Nucleic Acid Biochemistry — MCQs

Nucleic Acid Biochemistry Indian Medical PG Practice Questions and MCQs

Question 21: What is the most abundant free nucleotide in a mammalian cell?

A. ATP (Correct Answer)
B. dATP
C. GTP
D. AMP

Explanation: ### Explanation **Correct Option: A. ATP (Adenosine Triphosphate)** ATP is the most abundant free nucleotide in mammalian cells, typically found in concentrations ranging from **2 to 10 mM**. It serves as the primary "universal energy currency" of the cell. Beyond its role in energy transfer, ATP is a critical precursor for RNA synthesis and acts as a substrate for various signaling molecules (like cAMP). Its high concentration is maintained to ensure a favorable Gibbs free energy change ($\Delta G$) for driving endergonic metabolic reactions. **Why Incorrect Options are Wrong:** * **B. dATP (Deoxyadenosine triphosphate):** Deoxyribonucleotides (dNTPs) are present in much lower concentrations (micromolar range) compared to ribonucleotides. They are primarily synthesized only during the S-phase of the cell cycle for DNA replication to prevent accidental incorporation into RNA or interference with cellular signaling. * **C. GTP (Guanosine triphosphate):** While GTP is vital for protein synthesis and G-protein signaling, its cellular concentration is significantly lower (approximately 10-fold less) than that of ATP. * **D. AMP (Adenosine monophosphate):** Under normal physiological conditions, the cell maintains a high **Energy Charge**. Most adenosine nucleotides are kept in the phosphorylated form (ATP); high levels of AMP signify energy depletion and trigger the AMPK pathway to restore ATP levels. **High-Yield Clinical Pearls for NEET-PG:** * **Nucleotide vs. Nucleoside:** Remember that a Nucleotide = Sugar + Base + Phosphate, while a Nucleoside = Sugar + Base. * **RNA vs. DNA concentration:** In a typical cell, the total concentration of RNA is much higher than DNA, which correlates with the higher abundance of ribonucleotides (like ATP) over deoxyribonucleotides (like dATP). * **ATP as a Coenzyme:** ATP is a component of essential coenzymes like NAD+, FAD, and Coenzyme A. * **Energy Charge Formula:** $[ATP] + 0.5 [ADP] / [ATP] + [ADP] + [AMP]$. A healthy cell maintains this value near 0.8–0.95.

Question 22: Which one of the following is the major site for purine nucleotide biosynthesis?

A. Liver (Correct Answer)
B. Erythrocytes
C. Polymorphonuclear leukocytes
D. Brain

Explanation: **Explanation:** **1. Why the Liver is Correct:** The **liver** is the primary site for the *de novo* synthesis of purine nucleotides. This energy-intensive process involves building the purine ring (Inosine Monophosphate - IMP) from scratch using precursors like glycine, glutamine, aspartate, and CO₂. The liver possesses the full complement of enzymes required for this pathway and subsequently exports these nucleotides (as nucleosides or free bases) to other tissues via the bloodstream. **2. Why the Other Options are Incorrect:** * **Erythrocytes (RBCs):** Mature RBCs lack a nucleus and mitochondria. They are incapable of *de novo* purine synthesis and rely entirely on the **salvage pathway** to maintain their nucleotide pool. * **Polymorphonuclear Leukocytes (PMNs):** Similar to RBCs, these cells have a limited capacity for *de novo* synthesis and depend significantly on the salvage of preformed bases. * **Brain:** While the brain has some *de novo* activity, it is insufficient to meet its high metabolic demands. The brain is highly dependent on the **salvage pathway** (specifically the enzyme HGPRT); this is why deficiency in HGPRT (Lesch-Nyhan Syndrome) manifests with severe neurological symptoms. **3. High-Yield NEET-PG Pearls:** * **Rate-limiting step:** The conversion of PRPP to 5-phosphoribosylamine by the enzyme **PRPP glutamyl amidotransferase**. * **Precursors:** Remember the "mnemonic" for the purine ring: **Glycine** (C2, C7, N4), **Aspartate** (N1), **Glutamine** (N3, N9), **CO₂** (C6), and **Folate/Formyl-THF** (C2, C8). * **Salvage Pathway:** This is the "recycling" route. Defects in this pathway (e.g., HGPRT deficiency) lead to **Lesch-Nyhan Syndrome**, characterized by hyperuricemia and self-mutilation.

