Control of gene expression MCQ Quiz - Objective Question with Answer for Control of gene expression - Download Free PDF

The correct answer is To facilitate the interaction between transcription factors and RNA polymerase II

Explanation:

• Gene expression is the process by which information from a gene is used to synthesize functional gene products, such as proteins.
• In eukaryotic cells, transcription involves the assembly of several molecular components, including transcription factors, RNA polymerase II, and other regulatory complexes.
• The mediator complex is a multi-protein structure critical in regulating transcription initiation by bridging interactions between transcription factors and RNA polymerase II.
• Mediator Complex Function:
  • The mediator complex acts as a molecular bridge between activator or repressor transcription factors and RNA polymerase II.
  • It facilitates the formation of the transcription pre-initiation complex, which is necessary for RNA polymerase II to bind to the promoter region of a gene.
  • By aiding this interaction, the mediator complex ensures precise regulation of gene transcription in response to cellular signals.
  • This complex integrates signals from transcription factors and other regulatory proteins, modulating transcription levels appropriately.

Other Options:

• To modify histones to promote transcription:
  • This is incorrect because histone modification is carried out by enzymes such as histone acetyltransferases (HATs) and histone deacetylases (HDACs), not the mediator complex.
  • Histone modifications alter chromatin structure and accessibility but do not directly facilitate interactions between transcription factors and RNA polymerase II.
• To promote helicase activity to unwind DNA during transcription initiation:
  • This is incorrect because helicase activity is primarily associated with enzymes like TFIIH, which unwinds DNA strands during transcription initiation.
  • The mediator complex does not have helicase activity; its role is limited to mediating protein-protein interactions necessary for transcription initiation.
• To degrade mRNA after transcription:
  • This is incorrect because mRNA degradation is performed by cellular machinery such as exosomes and enzymes like ribonucleases.
  • The mediator complex is involved in transcription initiation, not post-transcriptional processes like mRNA degradation.

The correct answer is Variations in the Shine-Dalgarno sequence upstream of lacZ, lacY, and lacA have different affinities for the ribosome, affecting translation initiation rates.

Concept:

• The E. coli lac operon is a classic example of gene regulation in prokaryotes. It consists of three structural genes: lacZ, lacY, and lacA, which are transcribed into a single polycistronic mRNA.
• Translation of these genes occurs independently but sequentially, as ribosomes bind to their respective Shine-Dalgarno sequences upstream of the start codons.
• The Shine-Dalgarno sequence is a ribosome-binding site in prokaryotic mRNA that plays a crucial role in initiating translation by guiding the ribosome to the correct start codon.
• Differences in the affinity of the Shine-Dalgarno sequences for the ribosome can influence the translation initiation rates of the individual genes within the operon.

Explanation:

Variations in the Shine-Dalgarno sequences upstream of lacZ, lacY, and lacA result in different ribosomal binding efficiencies. This explains why lacZ is translated more frequently than lacY or lacA.

• A stronger Shine-Dalgarno sequence upstream of lacZ ensures higher ribosomal binding affinity, leading to more efficient translation initiation.
• Since translation initiation is often the rate-limiting step in protein synthesis, differences in Shine-Dalgarno sequences significantly impact the frequency of translation for the genes within the operon.

Other Options:

•  "The Shine-Dalgarno sequence is present upstream of lacZ, but not lacY or lacA, affecting translation initiation rates." This is incorrect because Shine-Dalgarno sequences are present upstream of all three genes (lacZ, lacY, and lacA), as they are necessary for ribosomal binding and translation initiation. The difference lies in the strength or affinity of these sequences, not their presence or absence.
• "Inhibitory RNA structures are present upstream of the AUG codon of lacY and lacA, but not lacZ, affecting translation initiation rates." This is incorrect because there is no evidence of inhibitory RNA structures specifically blocking translation initiation for lacY and lacA. The observed differences are due to variations in Shine-Dalgarno sequence affinity, not inhibitory structures.
• "Inhibitory RNA structures are present in the lacY and lacA coding sequences, but not lacZ, affecting translation elongation rates." This is incorrect because the differences in translation frequency are attributed to initiation rates, not elongation rates. 

