Ultimate Molecular Biology Quiz

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Ultimate Molecular Biology Quiz - Quiz

Do you understand molecular biology well? Take this molecular biology quiz, and see how much you know about this topic. Molecular biology studies the chemical and physical structure of biological macromolecules. Go for this quiz, and test your knowledge. Even if you miss out on some questions, we will help you with the correct answer. Try to get 100% on this quiz. If you find this quiz informative, share it with others interested in testing their knowledge on this topic.


Questions and Answers
  • 1. 

    Which of the following is not present in the core RNA polymerase? 

    • A.

      B (Beta)

    • B.

      B' (Beta prime)

    • C.

      Sigma (o)

    • D.

      Both (Beta) and (sigma)

    Correct Answer
    C. Sigma (o)
    Explanation
    The correct answer is sigma (o). Sigma (o) is not present in the core RNA polymerase. The core RNA polymerase consists of two subunits, B (Beta) and B' (Beta prime), which are responsible for the catalytic activity of RNA synthesis. Sigma (o) is a separate subunit that associates with the core enzyme to form the holoenzyme, which is responsible for promoter recognition and initiation of transcription. Therefore, sigma (o) is not a component of the core RNA polymerase.

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  • 2. 

    An experiment was designed to obtain nonspecific transcription from both strands of a DNA molecule. Which of the following strategies would be most effective in achieving this? 

    • A.

      Include the RNA holoenzyme in the reaction.

    • B.

      Use the core enzyme of RNA polymerase.

    • C.

      Enrich the preparation with sigma subunit

    • D.

      Use intact DNA.

    • E.

      Include the RNA holoenzyme in the reaction and using the core enzyme of RNA polymerase are both effective.

    Correct Answer
    B. Use the core enzyme of RNA polymerase.
    Explanation
    Using the core enzyme of RNA polymerase would be the most effective strategy in obtaining nonspecific transcription from both strands of a DNA molecule. The core enzyme is capable of binding to DNA and initiating transcription without the need for additional factors or subunits. This allows for transcription to occur on both strands of the DNA molecule, resulting in nonspecific transcription. Including the RNA holoenzyme in the reaction or enriching the preparation with the sigma subunit may enhance transcription efficiency, but they are not necessary for achieving nonspecific transcription.

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  • 3. 

    Which of the following techniques would be most useful in testing the ability of the core RNA polymerase to bind to the promoter? 

    • A.

      Gel filtration

    • B.

      S1 mapping

    • C.

      Immunoblotting

    • D.

      DNase footprinting

    • E.

      Northern blotting

    Correct Answer
    D. DNase footprinting
    Explanation
    DNase footprinting would be the most useful technique in testing the ability of the core RNA polymerase to bind to the promoter. DNase footprinting is a method used to identify the specific DNA sequence that is bound by a protein. In this case, by using DNase footprinting, we can determine if the core RNA polymerase is binding to the promoter region of the DNA. This technique involves treating the DNA with DNase, which cuts the DNA at unprotected regions. If the core RNA polymerase is bound to the promoter, it will protect that region from being cut by DNase, creating a "footprint" that can be visualized and analyzed.

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  • 4. 

    Which of the following statements is false regarding the sigma factor? 

    • A.

      It does not have a DNA-binding domain.

    • B.

      Interaction with the core enzyme unmasks the DNA-binding region.

    • C.

      Subregions 2.4 and 4.4 are involved in promoter recognition.

    • D.

      It can also bind to the nontemplate strand.

    • E.

      It can bind to the -10 box.

    Correct Answer
    A. It does not have a DNA-binding domain.
    Explanation
    The sigma factor is a protein subunit that is involved in the initiation of transcription in bacteria. It is responsible for recognizing and binding to specific DNA sequences called promoter regions. Once bound, the sigma factor helps to recruit the RNA polymerase core enzyme to the promoter, allowing for the initiation of transcription. Therefore, the statement "It does not have a DNA-binding domain" is false because the sigma factor does indeed have a DNA-binding domain, which is essential for its function in transcription initiation.

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  • 5. 

    Which of the following is true about the holoenzyme of RNA pol in an open complex? 

    • A.

      The DNA is bound mainly to the sigma-subunit

    • B.

      There is no interaction between the sigma factor and the -10 region.

    • C.

      There are two Na+ ions in the core enzyme.

    • D.

      Region 2.4 of the sigma-factor binds the -35 region.

    • E.

      The DNA is bound mainly to the sigma-subunit, and there is no interaction between the sigma factor and the -10 region.

    Correct Answer
    A. The DNA is bound mainly to the sigma-subunit
    Explanation
    In an open complex, the holoenzyme of RNA pol binds mainly to the sigma-subunit. This means that the DNA is primarily interacting with the sigma-subunit of the holoenzyme. Additionally, the answer states that there is no interaction between the sigma factor and the -10 region. This means that the sigma factor, which is responsible for recognizing the promoter sequence, does not directly interact with the -10 region of the DNA.

