Final Molecular Exam II
Exam II answers
- 1.
Which of the conditions does NOt lead to a change in gene expression patterns in bacterial cells?
- A.
Heat shock
- B.
Nutrient availability
- C.
Sporultaion
- D.
Nitrogen deprivation
- E.
None of the choices are correct
Correct Answer
E. None of the choices are correctExplanation
All of the conditions listed in the question can lead to a change in gene expression patterns in bacterial cells. Heat shock, nutrient availability, sporulation, and nitrogen deprivation are all known to affect gene expression in bacteria. Therefore, none of the choices are correct.Rate this question:
-
- 2.
Which of the following is involved in maintaining lysogeny in phage?
- A.
P RE
- B.
P RO
- C.
P RL
- D.
P RM
- E.
All choices correct
Correct Answer
D. P RMExplanation
P RM is involved in maintaining lysogeny in phage.Rate this question:
-
- 3.
One method used by researchers to demonstrate the existence of nucleosome free DNA regions is
- A.
Nuclear run on experiments
- B.
Reporter gene experiments
- C.
DNase hypersensitivity experiment
- D.
RNA polymerase run off experiments
- E.
Anion exchange chromotography
Correct Answer
C. DNase hypersensitivity experimentExplanation
DNase hypersensitivity experiment is a method used by researchers to demonstrate the existence of nucleosome free DNA regions. In this experiment, DNase I enzyme is used to digest the DNA, and the regions that are more accessible and sensitive to DNase I digestion indicate the presence of nucleosome-free DNA. This experiment helps in identifying regulatory regions and active chromatin regions, as these regions are more likely to be nucleosome-free and accessible for DNase I digestion.Rate this question:
-
- 4.
The histone code states that the
- A.
Lysines are the only amino acids found in histones that can be acetylated and deacylated
- B.
Combination of histone modification on a given nucleosome near a gene's control region affects the efficiency of transcription of that gene
- C.
Combination of histone modification on a given nucleosome near a gene's control region affects the efficiency of transcription of all the nearby genes
- D.
Histones are found in all living cells
Correct Answer
B. Combination of histone modification on a given nucleosome near a gene's control region affects the efficiency of transcription of that geneExplanation
The histone code refers to the combination of histone modifications on a specific nucleosome near a gene's control region. This combination of modifications can either enhance or inhibit the efficiency of transcription of that particular gene. It does not necessarily affect the transcription of all nearby genes, but rather specifically influences the transcription of the gene associated with the modified nucleosome.Rate this question:
-
- 5.
Following are typical features of transcription activators in eukaryotes?
- A.
Kinase domain
- B.
Transcription-activation domain and kinase domain
- C.
Trasncription- activation domain
- D.
Transcription- activation domain and DNA- binding domain
- E.
DNA-binding domain
Correct Answer
D. Transcription- activation domain and DNA- binding domainExplanation
Transcription activators in eukaryotes typically possess a transcription-activation domain and a DNA-binding domain. The transcription-activation domain is responsible for interacting with other proteins and components of the transcription machinery to enhance the initiation of transcription. The DNA-binding domain allows the activator to bind to specific DNA sequences in the promoter or enhancer regions of target genes, thereby recruiting the transcription machinery to initiate transcription. Therefore, the presence of both the transcription-activation domain and the DNA-binding domain is necessary for the proper functioning of transcription activators in eukaryotes.Rate this question:
-
- 6.
The glucocortoicoid receptor is activated by binding to
- A.
Its hormone partner
- B.
DNA and then to its hormone ligand
- C.
Its hormone ligand then moving to the nucleus
- D.
None of the choices are correct
Correct Answer
C. Its hormone ligand then moving to the nucleusExplanation
The correct answer is "its hormone ligand then moving to the nucleus". The glucocorticoid receptor is activated when it binds to its hormone ligand. Once bound, the receptor-hormone complex translocates to the nucleus, where it can interact with DNA and regulate gene expression. This process allows the receptor to initiate cellular responses to glucocorticoid hormones.Rate this question:
-
- 7.
The following is not present in the core RNA polymerase of bacteria
- A.
B
- B.
Sigma
- C.
'B
- D.
None of the choices are correct
Correct Answer
B. SigmaExplanation
The presence of sigma is not a characteristic of the core RNA polymerase in bacteria. Sigma is a subunit that associates with the core enzyme to form the holoenzyme, which is responsible for recognizing specific promoter sequences on DNA and initiating transcription. Therefore, the correct answer is sigma.Rate this question:
-
- 8.
Exp designed to obtain nonspecific transcription from both strands of a DNA molecule in bacteria. which of the following strategies would bee most effective in achieving this?
- A.
Use the core enzyme of RNA polymerase
- B.
Enrich the preparation with sigma subunit
- C.
