Control of gene expression MCQ Quiz in தமிழ் - Objective Question with Answer for Control of gene expression - இலவச PDF ஐப் பதிவிறக்கவும்

The correct answer is Option 4 i.e.- Glucose, + Lactose

Key Points

• The lac operon is a genetic system found in the bacteria Escherichia coli (E. coli) that controls the expression of genes involved in the metabolism of lactose.
• The lac operon consists of three structural genes (lacZ, lacY, and lacA) that encode proteins necessary for the metabolism of lactose, and a regulatory region that controls the expression of these genes.
• The regulatory region of the lac operon includes a promoter sequence where RNA polymerase binds to initiate transcription of the genes, an operator sequence where a regulatory protein called the Lac repressor can bind to prevent transcription, and a regulatory gene called lacI that encodes the Lac repressor protein.

Explanation:

• The Lac operon in E. coli is regulated by the presence or absence of glucose and lactose in the environment.
• The Lac operon is switched on when there is a low level of glucose and a high level of lactose in the environment.
• In the absence of glucose, the Lac repressor protein cannot bind to the operator region of the Lac operon, allowing RNA polymerase to transcribe the structural genes.
• The presence of lactose induces the expression of the Lac operon by binding to the Lac repressor protein and causing it to release from the operator region.
• This allows RNA polymerase to transcribe the structural genes, resulting in the production of beta-galactosidase (encoded by lacZ), lactose permease (encoded by lacY), and transacetylase (encoded by lacA), which are involved in lactose metabolism.

Concept:

• Bacteriophage is a virus that infects a bacteria.
• Upon infection, the phage can propagate in either of two ways: lytically or lysogenically.
•  Lytic growth requires replication of the phage DNA and synthesis of new coat proteins.
• These components combine to form new phage particles that are released by lysis of the host cell.
• Lysogeny—the alternative propagation pathway—involves the integration of the phage DNA into the bacterial chromosome where it is passively replicated at each cell division, as though it were a legitimate part of the bacterial genome.

Explanation:

• Viral genes are transcribed in three temporally distinct waves: early, intermediate, and late gene transcription.
• Early genes code for intermediate transcription factors, intermediate genes code for late transcription factors.
• Late genes produce early transcription factors which are packaged into the virion.
• Since virions package a comprehensive set of transcription machinery, early gene transcription can begin almost immediately after core entry before uncoating occurs.

​Fig 1: Listed factors for late gene expression

• Late transcription normally also requires DNA replication and is, in fact, coupled to concurrent DNA synthesis.
• The coupling of late transcription to DNA replication is enforced by the action of gp30, the  DNA ligase
• Single-strand breaks make phage DNA subject to nucleolytic attack, but protecting against that degradation by knocking out the exonuclease function encoded by gene 46 generates a situation in which late transcription occurs in the absence of DNA replication (e.g., in the absence of T4 DNA polymerase (gp43) function)
• Research experiments clearly identify the involvement of gp45, the sliding clamp processivity factor of the phage DNA polymerase holoenzyme, in  late transcription (That this approach does not equally clearly identify the involvement of the gp44/62 clamp loader complex in T4 late transcription is puzzling)
• In summary, the primary direct roles in phage late transcription are played by three proteins--gp55, gp33 and gp45--and by a transient form of the phage DNA template that is generated in the process of replication.
• Gp55 is a very small, highly diverged σ 70-family protein.
• The σ 70/ σ A subunits of the bacterial RNAPs comprise 4 globular domains (σ 1, σ 2, σ 3 and σ4) that are widely separated on the surface of the RNAP holoenzymes.
• When σ detaches from the RNAP core, these domains swap their sites of interaction with the b and b” RNAP subunits for internal contacts and assume a compact structure.
• The 112-residue gp33 binds to the flap tip of the RNAP β subunit.
• This is also the RNAP core attachment site of σ domain 4, which recognizes the -35 promoter element.
• Gp45 is the phage representative of the sliding clamp proteins.
• Sliding clamps are six-domain rings with a central hole large enough to accommodate a DNA helix: head-to-tail homodimers of 3-domain subunits in the case of the E. coli replisome’s b protein; homo trimers of 2-domain subunits in the case of gp45 and the eukaryotic PCNA (proliferating cell nuclear antigen); homo-  or hetero trimers of 2-domain subunits in the case of archaeal PCNA .

