Final Exam Flashcards
- The direct analysis of the expression levels and post-translational status of an enzyme biomarker involved in a human disease would most likely involve which of the following tools?
A. An antibody microarray.
B. A DNA microarray.
C. A transcript microarray.
D. A gene microarray
A
- Which of the following limits the ability to accurately evaluate an individual’s genomic or proteomic profile in the context of a disease?
A. The quality of the isolated mRNA.
B. The availability of highly specific antibodies.
C. The ability to evaluate all combinations of post-translational modifications.
D. All of the above are correct.
D
- Which of the following molecules are found in the DNA double helix structure?
A. deoxyribose molecules, purine or pyrimidine bases, and phosphates.
B. amino acids, ribose molecules, and phosphates.
C. ribose molecules, peptide bonds, and phosphates.
D. hydrophobic carbon chains, acetyl groups, and glycosidic bonds.
A
Mitotic DNA may contain which of the following components?
A. centromeres
B. telomeres
C. proteins
D. All of the above.
D
Reactivating telomerase activity in a quiescent (non-dividing) cell may:
A. shorten telomeric DNA and cause the cell to start dividing.
B. lengthen telomeric DNA and cause the cell to start dividing.
C. shorten telomeric DNA and keep the cell in quiescence.
D. lengthen telomeric DNA and keep the cell in quiescence.
B
- DNA replication occurs:
A. in a 3’ to 5’ direction on only one strand of the DNA molecule.
B. in a 3’ to 5’ direction on both strands of the DNA molecule.
C. in a 5’ to 3’ direction on both strands of the DNA molecule
D. in a 5’ to 5’ direction on only one strand of the DNA molecule.
C
- Inhibition of DNA synthesis by targeting phosphodiester bond formation would involve which enzyme?
A. DNA primase
B. DNA ligase
C. DNA polymerase
D. DNA helicase
B
- A type II topoisomerase inhibitor such as Etoposide (brand names: Eposin, Etopophos, or Vepesid) would:
A. enhance relaxation of supercoiled DNA.
B. enhance cell proliferation.
C. inhibit double strand breaks in DNA.
D. inhibit single strand breaks in DNA.
C
- Conservative site specific recombination is used to alter gene sequences or regulate gene expression by:
A. excising a specific DNA sequences.
B. inverting a DNA sequences.
C. inserting a DNA sequences into other regions of the chromosome.
D. All of the above are correct.
D
- Deamination of cytosine is a common spontaneous mutation that:
A. results in loss of the base component of the nucleotide.
B. results in a deletion mutant if not repaired.
C. results in a substitution mutant if not repaired.
D. changes cytosine to guanine.
C
- A depurinated nucleotide, if not repaired, would result in which type of mutation:
A. deletion
B. substitution.
C. insertion.
D. All of the above.
A
- Which of the following is true about nucleotide excision repair that repairs a pyrimidine dimer?
A. A DNA glycosylase will remove the incorrect base.
B. DNA nucleases and helicases remove a nucleotide sequence surrounding the pyrimidine dimer.
C. An apurinic endonuclease facilitates the removal of the incorrect sugar phosphate.
D. All of the above.
B
- Which of the following is true about RNA molecules?
A. RNA is always a linear single-stranded sequence.
B. RNA synthesis occurs at the ribosomes in the cytoplasm.
C. RNA can form double-stranded secondary structures.
D. RNA requires 5’ capping for transport into the nucleus.
C
- Which of the following occurs during RNA splicing?
A. the exon regions are removed and intron regions are spliced together.
B. the intron regions are removed and exon regions are spliced together.
C. the exon and intron regions are removed.
D. the exon and intron regions are spliced together.
B
- An operon controls the expression of:
A. A single gene
B. Function-unrelated different genes
C. Function-related different genes
D. None of the above
C
- Why is eukaryotic gene regulation more complex than prokaryotic?
A. Eukaryotic genomes are larger.
B. Eukaryotic organisms are more complex.
C. Cells are differentiated in eukaryotic organisms.
D. All of the above.
D
- What is the nucleosome?
A. The entire complex of the cell’s DNA and associated proteins.
B. Five histones wound together as dimers and tetramers.
C. A unit of chromatin made up of 200 bp of DNA and two copies of four of the histones.
D. 145 bp of DNA and two copies of four of the histones.
C
- The leucine zipper motif contains:
A. Two alpha helices in a coiled-coil fold.
B. Two beta sheets in an anti-parallel beta sheet
C. Both alpha helices and beta sheets.
D. Z-form DNA.
A
- The coactivators function through the following mechanisms.
