been found to decrease receptor function. Hereditary non-polyposis colorectal cancer (HNPCC) HNPCC is an autosomal dominant form of colorectal cancer. Analysis of several polymorphic markers from the DNA of the tumors revealed microsatellite instability. which is an autosomal dominant myopathy. Question 32 A 44-year-old woman is found to have a small number of cancerous polyps along the proximal (right side) of the colon. HNPCC is associated with mutations in a set of mismatch repair genes. Question 31 Alexander has congenital myotonic dystrophy. Alex’s mother is mildly affected with moderate facial weakness and myotonia. Familial adenomatous polyposis (FAP) C. The insertion of additional polyglutamine residues in the DMPK coding region D. if the condition is more common among individuals of one specific gender. These cases are almost without exception inherited maternally (see Chapter 19). A splice-site mutation near the 5′ end of the DMPK gene B. Question 37 Which of the following statements is true concerning the function of proto-oncogenes: A. B. Proto-oncogenes scan the genome for DNA damage E. D. The adjusted odds ratio for heart failure among persons who were homozygous for the del322-325 allele was 0. Compared to the risk for heart failure among an African-American with other alleles. Question 35 Janice was diagnosed with breast cancer at the age of 42. leading to an overexpression of the MYC gene (see Chapter 14). Proto-oncogenes serve as a signal for cellular apoptosis B. Proto-oncogenes repair DNA damage across the genome Proto-oncogenes are components of cell growth pathways. Individuals in this family who inherit the BRCA1 mutation have an increased risk of colorectal cancer. followed by the acquisition of a second. Amplification of the BCR locus on homogeneously staining regions C. the risk for an African-American who is homozygous for the del322-325 allele will be which of the following? A. Premenopausal breast cancer was a common occurrence in her father’s family and genetic testing showed that Janice possesses a mutated version of BRCA1. The males in this family who inherit the BRCA1 mutation have an increased risk of breast cancer. C. Proto-oncogenes are components of cell growth pathways C. Assume these results can be replicated and validated. Increased B. Janice’s father likely has an increased risk of prostate cancer. Identical D. As the mutation likely passed to Janice from her father.38 compared with those who had other alleles. Promoter hypomethylation of both copies of the TP53 gene Burkitt lymphoma is generally the result of a translocation of the MYC oncogene from chromosome 8 to the heavy chain immunoglobulin locus on chromosome 14. Question 38 Examine the following pedigree.unrelated African-American patients with heart failure and 200 unrelated African-Americans without heart failure were genotyped at this locus. Proto-oncogenes are cell checkpoint regulators D. Janice’s daughters have a 1 in 4 risk of developing breast cancer. Inheritance of a mutation in the KIT locus. he probably carries the BRCA1 mutation himself (see Chapter 14). Male’s with a BRCA1 mutation have an increased risk of developing prostate cancer. Improper gene conversion. Translocation of the MYC proto-oncogene to a position downstream of the IgH locus D. leading to a deletion event in the ABL gene region B. a B-cell tumor of the jaw. Question 36 A 6-year-old Kenyan child develops Burkitt lymphoma. MELAS (mitochondrial encephalomyopathy with lactic acidosis and stroke-like episodes) . Which of the following disorders most likely affects this family? A. Decreased C. Which statement best describes the risk of developing cancer for Janice’s family members? A. generally involved in relaying a ‘pro-growth’ message to the cell and responding by stimulating the transcription of growth factors (see Chapter 14). somatic mutation at this locus E. This cancer is usually caused by which mechanism listed below? A. No risk assessment can be made The finding of an odds ratio significantly less than 1 suggests that individuals homozygous for the specific variant (the del322-325 allele) have a lower risk of heart failure. This ultimately impacts the translation of all mitochondrially-encoded genes (see Chapter 11). Li-Fraumeni syndrome E. The variant is in linkage disequilibrium with a nearby gene or other sequence variation involved in the . a range of symptoms becomes apparent. Parkinson’s disease D. For a small fraction of the population. To compare the frequency of specific alleles among populations with and without the disease of interest Affected sib pair analysis is a model-free method that attempts to identify alleles or chromosomal regions shared by affected relatives more ofen than expected by chance (see Chapter 9). Recently. the number of repeats is unstable from generation to generation and as the number of repeats expands. Neither the normal nor the variant allele shows an appreciable difference in structure or function of the transporter. Fragile X syndrome C. The variant identified in this receptor is an A/C nucleotide change in the 3′UTR of the gene. Mutations in the genes responsible for maintaining mitochondrial replication Mutations in MELAS and MERRF are generally due to mutations that impair the functioning of the mitochondrial- specific tRNAs. a CGG trinucleotide repeat disorder. clinical symptoms have also been described for individuals with a repeat size in the premutation range. To calculate the amount of recombination that occurs between a specific candidate gene and the disease of interest C. To determine the most appropriate mendelian inheritance patterns for a disease of interest B.B. resulting in a very early menopause (see Chapter 18). each composed of closely matched groups of cases and controls. Full-mutation individuals are mentally impaired and have a characteristic facial appearance. Question 41 A positive association has been identified between the dopamine transporter and individuals with attention deficit hyperactivity disorder (ADHD). Single base mutations in mitochondrial tRNA genes B. Question 40 Affected sib pair studies are primarily used for what purpose? A. this is ofen observed as a tremor/ataxia phenotype. The identified variant is the functional cause of the disorder C. Large nonspecific deletions in the mitochondrial genome C. To identify alleles or chromosomal regions shared between affected relatives more ofen than expected by chance D. nor is a difference observed in regulation of the gene’s transcription or translation. What is the most likely explanation for the positive association between the variant allele and ADHD? A. Mutations in nuclear-encoding genes required for function of the OXPHOS pathway D. This variant has been identified in multiple studies from around the world. Premutation females ofen undergo premature ovarian failure. This is a false positive. Question 39 The mitochondrial disorders MELAS (mitochondrial encephalomyopathy with lactic acidosis and stroke-like episodes) and MERRF (myoclonic epilepsy with ragged red fibers) are generally both caused by what type of mutation? A. Leigh’s syndrome This is the pedigree of a typical family affected with fragile X syndrome. For males. due to population stratification between the case and control groups gathered in each study B. O) at the blood group locus. If a disease were due solely to genes. The A and B alleles function in a codominant manner. Retinoblastoma B. The maternal inheritance pattern is the first clue that this patient is likely affected with a mitochondrial disease (see Chapter 11). The numerous studies identifying a positive association with this allele suggest the finding is not a false positive. The pedigree you construct of the proband’s family reveals vision loss in the patient’s mother. For both types of diabetes. Leigh’s syndrome Leber’s hereditary neuropathy is a mitochondrial disorder that leads to a rapid onset of vision loss. MELAS syndrome C. Concordance rates of 80% were observed for NIDDM. researchers calculated concordance rates for monozygotic (MZ) twins. Based on this information. NIDDM appears to due to mutations in a single. The concordance rates are lower than those values for this study. what is the genotype of Michael at the ABO blood group locus? A. Question 44 The ABO blood group is coordinated by three alleles (A. However. E. BO . AO D. The most likely explanation is that the variant is in linkage disequilibrium with a nearby functional change that increases the risk of ADHD (see Chapter 9). Genetic influences exert a larger role in NIDDM than in IDDM. there is a substantial concordance for NIDDM among MZ twins. Question 42 A 19-year-old male patient complains of a loss of central vision in both eyes. suggesting that genetic factors play a potentially larger role in the etiology of this disorder (see Chapter 15). NIDDM is primarily determined by non-genetic factors D. all her five siblings. etiology of the disorder The variant in the dopamine transporter does not appear likely to be a functional cause of the disorder because neither allele appears to influence the transporter activity. Question 43 In an effort to identify the influence of genetic factors on both insulin-dependent (IDDM) and non-insulin- dependent (NIDDM) diabetes mellitus. Leber’s hereditary neuropathy D. B. AA E. The onset was less than 6 months ago and has progressed rapidly. dizygotic (DZ) concordance rates were 15%. and the proband’s maternal grandmother. You develop a differential diagnosis of likely disorders that includes which of the following? A. AB C. Concordance rates of 30–50% have been found for IDDM. Genetic factors have little or no role in the etiology of IDDM C. The following pedigree shows the blood groups from a three-generation family. OO B. gene Twin studies are ofen used in an attempt to identify the degree of genetic influence exerted on a disorder. indicating that there is also a role for non- genetic (environmental) factors in the etiology of NIDDM and IDDM. Kearns-Sayre syndrome E. and both are dominant to the O allele. What does this information suggest concerning the relative effect of genetic and environmental factors for each type of diabetes? A. No conclusions can be drawn from this study B. as yet unidentified. concordance rates would approach 100% for MZ twins and 50% for DZ twins. The code is overlapping C. Other sequences (like introns) are transcribed. Ofen. leading to incorporation of incorrect amino acids E. The code is degenerate B. The first child has type A blood. it cannot be added to the polycystin transcript B. Depending on their specific genotypes. these are removed during RNA processing E. the genetic code is said to be degenerate (see Chapter 2). DNA contains non-coding sequences. AO or OO.g. Question 47 Anthony has a DNA change in the dystrophin gene at nucleotide position 486. resulting in improper or failed splicing of the transcript C. Question 46 Cheryl has polycystic kidney disease (PKD). What is the likely effect of this change on the polycystin mRNA? A. Question 45 Which molecular property best distinguishes DNA from RNA? A. The ‘capless’ transcript is ubiquitinated by the repair machinery of the cell. we can infer that Michael’s wife is BO at the ABO locus – she inherits a B from her father and an O from her mother. the child must have inherited an O from his mother and an A allele from Michael. Michael could either be AA. RNA replication is semi-conservative DNA contains several sequences that are not present in processed RNA. The code is universal D. CCC and CCG both code for proline). more than one codon represents the same amino acid (e. The code is non-overlapping In many cases. Michael’s wife has type B blood and is either BB or BO at the blood group locus. Based on her mother’s type A blood. The introns of the polycystin mRNA will not be excised. The polycystin mRNA will undergo premature degradation in the cytoplasm. Because the poly(A) tail is normally attached at the end of the 5′ cap. The third child has type O blood.Michael’s parents have type A blood. Michael’s wife can