Hence the correct answer is option 2 (A), (C), (D)

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.

The correct answer is Option 2 i.e.Transferrin receptor is post‐transcriptionally regulated, and the 3’UTR is the regulatory site 

Key Points

• In eukaryotes, gene regulation is more complex and operates at different levels.
• Gene regulation can be divided into these five stages:

1. Epigenetic control
   • It is the first level of gene control, it involves changes in the genes that does not cause the change in the nucleotide sequence. 
   • It result in change in the chromosomal structure that cause the gene to turn "ON" and "OFF".
2. Transcriptional control
   • Whether mRNA of the particular gene is to be produced in  a particular cells is controlled in transcriptional regulation.
   • Activator/enhancers are the molecules that facilitate transcription of the genes while repressor/inhibitor are proteins molecules that stop transcription of the gene.
3. Post-transcriptional control
   • Post transcriptional control consist of splicing, 5'capping, 3'-adenylation, 
   • mRNA transcripts also contain 5'UTR and 3'-UTR regions that are not translated into proteins but they regulate translation of the gene. 
4. Translation control
   • Whether mRNA is to be translated into protein is controlled in translational regulation.
5. Post-translational control
   • This included various modifications tp the proteins so as it affects their activity.
   • For example, phosphorylation of certain tyrosine reside can turn on or turn off function of that protein.
   • Glycosylation, ubiquitylation, phosphorylation, sumoylation, acetylation, etc are few example of post-transcritption regulation of the protein.

Explanation:

Option 1:

• Since mutation in the 3'-UTR region of the transferrin receptor does not result in increased in the transferring receptors mRNA level. Hence, we can say that transferrin receptor is regulated post-transcriptionally regulated.
• But there is no experiment done to shows that it is not regulated at other levels.
• Hence, this is an incorrect option.

Option 2:

• Mutation in the 3'-UTR region of the transferrin receptors result in the decreases in receptors mRNA.
• So, we can say that it is regulated post-transcriptionally and 3'-UTR is the regulatory site.
• Hence, this is the correct option.

Option 3:

• Absence of iron is responsible for induction of transcription and not the presence of iron. 
• In the presence of iron, the level of transferrin receptors mRNA decreases.
• Hence, this is an incorrect option.

Option 4:

• Addition of iron to the cells result in decreases in mRNA of transferrin receptors.
• So, transferrin receptors regulation is sensitive iron concentration. 
• Hence, this is an incorrect option.

Hence, the correct answer is option 2.

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 Saturation of transcription factor binding sites.

Explanation:

The graph shows the transcription rate of Gene Y increasing rapidly with increasing concentrations of the transcription factor, but then reaching a plateau at higher concentrations. This plateauing effect can be explained by the saturation of transcription factor binding sites on the gene's promoter region.

At low concentrations of the transcription factor, more binding sites are available, leading to a linear increase in transcription rate. However, as the transcription factor concentration increases, the binding sites become saturated, meaning that additional transcription factors cannot further bind and activate transcription. This results in the observed plateau in transcription rate, even as the transcription factor concentration continues to rise.

The other options do not adequately explain the plateauing effect seen in the graph:

• A) Allosteric inhibition of the transcription factor: This would cause a decrease in transcription rate, not a plateau.
• C) Inactivation of RNA polymerase: This would lead to a decrease in transcription rate, not a plateau.
• D) Feedback inhibition by the gene product: This would require the gene product to inhibit its own transcription, which is not the mechanism described in the question.

Therefore, the best explanation for the plateauing effect observed in the graph is the saturation of transcription factor binding sites on the gene's promoter region.