Fundamental Processes MCQ Quiz - Objective Question with Answer for Fundamental Processes - Download Free PDF

The correct answer is Base-excision

Explanation:

• DNA repair systems are critical mechanisms that maintain the integrity of genetic material by correcting damage that occurs due to environmental factors, replication errors, or cellular processes.
• Base-excision repair (BER) is a specialized DNA repair pathway that fixes small, non-helix-distorting base lesions caused by oxidation, alkylation, deamination, or spontaneous hydrolysis.
  • BER is initiated when DNA glycosylases recognize and excise damaged bases, leaving behind an abasic site (also known as an AP site).
  • The AP site is then processed by an enzyme called AP endonuclease, which cuts the DNA backbone at the site.
  • A DNA polymerase fills in the missing nucleotide(s), and DNA ligase seals the nick in the DNA strand, completing the repair.

Other Options:

• Direct repair:
  • Direct repair is a mechanism that directly reverses specific types of DNA damage without removing the damaged nucleotide or disrupting the DNA backbone.
  • For example, photolyase enzymes repair UV-induced thymine dimers through photoreactivation, and O₆-methylguanine-DNA methyltransferase (MGMT) repairs alkylated guanine bases by transferring the alkyl group to itself.
• Mismatch repair (MMR):
  • MMR corrects replication errors such as mispaired bases or small insertion/deletion loops that escape proofreading by DNA polymerase.
  • It involves proteins like MutS and MutL that recognize mismatches, recruit endonucleases to excise the erroneous DNA segment, and allow DNA polymerase to resynthesize the correct sequence.
  • MMR does not involve DNA glycosylases, as it addresses replication errors rather than damaged bases.
• Nucleotide-excision repair (NER):
  • NER repairs bulky, helix-distorting lesions such as thymine dimers and large chemical adducts caused by UV radiation or chemical exposure.
  • The pathway involves the excision of a short single-stranded DNA segment containing the damage, followed by resynthesis using the complementary strand as a template.
  • NER does not require DNA glycosylases, as it removes lesions through a different mechanism involving excision and resynthesis.

The correct answer is A, B, and D

Concept:

• A promoter is a DNA sequence that facilitates the initiation of transcription by recruiting transcription machinery, such as RNA polymerase II and transcription factors.
• The TATA box is a highly conserved sequence found in many eukaryotic promoters. It is recognized and bound by the TATA-binding protein (TBP), which is a critical component of the transcription factor complex.
• If the TATA box is mutated and cannot bind TBP, the transcription process may still occur but with reduced efficiency, as other promoter elements and transcription factors may partially compensate for the loss of TBP binding.

Explanation:

• Option A: An mRNA will be generated with an alternate reading frame.
  This is incorrect. A mutation in the promoter or TATA box affects the transcription initiation process, not the reading frame of the mRNA. The reading frame is determined by the coding sequence of the gene and the proper start codon during translation, which remains unaffected by the promoter mutation.
• Option B: The mRNA will be transcribed by RNA polymerase I instead of RNA polymerase II.
  This is incorrect. RNA polymerase I is responsible for transcribing ribosomal RNA (rRNA) genes, while RNA polymerase II transcribes protein-coding genes. A mutation in the TATA box does not alter the specificity of RNA polymerase recruitment. The gene will still be transcribed by RNA polymerase II but with reduced efficiency.
• Option C: Transcription may occur with a reduced efficiency.
  This is correct. The TATA box plays a key role in recruiting TBP and the transcription machinery. If the TATA box is mutated, the binding of TBP is impaired, leading to decreased transcription efficiency. However, transcription may still occur through the activity of other promoter elements or compensatory mechanisms.
• Option D: Transcription may occur but will always result in the formation of a non-functional mRNA.
  This is incorrect. A mutation in the TATA box impacts transcription initiation but does not inherently affect the functionality of the mRNA. The mRNA produced may still be functional if it is correctly transcribed from the gene's coding sequence.

The correct answer is (A) - (II), (B) - (I), (C) - (IV), (D) - (III)

Explanation:

Bacterial gene expression regulation involves several sophisticated mechanisms to ensure proper cellular function in response to environmental changes. These mechanisms can operate at the transcriptional, translational, or post-translational level and are mediated by various factors such as proteins, small RNAs, and ligand-binding elements.