pass on an O or B allele to her child. Because of this principle. The ribosome will be unable to establish the correct reading frame. reducing protein levels in the cell D. the third position of the codon can differ without altering the corresponding amino acid. preventing its transport out of the nucleus The 5′ cap is believed to protect the transcript from degradation by cellular exonucleases. Mutation analysis of her polycystin 2 gene identifies a DNA change that prevents the addition of a methylated guanine cap at the 5′ end of the polycystin transcript. DNA contains uracil. it does not alter the identity of amino acid 162). but later removed during the processing steps (see Chapter 2). Their genotypes are either AA or AO. DNA replicates in a conservative manner. RNA resides mainly in the nucleus D.e. DNA contains a hydroxyl group at the 2′ position of the ribose sugar. So Michael must also have an O allele. That means Michael is AO at the blood group locus and has type A blood. In order to have type A blood. . RNA substitutes adenine nucleotides C. which can only occur when an OO genotype is present. DNA is present in the cytoplasm and nucleolus. Michael and his wife have three children. Absence of the cap leads to early degradation and a subsequent reduction in protein (see Chapter 2). Some of these sequences (such as promoter regions) are not transcribed. This change is classified as a silent mutation (i. an autosomal dominant disorder. This scenario illustrates which principle of the genetic code? A. which corresponds to the third position of the codon for amino acid 162. RNA does not B. XYY. Michael is found to be 47. nondisjunction at meiosis I produces a gamete with one homolog of each chromosome pair (XY) (see Chapter 3). Mutation in the elastin promoter region The FISH results confirm a deletion in the 7q11 region. two signals are obtained showing hybridization of the chromosome 7 centromeric probes. . Paternal nondisjunction at meiosis I D. A 10q21q reciprocal translation C. too small to be seen on the standard karyotype (see Chapter 3). Question 50 Which of the following chromosomal conditions would result only in the production of gametes that were either nullisomic or disomic for chromosome 21? A. Maternal nondisjunction at meiosis I B. confirming the diagnosis of Williams syndrome.Question 48 A 2-year-old girl with hypercalcemia. A standard G-banded karyotype reveals no visible abnormalities. Frameshif mutation in the elastin gene D. What is the most likely genetic cause of Williams syndrome in this girl? A. short stature and mental impairment is suspected of having Williams syndrome. The additional Y chromosome likely arose as a result of which of the following events? A. In contrast. using two chromosome 7 probes: one corresponds to the centromeric region of the chromosome and the second is for the elastin gene located at 7q11. supravalvular aortic stenosis. Paternal nondisjunction at meiosis II XYY males likely arise from a nondisjunction event during the second round of meiosis in spermatogenesis. In contrast. Because there are two copies of the Y chromosome. Nondisjunction at the second stage of meiosis results in the gamete receiving both copies of one homolog (in this case YY). the nondisjunction must have occurred in the father. A 21q21q Robertsonian translocation Individuals in whom the q arms of the two chromosomes 21 have fused are only capable of producing gametes that are nullisomic (lacking the translocation) or disomic (containing the translocation) for chromosome 21 (see Chapter 3). A paracentric inversion for chromosome 21 D. Fluorescence in situ hybridization (FISH) analysis is performed. the most likely cause of the syndrome is a microdeletion. only one signal is obtained for the locus-specific elastin probe. Nonsense mutation in the elastin gene B. Maternal nondisjunction at meiosis II C. Question 49 Upon karyotype analysis. In all the examined cells of this girl. Monosomy for chromosome 7 C. Given that no abnormality was seen by G-band analysis. Microdeletion of the elastin gene E. Monosomy for chromosome 21 B. Oligonucleotide ligation assay B. DNA sequencing D. centrally located eye (see Chapter 6). Renal-coloboma syndrome C. 46 chromosomes. Question 3 The Sonic hedgehog gene (SHH) induces cell proliferation during embryonic development in a highly regulated manner. These moles. Which method is most likely to be used for this purpose? A. 23 of paternal origin and 46 of maternal origin B. Mutations or deletions in this gene result in which of the following clinical features? A. 69 chromosomes. Sizing of polymerase chain reaction (PCR) fragments D.Practice Test 2 (1-50 questions) Question 1 Diagnostic testing of patients with fragile X syndrome is performed by digesting the sample DNA with restriction enzymes. 46 of paternal origin and 23 of maternal origin D. which have a high malignant potential. but no fetus is present. 23 of paternal origin and 23 of maternal origin E. all of maternal origin C. all of which are paternal in origin. G-banding of metaphase chromosomes B. denaturing the fragments. What finding would you expect from chromosome analysis and subsequent DNA testing of tissue from the complete mole? A. 69 chromosomes. Exon trapping FISH allows high resolution mapping of fragments that are separated by 2–3 kb and up to 700 kb (see Chapter 5). An ultrasound of her uterus identifies an enlarged and poorly organized placenta. Single-stranded conformational polymorphism (SSCP) E. fractionating the fragments by size using gel electrophoresis. There is ofen an absence of midline facial formation. which in severe cases may result in the formation of a single. This is a description of which of the following techniques? A. are thought to arise by fertilization of an empty egg (lacking any chromosomes) by either two sperm or by a sperm that undergoes replication of its chromosomes before fertilization (see Chapter 6). 46 chromosomes. DNA microarray analysis This is an example of a Southern blot (see Chapter 4). Holoprosencephaly E. Southern blotting C. 46 chromosomes. Fluorescence in-situ hybridization (FISH) C. transferring them to a nitrocellulose filter and hybridizing the filter with a radioactively labeled nucleic acid probe for the region of interest. the incomplete separation of the developing brain into distinct hemispheres and ventricles. Question 5 . Question 2 Cytological mapping techniques allow a gene to be physically assigned to a chromosome or chromosome region. A complete hydatidiform mole is diagnosed. Synpolydactyly D. Hirschsprung disease SHH mutations lead to holoprosencephaly. Question 4 A 34-year-old woman is 10 weeks pregnant. all of paternal origin Complete hydatidiform moles have 46 chromosomes. Ulnar-mammary syndrome B. This same mutation was transmitted to Sarah. Based on this result. leading to a very low level of factor XIII being produced and the clinical features of hemophilia A (see Chapter 7). what can you conclude about the location of marker D18S452 relative to the causative gene under study? A. This type of choice. In this case. Each always knew they would only marry someone who also had achondroplasia. The researchers coordinating the study report that for the polymorphic marker D18S452. Based upon these findings. The mutation on the maternal X chromosome is a dominant negative. given a recombination fraction of 0. Assortative mating E. Karyotype analysis of Sarah is unremarkable. Skewed X inactivation in Sarah C. Mutation analysis of the family’s factor XIII genes identifies a nonsense mutation in one of the two copies of the gene in Sarah’s mother. Neither her mother nor her father has hemophilia. is an example of which of the following? A. causing a gain of function that leads to hemophilia. but is otherwise clinically normal. Multiple studies of a given disorder among siblings have estimated the heritability of the disorder as equal to . which at the population level can disturb Hardy-Weinberg equilibrium. C. Question 6 Steven and Sarah are married. Marker D18S452 shows some evidence of being near the disease-causing gene. B. Consanguinity D. the characteristic extends to a single gene trait (see Chapter 8). Marker D18S452 is likely located very close to the disease-causing gene and in these families does not experience any intervening recombination.Sarah has hemophilia A. Selective dominance One of the assumptions underlying Hardy-Weinberg equilibrium is that random mating occurs across the population. with no aneuploidy or structural abnormalities identified. Gene pooling B. Sarah likely has a skewed X inactivation profile. Question 8 A. but far enough away that recombination occurs between the gene and marker some proportion of the time. No mutations are found in the factor XIII genes from Sarah’s father. Marker D18S452 does not appear to be located anywhere near the disease causing gene. an X-linked recessive disorder.85. Positive selection C. but additional families need to be recruited into the study before linkage can be firmly established. The copy of the gene transmitted to Sarah from her father also appears to harbor no mutations or DNA variants. As a result. Question 7 Several families with what appears to be an autosomal dominant disorder are recruited into a genetic linkage study in an attempt to identify the location of the causative gene. the majority of her cells can only transcribe the factor XIII nonsense allele. A de novo mutation on the paternal copy of the factor XIII gene B. Marker D18S452 is on the same chromosome as the disease causing gene. the LOD score for these families was +4. A LOD score of +3 or greater is sufficient for evidence of linkage between a marker and disease gene. Homozygosity for an inactivating mutation in the factor XIII gene D. D. with most cells inactivating the chromosome with the normal copy of the factor XIII gene. They both have achondroplasia. Assortative mating is the tendency for humans to choose partners who share specific characteristics. what is the most likely cause of Sarah’s hemophilia? A. Sarah has Turner syndrome E. 85. Delta beta thalassemia D. Impair the ability of cells to degrade lipids. Five Hb H disease results from the deletion of three of the four alpha globin genes. What does this finding reveal about the cause of this disorder? B. it would be expected to occur with greater frequency among the affected population. The presence of this tetramer leads to the clinical symptoms of Hb H disease (see Chapter 10). leading to the presence of a small level of normal Hb A. 0. Question 11 Ryan is diagnosed with Hb H disease. The majority of the variance in the phenotype of this disease is caused by genetic influences C. with no environmental influences D. Does hypertension occur more ofen than expected in extended family members when one member has the ins555-560 ACE allele? B. Is the ins555-560 allele shared between affected relatives more ofen than expected by chance? D. Heritability is measured on a scale from 0 to 1 and the higher the variance. This is likely caused by the deletion of how many of the alpha globin genes? A. an autosomal recessive disorder. This disorder is likely caused by a single gene. If the allele is involved in the etiology of the disorder. Two C. A small amount of alpha globin is produced by the remaining gene. Three D. This change affects the solubility of the resulting protein under certain conditions. Hereditary persistence of fetal hemoglobin C. One B. Question 9 An association study is undertaken to investigate the possible link between hypertension and a specific allele variant (ins555-560) of the ACE gene. What question will this study attempt to answer? A. the greater the role of genetic factors (see Chapter 9). Hb H. There is an overwhelming majority of beta globin. How much recombination occurs around the ACE gene among individuals with hypertension and the ins555- 560 allele? C. resulting in the production of the beta globin tetramer. Cooley’s anemia E. leading to toxicity . What is the effect of PKU causing mutations? A. Melissa is diagnosed with phenylketonuria (PKU). Sickle cell disease Sickle cell disease is caused by the valine to glutamic acid change in beta globin (see Chapter 10). Does the frequency of ins555-560 differ between populations with and without hypertension? Association studies compare the frequency of a particular allele in populations with and without the disease of interest. Hydrops fetalis B. Question 12 Shortly afer birth. Four E. Environmental factors exert a greater influence over disease likelihood than genetic factors Heritability is a measure of the total variance in a trait phenotype that is caused by additive genetic factors. Question 10 A single nucleotide change in the gene for beta globin substitutes the amino acid valine for glutamic acid at the sixth position. What is the corresponding disorder? A. Genetic involvement in the metabolism process is minimal B. The metabolic process is polygenic C. Collagen type I C. heterozygous. Blockage of this pathway leads to a buildup of phenylalanine. Liver enzymes necessary to degrade glycogen to release glucose D. which damages the development of the CNS and interferes with brain function (see Chapter 11). An enzyme necessary for the metabolism of monosaccharides such as fructose Mucopolysaccharidosis disorders are lysosomal storage diseases (see Chapter 11). Metabolism occurs in a codominant manner by the enzymes from two genes A trimodal discontinuous response suggests metabolism is under the control of a single gene with the three responses corresponding to the three genotypes at this gene: homozygous dominant. An enzyme in the urea cycle leading to hyperammonemia B. Lead to hyperphenylalaninemia. Prevent the transport of sodium and chloride ions across the membrane of cells The mutation for PKU is in the gene for phenylalanine hydroxylase (PAH). Disrupt the function of the enzyme tyrosinase D. Enzymes that cleave sphingolipids E. Branched chain ketoacid decarboxylase (BCKD) D. dislocation of the lenses. curvature of the spine. leading to hypopigmentation C. Increased levels of homocystine are identified in his urine. Such a finding suggests what about the metabolism of the drug in question? A. Question 16 In cancer. most ofen caused by a deficiency of cystathionine beta synthetase. Result in an increase in melanin formation. loss of heterozygosity among tumor cells suggests that the region ‘lost’ may contain which gene type? A. A receptor for a growth factor . Three genes metabolize the drug D. A single gene metabolizes the drug E. but elevated homocystine is not a feature of Marfan syndrome (see Chapter 11). The concentrations of the drug among the individuals are shown here. which is responsible for converting phenylalanine to tyrosine. Which of the following genes is most likely to contain the causative mutation for this child? A.B. Cystathionine beta synthetase E. It is ofen confused with Marfan syndrome due to an overlap in some symptoms. which damages the brain and CNS E. A tumor suppressor B. Question 14 Which enzyme is most likely deficient in individuals with a mucopolysaccharidosis disorder? A. homozygous recessive (see Chapter 12). Lysosomal enzymes involved in macromolecule degradation C. pectus excavatum and arachnodactyly. Connexin 43 Increased levels of homocystine is the hallmark of homocystinuria. Fibrillin B. Question 13 A 5-year-old is identified with mental impairment. Question 15 A pharmaceutical company carries out a dose–response test for a new drug – 300 individuals are given a standard dose of the drug and are tested 1 h later to determine the amount of the drug circulating in the bloodstream. 25% E. The extent to which CFTR activity is reduced by mutation correlates well with the clinical phenotype of the disorder. 50% Levels of CFTR activity between 8% and 12% are associated with the mildest cystic fibrosis phenotype. The screen correctly identifies one hundred and ninety (190) individuals with cystic fibrosis. An oncogene D. loss of heterozygosity indicates the location of a putative tumor suppressor. RAS Tp53 is a tumor suppressor gene. KIT E. Which of the following levels of CFTR activity would most likely be found among Martin’s cells? A. 1% B. gene deletion.X)? A. A cytoplasmic tyrosine kinase Ofen. A cell cycle kinase E. c-MYC D. RB1 B. in fact. Tp53 C. 6% C. The formation of a ring chromosome during maternal meiosis D. DNA analysis carried out as part of a fertility study identifies mutations in both copies of Martin’s cystic fibrosis transmembrane conductance regulator (CFTR) gene. leading to mosaicism B. which results in a sperm that lacks a sex chromosome. Isochromosome formation for the X chromosome during maternal meiosis Turner syndrome arises in 80% of all cases through a paternal nondisjunction event. have cystic fibrosis. The first copy of the mutation is usually inherited. CBAVD (see Chapter 19). nor does he have a history of recurrent lung infections or pulmonary involvement. Martin does not have pancreatic insufficiency. The second mutation occurs through a mechanism that removes a parental allele—ofen mitotic nondisjunction. Question 19 Martin has congenital bilateral absence of the vas deferens (CBAVD). A deletion of the long arm of the X chromosome afer fertilization C. Question 18 What mechanism causes most cases of Turner syndrome (45. Individuals with Li- Fraumeni syndrome inherit a mutation in one of the two copies of the gene. Loss of a sex chromosome through paternal meiosis E. The other mutation is acquired somatically in a specific tissue. 10 individuals . leading to sterility. 5000 individuals receive a positive screening result. Question 17 Individuals with Li-Fraumeni syndrome have an increased risk of developing a large number of cancers at relatively early ages. Mutations in which genes are responsible for this disorder? A. but diagnostic studies show that they do not. 10% D. Question 20 A screening test for cystic fibrosis has been established and data gathered on the first year’s findings. It is the most frequently mutated gene identified to date. Post-zygotic loss of a sex chromosome. leading to the cancer. etc. (see Chapter 14).C. gene conversion following DNA repair. Question 23 A patient with a family history of breast cancer comes to you seeking advice about genetic testing. In the example above. 95% C. You provide her with information related to options. 97% D. 98% C. Anita has a daughter. 48 of out 50 affected individuals received a positive screening result = 48/50 = 96% (see Chapter 20). In this example. Two additional individuals with PKU received ‘screen-negative’ scores. It is a measure of how well the test detects individuals who are truly affected. this gives a 1/8 total probability. 48% Sensitivity is determined by identifying the proportion of truly affected individuals identified by the screen.000/(5000 + 495. Assuming that neither Anita’s husband nor Carol’s husband carries the mutation.000 individuals receive a negative screening result and do not have cystic fibrosis. compared with those individuals who test screen positive. 5% E. An additional 6000 ‘screen-positive’ individuals are found to be unaffected by further diagnostic testing. 1/2 C.000 individuals received a negative screening result and are not affected with PKU.receive a negative screening result. 10% D. disease-negative individuals divided by the total number of disease-negative individuals = 495. 194. 99% B. For every 100 individuals who inherit a mutation in the disease gene. but do not have cystic fibrosis. costs. What is the specificity of this screening test? A. but are also unaffected. what is the probability that Carol’s first child will exhibit the clinical symptoms of the disease? A. Taken together. 1/4 D. 99% B.000) = 99% (see Chapter 20). What is the sensitivity of this screening method? A. Question 21 A screening test for phenylketonuria (PKU) identifies 48 individuals as ‘screen positive’ who are diagnostically confirmed to have PKU. What is the ethical principle relevant in this case? . 1/16 There is a 1/2 chance that Anita will pass along the mutation to Carol and a 1/2 chance that Carol will pass along the mutation to her child. but actually do have cystic fibrosis. 1% Specificity describes the proportion of unaffected individuals who are identified by a screen as unaffected. 96% E. limits of interpretation. etc. Specificity is calculated as the percentage of screen-negative. Question 22 Anita is affected by a rare autosomal dominant disorder that occurs among males and females with an equal frequency. An additional 495. Carol. 1/8 E. If Carol’s child receives the mutation. 3/4 B. 5000 individuals are ‘false positive’ – they receive a positive screen test. She asks you ‘should I get the test?’ You tell her that it is entirely her decision to make. there is a 1/2 chance he or she will actually show clinical symptoms. only 50 present with clinical symptoms of this disorder. 3/4 E. The recurrence risk for children of an affected male is less than the recurrence risk for children of an affected female According to the concepts outlined in the liability/threshold model. Prenatal testing is performed. The recurrence risk for the sons of an affected male is equal to the recurrence risk for his daughters B. who has no family history of Huntington’s disease. This puts the chance of Huntington’s disease closest to zero for this fetus (see Chapter 9). Confidentiality D. which statement best describes recurrence risk in this population? A. if the condition is more common among individuals of one gender. Nonmaleficence Autonomy refers to the right of the patient to be in charge and make his or her own decisions. he is concerned about the potential risks for his children. By testing the fetus for a linked marker. the relatives of an affected individual of the more frequently affected sex will be at lower risk than the relatives of an affected individual of the less frequently affected sex (see Chapter 9). at least to the extent that is possible given the circumstances. using the alleles of an STR marker completely linked (θ = 0. The allele results are shown below on the pedigree of Ron’s family. Using this information. Ron does not want to know if he has inherited the mutation. The goal of this type of approach is to determine whether the fetus has inherited the copy of huntingtin gene that originated in Ron’s father. 1/2 D. Array-based comparative genome hybridization (aCGH) . Although Ron does not want to know if he has inherited the mutation. About 60–80% of affected individuals are males. Question 26 Miniaturized spots of cDNA or oligonucleotides from a set of predetermined genes are robotically placed on a glass surface. What technique is described above? A. Ron’s wife. The recurrence risk for a sibling of an affected sister is smaller than the recurrence risk for a sibling of an affected brother C. Given this information. Question 24 Clef lip and palate is a multifactorial trait with a frequency in Caucasians of 1/1000. 0 B. The specific location of the emission is used to track the relative amount of RNA present for specific genes in the cells of interest. which is of normal size. In the example given above. Beneficence B. the fetus inherits the copy of huntingtin gene that originated in Ron’s mother. Ron’s mother has no family history of Huntington’s disease. 1/4 C. 1 Although Huntington’s disease can be tested for directly. RNA is obtained from the cells of interest. the chance that the fetus has inherited the mutated (expanded) copy of the huntingtin gene is closest to which of the following? A. Unhybridized RNA is washed away. Autonomy C.0) to the huntingtin gene. Justice E. is pregnant. this can be avoided. Note that we do not know if this is the expanded or normal copy.A. Question 25 Ron’s father has Huntington’s disease. the hybridized samples emit light. Upon exposure to laser light. labeled with fluorochromes and allowed to hybridize with the cDNA or oligonucleotides present on the glass. hepatomegaly and splenomegaly. Enzyme replacement therapy D. A: Karyotype found in eight cells from culture 3. chosen to represent the contribution from other genetic factors plus the environmental influence B. In counseling situations involving these disorders.B. Question 29 Martha is a 38-year-old woman in her 16th week of pregnancy. Gene therapy B. In a level 2 mosaicism. based on previous observations of disease recurrence among actual families D. how are recurrence risks identified? A. a lipid storage disorder that results from the accumulation of glycosylceramide. bone and joint pain. using modifications of the ‘prior’ risk calculation utilized among single-gene defects C. it is usually necessary to employ empiric risks to provide recurrence information. presents with anemia. 