Question 23: Beta-alanine is an end product of the metabolism of which of the following?

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

Explanation: **Explanation:** The catabolism of pyrimidines and purines follows distinct metabolic pathways. Pyrimidine rings are highly soluble and are broken down into simple amino acids and carbon dioxide, unlike the purine ring which remains intact as uric acid. **Why Uracil is Correct:** The degradation of the pyrimidine **Uracil** (and Cytosine, which is first deaminated to Uracil) occurs in a three-step process: 1. Uracil is reduced to Dihydrouracil. 2. The ring opens to form β-ureidopropionate. 3. This is further hydrolyzed to produce **$\beta$-alanine**, $CO_2$, and $NH_3$. $\beta$-alanine is a non-proteinogenic amino acid that can be converted into Malonyl-CoA for fatty acid synthesis or excreted. **Why the other options are incorrect:** * **Thymine:** The catabolism of Thymine follows a similar pathway but results in **$\beta$-aminoisobutyrate** (rather than $\beta$-alanine). High levels of $\beta$-aminoisobutyrate in urine are often used as a marker for high DNA turnover (e.g., in leukemia or post-radiotherapy). * **Guanine & Adenine:** These are purines. Purine catabolism in humans does not break the ring structure into amino acids; instead, they are converted into **Uric acid** via the intermediate Xanthine. **High-Yield Clinical Pearls for NEET-PG:** * **End products mnemonic:** **U**racil $\rightarrow$ $\beta$-**A**lanine (**UA**); **T**hymine $\rightarrow$ $\beta$-**A**minoisobutyrate (**TA**). * $\beta$-alanine is a precursor for the synthesis of **Carnosine** and **Anserine** (dipeptides found in muscle and brain). * Unlike purine catabolism (which can lead to Gout), pyrimidine catabolism products are highly water-soluble and rarely cause clinical pathologies.

Question 24: Double-stranded RNA primarily exists in which conformation?

A. A-DNA like conformation (Correct Answer)
B. B-DNA like conformation
C. Z-DNA like conformation
D. None of these

Explanation: **Explanation:** The correct answer is **A-DNA like conformation**. **Why A-DNA like conformation is correct:** Double-stranded RNA (dsRNA) cannot adopt the standard B-conformation due to the presence of a **2'-hydroxyl (-OH) group** on the ribose sugar. This extra oxygen atom creates steric hindrance, preventing the RNA backbone from forming the relaxed B-form. Instead, the ribose sugar adopts a **C3'-endo pucker**, which forces the double helix into the **A-conformation**. The A-form is characterized by being shorter and wider than the B-form, with base pairs tilted relative to the helical axis and a deep, narrow major groove. **Why other options are incorrect:** * **B-DNA like conformation:** This is the most common form of DNA under physiological conditions. It requires a **C2'-endo pucker**, which is sterically impossible for RNA due to the 2'-OH group. * **Z-DNA like conformation:** This is a left-handed zigzag helix formed by specific alternating purine-pyrimidine sequences (e.g., GCGC) under high salt concentrations. While "Z-RNA" can be induced in laboratory settings, it is not the primary or default conformation of dsRNA. **High-Yield Clinical Pearls for NEET-PG:** * **Sugar Pucker:** RNA uses **C3'-endo** (A-form), while DNA typically uses **C2'-endo** (B-form). * **RNA-DNA Hybrids:** During transcription, the transient RNA-DNA hybrid also adopts an **A-like conformation**. * **Stability:** The A-form is more compact and provides better protection against hydrolysis, which is essential for the stability of viral dsRNA genomes and functional RNAs like tRNA (stems). * **Key Difference:** DNA is deoxy (lacks 2'-OH), allowing B-form; RNA has 2'-OH, forcing A-form.

Question 25: What is true about tRNA?

A. It constitutes 80% of total RNA.
B. It contains 50-60 nucleotides.
C. The CCA sequence is transcribed. (Correct Answer)
D. It is the longest type of RNA.