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  • 6. 

    Dimethyl sulfate is used to 

    • A.

      Cleave DNA molecules

    • B.

      Denature or melt DNA promotes region

    • C.

      Remove methyl groups from RNA

    • D.

      Add methyl groups to DNA

    • E.

      Promote the binding of RNA polymerase to DNA

    Correct Answer
    D. Add methyl groups to DNA
    Explanation
    Dimethyl sulfate is a chemical compound that is commonly used to add methyl groups to DNA. Methyl groups are small chemical units that can be added to DNA molecules, and this process is known as methylation. Methylation of DNA can have various effects on gene expression and is an important mechanism in regulating gene activity. Dimethyl sulfate is often used in laboratory settings to specifically add methyl groups to DNA at specific locations, allowing researchers to study the effects of methylation on gene function. Therefore, the correct answer is that dimethyl sulfate is used to add methyl groups to DNA.

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  • 7. 

    Which of the following would most likely occur if an in vitro transcription assay is performed by using a cell extract containing an (alpha) subunit without the C-terminal domain? 

    • A.

      Polymerase would become tightly associated with the promoter.

    • B.

      Polymerase would become more susceptible to proteolysis.

    • C.

      Transcription will be accelerated.

    • D.

      Polymerase complex would be loosely associated with the promoter.

    • E.

      Recognition of the UP element would be faster.

    Correct Answer
    D. Polymerase complex would be loosely associated with the promoter.
    Explanation
    If an in vitro transcription assay is performed using a cell extract containing an (alpha) subunit without the C-terminal domain, the most likely occurrence would be that the polymerase complex would be loosely associated with the promoter. The C-terminal domain of the (alpha) subunit is responsible for stabilizing the interaction between the polymerase complex and the promoter. Without this domain, the interaction becomes weaker, resulting in a looser association between the polymerase complex and the promoter. This can affect the efficiency and accuracy of transcription.

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  • 8. 

    Place the following in the order that they occur during transcription initiation: 1- formation of open complex2- formation of closed complex3- promoter clearance4- synthesis of about 10 nucleotides 

    • A.

      1,2,3,4

    • B.

      2,1,4.3

    • C.

      2,1,2,4

    • D.

      3,1,2,4

    • E.

      3,2,1,4

    Correct Answer
    B. 2,1,4.3
    Explanation
    During transcription initiation, the first step is the formation of a closed complex, where the RNA polymerase binds to the promoter region of the DNA. Then, the closed complex transitions into an open complex, allowing the RNA polymerase to unwind the DNA and expose the template strand. Once the open complex is formed, the synthesis of about 10 nucleotides begins. Finally, promoter clearance occurs, where the RNA polymerase moves away from the promoter region and transcription continues. Therefore, the correct order is 2,1,4,3.

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  • 9. 

    Which of the following techniques is suitable for the study of DNA: RNA hybrid formation during transcription? 

    • A.

      DNA: protein crosslinking

    • B.

      DNAase I footprinting

    • C.

      DNA: RNA crosslinking

    • D.

      Two-dimensional PAGE

    • E.

      FRET analysis

    Correct Answer
    C. DNA: RNA crosslinking
    Explanation
    DNA: RNA crosslinking is a suitable technique for the study of DNA: RNA hybrid formation during transcription. This technique involves the covalent crosslinking of DNA and RNA molecules, allowing the identification and characterization of DNA: RNA hybrids. By crosslinking the two molecules together, researchers can analyze the interaction between DNA and RNA, providing insights into the process of transcription and the formation of DNA: RNA hybrids.

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  • 10. 

    Which of the following are characteristic of intrinsic terminator elements? 

    • A.

      They contain an inverted repeat.

    • B.

      They contain a hairpin loop.

    • C.

      They contain several T's in the non-template strand of DNA.

    • D.

      Only they contain an inverted repeat, and they contain a hairpin loop are correct.

    • E.

      They contain an inverted repeat, they contain a hairpin loop, and they contain several T's in the non-template strand of DNA.

    Correct Answer
    E. They contain an inverted repeat, they contain a hairpin loop, and they contain several T's in the non-template strand of DNA.
    Explanation
    Intrinsic terminator elements are DNA sequences that play a role in the termination of transcription. They contain an inverted repeat, which is a sequence that is repeated in the opposite orientation. This inverted repeat forms a hairpin loop structure, which helps in the termination process. Additionally, intrinsic terminator elements also contain several T's in the non-template strand of DNA. Therefore, the correct answer is that intrinsic terminator elements contain an inverted repeat, a hairpin loop, and several T's in the non-template strand of DNA.

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  • 11. 

    Which of the following is incorrect regarding rho-dependent terminators? 

    • A.

      They contain several T's in the non-template strand.

    • B.

      They consist of inverted repeats.

