Bio 99 Midterm 2 Tatiane

A merodiploid is: A. a haploid organism that contains a second copy of part of its genome. B. a haploid organism that is missing part of its genome. C. a diploid organism that is missing part of its genome. D. a diploid organism that contains a third copy of its genome.

A

How does the mismatch repair pathway identify the incorrect base in E. coli? A. The error is found on the unmethylated DNA strand. B. The error is found on the DNA strand identified by the presence of nicks. C. The error is found on the methylated DNA strand. D. The error is found on the DNA strand identified by the absence of nicks.

A

How is it possible that regions thousands of base pairs from a promoter can impact transcription of that gene? A. DNA is capable of looping, bringing distant regions in the same proximity. B. Proteins can be very large molecules, much larger than a nucleotide. C. Transcription factors work in complexes, creating large chains of proteins.

A

If a fork stalls at a lesion the usual first response is: A. fork regression B. formation of a Holliday intermediate C. strand invasion D. branch migration E. initiation of gap repair

A

In the absence of regulatory factors, eukaryotic gene transcription is generally low. Why is this the case? A. Because DNA is not naked, it is tightly condensed as chromatin. B. Because polymerases do not easily interact with DNA. C. No answer text provided. D. Because repressors are found on DNA in their default state.

A

Overexpression of proteins using bacterial vectors: A. requires a bacterial promoter to be present in the bacterial vector. B. makes subsequent purification of the recombinant DNA harder. C. always results in a properly folded, active protein. D. is so effective the expressed protein can represent as much as 80% of the total bacterial protein. E. requires eukaryotic ribosome binding sites to be present in the bacterial vector.

A

RNA polymerase does not contain a 3' to 5' proofreading exonuclease like DNA polymerase. Which answer best explains why this is acceptable? A. RNA molecules are transient compared to DNA molecules which need to last "forever". B. The DNA polymerase is involved in RNA proofreading. C. The sequence of RNA is not important since there are many examples of RNA codons that encode for the same amino acid.

A

Stem cells that can differentiate into any type of tissue but cannot differentiate into a complete organism are ______________ cells. A. pluripotent B. differentiated C. totipotent D. unipotent

A

The result obtained in Jacob and Monods merodiploid analysis that suggested the lacI region acts in trans to regulate the lac operon was: A. a non-functional lacI mutant is rescued by a wild type lacI allele. B. a non-functional lacI mutant cannot be rescued by a wild type lacI allele. C. a non-functional lacO mutant is rescued by a wild type lacO allele. D. a non-functional lacO mutant cannot be rescued by a wild type lacO allele.

A

Transcription factors generally read DNA sequences through amino acid-nucleotide contacts in the major groove because: A. there are more hydrogen bond donors and acceptors in the major groove. B. the minor groove displays only methyl functional groups for recognition. C. all minor groove functional groups are the same. D. there are no minor groove hydrogen bond acceptors available at A-T base pairs.

A

What is the correct order of general transcription factor assembly at an RNA polymerase II promoter? A. TFIID, TFIIB, TFIIF, TFIIH B. TFIIF, TFIID, TFIIB, TFIIH C. TFIIB, TFIIH, TFIID, TFIIF D. TFIID, TFIIH, TFIIB, TFIIF

A

Which method is used to visualize proteins separated by gel electrophoresis? Select all that apply. A. Western blotting B. Immunoprecipitation C. Southern blotting D. Tandem affinity purification E. Northern blotting

A

Which of the following is not true regarding double stranded breaks (DSB)? A. They usually arise due to exposure to UV light. B. DSB repair is handled by the same factors involved in homologous recombination during meiosis C. Mutations in genes involved in DSB repair can lead to cancer D. DSB repair pathways were likely essential before genome sizes could increase in an organism

A

You wish to determine the cellular location of a protein but unfortunately that protein is not particularly antigenic (can't make an antibody against it). How could you solve this problem? A. Create a fusion protein with an epitope tag that can be visualized by incubation with fluorescently tagged antibody that recognizes the epitope. B. Create a fusion protein with green fluorescent protein (GFP). The fusion protein will fluoresce when streptavidin is added. C. Express the protein as a fusion with biotin. The addition of GFP will cause fluorescence. D. None of the given answers will work.

A

What is a operon?

