Block 7 Michigan Genetic - Pt 2

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| By Rossstudent
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Block 7 Michigan Genetic - Pt 2 - Quiz

Questions and Answers
  • 1. 

    Sickle cell anemia is:

    • A.

      An autosomal dominant disease.

    • B.

      Always caused by the same point mutation in the beta-globin gene.

    • C.

      Rarely due to the same mutation in unrelated individuals.

    • D.

      Caused by mutations in either the alpha-globin gene or the beta-globin gene

    • E.

      An X-linked recessive disease

    Correct Answer
    B. Always caused by the same point mutation in the beta-globin gene.
    Explanation
    Sickle cell anemia is always caused by the same point mutation in the beta-globin gene: an A->T transversion in the second nucleotide of the sixth codon. This point mutation leads to the substitution of valine for glutamate at the corresponding position in the beta-globin protein. Note that valine, a hydrophobic amino acid, has very different properties from those of glutamic acid, a hydrophilic amino acid. The substitution of a more chemically similar residue, like aspartic acid, would probably not lead to the dramatic phenotype seen in sickle cell anemia: the aggregation of hemoglobin molecules under deoxygenating conditions.

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  • 2. 

    The pedigree below is from a family with cystic fibrosis, an autosomal recessive condition. What is the best estimate that individual I-3 is a carrier of cystic fibrosis?

    • A.

      1/4

    • B.

      1/3

    • C.

      1/2

    • D.

      2/3

    • E.

      3/4

    Correct Answer
    C. 1/2
    Explanation
    The chance that individual I-3 is a carrier of cystic fibrosis is approximately 1/2. Since II-3 is affected, I-2 must be a carrier. I-2 inherited the mutant allele from one of her parents, and thus her sister (I-3) had a parent who carried the mutant allele. I-3's chance of inheriting this mutant allele, as opposed to the probably normal allele on the other chromosome, is ½

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  • 3. 

    Which of the following observations is the strongest evidence for an important genetic component in the causation of type-1 diabetes mellitus (IDDM)?

    • A.

      Pancreatic B-cell autoantibodies are frequently present

    • B.

      Approximately 10% of affected individuals have an affected sibling.

    • C.

      Onset of disease is usually in childhood

    • D.

      The concordance rate in monozygotic twins is approximately 30%.

    • E.

      The concordance rate in monozygotic twins is five times that in dizygotic twins.

    Correct Answer
    E. The concordance rate in monozygotic twins is five times that in dizygotic twins.
    Explanation
    Remember that monozygotic (MZ) twins have identical genotypes and dizygotic (DZ) twins share only 50% of their genetic information. Therefore if a trait is non-genetically determined then there should be no difference in the concordance rate (how often members of a twin pair show the same phenotype) between MZ and DZ twin pairs. If a trait has a genetic component then the rate of concordance is expected to be higher in MZ twins then in DZ twins.

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  • 4. 

    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%

    Correct Answer
    C. 10%
    Explanation
    When looking at the inheritance of multifactorial traits, as the number of affected children within a single family increases, the recurrence risk also increases.
    In Mendelian traits the number of affected in a family did not change the recurrence risks. But multiple affected children DOES change the recurrence risk for MULTIFACTORIAL TRAITS. The presence of one affected child means the parents probably are midway between the mean for affected and the mean of the normal population, but the presence of a second affected child means they probably are closer to the threshold, and hence, have a higher recurrence risk should they choose to have another child.

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  • 5. 

    Ankylosing spondylitis is a chronic inflammatory arthritis affecting the spine and sacroiliac joints. 95% of Caucasian patients are positive for the HLA-B27 haplotype; whereas, 7% of all Caucasians are positive. This is evidence for:

    • A.

      Association of ankylosing spondylitis with the B27 allele of the HLA-B locus.

    • B.

      Linkage of ankylosing spondylitis to the HLA-B locus.

    • C.

      Neither.