    • C.

      They form hairpin loops.

    • D.

      Rho binds to the growing transcripts.

    • E.

      Rho has no helicase activity.

    Correct Answer
    E. Rho has no helicase activity.
    Explanation
    Rho-dependent terminators are a type of termination signal in bacterial transcription. They contain several T's in the non-template strand, consist of inverted repeats, and form hairpin loops. Rho, a protein, binds to the growing transcripts and acts as a helicase to unwind the RNA-DNA hybrid, facilitating termination. Therefore, the statement that Rho has no helicase activity is incorrect.

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  • 12. 

    Which of the following does not have a DNA-binding domain? 

    • A.

      Sigma-factor

    • B.

      UP region

    • C.

      Alpha-factor

    • D.

      Beta

    • E.

      Beta prime

    Correct Answer
    B. UP region
    Explanation
    The UP region does not have a DNA-binding domain. The UP region is a regulatory sequence found in bacterial promoters that helps to increase the binding of RNA polymerase to the promoter region, thereby enhancing transcription. It does not directly bind to DNA itself, but rather interacts with the alpha subunit of RNA polymerase to facilitate transcription initiation. In contrast, sigma-factors, alpha-factor, beta, and beta prime are all protein subunits of RNA polymerase that contain DNA-binding domains and are involved in the recognition and binding of specific DNA sequences during transcription initiation.

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  • 13. 

    Which of the following bond is lactose linked between glucose and galactose? 

    • A.

      It is a glycoside bond.

    • B.

      The linkage is through (beta)-1, 4 bond.

    • C.

      It can be broken by (beta)-galactosidase.

    • D.

      It is a covalent bond.

    • E.

      All of the choices are correct.

    Correct Answer
    E. All of the choices are correct.
    Explanation
    The given correct answer is "All of the choices are correct." This means that all of the statements provided in the question are true. The bond between glucose and galactose in lactose is a glycoside bond, specifically a (beta)-1, 4 bond. This bond can be broken by (beta)-galactosidase, and it is a covalent bond. Therefore, all of the statements are correct.

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  • 14. 

    If an operon is under negative control, it means that the operon is never operational.

    • A.

      True

    • B.

      False

    Correct Answer
    B. False
    Explanation
    If an operon is under negative control, it means that the operon is regulated by a repressor protein. This protein binds to the operator region of the operon and prevents RNA polymerase from transcribing the genes. Therefore, the operon is only operational when the repressor protein is not bound to the operator. Hence, the statement that an operon under negative control is never operational is false.

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  • 15. 

    In the absence of glucose, there will be a rapid accumulation of lactose in the following mutant: I- Oc Z+ Y-I- O+ Z+ Y-

    • A.

      True

    • B.

      False

    Correct Answer
    B. False
    Explanation
    In the absence of glucose, there will not be a rapid accumulation of lactose in the given mutant. This can be determined by analyzing the genotype of the mutant. The presence of "I-" indicates that the mutant lacks the lac repressor, which normally binds to the operator region and prevents transcription of the lac operon in the absence of glucose. Additionally, the presence of "Oc" indicates that the operator region is constitutively active, meaning that the lac operon will be transcribed regardless of the presence of glucose. Therefore, in the absence of glucose, the mutant will still accumulate lactose. Hence, the correct answer is False.

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  • 16. 

    The following is an example of a cis-dominant mutation: I-d O+ Z+ Y+ A-

    • A.

      True

    • B.

      False

    Correct Answer
    B. False
    Explanation
    The given sequence of blood types (I-d O+ Z+ Y+ A-) does not represent a cis-dominant mutation. A cis-dominant mutation occurs when a mutation affects only one copy of a gene, while the other copy remains normal. In this case, the blood types listed do not indicate any mutation or alteration in the gene responsible for blood type. Therefore, the correct answer is false.

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  • 17. 

    Which of the following statements is true about a lac operon with the following genotype? I- Oc Z+ Y-I- O+ Z- Y-

    • A.

      A repressor protein will be constitutively produced.

    • B.

      (Beta)- galactosidase will be constitutively produced.

    • C.

      A trans-dominant mutation is present.

    • D.

      Permease will be constitutively produced.

    • E.

      Both a repressor protein will be constitutively produced and a cis-dominant mutation is present.

    Correct Answer
    B. (Beta)- galactosidase will be constitutively produced.
    Explanation
    The lac operon with the given genotype I- Oc Z+ Y-I- O+ Z- Y- indicates that the operator region (O) is not functional, and the repressor protein cannot bind to it. As a result, the repressor protein will not be produced, leading to the constitutive production of (Beta)-galactosidase. The other options do not align with the given genotype and are therefore incorrect.

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  • 18. 

    Which of the following plasmids could be used to restore inducible regulation of (Beta)-galactosidase in this mutant: I+ Oc Z- Y+ A+? 

    • A.