• (A) Translated protein acts as a repressor (translational feedback) — (II) Ribosomal protein operon regulation:
  • Many ribosomal proteins regulate their own synthesis through translational feedback mechanisms.
  • When ribosomal proteins are in excess, they bind to the mRNA of their operon, preventing translation. This ensures balanced production of ribosomal components.
• (B) Production of ppGpp in response to amino acid starvation, which in turn regulates transcription by binding to β subunit of RNA polymerase — (I) Stringent response:
  • The stringent response is a survival mechanism in bacteria during amino acid starvation or other stress conditions. When amino acid levels are low, uncharged tRNAs accumulate, signaling the ribosome to synthesize ppGpp (guanosine tetraphosphate). ppGpp then binds to RNA polymerase, altering its affinity for promoters and globally reprogramming gene expression to conserve resources.
  • RelA or SpoT proteins produce ppGpp (guanosine tetraphosphate), which binds to RNA polymerase and alters transcription, prioritizing genes required for stress survival.
• (C) Regulation of bacterial mRNA translation in cis — (IV) Riboswitches that bind a ligand:
  • Riboswitches are regulatory elements located in the 5' untranslated region (UTR) of bacterial mRNAs.
  • They bind specific metabolites (ligands) and undergo conformational changes, affecting transcription or translation of the downstream gene.
• (D) Regulation of bacterial mRNA translation in trans — (III) sRNA (small RNA) and chaperone requires pairing with mRNA:
  • Small RNAs (sRNAs) regulate gene expression by binding to target mRNAs in a sequence-specific manner, often facilitated by chaperone proteins like Hfq.
  • This interaction can either promote or inhibit translation.

The correct answer is The chromatin region is a facultative heterochromatin

Concept:

• Chromatin is composed of DNA and histone proteins organized into nucleosomes. These structures regulate DNA accessibility and gene expression by either compacting the DNA or allowing it to be accessible.
• Facultative heterochromatin form of chromatin is dynamic and can switch between condensed (inactive) and decondensed (active) states depending on the developmental stage or environmental signals.
• Micrococcal nuclease hypersensitivity analysis: This technique is used to identify regions of chromatin that are more accessible or less compact, revealing patterns of nucleosome arrangement and DNA accessibility.
• Nucleosome positioning: Regularly spaced nucleosomes often indicate transcriptionally inactive chromatin, while irregularly spaced nucleosomes and nucleosome-free regions are typically associated with active chromatin and gene expression.

Explanation:

• In early developmental stages, the chromatin region analyzed had regularly spaced nucleosomes, suggesting that the region was likely transcriptionally inactive during this time.
• In later developmental stages, the chromatin showed irregularly spaced nucleosomes with nucleosome-free regions, indicating increased accessibility to transcriptional machinery and a transition to a transcriptionally active state.
• Such dynamic changes in chromatin structure are characteristic of facultative heterochromatin, which can switch between inactive and active states depending on developmental or environmental contexts.

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 Option 4 i.e.Nus A is necessary for intrinsic transcription termination.

Concept:

• Bacterial transcription termination serves two important purposes:
  • regulation of gene expression
  • recycling of RNA polymerase (RNAP)
• Bacteria have 2 major modes of termination of bacterial RNA polymerase (RNAP):
  • Intrinsic (Rho-independent)
  • Rho-dependent

Intrinsic Termination - 

• Intrinsic termination occurs by the specific sequences present in the mRNA sequence itself. 
• These RNA sequences form a stable secondary hairpin loop-type structure signaling for termination. 
• The base-paired region called the stable 'stem' consists of 8-9 'G' and 'C' rich sequences.
• The stem is followed by 6-8 ‘U’ rich sequences.
• Intrinsic transcription terminators consist of an RNA hairpin followed by Uridine-rich nucleotide sequences.
• Intrinsic termination needs two major interactions: 1) nucleic acid elements with 2) RNAP.
• Additional interacting factors like Nus A, could enhance the efficiency of termination but not necessary for intrinsic termination.

Rho-dependent Termination - 

• Rho-dependent termination on the other hand requires Rho protein which is an ATP-dependent RNA hexamer translocase (or helicase). 
• Rho protein binds with ribosome-free mRNA and 'C' rich sites on the mRNA (Rut site).

Explanation:

Option 1: Some terminator sequences require Rho protein for termination

• Since Rho protein is required for termination this option is correct. 

Option 2: Inverted repeat and ‘T’ rich non‐ template strand define intrinsic terminators.

