UMICH Study questions
The "triplet repeat" in Huntington Disease refers to: A A nucleic acid repeat consisting of: C-A-G B A nucleic acid repeat consisting of: T-A-G C An amino acid repeat consisting of: C-A-G D An amino acid repeat consisting of: Gly-X-Y E A nucleic acid repeat consisting of: G-A-T
A As per lecture this disease is characterized by expansion of the nucleic acid triplet repeat CAG at the 5' end of the gene.
Assuming Hardy-Weinberg equilibrium for alleles at the CFTR (cystic fibrosis) locus in the U.S. Caucasian population, and given that the mutant allele frequency, q, is 1/50, what fraction of this population are carriers of a CFTR mutation? A 2/50 B (49/50)2 C 1/100 D (1/50)2 E 1/50
A The normal allele frequency is calculated using the formula p+q=1; so given the mutant allele frequency of q=1/50, p=49/50. For a population in Hardy-Weinberg equilibrium, the heterozygote carrier frequency is 2pq = 2 x 49/50 x 1/50 = 98/2500. The closest answer of those given is 2/50.
Which of the following human diseases is least likely to be caused by aneuploidy? A Klinefelter syndrome B Fragile X syndrome C Down syndrome D Turner syndrome
B Aneuploidy is when the chromosome number is not an exact multiple of the haploid number. Down syndrome (trisomy 21), Turner syndrome (45, X), and Klinefelter syndrome (47, XXY) are all examples of aneuploidy. In contrast, Fragile X syndrome is characterized by expansion of a CGG triplet repeat in the first exon of the FMR-1 gene on the long arm of the X chromosome, with the total chromosome number not being affected (i.e. still euploid).
Myotonic dystrophy may show increasing severity and earlier age of onset in successive generations. This phenomenon is known as: A Compound heterozygosity B Anticipation C Incomplete penetrance D Locus heterogeneity E Variable expressivity
B Anticipation may be defined as "increasing severity and earlier age of onset in successive generations". Anticipation is most often due to the gradual expansion of a trinucleotide repeat element in the gene whose mutation causes the disease. Myotonic muscular dystrophy is caused by the expansion of a GCT repeat in the 3' untranslated region of the gene. Normal individuals have 5-35 copies of the repeat, while patients with MMD always have greater than 50 copies, and some patients have greater than 1000 copies. Expansion of these trinucleotide repeats occurs primarily when the gene goes through female meiosis. In fact, congential MMD, caused by trinucleotide expansions containing thousands of copies, is caused only by alleles that are inherited maternally. Other genetic disorders caused by the expansion of trinucleotide repeats include fragile X syndrome and Huntington disease.
A 26-year-old woman of Norwegian descent seeks genetic counseling. Her brother died at age eight of documented cystic fibrosis. Both of their parents are deceased. The woman undergoes DNA testing for 70 CF mutations which collectively detects approximately 90% of CF carriers of northern European descent. Testing reveals that she is negative for all 70 mutations. What is the probability that she is a heterozygous carrier of CF? A 2/3 B 1/6 C 1/15 D 1/25 E < 1%
B Bayes it
A woman has a brother affected with a rare autosomal recessive disorder (she herself is not affected). This disorder is 100% penetrant at birth. She undergoes a carrier screening test which detects carriers with 98% sensitivity, and 5% false positive rate; she tests positive. Which of the following is the best estimate of her risk for being a carrier? A 100% B 97.5% C 66.7% D 50% E 33.3%
B Bayes it
A nonsense mutation in exon 14, resulting in failure of the LDL receptor to reach the cell surface, is found in individuals affected with familial hypercholesterolemia in multiple Lebanese Christian families. Which is the most likely to explain the high frequency of the mutation in this population? A Lebanese eat a diet high in animal fat. B Founder effect. C Homozygotes enjoy a survival advantage. D Family size is large. E The mutation has no detrimental effect, and therefore is not selected against.
B Founder effect is the best explanation, because there is a single mutation which is found at high frequency in the Lebanese Christian population. This is due to a "rapidly expanding population founded by a small ancestral group when one or more of the founders was, by chance, a carrier of the mutant allele."