    Correct Answer
    A. Association of ankylosing spondylitis with the B27 allele of the HLA-B locus.
    Explanation
    Ankylosing spondylitis is associated with the B27 allele of the the HLA B locus. As this questions describes, the HLA-B27 haplotype is found with much higher frequency among affected indiviuals than unaffected individuals; this is the definition of association. The cause of ankylosing spondylitis is unknown and likely to be multifactorial. For ankylosing spondylitis to be linked to the HLA-B locus, a gene causing the disease would have to be located close enough to the HLA-B locus to have recombination events occur between the two loci less often than by chance (50%).

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  • 6. 

    Which karyotype would be MOST frequently seen in liveborn infants (as opposed to spontaneous abortions)?

    • A.

      47,XX,+21

    • B.

      47,XX,+3

    • C.

      69,XXX

    • D.

      46,XY,-11,+22

    • E.

      46,YY

    Correct Answer
    A. 47,XX,+21
    Explanation
    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.

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  • 7. 

    The family below is segregating a very rare autosomal recessive disease with 100% penetrance. The disease is present and recognizable at birth and does not decrease fitness. Assume no locus heterogeneity. What is the chance that the fetus (IV-1) is affected with this disease?

    • A.

      1/4

    • B.

      1/2

    • C.

      1/9

    • D.

      2/9

    • E.

      4/9

    Correct Answer
    C. 1/9
    Explanation
    This disease is characterized as being "very rare"; from this, one can assume that I-1 and I-4 are not carriers. I-2 and I-3 must each carry two mutant alleles, because this is an autosomal recessive disease. From this information about generation I, it can be deduced that everyone in generation II (II-1, II-2, II-3, II-4) is a carrier.
    Neither III-1 nor III-2 is affected. Therefore, they each have a 2/3 chance of being a carrier. To calculate the chance that the fetus is affected with the disease, there is a 2/3 chance of each parent being a carrier and a 1/2 chance of inheriting the mutant allele from each parent. Thus the probability of the fetus being affected is: 2/3 x 1/2 x 2/3 x 1/2 = 1/9.

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  • 8. 

    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

    Correct Answer
    B. False
    Explanation
    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.

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  • 9. 

    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.

      (49/50)2

    • B.

      (1/50)2

    • C.

      1/100

    • D.

      1/50

    • E.

      2/50

    Correct Answer
    E. 2/50
    Explanation
    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.

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  • 10. 

    The proband, III-1, has cystic fibrosis (CF). DNA analysis indicates that III-1 is homozygous for a particular CF mutant, but his grandmother (I-2) is homozygous normal. Assume that q=1/50 for this mutant allele, and that the population is in Hardy-Weinberg equilibrium. What is the chance of I-1 being a carrier of CF?

    • A.

      100%

    • B.

      50%

    • C.

      4%

    • D.

      2%

    • E.

      0.5%

    Correct Answer
    A. 100%
    Explanation
    Cystic fibrosis is an autosomal recessive disorder, so for III-1 to be affected he must have received a mutant allele from both of his parents, II-2 and II-3, who must be carriers. II-2 and II-3 must have received a mutant allele from one of their parents; since we know I-2 is homozygous normal, the mutant allele must have come from I-1. The correct answer is therefore 100%.

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  • 11. 

    The proband, III-1, has cystic fibrosis (CF). DNA analysis indicates that III-1 is homozygous for a particular CF mutant, but his grandmother (I-2) is homozygous normal. Assume that q=1/50 for this mutant allele, and that the population is in Hardy-Weinberg equilibrium. What is the chance of II-5 being a carrier?

    • A.

      100%

    • B.

      50%

    • C.

      4%

    • D.

      2%

    • E.

      0.5%

    Correct Answer
    C. 4%
    Explanation
    To calculate the chance that II-5 is a carrier, it is important to observe in the pedigree that II-5 is not a part of the family that has an affected member. Rather, II-5 is marrying into this family from the population at large, so the chance of her being a carrier must be calculated using the equations describing carrier frequency for a population in Hardy-Weinberg equilibrium. The heterozygote carrier frequency is 2pq = 2 x 49/50 x 1/50 = 98/2500 = approximately 4%.