A cluster of genes under the control of a single promoter. It contains a promoter, operator and structural genes

Homologous recombination can be used in the lab to replace a gene of interest with a selection marker. A cell is also capable of non-homologous end joining. Which of the following would you expect to be true if you are transforming the selection marker into a cell that can perform both homologous recombination and non-homologous end joining? A. The selection marker could still be transformed, but may be randomly inserted into the genome (not necessarily replacing the gene of interest). B. Transformed cells could still grow on the selective media. C. Transformed cells would no longer grow on the selective media. D. The selection marker can no longer be transformed into the cell. E. The gene of interest could still be knocked out.

A,B

Margulies and Clark identified recA mutants. These were cells that (select all that apply): A. A)Were incapable of recombination. B. produced colonies that could not grow on media that lacked adenine and leucine. C. A)produced colonies that could grow on media that lacked adenine and leucine. D. Were capable of recombination.

A,B

The Ames test is used to identify potential carcinogens. A compound tested resulted in a few scattered colonies growing on the media, similar to the control plate with no disk. What does this result imply? A. The few colonies that grew contain reversion mutations that allow them to grow on media without histidine. B. The compound in the disk is not mutagenic so there was no increase in the reversion rate. C. The few colonies that grew are resistant to the compound. D. The compound in the disk is so toxic that it inhibits almost all growth on the plate.

A,B

Which of the following is true of the coding strand of DNA for a given gene? Select all that apply. A. It may be located in either strand of a given chromosome. B. Its sequence is the complement of the template strand. C. Its sequence is the complement of the primary RNA transcript.

A,B

Which of the following is true regarding Illumina sequencing? Select all that apply. A. Each of the nucleotides has a different colored label. B. The blocking group is removable. C. The fluorescent label remains on the DNA strand as additional nucleotides are added. D. A blocking group is present on the 2' OH of the nucleotide.

A,B

Why is repairing a single-stranded break more efficient than repairing a double-stranded break? A. It doesn't result in a loss of nucleotides. B. It does not require another chromosome to repair. C. It does not require ligase. D. It isn't caused by phosphodiester bond hydrolysis.

A,B

Histone acetyltransferaes (select all that apply): A. are proteins. B. can lead to recruitment of bromodomain containing proteins when active. C. can produce chromatin decondensation. D. cause decreased gene expression when active.

A,B,C

Which of the following is used to study protein-protein interactions? Select all that apply. A. Immunoprecipitation B. Yeast two-hybrid analysis C. Tandem affinity purification

A,B,C

A plasmid vector typically has which of the following features? Select all that apply. A. A)An origin of replication. B. Small size to facilitate entry into the cell and biochemical manipulation of the DNA. C. A)Unique restriction sites for insertion of DNA. D. A)A selectable marker to ensure that the only cells containing the intact vector can grow.

A,B,C,D

Which of the following are part of the story regarding the use of GFP in the lab? A. A researcher found jellyfish specimens from Washington state. B. Cloning of the GFP gene by a researcher at Woods Hole Oceanographic Institution. C. Expression of the GFP gene in E. coli. D. Modification of the GFP gene to produce different color variants.

A,B,C,D

Which of the following is true of genomic and cDNA libraries? Select all that apply. A. cDNA libraries contain sequences from expressed protein coding genes only. B. A)Genomic libraries contain fragments of genomic DNA that are generated by partial digests with restriction enzymes. C. cDNA libraries contain cDNAs made from all types of RNA. D. Genomic libraries usually contain large fragments and are likely to be constructed in YAC or BAC vectors.

A,B,D

Which of the following sequences can be found in the core promoter of RNA polymerase II? Select all that apply. A. Inr B. BRE C. Enhancer D. DPE

A,B,D

Fragile X syndrome is caused by the expansion of CGG repeats in the FMR1 gene. Normally, there are roughly 40 of these repeats, but in Fragile X patients, this can expand to 200 repeats. Which of the following techniques could be used in a genetic test for the presence of the Fragile X mutation? A. A fluorescence hybridization experiment using a probe complementary to the repeat portion of the FMR1 gene. B. Digestion of the genomic DNA in a manner that keeps the FMR1 gene intact. The genomic DNA is then run on a gel and exposed to a radioactive probe complementary to a repeat portion of the FMR1 gene. C. A fluorescence hybridization experiment using a probe complementary to a non-repeat portion of the FMR1 gene. D. Digestion of the genomic DNA in a manner that keeps the FMR1 gene intact. The genomic DNA is then run on a gel and exposed to a radioactive probe complementary to a non-repeat portion of the FMR1 gene. E. PCR of the FMR1 gene and sequencing of the PCR product.