Which of the following conditions should not occur in a single individual: A Both a-thalassemia and hemoglobin SS disease. B Both b-thalassemia and hemoglobin SC disease. C A total of 5 copies of the a-globin gene sequence (aaa/aa). D Both a-thalassemia and b-thalassemia. E Both a-thalassemia and hemoglobin SC disease.
B Hemoglobin SC disease is caused by point mutations in the b-globin genes that create qualitative alterations in the function of the hemoglobin tetramer, without affecting the ratio of chains in the tetramer. b-thalassemias are quantitative deficiencies that generate abnormalities in b-globin production that can lead to an imbalance of the chains in the tetramer (i.e. b chains are not being produced or are being produced in reduced amounts).It is important to consider the fact that the mechanism for generating the proposed answer would require that in at least one of the b-chains two mutations would exist, i.e. one that is either S or C and the other that is a b-thalassemia mutation. Given the geographic specialization of these two categories of mutations it is rather unlikely that the two mutations would co-exist in a single globin chain. Granted this distinction is a bit confusing, however when one considers the other options, this is the best answer. (It highly unlikely but theoretically possible that a genotype for SC and beta-thalassemia could occur.) It is possible to simultaneously have both a- and b-thalassemia, and that combination does indeed occur in practice. The key here is that the alpha and beta genes are on different chromosomes, and the inheritance of one is independent of the other. For example, if ~20-30 % of African Americans carry an a-thalassemia allele, then ~20-30% of African Americans with b thalassemia would incidentally also have inherited an a-thalassemia allele. Coinheriting b-thalassemia or sickle cell with a-thalassemia results in a milder disease.
All of the following provide supporting evidence for the clonal origin of a tumor EXCEPT: A All cells from the tumor carry an identical rearrangement of the immunoglobin gene cluster. B All cells from the tumor carry exactly 46 chromosomes. C In a person heterozygous for a particular DNA polymorphism, all tumor cells show reduction to homozygosity for the same marker allele. D All cells from the tumor carry the same abnormal chromosome. E The tumor is from a female heterozygous for a G6PD polymorphism. All of the cells from the tumor express the same G6PD allele.
B If a group of cells is of clonal origin, it means that they all derived from a single cell. The cells would then be expected to be identical in a cellular characteristic that normally varies among cells (such as a polymorphism), or all be abnormal is a specific way (such as loss of one chromosome. All of the answers listed are examples of clonal origin EXCEPT: All cells from the tumor carry exactly 46 chromosomes. This statement is incorrect, because in a karyotypically normal person, all cells (except germ cells) have 46 chromosomes.
Patients with hemoglobin SC disease: A Could not have a child with sickle cell disease. B Have a different mutation in each copy of their b-globin gene, though both are in the same codon. C Exhibit an excess of embryonic z-chains in their adult red blood cells. D Frequently die in utero from complications of the hemoglobinopathy. E Never experience sickle cell crises.
B In Hemoglobin S, the mutation is GAG->GTG at codon 6 changing Glu->Val. In Hemoglobin C GAG->AAG also at codon 6 changes Glu->Lys. Because the hemoglobin tetramer is composed of 2 b-chains, it is possible to have mutations that originate in the 2 alleles of the b-globin gene and thus combine to form the SC variant within the tetramer.
Why is b-thalassemia major usually evident only after birth? A The product of the b-globin pseudogene is highly expressed only after birth. B The switch from g-gene to b-gene expression occurs around the time of birth. C The mother's normal red blood cells provide oxygen to the fetus in utero. D The fetal Hb persists after birth when it should have been shut off. E The Hb subunits encoded by the a cluster (which has two copies of the a-globin gene) are sufficient until after birth.
B b-thalassemia major occurs when either no beta-globin is made from either gene copy (b0), or when a small amount of defective beta-globin is made (b+). The clinical features of b-thalassemia major are not evident until after birth, because in the fetus the g gene is expressed from the b-globin cluster. It is not until after birth that the g gene is turned off and b-globin gene expression begins; at this time the defective b-globin begins to cause clinical symptoms in b-thalassemia major.