A. direct binding to target gene DNA
B. bridge the enhancer-bound activator and the promoter-bound RNA polymerase II
C. recruit histone acetylase lead to the modification of chromatin structure.
D. Both B and C
D
- How does tamoxifen act to inhibit the function of the estrogen receptor protein?
A. It binds at the zinc finger domain and prevents DNA binding.
B. It binds in the pocket normally occupied by estrogen and it prevents conformational changes in the protein structure.
C. It completely unfolds the estrogen receptor protein.
D. It dimerizes with estrogen to form an oligomeric complex that then binds to DNA.
B
- Which of the following is NOT considered chromatin remodeling?
A. Binding of remodeling engine complex to acetylated lysine residues
B. Recruitment of a co-activator
C. Lysine acetylation
D. Cleavage of polypeptide bonds.
D
- Which of the following statements about Eukaryotic Transcription is FALSE?
A. DNA from eukaryotic chromosomes is bare.
B. DNA from eukaryotic chromosomes is associated with histones.
C. Histones are small, alpha helical proteins.
D. There are five types of histones.
A
- A typical nuclear receptor protein contains:
A. the DNA-binding domain which determines which target gene to bind with.
B. the ligand-binding domain which interacts with specific ligands
C. the ligand-binding domain which interacts with coactivators after agonistic ligand binding.
D. All of the above
D
- Which of the following statements concerning positive regulation is FALSE?
A. Positive regulation refers to both activator and repressor proteins.
B. Inducer molecules can act to either switch the gene on or off.
C. In positive regulation, an activator protein binds to the DNA and promotes transcription.
D. In positive regulation the activator protein binds to a different region of the DNA than RNA polymerase.
A
- Which of the following post-translational modifications could affect protein interactions with other molecules?
A. Phosphorylation
B. Acetylation
C. Glycosylation
D. All of the above.
D
- Similarities among different protein kinases include:
A. a region that binds ATP.
B. a region that binds GTP.
C. a region that binds CTP.
D. All of the above.
A
- Protein folding is monitored by associated heat shock (Hsp) proteins. Misfolded proteins may likely undergo which of the following fates?
A. Acetylation followed by targeting to the proteasome.
B. Glycosylation followed by targeting to the endoplasmic reticulum.
C. ubiquitination followed by targeting to the ribosome.
D. ubiquitination followed by targeting to the proteasome.
D
- Which of the following is true in regards to protein glycosylation?
A. Glycosylation occurs in the endoplasmic reticulum (ER) and Golgi complex.
B. Glycosylation involves adding and removing sugar molecules.
C. Glycosylation modifications may alter the charge of the protein.
D. All of the above.
D
- Proteins that are found only in the nucleus would likely have which of the following features?
A. Exposed leucine-rich amino acid residues.
B. Exposed lysine and arginine-rich amino acid residues.
C. The ability to be translated on ribosomes in the nucleus.
B. Tight interactions with heat shock proteins that block access to lysine and arginine-rich sequences
B
- I-cell disease is associated with defects in GlcNac (N-acetylglucosamine) transferases that lead to the accumulation of metabolic waste products in the lysosome. Which of the following is true about the cause of I-cell disease?
A. Decreased GlcNac transferase activity increases the pH of the lysosome.
B. Decreased GlcNac transferase activity in the lysosome causes a decreased pH in the lysosome.
C. Decreased GlcNac transferase activity in the Golgi causes a decrease in glysosylated lysosome enzymes.
D. Decreased GlcNac transferase activity increases enzyme transport to the lysosome.
C
- I-cell disease is associated with defects in GlcNac (N-acetylglucosamine) transferases that lead to the accumulation of metabolic waste products in the lysosome. Which of the following is true about the cause of I-cell disease?
A. Decreased GlcNac transferase activity increases the pH of the lysosome.
B. Decreased GlcNac transferase activity in the lysosome causes a decreased pH in the lysosome.
C. Decreased GlcNac transferase activity in the Golgi causes a decrease in glysosylated lysosome enzymes.
D. Decreased GlcNac transferase activity increases enzyme transport to the lysosome.
C