      I+ Oc Z- Y+ A+

    • B.

      I+Oc Z- Y- A-

    • C.

      I- O+ Z+ Y+ A+

    • D.

      I- Oc Z+ Y+ A+

    • E.

      Both I+ Oc Z- Y+ A+ and I+ Oc Z- Y- A- are correct

    Correct Answer
    C. I- O+ Z+ Y+ A+
  • 19. 

    Which of the following organisms would produce functional lac products in the presence of IPTG? 

    • A.

      Is Oc Z- Y+ A+

    • B.

      Is Oc Z- Y- A-

    • C.

      I+ O+ Z+ Y+ A+

    • D.

      I- Oc Z+ Y+ A+

    • E.

      Both Is Oc Z- Y+ A+ and Is Oc Z- Y+ A+ are correct.

    Correct Answer
    C. I+ O+ Z+ Y+ A+
    Explanation
    The organisms that would produce functional lac products in the presence of IPTG are those that have all the genes necessary for lactose metabolism (lac operon) present and functional. In this case, the answer is I+ O+ Z+ Y+ A+, which indicates that all the necessary genes (I, O, Z, Y, A) are present and functional in the organism.

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  • 20. 

    Which of the following statements is true about a lac operon with this genotype? I-d O+ Z+ Y- A- 

    • A.

      The operon is repressible.

    • B.

      The operon is non-repressible.

    • C.

      The mutation is cis-dominant.

    • D.

      The operon is uninducible.

    • E.

      None of the choices is correct.

    Correct Answer
    B. The operon is non-repressible.
    Explanation
    The lac operon with the given genotype (I-d O+ Z+ Y- A-) is non-repressible because the presence of the I-d mutation prevents the repressor protein from binding to the operator region, allowing the lac operon to be transcribed and expressed continuously. This means that even in the absence of lactose, the operon will still be active and producing the enzymes necessary for lactose metabolism.

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  • 21. 

    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.

      DNA fingerprinting

    • C.

      Run-off transcription assay

    • D.

      DNA sequencing

    • E.

      RACE

    Correct Answer
    C. Run-off transcription assay
    Explanation
    A 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 allowing RNA polymerase to transcribe the lac DNA template in the presence of labeled nucleotides, and then isolating and analyzing the newly synthesized RNA. By comparing the amount of labeled RNA produced 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 evidence of transcription initiation.

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  • 22. 

    In an in vitro transcription assay to detect the level of transcription from a lac operon, one of the RNA precursors was tagged with a fluorescent label on the y-phosphate. Which of the following is a likely outcome if transcription occurred? 

    • A.

      The level of fluorescence would increase along the new RNA strand.

    • B.

      Each RNA would have a fluorescent tag on every other nucleotide.

    • C.

      The level of fluorescence would be very high in the assay buffer.

    • D.

      No fluorescence would be detected in the buffer at the end of the assay.

    • E.

      All of the choices are possible.

    Correct Answer
    C. The level of fluorescence would be very high in the assay buffer.
    Explanation
    If transcription occurred in the in vitro transcription assay, the RNA precursors with a fluorescent label on the y-phosphate would be synthesized. These labeled RNA molecules would contribute to the level of fluorescence in the assay buffer, resulting in a high level of fluorescence. Therefore, the correct answer is that the level of fluorescence would be very high in the assay buffer.

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  • 23. 

    Which of the following is most likely to occur if the level of glucose is low in a bacterial cell? 

    • A.

      Cyclic-AMP levels will be depressed.

    • B.

      CAP will assist in stimulating transcription of the lac operon if lactose is added

    • C.

      CRP activity will be inhibited

    • D.

      Both Cyclic-AMP levels will be depressed and CRP activity will be inhibited.

    • E.

      None of the choices is correct.

    Correct Answer
    B. CAP will assist in stimulating transcription of the lac operon if lactose is added
    Explanation
    If the level of glucose is low in a bacterial cell, it indicates that the cell is in need of an alternative energy source. In this situation, CAP (catabolite activator protein) will assist in stimulating transcription of the lac operon if lactose is added. This is because the lac operon is responsible for the metabolism of lactose as an alternative energy source. When glucose is scarce, CAP helps to activate the lac operon, allowing the cell to utilize lactose for energy production.

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  • 24. 

    Which of the following is true about the action of CAP at the lac promoter? 

    • A.

      CAP monomer binds directly to the promoter, stimulating polymerase to bind

    • B.

      CAP-AMP blocks the recruitment of polymerase to the promoter

    • C.

      CAP blocks the (alpha) CTD of RNA polymerase

    • D.

      Binding of the CAP-cAMP to the lac activator-binding site recruits RNA polymerase.

    • E.