Each normal polypeptide chain that makes up collagen is generally characterized by: A triplet repeats of the nucleic acid sequence: C-A-G B triplet repeats of the nucleic acid sequence: G-X-Y C triplet repeats of the amino acid sequence: Gly-X-Y D triplet repeats of the amino acid sequence: Cys-X-Y E back to back ALU repeats
C
Linkage analysis is performed in a large family with an autosomal dominant hemolytic anemia, using a polymorphic marker within the b-globin locus. The LOD score at q=0 is negative infinity. The LOD score at q=0.01 is -4.5. You conclude that the disorder in this family: A is due to a mutation in a gene on chromosome 11, 10 cM centromeric of b-globin B is due to a b-globin gene mutation C is not due to a b-globin gene mutation D is due to a mutation in the a-globin gene E is an acquired disorder, due to a somatic gene mutation
C A negative LOD score of <= -2 is evidence that linkage is excluded for this value of theta (page 206 of the book). Thus it is very unlikely that the gene for hemolytic anemia is linked to a region within or close to the beta-globin locus.
A large percentage of individuals with a1-antitrypsin deficiency will develop chronic obstructive pulmonary disease (COPD) or emphysema. The severity of this disease will be significantly increased if the patient is: A heterozygous for the mutation, with one normal copy of the a1-antitrypsin gene. B homozygous for null alleles of the elastase gene. C a cigarette smoker. D a woman. E neutropenic.
C About 80% of individuals homozygous for the Z allele of the Pi (protease inhibitor) locus develop pulmonary emphysema. This observation lends support to the protease-antiprotease theory of emphysema, which holds that emphysema results from an imbalance between proteases like neutrophil elastase and protease inhibitors like a1-antitrypsin in the lung. The Pi locus codes for the a1-antitrypsin protein, which is produced predominantly in the liver and secreted into the serum. The Z allele results in defective transport of the a1-antitrypsin protein from the endoplasmic reticulum to the Golgi apparatus and consequent reduced serum concentrations. PiZZ homozygotes have serum a1-antitrypsin levels that are about 10% of normal. Pulmonary emphysema in these patients occurs with greater severity and at an earlier age if the patient is a smoker. The protease-antiprotease theory of emphysema can explain the tendency of even normal smokers to develop emphysema, since cigarette smoke both increases the amount of protease in the lung and decreases the activity of a1-antitrypsin. In PiZZ individuals this effect is magnified.
You would appropriately suspect a possible collagen disorder in a patient who has any of the following symptoms except: A rupture of the bowel B multiple fractures C progressive neurological deterioration D joint hypermobility or laxity E blue sclerae
C Correct. Collagen disorders affect the integrity of connective tissue and would not be characterized by a progressive neurological deterioration.
A phenotypically normal woman with a 45, XX, -14, -21, +t(14q,21q) karyotype has a karyotypically normal husband. Among their liveborn offsping, which of the following is LEAST likely? A 46, XX B 46, XY, -14, +t(14q,21q) C 46, XY, -21, +t(14q,21q) D 45, XY, -14, -21, +t(14q,21q) E 45, XX, -14, -21, +t(14q,21q)
C First note that all of the listed karyotypes are possible when this woman mates with a karyotypically normal male. However, 46, XY, -21, +t(14q,21q) is trisomy 14, which is not viable and therefore would not be seen in liveborn offsping of this couple. Of viable offpring in such a mating, 50% have a balanced translocation, 40% are karyotypically normal, and 10% have an unbalanced translocation and Down syndrome.
major abnormal form of hemoglobin that accumulates in a fetus with the severe form of a-thalassemia (hydrops fetalis) is composed of: A A tetramer of 4 a-subunits (a4) B A tetramer containing 2 d-subunits and 2 b-subunits (d2b2) C A tetramer of 4 g-subunits (g4) D A tetramer of 4 b subunits (b4) E A tetramer containing 2 z-subunits and 2 e-subunits (z2e2)
C In hydrops fetalis, no alpha globin is produced. Since the fetus produces only gamma chains, they are the only components that are available to form the tetramer. NOTE: We are only speaking about the fetus, not the newborn.
