Ultimate Molecular Biology Quiz

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Quizzes Created: 16 | Total Attempts: 13,090

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1/80 Questions

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.
- True
- False

About This 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.

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2.

Epitope tagging is an efficient way to immunoprecipitate to isolate an entire polymerase complex.
- True
- False
Correct Answer

A. True

Explanation

Epitope tagging involves attaching a small protein sequence, known as an epitope tag, to a target protein. This tag allows for easy detection and purification of the tagged protein using specific antibodies. In the given statement, it is stated that epitope tagging is an efficient way to immunoprecipitate and isolate an entire polymerase complex. This is true because by tagging a specific subunit of the polymerase complex with an epitope tag, the entire complex can be precipitated and isolated using antibodies that recognize the epitope tag.

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3.

Arabinose can derepress the ara operon by causing AraC to loosen its attachment to araO2 and bind to araI2 instead.
- True
- 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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4.

Which of the following is not present in the core RNA polymerase?
- B (Beta)
- B' (Beta prime)
- Sigma (o)
- Both (Beta) and (sigma)
Correct Answer

A. 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
5.

Which of the following is absent in an operon?
- Operator
- Promoter
- Intron
- Both operator and promoter
- Both promoter and intron
Correct Answer

A. 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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6.

TFIID collaborates with TFIIA to bind to the TATA box.
- True
- False
Correct Answer

A. True

Explanation

TFIID and TFIIA are transcription factors that are involved in the initiation of transcription in eukaryotic cells. TFIID is responsible for recognizing and binding to the TATA box, a DNA sequence located in the promoter region of genes. TFIIA, on the other hand, helps stabilize the binding of TFIID to the TATA box. Therefore, TFIID and TFIIA work together to bind to the TATA box, making the statement true.

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7.

If an operon is under negative control, it means that the operon is never operational.
- True
- False
Correct Answer

A. 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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8.

Attenuation causes the length of the RNA molecules to be significantly extended.
- True
- False
Correct Answer

A. 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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9.

Eukaryotic ribosomal RNA genes are transcribed by RNA polymerase II.
- True
- False
Correct Answer

A. False

Explanation

Eukaryotic ribosomal RNA genes are transcribed by RNA polymerase I, not RNA polymerase II. RNA polymerase I is responsible for transcribing the genes that encode for the large rRNA molecules found in the ribosomes of eukaryotic cells. RNA polymerase II, on the other hand, transcribes protein-coding genes and some non-coding RNA genes. Therefore, the correct answer is False.

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10.

RNA polymerase prevents the binding of TFIIF to the promoter region
- True
- False
Correct Answer

A. False

Explanation

RNA polymerase does not prevent the binding of TFIIF to the promoter region. In fact, TFIIF is one of the general transcription factors that helps RNA polymerase bind to the promoter region and initiate transcription. TFIIF stabilizes the binding of RNA polymerase to the promoter and also plays a role in transcription initiation. Therefore, the correct answer is False.

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11.

CAMP level is increased by the presence of a high level of glucose
- True
- False
Correct Answer

A. 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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12.

Recent experimental evidence has shown that only one operator sequence is needed for maximal expression.
- True
- False
Correct Answer

A. 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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13.

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
- 1,2,3,4
- 2,1,4.3
- 2,1,2,4
- 3,1,2,4
- 3,2,1,4
Correct Answer

A. 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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14.

Roeder and Rutter showed that there are four polymerases operating in the eukaryotic cell.
- True
- False
Correct Answer

A. False

Explanation

Roeder and Rutter did not show that there are four polymerases operating in the eukaryotic cell. Therefore, the statement is false.

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15.

Which of the following techniques is suitable for the study of DNA: RNA hybrid formation during transcription?
- DNA: protein crosslinking
- DNAase I footprinting
- DNA: RNA crosslinking
- Two-dimensional PAGE
- FRET analysis
Correct Answer

A. 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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16.

Which of the following statements is false regarding the sigma factor?
- It does not have a DNA-binding domain.
- Interaction with the core enzyme unmasks the DNA-binding region.
- Subregions 2.4 and 4.4 are involved in promoter recognition.
- It can also bind to the nontemplate strand.
- 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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17.

Which of the following is most likely to occur if the level of glucose is low in a bacterial cell?
- Cyclic-AMP levels will be depressed.
- CAP will assist in stimulating transcription of the lac operon if lactose is added
- CRP activity will be inhibited
- Both Cyclic-AMP levels will be depressed and CRP activity will be inhibited.
- None of the choices is correct.
Correct Answer

A. 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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18.

Which of the following is true about the holoenzyme of RNA pol in an open complex?
- The DNA is bound mainly to the sigma-subunit
- There is no interaction between the sigma factor and the -10 region.
- There are two Na+ ions in the core enzyme.
- Region 2.4 of the sigma-factor binds the -35 region.
- 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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19.

