Reading...
Play button
Play button
Use LEFT and RIGHT arrow keys to navigate between flashcards;
Use UP and DOWN arrow keys to flip the card;
H to show hint;
A reads text to speech;
74 Cards in this Set
- Front
- Back
|
The best definition of an endonuclease is an enzyme that hydrolyzes:
A. A nucleotide from only the 3’-end of an oligonucleotide. B. A nucleotide from either terminal of an oligonucleotide. C. A phosphodiester bond located in the interior of a polynucleotide. D. A bond only in a specific sequence of nucleotides. E. A bond that is distal to the base that occupies the 5’-position of the bond. |
C. A phosphodiester bond located in the interior of a polynucleotide.
|
|
|
A palindrome is a sequence of nucleotides in DNA that:
A. Is highly reiterated. B. Is part of the introns of eukaryotic genes. C. Is a structural gene. D. Has local symmetry and may serve as a recognition site for various proteins. E. Has the information necessary to confer antibiotic resistance in bacteria. |
D. Has local symmetry and may serve as a recognition site for various proteins.
|
|
|
The Z DNA helix:
A. Has fewer base pairs per turn than the B DNA. B. Is favored by an alternating GC sequence. C. Tends to be found at the 3’-end of genes. D. Is inhibited by methylation of the bases. E. Is a permanent conformation of DNA. |
B. Is favored by an alternating GC sequence.
|
|
|
A nucleosome:
A. Is a regulatory repeating structure of DNA and histone proteins. B. Has a core of DNA with proteins wrapped around the outside. C. Uses only one type of histone per nucleosome. D. Is separated by a second nucleosome by nonhistone proteins. E. Has histones in contact with the major groove of the DNA. |
A. Is a regulatory repeating structure of DNA and histone proteins.
|
|
|
RNA:
A. Incorporates both modified and unmodified bases during transcription. B. Does not exhibit any double helical structure. C. Structures exhibit base-stacking and hydrogen-bonded base-pairing. D. Usually contains about 65-100 nucleosides. E. Does not exhibit Watson-Crick base-pairing. |
C. Structures exhibit base-stacking and hydrogen-bonded base-pairing.
|
|
|
Ribozymes:
A. Are any ribonucleoprotein particles. B. Are enzymes whose catalytic function resides in RNA subunits. C. Carry out self-processing reactions but cannot be considered true catalysts. D. Require a protein cofactor to form a peptide bond. E. Function only in the processing of mRNA. |
B. Are enzymes whose catalytic function resides in RNA subunits.
|
|
|
Nearly every process in which DNA participates requires that the DNA interact with proteins. Interaction with proteins depends on the local three-dimensional structure of DNA. One antitumor drug, cisplatin, is a tetracoordinate platinum complex. It forms intrastrand cross-links with DNA, causing the double helix to strongly bend toward the major groove. Cellular processes such as transcription and programmed cell death are affected.
Bent DNA: A. Occurs only in the presence of external agents like the antitumor drugs. B. May be a fundamental element in the interaction between DNA sequences and proteins. C. Occurs primarily in the presence of triple-stranded DNA. D. Requires the presence of inverted repeats. E. Occurs only in DNA that is in the Z-form. |
B. May be a fundamental element in the interaction between DNA sequences and proteins.
|
|
|
Nearly every process in which DNA participates requires that the DNA interact with proteins. Interaction with proteins depends on the local three-dimensional structure of DNA. One antitumor drug, cisplatin, is a tetracoordinate platinum complex. It forms intrastrand cross-links with DNA, causing the double helix to strongly bend toward the major groove. Cellular processes such as transcription and programmed cell death are affected.
Another unusual form of DNA is a triple-stranded complex. Triple-stranded DNA: A. Generally occurs in DNA in regions that play no role in transcription. B. May involve Hoogsteen hydrogen bonding. C. Is characterized by the presence of a string of alternating purine-pyrimidine bases. D. Forms only intermolecularly. E. Assumes a cruciform conformation. |
B. May involve Hoogsteen hydrogen bonding.
|
|
|
Telomeres are guanine-rich repetitive sequences at the ends of linear eukaryotic chromosomes. They are progressively shortened during each cycle of cell division until a critical length is reached and apoptosis occurs. The ribonucleoprotein, teloisomerase, maintains the length of telomeres for a finite number of cell divisions and then it is turned off, but some of the tumor cells maintain telomerase activity. The higher the tumor telomerase activity, the poorer the clinical prognosis is. A current chemotherapeutic approach being studied is selective inhibition of telomerase. Certain drugs selectively bind to G-quadruplex DNA structure in telomerase and have been shown to inhibit telomerase activity. These drugs are now being studied for antitumor activity.
9. G-quadrulpex DNA: A. Is a stacked structure of guanine tetramers. B. Is stabilized by metal cations. C. Involves bases held together by Hoogsteen hydrogen bonds. D. Can form either parallel or antiparallel arrangements. E. All of the above are correct. |
E. All of the above are correct.
|
|
|
Telomeres are guanine-rich repetitive sequences at the ends of linear eukaryotic chromosomes. They are progressively shortened during each cycle of cell division until a critical length is reached and apoptosis occurs. The ribonucleoprotein, teloisomerase, maintains the length of telomeres for a finite number of cell divisions and then it is turned off, but some of the tumor cells maintain telomerase activity. The higher the tumor telomerase activity, the poorer the clinical prognosis is. A current chemotherapeutic approach being studied is selective inhibition of telomerase. Certain drugs selectively bind to G-quadruplex DNA structure in telomerase and have been shown to inhibit telomerase activity. These drugs are now being studied for antitumor activity.