Include the RNA holoenzyme in the reaction
- D.
Use intact DNA
Correct Answer
A. Use the core enzyme of RNA polymeraseExplanation
Using the core enzyme of RNA polymerase would be the most effective strategy in obtaining nonspecific transcription from both strands of a DNA molecule in bacteria. The core enzyme is responsible for the elongation of the RNA transcript and does not require any additional factors or subunits for activity. It can initiate transcription from a DNA template without the need for specific promoter sequences, allowing for transcription from both strands of the DNA molecule. Enriching the preparation with sigma subunit or including the RNA holoenzyme in the reaction may enhance specific transcription from certain promoters, but they may not be effective in achieving nonspecific transcription from both strands. Using intact DNA alone may not be sufficient for transcription to occur without the core enzyme.Rate this question:
-
- 9.
The following is the correct order of addition of general transcription factors to the initiation complex in eukaryotes
- A.
TFIIIA+TFIID, TFIIIIB, TFIIF+pol II, TFIIE, TFIIH
- B.
TFIIA+TFIID, TFIIB, TFIIH,TFIIF +pol II, TFIIE
- C.
TFIIA+TFIID, TFIIB, TFIIE, TFIIF +pol II, TFIIH
- D.
TFIIA+TFIID, TFIIH, TFIIIIB, TFIIF +pol II, TFIIE
Correct Answer
A. TFIIIA+TFIID, TFIIIIB, TFIIF+pol II, TFIIE, TFIIHExplanation
The correct order of addition of general transcription factors to the initiation complex in eukaryotes is TFIIIA+TFIID, TFIIIIB, TFIIF+pol II, TFIIE, TFIIH. This order is supported by experimental evidence and is consistent with the known functions of these factors in the transcription process. TFIIA and TFIID are the first factors to bind to the promoter region, followed by TFIIIIB which helps in the recruitment of RNA polymerase II. TFIIF and pol II then join the complex, followed by TFIIE which helps in the recruitment of TFIIH. TFIIH is the final factor to join the complex and is responsible for unwinding the DNA and initiating transcription.Rate this question:
-
- 10.
THE FOLLOWING PLASMIDS COULD BE USED TO RESTORE INDUCIBLE regulation of B galactosidase in this mutant I+Oc Z- Y+ A+
- A.
I+ Oc Z- Y+ A+
- B.
I- O+ Z+ Y+ A+
- C.
I- Oc Z+ Y+ A+
- D.
I+ Oc Z- Y+ A+
Correct Answer
B. I- O+ Z+ Y+ A+Explanation
The correct answer is "I- O+ Z+ Y+ A+". This plasmid can be used to restore inducible regulation of B galactosidase in the given mutant because it lacks the I gene, which is responsible for repressing the lac operon. Additionally, it contains the O, Z, Y, and A genes, which are necessary for the proper functioning of the lac operon and the production of B galactosidase. Therefore, this plasmid can provide the necessary components for the restoration of inducible regulation in the mutant.Rate this question:
-
- 11.
Which statements is true about a lac operon with this genotype I- d O+ Z+ Y- A-
- A.
Operon is repressible
- B.
Mutation is only cis- dominant
- C.
Operon is uninducible
- D.
Operon is nonrepressible
- E.
None of the choices is correct
Correct Answer
D. Operon is nonrepressibleExplanation
The given genotype I- d O+ Z+ Y- A- indicates that the operon is nonrepressible. This is because the presence of the I- mutation means that the repressor cannot bind to the operator, resulting in the inability to shut off the lac operon. Additionally, the presence of the O+ mutation indicates that the operator is functional, allowing for transcription to occur even in the absence of lactose. Therefore, the operon is constitutively active and cannot be repressed.Rate this question:
-
- 12.
You are planning an experiment to look at the specific binding of the TATA box binding proteins to the promoter. which would be a suitable negative control?
- A.
DNA sequence made with multiple cytosines and a DNA strand with TATA box linked to multiple adenine
- B.
DNA strand containing the TATA box linked to multiple guanines
- C.
DNA strand containing the TATA box linked to multiple adenine
- D.
A DNA sequence made with multiple cytosines
- E.
All likely
Correct Answer
D. A DNA sequence made with multiple cytosinesExplanation
A suitable negative control in this experiment would be a DNA sequence made with multiple cytosines. This control would allow the researcher to observe if the TATA box binding proteins are specifically binding to the TATA box or if they are binding nonspecifically to the cytosines. By comparing the binding of the TATA box binding proteins to this control with the binding to the other DNA sequences, the researcher can determine if the binding is specific to the TATA box or if it is occurring randomly.Rate this question:
-
- 13.
Which of the following is true about the holoenzyme of bacterial RNA polymerase in an open complex of bacteria?
- A.
