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1

Adjacent deoxyribonucleotides on one strand of a double-stranded DNA molecule are connected by which of the following types of chemical bonds?

Phosphodiester Bonds

2

p16 is a member of the INK family of CDK inhibitor proteins (CIP). It functions by binding to CDK4 and CDK6. What effect does the action of p16 have on the cell cycle?

A. p16 promotes the associations between CDK4/6 and cyclin D.
B. p16 triggers the release of E2F from RB.
C. p16 blocks passage of cell cycle through restriction point in late G1.
D. p16 promotes the hyperphosphorylation of RB.

C − Binding of p16 to CDK4 and CDK6 prevents them from associating with cyclin D. Without active cyclin D-CDK4/6 complexes, cells cannot pass the restriction point in late G1.

3

A mutation resulting in a nucleotide substitution was observed within the open reading frame of a gene. When the protein produced from this gene was analyzed, it was found that no changes in amino acid sequence were present. Which of the following characteristics of the genetic code accounts for these observations?

A The genetic code is nonoverlapping.
B The genetic code is not punctuated.
C The genetic code is degenerate.
D The genetic code is a triplet code.
E The genetic code is essentially universal among species.

C − The degenerate nature of the genetic code means that some amino acids are represented by multiple codons. For example, phenylalanine is coded for by TTT and TTC. If a mutation alters the third T in the codon to C, then no amino acid change would be observed. This is known as a silent mutation.

4

A strain of bacteria resistant to a specific antibiotic was isolated in a hospital laboratory. Biochemical analysis demonstrated that treatment of the sensitive strain with the antibiotic resulted in a block in bacterial RNA synthesis, whereas in the resistant strain, RNA synthesis was unaffected. The resistant strain is unaffected by which of the following drugs?

A Chloramphenicol
B Tetracycline
C Rifampicin
D Acyclovir
E Amoxicillin

C − Rifampicin inhibits initiation of RNA synthesis by RNA polymerase holoenzyme in bacteria.





A Chloramphenicol is a broad-spectrum antibiotic that inhibits protein synthesis at the elongation phase by inhibiting peptidyl transferase activity of the bacterial ribosome.

B Tetracyclines bind to the small 30S subunit of bacterial ribosomes and inhibit translational elongation.

D Acyclovir is an inhibitor of viral DNA replication; it is not active against bacteria.

E Amoxicillin is a beta-lactum antibiotic that inhibits bacterial cell wall synthesis by blocking the cross-linking of peptidoglycan polymers.

5

A recent report identified a G --> C transversion in the gene encoding the hematopoietic transcription factor GATA1 in the genome of a patient with the congenital syndrome Diamond-Blackfan anemia. The deletion caused a loss of exon 2 in the transcribed mRNA. Biochemical analysis of cells homozygous for this mutation indicated that no functional GATA1 protein was produced. What is the likely location of the nucleotide substitution in the GATA1 genes of this patient?

A 5'-UTR
B 3'-UTR
C Splice donor site
D Promoter
E Enhancer

C − Mutations in intron-exon boundaries that affect splicing tend to cause loss of exons in the transcribed mRNA. In this case, the loss of exon 2 leads to the production of GATA1 that lacks the transactivation domain.

6

A novel drug that binds the 30S subunit of prokaryotic ribosomes was found to have bactericidal activity in laboratory tests and is proposed for clinical trials. Studies were performed in which bacterial cultures were treated with high doses of this drug, followed by biochemical analysis. Drug treatment was found to cause a significant reduction in the initiation phase of protein synthesis. The tested drug has a mechanism of action similar to which of the following antibiotics?

A Cycloheximide
B Puromycin
C Chloramphenicol
D Dihydrostreptomycin
E Tetracycline

D − Streptomycin inhibits protein synthesis by binding to the 16S rRNA of the 30S subunit to disrupt the assembly of the 70S ribosome, mRNA and aminoacyl-tRNA to form the initiation complex.

7

During protein synthesis, the energy for peptide bond formation comes from which of the following chemical bonds?