      CAP monomer binds directly to the promoter, stimulating polymerase to bind, and CAP blocks the (alpha) CTD of RNA polymerase

    Correct Answer
    D. Binding of the CAP-cAMP to the lac activator-binding site recruits RNA polymerase.
    Explanation
    The correct answer states that the binding of CAP-cAMP to the lac activator-binding site recruits RNA polymerase. This means that CAP-cAMP plays a role in bringing RNA polymerase to the lac promoter, which is necessary for gene expression. This explanation is supported by the fact that CAP is known to act as an activator of transcription in the lac operon system, and its binding to the activator-binding site is necessary for the recruitment of RNA polymerase.

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  • 25. 

    Which of the following is true about the regulation of the trp operon? 

    • A.

      An aporepressor is involved.

    • B.

      A corepressoris involved.

    • C.

      Attenuation is one of the control mechanisms.

    • D.

      A negative control is involved.

    • E.

      All of the choices are true.

    Correct Answer
    E. All of the choices are true.
    Explanation
    All of the choices are true. The regulation of the trp operon involves an aporepressor, which is a protein that is inactive until it binds to a corepressor molecule. Attenuation is also a control mechanism in the regulation of the trp operon, where the transcription of the operon is prematurely terminated under conditions of high tryptophan levels. Additionally, the regulation of the trp operon is a negative control mechanism, as it involves the repression of gene expression.

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  • 26. 

    Which of the following is absent in an operon? 

    • A.

      Operator

    • B.

      Promoter

    • C.

      Intron

    • D.

      Both operator and promoter

    • E.

      Both promoter and intron

    Correct Answer
    C. Intron
    Explanation
    An operon is a functional unit of DNA that consists of a promoter, operator, and one or more genes. The promoter is responsible for initiating the transcription of the genes, while the operator controls the access of RNA polymerase to the genes. In an operon, introns are absent. Introns are non-coding sequences of DNA that are present in eukaryotic genes but are spliced out during mRNA processing. In prokaryotes, which have operons, the genes are typically organized in a continuous sequence without introns. Therefore, the correct answer is intron.

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  • 27. 

    Recent experimental evidence has shown that only one operator sequence is needed for maximal expression.

    • A.

      True

    • B.

      False

    Correct Answer
    B. False
    Explanation
    The recent experimental evidence suggests that more than one operator sequence is needed for maximal expression, which contradicts the statement. Therefore, the correct answer is False.

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  • 28. 

    One hypothesis explaining the mechanism of repression of the lac operon states that the repressor binds the operator and blocks access by the polymerase to the adjacent promoter.

    • A.

      True

    • B.

      False

    Correct Answer
    A. True
    Explanation
    This statement is true because the lac operon is a system in bacteria that controls the expression of genes involved in lactose metabolism. The lac repressor protein can bind to the operator region of the operon, preventing RNA polymerase from accessing the promoter and initiating transcription. This mechanism allows the bacteria to regulate the expression of the lac genes in response to the presence or absence of lactose in the environment.

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  • 29. 

    CAMP level is increased by the presence of a high level of glucose 

    • A.

      True

    • B.

      False

    Correct Answer
    B. False
    Explanation
    The statement is false because the presence of a high level of glucose does not increase cAMP levels. cAMP (cyclic adenosine monophosphate) is a molecule involved in cellular signaling and is typically increased by the activation of certain receptors or enzymes, not by the presence of glucose.

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  • 30. 

    Arabinose can derepress the ara operon by causing AraC to loosen its attachment to araO2 and bind to araI2 instead. 

    • A.

      True

    • B.

      False

    Correct Answer
    A. True
    Explanation
    Arabinose can derepress the ara operon by causing AraC to loosen its attachment to araO2 and bind to araI2 instead. This means that when arabinose is present, it triggers a change in the binding of AraC protein, allowing it to activate gene expression from the araI2 site. This process leads to the production of enzymes necessary for the metabolism of arabinose. Therefore, the statement is true.

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  • 31. 

    Attenuation causes the length of the RNA molecules to be significantly extended. 

    • A.

      True

    • B.

      False

    Correct Answer
    B. False
    Explanation
    Attenuation does not cause the length of RNA molecules to be significantly extended. Attenuation is a regulatory mechanism in gene expression where the transcription of certain genes is prematurely terminated. It involves the formation of a hairpin structure in the RNA molecule, which leads to the termination of transcription. This process does not result in the extension of RNA molecules. Therefore, the statement is false.

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  • 32. 

    The following conditions lead to a change in gene expression patterns in bacterial cells.

    • A.

      Sporulation

    • B.

      Heat shock

    • C.

      Nutrient availability

    • D.

      Nitrogen deprivation

    • E.

      Phage infection

    Correct Answer
    A. Sporulation
    Explanation
    Sporulation is a process in which bacteria form endospores as a survival mechanism under unfavorable conditions. During sporulation, there is a significant change in gene expression patterns in bacterial cells. This is because the bacteria need to activate specific genes that are involved in the formation of endospores and deactivate genes that are not necessary for survival in the dormant state. Therefore, sporulation leads to a change in gene expression patterns in bacterial cells.