Region 2.4 of the sigma factor binds to the 35 region
- B.
There no interaction between the sigma factor and the 10 region
- C.
DNA is bound mainly to the sigma subunit
- D.
There are two Na+ ions in the core enzyme
Correct Answer
C. DNA is bound mainly to the sigma subunitExplanation
In an open complex of bacteria, the DNA is mainly bound to the sigma subunit of the holoenzyme of bacterial RNA polymerase. This means that the sigma subunit plays a crucial role in initiating transcription by binding to the DNA at specific regions, such as the -10 and -35 regions. The sigma subunit helps in positioning the RNA polymerase at the correct location on the DNA template, allowing for the efficient initiation of transcription.Rate this question:
-
- 14.
Which of the following would most likely occur in bacteria if an in vitro transcription assay is performed using cell extract containing an alpha subunit without the C-terminal domain
- A.
RNA polymerase would become tightly associated with the promoter.
- B.
RNA polymerase complex would be loosely associated with the promoter
- C.
RNA polymerase would become more susceptible to proteolysis
- D.
RNA polymerase complex would be loosely associated with the operon
Correct Answer
B. RNA polymerase complex would be loosely associated with the promoterExplanation
If an in vitro transcription assay is performed using cell extract containing an alpha subunit without the C-terminal domain, it is likely that the RNA polymerase complex would be loosely associated with the promoter. The C-terminal domain of the alpha subunit is responsible for the interaction between RNA polymerase and the promoter region of DNA. Without this domain, the binding between RNA polymerase and the promoter would be weakened, leading to a loose association between the complex and the promoter.Rate this question:
-
- 15.
Determine which of the following would most likely lead to a dramatic reduction in eukaryotic transcription
- A.
Removal of some TAFs
- B.
Removal of TAFII250
- C.
Absence of the TATA box
- D.
Absence of the TATA box removal of some TAFs
- E.
None are correct
Correct Answer
E. None are correctExplanation
None of the given options would most likely lead to a dramatic reduction in eukaryotic transcription. The removal of some TAFs or TAFII250 may affect the efficiency or specificity of transcription, but it would not necessarily cause a dramatic reduction. Similarly, the absence of the TATA box may affect the initiation of transcription, but it would not necessarily result in a dramatic reduction. Therefore, none of the options provided can be considered as the most likely cause for a dramatic reduction in eukaryotic transcription.Rate this question:
-
- 16.
New mutant cell line was developed and was found to be defective in ploymerase III activity of eukaryotes. which is likely to observe the cell line?
- A.
There will be an overabundance of secreted proteins
- B.
There will be overproduction of 7 SL RNA
- C.
Splicing function is impaired
- D.
All likely
Correct Answer
C. Splicing function is impairedExplanation
In eukaryotes, polymerase III is responsible for transcribing small structural RNA molecules, including 7 SL RNA, which is involved in splicing. Therefore, if the mutant cell line is defective in polymerase III activity, it is likely to observe impaired splicing function. This is because the production of 7 SL RNA, which is required for proper splicing, would be reduced or absent. The other options, such as an overabundance of secreted proteins or overproduction of 7 SL RNA, are not directly related to the defect in polymerase III activity.Rate this question:
-
- 17.
Which would you choose ti study the role of polymerase subunits in elongation of eukaryotes
- A.
Epitope tagging
- B.
Western blotting
- C.
In vitro transcription
- D.
DNase footprinting
Correct Answer
C. In vitro transcriptionExplanation
In vitro transcription would be the best choice to study the role of polymerase subunits in elongation of eukaryotes. In vitro transcription involves the synthesis of RNA molecules in a test tube using purified components, including polymerase subunits. This method allows researchers to specifically analyze the activity and function of the polymerase subunits during the elongation process without the complexity of the cellular environment. Epitope tagging, western blotting, and DNase footprinting are techniques that can provide valuable information in other aspects of gene expression, but they may not be as suitable for directly studying the role of polymerase subunits in elongation.Rate this question:
-
- 18.
Which technique would you use to study role of BRE in eukayrotic promoter region
- A.
Linker scanning mutations
- B.
S1 nuclease protection analysis
- C.
X-Ray crystallography
- D.
Linker scanning mutations and X ray crystallography
Correct Answer
A. Linker scanning mutationsExplanation
Linker scanning mutations would be the appropriate technique to study the role of BRE (TFIIB recognition element) in the eukaryotic promoter region. This technique involves systematically introducing mutations in the DNA sequence, specifically in the region of interest, to determine the functional significance of different elements. By creating mutations in the BRE sequence and observing the effects on transcription initiation, researchers can gain insights into the role of BRE in regulating gene expression. S1 nuclease protection analysis and X-Ray crystallography are not directly related to studying the role of BRE in the promoter region.Rate this question:
-
- 19.
