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

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 It functions as an activator complex, facilitating RNA polymerase binding.

Explanation:

A researcher observes that gene expression is upregulated when a protein complex binds to a specific DNA sequence near the gene's promoter. This suggests that the protein complex plays a positive role in gene regulation, likely by enhancing the transcription process. Let’s examine the potential roles:

1. Histone deacetylase:

   • Histone deacetylases remove acetyl groups from histones, leading to tighter DNA-histone interactions. This usually results in repression of gene expression, not upregulation.

2. Activator complex:

   • Activator complexes are proteins that bind to specific DNA sequences and enhance the transcription of associated genes. They facilitate the binding of RNA polymerase to the promoter, thereby upregulating gene expression.

3. Repressor complex:

   • Repressor complexes bind to DNA sequences and prevent the binding of transcription factors or RNA polymerase, thereby downregulating gene expression.

4. Methyltransferase:

   • Methyltransferases add methyl groups to DNA or histones, usually leading to gene silencing. This process is typically associated with the repression of gene expression rather than activation.

The correct answer is A-1, B-2, C-3, D-4.

Explanation:

Here's the detailed matching of the terms in Column I with their respective descriptions in Column II:

1. A. Lac Operon - 1. Repressor binding site:
   • The Lac Operon is a well-studied example of gene regulation in prokaryotes, particularly in E. coli. It consists of genes involved in lactose metabolism.
   • The repressor protein (LacI) binds to the operator region of the Lac Operon in the absence of lactose, preventing transcription of the operon's genes.
   • When lactose is present, it binds to the repressor, causing a conformational change that releases it from the operator, allowing transcription to proceed.
2. B. Enhancer - 2. Binding of activators:
   • Enhancers are regulatory DNA sequences that can be located far from the gene they control (either upstream, downstream, or within introns).
   • They increase the likelihood of transcription by providing binding sites for transcription activators and coactivators.
   • Enhancers help in forming a loop structure in the DNA, bringing the enhancer-bound activators in close proximity to the promoter and the transcription machinery.
3. C. Silencer - 3. Downregulation of genes:
   • Silencers are regulatory elements that can inhibit gene expression by binding repressor proteins.
   • These elements can function by blocking the assembly of the transcription machinery or by inducing a more compact chromatin structure to make the DNA less accessible 
   • Silencers are involved in processes like cell differentiation and development where selective gene expression is crucial.
4. D. Promoter - 4. Transcription initiation:
   • Promoters are regions of DNA that initiate transcription of a particular gene.
   • Promoters also contain other regulatory elements that assist in the binding of transcription factors and RNA polymerase.

The correct answer is The lac operon will not be expressed at all.

Explanation:

In a bacterial cell, the lac operon is a well-studied example of gene regulation involving the control of lactose metabolism. The promoter region in the lac operon is crucial for the binding of RNA polymerase, which initiates transcription of the downstream genes (lacZ, lacY, and lacA).

1. Mutation in the promoter region:

   • A mutation in the promoter region that causes a complete loss of RNA polymerase binding would prevent the transcription of the lac operon.
   • RNA polymerase must bind to the promoter to begin the transcription of the operon's genes.

2. Effect on lac operon expression:

   • Without RNA polymerase binding, no mRNA will be produced from the lac operon.
   • Consequently, the genes responsible for lactose metabolism (lacZ, lacY, and lacA) will not be expressed, even in the presence of lactose.

Key Points

• Promoter mutations that inhibit RNA polymerase binding result in a loss of transcription.
• The presence of lactose alone cannot induce expression if the promoter is non-functional.

The correct answer is A - 3, B - 2, C - 4, D - 1

Solution:

• A. lac operon inducer — 3. Allolactose: Allolactose binds to the lac repressor, causing a conformational change that releases the repressor from the operator, thus inducing the operon.
• B. lac repressor binding site — 2. Operator: The operator is a region of DNA where the lac repressor protein binds to prevent transcription.
• C. Attenuation in the trp operon — 4. Leader peptide: The leader peptide sequence is involved in the attenuation mechanism, which regulates the trp operon based on tryptophan levels through ribosome stalling.
• D. trp operon repressor activation — 1. Tryptophan: Tryptophan binds to the trp repressor protein, activating it to bind to the operator and repress transcription of the trp operon.

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.