Another approach to cancer treatment is drugs that inhibit topoisomerases. Topoisomerases: A. Regulate the level of superhelicity of DNA in the cells. B. Always break one strand of DNA. C. Can create but not remove supercoils. D. Must hydrolyze ATP for their action. E. Of the subclass gyrases, introduce negative superhelices in eukaryotic DNA. |
A. Regulate the level of superhelicity of DNA in the cells.
|
|
|
Eukaryotic DNA contains many sequences that can be repeated a few times, for certain coding genes, to millions of times for some relatively short sequences. Abnormal expansion of reiterated 3-bp DNA sequences can lead to a number of diseases. Fragile X syndrome, which causes mental retardation, is characterized by expansion of a GCC triplet near the FMR-1 gene, from a normal 30 copies to thousands of copies. Diseases of this type generally increase in severity with each successive generation. One possible mechanism of expansion involves slippage in DNA during synthesis of the lagging strand.
Slipped, mispaired DNA occurs when the DNA region has: A. Direct repeats. B. Homopurine-homopyrimidine sequences. C. Inverted repeats D. Mirror repeats. E. Palindromes. |
A. Direct repeats.
|
|
|
Eukaryotic DNA contains many sequences that can be repeated a few times, for certain coding genes, to millions of times for some relatively short sequences. Abnormal expansion of reiterated 3-bp DNA sequences can lead to a number of diseases. Fragile X syndrome, which causes mental retardation, is characterized by expansion of a GCC triplet near the FMR-1 gene, from a normal 30 copies to thousands of copies. Diseases of this type generally increase in severity with each successive generation. One possible mechanism of expansion involves slippage in DNA during synthesis of the lagging strand.
Normally, certain kinds of reiterated sequences occur in a chromosome as an interspersion pattern that is: A. Highly repetitive DNA sequences. B. The portion of DNA composed of single copy DNA. C. Alu sequences. D. Alternating blocks of single copy DNA and moderately repetitive DNA. E. Alternating blocks of short interspersed repeats and long interspersed repeats. |
D. Alternating blocks of single copy DNA and moderately repetitive DNA.
|
|
|
Replication:
a. Is semiconservative b. Requires only proteins with DNA polymerase activity c. Uses 5’ to 3’ polymerase activity to synthesize one strand and 3’ to 5’ polymerase activity to synthesize the complementary strand d. Requires a primer in eukaryotes but not in prokaryotes e. Must begin with an excision step |
a. Is semiconservative
|
|
|
In eukaryotic DNA replication:
a. Only one replisome forms because there is a single origin of replication b. The Okazaki fragments are 1000 to 2000 nucleotides in length c. Helicase dissociates from DNA as soon as the initiation bubble forms d. FEN1 (flap endonuclease 1) is involved in removing the primer e. The process occurs throughout the cell cycle |
d. FEN1 (flap endonuclease 1) is involved in removing the primer
|
|
|
All of the following statements about telomerase are correct except:
a. The RNA component acts as a template for the synthesis of a segment of DNA b. It adds telomerase to the 5’ ends of the DNA strands c. It provides a mechanism for replicating the ends of linear chromosomes d. It recognizes a G-rich single strand of DNA e. It is a reversible transcriptase |
b. It adds telomerase to the 5’ ends of the DNA strands
|
|
|
A transition mutation:
a. Occurs when a purine is substituted for a pyrimidine, or vice-versa b. Results from the insertion of one or two bases into the DNA strand c. Decreases in frequency in the presence of base analogs d. Results from substitution of one purine for another or of one pyrimidine for another e. Always is a missense mutation |
d. Results from substitution of one purine for another or of one pyrimidine for another
|
|
|
Homologous recombination:
a. Occurs only between two segments from the same DNA molecule b. Requires that a specific DNA sequence be present c. Requires one of the duplexes undergoing recombination be nicked in both strands d. May result in strand exchange by branch migration e. Is catalyzed by transposases |
d. May result in strand exchange by branch migration
|
|
|
All of the following are true about transpositions except:
a. Transposons move from one location to a different one within a chromosome b. Both the donor and target sites must be homologous c. Transposons have insertion sequences that are recognized by transposases d. The transposon may either be excised and moved, or be replicated with the replicated piece moving e. They may either activate or inactivate a gene |
b. Both the donor and target sites must be homologous
|
|
|
Retroviruses, like HIV which causes AIDS, have their genetic information in the form of RNA. Reverse transcriptase synthesizes a DNA copy of the viral genome. One drug used in treating AIDS is AZT, an analog of deoxythymidine, which has an azido group at the 3’ position of the sugar. It can be phosphorylated and competes with dTTP for incorporation into the reverse transcript. Once incorporated, its presence terminates chain elongation.
The growing chain is terminated because: a. The analog cannot hydrogen bond to RNA b. The presence of the AZT analog inhibits the proofreading ability of reverse transcriptase c. AZT does not have a free 3’ – OH d. The analog causes distortion of the growing chain inhibiting reverse transcriptase e. dTTP can no longer be added to the growing chain |
c. AZT does not have a free 3’ – OH
|
|
|
Retroviruses, like HIV which causes AIDS, have their genetic information in the form of RNA. Reverse transcriptase synthesizes a DNA copy of the viral genome. One drug used in treating AIDS is AZT, an analog of deoxythymidine, which has an azido group at the 3’ position of the sugar. It can be phosphorylated and competes with dTTP for incorporation into the reverse transcript. Once incorporated, its presence terminates chain elongation.
