Nucleic Acid Biochemistry — MCQs

Nucleic Acid Biochemistry Indian Medical PG Practice Questions and MCQs

Question 191: Which type of DNA is characterized by a left-handed helix?

A. B-DNA
B. A-DNA
C. Z-DNA (Correct Answer)
D. F-DNA

Explanation: ***Z-DNA*** - **Z-DNA** is a unique form of DNA characterized by its distinctive **left-handed helical structure**, as opposed to the right-handed helices of A-DNA and B-DNA. - This structure forms under specific conditions, often in regions with alternating **purine-pyrimidine sequences** (e.g., GCGCGC) and is thought to play roles in **gene regulation** and chromatin structure. - It has a **zigzag backbone** appearance, which gives it its name. *B-DNA* - **B-DNA** is the most common and **biologically significant form of DNA** found under physiological conditions in living cells. - It exhibits a **right-handed helical structure**, with approximately 10-10.5 base pairs per turn. - This is the predominant form in aqueous environments at neutral pH. *A-DNA* - **A-DNA** is a **right-handed helical DNA** form that is usually found under dehydrating conditions or when DNA is complexed with certain proteins. - It has a wider and shorter helix than B-DNA, with about 11 base pairs per turn. - The bases are tilted relative to the helix axis. *F-DNA* - **F-DNA** is not a recognized standard DNA conformation in biochemistry. - The three well-established DNA forms are **A-DNA, B-DNA, and Z-DNA**, based on their structural characteristics and biological relevance.

Question 192: What is the primary functional mechanism of small RNAs in cellular processes?

A. Typically less than 200 nucleotides in length
B. Involved in post-transcriptional regulation of gene expression
C. Always synthesized endogenously
D. Regulate gene expression through RNA interference mechanisms (Correct Answer)

Explanation: ***Regulate gene expression through RNA interference mechanisms*** - Small RNAs, such as **miRNAs (microRNAs)** and **siRNAs (small interfering RNAs)**, primarily function by guiding **RNA-induced silencing complexes (RISC)** to target mRNA molecules. - This interaction leads to either the **degradation of mRNA** or the **inhibition of its translation**, thereby regulating gene expression at the post-transcriptional level. *Typically less than 200 nucleotides in length* - While small RNAs are indeed short, this statement describes a **physical characteristic** rather than their primary cellular function. - Their function is linked to specific molecular interactions, not simply their size. *Involved in post-transcriptional regulation of gene expression* - This is a correct statement about their function but is **less specific** than regulating gene expression via RNA interference. - RNA interference is the specific mechanism by which many small RNAs achieve post-transcriptional regulation. *Always synthesized endogenously* - While many small RNAs like miRNAs are **endogenously synthesized**, some siRNAs can be **exogenously introduced** (e.g., in research) or derived from viral infections. - Therefore, stating they are *always* synthesized endogenously is inaccurate.

Question 193: Which of the following statements about purine synthesis is true?

A. Glutamine PRPP amidotransferase is the rate-limiting enzyme in purine synthesis.
B. THFA derivatives are required for the formation of C2 and C8 in the purine ring.
C. Glutamine donates the amino group for N1 of the purine ring.
D. IMP is the first nucleotide synthesized during purine synthesis. (Correct Answer)

Explanation: ***IMP is the first nucleotide synthesized during purine synthesis.*** - **Inosine monophosphate (IMP)** is the first complete purine nucleotide formed in de novo purine synthesis. - It serves as the precursor for both **AMP** and **GMP**, making it the foundational molecule in the purine biosynthesis pathway. - The synthesis pathway converges at IMP before branching to form the specific adenine and guanine nucleotides. *Glutamine PRPP amidotransferase is the rate-limiting enzyme in purine synthesis.* - While **glutamine PRPP amidotransferase** catalyzes the committed step in de novo purine synthesis, it is often considered a key regulatory enzyme. - However, the statement is marked incorrect in this context because **PRPP synthetase** (which forms PRPP from ribose-5-phosphate) can also be considered rate-limiting depending on PRPP availability. - The first committed step specific to purines is the glutamine PRPP amidotransferase reaction, making this a debatable but commonly accepted alternative answer. *THFA derivatives are required for the formation of C2 and C8 in the purine ring.* - **Tetrahydrofolate (THF) derivatives** do provide one-carbon units to positions **C2 and C8** of the purine ring. - Specifically, **N10-formyl-THF** donates the formyl group for C2, and **N5,N10-methenyl-THF** (which converts to N10-formyl-THF) provides C8. - This statement is technically correct, but may be marked incorrect if the question seeks a more fundamental defining feature of purine synthesis (such as IMP being the first nucleotide). *Glutamine donates the amino group for N1 of the purine ring.* - **Glutamine** provides nitrogen atoms at positions **N3 and N9** of the purine ring. - The **N1 nitrogen** is derived from the amino group of **aspartate**, not glutamine. - This statement is clearly incorrect.