Predict possible effect of deleting the enhancers region of the polymerase I gene in eukaryotes
- A.
Reduction in production of Rpb 1
- B.
Reduction in amount of rRNA made
- C.
Reduction in production of Rpb2
- D.
Reduced transcription of ribosomes
Correct Answer
B. Reduction in amount of rRNA madeExplanation
Deleting the enhancers region of the polymerase I gene in eukaryotes would lead to a reduction in the amount of rRNA made. Enhancers are DNA sequences that enhance the transcription of specific genes. In this case, the enhancers region of the polymerase I gene is responsible for enhancing the transcription of rRNA. Therefore, deleting this region would result in a decrease in the transcription of rRNA, ultimately leading to a reduction in the amount of rRNA produced.Rate this question:
-
- 20.
Which of the followin is the correct order chromatin folding
- A.
Nucleosome formation, 30 nM fiber formation, radial loop structure
- B.
Nucelosome formation, radial loop formation, nucleosome formation
- C.
Nucleosome formation, radial loop structure, 30 nM fiber formation
- D.
30 nM fiber formation, nucleosome formation, radial loop formation
Correct Answer
A. Nucleosome formation, 30 nM fiber formation, radial loop structureExplanation
The correct order of chromatin folding is nucleosome formation, followed by the formation of 30 nM fiber, and finally the formation of radial loop structure. This is because nucleosome formation involves the wrapping of DNA around histone proteins, which forms the basic unit of chromatin. The nucleosomes then further condense to form a 30 nM fiber, where the nucleosomes are organized into a more compact structure. Finally, the 30 nM fiber is folded into radial loop structures, which help to further compact the chromatin and facilitate gene regulation.Rate this question:
-
- 21.
Studying Xenopus transcription of somatic and oocyte 5S rRNA genes patterns in two types of cells could be seen are between
- A.
RNA Pol III and repressors
- B.
Histones and enhancer elements
- C.
Histones and DNA gyrase
- D.
Histones and transcription factors
Correct Answer
D. Histones and transcription factorsExplanation
The correct answer is histones and transcription factors. Histones are proteins that help organize DNA into a compact structure called chromatin. Transcription factors are proteins that bind to specific DNA sequences and regulate the transcription of genes. In the context of studying Xenopus transcription of somatic and oocyte 5S rRNA genes, histones and transcription factors are likely involved in regulating the expression of these genes. The interaction between histones and transcription factors is important for controlling gene expression and determining the patterns of transcription in different types of cells.Rate this question:
-
- 22.
Which sigma specificity factors is involved in middle gene transcription during SPO I phage infection of bacteria
- A.
Gp33
- B.
Gp34
- C.
Gp43
- D.
Gp24
- E.
Gp28
Correct Answer
E. Gp28Explanation
The correct answer is gp28. Gp28 is involved in middle gene transcription during SPO I phage infection of bacteria.Rate this question:
-
- 23.
Which is true of early phase of T7 infection of bacteria
- A.
Class I genes are the first to be transcribed
- B.
Class II genes are the first to be transcribed
- C.
Host polymerase transcribes class II genes
- D.
All are correct
Correct Answer
A. Class I genes are the first to be transcribedExplanation
In the early phase of T7 infection of bacteria, Class I genes are the first to be transcribed. This means that these genes are the first to be copied into RNA molecules, which can then be used to produce proteins. This early transcription of Class I genes is an important step in the infection process, as it allows the virus to quickly take control of the host cell's machinery and begin replicating itself.Rate this question:
-
- 24.
Which operators is a target of cl gene production of the phage
- A.
O R1
- B.
O L1
- C.
O L2
- D.
O R2
- E.
All are correct
Correct Answer
E. All are correctExplanation
The correct answer is "all are correct" because all of the operators mentioned (O R1, O L1, O L2, O R2) can be targets of cl gene production of the phage.Rate this question:
-
- 25.
Which is true about interaction with repressor and RNA polymerase at P RM?
- A.
Interaction between repressor and polymerase is essential for activation of transcription and interaction involves region of sigma factor
- B.
Interaction involves region 4 sigma factor
- C.
Interaction can be disrupted by binding of CIII
- D.
Interaction between repressor and polymerase is essential for activation of transcription
Correct Answer
A. Interaction between repressor and polymerase is essential for activation of transcription and interaction involves region of sigma factorExplanation
The interaction between the repressor and RNA polymerase at P RM is essential for the activation of transcription. This interaction also involves the region of the sigma factor. The sigma factor is a subunit of RNA polymerase that is responsible for recognizing and binding to specific promoter sequences. Therefore, the interaction between the repressor and polymerase, specifically involving the sigma factor, is necessary for the activation of transcription at the P RM.Rate this question:
-
- 26.
Which statement is true about the cro gene of the phage
- A.
