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

The correct answer is P

Explanation:

• The lac operon is an example of an inducible operon in prokaryotes, specifically in Escherichia coli. It consists of three structural genes: lacZ (encodes β-galactosidase), lacY (encodes permease), and lacA (encodes transacetylase).
• In the absence of lactose, the repressor protein binds to the operator region, preventing the RNA polymerase from transcribing the genes (lacZ, lacY, lacA).
• When lactose (or an analog such as IPTG) is present, it acts as an inducer by binding to the repressor protein. This binding inactivates the repressor, allowing RNA polymerase to transcribe the genes.
• As a result, lac mRNA, β-galactosidase, and permease are synthesized, enabling the cell to metabolize lactose.
• The response of the lac operon to lactose is typically an "on-off" switch: in the absence of lactose, the operon is "off," and in the presence of lactose, the operon is "on."

Profile P:

• Lactose added: lac mRNA (dashed line) rises first, followed by PE (solid line) and BG (dotted line). This is correct in terms of order.
• Lactose removed: lac mRNA drops rapidly. PE and BG drop more slowly. This is also correct.

Profile Q:

• Lactose added: PE (solid line) rises before BG (dotted line) and lac mRNA (dashed line) is shown rising with or after PE. This is incorrect. lac mRNA must rise first, followed by both enzymes.

Profile R:

• Lactose added: PE (solid line) rises before BG (dotted line), and lac mRNA (dashed line) also seems to rise later or at the same time as enzymes. This is incorrect. mRNA must precede enzyme synthesis.

Profile S:

• Lactose added: BG (dotted line) and PE (solid line) seem to rise simultaneously or even before lac mRNA (dashed line). This is incorrect.

The correct answer is Histone acetyl transferases and Histone methyl transferases 

Concept:

• Histone writers are enzymes that add specific chemical modifications to histone proteins. These modifications can alter chromatin structure and regulate gene expression.
• Histone readers are proteins that recognize and bind to these specific histone modifications.
• Histone erasers are enzymes that remove these specific chemical modifications from histone proteins.

Explanation:

• Histone acetyl transferases (HATs): These enzymes add acetyl groups to lysine residues on histone tails. This acetylation typically loosens chromatin structure, making DNA more accessible for transcription. Since they add a modification, they are histone writers.
• Histone methyl transferases (HMTs): These enzymes add methyl groups to lysine or arginine residues on histone tails. Depending on the specific residue and the degree of methylation, this can lead to either activation or repression of gene expression. Since they add a modification, they are histone writers.
• Histone deacetylases (HDACs): These enzymes remove acetyl groups from histone tails. This deacetylation generally leads to a more condensed chromatin structure, making DNA less accessible and thus repressing gene expression. Since they remove a modification, they are histone erasers.
• DNA methyl transferases (DNMTs): These enzymes add methyl groups directly to DNA bases (specifically cytosines, often in CpG dinucleotides). While DNA methylation is a crucial epigenetic modification affecting gene expression and chromatin, it is a modification of DNA itself, not histones. Therefore, DNMTs are not classified as histone writers, readers, or erasers.

The correct answer is P−4 ; Q−1 ; R−2 ; S−3

Explanation:

• RNase P (P−4):
  • RNase P is a ribozyme, which is an RNA molecule that catalyzes chemical reactions, similar to enzymes. Its primary function is in the processing of precursor tRNA molecules, where it cleaves the 5' leader sequence of pre-tRNA to produce mature tRNA. This makes RNase P an important example of a catalytic RNA molecule.
• RNA Polymerase I (Q−1):
  • RNA Polymerase I is responsible for transcribing rRNA genes, which are essential for ribosome assembly. It is a eukaryotic enzyme that synthesizes large ribosomal RNA precursors in the nucleolus. This transcription process is key to producing the ribosomal RNA components of the ribosome.
• siRNA (R−2):
  • Small interfering RNA (siRNA) is involved in the process of gene silencing through RNA interference (RNAi). siRNA binds to complementary mRNA sequences, leading to their degradation and preventing translation into proteins. This mechanism is utilized in research and therapeutic applications to silence specific genes.
• Guide RNA (S−3):
  • Guide RNA (gRNA) is an essential component of the CRISPR-Cas9 genome-editing system. It directs the Cas9 protein to a specific DNA target sequence by complementary base pairing. Cas9 then introduces double-strand breaks at the target site, allowing for genome editing or repair.

The correct answer is 0.14

Explanation:

Let's track the type of DNA molecules over four generations, starting with DNA containing only heavy nitrogen (¹⁵N¹⁵N) and then shifting to a medium with light nitrogen (¹⁴N).