• The image given below represents a pre-requisite template for the intrinsic terminator.
• We can find a T-rich sequence on the non-template DNA strand.
• The inverted repeat sequence is also present and helps in the formation of the hairpin loop (as shown in the image).
• Hence, the statement is correct.

Option 3: Rho-dependent terminators may possess inverted repeat elements.

• In some cases, Rho-dependent terminators could possess inverted repeat elements, but Rho proteins do not rely on these inverted repeat elements for their action.
• Hence the statement is correct.

Option 4: Nus A is necessary for intrinsic transcription termination.

• NusA is not a necessary element for intrinsic transcription termination.
• It might enhance transcription termination in some cases but only as an accessory element.
• Hence, this option is incorrect.

Additional Information

Other mode of termination -

• It is reported in bacteria and is Mfd dependent.
• Mfd-dependent termination occurs with the help of Mfd protein which is a type of DNA translocase and requires ATP for its action just like Rho.

Hence, the correct answer is option 4.

The correct answer is Option 3 i.e.Cdc45

Concept:

• DNA replication in eukaryotes could be divided into three major parts:
  • Initiation
  • Elongation
  • Termination.
• DNA replication initiation could be divided into:
  •  pre-replicative complex
  •  initiation complex.
• Pre-replicative complex majorly consists of
  • ORC (origin recognition complex)+ Cdc6 + Cdt + MCM complex (mini-chromosome maintenance complex)
• Initiation complex consists of 
  • Cdc45 + MCM 10 + GINS + DDK and CDK kinase + Dpb11, Sld3, Sld2 protein complex.

Explanation:

• All the proteins given in the options belong to eukaryotic cells and so we must consider only eukaryotic DNA replication here.

Option 1: ORC - INCORRECT

• DNA replication is initiated from the origin of replication, having specific sequences to initiate replication.
• The ORC is a hexameric DNA binding complex that binds with the origin of replication followed by the recruitment of Cdc6 protein followed by Cdt1.
• ORC dephosphorylates and becomes inactivated before the elongation process.
• Hence, this option is incorrect. 

Option 2: Geminin - INCORRECT

• It binds to cdt1 to prevent the re-initiation of DNA replication and hence it works as a regulator/inhibitor rather than an initiator of DNA replication.
• It is an inhibitor of Cdt1.

Option 3: Cdc45 - CORRECT

• Cdc refers to the cell division control proteins that are involved in various steps of DNA replication process.
• Cdc45 remains with MCM complex and GINS to work as a helicase.
• Thus, it helps in the initiation of DNA replication as well as advancement of the replication fork.

​Option 4: Cdc6 - INCORRECT

• It helps in the assembling of the pre-replicative complexes and interacts with the ORC.
• Cdc6 degrades before initiation of the replication fork.
• The concentration of both cdc6 and cdt1 declines before DNA elongation starts.

​Hence, the correct option is option 3.

The correct answer is Option 2 i.e. CCA sequence is added to these tRNA transcripts in a DNA template independent manner.

Concept:

• Transfer RNA(tRNA) is a small RNA molecule that plays an important role in protein synthesis. It is also known as an adaptor molecule.
• It provides an interface between nucleic acid language and protein language.
• It folds into a cloverleaf-like secondary structure with stems and loops.

Important Points

• Stems and loops in tRNA consist of the acceptor arm, D arm, anticodon arm, and TψC arm.
• Acceptor’s arm has 7 base pairs and 4 unpaired nucleotide sequences along with conserved CCA sequences. Many genes that encode tRNAs do not encode the CCA end
• A specialized RNA polymerase known as CCA-adding enzyme or tRNA nucleotidyltransferase adds it post-transcriptionally. 
• This enzyme adds the terminal CCA to tRNAs that initially lack this sequence without an RNA or DNA template.
• D arm has 5 to 7 nucleotides loop that contains modified dihydrouridine.
• Anticodon arm contains anticodon for recognizing the base pairing with the mRNA.
• TψC contains unusual base pseudouridine.
• Variable arm contains 4 to 5 nucleotides and can contain up to 24 nucleotides as well.

So, the correct answer is option 2.

The correct answer is create DNase I hypersensitive sites.

Concept:

Nucleosome remodelers are enzymes that use ATP to reposition or modify nucleosomes, the structural units of chromatin in eukaryotes. These enzymes help in regulating the accessibility of DNA to transcription factors, RNA polymerases, and other DNA-binding proteins by changing the arrangement of histones on DNA.