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  • 12. 

    The proband, III-1, has cystic fibrosis (CF). DNA analysis indicates that III-1 is homozygous for a particular CF mutant, but his grandmother (I-2) is homozygous normal. Assume that q=1/50 for this mutant allele, and that the population is in Hardy-Weinberg equilibrium. What is the chance of III-2 being affected with CF

    • A.

      100%

    • B.

      50%

    • C.

      4%

    • D.

      2%

    • E.

      0.5%

    Correct Answer
    E. 0.5%
    Explanation
    To calculate the chance that III-2 is affected, one must determine the probability that each parent is a carrier and then the chance of inheriting both mutant alleles. II-4 has a 1/2 chance of being a carrier, because his father is a carrier and there is a 1/2 chance of inheriting that mutant allele from him (note: we know is mother is homozygous normal). As calculated by the Hardy-Weinberg equilibrium equation, there is approximately a 2/50 chance that II-5 is a carrier. To calculate III-2's chance of being affected, we must multiple the probabilities of his parents being carriers but 1/2 probability of inheriting the mutant allele from each parent (1/2 x 1/2 or 1/4 probability of inheriting both mutant alleles). So, overall we have 1/2 x 1/2 x 2/50 x 1/2 = 1/200 = 0.5%

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  • 13. 

    Prader-Willi syndrome (PWS) can result from either an interstitial deletion involving the paternal copy of chromosome subregion 15q1-q13 or from maternal uniparental disomy of chromosome 15. The reason for this is:

    • A.

      The maternal copy of the gene(s) responsible for Angelman syndrome (AS) is imprinted and is not expressed.

    • B.

      PWS results from an anomaly of X-chromosome inactivation

    • C.

      The paternal copy of the gene(s) responsible for PWS is imprinted and is not expressed.

    • D.

      The maternal copy of the gene(s) responsible for PWS is imprinted and is not expressed

    • E.

      The maternal copy of the gene(s) responsible for PWS exerts a dominant negative effect of the paternal allele

    Correct Answer
    D. The maternal copy of the gene(s) responsible for PWS is imprinted and is not expressed
    Explanation
    Prader-Willi Syndrome (PWS) is caused by genomic imprinting, defined as the differential expression of maternal and paternal alleles. When the information in region 15q11-q13 is derived only from a mother (either via uniparental disomy (both chromosomes from a single parent) or deletion on the paternal chromosome), the maternally imprinted chromosome is unable to express its genetic information and the result is PWS.

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  • 14. 

     Familial retinoblastoma (FRB) is an autosomal dominant cancer predisposition syndrome, due to a defect in the RB tumor suppressor gene. An affected individual typically inherits a single defective copy from one parent, and a normal copy from the other parent. Below are Southern blots for an individual with FRB; both normal and tumor cells were typed for three RFLP markers flanking the RB locus on chromosome 13 (N = normal cells, T = tumor cells). Based upon these data, which of the following is the most likely explanation for the loss of heterozygosity in this individual?

    • A.

      Mitotic crossover

    • B.

      Loss of the normal chromosome 13

    • C.

      Independent second mutation

    • D.

      Loss of the normal chromosome 13 and reduplication of the mutant chromosome 13

    Correct Answer
    A. Mitotic crossover
    Explanation
    The Southern blot results show a single band of double intensity for Marker 1 and Marker 2. For Marker 3, the normal and tumor cells have the same banding pattern. These experimental results suggest a mitotic crossover event occurred in the region containing Marker 1 and Marker 2, but not Marker 3.

    Southern blots for an independent second mutation mechanism would show the same bands for both the normal and tumor cells for all three markers. Loss of the normal chromosome 13 would result in a single band of weak intensity for all three markers. Finally, loss of the normal chromosome 13 and reduplication of the mutant chromosone 13 would result in a single band of strong intensity for all three markers.

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  • Current Version
  • Mar 21, 2023
    Quiz Edited by
    ProProfs Editorial Team
  • Mar 30, 2012
    Quiz Created by
    Rossstudent
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