There is a window in which the effect is primarily on viral replication since AZT is much less effective at competing with dTTP for incorporation by cellular DNA polymerases because of the proofreading ability of DNA polymerases. Proofreading activity to maintain the fidelity of DNA synthesis: a. Occurs after the synthesis has been completed b. Is a function of 3’ to 5’ exonuclease activity intrinsic to or associated with DNA polymerases c. Requires the presence of an enzyme separate from the DNA polymerases d. Occurs in prokaryotes but not eukaryotes e. Is independent of the polymerase activity in prokaryotes |
b. Is a function of 3’ to 5’ exonuclease activity intrinsic to or associated with DNA polymerases
|
|
|
: Patients with the rare disease xeroderma pigmentosum (XP) are very sensitive to light and are highly susceptible to skin cancers. Study of such patients has enhanced our knowledge of DNA repair because XP is caused by defective DNA repair – nucleotide excision repair. (A variant, XP-V, is deficient in post-replication repair.)
All of the following are true about nucleotide excision repair except: a. Removal of the damaged bases occurs on only one strand of the DNA b. It removes thymine dimmers generated by UV light c. It involves the activity of an excision nuclease, which is an endonuclease d. It requires a polymerase and a ligase e. Only the damaged nucleotides are removed |
e. Only the damaged nucleotides are removed
|
|
|
Patients with the rare disease xeroderma pigmentosum (XP) are very sensitive to light and are highly susceptible to skin cancers. Study of such patients has enhanced our knowledge of DNA repair because XP is caused by defective DNA repair – nucleotide excision repair. (A variant, XP-V, is deficient in post-replication repair.)
Another type of DNA repair is base excision repair. Base excision repair: a. Is used only for bases that have been deaminated b. Uses enzymes called DNA glycosylases to generate an abasic sugar site c. Removes about 10 to 15 nucleotides d. Does not require an endonuclease e. Recognizes a bulky lesion |
b. Uses enzymes called DNA glycosylases to generate an abasic sugar site
|
|
|
Interfering with topoisomerases is one way of inhibiting DNA replication. Certain antibiotics target DNA gyrase (type II topoisomerase) of E. coli inhibiting catalytic activity. Topoisomerase poisons prevent resealing of the phosphodiester bond, leaving covalent protein-DNA junctions. These compounds are used in treating infections and as chemotherapeutic agents.
DNA gyrase: a. ATP subunits hydrolyze ATP to form new phosphodiester bonds b. Removes negative supercoils c. Swivelase subunits create and reseal transient nicks on both strands d. Increase the linking number e. Occurs in eukaryotes as well as prokaryotes |
c. Swivelase subunits create and reseal transient nicks on both strands
|
|
|
Interfering with topoisomerases is one way of inhibiting DNA replication. Certain antibiotics target DNA gyrase (type II topoisomerase) of E. coli inhibiting catalytic activity. Topoisomerase poisons prevent resealing of the phosphodiester bond, leaving covalent protein-DNA junctions. These compounds are used in treating infections and as chemotherapeutic agents.
All of the following are correct about double strand break in DNA except they: a. Can lead to loss of genetic material b. Are always involved in homologous recombination c. Are involved in nonhomologous recombination d. Are associated with a heterodimer (Ku) in mammals e. Can lead to mutations or improper regulation of gene expression |
b. Are always involved in homologous recombination
|
|
|
In Eukaryotic transcription:
a. RNA polymerase does not require a template b. All RNA is synthesized in the nucleolus c. Consensus sequences are the only known promoter elements d. Phosphodiester bond formation is favored because there is pyrophosphate hydrolysis e. RNA polymerase requires a primer |
d. Phosphodiester bond formation is favored because there is pyrophosphate hydrolysis
|
|
|
The Sigma subunit of prokaryotic RNA polymerase:
a. Is part of the core enzyme b. Binds the antibiotic rifampicin c. Is inhibited by α-amanitin d. Must be present for transcription to occur e. Specifically recognizes promoter sites |
e. Specifically recognizes promoter sites
|
|
|
Termination of a prokaryotic transcript:
a. Is a random process b. Requires the presence of the rho subunit of the holoenzyme c. Does not require rho factor if the end of the gene contains a G-C rich palindrome d. Is most efficient if there is an A-T rich segment at the end of the gene e. Requires an ATPase in addition to the rho factor |
c. Does not require rho factor if the end of the gene contains a G-C rich palindrome
|
|
|
Eukaryotic transcription:
a. Is independent of the presence of upstream consensus sequences b. May involve a promoter located within the region transcribed rather than upstream c. Requires a separate promoter region for each of the three ribosomal RNAs transcribed d. Requires that the entire gene be in the nucleosome form of chromatin e. Is affected by enhancer sequences only if they are adjacent to the promoter |
b. May involve a promoter located within the region transcribed rather than upstream
|
|
|
All of the following are correct about a primary transcript in eukaryotes except that it:
a. Is usually longer than the functional RNA b. May contain nucleotide sequences that are not present in functional RNA c. Will contain no modified bases d. Could contain information for more than one RNA molecule e. Contains a TATA box |
e. Contains a TATA box
|
|
|
In the cellular turnover of RNA:
a. Repair is more active than degradation. b. Regions of extensive base pairing are more susceptible to cleavage c. Endonucleases may cleave the molecule starting at either the 5’-or 3’end d. The products are nucleotides with a phosphate at either the 3’ or 5’ OH group e. All species except rRNA are cleaved. |
d. The products are nucleotides with a phosphate at either the 3’ or 5’ OH group
|
|
|
Fragile X syndrome is a common form of inherited mental retardation. The mutation in the disease allows the increase of CGG repeat in a particular gene from a normal of about 30 repeats to 200-1000 repeats. This repeat is normally found in the 5’-untranslated region of a gene for the protein FMR1. FMR1 might be involved in the translation of brain-specific mRNAs during brain development. The consequence of the very large number of CGG repeats in the DNA is extensive methylation of the entire promoter region of the FMR1 gene.