A,B,D,E

Restriction enzymes (select all that apply): A. Typically cleave at palindromic sequences 4-8bp long. B. Are necessary for joining two DNA fragments together. C. Cleave invading viral DNAs. D. Always produce sticky ends. E. Are found in a wide variety of mammalian cell types.

A,C

Which of the following is a conclusion that Bill Dynan and Bob Tijan made about Sp1 binding to the SV40 promoter based on their DNA footprinting experiment? A. Sp1 binding to the SV40 promoter is location specific. B. Increasing concentration of Sp1 had no impact on the experiment. C. Sp1 bound to both the -21 binding site and -42 binding site.

A,C

Which of the following is true of the lac operon? Select all that apply. A. Transcription produces a single polycistronic mRNA containing the lacZ, lacY and lacA genes. B. The lacI gene is controlled by the same promoter as the lacZ gene. C. The operator region regulates transcription through interaction with the lac repressor protein. D. It is controlled by a negative regulator but not a positive regulator.

A,C

Which of the following is true of the Sanger sequencing method? Select all that apply. A. Requires a primer for the polymerase to extend from B. Requires an RNA template to copy C. Requires chain terminating ddNTPs which are lacking a 2' OH and are unable to form a bond with the next nucleotide. D. The sequencing reaction products are separated by affinity chromatography. E. Incorporates dNTPs using a DNA polymerase

A,E

How does the cAMP affect the expression of the lac operon? A. cAMP binds to the CRP, decreasing its affinity for a DNA site near the promoter. B. cAMP binds to the CRP, increasing its affinity for a DNA site near the promoter. C. cAMP binds to the lac repressor, decreasing its affinity for a DNA site near the promoter. D. cAMP binds to the lac repressor, increasing its affinity for a DNA site near the promoter.

B

Prior to the identification and isolation of restriction enzymes, what was one method researchers used to join together two strands of DNA? A. Doing a PCR reaction with two separate template molecules. B. Adding poly A tails to one strand of DNA and poly T tails to another strand of DNA. C. Adding DNA polymerase and dNTPs to a tube containing two strands of DNA.

B

The activator for the lac operon is: A. CRP B. cAMP C. glucose D. lactose E. allolactose

B

The effector for the lac repressor is: A. lactose B. allolactose C. glucose D. beta-galactosidase E. cAMP

B

The sequence of a promoter constitutes the most basic mechanism of transcription regulation because: A. promoters are always bound by repressors. B. RNA polymerase has differential affinities for different sequences that correlate with the efficiency of transcription. C. promoters are always bound by activators. D. the expression levels of different housekeeping genes are always identical.

B

Which of the following is not true regarding kinetic proofreading during transcription? A. It occurs because the RNA polymerase stalls as it moves along the template. B. The goal is to improve the accuracy of DNA synthesis. C. It involves a pyrophosphorolysis reaction. D. All of the given answers are true.

B

Which of the following is the least likely mechanism for reducing the rate of gene transcription by a repressor? A. The repressor binds directly to the RNA polymerase to block the closed-to-open transition at initiation. B. The repressor induces a conformational change in the polymerase that accelerates the closed-to-open transition. C. The binding of the repressor sterically occludes binding of the RNA polymerase to the promoter. D. The repressor binds directly to the DNA to stabilize the closed complex over the open complex.

B

Which of the following conclusions were made from the Modrich experiment regarding mismatch repair and methylation? Select all that apply. A. Methylated strands were cleaved by the repair machinery.. B. Unmethylated strands were cleaved by the repair machinery.. C. Cleavage requires MutL, MutS, and MutH. D. Cleavage requires only MutL and MutS.

B,C

Base excision repair is responsible for fixing (select all that apply): A. double stranded breaks. B. individual damaged nucleotides that minimally impact the helix structure. C. individual damaged nucleotides that cause a major impact to the helix structure. D. single stranded breaks.

B,D

You would like to set up a test similar to the Ames test using a species of fungus. You start with a methionine auxotroph that is resistant to the antibiotic kanamycin. You then test the potential fungicidal compound X and grow your cells on a plate lacking methionine. Which of the following results could you lead to conclude that compound X is a mutagen? A. Compound X causes the cells to produce kanamycin. B. There are more cells on the compound X containing plate compared to a control plate without compound X. C. There are more cells if the fungus is moved to a plate lacking histidine. D. There are fewer cells right around the compound X disk compared to a control plate without compound X.