Question 194: For which tissue/organ is the salvage pathway of purine biosynthesis particularly important?

A. Liver
B. RBCs (Correct Answer)
C. Kidney
D. Lung

Explanation: ***RBCs*** - **Red blood cells (RBCs)** lack a nucleus and the machinery for *de novo* purine synthesis, making them entirely dependent on the **salvage pathway** to acquire purines. - The **salvage pathway** reuses pre-existing purine bases and nucleosides to synthesize new purine nucleotides via enzymes like **HGPRT** (hypoxanthine-guanine phosphoribosyltransferase), which is crucial for RBC function. - **Brain tissue** is another organ critically dependent on salvage pathways, but among the given options, RBCs represent the classic example of absolute salvage pathway dependence. *Liver* - The liver is a major site of **_de novo_ purine synthesis** and is not primarily dependent on the salvage pathway for its purine requirements. - While the liver does utilize the salvage pathway, it also has robust **_de novo_ synthesis** capabilities, making it less critical than for RBCs. *Kidney* - The kidney performs both **_de novo_ purine synthesis** and utilizes the salvage pathway, similar to most other metabolically active tissues. - It is not uniquely or predominantly reliant on the salvage pathway for its purine needs compared to _de novo_ synthesis. *Lung* - The lung tissue, like most tissues with active metabolism and cell division, has the capacity for both **_de novo_ purine synthesis** and the salvage pathway. - It does not have a specific or heightened dependence on the salvage pathway that would make it particularly important compared to other tissues.

Question 195: Which of the following is a nucleoside?

A. Adenine
B. Uridine (Correct Answer)
C. Thymine
D. Guanine

Explanation: ***Uridine*** - Uridine is a **nucleoside** composed of the nitrogenous base **uracil** covalently attached to the sugar **ribose** via a β-N1-glycosidic bond. - Nucleosides consist of a **nitrogenous base** (purine or pyrimidine) linked to a **pentose sugar** (ribose or deoxyribose). *Adenine* - Adenine is a **purine nitrogenous base**, not a nucleoside. - It is a component of both DNA and RNA, and when combined with ribose, it forms the nucleoside **adenosine**. *Thymine* - Thymine is a **pyrimidine nitrogenous base**, not a nucleoside. - When combined with deoxyribose, it forms the deoxynucleoside **thymidine**, which is found in DNA. *Guanine* - Guanine is a **purine nitrogenous base**, not a nucleoside. - When combined with ribose, it forms the nucleoside **guanosine**.

Question 196: Which enzyme is responsible for synthesizing RNA during transcription?

A. RNA polymerase (Correct Answer)
B. Primase
C. Ligase
D. Topoisomerase

Explanation: ***RNA polymerase (Correct Answer)*** - **RNA polymerase** is the central enzyme responsible for synthesizing an **RNA strand** from a DNA template during transcription. - It unwinds the DNA helix, reads the nucleotide sequence, and adds complementary RNA nucleotides to form a new RNA molecule. - This is a fundamental process in **gene expression**, occurring in the nucleus of eukaryotic cells. *Primase (Incorrect)* - **Primase** is an enzyme involved in **DNA replication**, not transcription. - Its function is to synthesize short **RNA primers** (8-12 nucleotides) that provide a starting point for DNA polymerase. *Ligase (Incorrect)* - **Ligase** is an enzyme that joins DNA fragments together by forming **phosphodiester bonds**. - It is primarily involved in **DNA replication** and repair processes, connecting Okazaki fragments on the lagging strand or repairing nicks in DNA strands. *Topoisomerase (Incorrect)* - **Topoisomerases** are enzymes that regulate the **supercoiling** of DNA. - They relieve **torsional stress** that builds up during DNA replication and transcription by cutting and rejoining DNA strands, preventing tangling.

Question 197: If a sample of DNA has adenine at 28%, what will be the amount of Cytosine present?

A. 20%
B. 24%
C. 22% (Correct Answer)
D. 26%

Explanation: ***22%*** - According to **Chargaff's rules**, in a double-stranded DNA molecule, the amount of **adenine (A)** is approximately equal to the amount of **thymine (T)**, and the amount of **guanine (G)** is approximately equal to the amount of **cytosine (C)**. - If adenine (A) is 28%, then thymine (T) is also 28%. The total percentage of A and T is 28% + 28% = 56%. The remaining percentage for G and C is 100% - 56% = 44%. Since G = C, cytosine (C) will be 44% / 2 = 22%. *20%* - This value is not consistent with the given **adenine percentage** when applying **Chargaff's rules** for DNA base pairing. - If cytosine were 20%, then guanine would also be 20%, making the total G+C content 40%. This would leave 60% for A+T, meaning A would be 30%, not 28%. *24%* - This percentage does not align with the fundamental **base-pairing rules** of DNA. - If cytosine were 24%, then guanine would also be 24%, totaling 48% for G+C. This would imply 52% for A+T, meaning adenine would be 26%, which contradicts the given 28%. *26%* - This would only be correct if the **adenine percentage** was lower, as it suggests a different **G+C content**. - If cytosine were 26%, then guanine would also be 26%, making the total G+C content 52%. This would imply 48% for A+T, meaning adenine would be 24%, not 28%.