Methylation of bases in DNA usually: a. facilitates the binding of transcription factors to the DNA. b. makes a differences in activity only if it occurs in an enhancer region. c. prevents chromatin from unwinding. d. inactivates DNA for transcription e. results in increased production of the product of whatever gene is methylated. |
d. inactivates DNA for transcription
|
|
|
Fragile X syndrome is a common form of inherited mental retardation. The mutation in the disease allows the increase of CGG repeat in a particular gene from a normal of about 30 repeats to 200-1000 repeats. This repeat is normally found in the 5’-untranslated region of a gene for the protein FMR1. FMR1 might be involved in the translation of brain-specific mRNAs during brain development. The consequence of the very large number of CGG repeats in the DNA is extensive methylation of the entire promoter region of the FMR1 gene.
Transcription of eukaryotic genes requires the presence of a promoter and usually the presence of enhancers. An enhancer: a. is a consensus sequence in DNA located where RNA polymerase first binds. b. may be located in various places in different genes. c. may be on either strand of DNA in the region of the gene. d. functions by binding RNA polymerase. e. stimulates transcription in al prokaryotes and eukaryotes. |
b. may be located in various places in different genes.
|
|
|
The synthesis of normal adult hemoglobin (HbA) requires the coordinated synthesis of B-globin and B-globin. B-Thalassemia is a genetic disease leading to a deficiency of B-globin and an inability of the blood to deliver oxygen properly. B-Thalassemia can result from a wide variety of mutations.
All of the following could lead to lack of a or nonfunctional B-globin except: a. a frameshift mutation leading to premature termination of protein synthesis. b. mutation in the promoter region of the B-globin gene. c. mutation toward the 3’-end of the B-globin gene that codes for the polyadenylation site. d. mutation in the middle of the intron that is not an A base. e. all of the above could lead to this result. |
d. mutation in the middle of the intron that is not an A base.
|
|
|
The synthesis of normal adult hemoglobin (HbA) requires the coordinated synthesis of B-globin and B-globin. B-Thalassemia is a genetic disease leading to a deficiency of B-globin and an inability of the blood to deliver oxygen properly. B-Thalassemia can result from a wide variety of mutations.
One mutation leading to the B-Thalassemia occurs at a splice junction. Which of the following statements about removing introns is/are correct? a. small nuclear ribonucleoproteins (snRNP) are necessary for removing introns. b. the consensus sequences at the 5’- and 3’-ends of the introns are identical. c. removal of an intron does not require metabolic energy. d. the exon at one end of an intron must always be joined to the exon at its other end. e. the nucleoside at the end of the intron first released forms a bond with a 3’-OH group on one of the nucleosides within the intron. |
a. small nuclear ribonucleoproteins (snRNP) are necessary for removing introns.
|
|
|
Protooncogenes produces products that have specific roles in regulation growth and differentiation of normal cells. Mutations can turn these genes into oncogenes whose products are less responsive to normal control. Unmutated protein p53, a tumor suppressor, is a transcription factor, inhibiting some genes and activation others. P53 inhibits genes with TAT sequences and activates genes for DNA repair.
The TATA sequence: a. occurs about 25bp downstream from the start of transcription. b. binds directly to RNA polymerase. c. binds transcription factors which bind RNA polymerase. d. bind p53. e. is an enhancer sequence. |
c. binds transcription factors which bind RNA polymerase.
|
|
|
Protooncogenes produces products that have specific roles in regulation growth and differentiation of normal cells. Mutations can turn these genes into oncogenes whose products are less responsive to normal control. Unmutated protein p53, a tumor suppressor, is a transcription factor, inhibiting some genes and activation others. P53 inhibits genes with TAT sequences and activates genes for DNA repair.
Transcription-coupled DNA repair: a. occurs because a DNA lesion causes the RNA polymerase to stall during transcription. b. occurs when a DNA lesion forms in a region of DNA which is highly compacted. c. leads to a normal transcript because transcription continues after the repair is made. d. occurs because both the template and coding strands of the DNA have lesions. e. all of the above are correct. |
a. occurs because a DNA lesion causes the RNA polymerase to stall during transcription.
|
|
|
Degeneracy of the genetic code denotes the existence of:
A. multiple codons for a single amino acid. B. codons consisting of only two bases. C. base triplets that do not code for any amino acid. D. different systems in which a given triplet codes for different amino acids. E. Codons that include one or more of the “unusual” bases. |
A. multiple codons for a single amino acid.
|
|
|
In the formation of an aminoacyl-tRNA:
A. ADP and Pi are products of the reaction. B. aminoacyl adenylate appears in solution as a free intermediate. C. aminoacyl-tRNA synthetase is believed to recognize and hydrolyze incorrect aminoacyl-tRNA’s it may have produced. D. separate aminoacyl-tRNA synthetases exist for every amino acid in the functional protein. E. there is a separate aminoacyl-tRNA synthetase for every tRNA species. |
C. aminoacyl-tRNA synthetase is believed to recognize and hydrolyze incorrect aminoacyl-tRNA’s it may have produced.
|
|
|
During initiation of protein synthesis:
A. methionyl-tRNA appears at the A site of the 80 S initiaton complex. B. eIF-3 and the 40 S ribosomal subunit participate in forming a preinitiation complex. C. eIF-2 is phosphorylated by GTP. D. the same methionyl-tRNA is used as is used during elongation. E. a complex of mRNA, 60 S ribosomal subunit, and certain initiation factors is formed. |