• Initial State (Generation 0):
  • All E. coli DNA is grown in15NH4Cl. So, we begin with 1 molecule of "heavy" DNA 15N15N
  • Total DNA molecules = 1 (Heavy)
  • Hybrid DNA = 0
  • Light DNA = 0
• After 1st Generation of Replication (in ¹⁴N medium):
  • The 1 heavy DNA molecule replicates semi-conservatively. Each original 15N strand serves as a template, and new complementary strands are synthesized using 14N
  • Result: 2 hybrid DNA molecules 15N -14N
  • Total DNA molecules = 2¹ = 2
  • Hybrid DNA = 2 molecules.
  • Light DNA = 0 molecules.
• After 2nd Generation of Replication (in 14N medium):
  • The 2 hybrid DNA molecules replicate. Each hybrid molecule separates into one ¹⁵N strand and one ¹⁴N strand.
  • The 15N strands (2 of them) serve as templates for new 14N strands, forming 2 new hybrid (15N -14N) molecules.
  • The 14N strands (2 of them) serve as templates for new 14N strands, forming 2 new light (14N -14N) molecules.
  • Result: 2 hybrid DNA molecules and 2 light DNA molecules.
  • Total DNA molecules = 2² = 4.
  • Hybrid DNA = 2 molecules.
  • Light DNA = 2 molecules.
• After 3rd Generation of Replication (in 14N medium):
  • The existing 2 hybrid and 2 light molecules replicate.
  • From the 2 hybrid molecules: 2 hybrid + 2 light DNA molecules.
  • From the 2 light molecules: 4 light DNA molecules (each light molecule produces 2 new light molecules).
  • Result: 2 hybrid DNA molecules and 2 + 4 = 6 light DNA molecules.
  • Total DNA molecules = 2³ = 8.
  • Hybrid DNA = 2 molecules.
  • Light DNA = 6 molecules.
• After 4th Generation of Replication (in 14N medium):
  • The existing 2 hybrid and 6 light molecules replicate.
  • From the 2 hybrid molecules: 2 hybrid + 2 light DNA molecules.
  • From the 6 light molecules: 12 light DNA molecules.
  • Result: 2 hybrid DNA molecules and 2 + 12 = 14 light DNA molecules.
  • Total DNA molecules = 2⁴ = 16.
  • Hybrid DNA = 2 molecules.
  • Light DNA = 14 molecules.
• Calculate the Molar Ratio:
  • The molar ratio of hybrid DNA (15N -14N) to light DNA (14N -14N) is:
  • Ratio = $\frac{\text{Number of Hybrid DNA molecules}}{\text{Number of Light DNA molecules}}=\frac{2}{14}=\frac{1}{7}$ = 0.14

The correct answer is 146

Explanation:

• In eukaryotic cells, DNA is meticulously packaged to fit within the confined space of the nucleus. The primary level of this packaging involves wrapping DNA around specialized proteins called histones to form structures called nucleosomes. Nucleosomes are the fundamental repeating units of chromatin.
• This is the core component of a nucleosome, consisting of approximately 146 to 147 base pairs of DNA wrapped nearly two times (1.65 turns) around an octamer of histone proteins (two molecules each of H2A, H2B, H3, and H4).
• The assembly of nucleosomes forms a "beads-on-a-string" structure, which is further compacted into higher-order structures to form chromosomes.

Given:

• The length of DNA per base pair (bp) is 0.34 nm/bp.
• The diameter of a nucleosome core particle is 11 nm.
• The ratio of wrapped DNA length to nucleosome diameter is 4.51.

Calculate the Length of DNA Wrapped Around the Nucleosome (in nanometers):

• $\text{Ratio}=\frac{\text{Wrapped DNA Length (nm)}}{\text{Nucleosome Diameter (nm)}}$

Rearranging this formula to solve for the wrapped DNA length:

• Wrapped DNA Length (nm) = Ratio x Nucleosome Diameter (nm)
• Wrapped DNA Length (nm) = 4.51 x 11 nm
• Wrapped DNA Length (nm) = 49.61 nm

Convert the Wrapped DNA Length from Nanometers to Base Pairs (bp):

• Length of DNA in bp = $\frac{\text{Wrapped DNA Length (nm)}}{\text{DNA length per bp (nm/bp)}}$
• Length of DNA in bp = $\frac{49.61\text{ nm}}{0.34\text{ nm/bp}}$
• Length of DNA in bp = 145.91176 bp.

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.