Its product promotes repressor activity
- B.
It is adjacent to the cIII gene
- C.
Cro repression is important to lysogeny
- D.
It must be stimulated during lysogeny and its product
Correct Answer
C. Cro repression is important to lysogenyExplanation
The statement "cro repression is important to lysogeny" is true about the cro gene of the phage. Lysogeny is the process by which a phage integrates its DNA into the host cell's genome and remains dormant. The cro gene plays a crucial role in this process by repressing the expression of other phage genes, allowing the phage to establish a stable lysogenic state within the host cell.Rate this question:
-
- 27.
Predict the outcome of the removal of the N from the DNA sequence of the phage
- A.
Antitermination will be affected and transcription from the left promoter will be interrupted
- B.
Antitermination will be affected
- C.
Transcription from the left promoter will be interrupted
- D.
Lysogenic cycle will be induced
Correct Answer
A. Antitermination will be affected and transcription from the left promoter will be interruptedExplanation
The removal of the N from the DNA sequence of the phage will affect antitermination and interrupt transcription from the left promoter. This is because N is a phage protein that plays a role in antitermination, which allows the continuation of transcription past termination sites. Without N, antitermination cannot occur, leading to the interruption of transcription from the left promoter.Rate this question:
-
- 28.
Which of following is possible mechanism for action of Nus A in transcription termination of phage
- A.
It binds to the RNA polymerase causing it to stall
- B.
It interacts with S10 to promote detachment of the RNA polymerase for the DNA template
- C.
It stimulates termination at intrinsic terminator by facilitating hairpin loop formation
- D.
It binds to NusB to promote detachment of the RNA polymerase
Correct Answer
C. It stimulates termination at intrinsic terminator by facilitating hairpin loop formationExplanation
Nus A stimulates termination at intrinsic terminator by facilitating hairpin loop formation. This suggests that Nus A plays a role in the termination of transcription by promoting the formation of a hairpin loop structure at the intrinsic terminator region of the DNA template. This hairpin loop formation signals the RNA polymerase to dissociate from the DNA template, effectively terminating transcription.Rate this question:
-
- 29.
During experiment to study the rate of infection of bacteria with T4, the bacterial cultures were accidentally exposed to a UV light source, which would be expected for this exposure
- A.
SOS response is induced
- B.
Corpotease activity in Rec A protein is activated and SOS response is induced
- C.
RecA gene is turned off
- D.
Coprotease activity in Rec A protein is activated
Correct Answer
B. Corpotease activity in Rec A protein is activated and SOS response is inducedExplanation
When bacterial cultures are exposed to a UV light source, it can cause DNA damage. In response to this damage, the RecA protein is activated and its coprotease activity is increased. This activation of coprotease activity in RecA protein leads to the induction of the SOS response. The SOS response is a DNA repair mechanism that helps the bacteria to repair the damaged DNA and survive the UV exposure. Therefore, the activation of coprotease activity in RecA protein and induction of the SOS response would be expected as a result of the accidental exposure to UV light.Rate this question:
-
- 30.
Which statements is true about a lac operon with the following genotypeI- Oc Z+ Y-I- O+ Z- Y-
- A.
A repressor protein will be constitutively produced
- B.
A trans-dominant mutation is present
- C.
B galactosidase will be constitutively produced
- D.
Permease will be constitutively produced
Correct Answer
C. B galactosidase will be constitutively producedExplanation
Based on the given genotype (I- Oc Z+ Y-I- O+ Z- Y-), we can determine that the operator (Oc) is mutated and the lacI gene is non-functional (I-). This means that the repressor protein will not be produced, leading to the constitutive production of the lac operon enzymes. Therefore, the statement "B galactosidase will be constitutively produced" is true.Rate this question:
-
- 31.
Following can be found in DNA binding domains of transcription activators in eukaryotes
- A.
Zinc fingers
- B.
BZIP motifs
- C.
BHLH motif
- D.
Glutamine- rich regions
- E.
All of hte choices are correct
Correct Answer
E. All of hte choices are correctExplanation
The correct answer is that all of the choices are correct. DNA binding domains of transcription activators in eukaryotes can contain zinc fingers, bZIP motifs, bHLH motifs, and glutamine-rich regions. These domains are responsible for binding to specific DNA sequences and regulating gene expression. The presence of all these motifs suggests that transcription activators have multiple mechanisms for binding to DNA and controlling gene transcription.Rate this question:
-
- 32.
One effect of mutating or deleting the activation domain of a transcription activaor in eukaryotes would be
- A.
Reduction in the binding of TFIIA
- B.
Reduction in the binding TFIID
- C.
Reduction in the binding of TFIIA and reduction in the binding TFIID
- D.