A man and a woman, each with a normal karyotype, have a child with Down Syndrome due to trisomy for chromosome 21. Shown below is the haplotype, based on a polymorphic marker locus near the centromere of chromosome 21, for the child, her mother, and her father. Based on this information, in which meiotic division did the non-disjunction event occur? Haplotype: Mom (CD); Father (AB); Affected fetus: (BCD) A Paternal meiosis I B Paternal meiosis II C Maternal meiosis I D Maternal meiosis II E Maternal meiosis I or paternal meiosis I
C In meiosis I, each chromosome replicates into sister chromatids and then two homologous chromosomes separate to opposite poles of the cell, with each daughter cell retaining only one of the homologous pair. In meioisis II, the sister chromatids then separate. The affected child received both of the maternal alleles at this polymorphic locus (in addition to one paternal allele), so the non-disjunction event must be maternal. The child has both C and D alleles, so it must be maternal meiosis I. If the non-disjunction event occurred during maternal meiosis II (and assuming a B allele from the father), the child's haplotype would be either B,C,C or B,D,D.
A 26-year-old male medical student who has no neurological or psychiatric problems undergoes genetic testing to determine if he has a mutation in his Huntington Disease (HD) gene. Several of his maternal relatives, including his mother, have well documented HD. His test results are: HD allele 1: 17 repeats HD allele 2: 43 repeats The most accurate interpretation of this test is: A Luckily, he has one normal allele. Therefore he is only a gene carrier and will never develop the disease. B He has inherited one HD gene mutation, but given that he currently has no symptoms by the age of 26, it is most likely that he will never develop the disease. C He is an asymptomatic individual who inherited an HD gene mutation and will develop symptoms of the disease if he lives long enough. D He has HD and should seriously consider another occupation as he will never be able to practice medicine competently. E These test results are inconclusive and therefore his 50/50 risk of inheriting the disease from his mother has not been modified by the test.
C This individual has had presymptomatic testing for an autosomal dominant condition. This test has 100% predictive value.
The average recurrence risk for a couple that has had a child with cleft lip, a multifactorial birth defect, is approximately 4%. What is the recurrence risk if the couple has two affected children? A 2% B 4% C 10% D 25% E 50%
C When looking at the inheritance of multifactorial traits, as the number of affected children within a single family increases, the recurrence risk also increases. This specific empiric risk was quoted on p.62 of the text.
A phenotypically normal woman with a 45, XX, -14, -21, +t(14q, 21q) karyotype has a karyotypically normal (46, XY) husband. Among their liveborn offspring, the most likely karyotypes are 46, XY and 46, XX. Which of the following karyotypes is the next most likely among liveborn offspring? A 47, XY, +21 B 46, XY, -14, +t(14q, 21q) C 46, XY, -21, +t(14q, 21q) D 45, XY, -14, -21, +t(14q, 21q) E 45, XY, -14, -21, +16
C XY, -14, -21, +16 and 47, XY, +21 are not possible karyotypes for offspring of this couple. Of the remaining three answers, 46, XY, -21, +t(14q, 21q), which is trisomy 14, is not viable. 45, XY, -14, -21, +t(14q, 21q) and 46, XY, -14, +t(14q, 21q) (which is a trisomy 21) both result in liveborn offspring. Of these two, 45, XY, -14, -21, +t(14q, 21q) is much more common.
A young woman of northern European descent is the single parent of a child with autosomal recessive cystic fibrosis (CF). She marries a genetically unrelated man of northern European descent and wishes to have more children. What is the risk that he is a carrier of CF? Assume the frequency of CF is 1/2500 in this population. A 1/2500 B 1/1250 C 1/50 D 1/25 E 1/4
D Cystic fibrosis is an autosomal recessive disease. According to Hardy Weinberg Equilibrium , the frequency of an autosomal recessive disease in a population is q2, which in this case = 1/2500 and thus q = 1/50. Because the frequency of the two alleles (p & q ) must equal 1, p = 49/50 ~ 1. Finally according to Hardy Weinberg Equilibrium, the frequency of the heterozygous genotype (a carrier in this case ) is 2pq. This carrier frequency is 2 x 1/50 x 1 = 1/25.