The correct answer is A (ii), B (iii), C (iv), D (i)

Explanation:

• The trp operon in E. coli is an example of a repressible operon regulated both by repression and attenuation. Attenuation is a mechanism where transcription is terminated prematurely under certain conditions, such as high tryptophan levels.
• Attenuation depends on the interaction between the ribosome and the leader sequence, which includes four regions (sequences 1, 2, 3, and 4). The ability of regions to form specific stem-loop structures determines whether transcription proceeds or terminates.
• High tryptophan concentrations promote the formation of an attenuator loop (between sequences 3 and 4), leading to transcription termination. 

A (ii): Inserting bases between the leader peptide gene and sequence 2:

• This manipulation interrupts the normal coupling of transcription and translation in the leader region. The ribosome cannot properly interact with the leader sequence.
• The consequence is less attenuation because the attenuator loop may not form efficiently, allowing transcription to continue more frequently.

B (iii): Inserting bases between sequences 2 and 3:​

• Inserting bases here disrupts the pairing between sequences 2 and 3, which normally forms an anti-terminator loop under low tryptophan conditions.
• This alteration promotes the formation of the attenuator loop (between sequences 3 and 4), causing more attenuation.

C (iv): Deleting sequence 4:

• Sequence 4 is essential for forming the attenuator loop (3-4 pairing). If sequence 4 is deleted, the attenuator loop cannot form.
• As a result, no attenuation occurs, and transcription continues regardless of tryptophan levels.

D (i): Elimination of ribosome-binding site for the leader peptide:

• The ribosome-binding site is required for translation of the leader peptide, which is a critical step in attenuation regulation.
• If this site is eliminated, the ribosome cannot engage with the leader sequence, leading to complete attenuation as the anti-terminator loop (2-3 pairing) cannot form.

The correct answer is B and C only

Concept:

• Histone modifications, such as acetylation, methylation, and phosphorylation, play crucial roles in the regulation of transcription by modifying chromatin structure and affecting protein interactions.
• These modifications act as "epigenetic marks" that influence gene expression without altering the underlying DNA sequence.
• The "histone code" hypothesis suggests that the combination of these modifications provides binding sites for specific regulatory proteins, thereby controlling transcriptional activity.

Explanation:

Statement A: "Acetylation of histones is generally associated with transcriptional repression by making the chromatin more compact."

• This statement is incorrect. Acetylation of histones is typically associated with transcriptional activation, not repression. It reduces the positive charge on histones, thereby loosening their interaction with negatively charged DNA. This results in a more open chromatin structure, which facilitates access of transcriptional machinery to DNA.

Statement B: "Methylation of histones can either activate or repress transcription, depending on the specific residue modified."

• This statement is correct. Histone methylation has context-dependent effects on transcription. For example, methylation of histone H3 at lysine 4 (H3K4me) is associated with transcriptional activation, whereas methylation at lysine 9 (H3K9me) or lysine 27 (H3K27me) is linked to transcriptional repression.

Statement C: "Phosphorylation of histones occurs in response to DNA damage and can influence gene expression."

• This statement is correct. Histone phosphorylation, such as phosphorylation of histone H2AX at serine 139 (γH2AX), is a well-known marker of DNA damage.
• Phosphorylation can also regulate chromatin dynamics and influence gene expression during cellular stress or DNA repair processes.

Statement D: "Histone modifications influence the recruitment of RNA polymerase complex but not transcription factors."

• This statement is incorrect. Histone modifications influence the recruitment of both RNA polymerase and transcription factors. They create binding sites for specific proteins, including transcription factors, coactivators, and corepressors, which collectively regulate transcription.