80% E. two or more cells in . He is diagnosed with Gaucher disease. 50% D. an 8-year-old of Ashkenazi Jewish ancestry. Northern blotting E. slight modification to the enzyme targets it to macrophage lysosomes. Bayes’ theorem is applied. The risks are determined empirically. The resulting karyotypes are shown below. 100% This is an example of a level 2 mosaicism for trisomy 21 in a male fetus. Virtually 0% B. B: Karyotype found in the remaining cells of culture 3 and among all cells examined from cultures 1 and 2. These are based on the observed recurrence among families studied to date. 20% C. Recurrence risks cannot be determined for non-Mendelian disorders In the situation described above. Serial analysis of gene expression (SAGE) Microarray gene expression analysis utilizes multiplex hybridization assays to measure the relative abundance of thousands of gene transcripts simultaneously. She is referred for amniocentesis on the basis of advanced maternal age. Microarray gene expression analysis C. but are multifactorial. The amniotic sample received by the cytogenetics lab is split into three separate aliquots and cell cultures are established from each aliquot. Question 27 Aaron. leading to dramatic alleviation of symptoms in affected individuals (see Chapter 11). you need to assess the risk for mosaicism in the fetus. Liver transplant C. Reverse-transcriptase polymerase chain reaction (RT-PCR) D. Palliative care Although initial attempts at enzyme replacement therapy showed little success. Single gene inheritance risks are multiplied by a multifactorial fraction. What is the best option to manage Aaron’s disorder? A. As the lab director. What will you report? A. Question 28 Many disorders are not caused by single gene inheritance. Diagnostic tests measure the size of the CAG repeat tract: allele sizes of 40 or more CAG repeats are invariably associated with the disease (see Chapter 19). Continued progression of muscle weakness affected both swallowing and respiratory function. two of whom are affected with type 1 neurofibromatosis (NF1). The presence of single-base changes in the coding region of the huntingtin gene Nearly all individuals with Huntington’s disease have an expansion of a CAG polyglutamine repeat in the 5′ region of the huntingtin gene. The presence and amount of the huntingtin protein B.only one culture have an alternate karyotype. What is the most likely explanation for the presence of NF1 in the children? A. Independent de-novo mutations in affected children C. Question 32 Population screening for carriers of cystic fibrosis is now commonly offered to all pregnant women. multiple affected children from unaffected healthy parents are more likely due to parental gonadal mosaicism. The girl’s mother remembered fetal movements were both faint and sporadic during pregnancy. Question 33 An infant girl was born with severe hypotonia and lack of spontaneous movement. Environmental exposure limited to affected children E. What is the most likely diagnosis for this disorder? . Genetic testing identified a deletion at chromosome 5q13. Cascade screening B. The number of polyglutamine residues in the 3′ UTR of the huntingtin gene D. If the woman is found to carry a mutation in CFTR. There is no family history of NF1 and neither Elizabeth nor Robert has any of the clinical findings associated with NF1. Family screening D. He has asked to undergo genetic testing to determine whether he is likely to develop Huntington’s. completely penetrant disorder. The presence of deletions spanning one or more exons of the huntingtin gene C. prenatal diagnosis is then offered. testing is offered to the father (see Chapter 20). Question 31 Elizabeth and Robert have five children. Couple screening C. Incomplete penetration D. Follow-up fetal blood sampling is sometimes used in an attempt to resolve these types of findings. This is most likely a culture artifact. The number of CAG repeats in the 5′ region of the huntingtin gene E. but empiric studies have shown that there is up to a 20% chance the mosaicism is real and will be identified in the fetus. she is invited to bring in her partner for testing. What is this screening approach called? A. Two-step screening This is a description of two-step screening: first the mother is tested and if positive. encompassing both the survival motor neurone gene and the neuronal apoptosis inhibitory protein gene. Question 30 Michael is the grandson of a woman affected with Huntington’s disease. where a proportion of the gametes from one parent carry the NF1 mutation (see Chapter 19). If the partner is also a carrier. 25% recurrence risk for each child NF1 is an autosomal dominant. Parental gonadal mosaicism B. You tell him that the diagnostic test for Huntington’s disease specifically measures what? A. leading to death at 18 months. Targeted screening E. He’d like more information about this genetic test. Although about 1/2 of all cases are due to new mutations. The disease is characterized by degeneration of the anterior horn cells of the spinal cord leading to progressive muscle weakness (see Chapter 19).A. Heterokaryon C. Morphogen This is a classic example of a teratogen – an agent that causes birth defects by affecting embryonic or fetal . an amniotic band is discovered to encircle the lef hand of the fetus. What is the term describing the mechanism that led to this single congenital abnormality? A. Which chromosomal breakage syndrome is most consistent with this clinical description? A. One has short stature. Chimera B. She gives birth to a son who is affected with hydrocephaly. ICF syndrome This is a description of Fanconi anemia. Malformation E. a 26-year-old woman. The association with upper limb abnormalities is unique among the chromosomal breakage syndromes (see Chapter 18). Dysplasia C. At birth. developmental delay and bilateral radial aplasia (an upper limb abnormality that involves the radius). Question 36 Angie. Juvenile Parkinson’s disease C. several of the fingers on the infant’s lef hand are missing and the remainder are poorly formed. Question 34 David and Sarah have four children. An examination of cultured cells from this child revealed multiple chromosomal breaks. Question 35 During prenatal ultrasound. leading to reduced levels of all blood cell types. Ataxia telangiectasia D. Amyotrophic lateral sclerosis D. due to the constriction of the amniotic band. retinoic acid is a(n): A. has been taking a medicine prescribed by her doctor to treat severe acne. These malformations can be traced to the effect of the retinoic acid. an autosomal recessive disorder characterized at the cytogenetic level by chromosome breakage and gaps. Deformation D. Spinal muscular atrophy (SMA) E. Teratogen E. In this example. Sequence B. While taking the medication. This child was born without thumbs and later developed bone marrow failure. Duchenne muscular dystrophy B. The active ingredient in this medication is retinoic acid. Disruption A disruption occurs when some outside factor (usually not genetic) disrupts the normal developmental pathway (see Chapter 16). Bloom’s syndrome C. Angie becomes pregnant. Fanconi anemia B. Xeroderma pigmentosa E. Enhancer D. clef palate and low-set deformed ears. Myotonic dystrophy This is a description of SMA type I. Question 38 Emily is an 18-year-old who has recently been diagnosed with chronic myeloid leukemia.22)(q34.q11) B. Multiple genes . Recurrence risk increases with increased number of affected first-degree relatives E. del(13q) The ‘Philadelphia chromosome’. Double minute for MYCN E.14)(q24. Migrant populations show no change in disease prevalence B.q32) D. del(17p13. observed in 90% of individuals with this form of leukemia. t(8. one with normal and one with elevated systolic pressures D. When plotted. the concentration of the drug circulating in the blood was measured. Question 40 The metabolism of a certain drug is measured using a dose–response test in 1000 individuals. adopted children would show lower correlations with their biological parents (who do not share their environment) than would children who stay with (and thus share the environment of) biological parents (see Chapter 15). A single dominant gene B. t(9. Chromosome analysis of her white blood cells would reveal which abnormality? A. C. Ovarian cancer C. Two hypertensive groups exist. Breast cancer B. Thyroid cancer E. Li-Fraumeni syndrome The Amsterdam criteria are used to identify individuals at highest risk of colorectal cancer (see Chapter 14). at least two successive generations affected. A single recessive gene C. Colorectal cancer D. is the product of a reciprocal translocation between the long arms of chromosomes 22 and 9 (see Chapter 14). Question 37 Which finding would suggest that environmental factors play a major role in the etiology of essential hypertension? A. All individuals were given a standard dose and afer 30 minutes. cancer diagnosed before age 50 in at least one relative) is used to select high-risk individuals for what form of inherited cancer? A.1) C. Incidence of hypertension is increased in identical twins compared to non-identical twins If a disorder is driven in large part by environmental factors. What action is the metabolism of this drug most likely controlled by? A. Question 39 The ‘Amsterdam criteria’ (at least three affected relatives. the overall concentrations formed the following distribution. The correlation for blood pressure between adoptive children and their biological parents is lower than the correlation for children who remain with their biological parents.development (see Chapter 16). fully translucent irises. has become difficult to rouse and is now having seizures. Evelyn has a low core temperature. leading to the absence of melanin pigmentation (see Chapter 11). Camptomelic dysplasia Congenital adrenal hyperplasia is the result of mutations in 21-hydroxylase. her parents report she has not been feeding well. Congenital adrenal hyperplasia B. Pompe’s disease C. 21-hydroxylase Andrew presents with the symptoms of type I oculocutaneous albinism (OCA). Ferrochelatase B. Hexosaminidase-A C. Cannot be determined with the available data A unimodal distribution suggests that the metabolism is under the control of multiple genes (i. When screening individuals in this family. see Chapter 12). white skin that does not tan. and blue. Appearance of a ‘cherry-red’ retinal spot B. These symptoms are likely to be caused by a deficiency in which one of the following enzymes? A. It leads to an accumulation of ammonia which is toxic to the nervous system (see Chapter 11). Question 44 Miguel’s family has familial hypercholesterolemia (FH) and is at increased risk of early coronary artery disease. She is diagnosed with ornithine transcarbamylase deficiency. Uric acid B. The internal genitalia are female and chromosome analysis is 46. Biochemical studies reveal a deficiency of 21-hydroxylase. a 4-year-old male. cerebral edema and has slipped into a coma. Phenylalanine Ornithine transcarbamylase deficiency is an inborn error of the urea cycle. Question 41 Andrew. Ammonia C. what early clinical sign will you look for? A. Analysis of Evelyn’s blood will most likely show an elevation of which metabolic product? A.D.XX. Hurler’s syndrome E. Although she appeared normal at birth. It is the most common cause of ambiguous genitalia in females (see Chapter 11). Androgen insensitivity D. Arginase D. Question 43 Marcus and Carmen have a child with ambiguous external