B,D

How are recombinat DNAs introduced into bacterial cells? A. Most bacterial cells are naturally competent so DNA just needs to be mixed with cells. B. Transformation can occur if DNA is mixed with bacterial cells that have already been heat shocked. C. Many bacterial cells can be transformed after being made competent by treatment with cold CaCl followed by a brief heat shock. D. Many bacterial cells can be transformed by applying high voltage to a mixture of cells and DNA for a sustained period of time.

C

In a yeast two hybrid experiment, two fusion proteins are created. Each fusion protein contains a protein from the target organism's genome (Protein X and Y). The fusion proteins also contain a DNA binding domain and an activation domain of a transcription factor. When is this transcription factor functional? A. When protein X and protein Y are expressed. B. When protein Y is functionally related to the transcription factor. C. When protein X and protein Y interact. D. When protein X is functionally related to the transcription factor.

C

Repair of double stranded breaks by non-homologous end joining is more likely to occur outside of S phase and G2 than homologous recombination. Why is this? A. Double stranded breaks are unlikely to occur during S phase and G2. B. Proteins for non homologous end joining cannot be transcribed during S phase and G2. C. Outside of S phase and G2, there is no homologous chromosome that the homologous recombination pathway can use for repair. D. DNA ligase, which seals two broken chromosomes in non homologous end joining is not present during S phase or G2.

C

Which of the following characteristics of DNA-dependent DNA synthesis is not the same for DNA-dependent RNA synthesis? A. Synthesis requires a template. B. The process results in phosphodiester bond formation. C. Initiation of synthesis requires a primer. D. All of the given answers are the same for both. E. Synthesis is catalyzed in the 5' to 3' direction.

C

Which of the following characteristics of bacterial transcription is not true of eukaryotic transcription? A. Abortive transcripts can be formed prior to the elongation phase. B. Specific factors are used to direct the polymerase to bind a specific type of promoter sequence. C. RNA polymerase interacts with a single transcription factor during initiation. D. RNA polymerase generally interacts with DNA upstream from the coding sequence in the template.

C

Which of the following conclusions were made from the Kelner experiment regarding DNA repair and sunlight? Select all that apply. A. Cell survival decreases with more exposure to visible light. B. No answer text provided. C. Cell survival increases with more exposure to visible light. D. The impact of sunlight on DNA repair is mediated primarily by sunlight's ability to increase temperature.

C

Which of the following describes translesion synthesis? A. It is the process by which double stranded breaks are repaired by replacing damaged DNA. B. It is the process by which the DNA polymerase stalls at a lesion until the repair machinery fixes it. C. It is the process by which DNA polymerase continues replication over a lesion.

C

Which of the following is true of the structure of typical transcriptional activators? A. The DNA-binding domain always binds to the coactivator. B. They contain only one type of motif. C. They contain a regulatory domain that is still functional when removed from the DNA-binding domain. D. The regulatory domain contains the DNA-binding motif.

C

You would like to set up a test similar to the Ames test using a species of fungus. You start with a methionine auxotroph that is resistant to the antibiotic kanamycin. You then test the potential fungicidal compound X and grow your cells on a plate lacking methionine. Instead of growing cells on a plate lacking methionine to test compound X, you grow them on the plate with kanamycin. Which of the following results could imply that compound X is a mutagen? A. More colonies are observed on the kanamycin plate after exposure to compound X compared to the no compound X control plate. B. Growth is observed on the kanamycin plate after exposure to compound X. C. Fewer colonies are observed on the kanamycin plate after exposure to compound X compared to the no compound X control plate.

C

A DNA sequence of interest was digested and ligated into the middle of the AmpR gene region of pBR322 vector. The vector also contains a TetR gene which is not interrupted. After introducing the recombinat DNA into bacterial cells, the resulting transformants were tested on a variety of media. Which of the following results would be expected? A. Growth on ampicillin containing media. B. Growth on media containing both tetracyclin and ampicillin. C. Death on media containing both tetracyclin and ampicillin. D. A)Growth on tetracyclin containing media.

C,D

DNA footprinting is a technique that can be used to identify (select all that apply): A. A region of DNA that has been damaged by mutation. B. The position of a particular gene on a chromosome. C. Whether a protein binds to a specific DNA sequence. D. The binding site of a repressor, polymerase, or other protein on a DNA sequence.

C,D

Homologous recombination can be used in the lab to replace a gene of interest with a selection marker. What is required for the selection marker to specifically replace the gene of interest? A. Specific selection markers are only capable of replacing specific genes. B. The sequences surrounding the selection marker will be placed around the gene of interest. C. The sequences surrounding the gene of interest will be placed around the selection marker. D. It depends on the media used to test for incorporation of the selection marker.

