Block 7 Michigan Genetic
.JPG "Block 7 Michigan Genetic - Quiz")
Questions and Answers
- 1.
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
Correct Answer
B. Fragile X syndromeExplanation
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).Rate this question:
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- 2.
Myotonic dystrophy may show increasing severity and earlier age of onset in successive generations. This phenomenon is known as:
- A.
Variable expressivity
- B.
Compound heterozygosity
- C.
Locus heterogeneity
- D.
Incomplete penetrance
- E.
Anticipation
Correct Answer
E. AnticipationExplanation
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.Rate this question:
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- 3.
To establish a successful and cost-effective screening program for detecting heterozygous carriers of an autosomal recessive disease, all of the following are essential, EXCEPT:
- A.
The screening test has a positive predictive value of 100%
- B.
A high-risk population can be identified
- C.
Reproductive options are available
- D.
Genetic counseling is provided with the testing
- E.
The disease is severe enough to be clinically significant
Correct Answer
A. The screening test has a positive predictive value of 100%Explanation
Screening tests are aimed at identifying a subset of the population for which more specific diagnostic tests are applicable. Therefore a successful "screening" program does not need to have 100% positive predictive value.Rate this question:
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- 4.
True or False: Presymptomatic diagnosis for Huntington's disease requires DNA analysis of at least one affected relative.
- A.
True
- B.
False
Correct Answer
B. FalseExplanation
Since Huntington's disease (HD) is caused by a single, known mutation, a PCR test can provide direct diagnosis without the need to test other family members. HD is caused by the expansion of a CAG repeat in the coding region of the HD gene, huntingtin. The CAG repeat ranges from 10-30 copies in length in normal chromosomes and from 36-121 copies in length on HD chromosomes.Rate this question:
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- 5.
Below is a family with a normal son, and a daughter with trisomy 13. DNA samples from all four individuals were analyzed for an RFLP on chromosome 13. The results are shown below the pedigree. This RFLP has alleles of 10kb, 8kb, 6kb, and 4kb. Note that the 6kb band on the daughter (II-2) and the 4kb band on the mother (I-2) are of double intensity. Based on these data, in which meiotic division did the non-disjunction event occur?
- A.
Paternal meiosis I
- B.
Paternal meiosis II
- C.
Maternal meiosis I
- D.
Maternal meiosis II
- E.
Maternal meiosis I or maternal meiosis II
Correct Answer
B. Paternal meiosis IIExplanation
The father (I-1) has 2 different alleles at the RFLP analyzed, one 10kb in size and one 6kb in size. The mother (I-2) has the same 4kb allele for this RFLP on each copy of her chromosome 13. The affected daughter (II-2) received two copies of the 6kb allele from her father and one copy of the 4kb allele from her mother. The non-disjunction event must have occurred in paternal meiosis II, as this is the stage when the identical sister chromatids separate.
NOTE: If an affected daughter had received 2 copies of the maternal 4kb allele, it would be impossible to determine if the non-disjunction event had occurred during maternal meiosis I or maternal meiosis II.Rate this question:
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- 6.
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 the a-globin gene
- B.
Is due to a B-globin gene mutation
- C.
Is an acquired disorder, due to a somatic gene mutation
- D.
Is due to a mutation in a gene on chromosome 11, 10 cM centromeric of B-globin
- E.
Is not due to a B-globin gene mutation
Correct Answer
E. Is not due to a B-globin gene mutationExplanation
A negative LOD score ofRate this question:
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- 7.
True or False: In X chromosome inactivation, all of the X chromosome genes are inactivated.
- A.
True
- B.
False
Correct Answer
B. FalseExplanation
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.Rate this question:
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- 8.
The concept that genetic counseling must be nondirective arises directly from respect for which of the following principles of medical ethics?
- A.
Autonomy
- B.
Justice
- C.
Beneficence
- D.
Paternalism
- E.
None of the above
Correct Answer
A. AutonomyExplanation
Genetic counseling must be nondirective to respect the autonomy of the patient. Autonomy is synonymous with self-determination, self-rule, or self-governance. In genetic counseling, the patient should be given as much information as possible so that he/she can make the best decision possible for himself/herself (which will not be the same decision for all patients).Rate this question:
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- 9.
A young woman of northern European descent is the single parent of a child with autosomal recessive cystic fibrosis. She marries her first cousin and becomes pregnant. What is the probability that her child will have cystic fibrosis?
- A.
1/2500
- B.
1/100
- C.
1/32
- D.
1/8
- E.