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-
- True
- False
Correct Answer

A. 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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20.

Which of the following sigma specificity factor is involved in middle gene transcription during SPO1 phage infection of bacteria?
- Gp33
- Gp28
- Gp34
- Gp43
- Host sigma
Correct Answer

A. Gp28

Explanation

During SPO1 phage infection of bacteria, gp28 is the sigma specificity factor involved in middle gene transcription.

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21.

Which of the following techniques would be most useful in testing the ability of the core RNA polymerase to bind to the promoter?
- Gel filtration
- S1 mapping
- Immunoblotting
- DNase footprinting
- Northern blotting
Correct Answer

A. 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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1
22.

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.
- Rough endoplasmic reticulum
- Nucleolus
- Endoplasmic reticulum
- Ribosomes
Correct Answer

A. 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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23.

Which of the following can be found in DNA-binding domains?
- Zinc fingers
- Glutamine-rich regions
- BZIP motifs
- BHLH motif
- All of the choices are correct
Correct Answer

A. All of the choices are correct

Explanation

DNA-binding domains can contain various motifs and regions that allow them to bind to DNA. Zinc fingers are one type of DNA-binding motif that consists of a zinc ion coordinated by cysteine and histidine residues. Glutamine-rich regions are another type of DNA-binding domain that contains an abundance of glutamine amino acids and can interact with DNA. bZIP motifs, short for basic leucine zipper motifs, are DNA-binding domains that consist of a basic region and a leucine zipper region. bHLH motifs, short for basic helix-loop-helix motifs, are also DNA-binding domains that contain a basic region and two helix-loop-helix regions. Therefore, all of the choices are correct.

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24.

Dimethyl sulfate is used to
- Cleave DNA molecules
- Denature or melt DNA promotes region
- Remove methyl groups from RNA
- Add methyl groups to DNA
- Promote the binding of RNA polymerase to DNA
Correct Answer

A. 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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25.

Which of the following does not have a DNA-binding domain?
- Sigma-factor
- UP region
- Alpha-factor
- Beta
- Beta prime
Correct Answer

A. 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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26.

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?
- Include the RNA holoenzyme in the reaction.
- Use the core enzyme of RNA polymerase.
- Enrich the preparation with sigma subunit
- Use intact DNA.
- Include the RNA holoenzyme in the reaction and using the core enzyme of RNA polymerase are both effective.
Correct Answer

A. 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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2
27.

Differences in the transcription of T7 and SPO1 genes are
- T7 phage encodes a new RNA polymerase for phage genes.
- SPO1 phage uses the new sigma factor for phage gene.
- Both correct
- Both wrong
- They have the same factors.
Correct Answer

A. 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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28.

Which of the following is true about the action of CAP at the lac promoter?
- CAP monomer binds directly to the promoter, stimulating polymerase to bind
- CAP-AMP blocks the recruitment of polymerase to the promoter
- CAP blocks the (alpha) CTD of RNA polymerase
- Binding of the CAP-cAMP to the lac activator-binding site recruits RNA polymerase.
- CAP monomer binds directly to the promoter, stimulating polymerase to bind, and CAP blocks the (alpha) CTD of RNA polymerase
Correct Answer

A. 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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29.

Why three RNA polymerases? How many types of RNA are in the cells?
- RRNA: 28S, 18s, 5.8S, 5s
- MRNA
- TRNA
- Small nuclear RNA or snRNA
- 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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30.

Which of the following bond is lactose linked between glucose and galactose?
- It is a glycoside bond.
- The linkage is through (beta)-1, 4 bond.
- It can be broken by (beta)-galactosidase.
- It is a covalent bond.
- All of the choices are correct.
Correct Answer

A. 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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31.

Which of the following statements is true about a lac operon with this genotype? I-d O+ Z+ Y- A-
- The operon is repressible.
- The operon is non-repressible.
- The mutation is cis-dominant.
- The operon is uninducible.
- None of the choices is correct.
Correct Answer

A. 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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32.

Which of the following products are made by RNA polymerase III?
- 7SL RNA
- 5SrRNA
- SnRNA
- 7SL RNA and 5S rRNA and U6 snRNA
- 7SL RNA, 5S rRNA and snRNA
Correct Answer