genitalia. the metabolism is polygenic. Individuals with OCA type 1 have deficiencies in tyrosinase. What is the most likely diagnosis? A. Urea E. Folate D.e. Question 42 Evelyn is a 6-day-old newborn brought to the emergency room by her parents. High cholesterol levels in persons with FH are due to defective or deficient low-density lipoprotein receptors. He also presents with poor visual acuity and nystagmus. Tyrosinase E. has white hair. Hirsutism . Formation of Heinz bodies in the red blood cells D. Upon clinical examination. Maria is noted to have a Kayser-Fleischer ring at the corneal margin of her eyes. weakness and lassitude. Because the gamma and delta globin chains are intact. Given the information shown in this pedigree. growth retardation and hepatosplenomegaly. Galactosemia Wilson’s disease is a disorder of copper metabolism. what disorder should be strongly considered? A. McArdle’s disease E. Maple syrup urine disease B. Her parents have also noticed sharp personality changes in Maria. An elevation of serum creatine E. Penetrance D. what is the chance that the firstborn child of Jin (Peng’s son) will also be homozygous for the factor V Leiden mutation? . Question 45 Maria is a 23-year-old woman who has recently noticed loss of coordination. Question 46 Rehan has sickle cell anemia. caused by a single nucleotide change in the beta-globin chain of hemoglobin. The effect of this single nucleotide change on multiple organ systems is an example of: A. early renal failure. Based upon Maria’s symptoms. and difficulty in speaking and swallowing. Wilson’s disease D. Locus heterogeneity E. Hydrops fetalis D. HbA2 and HbF production continues (see Chapter 10). Peng is homozygous for the mutation. Sickle cell disease C. Question 48 The pedigree shown here follows the inheritance of the factor V Leiden mutation. In general. HbH disease Beta thalassemia results from the absence of functional beta globin (and the consequent absence of HbA). which increases the risk of thrombophilia. Pleiotropy C.4) gives the following results. Hemoglobin electrophoresis on cellulose acetate (pH 8. splenomegaly. Propionic acidemia C. It ofen appears during childhood or adolescence in individuals with FH (see Chapter 11). Question 47 Mikklos is a 9-month-old boy of Greek origin who presents with severe anemia. Beta thalassemia B.C. Which disorder is most consistent with this clinical profile? A. It ofen causes both neurological and behavioral changes in affected individuals and is recognized by the presence of a Kayser-Fleischer ring in the eye (see Chapter 11). Allelic heterogeneity B. Formation of multiple xanthomas A xanthoma is a subcutaneous deposit of lipid. limb pain. leading to the symptoms described above. He has splenic infarctions. these features do not appear until a few months afer birth because gamma globin is the major non-alpha globin chain during fetal development and early postnatal life. Anticipation A mutation that affects numerous organ or tissue types is said to exhibit pleiotropy (see Chapter 7). Methemoglobinemia E. but had lost the use of her hands for other purposes and had severe impairment in expressive and receptive language. 100% Jin is heterozygous for the mutation. Congenital hypertrophy of the retinal pigment epithelium (pigmented areas of the retina) is a finding ofen associated with FAP (see Chapter 14). 75% E. It is caused by mutations in the MECP2 gene. APC D. Rett syndrome E. TP53 E. DiGeorge syndrome B. RET Mutations in both copies of the APC gene are a hallmark of familial adenomatous polyposis (FAP). she had developed a pattern of hand wringing. By 24 months. Patau syndrome Rett syndrome is a neurodevelopmental X-linked dominant disorder that occurs almost exclusively among females. a mediator of gene expression and transcriptional silencing. Question 49 Kirsten is a 5-year-old girl with ataxia and seizures. Question 50 Benjamin is a 38-year-old man who has just undergone a total colectomy with ileoanal pull-through. The result of an apparently normal pregnancy and birth.A. The probability that they both pass the mutation on to a child is therefore 25%. 50% D. Kirsten would be most likely to be affected with which disorder? A. It is characterized by hundreds to thousands of adenomatous polyps throughout the colon. in which gene is Benjamin most likely to have a mutation? A. an autosomal dominant disorder accounting for approximately 1% of colorectal cancer. MLH1 C. Edward syndrome C. Given this set of clinical findings. Miller-Dieker syndrome D. BRCA2 B. Kirsten progressed normally until the middle of her second year. . Virtually 0% B. His colon contained more than 2400 polyps. as is his wife. 25% C. when she began losing speech and motor skills. Benjamin also has congenital hypertrophy of the retinal pigment epithelium. Based upon these findings. Benjamin’s mother and a maternal uncle died of cancer before the age of 55. There was a marked deceleration in the growth of her head during this same period. Becker muscular dystrophy E. Ellis-van Creveld syndrome. Question 3 Roger is a 26-year-old man who complains of a decrease in strength and endurance and a reduced ability to walk and run. most likely accounts for this elevated frequency? A. Neurofibromatosis D. Which of the following factors. The test was positive. Upon examination. A founder effect among the Amish population B. Genetic drif Heterozygote advantage is an example of positive selection. the cells sickle and are destroyed. A reduced mutation rate among the Amish C. An environmental risk factor present at increased frequency among the Amish The most likely explanation for this observation is that one or two of the original founders of the Old Order Amish were carriers for the Ellis-van Creveld mutation. the neurologist requested molecular testing for a duplication of the peripheral myelin protein 22 gene (PMP22). indicating increased dosage of this gene. compared with individuals who have two normal copies of the beta globin gene. resulting in an increase in the number of marriages between two carriers and a subsequent increase in disease frequency (see Chapter 8). Leigh disease Charcot-Marie Tooth disease. coupled with the social and religious isolation of this population. Negative selection C. characterized by chronic motor and sensory polyneuropathy. is present at a much higher frequency than that observed among other groups of European descent. Allelic diffusion B. He also notes frequent leg cramping and difficulty climbing stairs and stepping over objects. Reduced fitness observed for Amish affected with this disease D. Consequently. an autosomal recessive disorder. A neurological evaluation identifies a reduction in nerve conduction velocity. but segmental demyelination was observed. Hardy- Weinberg equilibrium is disturbed for the beta globin mutation. In these populations. When carriers’ red blood cells are invaded by the malaria parasite. is most ofen caused by increased dosage of PMP22 as a result of a duplication of the chromosome 17 gene (see Chapter 19). A subsequent nerve biopsy showed no evidence of inflammation. Heterozygote advantage E. Question 2 Among the Old Order Amish in the US. Due to the restricted number of marriage partners available to members of this group. where the presence of one copy of the sickle cell mutation increases the fitness of the individual (see Chapter 8). Increased reproductive fitness of the affected males among the Amish E. Inflammatory peripheral neuropathy B. carriers of sickle cell anemia have greater resistance to infection. . What is the most likely diagnosis? A. Charcot-Marie-Tooth disease C. There are no other family members with similar problems and no family history of neuromuscular disorders. Assortative mating D. the frequency of the mutation increased to a relatively high frequency. atrophy of his distal leg muscles and reduced ankle and patellar reflexes are noted. This disturbance of genetic inheritance in populations is referred to as what? A.Question 1 In areas where Plasmodium falciparum malaria occurs. there is an 80% chance that he will develop the clinical symptoms of the disorder. Digenic inheritance D. XX individuals with ambiguous genitalia B. but infertile The SRY gene is the primary factor that determines maleness. many of the genes that control sperm development and fertility reside on the Y chromosome. translocating the SRY gene from the Y chromosome onto the X. leading to the formation of masculine internal and external genitalia. Marsha. 40% C. only patients heterozygous for mutations in both genes develop the disease. 80% E. Genetic heterogeneity Digenic inheritance describes a situation where the additive effects of heterozygous mutations at two different genes leads to the occurrence of a disorder. The probability that Marcus will have the clinical signs associated with the disorder is closest to which of the following choices? A. They are genetically equivalent to siblings. However. There is a 50% that Marsha passed along the mutation to Marcus. It is theorized that the individual effect of each mutation is not sufficient by itself to cause the disorder. leading to symptoms (see Chapter 7). This form of the disease only occurs when individuals are heterozygous for mutations in two different unlinked genes. If such a translocation event takes place and the resulting sperm fertilizes a normally developed oocyte. Triallelic inheritance E. The affection status of Michael and Marcus are independent. XY individuals who are phenotypically male. Question 6 The following pedigree shows your patient. Michael and Marcus. 20% B. but infertile D. SRY expression triggers a series of downstream gene activation. there is a 40% overall chance that Marcus will show signs of the disease (see Chapter 5). who is affected with a single-gene. Rarely. and XX males who lack these genes are therefore infertile (see Chapter 6). both coding for photoreceptor membrane proteins. Question 5 In sperm. This is an example of what type of complex inheritance pattern? A. which combination of karyotype and phenotype is most likely to be observed? A. autosomal dominant disorder that shows 80% penetrance. ROM1 and peripherin. and her two children. If Marcus received the mutation. aberrant recombinations take place outside this region. 100% Marsha’s dizygotic twins arose from the fertilization of two oocytes by two sperm.Question 4 Anne-Elizabeth is affected with a rare form of retinitis pigmentosa (a progressive retinal degeneration). sharing 50% of the genes in common on average. 50% D. XX individuals who are phenotypically male. XX individuals who are phenotypically female. but infertile E. XY individuals who are phenotypically female and are fertile C. most meiotic recombination between the X and Y chromosomes occurs in the pseudoautosomal region. Michael is also clinically affected with the disorder. but the joint presence crosses some threshold of cell damage or injury. Translocation of SRY onto the X chromosome results in XX males. As a result. Pleiotropy B. Individuals who carry a mutation in only one of these genes are not affected. . Co-dominant inheritance C. Ordering which cytogenetic analysis results in the fastest clinical report? A. Large amounts of starting DNA are needed C. CTACAC B. Fluorescent in-situ hybridization (FISH) C. for this reason. Of the choices provided. DNA microarray for resequencing purposes C. Multiple regions of DNA cannot be coamplified in the same reaction D. The test you want performed involves segmental amplification of the promoter and exonic regions of both collagen type I genes. Ephraim. ATCTCAG E. use of the PCR is limited by which of the following experimental conditions? A. Which of the following laboratory tests will you order? A. which is ofen associated with trisomy 21. Question 8 The polymerase chain reaction (PCR) is a powerful tool and is commonly used for diagnostic studies. only option C contains a palindrome (see Chapter 4). Nirupama. is suspected of having osteogenesis imperfecta. looking for variations from the normal electrophoretic mobility of a fragment. High-resolution Giemsa banding B. Question 9 Your patient. GCTGG C. on their specific DNA sequence (see Chapter 4).Question 7 The following sequence is found in the promoter region of the insulin gene: 5′- CACTACACGCTGCTGGGATCCTGGATCTCAGCTCCCTGGCCGACAAC-3′ Which of the subsets of this sequence shown below represent the most likely target region for recognition and cleavage by a restriction endonuclease? A. GGATCC D. You seek molecular confirmation of this diagnosis. Ultrasound of the fetus showed an increase in the nuchal thickness. Oligonucleotide ligation assay (OLA) SSCP. Question 10 Your patient. PCR is carried out under strict guidelines in areas free of contaminating sequence (see Chapter 4). GCCGACA Restriction endonucleases cleave double-stranded DNA around palindromic nucleotide sequences (where the same sequence of nucleotides occurs on each of the two DNA strands when read in the same direction of polarity). is 16 weeks’ pregnant. in part. Denaturing high-performance liquid chromatography (DHPLC) D. D. and both you and your patient are anxious for a rapid diagnosis. However. 