1/4
Correct Answer
C. 1/32Explanation
To work this problem draw out the family lineage, calculate the risk of having the CF gene at each branch and multiply each risk. The parent of this woman through whom she is related to her first cousin has a 1/2 chance of carrying the mutant CF, this parental sib has a 1/2 chance of carrying the allele and this person's child (the first cousin) also has a 1/2 chance of carrying the mutant allele. Cumulative risk = 1/2 x 1/2 x 1/2 = 1/8 that the first cousin has a mutant allele. Then the risk to have an affected child is 1 x 1/2 x 1/8 x 1/2 = 1/32.Rate this question:
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- 10.
A, B, C, and D are defined in the table below. For each question, match the term with its correct definition. Affected Unaffected Positive Test Result A B Negative Test Result C D Specificity
- A.
C/(C + D)
- B.
B/(B + D)
- C.
A/(A + C)
- D.
D/(B + D)
- E.
B/(A + B)
Correct Answer
D. D/(B + D)Explanation
Specificity is defined as the number of unaffected people who have a negative test result out of the total number of unaffected people. For the table shown, this is D/(B + D).
Specificity can be remembered as "negative in health" or NIH.Rate this question:
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- 11.
A, B, C, and D are defined in the table below. For each question, match the term with its correct definition. Affected Unaffected Positive Test Result A B Negative Test Result C D Sensitivity
- A.
C/(C + D)
- B.
B/(B + D)
- C.
A/(A + C)
- D.
D/(B + D)
- E.
B/(A + B)
Correct Answer
C. A/(A + C)Explanation
Sensitivity if defined as the number of affected individuals who test positive for the disease out of the total number of affected individuals. For the table shown, this is equal to A/(A + C).
A way to remember this is "positive in disease" or PID, because PID (pelvic inflammatory disease) is a sensitive topic.Rate this question:
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- 12.
A, B, C, and D are defined in the table below. For each question, match the term with its correct definition. Affected Unaffected Positive Test Result A B Negative Test Result C D False positive rate
- A.
C/(C + D)
- B.
B/(B + D)
- C.
A/(A + C)
- D.
D/(B + D)
- E.
B/(A + B)
Correct Answer
E. B/(A + B)Explanation
The false positive rate is the number of unaffected individuals who test positive out of the total number of positive test results. For the table shown, the false positive rate is equal to B/(A + B).Rate this question:
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- 13.
A, B, C, and D are defined in the table below. For each question, match the term with its correct definition. Affected Unaffected Positive Test Result A B Negative Test Result C D False negative rate
- A.
C/(C + D)
- B.
B/(B + D)
- C.
A/(A + C)
- D.
D/(B + D)
- E.
B/(A + B)
Correct Answer
A. C/(C + D)Explanation
The false negative rate is the number of affected individuals who test negative divided by the total number of negative test results. This is C/(C + D) for the table shown.Rate this question:
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- 14.
Each of the following have been observed as mechanisms resulting in the activation of a proto-oncogene except:
- A.
A chromosome translocation fusing portions of the oncogene and another cellular gene
- B.
Capture of the oncogene sequence by a retrovirus
- C.
Amplification of an oncogene as small, sub-chromosomal fragments (double minutes)
- D.
Inactivation of an oncogene by telomerase activity
- E.
A point mutation altering the function of the oncogene protein product
Correct Answer
D. Inactivation of an oncogene by telomerase activityExplanation
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.Rate this question:
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- 15.
Which of the following would NOT benefit a patient with B-thalassemia major?
- A.
Increasing fetal Hb production.
- B.
Increasing B-globin production
- C.
Decreasing a-globin production.
- D.
Increasing a-globin production.
- E.
Transfusion and iron chelation therapy
Correct Answer
D. Increasing a-globin production.Explanation
B-thalassemia major is characterized by very little or no -globin. This results in the formation of homotetramers (a4) which precipitate in the RBCs causing RBC destruction in the bone marrow and spleen. Increasing a-globin production would only exacerbate homotetramer formation and worsen the patient's condition.Rate this question:
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- 16.
The results of linkage analysis for marker A and the NF-1 gene are shown below. What is the approximate genetic distance between marker A and the NF-1 gene?
- A.
0.1 cM
- B.
0.5 cM
- C.
1 cM
- D.
10 cM
- E.
30 cM
Correct Answer
D. 10 cMExplanation
The most likely recombination fraction (), is that which gives the highest positive lod score (). For this question, the highest lod score is when =0.10. To change this to centimorgans, use the conversion that 1 cM is equal to a 1% or 0.01 recombination frequency between the two loci. Therefore, =0.10 means that the two loci are separated by a distance of approximately 10 cM.Rate this question:
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.JPG "The results of linkage analysis for marker A and the NF-1 gene are shown below. What is the approximate genetic distance between marker A and the NF-1 gene?")
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