C,D

Which of the following are true in regards to the use of double stranded break (DSB) mechanisms for genome editing? Select all that apply. A. These mechanisms are specific for prokaryotes. B. One of the main advantages of these systems is that it is possible to generate mutations in a sequence-independent manner. C. The addition of foreign DNA can lead to specific gene modifications. D. Repair of breaks by NHEJ can lead to gene inactivation.

C,D

Which of the following statements about regulation of the lac operon is true? Select all that apply. A. When glucose is present but lactose is absent, gene expression is activated at the lac operon. B. The presence of glucose in the growth medium does not affect the level of gene expression activated by the presence of lactose. C. The lac repressor can dissociate from the operator in the presence of glucose. D. Gene expression is high when the lac repressor is not bound to the operator and glucose is unavailable. E. Gene expression is high when the lac repressor is not bound to the operator and glucose and lactose are available.

C,D

A polylinker is: A. A synthetic fragment inserted into a vector that allows replication of the vector in the host cell. B. A synthetic fragment used in colony hybridization to visualize colonies that contain the DNA of interest. C. A sequence that allows for selection of cells that contain the vector. D. A synthetic fragment that contains multiple sites for restriction enzymes to cut the DNA. E. An enzyme used to connect DNA fragments together.

D

Homologous recombination can be used in the lab to replace a gene of interest with a selection marker. To confirm that the selection marker is integrated into the correct place in the genome, sometimes a negative selection is used in addition to the standard positive selection marker. Which of the following would act as a negative selection marker? A. A second marker inserted into the transformed cassette is now required to grow on the selective media. B. Transformation of a different gene, which does not act as a selection marker, to replace the gene of interest. C. Transforming the cells with two different knock-out cassettes, each containing the same selection marker. D. A second marker in the knock-out cassette, that if inserted into the genome results in cell death when plated on the selective media.

D

In a typical Sanger sequencing reaction, the amount of dNTPs is much greater than of the ddNTPs. If equal amounts of both types of nucleotides were used in each reaction, how might that affect the outcome? A. After gel electrophoresis and visualization of the results, mostly long fragments would be present and no sequence could be read close to the primer. B. After gel electrophoresis and visualization of the results, the sequence could be read normally. The dNTP/ddNTP ratio does not affect the products of the reaction. C. After gel electrophoresis and visualization of the results, only the purine reactions would show termination and be read from the gel. D. After gel electrophoresis and visualization of the results, mostly short fragments would be present and only sequences very close to the primer could be read. E. After gel electrophoresis and visualization of the results, no bands would be visible because the high concentration of ddNTPs would inhibit the activity of the polymerase.

D

Nucleotide excision repair is responsible for fixing (select all that apply): A. individual damaged nucleotides that minimally impact the helix structure. B. single stranded breaks. C. double stranded breaks. D. individual damaged nucleotides that cause a major impact to the helix structure.

D

When performing automated DNA sequencing, why can all four sequencing reactions be analyzed together in a capillary gel? A. The four reactions are carried out separately and then run together in the capillary gel. B. Each primer is labeled with a different fluroescent tag.. C. Each dNTP is labeled with a different radioisotope. D. Each ddNTP has a different fluorescent tag.

D

Which of the following damaged templates encountered by a replication fork is repaired by double-stranded break repair? A. The replication fork encounters a lesion and stalls. B. The replication fork encounters a lesion and bypasses it restarting synthesis on the other side of the lesion. C. The replication fork encounters a lesion and translesion synthesis occurs. D. The replication fork encounters a lesion at which repair has been initiated and the fork collapses.

D

Which of the following is not true regarding patients with xeroderma pigmentosum? A. All of the answers are true. B. They have an excess of T-T dimers. C. The have higher incidence of skin cancer. D. They have a faulty base excision repair pathway. E. Proteins extracts from a single XP patient could not repair a UV lesion in vitro.

D

Which of the following is not true regarding the RNA polymerase II CTD? A. It gets phosphorylated after transcription is initiated. B. It is highly conserved. C. It is involved in mRNA processing. D. All of the given answers are true. E. It consists of a seven nucleotide repeated sequence.

E

How to use EMSA to identify transcription factors

Free floating DNA stays at the bottom and MW increases as you go up the assay. AS DNA & protein Concentration increases going to the right. The lower band of free floating DNA decreases going to the right