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  • 33. 

    Which of the following sigma specificity factor is involved in middle gene transcription during SPO1 phage infection of bacteria?

    • A.

      Gp33

    • B.

      Gp28

    • C.

      Gp34

    • D.

      Gp43

    • E.

      Host sigma

    Correct Answer
    B. Gp28
    Explanation
    During SPO1 phage infection of bacteria, gp28 is the sigma specificity factor involved in middle gene transcription.

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  • 34. 

    Which of the following is true about the early phase of T7 infection of bacteria?

    • A.

      Class II genes are the first to be transcribed.

    • B.

      The host polymerase transcribes class II genes.

    • C.

      The phage-encoded polymerase transcribes the class I phage genes.

    • D.

      Class I genes are the first to be transcribed.

    • E.

      All of the choices are correct.

    Correct Answer
    D. Class I genes are the first to be transcribed.
    Explanation
    In the early phase of T7 infection of bacteria, the first genes to be transcribed are the Class I genes. This means that the phage-encoded polymerase transcribes the Class I phage genes, indicating that the correct answer is "Class I genes are the first to be transcribed." This suggests that the phage takes control of the transcription process early on in the infection, utilizing its own polymerase to transcribe its own genes.

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  • 35. 

    Differences in the transcription of T7 and SPO1 genes are 

    • A.

      T7 phage encodes a new RNA polymerase for phage genes.

    • B.

      SPO1 phage uses the new sigma factor for phage gene.

    • C.

      Both correct

    • D.

      Both wrong

    • E.

      They have the same factors.

    Correct Answer
    C. Both correct
    Explanation
    The correct answer is "Both correct". This is because the given information states that T7 phage encodes a new RNA polymerase for phage genes, while SPO1 phage uses the new sigma factor for phage genes. This implies that there are differences in the transcription of T7 and SPO1 genes, supporting the statement that both options are correct.

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  • 36. 

    Which of the following conditions does not lead to a change in gene expression patterns in bacteria cells? 

    • A.

      Sporulation

    • B.

      Heat shock

    • C.

      Nutrient availability

    • D.

      Nitrogen deprivation

    • E.

      None of the choices are correct

    Correct Answer
    E. None of the choices are correct
    Explanation
    All of the conditions listed in the question can lead to a change in gene expression patterns in bacteria cells. Sporulation, heat shock, nutrient availability, and nitrogen deprivation are all known to trigger changes in gene expression. Therefore, none of the choices are correct in terms of not leading to a change in gene expression patterns in bacteria cells.

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  • 37. 

    Predict the outcome of the removal of the N from the DNA sequence of the (lambda) phage:

    • A.

      Lysogenic cycle will be induced

    • B.

      Entry into the bacterial host will be blocked

    • C.

      Antitermination will be affected

    • D.

      Transcription from the left promoter will be interrupted

    • E.

      Antitermination will be affected and transcription from the left promoter will be interrupted

    Correct Answer
    E. Antitermination will be affected and transcription from the left promoter will be interrupted
    Explanation
    The removal of the N from the DNA sequence of the (lambda) phage will affect antitermination and interrupt transcription from the left promoter. The N protein is responsible for preventing transcription termination and allowing continuous transcription of the phage genes. Without the N protein, antitermination will be affected, leading to premature termination of transcription. Additionally, the left promoter is dependent on the N protein for efficient transcription initiation. Therefore, the removal of N will interrupt transcription from the left promoter.

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  • 38. 

    Which of the following is a possible mechanism for the action of NusA in termination? 

    • A.

      It binds to the RNA polymerase causing it to stall.

    • B.

      It stimulates termination at the intrinsic terminator by facilitating hairpin loop formation.

    • C.

      It binds to NusB to promote detachment of the RNA polymerase for the DNA template.

    • D.

      It interacts with S10 to promote dissociation of the RNA polymerase.

    • E.

      It binds to the RNA polymerase causing it to stall, and it stimulates termination at the intrinsic terminator by facilitating hairpin loop formation.

    Correct Answer
    B. It stimulates termination at the intrinsic terminator by facilitating hairpin loop formation.
    Explanation
    NusA is a protein that plays a role in transcription termination in bacteria. It interacts with the RNA polymerase and aids in the formation of a hairpin loop structure at the termination site, which leads to the release of the RNA transcript. This mechanism is known as intrinsic termination. The other options mentioned in the question, such as binding to the RNA polymerase causing it to stall or promoting detachment of the RNA polymerase, are not directly associated with the action of NusA in termination. Therefore, the correct answer is that NusA stimulates termination at the intrinsic terminator by facilitating hairpin loop formation.

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  • 39. 

    Which of the following is involved in maintaining lysogeny in (lambda) phage? 

    • A.

      P (RE)

    • B.

      P (RM)

    • C.

      P (RL)

    • D.