B. eIF-3 and the 40 S ribosomal subunit participate in forming a preinitiation complex.
|
|
|
During the elongation stage of eukaryotic protein synthesis:
A. the incoming aminoacyl-tRNA binds to the P site. B. a new bond synthesized by peptidyl transferase requires GTP hydrolysis. C. the peptidyl-tRNA is translocated to a different site on the ribosome. D. streptomycin can cause premature release of the incomplete peptide. E. peptide bond formation occurs by the attack of the carboxyl group of the incoming amino acyl-tRNA on the amino group of the growing peptide chain. |
C. the peptidyl-tRNA is translocated to a different site on the ribosome.
|
|
|
Formation of mature insulin includes all of the following except:
A. removal of a signal peptide. B. folding in to a three-dimensional structure. C. disulfide bond formation. D. removal of a peptide from an internal region. E. γ-carboxylation of glutamate residues. |
E. γ-carboxylation of glutamate residues.
|
|
|
Chaperones:
A. are always required to direct the folding of proteins. B. when bound to protein increase the rate of protein degradation. C. usually bind to strongly hydrophilic regions of unfolded proteins. D. sometimes maintain proteins in an unfolded state to allow passage through membranes. E. foster aggregation of proteins into plaques. |
D. sometimes maintain proteins in an unfolded state to allow passage through membranes.
|
|
|
Cystic fibrosis is a frequent genetic disease of Caucasians. The CF gene codes for a protein called the cystic fibrosis transmembrane conductance regulator (CFTR) which functions as a cAMP-regulated chloride channel. The protein has two membrane-spanning domains, two domains that interact with ATP and one regulatory domain. The most common defect is deletion of three consecutive bases in the gene for one of the ATP binding domains. The result is a protein that does not fold correctly in the endoplasmic reticulum, is not properly glycosylated, and is not transported to the cell surface. Rather, it is degraded in the cytosol within proteasomes.
The particular mutation described above most likely leads to a protein that: A. must be longer than it should be. B. must be several amino acids shorter than it should be. C. would always have a different amino acid sequence from the point of mutation to the end of the protein compared to the unmutated situation. D. has a substitution of one amino acid for another compared to the protein from the unmutated gene. E. has the deletion of one amino acid compared to the protein from the unmutated gene. |
E. has the deletion of one amino acid compared to the protein from the unmutated gene.
|
|
|
Cystic fibrosis is a frequent genetic disease of Caucasians. The CF gene codes for a protein called the cystic fibrosis transmembrane conductance regulator (CFTR) which functions as a cAMP-regulated chloride channel. The protein has two membrane-spanning domains, two domains that interact with ATP and one regulatory domain. The most common defect is deletion of three consecutive bases in the gene for one of the ATP binding domains. The result is a protein that does not fold correctly in the endoplasmic reticulum, is not properly glycosylated, and is not transported to the cell surface. Rather, it is degraded in the cytosol within proteasomes.
Targeting a protein to be degraded within proteasomes usually requires ubiquitin. In the function of ubiquitin all of the following are true except: A. ATP is required for activation of ubiquitin. B. a peptide bond forms between the carboxyl terminal of ubiquitin and an Є-amino group of a lysine. C. linkage of a protein to ubiquitin does not always mark it for degradation. D. the N-terminal amino acid is one determinant of selection for degradation. E. ATP is required by the enzyme that transfers the ubiquitin to the protein to be degraded. |
E. ATP is required by the enzyme that transfers the ubiquitin to the protein to be degraded.
|
|
|
Collagen is unusual in its amino acid composition and requires a wide variety of posttranslational modifications to convert it to a functional molecule. Because of the complexity of collagen synthesis, there are many diseases, resulting in structural weaknesses in connective tissue, caused by defects in the process. Scurvy leads to a less stable collagen lacking sufficient hydroxyproline.
9. 4-Hydroxylation of specific prolyl residues during collagen synthesis requires all of the following except: A. Fe2+. B. a specific amino acid sequence at the site of hydroxylation. C. ascorbic acid. D. co-hydroxylation of lysine. E. individual α chains, not yet assembled into a triple helix. |
D. co-hydroxylation of lysine.
|
|
|
Collagen is unusual in its amino acid composition and requires a wide variety of posttranslational modifications to convert it to a functional molecule. Because of the complexity of collagen synthesis, there are many diseases, resulting in structural weaknesses in connective tissue, caused by defects in the process. Scurvy leads to a less stable collagen lacking sufficient hydroxyproline.
Much of procollagen formation occurs in the endoplasmic reticulum and Golgi apparatus which requires signal peptide. All of the following statements about targeting a protein for the ER are true except: A. signal peptide usually has a positively charged N-terminus and a stretch of hydrophobic amino acids. B. signal peptide emerging from a free ribosome, binds signal recognition particle (SRP). C. signal peptide is usually cleaved from the protein before the protein is inserted into the ER membrane. D. docking protein is actually an SRP receptor and serves to bind the SRP to the ER. E. SRP and docking protein do not enter the ER lumen but are recycled. |
C. signal peptide is usually cleaved from the protein before the protein is inserted into the ER membrane.
|
|
|
Since protein synthesis is necessary for cells to survive and reproduce, any interference with this will have profound effects. Many antibiotics and toxins function by interfering with protein synthesis. Antibiotics selective for prokaryotic protein synthesis are useful clinically. Agents affecting eukaryotic synthesis are not.
Streptomycin binds the small subunit of prokaryotic ribosomes and: A. causes premature release of the incomplete peptide. B. prevents binding of the 40S and 60S subunits. C. interferes with initiation of protein synthesis. D. inhibits peptidyl transferase activity. E. acts as an N-glycosidase. |
C. interferes with initiation of protein synthesis.
|
|
|
Since protein synthesis is necessary for cells to survive and reproduce, any interference with this will have profound effects. Many antibiotics and toxins function by interfering with protein synthesis. Antibiotics selective for prokaryotic protein synthesis are useful clinically. Agents affecting eukaryotic synthesis are not.