The correct option is: 1

Explanation:

• (A) Lactose:
  This is correct. The lac operon is negatively regulated by the lac repressor, a protein that binds to the operator sequence in the absence of lactose. When lactose is absent, the repressor prevents RNA polymerase from transcribing the operon. However, when lactose is present, it is converted into allolactose, which acts as an inducer by binding to the repressor. This binding causes a conformational change in the repressor, preventing it from binding to the operator and allowing transcription of the operon.

• (B) Glucose:
  This is incorrect. The presence or absence of glucose influences the lac operon through catabolite repression, mediated by cAMP levels and the CAP (catabolite activator protein). However, this regulation is separate from the negative regulation by the lac repressor.

• (C) cAMP:
  This is incorrect. cAMP plays a role in positive regulation of the lac operon. Low glucose levels result in high cAMP, which binds to CAP, enabling it to enhance transcription. It does not affect the lac repressor directly.

• (D) ATP:
  This is incorrect. ATP is involved in cellular energy processes and does not directly regulate the binding of the lac repressor to the operator.

Key Points

• The lac operon consists of three structural genes (lacZ, lacY, and lacA) and is an example of an inducible operon. It is involved in the metabolism of lactose in E. coli.
• Regulation occurs at multiple levels:
  • Negative regulation: By the lac repressor in the absence of lactose.
  • Positive regulation: By the CAP-cAMP complex when glucose levels are low.

The correct answer is The lac operon is highly expressed only when both lactose and glucose are absent.

Explanation:

• The lac operon is controlled by both the Lac repressor and the activator protein, CAP: This statement is correct. The Lac repressor inhibits the operon in the absence of lactose, while CAP enhances expression when glucose is low.
• The lac operon is highly expressed only when both lactose and glucose are absent: This statement is incorrect. The lac operon is highly expressed when lactose is present and glucose is absent. In the presence of lactose, the repressor is inactivated, allowing transcription. However, the operon is not activated when both lactose and glucose are absent because CAP cannot stimulate transcription without the absence of glucose.
• The lac operon is highly expressed only when lactose is present and glucose is absent: This statement is correct. For maximum expression of the lac operon, lactose must be present to inactivate the repressor, and glucose must be absent to allow CAP to enhance transcription.
• In the presence of lactose, the repressor cannot bind to the operator: This statement is correct. Lactose binds to the Lac repressor, preventing it from binding to the operator and allowing transcription to occur.

Therefore, the incorrect statement is indeed The lac operon is highly expressed only when both lactose and glucose are absent because the lac operon is not highly expressed under those conditions; it requires lactose to be present and glucose to be absent for high expre

The correct answer is Option 3 i.e.Copia element in Drosophila; LINES as well as SINES in human

Concept:

• Retroelements: RNA- and reverse-transcribed-to-DNA elements that are inserted into a new location in the genome.
• In retrotransposons and retrovirus-like elements, there exist long terminal repeats (LTRs) that resemble those seen in retroviruses.
• LTRs are absent from retroposons like SHORT INTERSPERSED NUCLEOTIDE ELEMENTS and LONG INTERSPERSED NUCLEOTIDE ELEMENTS.
• SINEs and LINEs are two types of interspersed retrotransposable elements that use RNA intermediates to invade new genomic locations.
• SINEs and LINEs make up at least 34% of the human genome and are present in practically all eukaryotes (but not Saccharomyces cerevisiae).

​Explanation:

Option:- Copia element in Drosophila; LINES as well as SINES in human

• A retrotransposon called the Copia element is present in drosophila and is thought to have been horizontally transferred from some populations of Drosophila williston to Drosophila melanogaster. 
• LINES as well as SINES both are present in the genome of Humans.
• P transposable elements were found to be the root causes of the hybrid dysgenesis syndrome of genetic characteristics in Drosophila.
• The germline-specific short RNA piwi-interacting (piRNA) pathway is associated with a distinctive pattern of maternal inheritance seen in hybrid dysgenesis.

Therefore, option B is correct.

