Block 7 Michigan Genetic - Pt 2
.JPG "Block 7 Michigan Genetic - Pt 2")
- 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
- 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?
- ProProfs]( "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?")
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
- 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.
- 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%
- 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.
- 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
- 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?
- ProProfs]( "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?")
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
- 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
- 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
- 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?
- ProProfs]( "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?")
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%
- 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? - ProProfs]( "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?")
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%
- 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 - ProProfs]( "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")
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%
- 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
- 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? - ProProfs]( "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?")
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
.JPG "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?")
.JPG "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?")
.JPG "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?")
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