• DNase I hypersensitive sites are regions of chromatin that are more accessible to the enzyme DNase I because the DNA is less tightly packed. Nucleosome remodelers can make chromatin more open or "relaxed," which exposes the DNA and makes it hypersensitive to DNase I. This allows regulatory proteins to access these sites more easily, facilitating processes like transcription.

Explanation:

• 1) Methylate histone H3: Histone methylation is performed by histone methyltransferases, not nucleosome remodelers. Methylation of histone tails can either activate or repress gene expression depending on the specific amino acid that is methylated.
• 2) Acetylate histone H3 and H4: Acetylation is done by histone acetyltransferases (HATs), not nucleosome remodelers. Acetylation reduces the positive charge of histones, loosening their interaction with negatively charged DNA, which can lead to increased transcription.
• 4) Degrade histone subunits: Nucleosome remodelers do not degrade histones. Their function is to reposition or remove nucleosomes, not degrade their subunits.

Thus, nucleosome remodelers primarily create DNase I hypersensitive sites by modifying chromatin structure to make DNA more accessible.

Concept:

• DNA polymerase proofreading activity rectifies errors that occurred during the incorporation of nucleotides in the process of new DNA synthesis.
• In PCR, DNA polymerases remove mismatched nucleotides from 3'-end.
• It exhibits 3'→5' exonuclease activity.

Important Points

Option 1:- CORRECT

• The chain termination reaction of PCR is almost similar to that of the Termination reaction in DNA replication.
• The only difference is that ddNTPs are used in PCR in place of dNTPs.
• 3' OH group is absent in ddNTPs that is required for phosphodiester bond formation.
• Therefore if DNA polymerase is incorporated at random sites, an extension of DNA is ceased.

Option 2:- CORRECT

• The Southwestern Blotting technique determines DNA-protein interactions.

Option 3:- INCORRECT

• In PCR (polymerase chain reaction), DNA polymerase enzyme synthesizes new DNA strands based on a template DNA.
• DNA polymerases used in PCR is able to proofread and correct errors during DNA synthesis.
• This proofreading activity is carried out by the 3' - 5' exonuclease activity of the DNA polymerase.
• The 3' - 5' exonuclease activity allows the DNA polymerase to detect and remove incorrectly incorporated nucleotides by excising them from the growing DNA strand, enhancing the fidelity and accuracy of DNA replication in PCR.
• On the other hand, the 5' - 3' exonuclease activity is associated with a different type of enzyme called exonucleases.
• These enzymes degrade or remove nucleotides from the end of a DNA or RNA molecule in a 5' to 3' direction.
• They are not directly involved in proofreading during DNA synthesis.

Option 4:- CORRECT

• Yeast-2-Hybrid (Y2H) is a molecular technique that is used to detect protein-protein interactions as well as protein-DNA interactions.

Conclusion:-

So, the 5' - 3' exonuclease activity ∶ Proof reading polymerase for PCR represents a combination of terms that are matched INCORRECTLY.

The correct answer is Option 4 i.e. rRNA, tRNA and mRNA encoding genes. 

Concept:

• Introns are non-coding regions of DNA found within eukaryotic genes.
• They are present in genes that encode for various types of RNA molecule, including rRNA, tRNA, and mRNA.
• The presence of introns in these RNA genes due to the complex nature of eukaryotic gene regulation and RNA processing.
• Introns play important role in gene expression and RNA maturation.
• Before the pre-mRNA can be used to produce functional RNA molecules (such as rRNA, tRNA or mature mRNA), the introns must be removed through a process called splicing.
• Splicing involves the precise removal of introns and the joining together of exons to form the mature RNA molecule.
• The removal of introns and the joining of exons is a critical step in generating functional RNA molecule.

Steps of intron splicing -

1. Recognition:
   • The spliceosome identifies the 5' and 3' splice sites, at the ends of the intron and the branch point site within the intron.
2. Cleavage:
   • The spliceosome cuts the pre-mRNA at the 5' splice site, releasing the intron as a lariat-shaped structure.
3. Formation of the spliceosome: 
   • The 5' end of the intron is connected to the branch point site, forming a loop, while the 3' end of the intron is connected to the 5' end of the next exon.
4. Exon ligation:
   • The spliceosome catalyzes the joining of the two exons, releasing the intron lariat and forming the mature mRNA.

Hence, the correct answer is option 4.

Additional Information

• In prokaryotic genes, which lack introns, the transcription and translation processes occur simultaneously, as there no need for splicing. 