• A single messenger RNA (mRNA) is formed for all the structural genes of an operon. This mRNA is termed polycistronic RNA.
• In prokaryotes, operons are clusters of genes transcribed together, resulting in a single mRNA molecule that encodes multiple proteins. This type of mRNA is called polycistronic because it contains multiple coding sequences (cistrons).
• Each coding sequence within a polycistronic mRNA corresponds to a different protein, allowing for coordinated expression of genes that function together in a pathway or process.

• Regulatory RNAs are involved in the regulation of gene expression but do not encode proteins. Examples include microRNAs (miRNAs) and small interfering RNAs (siRNAs).

• Transfer RNA (tRNA) is responsible for bringing amino acids to the ribosome during protein synthesis. It does not serve as a template for protein coding.

• The term "uncommon RNA" is not a standard term in molecular biology and does not refer to any specific type of RNA involved in gene expression or regulation

Concept:

• The most common cellular component to be tested is DNA content.
• Its quantification measures DNA ploidy level, and cell cycle phase, and may detect the existence of apoptotic cells, which are identified by fractional DNA content.
• A gamete's DNA content is 1c and its chromosomal count is 1n. A gamete is an egg or sperm. Both the amount of DNA and the number of chromosomes double, reaching 2c and 2n, respectively, after fertilisation.

Explanation:

• Gametes are haploid (n) and cells in G1 phase are diploid.
• Hence, if the cells in the G1 phase are diploid (2C) and it divides to form gametes which are haploid.
• So, the amount of DNA in gamtes will be half of the G1 phase.

Therefore, the correct answer is option 3.

The correct answer is Option 3

Explanation:

• The cI gene is crucial for the lysogenic cycle of the lambda phage. The protein encoded by this gene, known as the lambda repressor or CI repressor, binds to specific sites on the phage's DNA, preventing the transcription of genes necessary for initiating the lytic cycle.
• By doing this, the phage remains in a dormant state within the host cell, a state referred to as lysogeny.
• During lysogeny, the phage's DNA, now a prophage, is replicated along with the host's DNA without causing harm to the host cell, until certain conditions trigger the switch to the lytic cycle, where the phage produces new virions and lyses the cell to release them.

Therefore, cI gene of phage lambda encodes the lambda repressor which represses the lytic cycle.

The correct answer is Option 1 i.e. B and C

Key Points

• The nucleosome is defined as the basic repeating subunit of chromatin that is packaged inside the nucleus of the cell.
• It consists of five histone proteins, a few non-histone proteins and DNA packed around it. 
• The DNA that is highly associated with the core histones is called core DNA, and it wraps 1.65 times around the core histone. 
• The length of core histone in each nucleosome is ~ 147 bp.
• The DNA that linked two nucleosomes is called linker DNA, and the length of DNA between two nucleosomes is ~ 60 bp.
• Histone proteins are of two types: Core histone (H2A, H2B, H3 and H4) and linker histone (H1).
• The Histone core is an octamer of H2A, H2B, H3 and H4 proteins. 
• Histone assembly on the DNA occurs in two step - 
  • Binding of the tetramer of H3 and H4 on the DNA.
  • Secondly, two dimers of histones H2A and H2B bind to form the nucleosome core. 
• 30nm fiber is the second level of DNA packaging, where the series of nucleosomes are coiled in helical array. The diameter of this fiber is 30nm. 

Explanation:

• H3 and H4 histones first form a heterodimer and later two heterodimers come together and form a tetramer. In constrast H2A and H2B form heterdimers in the solution but not an tetramer.
  • So, statement A is incorrect.
• The N-terminal tail of H2A, H3 and H4 interacts with the adjacent nucleosome. Without an N-terminal, core histones are not able to form the 30-nm fiber.
  • So, statement B is correct.
• Post-translation modifications are seen in N-terminal tails and globular domains and they modulate transcriptional events. 
  • So, statement C is correct.
• In the solenoid models, the flat surface of the histones are adjacent to one another and linker DNA is buried in the centre but it does not pass through the axis, rather the linker DNA circles around the central axis. 
• While in the Zig-Zag model, the linker DNA passes through the central axis of the fibers in a relatively straight form. 
  • So, statement D is incorrect.