      P (RO)

    • E.

      All of the choices are correct

    Correct Answer
    B. P (RM)
    Explanation
    P (RM) is involved in maintaining lysogeny in (lambda) phage.

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  • 40. 

    Which of the following operators is a target of CI gene production? 

    • A.

      O RI

    • B.

      O R2

    • C.

      O L1

    • D.

      O L2

    • E.

      All of the choices are correct

    Correct Answer
    E. All of the choices are correct
    Explanation
    All of the operators mentioned in the options (O RI, O R2, O L1, O L2) are targets of CI gene production.

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  • 41. 

    Which of the following is true about the interaction with the (lambda) repressor and RNA polymerase at P (RM)? 

    • A.

      Interaction b/w repressor and polymerase is essential for the activation of transcription.

    • B.

      Interaction involves region 4 of sigma-factor.

    • C.

      Interaction can be disrupted by the binding of CIII.

    • D.

      Interaction between repressor and polymerase is essential for activation of transcription, and interaction involves region 4 of the sigma-factor.

    • E.

      Interaction between repressor and polymerase is essential for activation of transcription, and interaction can be disrupted by the binding of CIII.

    Correct Answer
    D. Interaction between repressor and polymerase is essential for activation of transcription, and interaction involves region 4 of the sigma-factor.
    Explanation
    The correct answer states that the interaction between the repressor and polymerase is essential for the activation of transcription, and this interaction specifically involves region 4 of the sigma-factor. This means that in order for transcription to occur, the repressor protein must interact with the RNA polymerase enzyme, and this interaction is mediated by region 4 of the sigma-factor. This interaction is crucial for the proper functioning of the transcription process.

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  • 42. 

    Which of the following is true about the cro gene? 

    • A.

      It must be stimulated during lysogeny.

    • B.

      Its product promotes repressor activity.

    • C.

      Cro repression is important to lysogeny.

    • D.

      It is adjacent to the CIII gene.

    • E.

      It must be stimulated during lysogeny, and its products promote repressor activity.

    Correct Answer
    C. Cro repression is important to lysogeny.
    Explanation
    The correct answer is "cro repression is important to lysogeny." This means that the repression of the cro gene plays a crucial role in the process of lysogeny. Lysogeny is a state in which a bacteriophage's DNA is integrated into the host cell's genome and remains dormant. The cro gene codes for a protein that acts as a repressor, preventing the expression of other genes involved in the lytic cycle of the bacteriophage. This repression is necessary for the establishment and maintenance of lysogeny.

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  • 43. 

    During an experiment to study the rate of infection of bacteria with T4, bacteria with T4, bacterial cultures were accidentally exposed to a UV light source. Which of the following would be expected for this exposure? 

    • A.

      RecA gene is turned off

    • B.

      Coprotease activity in RecA protein is activated.

    • C.

      SOS response is induced.

    • D.

      Coprotease activity in RecA protein is activated, and SOS response is induced.

    • E.

      RecA gene is turned off, and coprotease activity in RecA protein is activated are correct

    Correct Answer
    D. Coprotease activity in RecA protein is activated, and SOS response is induced.
    Explanation
    When bacterial cultures are exposed to UV light, it can cause DNA damage. The activation of coprotease activity in the RecA protein helps to repair this DNA damage by promoting the cleavage of LexA repressor, which leads to the induction of the SOS response. The SOS response is a cellular response to DNA damage, where a variety of repair mechanisms are activated. Therefore, the correct answer is that coprotease activity in RecA protein is activated, and SOS response is induced.

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  • 44. 

    You were asked to choose a method to separate a mixture of eukaryotic polymerases. Which of the following methods would you choose? 

    • A.

      Chromatography

    • B.

      Western blotting

    • C.

      S1 nuclease assay

    • D.

      DNASE 1 footprinting

    • E.

      Chromatography and UMP incorporation assay

    Correct Answer
    E. Chromatography and UMP incorporation assay
    Explanation
    Chromatography is a commonly used method to separate mixtures based on their different physical and chemical properties. In the context of separating eukaryotic polymerases, chromatography can be used to separate the different types of polymerases based on their size, charge, or affinity for certain molecules. UMP incorporation assay, on the other hand, is a technique used to measure the activity of polymerases by monitoring the incorporation of UMP molecules into a growing DNA or RNA chain. By combining chromatography with UMP incorporation assay, one can not only separate the polymerases but also determine their activity levels, providing a more comprehensive analysis of the mixture.

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  • 45. 

    Why three RNA polymerases? How many types of RNA are in the cells? 

    • A.

      RRNA: 28S, 18s, 5.8S, 5s

    • B.

      MRNA

    • C.

      TRNA

    • D.

      Small nuclear RNA or snRNA

    • E.