A The ester bond joining the amino acid and the terminal ribose of tRNA, broken during peptide synthesis

B The phosphodiester bond of adenosine triphosphate (ATP), hydrolyzed by the catalytic rRNA.

C The phosphodiester bond of guanosine triphosphate (GTP), hydrolyzed by an elongation factor

D The phosphodiester bond of GTP, hydrolyzed by an initiation factor

E The phosphodiester bond of GTP, hydrolyzed by a release factor

A − During the charging of tRNAs by aminoacyl tRNA synthetases, ATP is consumed to form an ester bond between the 3'-OH of the terminal adenosine of tRNA. The energy stored in this bond is then consumed to form a peptide bond during protein synthesis.

8

Which type of mutation in a codon would lead to recruitment of release factor to the ribosome when the codon is present in the A site?

D − Nonsense mutations are those that result in the generation of a stop codon in place of an amino acid codon. When a stop codon is present in the A site of a ribosome, it leads to recruitment of release factor, which triggers peptidyl transferase to cleave the ester bond between the peptide's C terminus and the tRNA in the P site.

9

The overexpression of BCL-2 observed in certain types of B cell lymphoma serves as one of the first and strongest forms of evidence that failure of cell death contributes to cancer. Tumors expressing high levels of BCL-2 (and other antiapoptotic family members) are frequently found to be resistant to radiation therapy and chemotherapeutic agents. Early advances in BCL-2 targeted therapy involved the use of antisense technology that aimed at inhibiting BCL-2 expression. One promising antisense oligonucleotide was G3139 (oblimersen). Which experimental result would provide evidence that G3139 is effective at overcoming the antiapoptotic effects of elevated BCL-2 in B cell lymphoma?

A. Organization of the cytoskeleton remains unchanged
B. A decrease in the cleaved form of caspase-3
C. An increase in the cleaved form of caspase-9
D. An increase in BCL-2 mRNA
E. An increase in the number of cells that swell and burst

C − Overexpression of BCL-2 would prevent the cleavage of caspase-9 into its active form. Thus, G3139 may be deemed effective if its use results in an increase in the cleaved form of caspase-9.

10

A 31-year-old man who emigrated from Vietnam and recently returned from a visit to his home country presents in your office with a chief complaint of sore throat, fever, and malaise. Upon physical examination, you observe a gray membrane that covers his tonsils and pharynx. Upon questioning, you find that he has never been vaccinated for diphtheria. In cells exposed to the toxin, which of the following processes related to gene expression are affected as a direct result of the action of the diphtheria toxin?

A Transcriptional initiation
B Transcriptional elongation
C Transcriptional termination
D Translational initiation
E Translational elongation

E − Diphtheria toxin catalyzes the covalent attachment of ADP to the catalytic unit of eEF-G, irreversibly inactivating it. Translocation, and therefore elongation, is blocked in eukaryotic cells.

11

A mutation in a gene coding for a specific type of RNA molecule from which class of eukaryotic RNAs could potentially cause amino acid misincorporations in a wide variety of proteins?

A 18S ribosomal RNA (rRNA)
B mRNA
C Micro RNA (miRNA)
D snRNA
E tRNA

E − tRNA "charging" is governed by "acceptor identity," which constitutes in essence a second genetic code. Specifically, the tRNA synthetase enzymes recognize both the amino acid and its cognate tRNA by sensing a particular three-dimensional conformation. As such, a mutation in a tRNA could potentially alter its structure such that the mutant tRNA is recognized by an incorrect tRNA synthetase that "charges" the mutant tRNA with the wrong amino acid.

12

A physician is exploring the best treatment options for his 56-year-old female patient who has been diagnosed with breast cancer. To show that his patient may be a good candidate for Herceptin, he plans to examine a biopsy of her cancerous tissue for expression levels of which of the following proteins?

A BCR-ABL tyrosine kinase
B HER2/NEU family of receptors
C Insulin receptors
D FAS receptors

B − Herceptin (trastuzumab) is a monoclonal antibody directed against HER2/NEU receptors (which are members of the EGF family of receptors).