A. 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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33.

Which of the following are typical features of transcriptional activators?
- Transcriptional activators
- DNA-binding domain
- Kinase domain
- Transcription-activation domain and DNA-binding domain
- Transcription activation domain and kinase domain
Correct Answer

A. Transcription-activation domain and DNA-binding domain

Explanation

Transcriptional activators are proteins that bind to specific DNA sequences and enhance the transcription of genes. They typically have two key features: a transcription-activation domain and a DNA-binding domain. The transcription-activation domain interacts with other proteins involved in the transcription process, such as the RNA polymerase, to promote gene expression. The DNA-binding domain allows the activator to recognize and bind to specific DNA sequences, positioning it in the correct location to activate transcription. Therefore, the correct answer is transcription-activation domain and DNA-binding domain.

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34.

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

A. 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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35.

Which of the following organisms would produce functional lac products in the presence of IPTG?
- Is Oc Z- Y+ A+
- Is Oc Z- Y- A-
- I+ O+ Z+ Y+ A+
- I- Oc Z+ Y+ A+
- Both Is Oc Z- Y+ A+ and Is Oc Z- Y+ A+ are correct.
Correct Answer

A. 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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36.

An experiment is planned to study the effect of reducing the transcription of the myosin gene. Which of the following inhibitors would be useful in this study?
- (alpha)-amanitin
- DMS
- Heparin
- Ammonium sulfate
- None of the choices are correct
Correct Answer

A. (alpha)-amanitin

Explanation

(alpha)-amanitin is a potent inhibitor of RNA polymerase II, which is responsible for transcribing the myosin gene. By using (alpha)-amanitin, the transcription of the myosin gene can be reduced, allowing the researchers to study the effect of this reduction on various cellular processes. DMS, heparin, and ammonium sulfate are not known to specifically inhibit transcription of the myosin gene, so they would not be useful in this study. Therefore, the correct choice is (alpha)-amanitin.

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37.

A promoter region replacing the TATA box with the ICIC box will not bind to TFIID.
- True
- False
Correct Answer

A. False

Explanation

The statement is false because the TATA box is a crucial element in the promoter region that helps in binding to TFIID, which is a protein complex involved in the initiation of transcription. If the TATA box is replaced with the ICIC box, it is expected that TFIID will still be able to bind to the promoter region.

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38.

Which of the following are characteristic of intrinsic terminator elements?
- They contain an inverted repeat.
- They contain a hairpin loop.
- They contain several T's in the non-template strand of DNA.
- Only they contain an inverted repeat, and they contain a hairpin loop are correct.
- 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

A. 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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39.

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

A. I- O+ Z+ Y+ A+
40.

Which of the following is true about the cro gene?
- It must be stimulated during lysogeny.
- Its product promotes repressor activity.
- Cro repression is important to lysogeny.
- It is adjacent to the CIII gene.
- It must be stimulated during lysogeny, and its products promote repressor activity.
Correct Answer

A. 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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41.

The glucocorticoid receptor is activated by binding to
- Its hormone ligand then moving to the nucleus
- Its protein partner and then moving to the nucleus
- Its partner in the nucleus
- DNA and then its hormone ligand.
- None of the choices are correct.
Correct Answer

A. Its hormone ligand then moving to the nucleus

Explanation

The glucocorticoid receptor is activated by binding to its hormone ligand, which triggers a conformational change in the receptor. This activated receptor then translocates to the nucleus, where it binds to specific DNA sequences known as glucocorticoid response elements (GREs). This receptor-DNA interaction leads to the regulation of target gene expression, resulting in various physiological responses.

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42.

Which of the following techniques is most useful in determining if RNA polymerase has initiated transcription from the lac DNA template?
- Southern analysis
- DNA fingerprinting
- Run-off transcription assay
- DNA sequencing
- RACE
Correct Answer

A. 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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43.

Which of the following would you choose to specifically collect polymerase subunits in cell extracts?
- Epitope tagging
- Immunoprecipitation
- In vitro transcription
- Western blotting
- DNAase footprinting
Correct Answer

A. Epitope tagging

Explanation

Epitope tagging is a technique used to specifically collect polymerase subunits in cell extracts. This involves adding a short amino acid sequence, known as an epitope tag, to the protein of interest. Antibodies that specifically recognize the epitope tag can then be used to isolate and purify the tagged protein from the cell extract. This method allows for the specific collection of polymerase subunits and can be used to study their function and interactions. Immunoprecipitation, in vitro transcription, western blotting, and DNAase footprinting are not specifically designed for collecting polymerase subunits in cell extracts.