Which karyotype would be MOST frequently seen in liveborn infants (as opposed to spontaneous abortions)? A 46,XY,-11,+22 B 69,XXX C 46,YY D 47,XX,+21 E 47,XX,+3
D Down syndrome is the most common serious chromosomal disorder, seen in about 1/900 newborn infants. Since 47,XX,+21, or trisomy 21, is the usual cause of Down syndrome (95%), this is the correct answer. Most of the other trisomies, including 47,XX,+3, rarely lead to live infants. Trisomy 13 (Patau syndrome) and trisomy 18 (Edwards syndrome) occur in about 1/10,000 births. Triploidy, or 69,XXX, is seen in about 20% of spontaneously aborted fetuses but never in newborns. Monosomies generally involve the loss of too much genetic material to be compatible with live birth; monosomy is part of 46,XY,-11,+22. The 46,YY karyotype is also not seen in newborns, for similar reasons.
Alkaptonuria is caused by a deficiency of: A phenylalanine hydroxylase B homogentisic acid C normal colored urine D homogentisic acid oxidase E phenylalanine oxidase
D In individuals affected with alkaptonuria, there is a build-up of homogentisic acid because homogentisic acid oxidase is unable to metabolize this protein to maleylacetoacetic acid. This enzymatic step is part of the pathway involved in the catabolism of phenylalanine and tyrosine to fumaric acid and acetoacetic acid. See Fig. 7.2 in the book for the steps in this pathway.
Hemophilia A and hemophilia B have nearly identical phenotypes, but they result from mutations in different genes on the X chromosome. This is an example of: A Variable expressivity B Compound heterozygosity C Double heterozygosity D Locus heterogeneity E Allelic heterogeneity
D Locus heterogeneity refers to mutations at different genetic loci (genes) that can cause the same or a similar phenotype. This situation fits the definition perfectly, since mutations at either of two loci on the X chromosome can both cause the hemophiliac phenotype.
The cellular phenotype of in-frame insertion or deletion mutations in the ligand binding domain of the LDL-R gene is most likely to be: A Failure to synthesize any LDL-receptor protein B Failure to transport LDL-receptor through the endoplasmic reticulum C Failure of the LDL-receptor to localize in clathrin-coated pits D Failure of the LDL-receptor to bind LDL E Failure of the LDL-receptor to be internalized after binding LDL
D This is a class 3 mutation. The receptor is synthesized, reaches the cell surface but because this class of mutation alters the conformation of the protein, it is unable to bind LDL properly. In this group of proteins, the binding capacity of this mutant protein is 2-30% of the normal binding capacity.
Each of the following have been observed as mechanisms resulting in the activation of a proto-oncogene except: A Capture of the oncogene sequence by a retrovirus B a point mutation altering the function of the oncogene protein product C a chromosome translocation fusing portions of the oncogene and another cellular gene D amplification of an oncogene as small, sub-chromosomal fragments (double minutes) E inactivation of an oncogene by telomerase activity
E Conversion of a proto-oncogene into an oncogene is characterized by activation of the gene so that it loses normal control and functions inappropriately in cell growth. Thus an event that inactivates the gene, like the action of telomerase, would not be expected to activate it.
Confidentially in the physician-patient relationship should be broken in which of the following clinical scenarios? A A young woman has just tested positive for HIV. She begs you not to tell the test result to her boyfriend of 4 years. B Your 23 year-old patient is starting a new job and her employer requests the results of her genetic test for expansion of triplet repeats in the Huntington Disease gene. C You see a patient in the psychiatry outpatient clinic who tells you he has recurring thoughts of murdering his wife, and he is now devising a realistic plan to do so. D All of the above E A and C only
E Genetic testing information should be kept in confidence and not disclosed to a third party such as an employer. The other two examples, though, are scenarios in which physician-patient confidentiality should be broken due to the realistic potential of harm to another person. The correct answer is A and C only.
A chromosomal analysis is obtained on a young woman with mild signs of Turner syndrome and reveals a 46,XX/45,X karyotype. Nondisjunction is most likely to have occurred in: A maternal meiosis I B maternal meiosis II C paternal meiosis I D paternal meiosis II E mitosis after fertilization
E Since there is a cell line in this young woman that has a complete chromosome complement (46,XX) she must have received the appropriate amount of maternal and paternal genetic information at the time of fertilization. The loss would have occurred post-fertilization to generate the 45,X cell line. See p.184 in the book.