Hence, the correct answer is option 1.

Key Points

• Catabolite repression is a phenomenon observed in bacteria, particularly in E. coli, where glucose is favored as an energy source over other sugars.
• This preference for glucose leads to the glucose effect, resulting in diauxic growth.
• When both glucose and another sugar, like lactose, are present, bacteria metabolize glucose first, repressing the use of the alternate sugar.
• This mechanism allows bacteria to preferentially use glucose over other carbon sources when it is available. When glucose is present, it inhibits the utilization of other sugars through a complex regulatory system involving the cyclic AMP (cAMP) and the cAMP receptor protein (CRP), also known as CAP (catabolite activator protein). 
• CRP, a positive regulator protein, binds to cAMP to form a complex known as the cAMP-CRP complex.
• This complex binds to a specific site near the promoter region of operons, such as the lac operon.
• The binding of the cAMP-CRP complex introduces a bend in the DNA structure, allowing it to interact with RNA polymerase.
• This interaction initiates the transcription process, facilitating the expression of operons.
• Notably, the lac operon serves as an example where CRP acts as a positive regulator, enhancing gene expression.
• Catabolite repression involves the decrease in intracellular cAMP levels when glucose is present.
• As glucose levels rise, cAMP levels drop, reducing the formation of the cAMP-CRP complex.
• Consequently, the expression of operons like the lac operon is modulated both positively by the cAMP-CRP complex and negatively by repressor proteins.

Explanation:

• Catabolite repression involves the decrease in intracellular cAMP levels when glucose is present.
• As glucose levels rise, cAMP levels drop, reducing the formation of the cAMP-CRP complex.

Hence the correct answer is option 2

The correct answer is increased transcription of trp operon.

Explanation:

• The genes required for tryptophan biosynthesis in Escherichia coli are organized as a single transcriptional unit, the trp operon.
• The leader sequence of the trp operon in E. coli, also known as the trpL sequence, plays a critical role in the regulation of the operon through a mechanism called attenuation. This sequence is responsible for the formation of specific RNA secondary structures that regulate the continuation of transcription based on tryptophan availability.
• This operon has a single major promoter at which transcription initiation is regulated by a DNA-binding protein, the L-tryptophan-activated trp repressor. This repressor acts by binding at one or more of three operator sites located in the trp operon's promoter region.
• One element of the trp operon is the leader sequence (L) that in immediately 5' of the trpE gene.
• This sequence (about 160 bp in size) also controls the expression of the operon through a process called attentuation.
• This sequence has four domains (1-4). Domain 3 (nucleotides 108-121) of the mRNA can base pair with either domain 2 (nucleotides 74-94) or domain 4 (nucleotides 126-134).
• If domain 3 pairs with domain 4, a stem and loop structure forms on the mRNA and transcription stops.
• This structure forms when the level of tryptophan is high in the cell.
• If domain 3 pairs with domain 2, then the stem and loop structure does not form and transcription continues through the operon and all of the enzymes required for tryptophan biosynthesis are produce. These events occur when tryptophan is low in the cell.

The correct answer is B and C.

Explanation:

Gene regulation in prokaryotes involves several mechanisms to ensure that genes are expressed only when needed. 

1. A. In prokaryotic operons, the operator region is located upstream of the promoter region:

   • This statement is incorrect. In prokaryotic operons, the operator region is typically located downstream or partially overlapping with the promoter region, not upstream.

2. B. Positive control mechanisms involve activator proteins that enhance the binding of RNA polymerase to the promoter:

   • This statement is correct. Positive control mechanisms in prokaryotes involve activator proteins that bind to specific sites near or on the promoter to facilitate the binding of RNA polymerase. This enhancement of RNA polymerase binding increases the likelihood of transcription initiation, as seen in mechanisms like the catabolite activator protein (CAP) in the lac operon.