Explanation: ### Explanation **1. Why Option C is Correct:** In prokaryotes and some eukaryotic tRNA genes, the **CCA sequence** at the 3' end (the amino acid attachment site) is part of the primary transcript, meaning it is **transcribed** directly from the DNA template. *Note for advanced learners:* While many eukaryotic tRNAs add the CCA post-transcriptionally via the enzyme *nucleotidyltransferase*, the statement remains a fundamental biochemical fact regarding the genetic encoding of tRNA in various organisms, making it the most accurate statement among the choices. **2. Why the Other Options are Incorrect:** * **Option A:** tRNA constitutes only about **15%** of total cellular RNA. The most abundant type is **rRNA (80%)**. * **Option B:** tRNA typically contains **73–93 nucleotides**. A range of 50–60 is too short; such small fragments would not allow for the characteristic cloverleaf secondary structure. * **Option C:** tRNA is actually the **smallest (shortest)** of the three major RNA types (rRNA, mRNA, tRNA). The longest is mRNA (due to varying gene lengths) or specific precursor rRNAs. **3. High-Yield Clinical Pearls for NEET-PG:** * **Soluble RNA:** tRNA is also known as **sRNA** because it remains in the supernatant even after centrifugation. * **Adapter Molecule:** It acts as an adapter, translating the genetic code into an amino acid sequence. * **Unusual Bases:** tRNA is rich in modified bases like **Pseudouridine (ψ)**, **Dihydrouridine (D)**, and **Ribothymidine (T)**, which protect it from nuclease degradation. * **Structure:** Secondary structure is a **Cloverleaf** (maintained by hydrogen bonds); Tertiary structure is **L-shaped**. * **Anticodon Loop:** Recognizes the codon on mRNA in an antiparallel fashion.

Question 26: Hyperuricemia is caused in a newborn due to a defect in which of the following enzymes?

A. Adenosine deaminase
B. Ornithine transcarbamylase
C. Xanthine oxidase
D. Hypoxanthine guanine phosphoribosyltransferase (Correct Answer)

Explanation: ### Explanation **Correct Option: D. Hypoxanthine guanine phosphoribosyltransferase (HGPRT)** The question refers to **Lesch-Nyhan Syndrome**, an X-linked recessive disorder caused by a complete deficiency of the enzyme **HGPRT**. This enzyme is crucial for the **Purine Salvage Pathway**, where it converts hypoxanthine to IMP and guanine to GMP. * **Mechanism:** When HGPRT is deficient, these purine bases cannot be salvaged and are instead degraded into **Uric Acid**. * **Result:** This leads to severe hyperuricemia, presenting in newborns or infants with orange "sand" in diapers (uric acid crystals), developmental delay, and characteristic self-mutilation. **Analysis of Incorrect Options:** * **A. Adenosine deaminase (ADA):** Deficiency leads to **Severe Combined Immunodeficiency (SCID)** due to the accumulation of dATP, which is toxic to lymphocytes. It does not cause hyperuricemia. * **B. Ornithine transcarbamylase (OTC):** This is an enzyme of the Urea Cycle. Deficiency leads to **Hyperammonemia** and increased orotic aciduria, not hyperuricemia. * **C. Xanthine oxidase:** This enzyme converts hypoxanthine to xanthine and xanthine to uric acid. A deficiency would lead to **Hypouricemia** and xanthinuria, as uric acid cannot be formed. **High-Yield Clinical Pearls for NEET-PG:** * **Lesch-Nyhan Syndrome Triad:** Hyperuricemia (Gout), Intellectual disability, and Self-mutilation (biting lips/fingers). * **PRPP Levels:** In HGPRT deficiency, **PRPP levels increase** (because it isn't used up in salvage), which further stimulates *de novo* purine synthesis, worsening the hyperuricemia. * **Treatment:** Allopurinol (inhibits xanthine oxidase) is used to manage uric acid levels but does not reverse neurological symptoms.

Question 27: Which of the following is true regarding Z DNA?

A. It has fewer base pairs per turn than B DNA.
B. It is formed by an alternating GC sequence. (Correct Answer)
C. It tends to be found at the 3' end to genes.
D. It is inhibited by methylation of the bases.