Type III collagen is a homo-trimer of a1 (III) chains. If one of the two copies of the COL3A1 gene (which codes for this protein) contains a missense mutation which produces a mutated (but stable) protein, what fraction of type III collagen molecules will contain a mutant subunit and thus be abnormal? A 1/8 B 1/4 C 1/2 D 3/4 E 7/8
E The easy way to find out what fraction of molecules will be abnormal is to find out what fraction of molecules will be normal and subtract from one. This is because there are many abnormal combinations (-++, --+, ---, etc.) but only one normal combination: +++. The fraction of molecules that will be normal is the fraction that gets a molecule derived from the normal copy of the gene three times in a row, or 1/2 x 1/2 x 1/2 = 1/8. The fraction of molecules that will be abnormal is thus 1 - 1/8 or 7/8.
An alpha-thal silent carrier man (aa/a-) mates with an alpha-thal trait woman (aa/--). Assuming no genetic recombination, which of the following is NOT a possible genotype of their offspring? A aa/a- B aa/-- C a-/-- D aa/aa E a-/a-
E a-/a- is not a possible genotype of the offspring. Since a child inherits one allele from each parent, this combination would imply that each parent possessed an a- allele. In fact, only the child's father has such an allele.
Each RNA nucleotide contains an additional 4' hydroxyl (-OH) group not present in DNA. A False B True
False Each RNA nucleotide contains an additional 2' hydroxyl group not present in DNA.
The G8 RFLP marker is closely linked to the Huntington disease (HD) locus, and it is useful for linkage analysis in HD families. True or False: The base changes responsible for the G8 polymorphism are also responsible for HD. A True B False
False Huntington disease is caused by triplet expansion in the coding region of the Huntington disease gene. While the G8 polymorphism is close to the HD gene (because it is linked), the sequence at the G8 locus is independent of whether or not a patient will have HD and can only be used to follow chromosome inheritance in families.
True or False: Huntington's disease results from a trinucleotide repeat expansion in intron 6 of the huntingtin gene. A True B False
False Huntington's disease results from a trinucleotide (CAG) repeat expansion in the 5' end of the huntingtin coding sequence, not in an intron. This leads to a polyglutamine tract in the protein. It is unkown how the expansion leads to disease.
In X chromosome inactivation, all of the X chromosome genes are inactivated. A True B False
False Not all genes on the X chromosome are inactivated during Lyonization. Two genes on the short arm of the X chromosome which are not inactivated are the Xga antigen gene and the steroid sulfatase gene. There are also genes on the long arm of the X chromosome which are not inactivated, because they are needed to maintain ovarian function until menopause.
Robertsonian translocations usually involve metacentric chromosomes. A True B False
False Robertsonian translocations usually involve acrocentric chromosomes, which cluster together during meiosis. Another name for Robertsonian translocation is centric fusion, because the chromosomes fuse at the centomere with loss of the short arms of the chromosomes. This loss is not clinically significant, although there is a signifiant risk of offspring with an unbalanced translocation.
You analyze the sequence for part of the Factor VIII gene in a male patient with hemophilia A, and find a missense base change substituting a Leu for Ile at codon 904. Analysis of the Factor VIII gene of another unrelated individual identifies the exact same base change. True or False: This second individual must also have hemophilia A. A True B False
False The given statement is false. The mutation you identified in the first individual may not be the cause of the hemophilia. In fact, given the conservative nature of the identified change (leucine and isoleucine both have hydrophobic side chains), it would not be surprising to learn that the mutation was unrelated to the hemophilia.
You are doing an emergency medicine rotation as a third year medical student. A 4 year-old boy who has just been in a car accident is rushed in because he has severe blood loss requiring transfusion. The boy's parents refuse to give consent for the transfusion because they are Jehovahs Witnesses. You go to the judge and get a court order for the transfusion citing which of the following principles of medical ethics? A Justice B Nonmaleficence C Paternalism D Confidentiality
Nonmaleficence You are acting out of nonmaleficence. Nonmaleficence is the duty to "first do no harm." Clearly it is harmful to this boy to not receive the blood transfusion. Although the parents object to the transfusion on religious grounds, this patient is to young to make his own decision. You should get the court order to insure that the patient is not harmed.