      Other RNAs

    Correct Answer
    A. RRNA: 28S, 18s, 5.8S, 5s
    Explanation
    The three RNA polymerases are necessary because each one is responsible for transcribing a specific type of RNA. RNA polymerase I transcribes the genes that encode for ribosomal RNA (rRNA), which are essential components of the ribosomes. The rRNA molecules produced by RNA polymerase I include the 28S, 18S, 5.8S, and 5S rRNA. RNA polymerase II transcribes the genes that encode for messenger RNA (mRNA), which carries the genetic information from DNA to the ribosomes for protein synthesis. RNA polymerase III transcribes the genes that encode for transfer RNA (tRNA) and other small non-coding RNAs, including small nuclear RNA (snRNA). Therefore, the presence of three RNA polymerases allows for the transcription of different types of RNA molecules in the cell.

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  • 46. 

    Which of the following are products of RNA polymerase II activity? 

    • A.

      TRNA

    • B.

      SnRNA

    • C.

      HnRNA

    • D.

      TRNA and snRNA

    • E.

      SnRNA and hnRNA

    Correct Answer
    E. SnRNA and hnRNA
    Explanation
    RNA polymerase II is responsible for transcribing protein-coding genes in eukaryotic cells. It synthesizes heterogeneous nuclear RNA (hnRNA), which is the precursor to messenger RNA (mRNA). hnRNA undergoes further processing to become mature mRNA. Additionally, RNA polymerase II also transcribes small nuclear RNA (snRNA), which is involved in the splicing of pre-mRNA. Therefore, the correct answer is snRNA and hnRNA.

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  • 47. 

    Which of the following products are made by RNA polymerase III? 

    • A.

      7SL RNA

    • B.

      5SrRNA

    • C.

      SnRNA

    • D.

      7SL RNA and 5S rRNA and U6 snRNA

    • E.

      7SL RNA, 5S rRNA and snRNA

    Correct Answer
    D. 7SL RNA and 5S rRNA and U6 snRNA
    Explanation
    RNA polymerase III is responsible for transcribing small non-coding RNAs, including 7SL RNA, 5S rRNA, and U6 snRNA. These RNA molecules play important roles in various cellular processes. 7SL RNA is involved in protein targeting and secretion, 5S rRNA is a component of the ribosome, and U6 snRNA is involved in splicing of pre-mRNA. Therefore, the correct answer is 7SL RNA and 5S rRNA and U6 snRNA.

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  • 48. 

    Which of the following statements about ribosomal genes is incorrect? 

    • A.

      They have a higher GC content compared to other nuclear genes.

    • B.

      They have a different base composition compared to other nuclear genes.

    • C.

      There are no repetitive sequences.

    • D.

      They are found in the nucleolus.

    • E.

      They are transcribed by RNA polymerases I and III.

    Correct Answer
    C. There are no repetitive sequences.
    Explanation
    Ribosomal genes do have repetitive sequences, so the statement "There are no repetitive sequences" is incorrect. These genes contain multiple copies of the same sequence, which are repeated in tandem. This repetition allows for the production of multiple ribosomal RNA molecules, which are essential for protein synthesis.

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  • 49. 

    An antibody was produced and used in an immunohistochemical assay to detect the location of RNA polymerase I in a thin section of a liver biopsy. The section was antibody-labeled with a fluorescent tag, then viewed under a fluorescent microscope. State where you would expect to find positive staining if the enzymes are present. 

    • A.

      Rough endoplasmic reticulum

    • B.

      Nucleolus

    • C.

      Endoplasmic reticulum

    • D.

      Ribosomes

    Correct Answer
    B. Nucleolus
    Explanation
    The correct answer is nucleolus. RNA polymerase I is responsible for transcribing ribosomal RNA (rRNA) in the nucleolus, which is a distinct subcompartment within the nucleus. Therefore, positive staining would be expected in the nucleolus if the enzymes are present. The other options, rough endoplasmic reticulum, endoplasmic reticulum, and ribosomes, are not directly involved in the transcription of rRNA.

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  • 50. 

    A new mutant cell line was developed and was found to be defective in polymerase III activity. Which of the following is likely to be observed in this cell line? 

    • A.

      There will be an overabundance of secreted proteins.

    • B.

      Splicing function is impaired.

    • C.

      There will be an overproduction of 7 SL RNA

    • D.

      There will be an overabundance of secreted proteins, and splicing function will be impaired correctly.

    • E.

      There will be an overabundance of secreted proteins, splicing function will be impaired, and there will be an overproduction of 7 SL RNA correct.

    Correct Answer
    B. Splicing function is impaired.
    Explanation
    If the mutant cell line is defective in polymerase III activity, it is likely to affect the splicing function. Polymerase III is responsible for transcribing small nuclear RNAs (snRNAs), which are essential for the splicing process. Therefore, if polymerase III activity is impaired, it would result in a defective splicing function. This explains why the correct answer is "Splicing function is impaired."

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  • Sep 02, 2023
    Quiz Edited by
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