Diphtheria toxin: A. acts catalytically. B. releases incomplete polypeptide chains from the ribosome. C. activates translocase. D. prevents release factor from recognizing termination signals. E. attacks the RNA of the large subunit. |
A. acts catalytically.
|
|
|
Development of recombinant DNA methodologies is based upon discovery of:
A. the polymerase chain reaction (PCR). B. restriction endonucleases. C. plasmids. D. complementary DNA (cDNA). E. yeast artificial chromosomes (YACs). |
(B)
The ability to cleave DNA predictably at specific sites is essential to recombinant DNA technology. A, C, D, and E. These techniques or structures are involved in recombinant DNA technology but came later. |
|
|
Construction of a restriction map of DNA requires all of the following except:
A. partial hydrolysis of DNA. B. complete hydrolysis of DNA. C. electrophoretic separation of fragments on a gel. D. staining of an electrophoretic gel to locate DNA. E. cyclic heating and cooling of the reaction mixture. |
(E)
Cyclic heating and cooling is part of the PCR process, not of restriction mapping. A, B: Restriction mapping involves all degrees of hydrolysis. Partial hydrolysis gives fragments of varying sizes, and complete hydrolysis gives the smallest possible fragments. C, D: Fragments are electrophoretically separated by size on agarose gel, which is stained to reveal the DNA. |
|
|
Preparation of recombinant DNA requires:
A. restriction endonucleases that cut in a staggered fashion. B. restriction endonucleases that cleave to yield blunt-ended fragments. C. poly(dT). D. DNA ligase. E. cDNA. |
(D)
DNA ligase covalently connects fragments held together by interaction of cohesive ends. A: This is the most desirable type of restriction endonuclease to use, but it is not essential. B: Restriction nucleases that make blunt cuts can also be used if necessary. C: This is used in conjunction with poly dA if restriction endonucleases that make blunt cuts are employed, but it is not essential to all of recombinant DNA preparation. E: cDNA is only one type that can be used. |
|
|
In the selection of bacterial colonies that carry cloned DNA in plasmids, such as pBR322, that contain two antibiotic resistance genes:
A. one antibiotic resistance gene is nonfunctional in the desired bacterial colonies. B. untransformed bacteria are antibiotic resistant. C. both antibiotic resistance genes are functional in the desired bacterial colonies. D. radiolabeled DNA or RNA probes play a role. E. none of the above. |
(A)
The foreign DNA is inserted into one antibiotic resistance gene, thus destroying it. B: Resistance is due to the plasmids. C: See the comment for A above. D: Radiolabeling detects the DNA of interest, not the colonies that contain cloned DNA. |
|
|
A technique for defining gene arrangement in very long stretches of DNA (50-100 kb) is:
A. RFLP. B. chromosome walking. C. nick translation. D. Southern blotting. E. SSCP. |
(B)
A: Restriction fragment length polymorphism (RFLP) is a characteristic of DNA, not a technique C: Nick translation is used to label DNA during chromosome walking. D: Southern blotting is a method for analyzing DNA. E: Single-strand conformation polymorphism (SSCP) is a method for detecting base changes in DNA that do not alter restriction endonuclease sites. |
|
|
Antisense nucleic acids:
A. complementary to mRNA would enhance translation. B. could result if a gene is inserted downstream of a promoter but in opposite direction to normal. C. can have no clinical uses. D. react only with DNA. E. are necessary for recombinant DNA technology. |
(B)
This would put the antisense strand of DNA under control of the promoter with subsequent transcription of antisense mRNA. A: Antisense nucleic acid binds mRNA blocking translation. C: If not now, in the future. Antisense DNA nucleotides have been produced that inhibit viral infection. D: React with both DNA and RNA. |
|
|
In this country, a major cause of death of babies during the first year is sudden infant death syndrome (SIDS). One study showed a strong correlation for an increased risk of SIDS with a prolonged QT interval in their electrocardiograms. In one child, a gene associated with the Long QT syndrome had a substitution of AAC for TCC. This gene codes for a protein associated with the sodium channel. The mutation was detected by a single-strand conformation polymorphism (SSCP).
Which of the following statements about SSCP is/are correct? A. The electrophoretic mobility on polyacrylamide gel of small, single-stranded DNA during the SSCP technique depends partly on the secondary conformation. B. There must be a restriction endonuclease site in the region studied for SSCP to work. C. Radiolabeling is not used in this technique. D. It is not necessary to know the sequence of the DNA to be studied. E. All of the above are correct. |
(A)
A single base substitution usually modifies the conformation enough to shift the mobility as detected by SSCP. B: SSCP is the method of choice if there is no restriction endonuclease site. C: DNA is amplified by PCR in the presence of radiolabeled nucleotides. There has to be a detection method for the bands. D: SSCP requires prior knowledge of the sequence. |
|
|
In this country, a major cause of death of babies during the first year is sudden infant death syndrome (SIDS). One study showed a strong correlation for an increased risk of SIDS with a prolonged QT interval in their electrocardiograms. In one child, a gene associated with the Long QT syndrome had a substitution of AAC for TCC. This gene codes for a protein associated with the sodium channel. The mutation was detected by a single-strand conformation polymorphism (SSCP).