24–48 hours is required to complete the assay B. The process is prone to amplify contaminating DNA. Real-time PCR E. Single strand conformational polymorphism (SSCP) B. Reverse banding (R-banding). You order an amniocentesis. Flow cytometry . Single stranded DNA fragments assume a specific three-dimensional structure based. Contaminating DNA can readily be coamplified The PCR is a very robust process that will amplify any DNA fragment present in a sample that is homologous to the forward and reverse primers used in the reaction. The resulting double-stranded products are denatured and one of the strands is subjected to gel electrophoresis. A locus for paroxysmal kinesigenic dyskinesia maps to human chromosome 16.E. What beta-globin regulatory element is removed by this deletion? A. A. Polyadenylation signal B. Question 11 Paroxysmal kinesigenic dyskinesia (PKD) is a rare autosomal dominant disorder characterized by episodic choreiform or dystonic movements that are brought on or exacerbated by voluntary movement. Virtually 0% B. coded by a mitochondrial gene. Sperm mitochondria are generally destroyed so that all mitochondria are derived from the mother. The allele of marker D16S419 that was co-inherited with the clinical phenotype is the PKD-causing mutation D. It is likely the location of the PKD gene will be found somewhere in this interval. Approximately 25% C. D16S3396 and D16S419 achieve this level. markers D16S753. Question 13 Anna-Marie is affected with beta-thalassemia. with correspondingly small recombination fractions. The physical location of the PKD gene is likely found between markers D16S753 and D16S419 LOD scores of 3. The two-point LOD scores from a portion of markers along chromosome 16 are shown here. D16S419 is contained within the gene boundaries of the putative PKD locus B. are accepted as significant evidence for linkage with a specific marker. In this example. What is the probability that both daughters have inherited this mutation? A. Roach ES.0 or greater. 3′ intron/exon boundary for intron 2 . Neurology 2000. Consequently. Approximately 50% E. Ryan has two daughters. none of the offspring of a male carrying a mitochondrial mutation will inherit the mutation (see Chapter 2). Bowcock AM. Question 12 Ryan is affected with Leber’s hereditary optic neuropathy (LHON). Molecular diagnosis of her beta-globin genes identifies a mutation that removes the 100 nucleotides immediately preceding the start site for transcription. leading to bilateral blindness. Approximately 38% D. A genome-wide linkage study was conducted using a large African-American kindred segregating the PKD phenotype in an attempt to identify the most likely location of the PKD gene (Bennett LB. Approximately 75% A defining characteristic of mitochondrial DNA is inheritance through only the maternal lineage. The strongest evidence for linkage to the PKD locus is found at marker D16S769 C. characterized by rapid optic nerve death. 54(1):125–130). FISH can be used to examine interphase cells (as well as metaphase cells) with a turnaround time as rapid as 24 hours (see Chapter 3). The descending order in the table corresponds to their physical order (from centromere to telomere) on the q arm of the chromosome. TATA box C. What conclusions can reasonably be drawn from these data? Table 1: Two point LOD scores for chromosome 16 loci and PKD. Subsequent molecular studies reveal the causative mutation is a single base substitution in NADH dehydrogenase subunit four. there are 200 chromosomes. Nonsense B. Homoplasmy D. The frequency of HbA is therefore 150/200 = 0. The corresponding normal sequence is also shown.75.32 B. These include the TATA box (approximately 25 base pairs upstream of the transcription start site) and the GC and CAAT boxes (usually within the 80 base pairs upstream of the start site) (see Chapter 2). Quantitative analysis of the electrophoresis results is shown below. only one mutation is needed to produce the clinical effects of OI. Of those 200.60 D. 0. This case illustrates which of the following genetic concepts? A.D. Missense C. Shown below is her DNA sequence for a portion of exon 6 for the collagen type 1 gene. Frameshif This is an example of a frameshif mutation.75 Among this sample of 100 individuals.68 E. Substitution D. HbA/HbL 30%. Heterogeneity B. Insertion E. Question 17 . What is the allele frequency of HbA in this sample? A. 0. 0. 0. HbL/HbL 10%. Question 14 May-lin has type 1 osteogenesis imperfecta (OI).56 C. Termination codon Sequences immediately upstream of the transcription start site generally contain conserved sequences important for the initiation of transcription. 150 are HbA (120 from the HbA homozygotes plus 30 contributed by the HbA/HbL heterozygotes). May-lin: 5′ AAACTCCACTTCTTCACGTAC – 3′ Normal: 5′ –AAACTCACTTCTTCACGTAC – 3′ May-lin has what type of mutation? A. Hypermorph E. an autosomal dominant disorder resulting from defective type I collagen. Haploinsufficiency C. which contains the mutation responsible for type 1 OI. A single base deletion has disrupted the reading frame for translation (see Chapter 2). Heteroplasmy Haploinsufficiency describes the situation where 1/2 of the normal levels of the gene product leads to phenotypic effects (see Chapter 2). Although Evann’s OI is caused by a loss of function mutation in the pro alpha I gene. Question 16 Hemoglobin protein electrophoresis was used to determine whether 100 individuals carried genes for normal hemoglobin (HbA) or the Lepore variant (HbL). HbA/HbA 60%. 3′ UTR (untranslated region) E. 0. Question 15 Evann is affected by osteogenesis imperfecta (OI). 0. 1/16 C. 1/512 B. What are the chances that both Sarah’s grandchildren have inherited a copy of the MSUD mutation? A. The frequency of this mutation in the population is negligible. Two of these three would result in a heterozygous carrier for the mutation. 2) he inherited a mutation from his mother and a normal copy from his father. an autosomal recessive disorder that results in the inability to metabolize branched chain amino acids. 1/2 will inherit one copy of the normal gene and one copy of the mutation.25 B. the probability that Roxanne and Richard will give birth to a child with hemochromatosis (an autosomal recessive disorder) is closest to which of the following options? A. In the same way. Question 19 Roxanne and Richard are married. So there is a 1/4 chance that one grandchild will be a carrier for the MSUD mutation. 0. The chance that both grandchildren will carry the mutation is 1/4 × 1/4 = 1/16. That leaves three equally likely possibilities: 1) he inherited both normal copies from his parents. 0. At birth. and 3) he inherited a normal copy from his mother and a mutation from his father. 1/256 C. 0. followed by a 1/2 chance that the daughter in turn passes the mutation along to her child. as Evan is 28 years old. 1/32 B. 1/4 D. 1/4 of their offspring will inherit two normal copies of the corresponding gene. an autosomal recessive lysosomal storage disorder that is fatal by the age of 2 years.33 C. 3/8 E. 0. because these individuals die before the age of 2 years. there is a 1/4 chance that the other grandchild will carry the MSUD mutation. 1/32 As hemochromatosis is an autosomal recessive disorder. 1/128 D. 1/2 There is a 1/2 chance that Sarah passes along the mutation to one of her daughters. and 1/4 will inherit both copies of the mutation (and have Gaucher disease). However.Evan is a 28-year-old man with a younger brother who died from type 2 Gaucher disease. Question 18 Sarah is a known heterozygous carrier of a mutation that causes maple syrup urine disease (MSUD). the great grandfather of Roxanne and the great grandmother of Richard are obligate .67 E. She has two daughters who each have one child. giving Evan a 2/3 probability.50 D. If all individuals who married into this family are not carriers for the mutation. What is the probability that Evan is a heterozygous carrier for the Gaucher mutation? A. Through a family history you establish the pedigree shown here. 1/64 E. he cannot possibly be affected with Gaucher disease.75 Evan’s parents were each carriers of the Gaucher mutation. In order for Richard and Roxanne to have a child with hemochromatosis they must both be carriers (1/8 × 1/8 = 1/64) and then each pass the mutation to that child (which would occur on average 1/4 of the time). 0.04. .20 for the second microsatellite. The genotype frequencies are determined using the Hardy-Weinberg equation.50 for the first microsatellite and 0.04. In the situation described above. 0. The frequency for the heterozygous genotype at the third microsatellite is 2pq. The matching allele frequencies among the general Caucasian population are 0. who also has Becker muscular dystrophy.0004. Helene has just given birth to a daughter. The genotype frequency of each homozygous microsatellite is p². In the same way.0002 B. Virtually 0 B. What is the probability that a random individual drawn from the Caucasian population would match the alleles present on the male fraction obtained from the rape kit? A.25. He is a Caucasian of European descent.20 × 0. Richard has a 1/8 chance of being a carrier.0016 E. Taken together.carriers. For a female to be affected with Becker muscular dystrophy. So there is a 1/2 × 1/2 = 1/4 chance that a child will inherit a maternal copy of the mutation.0010 D.10.10 = 0. Question 20 A suspect in a rape and assault case has been genotyped at three independent microsatellite repeat markers. 1/4 C. What is the probability that this child will be affected with Becker muscular dystrophy? A. 0. it is likely that one of her parents carried the mutation. a 1/2 chance he then passed it to his daughter (Roxanne’s mother) and a 1/2 chance she passed it to Roxanne.04 × 0. Question 21 Helene has two brothers with Becker muscular dystrophy. 1 As both Helene’s brothers have Becker muscular dystrophy (an X-linked recessive disorder).50² = 0. His alleles at these three markers match the alleles recovered from the male fraction of the rape kit collected from the victim shortly afer the attack. There is a 1/2 chance that Roxanne’s great grandfather passed the mutation to his son (Roxanne’s grandfather). the probability of this occurring is 1/64 × 1/4 = 1/256. she must inherit a mutation from both her father and her mother.20 and 0.0004 C. Gary will pass the mutation for Becker to all his daughters. 1/2 D. The matching allele frequencies for both alleles of the third microsatellite are 0. so the genotype frequency for the first microsatellite is 0. If Helene carries the mutation. This gives Roxanne a 1/2 × 1/2 × 1/2 = 1/8 chance of carrying the hemochromatosis mutation. there is a 1/2 chance she will pass it to a child. Multiplying the three genotype frequencies together gives us the probability that a randomly chosen individual in the Caucasian population would have the same genotypes as those identified on the rape kit. Helene has a 1/2 chance of being a carrier. 3/4 E. the genotype frequency for the homozygote at the second microsatellite is 0. 