Binding of the activator to DNA without activating transcription
Correct Answer
D. Binding of the activator to DNA without activating transcriptionExplanation
If the activation domain of a transcription activator is mutated or deleted in eukaryotes, it would result in the binding of the activator to DNA without activating transcription. This means that even though the activator is able to bind to the DNA, it would not be able to initiate the transcription process. This could be due to the fact that the activation domain is responsible for interacting with other transcription factors and components of the transcription machinery, which are necessary for the activation of transcription. Without a functional activation domain, the activator would be unable to activate transcription despite its binding to the DNA.Rate this question:
-
- 33.
Which of the following techniques is most useful in determining if RNA polymerase has initiated transcription from the lac DNA template
- A.
Southern analysis
- B.
RACE
- C.
Run off transcription assay
- D.
DNA sequencing
- E.
DNA fingerprinting
Correct Answer
C. Run off transcription assayExplanation
The run off transcription assay is the most useful technique for determining if RNA polymerase has initiated transcription from the lac DNA template. This assay involves isolating and labeling newly synthesized RNA molecules and then separating them from the DNA template. By comparing the amount of labeled RNA in the presence and absence of RNA polymerase, it can be determined if transcription has occurred. This technique directly measures the activity of RNA polymerase and provides a reliable indication of transcription initiation. Southern analysis, RACE, DNA sequencing, and DNA fingerprinting are not specifically designed to assess transcription initiation and may not provide the same level of information.Rate this question:
-
- 34.
HMG I (Y) plays a role in IFN-B transcription by
- A.
Bending the DNA causing more activators to bind and to interact with each other to enhance transcription
- B.
Suppressing IFN-B expresssion during a viral infection
- C.
Suppressing IFN-B expression during a viral infection and binding to the HMG domain of LEF
- D.
Dending the DNA
Correct Answer
A. Bending the DNA causing more activators to bind and to interact with each other to enhance transcriptionExplanation
HMG I (Y) plays a role in IFN-B transcription by bending the DNA, which leads to an increased ability for activators to bind and interact with each other. This enhanced interaction between activators ultimately enhances transcription of IFN-B.Rate this question:
-
- 35.
An electrophoretic mobility shift assay (EMSA) was conducted to check for the binding of TFIIB to a segment of a DNA promoter region. which approaches could be used to confirm that TFIIB is indeed binding?
- A.
Complete TFIIB binding with THIID
- B.
Use an antibody to THIIB and check for a supershift by EMSA
- C.
Use an antibody to TFIIB and check for a supershift by EMSA and complete TFIIB binding with TFIID are correct
- D.
Conduct a nuclear run off assay
Correct Answer
B. Use an antibody to THIIB and check for a supershift by EMSA -
- 36.
In an in vitro transcription assay to detect the level of transcription from a lac operon
- A.
The level of fluorescence would increase along the new RNA strand
- B.
The level of fluorescence would be very high in the assay buffer
- C.
Each RNA would have a fluorescent tag on every other nucleotide
- D.
All of the choices are possible
Correct Answer
B. The level of fluorescence would be very high in the assay bufferExplanation
In an in vitro transcription assay, the level of fluorescence would be very high in the assay buffer because the assay buffer contains the necessary components and conditions for transcription to occur efficiently. This includes the presence of RNA polymerase, nucleotides, and other factors required for transcription. As a result, the transcription process would be highly active in the assay buffer, leading to a high level of fluorescence. The other options are not correct because they do not explain the reason for the high level of fluorescence in the assay buffer.Rate this question:
-
- 37.
Which of the following is most likely to occur if the level of glucose is low in bacterial
- A.
Both cyclic AMP levels will be depressed and CRP activity will be inhibited are correct
- B.
CAP will assist in stimulating transcription of the lac operon if lactose is added
- C.
CRP activity will be inhibited
- D.
Cyclic AMP blocks recruitment of polymerase to the promoter
Correct Answer
B. CAP will assist in stimulating transcription of the lac operon if lactose is addedExplanation
When the level of glucose is low in bacteria, cyclic AMP (cAMP) levels increase. cAMP then binds to the catabolite activator protein (CAP), forming a complex. This complex binds to the promoter region of the lac operon in the presence of lactose, stimulating transcription of the lac operon. Therefore, CAP assists in stimulating transcription of the lac operon when lactose is added.Rate this question:
-
- 38.
Which of the following is true about the action of CAP at the lac promoter?
- A.
Binding of the CAP cAMP to the lac activator binding site recruits RNA polymerase
- B.
CAP blocks the a CTP of RNA polymerase
- C.
CAP monomer binds directly to the promoter stimulating polymerase to bind
- D.