The correct answer is Option 1

Concept:-

• Mammalian cells contain two forms of RNA polymerase II, designated IIO and IIA, that differ in the extent of phosphorylation within the C-terminal domain of their largest subunit.
• Phosphorylation of this domain, which results in the conversion of RNA polymerase IIA to IIO.
• Kinases help is phosphorylation.
• So IIA can be converted into IIO by kinases. The reverse conversion can be done by phosphatases that remove the phosphate group.
• A third form of the enzyme, RNA polymerase IIB, is found in vitro and lacks the repetitive C-terminal domain.
• So IIB can be made from IIA or IIO by the action of a protease that will remove the C-terminal domain. 
• Kinase: Responsible for protein phosphate group attachment. 
• Phosphatase:- Takes a phosphate group out of a protein.
• Together, these two groups of enzymes control how a cell's proteins behave, frequently in response to outside stimuli.
• The hydrolysis of peptide bonds is a typical chemical reaction that is effectively carried out by proteases.

Explanation:-

Optiion:-​

• Kinases add phosphates to IIa which then gets converted into IIo and it can be converted back to IIa by phosphatase which removes the phosphate. IIa and IIo are converted into IIb with the help of protease.

Hence, this option is correct.

The correct answer is syn A base-paired with anti T

Explanation:

Classical Hoogsteen base pairing involves an alternative hydrogen bonding pattern compared to the standard Watson-Crick base pairing.

• In the Watson-Crick model, adenine (A) pairs with thymine (T) using two hydrogen bonds, and guanine (G) pairs with cytosine (C) using three hydrogen bonds.
• Hoogsteen base pairing, on the other hand, arises when purine bases (adenine and guanine) form hydrogen bonds with their complementary pyrimidine bases (thymine and cytosine) involving different atoms or additional edge of the base, leading to alternative hydrogen bonding patterns.

• Syn Conformation: In this conformation, the base is positioned such that it is over the sugar ring. For adenine, this means the bulkier parts of the adenine ring (like the amino group at position 6) are closer to the deoxyribose (sugar).
• Anti Conformation: In this conformation, the base is flipped away from the sugar, so that it is extended outward, away from the deoxyribose.
• In Watson-Crick base pairing, both bases are usually in the anti conformation.

In Hoogsteen base pairs:

• Adenine (A) pairs with thymine (T) in such a way that adenine uses its N7 and amino group (rather than N1 and 6-amino group in a standard Watson-Crick base pair) to form hydrogen bonds with the O4 and N3 of thymine.
• Guanine (G) pairs with cytosine (C) such that guanine uses its N7 and amino group to form hydrogen bonds with cytosine’s N3 and amino group.

Classical Hoogsteen base pairing can be represented as:

• A Hoogsteen base pair between adenine (A) and thymine (T): In this pair, adenine uses its N7 position to bond with thymine's N3, and using its 6-amino group (NH₂) to bond with thymine's O4.

The classic Hoogsteen base pair would be syn Adenine (A) and syn Thymine (T) in a Hoogsteen configuration

Therefore, if you are looking at possible pairings for classical Hoogsteen interactions: syn A base-paired with anti T

The correct answer is G = 18.5%, C = 24.1%, A = 32.8%, U = 24.6%

Explanation:

Given Base Composition of the DNA Template Strand:

• G (Guanine) = 24.1%
• C (Cytosine) = 18.5%
• A (Adenine) = 24.6%
• T (Thymine) = 32.8%

RNA Transcription:

During transcription, the RNA polymerase synthesizes RNA based on the DNA template strand, following specific base pairing rules:

• Adenine (A) in DNA pairs with Uracil (U) in RNA.
• Thymine (T) in DNA pairs with Adenine (A) in RNA.
• Cytosine (C) in DNA pairs with Guanine (G) in RNA.
• Guanine (G) in DNA pairs with Cytosine (C) in RNA.

Corresponding RNA Base Composition:

1. From G (24.1% in DNA) → C in RNA: C = 24.1%
2. From C (18.5% in DNA) → G in RNA: G = 18.5%
3. From A (24.6% in DNA) → U in RNA: U = 24.6%
4. From T (32.8% in DNA) → A in RNA: A = 32.8%

Final RNA Base Composition:

• G = 18.5%
• C = 24.1%
• A = 32.8%
• U = 24.6%

Conclusion: The correct answer isG = 18.5%, C = 24.1%, A = 32.8%, U = 24.6%, based on the accurate transcription rules from the DNA template strand.