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(MB) ASCP Practice Exam/155 Questions And Answers 2024
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(MB) ASCP Practice Exam/155Questions And Answers 2024
Which of the following is not a component of a nucleotide?
Phosphate group
Anti-codon
Ribose sugar
Nitrogen base - -Anti-codon
-According to Chargaff's rule of base pairing, adenine pairs with: - -Thymine
-What genes would be screened in a breast cancer panel? - -HER2, ERBB2,
BRCA1
-Next Generation Sequencing uses: - -Short sequence reads
-Purines and pyrimidines differ from each other in that: - -Purines have two
rings; pyrimidines have one ring
-The purines are:
Cytosine and uracil
Adenine and thymine
Thymine and cytosine
Adenine and guanine - -Adenine and guanine
-What is the rate of mutation per round of DNA replication?
1 in 1,000 base pairs
1 in 10,000 base pairs
1 in 1,000,000 base pairs
1 in 1,000,000,000 base pairs - -1 in 1,000,000,000 base pairs
-The rate of DNA migration through an agarose gel during electrophoresis
does not depend on which of the following factors?
Net charge of the molecule
Size of the molecule
Shape of the molecule
Nucleotide sequence of the molecule - -Nucleotide sequence of the molecule
-What are the phases in a qPCR Amplification Plot?
Initiation, exponential, plateau
Baseline, exponential, plateau
Baseline, threshold, exponential, plateau
Baseline, initiation, threshold, exponential, plateau - -Initiation, exponential,
plateau
,-Find the palindrome in this restriction enzyme site: 5'-CTGCAG-3'?
5'-GAC
3'-GAC
3'-CTG
5'-GTC - -3'-GAC
-A patient with impaired judgment, personality changes, signs of abnormal
body movements and depression comes to the physician's office for a follow-
up visit. The physician suspects a single-gene disorder may be the cause of
those clinical manifestations. A blood specimen was then sent to your clinical
laboratory for mutation screening in the Huntington gene. Testing with
standard PCR indicates that patient has Huntington Disease, HD. Which of
the following would be consistent with this diagnosis?
25 CAG repeats in the Huntington gene
85 CAG repeats in the Huntington gene
25 CGA repeats in the Huntington gene
85 CGA repeats in the Huntington gene - -85 CAG repeats in the Huntington
gene
-Which two HPV types are responsible for most cases of cervical cancer?
16 and 18
31 and 59
16 and 58
44 and 59 - -16 and 18
-Replication forks, known as origins of DNA replication, are created by this
enzyme:
Ligase
Taq Polymerase
Primase
Helicase - -Helicase
-Mutation in what gene is associated with Fragile X syndrome? - -FMR1
-Mantle cell lymphoma (MCL) is caused by what translocation? - -t(11;14)
-This polymerase is involved in "initiation of DNA replication and has
primase activity": - -Pol α
-Its discovery shed light on why there is simultaneous, though not
continuous, synthesis of DNA on both leading and lagging strands of DNA:
Klenow fragment of DNA polymerase
Okazaki fragments
Sanger fragments
, RNA fragments - -Okazaki fragments
-What gene is measured following treatment with imatinib (Gleevec)?
FLT3
BCR/Abl
Jak2
MAPK - -BCR/Abl
-What is the rate of mammalian DNA replication?
500 nucleotides per second
100 nucleotides per second
50 nucleotides per second
10 nucleotides per second - -50 nucleotides per second
-This polymerase is involved in "replicates mitochondrial DNA": - -Pol γ
-A patient with impaired judgment, personality changes, signs of abnormal
body movements and depression comes to the physician's office for a follow-
up visit. The physician suspects a single-gene disorder may be the cause of
those clinical manifestations. A blood specimen was then sent to your clinical
laboratory for mutation screening in the Huntingtin gene. Which of these
methods would best accomplish this task?
Methylation-specific PCR
Standard PCR
PFGE
RAPD PCR - -Standard PCR
-Which of the following storage options is optimal for storing isolated DNA
for a period greater than seven years?
22-25ºC
2-8ºC
-20ºC
-70ºC - --70ºC
-Consider a hypothetical mutation involving gene X. Let's say you amplify a
specific exon, say exon 11, of that gene then you cut it with restriction
enzyme W. In a person without the mutation, cutting the gene with
restriction enzyme W generates two fragments of sizes, 100 bp and 250 bp.
Suppose a C>T mutation in gene X deletes a restriction site, yielding a
fragment of 350 bp. You would expect a heterozygous person for gene X to
have these fragments on a restriction gel:
+/+ = 350 bp; 250 bp; 100 bp;
m/+ = Only the 350 bp
m/+ = 350 bp; 250 bp; 100 bp
m/m = 350 bp; 250 bp - -m/+ = 350 bp; 250 bp; 100 bp
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