13

Six weeks after possible exposure to the human immunodeficiency virus (HIV), a 28-year-old man visits his doctor to request an HIV test. Blood was drawn, and an enzyme-linked immunosorbent assay (ELISA) for specific HIV antigens was positive. What additional antigen-antibody test must be performed to rule out the possibility that the ELISA result was false-positive?

A Southern blot
B Western blot
C DNA microarray
D Fluorescence in situ hybridization (FISH)
E Northern blot

B − The Western blot (immunoblot) is the gold standard for validation of HIV infection and is used to confirm the results of an ELISA. It allows for the detection of some circulating HIV proteins whose levels increase prior to the appearance of anti-HIV antibodies (which are detected in HIV ELISAs).

14

During the elongation phase of prokaryotic translation, for each amino acid added to the growing peptide chain, two GTP molecules are hydrolyzed to GDP. Which of the following aspects of ribosome function require the hydrolysis of one of these GTP molecules?

A Peptide bond formation
B Aminoacyl-tRNA delivery to A site
C Binding of CAP-binding complex to the 7-methyl-guanosine cap of mRNA
D Assembly of the small and large ribosomal subunits

B − One GTP molecule is hydrolyzed by the G protein elongation factor-1 (EF-1 in eukaryotes) when an amino acyl tRNA is inserted into the A site. A second is expended during the translocation process.

15

A 65-year-old woman presents to her physician complaining of fatigue, lack of energy, and weight loss. Physical exam reveals splenomegaly, and laboratory tests indicate an elevation in white blood cell count. Bone marrow analysis reveals the presence of a shortened version of chromosome 22 caused by a reciprocal translocation of DNA between chromosomes 9 and 22.Which of the listed drugs is a suitable first-line therapeutic agent to treat this patient's condition?

A Trastuzumab (Herceptin)
B Imatinib (Gleevec)
C Cetuximab (Erbitux)
D Infliximab (Remicade)
E Basiliximab (Simulect)

B − The patient has chronic myelogenous leukemia (CML), which is driven by the unregulated tyrosine kinase activity of the BCR-ABL protein product of the Philadelphia chromosome (der 22). Imatinib binds to the active site of the ABL tyrosine kinase and inhibits its activity.

16

A 7-methylguanosine residue is found on the 5' end of the product of which of the following enzymes?

A RNA polymerase I
B RNA polymerase II
C RNA polymerase III
D DNA polymerase alpha
E DNA polymerase delta

B − A 7-methylguanosine cap is added to the 5' end of mRNAs following their synthesis. RNA polymerase II synthesizes mRNA in eukaryotic cells.

17

While working in a medical mission in a small village in Honduras, a 7-year-old male child is brought to you by his parents. The parents tell you that the child has always had "very bad skin and been very sick," and whenever they can afford it, they take the boy to a hospital in the nearest city for treatment of his skin lesions. While the child was still an infant, the doctors told the parents that he cannot ever be exposed to sunlight, a restriction that is difficult to impose as both parents work in the fields all day. Upon examination of the boy, you find significant patches of discolored skin with areas of severe blistering on his face, arms, and neck that are oozing and raw (indicative of xeroderma pigmentosum). Molecular analysis of the DNA from the boy's skin keratinocytes would reveal an abundance of which of the following types of lesions?

A Double-stranded breaks
B Pyrimidine dimers
C Chromosomal translocations
D Nucleotide deletions
E Nucleotide substitutions

B − The boy has the genetic condition xeroderma pigmentosum due to a defect in nucleotide excision repair. The most common form of DNA damage resulting from sunlight exposure is the production of thymidine dimers. Nucleotide excision repair recognizes and removes bulky adducts that alter or distort the normal shape of DNA, which includes thymidine dimers.

18

What effect does damage to DNA have on p53 protein?

A Inactivates p53
B Stabilizes p53
C Degrades p53
D Dephosphorylates p53

B − DNA damage triggers the phosphorylation and thus stabilization of p53.