In order to get enough DNA to analyze, DNA is amplified by a polymerase chain reaction (PCR). The essential property of the DNA polymerase employed is that it: A. does not require a primer. B. is unusually active. C. is thermostable. D. replicates double-stranded DNA. E. can replicate both eucaryotic and procaryotic DNA. |
(C)
PCR requires cycling between low temperatures, where hybridization of template DNA and oligomeric primers occurs, and high temperatures, where DNA melts. The Taq DNA polymerase, isolated from a thermophilic organism discovered in a hot spring, is stable at high temperatures, and it makes the cycling possible with no addition of fresh polymerase after each cycle. |
|
|
The purpose of gene therapy is to introduce a normal gene into cells containing a defective gene. The first authorized human gene therapy was given to a 4-year-old girl with adenosine deaminase deficiency (ADA). ADA children have a severe combined immune deficiency (SCID) and usually die within the first few years from overwhelming infections. A modified retrovirus was constructed to contain the human ADA gene which could be expressed in human cells without replication of the virus. The child’s isolated T cells were “infected” with the retrovirus to transfer the normal gene to them. The modified T cells were reintroduced into the patient. The girl is now 13 years old and doing well.
Expression of recombinant genes in mammalian cells: A. will not occur if the gene contains an intron. B. occurs most efficiently if cDNA is used in the vector. C. does not require directional insertion of the gene into the vector. D. requires that the vector have enhancer and promoter elements engineered into the vector. E. does not require that the vector have an origin of replication (ori). |
(D)
Eukaryotic systems require controlling elements, which are not necessary in bacterial systems. A: Expression may be improved if an intron is present. B: cDNA does not possess the required controlling elements. C: The gene must be inserted in the proper orientation relative to the control elements. E: The vector must replicate, so it needs the sequence to promote replication. |
|
|
The purpose of gene therapy is to introduce a normal gene into cells containing a defective gene. The first authorized human gene therapy was given to a 4-year-old girl with adenosine deaminase deficiency (ADA). ADA children have a severe combined immune deficiency (SCID) and usually die within the first few years from overwhelming infections. A modified retrovirus was constructed to contain the human ADA gene which could be expressed in human cells without replication of the virus. The child’s isolated T cells were “infected” with the retrovirus to transfer the normal gene to them. The modified T cells were reintroduced into the patient. The girl is now 13 years old and doing well.
Another clinical use for recombinant DNA technology is to have rapidly replicating bacteria produce large amounts of specific proteins (e.g., hormones). Expression of a eucaryotic gene in prokaryotes involves: A. a Shine-Delgarno (SD) sequence in mRNA. B. absence of introns. C. regulatory elements upstream of the gene. D. a fusion protein. E. all of the above. |
(E)
A: the SD sequence is necessary for the bacterial ribosome to recognize the mRNA. B: Bacteria do not have the intracellular machinery to remove introns from mRNA. C: Appropriate regulatory elements are necessary to allow the DNA to be transcribed. D: A fusion protein may be a product of the reaction. |
|
|
Whole genomic DNA was isolated from three individuals, digested separately with restriction endonuclease to fragments and the fragments separated on agarose gel in an electric field. The gene of interest was isolated and analyzed using Southern blot technique. Each individual sample showed two bands. The bands were identical for two of the individuals. For the third, one band was identical to one band of the other two, but the second band was of lower molecular weight than the others. This is an example of restriction fragment length polymorphism (RFLP).
A reasonable explanation for this RFLP is that the gene in the third individual: A. lacked the recognition sequence for the restriction endonuclease. B. had a mutation at the cleavage site. C. had an additional recognition site for the endonuclease. D. had a deletion of a segment of the gene. E. underwent a random cleavage. |
(D)
This would account for one band being of lower molecular weight. A, B, E: The cleavage site has a specific recognition sequence C: This would yield more than two bands. |
|
|
Whole genomic DNA was isolated from three individuals, digested separately with restriction endonuclease to fragments and the fragments separated on agarose gel in an electric field. The gene of interest was isolated and analyzed using Southern blot technique. Each individual sample showed two bands. The bands were identical for two of the individuals. For the third, one band was identical to one band of the other two, but the second band was of lower molecular weight than the others. This is an example of restriction fragment length polymorphism (RFLP).
The Southern blot technique: A. transfers DNA fragments from agarose gel to a nitrocellulose filter. B. requires that the DNA fragments remain double-stranded. C. requires that the DNA is radiolabeled prior to addition to agarose gel. D. alters the position of the DNA fragments during the process. E. amplifies the amount of DNA material. |
(A)
This occurs by elution from the gel by a high salt solution. B: DNA is hydrolyzed by alkali and single-strand DNA binds to the filter. C: The probe to identify the bands on the filter is radiolabeled. D: The positions on the filter are identical to the gel. E: Doesn’t happen. |
|
|
The X-ray autoradiogram of one strand of a fragment of DNA sequenced by the Sanger method was obtained. The 5’-end of the nucleotide was labeled with 32P. Numbering from bottom to top of the autoradiogram, the lane from the ddG tube showed bands at position 4 and 11; ddC tube had bands as 1,2,7,8,9, and 13; ddA at 5,10, and 15; ddT at 3,6,12, and 14. What is the sequence of the fragment? Construct the autoradiogram pattern of the complementary sequence.
|
The sequence is C C T G A T C C C A G T C T A reading 5’ to 3’. The autoradiograph of the complementary strand should have bands for ddG at 1, 2, 7, 8, 9, and 13; ddC at 4 and 11; ddA at 3, 6, 12, and 14; ddT at 5, 10, 15.
|
|
|
What would be the best vector to use to carry a segment of DNA that is 250 kb in size? Would any other type of vector work? Why?
|
YACs (yeast artificial chromosomes) would be the best vector since they can accept DNA between 200 and 500 kb. Cosmids accept the next largest piece, but it is only 45 kb. Plasmid and bacteriophages accept much smaller pieces.