Question 198: What is the mechanism responsible for the intestine-specific expression of apoprotein B-48?

A. DNA rearrangement
B. RNA alternative splicing
C. RNA editing (Correct Answer)
D. Protein synthesis

Explanation: ***RNA editing*** - In the intestine, a **cytidine deaminase enzyme (APOBEC-1)** deaminates a specific **cytidine to uridine** at position 6666 in the apoB mRNA. - This C-to-U change creates a **premature stop codon (UAA)**, resulting in the truncated **apoB-48 protein** (48% of the full-length apoB-100). *DNA rearrangement* - This mechanism involves permanent changes in the **genomic DNA sequence**, often seen in immune gene diversification (e.g., V(D)J recombination). - It would lead to a different gene product at the DNA level, which is not how apoB-48 is generated. *RNA alternative splicing* - This process involves the selective inclusion or exclusion of **exons** during mRNA processing, leading to different protein isoforms from a single gene. - While it generates multiple protein products, it does not involve a nucleotide change within the mRNA sequence to create a new stop codon. *Protein synthesis* - This is the process of translating mRNA into protein, directed by the codons in the mRNA sequence. - While apoB-48 is a product of protein synthesis, the mechanism for its *intestine-specific expression* lies in the modification of the mRNA *before* translation, not in the synthesis process itself.

Question 199: Which of the following statements about DNA polymerase I is correct?

A. Participates in the synthesis of Okazaki fragments.
B. Is the primary enzyme for DNA replication in bacteria
C. Involved in DNA repair processes. (Correct Answer)
D. Not essential for DNA replication in bacteria.

Explanation: ***Involved in DNA repair processes.*** - **DNA polymerase I** possesses **5' to 3' exonuclease activity**, which is crucial for removing **RNA primers** and damaged DNA segments during DNA repair. - Its **DNA repair function** is essential for maintaining genome integrity by excising incorrect nucleotides and filling the gaps. - DNA pol I plays a key role in **nick translation** and **gap filling** after primer removal during DNA replication. *Participates in the synthesis of Okazaki fragments.* - **DNA polymerase III** is the primary enzyme responsible for synthesizing **Okazaki fragments** on the lagging strand during bacterial DNA replication. - While DNA polymerase I does **process** Okazaki fragments by removing RNA primers and filling gaps, it does not *synthesize* them. *Is the primary enzyme for DNA replication in bacteria* - **DNA polymerase III** is the main enzyme responsible for the bulk of DNA synthesis during replication in **bacteria**. - DNA polymerase I plays a more specialized role in **primer removal** and **gap filling** rather than primary elongation. *Not essential for DNA replication in bacteria.* - **DNA polymerase I** is **essential** for bacterial viability despite not being the primary replicative polymerase. - Its crucial role in **primer removal** and **gap filling** after primer excision is indispensable for completing DNA replication and repair processes.

Question 200: If the content of adenine (A) is 15%, what is the percentage of guanine (G) in the DNA?

A. 15%
B. 85%
C. 70%
D. 35% (Correct Answer)

Explanation: ***35%*** - According to **Chargaff's rules**, in a DNA molecule, the amount of **adenine (A) is equal to the amount of thymine (T)**, and the amount of **guanine (G) is equal to the amount of cytosine (C)**. - If A = 15%, then T must also be 15%. This means A + T = 30%. Since the total percentage of all bases is 100%, G + C must be 100% - 30% = 70%. As G = C, then G = 70% / 2 = 35%. *15%* - This would only be correct if guanine paired with adenine, which it does not; guanine pairs with **cytosine**. - This answer incorrectly assumes that all four bases are present in equal proportions, or that G equals A, which violates **Chargaff's rules**. *85%* - This percentage would imply an incorrect base pairing or an imbalanced ratio of purines and pyrimidines, violating the fundamental structure of DNA. - An 85% guanine content would mean that G + C far exceeds 100% or that T is extremely low, which is biologically impossible. *70%* - This represents the combined percentage of **guanine and cytosine**, not guanine alone. - While it correctly acknowledges the remaining proportion of bases, it fails to divide this sum between the two equal components, **G and C**.

Practice by Chapter

Nucleotide Structure and Function

Practice Questions

DNA Structure and Replication

Practice Questions

RNA Structure and Types

Practice Questions

Transcription: RNA Synthesis

Practice Questions

Post-Transcriptional Modifications

Practice Questions

Translation: Protein Synthesis

Practice Questions

Genetic Code and Codon Usage

Practice Questions

Regulation of Gene Expression

Practice Questions

Mutations and DNA Repair

Practice Questions

Purine Metabolism and Disorders

Practice Questions

Pyrimidine Metabolism and Disorders

Practice Questions

Nucleotide Degradation and Salvage Pathways

Practice Questions

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