19

Which of the following mutations is an example of a "gain of function" mutation that converts a normal gene (proto-oncogene) into an oncogene that leads to the production of oncoproteins?

A A mutation in BRCA1/BRCA2 that disrupt homologous recombination

B A mutation in p53 that prevents its phosphorylation in response to DNA damage

C A mutation in NF-1 that prevents activated RAS protein from being turned off

D A point mutation in HER2 that allows the protein product to signal in absence of ligand

E A microdeletion in RB that results in loss of control of the G1 checkpoint

D − HER2 is a proto-oncogene that encodes HER2 (a member of the family of epidermal growth factor receptors). A point mutation in HER2 that changes a valine to glutamine converts it into an oncogene whose oncoprotein product is Neu. The dimerization of tyrosine kinase activity of Neu is triggered in the absence of ligand.

20

Valproic acid is a drug used clinically to treat seizures. Biochemically, it is an inhibitor of histone deacetylases. Treating cells with valproic acid will likely alter the charge of which of the following amino acids in histones?

A Aspartate
B Glutamate
C Histidine
D Arginine
E Lysine

E − Histones are rich in the basic amino acids arginine and lysine. Acetylation occurs only on the positively charged amino groups of lysine residues. This neutralizes the basic charge and reduces the affinity of DNA for nucleosomes, making it more accessible to the transcriptional machinery.

21

Which of the following events is characteristic of both the extrinsic and intrinsic pathways of apoptosis?

A Activation of caspase 8

B Cleavage of the proapoptotic factor, BID

C Involvement of Fas-associated death domain protein (FADD)

D Activation of caspases 3, 6, and 7

D − Caspases 3, 6, and 7 are activated by both the extrinsic and intrinsic pathways of apoptosis.

22

The public health department in your state recently reported the isolation of a tetracycline-resistant strain of Treponema pallidum in your area. Biochemical analysis has determined that the resistance is associated with a change in ribosome structure, which alters the ribosome'™s function. Which of the following functions in protein synthesis is most likely to be altered in the tetracycline-resistant strain of T. pallidum?

A Assembly of the 30S preinitiation complex

B Entry of the 50S ribosomal subunit to create the initiation complex

C Binding of aminoacyl-tRNAs to the A site

D Peptidyl transferase activity of the 70S ribosome

E Translocation of the 70S ribosome

C − Tetracylclines bind to the small 30S subunit of bacteria, blocking access of aminoacyl-tRNAs to the A site. An alteration in ribosome structure, perhaps affecting the A site, could potentially disrupt the binding of tetracycline to the ribosome, generating a resistant phenotype.

23

A colleague claims that inhibition of DNA primase would block initiation of new DNA synthesis at origins of replication but would not affect ongoing DNA synthesis. For which of the following reasons is your colleague incorrect?

A DNA synthesis is bidirectional.
B DNA synthesis is semidiscontinuous.
C DNA synthesis requires a primer for DNA polymerase I to extend.
D DNA synthesis requires strand separation by helicase.
E DNA synthesis is semiconservative.

B − DNA polymerase can synthesize DNA only in a 5'-->3' direction. On one strand of the DNA molecule (the lagging strand), the polymerase synthesizes a series of short DNA polymers (Okazaki fragments) that each require a primer. Therefore, new primers synthesized by primase are required for ongoing DNA synthesis, not just when DNA synthesis initiates at origins of replication.

24

Meningitis is a rare and very serious complication of tularemia, which is caused by Francisella tularensis infection. Antibiotic drugs inhibit protein translation in prokaryotes by interfering with the initiation, elongation, or termination phase. One treatment modality for the treatment of tularemia is the antibiotic chloramphenicol in combination with one other antibiotic that inhibits a different phase of protein translation. Which of the following drugs would be the appropriate choice to pair with chloramphenicol?

A Erythromycin
B Puromycin
C Tetracycline
D Streptomycin
E Cycloheximide

D − Streptomycin inhibits protein synthesis at the initiation phase by binding to the small ribosomal subunit and inhibiting assembly of the preinitiation complex. Because chloramphenicol inhibits the elongation phase (by blocking peptidyl transferase activity), the two drugs inhibit different aspects of translation and are synergistic.