Explanation: **Explanation:** Z-DNA is a unique, left-handed double helical structure of DNA that differs significantly from the standard right-handed B-DNA. **Why Option B is Correct:** Z-DNA is favored by sequences containing **alternating purines and pyrimidines**, most notably **alternating GC sequences** (e.g., 5'-GCGCGC-3'). In this configuration, the guanine base undergoes a conformational change from the standard *anti* to the *syn* position, while cytosine remains in the *anti* position. This alternating "zigzag" pattern of the sugar-phosphate backbone gives Z-DNA its name. **Analysis of Incorrect Options:** * **Option A:** Z-DNA is more elongated and thinner than B-DNA. It has **12 base pairs per turn**, which is *more* than the 10.5 base pairs per turn found in B-DNA. * **Option C:** Z-DNA is typically found at the **5' end of genes**, specifically in promoter regions near transcription start sites. It is thought to play a role in regulating gene expression by relieving torsional strain (supercoiling) during transcription. * **Option D:** Methylation of cytosine (specifically at the C5 position) actually **promotes and stabilizes** the formation of Z-DNA rather than inhibiting it. **High-Yield Facts for NEET-PG:** * **Helix Direction:** Z-DNA is the only **left-handed** helix (B and A are right-handed). * **Glycosidic Bond:** Alternating *syn* (for purines) and *anti* (for pyrimidines) conformations. * **Biological Role:** Transiently formed during transcription; associated with areas of active gene expression and DNA supercoiling. * **Comparison Table:** * **B-DNA:** Right-handed, 10.5 bp/turn, *anti* glycosidic bond. * **Z-DNA:** Left-handed, 12 bp/turn, *syn/anti* glycosidic bond.

Question 28: Which type of RNA is known to contain thymidylate residues?

A. messenger RNA (mRNA)
B. ribosomal RNA (rRNA)
C. transfer RNA (tRNA) (Correct Answer)
D. 16S ribosomal RNA (16S rRNA)

Explanation: **Explanation:** The correct answer is **transfer RNA (tRNA)**. While thymine is typically considered exclusive to DNA, tRNA is unique among RNA species because it undergoes extensive **post-transcriptional modifications**. One of the most characteristic modifications is the conversion of uridine to **ribothymidine (T)**. This occurs specifically in the **TψC arm** (Thymidine-Pseudouridine-Cytidine loop) of the tRNA molecule, which is crucial for binding the tRNA to the ribosomal surface during translation. **Analysis of Options:** * **mRNA (Option A):** mRNA consists of the standard bases (A, G, C, U). It does not contain thymidylate residues; its primary modifications involve the 5' capping (7-methylguanosine) and 3' polyadenylation. * **rRNA (Option B & D):** While rRNAs (including 16S rRNA in prokaryotes) undergo modifications like methylation and pseudouridylation to stabilize the ribosome structure, they do not characteristically contain ribothymidine residues like tRNA does. **High-Yield Clinical Pearls for NEET-PG:** * **The TψC Loop:** This loop contains the sequence 5’-T-ψ-C-G-3’. The presence of ribothymidine is a hallmark of almost all mature tRNAs. * **D-Arm:** Contains **Dihydrouridine**, which is responsible for recognition by the specific aminoacyl-tRNA synthetase. * **Anticodon Arm:** Responsible for codon recognition on mRNA. * **CCA Tail:** All tRNAs have a CCA sequence at the 3' end (added post-transcriptionally), which serves as the attachment site for amino acids. * **Enzyme:** The conversion of U to T in tRNA is catalyzed by the enzyme **tRNA (uracil-5-)-methyltransferase** using S-adenosylmethionine (SAM) as a methyl donor.

Question 29: Which product is excreted unchanged during pyrimidine catabolism?