0.25 × 0. 0. She marries Gary. there is a 1/4 chance the newborn daughter of Helene and Gary will inherit two mutations and be affected.0020 The probability is determined by multiplying together the frequencies of the three genotypes present on the rape kit.20² = 0. The suspect is homozygous for the first two microsatellites examined and heterozygous for the third. Similarly. or 2 × 0.04 = 0. 0. Jean. carried on the same chromosome as allele 5.dup(5)p15 for Margaret. from his father (individual I-1). Based only on this information. Under the hypothesis of linkage between the disease and the genotyped marker. More severe than Margaret B. This is the case for all but one of the offspring (III-7). He would inherit the normal copy of the gene. About the same severity as Margaret D. 1/4 D. 0 B. Based on this information alone. while those that inherit the ‘5’ allele would not. from his mother (individual I-2). One recombinant in eight meiotic events yields a recombinant fraction of 1/8. There is a maternal family history of this disorder. Less severe than Margaret C.del(5)p15 for Anita and a karyotype of 46. Impossible to tell from the information available In general. carried on the same chromosome as allele 3. 1/2 The mother of Chris and Todd is an obligate carrier. the boy’s maternal aunt. There is a 1/2 chance that the mutation was passed to her sister as well. which means that one of her parents most likely carried the mutation responsible for Duchenne muscular dystrophy. 1/8 C. A four allele marker has been genotyped and the corresponding alleles are shown below the symbol for each individual. a loss of genetic material produces more severe consequences than a gain of genetic material. What is the recombination frequency for the marker and disease locus? A. Virtually 0 B. how severe would you expect Anita’s clinical phenotype to be compared to Margaret’s clinical phenotype? A. What is the probability that she is a heterozygous carrier of the disease gene? A. this man would inherit the disease causing mutation. Thus. 3/8 D. and those of his unaffected mate (individual II-2). Question 23 Consider the pedigree shown here.XX. 1/8 C. Question 24 A cytogenetics laboratory reports a karyotype of 46. is interested in beginning a family. there are seven nonrecombinants and one recombinant among the offspring. 1/2 E. Based on these haplotypes.Question 22 Chris and his brother Todd both have Duchenne muscular dystrophy. you would expect the deletion patient (Anita) to be more severely affected than a . we would predict that the children who inherit the ‘3’ allele from their father would also be affected. 5/8 The genotypes of the individuals in generation I establish the linkage phase for individual II-1. This individual would be termed a recombinant. where an autosomal dominant disorder is being transmitted. 3/8 E.XX. a mature oocyte contained the centromeres labeled A and B.00. Chromosome 1 homologs are labeled at the centromeres as A. will be unaffected. 1/2 will have green blindness E. 10% C.duplication patient (Margaret). The normal allele ‘G’ is dominant to the other two alleles. The second polar body would contain sister chromatids that are identical to those that remain in the mature oocyte (see Chapter 3). Virtually 0% B. Question 27 A three-allele locus on the X chromosome controls the green-sensitive pigment for color vision. has been typed for all members of the family. Question 26 Consider chromosomes 1 and 2 at the beginning of meiosis. which results in ‘green blindness’. we can use the alleles of the microsatellite to predict the affection status of the family members. A and b C. Mutations at this locus have a penetrance of 90%. who inherited allele 4 from his father. 100% Because there is no recombination between the OI locus and the typed microsatellite. All will have green blindness For X-linked disorders. All will have green shif D. A man with normal vision marries a woman with green blindness. The g1 mutation leads to a condition called ‘green shif’ – the presence of a weakened green-sensitive pigment where only certain shades of green are indistinguishable from browns.a. Shaded individuals are affected with osteogenesis imperfecta (OI). Question 25 The pedigree of the Patel family is shown here. 90% E. where reds. The accompanying genotypes are shown in the electrophoresis image below the pedigree. A four allele microsatellite. At the second meiotic division. sister chromatids are separated. Based on these genotypes what is the approximate risk that individual 7 will have the clinical symptoms of OI? A. 50% D. a and B D. a and b would be found in the first polar body and A and B would remain in the oocyte. A and B During the first meiotic division. homologous chromosomes are separated. All will have normal green vision B. If at the time of fertilization. Individual 7. All sons . 1/2 will have green shif C. The OI causing mutation is on the same chromosome as allele 2 in the affected father. who are both affected with OI. both of which are mutations. This allele was inherited by individuals 3 and 6. linked to the OI locus at a recombination fraction of 0. Chromosome 2 homologs are labeled at the centromeres as B. 1/2 will have green shif. which combination would be found in the second polar body? A. At this point. greens and yellows cannot be distinguished. an autosomal dominant disorder.b. there is no contribution from the father (who provides the Y chromosome instead). 1/2 with have normal vision. The g1 mutation is dominant to the g2 mutation. a and b B. What type of green vision will the sons born to this couple have? A. Question 31 Unlike alpha thalassemia. composed of alpha and gamma subunits. .XX. 24. 45. The infant’s hands and feet appear normal. Beta globin C. 47. has just given birth to a daughter with severe developmental abnormalities. a 25-year-old woman. This results in an overabundance of alpha globin subunits. alpha-globin tetramers are capable of supplying oxygen to the fetus E.will receive one of the mother’s two X chromosomes. Question 30 Monique.X C. present until just afer birth. all oxygen requirements are provided by maternal red blood cells C. 46. Around the time of birth. This combination will result in green blindness. gamma expression is diminished as beta globin expression is increased. 46. der(13.Y B.XX Mature human gametes have 23 chromosomes. The infant dies 10 days afer birth. normal. Embryonic hemoglobin. low-set ears and a ventricular septal defect. Chromosome studies of this infant would likely reveal which of the following karyotypes? A. 45. Prenatally. Fetal hemoglobin is sufficient to meet oxygen needs prenatally B. meets fetal oxygen requirements Fetal hemoglobin. Gamma globin Individuals affected with beta thalassemia major produce little to no beta globin. microphthalmia with bilateral colobomas. Delta hemoglobin tetramers transport oxygen during fetal development D. which is ofen identifiable in utero.Y D. bilateral clef lip and palate. so all sons will have one g2 allele and a Y chromosome. –15q11. human sperm? A. Decreasing the levels of alpha globin would theoretically reduce the formation of alpha tetramers (see Chapter 10). 46. beta thalassemia usually presents only afer birth.XX.XX. but not both (see Chapter 3). The infant has microcephaly. both of the mother’s X chromosome carry the g2 mutation. +13 E. leading to the formation of alpha subunit tetramers. Fetal Hb B. Question 29 Which of the following karyotypes would most likely be present among mature. 23.XO D.21)(q10:q10) B.XY E. is the primary hemoglobin present during development. These insoluble tetramers are harmful to red blood cells and lead to their premature destruction. Alpha globin D. Question 28 Decreasing production of which form of hemoglobin would theoretically benefit a patient with beta thalassemia major? A. Sperm will have either an X or Y sex chromosome. 22. This is because: A. In utero. +18 C.XX. In this case. 47.2 –q13 The above physical description is consistent with a diagnosis of trisomy 13 (see Chapter 18). which is 1/50.Individuals with mutations that impair beta globin production begin to show symptoms as the proportion of fetal hemoglobin drops without a concomitant rise in adult (alpha. Question 32 An autosomal recessive disorder has a frequency of 1 per 2500 in the Caucasian population. 1/4 Cystic fibrosis is an autosomal recessive disorder. Approximately 4% C. the frequency of the disease allele (q) is the square root of the frequency of the disease (1/2500). if the disease frequency is 1/2500 (q²). She divorces her husband and marries Alex. All known cases of this disorder result from an identical mutation in the causative gene. Question 33 Anne. Since the frequency of the disease allele plus the frequency of the normal allele must equal 1. 1/250 C. 99% Before the screening test. a genetically unrelated man from France. 95% C. There is a 1/2 chance Anne will pass along her mutated allele to her next child. If Alex has the mutation. Question 35 Leevi. Approximately 90% D. According to Hardy-Weinberg equilibrium. Leevi. 1/100 D. Putting this all together gives a risk of 1/25 × 1/2 × 1/2 = 1/100. The screening test detects affected individuals with a 98% sensitivity and a 5% false positive rate. the chance he would pass it along to his child is 1/2. Question 34 Rebekah. Approximately what percentage of this population has two copies of the normal allele for this gene? A. Bayesian analysis can be used to incorporate the findings from the screening test to determine a new risk. What now is the best estimate of her risk of developing the disease? A. she is herself a carrier for a mutation that causes cystic fibrosis. a 27-year-old woman is screened for a fully penetrant autosomal dominant disorder. Rebekah tests positive. a young woman from the UK. Studies have shown that the frequency of cystic fibrosis among individuals of Northern European descent is 1/2500. The chance that Alex is a carrier for cystic fibrosis is 2pq ≈ 2q = 1/25. The frequency of the homozygous normal genotype is (p)x(p) = p² (49/50)² = 2401/2500 ≈ 96%. Clinical symptoms for this disease do not generally appear until the fourth decade of life. . the frequency of the normal allele (p) is 49/50. Rebekah’s mother had the disease and died in her early 40s.5% D. Approximately 2% B. 98% E. Approximately 98% According to Hardy-Weinberg equilibria. has a child with cystic fibrosis. Approximately 96% E. 1/25 E. there was an a priori risk of 50% that Rebekah would inherit the mutation from her mother and develop the disease. beta) hemoglobin (see Chapter 10). 1/2500 B. 97. As Anne has an affected child. What is the risk that Anne and Alex will have a child with cystic fibrosis? A. the frequency of the disease allele is q = 1/50. 90% B. a 32-year-old man of Finnish descent had a sibling who died at the age of 10 from cystic fibrosis (CF). The mutation is too large to be detected using standard DNA sequencing or DHPLC methods — the normal copy of the LIS1 gene would be amplified and screened. his new risk of being a carrier for CF is 1/6. FISH examination of the PWS 15q locus shows no evidence of an interstitial deletion in this area. Autosomal monosomy Uniparental disomy (the inheritance of two chromosomes from one parent) occurs in approximately 30% of Prader-Willi cases. who are both deceased. also be identified using a quantitative approach such as quantitative PCR. Denaturing high performance liquid chromatography (DHPLC) FISH is the best technique for the case described. Question 36 Robert is diagnosed with Prader-Willi syndrome (PWS). which molecular mechanism listed below is the most likely cause of Prader-Willi syndrome for Robert? A. characterized by short stature. 