25

A nucleotide sequence 5'-TCCGAT-3' within human genomic DNA underwent spontaneous mutation resulting in a nucleotide substitution, producing the sequence 5'-TTCGAT-3'. Which one of the following chemical reactions could account for the nucleotide substitution above?

A Conversion of a deoxyribosyl group to a ribosyl group

B Deamination of a pyrimidine

C Hydrolysis of the N-glycosidic bond in a purine

D Hydrolysis of a phosphodiester bond

E Methylation of a cytosine

B − Deamination of cytosine converts the base to a uracil. Uracil will be replaced by thymidine either by repair mechanisms or during the next round of DNA replication.

26

The DNA translocation generating the Philadelphia chromosome involves which proto-oncogene and is diagnostic for which disease?

A RAS G protein, colon carcinoma

B RAS G protein, chronic myelogenous leukemia

C c-MYC transcription factor, Burkitt'™s lymphoma

D ABL tyrosine kinase, chronic myelogenous leukemia

E p53 transcription factor, chronic myelogenous leukemia

D − The Philadelphia chromosome (der (22)) results from the reciprocal translocation of the abl proto-oncogene from chromosome 9 to chromosome 22 and the BCR region of chromosome 22 to 9. This translocation creates a BCR-ABL fusion gene (and an unregulated tyrosine kinase protein product) and is associated with chronic myelogenous leukemia (CML).

27

Cyclin D—dependent kinases (CDK4/6) are key targets for cancer chemotherapy. The compound palbociclib, or PD 0332991, is a selectively potent inhibitor of CDK4 and CDK6 in retinoblastoma protein (RB)—positive cancer cell lines. Studies show that oral administration of PD 0332991 to mice harboring human tumor xenografts resulted in a broad spectrum of anticancer activity and caused the regression of certain tumors. Which of the following effects would be a consequence of treatment with PD 0332991?

A Progression of tumor cell cycle through the G1-S phase

B Stimulation of RB phosphorylation

C Reduction in the transcription of genes under the control of E2F

D Arrests cells in the G2 phase

E Promotion of E2F release

C − Inhibition of CDK4/6 by PD 0332991 inhibits the phosphorylation of RB in tumor cells. Consequently, E2F remains sequestered by RB and unable to promote the transcription of target genes.

28

Which one of the following antibiotics will inhibit cytoplasmic protein synthesis in both prokaryotic and eukaryotic organisms?

A Erythromycin
B Tetracycline
C Streptomycin
D Chloramphenicol
E Puromycin

E − Puromycin acts upon both eukaryotic and prokaryotic peptidyl transferases, releasing small peptides ending in puromycin at their C-terminal ends.

29

Following a day spent at the beach, which of the following types of DNA damage would most commonly be found in melanocytes within the basal layer of the epidermis?

A Apurinic sites caused by depurination
B Intrastrand cross-links
C Double-strand breaks
D Pyrimidine dimers
E Nucleotide alkylation

D − Ultraviolet (UV) radiation in sunlight most commonly causes the formation of pyrimidine dimers.

30

Fabry disease is an X-linked recessive disorder caused by defects in the activity of the lysosomal enzyme alpha-galactosidase. This results in the accumulation of glycolipids in the plasma and cellular liposomes of the brain, heart, and kidneys. Patients affected by Fabry disease present with raphes, peripheral neuropathy, cardiac disease, and kidney failure. The pedigree below shows inheritance of the X-linked recessive Fabry disease. Which statement is a correct interpretation of this family's pedigree?

A Both males and females of all generations are affected.

B All female descendants of the affected male are carriers.

C Sons of heterozygous mothers have a 50% chance of being affected.

D The genetic mutation cannot be passed on to great-grandchildren (third generation).

C − For X-linked disorders, sons of heterozygous mothers (i.e., carriers) have a 50% chance of having the disorder.