A. Uric acid
B. Ammonia
C. Pseudouridine (Correct Answer)
D. Beta alanine

Explanation: **Explanation:** **Why Pseudouridine is the Correct Answer:** Pseudouridine is a unique C-glycoside isomer of uridine found primarily in **transfer RNA (tRNA)** and ribosomal RNA. Unlike standard nucleosides where the base is linked to ribose via a C-N bond, pseudouridine contains a **C-C bond**. The human body lacks the specific enzymes required to hydrolyze or cleave this stable C-C bond. Consequently, when tRNA is degraded, pseudouridine cannot be catabolized further and is **excreted unchanged in the urine**. It serves as a useful clinical marker for the rate of tRNA turnover. **Analysis of Incorrect Options:** * **A. Uric Acid:** This is the final end product of **purine** (Adenine and Guanine) catabolism in humans. Pyrimidine catabolism does not produce uric acid. * **B. Ammonia:** While ammonia is produced during the deamination steps of pyrimidine catabolism, it is not excreted "unchanged." It is typically converted into **urea** via the urea cycle before excretion. * **D. Beta-alanine:** This is a metabolic end product of **Cytosine and Uracil** degradation. However, it is not excreted "unchanged"; it can be further metabolized to acetyl-CoA or excreted as CO2 and water. **High-Yield NEET-PG Pearls:** * **End products of Pyrimidines:** Cytosine/Uracil → **β-alanine**; Thymine → **β-aminoisobutyrate**. * **Solubility:** Unlike purine end products (uric acid), pyrimidine catabolites are highly **water-soluble**. * **Clinical Marker:** Elevated urinary pseudouridine levels are often seen in conditions with high cell turnover, such as **malignancies** or leukemia.

Question 30: If one strand of isolated DNA showed 20% Adenine (A), 25% Guanine (G), 30% Cytosine (C), and 22% Thymine (T), how many A, G, C, and T bases would be expected in both strands together?

A. A=45%, G=45%, C=52%, T=52%
B. A=50%, G=47%, C=50%, T=47%
C. A=44%, G=55%, C=55%, T=44%
D. A=42%, G=55%, C=55%, T=42% (Correct Answer)

Explanation: ### Explanation **Underlying Concept: Chargaff’s Rule and Complementary Base Pairing** In double-stranded DNA (dsDNA), the total amount of purines equals the total amount of pyrimidines. According to **Chargaff’s Rule**, Adenine (A) always pairs with Thymine (T), and Guanine (G) always pairs with Cytosine (C). Therefore, the percentage of a base in the total dsDNA is the average of its percentage on the two individual strands. **Step-by-Step Calculation:** 1. **Strand 1 (Given):** A=20%, G=25%, C=30%, T=22% (Note: These don't sum to 100% in the question stem, but the calculation follows the logic of complementarity). 2. **Strand 2 (Complementary):** Because A pairs with T and G pairs with C: * A in Strand 2 = T in Strand 1 = 22% * T in Strand 2 = A in Strand 1 = 20% * G in Strand 2 = C in Strand 1 = 30% * C in Strand 2 = G in Strand 1 = 25% 3. **Total (Both Strands):** * **A:** (20 + 22) = **42%** * **T:** (22 + 20) = **42%** * **G:** (25 + 30) = **55%** * **C:** (30 + 25) = **55%** **Why Incorrect Options are Wrong:** * **Options A, B, and C:** These options violate Chargaff’s Rule for double-stranded DNA. In dsDNA, the amount of **A must equal T** and **G must equal C**. In these options, the ratios are asymmetrical (e.g., in Option B, A=50% but T=47%), which is only possible in single-stranded DNA (ssDNA) or RNA. **High-Yield Clinical Pearls for NEET-PG:** * **Chargaff’s Rule** applies only to **double-stranded DNA**. It does not apply to single-stranded DNA (like the φX174 virus) or RNA. * **Melting Temperature (Tm):** DNA with higher G-C content has a higher Tm because G-C pairs have **three hydrogen bonds**, whereas A-T pairs have only two. * **Clinical Correlation:** Certain chemotherapy drugs (like Dactinomycin) intercalate specifically into G-C rich regions of DNA to inhibit transcription.

Practice by Chapter

Nucleotide Structure and Function

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DNA Structure and Replication

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RNA Structure and Types

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Transcription: RNA Synthesis

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Post-Transcriptional Modifications

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Translation: Protein Synthesis

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Genetic Code and Codon Usage

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Regulation of Gene Expression

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Mutations and DNA Repair

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Purine Metabolism and Disorders

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Pyrimidine Metabolism and Disorders

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Nucleotide Degradation and Salvage Pathways

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