1/4 C. 1/15 E. 1/6 D. given that he tested negative for the CF panel. Question 38 Steven is a 24-hour-old. but is concerned about the risk of passing CF to his children. infant who is hypotonic. his new risk of being a carrier for CF is 1/6. DNA samples are not available from his brother or from his parents. Fluorescent in-situ hybridization (FISH) D. hypogonadism and learning disabilities. DNA sequencing B. lethargic and has shallow. He undergoes DNA testing for the 70 most common CFTR mutations – a panel that detects 90% of the mutations found among individuals of northern European descent. the risk he is a carrier is 2/3. Standard G-banding C. 2/3 B. Given this finding. full-term. Because he is not affected with the disorder. what is the probability that he is a heterozygous carrier for CF? A. Nucleotide substitution E. rapid breathing. What test should be ordered . So. masking the presence of the deletion. 1/25 Begin with the prior risk that Leevi is a carrier for CF. Then use Bayesian analysis to calculate the risk given his negative testing results. The difference in gene dosage at the LIS1 gene could. located at 17p13. however. The microdeletion is too small to be identified by standard G- banding. Question 37 Many cases of Miller-Dieker syndrome are caused by microdeletions within the region of the LIS1 gene. Chromosome translocation B. has recently married and would like to start a family. His mother gave birth 2 years ago to a son with similar symptoms who died 72 hours afer birth. Which diagnostic technique would most likely detect a 75 kilobase deletion in this gene region? A.3.who does not himself have CF. Chromosome duplication D. He tests negative for all 70 mutations. obesity. The two chromosomes 15 are inherited maternally. Given this information. Uniparental disomy C. disomic. Plasma phenylalanine level E. Question 42 Andrew is heterozygous for both the factor V Leiden mutation. Mike E. Plasma ammonia level C. 2:1:1 ratio of normal. What finding suggests the development of schizophrenia is. Hemoglobin electrophoresis Steven has the symptoms of a urea cycle disorder. Elevated ammonia levels are a hallmark of urea cycle disorders. The incidence of schizophrenia is increased among biological relatives of schizophrenic adoptees compared to the biological relatives of control adoptees C. genetically influenced? A. Equal number of disomic and nullisomic gametes D. is an obligate carrier. disomic and nullisomic gametes The first meiotic division separates homologous chromosomes. Schizophrenia is a psychotic illness characterized by disorganized thought. Question 39 A nondisjunction event for chromosome 2 occurs during the first meiotic division of a primary spermatocyte. which leads to a build-up of ammonia. More disomic than nullisomic gametes C. disomic and nullisomic gametes E. Question 40 The following pedigree shows the transmission of a very rare disorder. The concordance rate for monozygotic twins with schizophrenia is 15%. which renders the factor V resistant to cleavage. who gives birth to a son with the disorder. All disomic gametes B. which is equal to the concordance rate for dizygotic twins D. Which individual in the pedigree is an obligate carrier for the mutant allele of the disease-causing gene? A. Ann B. Sallie This represents a disorder with an X-linked recessive inheritance pattern. Nondisjunction at this stage results in two gametes with extra chromosomes and two gametes with missing chromosomes (see Chapter 3). Beth C. altered behavior and a decline in social functioning. at least in part. 1:1:2 ratio of normal. The concordance rate for schizophrenia for identical twins born to a schizophrenic parent is higher if the twins are reared together rather than reared separately Increased incidence among biological relatives provides support for genetic factors in the etiology of schizophrenia (see Chapter 15). Sweat chloride level D. and nullisomic gametes? A. Jim D. Sallie. At the end of meiosis what would the proportions be of normal. Chromosome analysis B. The comparison of birth month to incidence of schizophrenia reveals an excess of winter births among individuals with schizophrenia E. which reaches the brain and can cause irreversible damage and death.on Steven first? A. . The lifetime risk to develop schizophrenia is 1% for an individual in the general population and rises to 4% if the associated schizoid personality disorder is also considered B. and a daughter who is a carrier. There are over 80 distinct alleles identified for this gene. Venous thrombosis D. Inherited together. Question 44 Many cases of androgen insensitivity syndrome are caused by deletions within the androgen receptor gene located on the X chromosome. which increases prothrombin levels. If the other allele is a poor metabolizer. He is given standard dosage of oral codeine to relieve the cough. Fluorescent in-situ hybridization (FISH) E. Normal/normal B. Question 43 One of the genes involved in codeine metabolism is CYP2D6. fully functional extra copy of the CYP2D6 gene (‘ultrarapid’ alleles). Southern blotting B. On the other hand. A 500 band karyotype D. This enzyme mediates O-demethylation of codeine to produce morphine. Hemophilia B. a component of the cytochrome P-450 family. a deletion of this size is too large for detection by standard sequencing techniques. may experience exaggerated clinical consequences in response to standard doses. Other alleles inactivate CYP2D6 activity (‘poor metabolizer’ alleles). This patient most likely possesses which of the following CYP2D6 allele combinations? A. A small number of individuals have CYP2D6 genes where a gene duplication event has created an adjacent. A few days into treatment. Question 45 Individuals with hereditary non-polyposis colorectal cancer (HNPCC) ofen have mutations in what class of genes? A. Hypertension E. the overall metabolic effect will appear as if there are two normal alleles present (see Chapter 12). The deletion is certainly too small to detect by karyotype. Poor metabolizer/ultrarapid D. the patient becomes unresponsive and develops a life-threatening morphine intoxication. Apoptosis pathway B. the factor V Leiden mutation and prothrombin variant both increase the risk of deep vein thrombosis four. they confer a twentyfold increase in risk (see Chapter 15). DNA sequencing C.or fivefold. An 800 band karyotype Southern blots are most likely to detect this relatively small deletion. Normal/ultrarapid Individuals with ultrarapid CYP2D6 activity can produce greater amounts of morphine from codeine and as a result. and very likely too small for FISH detection as well. Many alleles (classified as ‘normal’ alleles) do not alter enzyme activity levels. A 62-year-old man with bilateral pneumonia has a cough that does not respond to over-the-counter medication. This is especially likely when the other CYP2D6 allele is either normal or also ultrarapid. Thalassemia Individually.and the prothrombin variant G20210A. Coronary artery disease C. Normal/poor metabolizer C. What diagnostic technique would most likely detect a 700 base pair deletion that completely removes exon 8 in this gene? A. Mismatch repair . What disease is he is at an increased risk of developing? A. Approximately 80% of classic Rett syndrome mutations are point mutations and 15% are microdeletions encompassing the Rett gene. the chromosomes of the parents are also examined to determine whether either parent is a balanced carrier of the translocation. Denaturing high-performance liquid chromatography (DHPLC) screening and DNA sequencing C. a key component of the nuclear membrane. progressive neurological disorder observed in girls who appear normal for the first few months of life. Recurrence risks are not influenced by the gender of the carrier B. Question 47 Classic Rett syndrome is an X-linked. A different amino acid is inserted in the protein at the site of the mutation C. Of the identified mutations. G-banding and quantitative PCR D. premature aging syndrome. What is the most likely effect of this mutation? A. 99. how does the gender of the carrier affect recurrence risks for additional children with Down syndrome? A. the protein will be 50 amino acids shorter. autosomal dominant. FISH or quantitative PCR will identify the microdeletions. Question 46 The LMNA gene on chromosome 1 encodes lamin protein. Rett syndrome is caused by mutations in the MECP2 gene and mutations that cause classic Rett syndrome are male lethal. decreasing protein levels Activation of a cryptic splice-site donor creates a new location for the exon/intron boundary. Southern blotting. The lamin gene is transcribed at a reduced rate. If one of the parents is identified as carrying the translocation.C. DNA sequencing and Southern blotting E. When the transcript undergoes processing. G-banding and MECP2-specific fluorescent in-situ hybridization (FISH) B.5% occur de novo. Most cases are caused by a silent base substitution. the last exon is spliced at this location. motor skills and eventually purposeful hand movements. but cannot detect the microdeletions (the non-deleted copy of the gene will amplify and provide normal sequence). The entire last exon is spliced out of the transcript during processing D. What combination of tests should be ordered to maximize the likelihood of identifying a causative mutation in a patient with the clinical symptoms of Rett syndrome? A. resulting in the loss of the final 150 nucleotides. A single point mutation in this gene is responsible for the majority of cases of Hutchinson-Gilford progeria. the additional chromosomal material is the result of a Robertsonian translocation between chromosome 21 and usually either chromosome 13 or 14. Cell adhesion HNPCC is due to mutations in a series of proofreading genes involved in the repair of DNA mismatches that arise due to errors in DNA replication (see Chapter 14). which are too small to be detected by a standard G-banded karyotype. Recurrence risks are higher if the mother is the carrier . The final 50 amino acids are deleted from the end of the protein B. but then experience loss of language. an extremely rare. which activates a cryptic splice-site donor in the last exon of the gene that is 150 nucleotides 5′ of the traditional GT donor site. Question 48 In approximately 4% of all cases of Down syndrome. Quantitative PCR and Southern blotting DNA sequencing or DHPLC-based screening will identify the point mutations. Consequently. When a translocation Down syndrome case is identified. Recurrence risks are higher if the father is the carrier C. Signal transduction D. All her children would be unaffected If this woman were to become pregnant. PKD1 and PKD2. ultimately leading to kidney failure. Her children could either be unaffected or have PWS C. Question 50 A woman is affected with Prader-Willi syndrome (PWS) due to a deletion that includes both the Angelman syndrome (AS) and PWS genes. the AS gene is inactive on the paternal copy and as this gene is missing from the maternal copy. they will be unaffected.The recurrence risks for Down syndrome are approximately 10% if the mother carries the Robertsonian translocation involving chromosome 21. . Locus heterogeneity C. If her children receive the normal copy. For male carriers. Although rare. Which of the following options describe the potential outcomes if this woman were to become pregnant? A. Germline mosaicism B. The PWS genes are active on the paternal copy. they will only have their father’s PWS and AS gene regions. but not both. the risk is only 1 –3% (see Chapter 18). She could have children who are unaffected. Individuals with ADPKD inherit a mutation in one of the genes. These cysts impair kidney function. Co-dominant inheritance D. so the child will not have PWS. Her children could either be unaffected or have AS B. There are two causative genes. However. If they receive the deleted copy. she would either pass to her children her normal copy of 15q or the copy with the deletion. this child would have Angelman syndrome. Question 49 Individuals with the autosomal dominant form of polycystic kidney disease (ADPKD) have numerous fluid-filled cysts within each kidney. fertility among PWS individuals has been documented. Clinical variability Locus heterogeneity occurs when a disorder can be caused by a mutation in more than one gene. have AS or have PWS D. Digenic inheritance E. Which genetic term best describes the inheritance of this disorder? A.