None are correct
Correct Answer
A. Binding of the CAP cAMP to the lac activator binding site recruits RNA polymeraseExplanation
The correct answer is that the binding of the CAP cAMP to the lac activator binding site recruits RNA polymerase. This means that CAP acts as an activator by binding to the lac promoter and facilitating the recruitment of RNA polymerase, which is necessary for the initiation of transcription. This binding enhances the efficiency of transcription initiation at the lac promoter.Rate this question:
-
- 39.
Which would be effective in vitro in blocking the transcription from the lac operon in the presence of permease
- A.
Antibody to the repressor protein
- B.
Antibody to RNA polymerase
- C.
Mutation in the operator
- D.
Both antibody to RNA polymerase and mutation in the operator
- E.
An antibody to B galactosidase
Correct Answer
D. Both antibody to RNA polymerase and mutation in the operatorExplanation
Both the antibody to RNA polymerase and mutation in the operator would be effective in vitro in blocking the transcription from the lac operon in the presence of permease. The antibody to RNA polymerase would inhibit the activity of RNA polymerase, preventing it from binding to the promoter region and initiating transcription. The mutation in the operator would disrupt the binding of the repressor protein, allowing RNA polymerase to bind and transcribe the lac operon. Therefore, both of these factors together would effectively block transcription from the lac operon.Rate this question:
-
- 40.
Which of the following is true about the regulation of the trp operon?
- A.
An aporepressor is involved
- B.
A corepressor is involved
- C.
Attenuation is one of the control mechanisms
- D.
Negative control is involved
- E.
All are true
Correct Answer
E. All are trueExplanation
All the statements are true regarding the regulation of the trp operon. The trp operon is a group of genes involved in the synthesis of tryptophan in bacteria. An aporepressor is involved in the regulation of the operon, which is a protein that requires binding with a co-repressor (tryptophan) to become an active repressor. Attenuation is another control mechanism in the trp operon, where the transcription of the operon is regulated by the presence of tryptophan in the cell. Lastly, negative control is involved, as the repressor protein binds to the operator region and prevents transcription when tryptophan levels are high.Rate this question:
-
- 41.
Which is absent in an operon
- A.
Operator
- B.
Promoter
- C.
Intron
- D.
Both operator and promoter
- E.
Both promoter and intron
Correct Answer
C. IntronExplanation
An operon is a unit of genetic material that consists of a promoter, operator, and structural genes. The promoter is the region of DNA where RNA polymerase binds to initiate transcription, while the operator is a regulatory region that controls the access of RNA polymerase to the structural genes. In contrast, an intron is a non-coding sequence of DNA that is present within a gene but is not transcribed into RNA. Therefore, the correct answer is intron, as it is not a component of an operon.Rate this question:
-
- 42.
Which technique would be most useful in testing the ability of the core RNA polymerase of bacteria to bind to the promoter?
- A.
Northern blotting
- B.
Gel filtration
- C.
S1 mapping
- D.
DNase footprinting
- E.
Immunoblotting
Correct Answer
D. DNase footprintingExplanation
DNase footprinting would be the most useful technique in testing the ability of the core RNA polymerase of bacteria to bind to the promoter. DNase footprinting is a method that allows the identification of specific DNA-protein interactions. It involves the digestion of DNA with DNase, which cleaves the DNA at sites where it is not bound to proteins. By comparing the digestion patterns of DNA in the presence and absence of the RNA polymerase, one can determine if the polymerase is binding to the promoter region and protecting it from DNase cleavage. This technique provides direct evidence of protein-DNA interactions and can therefore assess the ability of the RNA polymerase to bind to the promoter.Rate this question:
-
- 43.
Which statement is false for sigma factor of bacterial polymerase
- A.
It does not have a DNA binding domain
- B.
It can bind to the 10 box
- C.
It can also bind to the nontemplate strand
- D.
None are correct
Correct Answer
A. It does not have a DNA binding domainExplanation
The statement that the sigma factor of bacterial polymerase does not have a DNA binding domain is false. The sigma factor is a subunit of the bacterial RNA polymerase that is responsible for recognizing the promoter region of the DNA and initiating transcription. It binds to the -10 box (also known as the Pribnow box) in the promoter region, which is crucial for the initiation of transcription. Additionally, the sigma factor can also bind to the nontemplate strand of the DNA during transcription initiation. Therefore, all the statements provided are incorrect.Rate this question:
-
- 44.
Place following in order1-formation of open complex2- formation of closed complex3-promoter clearance 4-synthesis
- A.
3124
- B.
3214
- C.
2143
- D.
1234
Correct Answer
C. 2143Explanation
The correct answer, 2143, represents the correct sequence of events in transcription initiation. First, the RNA polymerase binds to the promoter region of the DNA, forming a closed complex (2). Then, the DNA strands separate, forming an open complex (1). Next, the RNA polymerase clears the promoter region and begins synthesizing the RNA molecule (4). Therefore, the correct order is 2-1-4-3.Rate this question:
-
- 45.