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44.

Which of the following is not a part of the core class II promoter?
- TATA box
- Upstream element
- BRE
- DPE
- Inr
Correct Answer

A. Upstream element

Explanation

The upstream element is not a part of the core class II promoter. The core class II promoter consists of the TATA box, BRE (TFIIB recognition element), DPE (downstream promoter element), and Inr (initiator element). The upstream element is a separate regulatory element located upstream of the core promoter region. It plays a role in the regulation of transcription initiation but is not considered a part of the core promoter itself.

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45.

Predict a possible effect of deleting the enhancers region of the polymerase I gene.
- Reduced transcription of ribosomes
- Reduction in the production of most hnRNAs
- Reduction in the amount of rRNA made
- Reduction in the production of Rpb1
- Reduction in the production of Rpb2
Correct Answer

A. Reduction in the amount of rRNA made

Explanation

Deleting the enhancers region of the polymerase I gene would likely result in a reduction in the amount of rRNA made. Enhancers are DNA sequences that enhance the transcription of specific genes. In the case of the polymerase I gene, enhancers play a crucial role in promoting the transcription of rRNA, which is essential for ribosome synthesis. Therefore, deleting the enhancers region would likely lead to a decrease in the transcription of rRNA and subsequently reduce the amount of rRNA produced.

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46.

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?
- Polymerase would become tightly associated with the promoter.
- Polymerase would become more susceptible to proteolysis.
- Transcription will be accelerated.
- Polymerase complex would be loosely associated with the promoter.
- Recognition of the UP element would be faster.
Correct Answer

A. 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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47.

Which of the following is true about the regulation of the trp operon?
- An aporepressor is involved.
- A corepressoris involved.
- Attenuation is one of the control mechanisms.
- A negative control is involved.
- All of the choices are true.
Correct Answer

A. 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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48.

Which statement is wrong regarding promoters in humans?
- Promoter lies before the gene.
- Promoter is recognized by RNA polymerase
- Promoter can be never within a gene
- Class III promoter in type I genes is recognized by RNA polymerase.
- Bacterial promoters have the same function as the humans'
Correct Answer

A. Promoter can be never within a gene

Explanation

The given statement "Promoter can be never within a gene" is incorrect. In humans, promoters can indeed be located within a gene. Promoters are DNA sequences that are responsible for initiating the process of transcription, where RNA polymerase binds and starts transcribing the gene. Promoters can be found upstream or downstream of the gene, as well as within the gene itself. Therefore, the statement that promoters can never be within a gene is incorrect.

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49.

Which of the following is the correct order of chromatin folding?
- Nucleosome formation, 30 nM fiber formation, radial loop structure
- Radial loop structure, 30 nM fiber formation, nucleosome formation
- 30 nM fiber formation, nucleosome formation, radial loop formation
- 30 nM fiber formation, radial loop formation, nucleosome formation
- Nucleosome formation, radial loop formation, nucleosome formation
Correct Answer

A. Nucleosome formation, 30 nM fiber formation, radial loop structure

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Ultimate Molecular Biology Quiz

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.

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Epitope tagging is an efficient way to immunoprecipitate to isolate an entire polymerase complex.

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

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

Which of the following is absent in an operon?

TFIID collaborates with TFIIA to bind to the TATA box.

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

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

Eukaryotic ribosomal RNA genes are transcribed by RNA polymerase II.

RNA polymerase prevents the binding of TFIIF to the promoter region

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

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

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

Roeder and Rutter showed that there are four polymerases operating in the eukaryotic cell.

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

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

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

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

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-

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

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

Which of the following can be found in DNA-binding domains?

Dimethyl sulfate is used to

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

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?

Differences in the transcription of T7 and SPO1 genes are

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

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

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

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

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

Which of the following are typical features of transcriptional activators?

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

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

An experiment is planned to study the effect of reducing the transcription of the myosin gene. Which of the following inhibitors would be useful in this study?

A promoter region replacing the TATA box with the ICIC box will not bind to TFIID.

Which of the following are characteristic of intrinsic terminator elements?

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

Which of the following is true about the cro gene?

The glucocorticoid receptor is activated by binding to

Which of the following techniques is most useful in determining if RNA polymerase has initiated transcription from the lac DNA template?

Which of the following would you choose to specifically collect polymerase subunits in cell extracts?

Which of the following is not a part of the core class II promoter?

Predict a possible effect of deleting the enhancers region of the polymerase I gene.

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?

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

Which statement is wrong regarding promoters in humans?

Which of the following is the correct order of chromatin folding?