Deck 5: Principles of Clinical Cytogenetics and Genome Analysis
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Deck 5: Principles of Clinical Cytogenetics and Genome Analysis
1
What is needed in determining to assess whether a nondisjunction event occurred at meiosis I or meiosis II?
A) The karyotype
B) Whether the abnormal gamete contributed two copies of the same chromosome or one of each homolog
C) The parent of origin of the nondisjunction
D) Whether or not recombination occurred between the homologs of the nondisjoined chromosome
E) Cannot be determined
A) The karyotype
B) Whether the abnormal gamete contributed two copies of the same chromosome or one of each homolog
C) The parent of origin of the nondisjunction
D) Whether or not recombination occurred between the homologs of the nondisjoined chromosome
E) Cannot be determined
Whether the abnormal gamete contributed two copies of the same chromosome or one of each homolog
2
Which of the following are possible causes of Angelman syndrome?
A) Uniparental disomy for chromosome 15
B) A maternal deletion on chromosome 15q
C) Defects in the imprinting center at chromosome 15q
D) A mutation in the gene for E6-AP ubiquitin protein ligase
E) All of the above
A) Uniparental disomy for chromosome 15
B) A maternal deletion on chromosome 15q
C) Defects in the imprinting center at chromosome 15q
D) A mutation in the gene for E6-AP ubiquitin protein ligase
E) All of the above
All of the above
3
To be stable, a rearranged chromosome must have which of the following?
A) No inverted segments
B) A centromere and two telomeres
C) A centromere
D) A balanced composition (no losses or gains of genetic material)
E) A similarly rearranged homolog in the embryo
A) No inverted segments
B) A centromere and two telomeres
C) A centromere
D) A balanced composition (no losses or gains of genetic material)
E) A similarly rearranged homolog in the embryo
A centromere and two telomeres
4
When two or more chromosome complements can be found within an individual, this is known as:
A) Heterozygosity
B) Mosaicism
C) Genetic heterogeneity
D) Genomic duplication
E) Trisomy
A) Heterozygosity
B) Mosaicism
C) Genetic heterogeneity
D) Genomic duplication
E) Trisomy
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5
From largest to smallest, put these techniques in order in terms of the size of the genetic abnormality each could detect. I. Comparative genome hybridization
II) G-banding
III) Allele specific oligonucleotide hybridization
IV) Fluorescent in situ hybridization (FISH)
A) I, II, III, IV
B) II, IV, III, I
C) IV, II, III, I
D) II, IV, I, III
E) I, II, IV, III
II) G-banding
III) Allele specific oligonucleotide hybridization
IV) Fluorescent in situ hybridization (FISH)
A) I, II, III, IV
B) II, IV, III, I
C) IV, II, III, I
D) II, IV, I, III
E) I, II, IV, III
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6
If a single, 23X sperm fertilizes an ovum that lacks a nucleus, and its chromosomes double to yield a 46XX karyotype, what is the outcome?
A) A normal female
B) A partial mole
C) An ovarian teratoma
D) A complete mole
E) A normal placenta but no fetus
A) A normal female
B) A partial mole
C) An ovarian teratoma
D) A complete mole
E) A normal placenta but no fetus
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7
The inability of one copy of a gene to perform the function of the normal two copies of the gene is known as:
A) Dominant negativity
B) Recessivity
C) Deletion sensitivity
D) Mutation threshold
E) Haploinsufficiency
A) Dominant negativity
B) Recessivity
C) Deletion sensitivity
D) Mutation threshold
E) Haploinsufficiency
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8
What type of chromosomal aberration is indicated by the following cytogenetic abbreviation: 46,XX, t(1;2)(q25;p23)?
A) Translocation
B) Transition
C) Transversion
D) Terminal deletion
E) Loss of telomeres
A) Translocation
B) Transition
C) Transversion
D) Terminal deletion
E) Loss of telomeres
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9
Which of the following is not seen in a normal human karyotype?
A) Acrocentric chromosome
B) Metacentric chromosome
C) Submetacentric chromosomes
D) Satellite sequences
E) Telocentric chromosomes
A) Acrocentric chromosome
B) Metacentric chromosome
C) Submetacentric chromosomes
D) Satellite sequences
E) Telocentric chromosomes
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10
Which of the following is the most common cause of chromosomal aneuploidy?
A) Unbalanced translocation
B) Nonhomologous recombination
C) Chromosomal nondisjunction
D) Sequence inversion
E) Meiotic arrest
A) Unbalanced translocation
B) Nonhomologous recombination
C) Chromosomal nondisjunction
D) Sequence inversion
E) Meiotic arrest
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11
Cells are arrested at which stage of the cell cycle for use in routine karyotyping?
A) Interphase
B) Prophase
C) Metaphase
D) Anaphase
E) Telophase
A) Interphase
B) Prophase
C) Metaphase
D) Anaphase
E) Telophase
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12
Which term is used to describe differences in gene expression based on the parent of origin of an allele?
A) Parental determination
B) Sex-determination
C) Imprinting
D) Sex-linked dominance
E) Heterodisomy
A) Parental determination
B) Sex-determination
C) Imprinting
D) Sex-linked dominance
E) Heterodisomy
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13
This type of chromosome has satellites attached to their short arms by stalks.
A) Acrocentric
B) Metacentric
C) Submetacentric
D) Sex chromosomes
E) Autosomes
A) Acrocentric
B) Metacentric
C) Submetacentric
D) Sex chromosomes
E) Autosomes
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14
To do high resolution banding of chromosomes, what change to the procedure must be made relative to normal G-banding?
A) Use more cells
B) Lengthen the staining procedure
C) Amplify the chromosomes
D) Use a more sensitive stain
E) Arrest the cells earlier in mitosis
A) Use more cells
B) Lengthen the staining procedure
C) Amplify the chromosomes
D) Use a more sensitive stain
E) Arrest the cells earlier in mitosis
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15
What is the term used for a chromosome in which one arm is deleted and the other arm is duplicated in a mirror image fashion?
A) Robertsonian translocation
B) Balanced translocation
C) Ring chromosome
D) Mirrored chromosome
E) Isochromosome
A) Robertsonian translocation
B) Balanced translocation
C) Ring chromosome
D) Mirrored chromosome
E) Isochromosome
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16
Without taking into account the likelihood of each type of segregation, what portion of the possible 2:2 segregation products resulting from meiosis in a carrier of a balanced translocation will be normal?
A) 1/2
B) 1/3
C) 1/4
D) 1/6
E) 1/12
A) 1/2
B) 1/3
C) 1/4
D) 1/6
E) 1/12
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