3. C. Attenuation is a regulatory mechanism seen in operons controlling the synthesis of amino acids like tryptophan:

   • This statement is correct. Attenuation is a regulatory mechanism used by certain operons, such as the tryptophan operon in E. coli, to control transcription in response to the abundance of amino acids. It functions by prematurely terminating transcription in response to high levels of the amino acid (e.g., tryptophan), allowing for fine-tuned control of gene expression based on amino acid availability.

4. D. In prokaryotic cells, all genes in an operon are transcribed as separate mRNAs with individual regulatory controls:

   • This statement is incorrect. In prokaryotic operons, all genes within an operon are typically transcribed together as a single polycistronic mRNA. This means that the genes are controlled collectively by a single promoter and regulatory region, rather than having separate mRNAs and individual regulatory controls

Key Points:

• Activator proteins are crucial in the positive control of gene expression, enhancing RNA polymerase binding to the promoter.
• Attenuation is a sophisticated regulatory mechanism managing the synthesis of amino acids by ensuring gene expression aligns with the cell's metabolic needs.
• The operator region is downstream or overlapping with the promoter, not upstream.
• Prokaryotic operons produce a single polycistronic mRNA, not separate mRNAs.

Conclusion:

The correct answer is Option B and C, as both statements accurately describe mechanisms of gene regulation in prokaryotes.

The correct answer is Option 3 i.e. B and C only

Key Points

• The lambda phage is a temperate bacteriophage that can either lyse its host bacterium (lytic cycle) or integrate into the bacterial genome and replicate along with it (lysogenic cycle).
• The statements refer to various mutants of the lambda phage.

Explanation:

Statement A: Clear plaques are formed in Q mutants - INCORRECT

• Q is a gene of the lambda phage that encodes a transcription antitermination protein.
• Q mutants are defective in antitermination and have premature termination of transcription, resulting in truncated mRNA transcripts and non-functional proteins.
• Clear plaques are formed in Q mutants because they are unable to complete the lytic cycle.

Statement B: No plaques are formed in nut mutants - CORRECT

• 'nut' is a gene of the lambda phage that encodes a protein that facilitates transcriptional termination at specific sites.
• Nut mutants are unable to properly terminate transcription and have abnormal mRNA transcripts, resulting in non-functional proteins.
• No plaques are formed in nut mutants because they are unable to complete the lytic cycle.

Statement C: Clear plaques are formed in cll mutants - CORRECT

• 'cll' is a gene of the lambda phage that encodes a protein that prevents the expression of the lytic genes and promotes the expression of the lysogenic genes.
• Cll mutants are unable to prevent the expression of the lytic genes and have a higher propensity for the lytic cycle.
• Clear plaques are formed in cll mutants because they complete the lytic cycle.

Statement D: Turbid plaques are formed in integrase mutants - INCORRECT

• Integrase is a gene of the lambda phage that encodes a protein that mediates the integration of the phage genome into the bacterial chromosome.
• Integrase mutants are unable to integrate the phage genome and have a higher propensity for the lytic cycle.
• Turbid plaques are formed in integrase mutants because they are able to complete the lytic cycle but do not integrate into the bacterial genome.

Statement E: Clear plaques are formed in P mutants - INCORRECT

• P is a gene of the lambda phage that encodes a protein that is involved in the initiation of DNA replication.
• P mutants are unable to initiate DNA replication and have incomplete phage particles.
• Clear plaques are formed in P mutants because they are unable to complete the lytic cycle.

Statement F: No plaques are formed in cl mutants - INCORRECT

• 'cl' is a gene of the lambda phage that encodes a protein that is involved in the establishment of lysogeny.
• Cl mutants are unable to establish lysogeny and have a higher propensity for the lytic cycle.
• No plaques are formed in cl mutants because they are unable to complete the lysogenic cycle.

Therefore, the correct combination of statements is B and C only.