Which following is most involved in elongation of the RNA transcript in bacteria
- A.
Core polymerase
- B.
Sigma factor
- C.
UP elements
- D.
Both sigma factor and UP elements
Correct Answer
B. Sigma factorExplanation
The sigma factor is most involved in the elongation of the RNA transcript in bacteria. The sigma factor is responsible for recognizing the promoter region on the DNA and initiating transcription by recruiting the RNA polymerase to the correct site. Once transcription is initiated, the core polymerase takes over and continues the elongation of the RNA transcript. UP elements are DNA sequences that can enhance transcription initiation but are not directly involved in elongation. Therefore, the correct answer is sigma factor.Rate this question:
-
- 46.
State where you would expect to find postitive staining if the enzyme is present
- A.
Nucleous
- B.
Ribosomes
- C.
Wendy's
- D.
KFC
Correct Answer
A. Nucleous -
- 47.
Which are products of RNA polymerase II activity in eukaryotes
- A.
SnRNA and hnRNA
- B.
TRNA
- C.
SnRNA
- D.
HnRNA
Correct Answer
A. SnRNA and hnRNAExplanation
snRNA (small nuclear RNA) and hnRNA (heterogeneous nuclear RNA) are both products of RNA polymerase II activity in eukaryotes. snRNA is involved in the processing of pre-mRNA and plays a crucial role in splicing, which removes introns and joins exons to form mature mRNA. hnRNA, on the other hand, is the primary transcript of a gene before it undergoes processing. It contains both introns and exons and is eventually modified and processed into mature mRNA. Therefore, snRNA and hnRNA are the correct products of RNA polymerase II activity in eukaryotes.Rate this question:
-
- 48.
Which of the following techniques is suitable for the study of DNA; RNA hybrid formaiton during transcription
- A.
DNA; protein crosslinking
- B.
DNAse I footprinting
- C.
DNA;RNA crosslinking
- D.
FRET analysis
Correct Answer
C. DNA;RNA crosslinkingExplanation
DNA;RNA crosslinking is a suitable technique for the study of DNA; RNA hybrid formation during transcription. This technique involves chemically crosslinking DNA and RNA molecules that are in close proximity to each other, allowing for the identification and characterization of DNA; RNA hybrids. By crosslinking the two molecules, researchers can stabilize and capture the transient DNA; RNA hybrids, providing valuable insights into the process of transcription and the interaction between DNA and RNA molecules.Rate this question:
-
- 49.
Is incorrect regarding rhodependent terminator in bacterial transcription?
- A.
They consist of inverted repeats
- B.
Rho binds to the growing transcripts
- C.
They form hairpin loops
- D.
Rho has no helicase
Correct Answer
D. Rho has no helicaseExplanation
The correct answer is that Rho has no helicase. Helicase is an enzyme that unwinds the DNA double helix during transcription. In bacterial transcription, the rho-dependent terminator is a sequence of DNA that signals the termination of transcription. Rho is a protein that binds to the growing transcripts and helps in the termination process. However, unlike other helicases, Rho does not possess helicase activity, which is the ability to unwind DNA strands.Rate this question:
-
- 50.
Which of the following does not have a DNA binding domain
- A.
Sigma factor
- B.
Alpha factor
- C.
Beta
- D.
UP region
- E.
Beta prime
Correct Answer
D. UP regionExplanation
The UP region does not have a DNA binding domain. Sigma factors, alpha factors, and beta prime are all proteins that bind to DNA and are involved in gene regulation. The UP region is a regulatory sequence located upstream of the promoter region in bacterial DNA. It helps in facilitating the binding of RNA polymerase to the promoter, but it does not directly bind to DNA itself. Therefore, the UP region does not have a DNA binding domain.Rate this question:
-
Quiz Review Timeline +
Our quizzes are rigorously reviewed, monitored and continuously updated by our expert board to maintain accuracy, relevance, and timeliness.
-
Current Version
-
Mar 21, 2023Quiz Edited by
ProProfs Editorial Team -
May 04, 2010Quiz Created by
Imasportydiva24
- Atomic Theory Quizzes
- Chemical Bonding Quizzes
- Chemical Compound Quizzes
- Chemical Element Quizzes
- Chemical Reaction Quizzes
- Chemistry Practice Quizzes
- Clinical Chemistry Quizzes
- Electrolysis Quizzes
- Functional Group Quizzes
- Gas Quizzes
- General Chemistry Quizzes
- IBSL Chemistry Quizzes
- Medicinal Chemistry Quizzes
- Periodic Table Quizzes
- Rate Of Reaction Quizzes
- Reaction Quizzes
- Soap Quizzes
- Solution And Mixture Quizzes
- State Of Matter Quizzes
- Stoichiometry Quizzes
Back to top