|
|
|
Full expression of the lac operpon requires:
A. lactose and cAMP B. allolactose and cAMP C. lactose alone D. allolactose alone E. absence or inactivation of the lac operon |
B. allolactose and cAMP
|
|
|
In an operon:
A. each gene of the operon is regulated independently B. control may be exerted via induction or via repression C. operator and promoter may be trans to the genes they regulate D. the structural genes are either not expressed at all or they are fully expressed E. control of gene expression consists exclusively of induction and repression |
B. control may be exerted via induction or via repression
|
|
|
The E.coli lacZYA region will be upregulated if:
A. there is a defect in binding of the inducer to the product of the lacI gene B. glucose and lactose are both present, but the cell cannot bind the CAP protein C. glucose and lactose are both readily available in the growth medium D. the operator has mutated so it can no longer bind repressor E. the lac corepressor is not present |
D. the operator has mutated so it can no longer bind repressor
|
|
|
All of the following describe an operon except:
A. control mechanism for eukaryote genes B. includes structural genes C. expected to code for polycistronic mRNA D. contains control sequences such as an operator E. can have multiple promoters |
A. control mechanism for eukaryote genes
|
|
|
In bacteria, amino acid starvation is associated with the production of guanosine tetraphosphate and guanosine pentaphosphate. This situation is referred to as:
A. attenuation B. corepression C. repression D. self-regulation E. stringent response |
E. stringent response
|
|
|
Tryptophan as a corepressor for the trp operon binds to the:
A. operator B. promoter C. repressor protein D. RNA polymerase E. upstream region from the promoter |
C. repressor protein
|
|
|
In eukaryotic transcription by RNA polymerase II, formation of a preinitiation complex:
A. begins with the binding of a protein (TRB) to the TATA box of the promoter B. involves the ordered addition of several transcription factors and the RNA polymerase C. allows an ATP-dependent opening of the two strands of DNA D. requires that the C-terminal domain of RNA polymerase II not be phosphorylated E. all of the above are correct |
E. all of the above are correct
|
|
|
Enhancers:
A. are sequences in the promoter that bind to hormone-transcription factor complexes B. are more effective the closer they are to the TATA box C. may be thousands of base pairs away from preinitiation complex assembly D. must be upstream of the site of the preinitiaton complex assembly E. bind transcription factors that act directly with the initiation complex |
C. may be thousands of base pairs away from preinitiation complex assembly
|
|
|
The problem of pathogenic bacteria becoming resistant to a large number of antibiotics is a serious public health concern. A bacterial strain in a patient being treated with an antibiotic may suddenly become resistant not only to that antibiotic but to others as well even though it has not been exposed to the other antibiotics. This occurs when the bacteria acquire a plasmid from another strain that contains several different transposons.
All of the following phrases describe transposons except: A. a means for the permanent incorporation of antibiotic resistance into the bacterial chromosome. B. contain short inverted terminal repeat sequences C. code for an enzyme that synthesizes guanosine tetraphosphate and guanosine pentaphosphate, which inhibit further transposition D. include at least one gene that codes for a transposase E. contain varying numbers of genes |
C. code for an enzyme that synthesizes guanosine tetraphosphate and guanosine pentaphosphate, which inhibit further transposition
|
|
|
The problem of pathogenic bacteria becoming resistant to a large number of antibiotics is a serious public health concern. A bacterial strain in a patient being treated with an antibiotic may suddenly become resistant not only to that antibiotic but to others as well even though it has not been exposed to the other antibiotics. This occurs when the bacteria acquire a plasmid from another strain that contains several different transposons.
In the operation of transposons: A. typically the transposon moves from its original site and relocates to a different site B. a duplicated transposon must be inserted into the same DNA molecule as the original C. all transposons are approximately the same size D. the insertion sites must be in a concensus sequence E. the transposase may recognize the repetitive ends of the transposon and participate in the cleavage of the recipient site |
E. the transposase may recognize the repetitive ends of the transposon and participate in the cleavage of the recipient site
|
|
|
Genes present in a region of DNA which is in the highly condensed heterochromatin form cannot be transcribed. To be transcribed, this region of DNA must change to a more open structure which may occur by acetylation of histones by an acetyltransferase like CBP, CREB binding protein. CREB is a transcription factor. Rubinstein-Taybi syndrome is caused by mutations of the CBP gene. Rubinstein-Taybi patients are mentally retarded and have other developmental abnormalities, the severity of which depends on the extent of the mutation.
In chromatin: A. a nucleosome consists of four molecules of histones surrounding a DNA core B. DNA positioned so that the major and minor grooves are on the outside of the nucleosome is more accessible for transcription than if the grooves face the interior C. DNA must be completely removed from the nucleosome structure for transcription D. linker DNA is the only DNA capable of binding transcription factors E. the histone octamer consists of eight different kinds of histone proteins |
B. DNA positioned so that the major and minor grooves are on the outside of the nucleosome is more accessible for transcription than if the grooves face the interior
|
|
|
Genes present in a region of DNA which is in the highly condensed heterochromatin form cannot be transcribed. To be transcribed, this region of DNA must change to a more open structure which may occur by acetylation of histones by an acetyltransferase like CBP, CREB binding protein. CREB is a transcription factor. Rubinstein-Taybi syndrome is caused by mutations of the CBP gene. Rubinstein-Taybi patients are mentally retarded and have other developmental abnormalities, the severity of which depends on the extent of the mutation.
Acetylation of histones can lead to a more open DNA structure by: A. weakening the electrostatic attraction between histones and DNA B. causing histones to interact with the C-terminal domain (CTD) of RNA polymerase C. causing electrostatic repulsion between histones and DNA D. facilitating methylation of DNA E. attracting transcription factors to DNA |
A. weakening the electrostatic attraction between histones and DNA
|
