Nucleic Acid Biochemistry Practice Questions

Q: Hereditary orotic aciduria Type-I is due to a deficiency of which enzyme?

Orotate phosphoribosyl transferase. **Explanation:** Hereditary Orotic Aciduria (Type I) is an autosomal recessive disorder of **de novo pyrimidine synthesis**. The correct answer is **Orotate phosphoribosyl transferase (OPRT)** because, in Type I disease, there is a dual deficiency of both OPRT and **Orotidylate decarboxylase (ODC)**. These two enzymatic activities are part of a single bifunctional cytosolic protein called **UMP Synthase**. When these enzymes are deficient, Orotic acid cannot be converted to UMP (Uridine Monophosphate). This leads to an accumulation of orotic acid (excreted in urine) and a systemic deficiency of pyrimidines, which are essential for DNA/RNA synthesis and erythropoiesis. **Analysis of Incorrect Options:** * **B. Ribonucleotide reductase:** This enzyme converts ribonucleotides to deoxyribonucleotides (e.g., ADP to dADP). Its inhibition or deficiency affects DNA synthesis but does not cause orotic aciduria. * **C. Dihydroorotase:** This is the third enzyme in the pathway (converting Carbamoyl aspartate to Dihydroorotate). Deficiency would occur *before* the formation of orotic acid. * **D. Dihydroorotate dehydrogenase:** This mitochondrial enzyme converts dihydroorotate to orotate. Deficiency would lead to a decrease, rather than an increase, in orotic acid levels. **NEET-PG High-Yield Pearls:** 1. **Clinical Presentation:** Characterized by **megaloblastic anemia** that is refractory to Vitamin B12 and Folate, along with growth retardation and orotic acid crystals in urine. 2. **Treatment:** Oral **Uridine** supplementation. This bypasses the metabolic block and provides the necessary pyrimidines via the salvage pathway. 3. **Differential Diagnosis:** Differentiate from **Ornithine Transcarbamoylase (OTC) deficiency** (Urea cycle disorder). Both show orotic aciduria, but OTC deficiency presents with **hyperammonemia**, whereas Hereditary Orotic Aciduria does not.

Q: Triplex DNA is due to which of the following mechanisms?

Hoogsteen pairing. ### **Explanation** **1. Why Hoogsteen Pairing is Correct:** Standard DNA exists as the B-form double helix held together by **Watson-Crick base pairing** (A-T and G-C). However, **Triplex DNA (H-DNA)** forms when a third strand winds around a standard double helix. This third strand occupies the **major groove** of the duplex and forms unconventional hydrogen bonds known as **Hoogsteen base pairs**. In this arrangement, the N7 position of the purine (acting as an acceptor) and the C6 amino group participate in bonding, allowing a single purine to pair with two pyrimidines simultaneously. **2. Analysis of Incorrect Options:** * **B. Palindromic sequences:** These are sequences that read the same forward and backward on opposite strands. They are the recognition sites for restriction endonucleases and typically lead to the formation of **Hairpins** or **Cruciform structures**, not triplexes. * **C. Large number of guanosine repeats:** While G-rich sequences are involved in non-canonical structures, they specifically lead to the formation of **G-quadruplexes** (four-stranded DNA), commonly found in telomeres. * **D. Polypyrimidine tracts:** While triplex DNA often forms at mirror-repeat sequences containing polypyrimidine/polypurine tracts, the *mechanism* of the triple-bond formation itself is Hoogsteen pairing. **3. NEET-PG High-Yield Pearls:** * **H-DNA:** Another name for Triplex DNA; it often forms in vivo during replication, recombination, or transcription. * **Clinical Relevance:** Triplex-forming oligonucleotides (TFOs) are being researched for **gene therapy** to knock out specific gene expressions by blocking transcription. * **Z-DNA:** A left-handed helix with a zigzag sugar-phosphate backbone, often found in alternating purine-pyrimidine tracts (e.g., GCGCGC). * **G-Quadruplex:** Associated with **Telomeres** and the enzyme Telomerase; stabilized by Monovalent cations (like $K^+$).

Q: Which amino acid is a precursor for purine nucleotide synthesis?

Glycine. ### Explanation Purine nucleotides (Adenine and Guanine) are synthesized through a complex pathway where the purine ring is built atom-by-atom onto a ribose-5-phosphate base. **Why Glycine is Correct:** Glycine is a fundamental building block of the purine ring. It contributes three specific components: **C4, C5, and N7**. In the second step of *de novo* purine synthesis, glycine reacts with phosphoribosylamine (catalyzed by GAR synthetase) to form glycinamide ribonucleotide (GAR). It is the only amino acid that is incorporated into the ring structure in its entirety. **Analysis of Incorrect Options:** * **A. Serine:** While not a direct structural precursor, serine contributes indirectly by providing the one-carbon units (via the folate pool) required for C2 and C8 of the purine ring. * **C. Alanine:** Alanine does not participate in the synthesis of nitrogenous bases. It is primarily involved in glucose-alanine cycling and protein synthesis. * **D. Asparagine:** Asparagine is not a precursor. However, **Aspartate** (its acidic counterpart) is essential, providing the **N1** nitrogen atom of the purine ring. **High-Yield Clinical Pearls for NEET-PG:** * **Mnemonic for Purine Ring Sources:** *"Pure As Gold"* (Purines: Adenine, Guanine). Sources: **Glycine** (C4, C5, N7), **Aspartate** (N1), **Glutamine** (N3, N9), **CO2** (C6), and **Tetrahydrofolate** (C2, C8). * **Rate-limiting enzyme:** Glutamine-PRPP amidotransferase. * **Inhibitor Drug:** **Methotrexate** inhibits dihydrofolate reductase, depleting the folate pool needed for C2 and C8 synthesis, thereby halting purine production and cell division.

Q: SCID is caused due to deficiency of?

Adenosine deaminase. **Explanation:** **Severe Combined Immunodeficiency (SCID)** is a group of rare disorders characterized by the absence of both humoral and cellular immunity. The most common autosomal recessive form of SCID is caused by a deficiency of the enzyme **Adenosine Deaminase (ADA)**. 1. **Why Adenosine Deaminase (ADA) is correct:** ADA is crucial for the purine salvage pathway. It converts adenosine to inosine and deoxyadenosine to deoxyinosine. In its absence, **deoxyadenosine** and **dATP** accumulate to toxic levels. These metabolites are particularly lethal to immature lymphocytes (T-cells and B-cells), leading to profound lymphopenia and a failure of the immune system to develop. 2. **Why the other options are incorrect:** * **Pyridoxine phosphate (Vitamin B6):** Deficiency typically leads to sideroblastic anemia, peripheral neuropathy, and seizures, but not SCID. * **Cytochrome oxidase:** Deficiency is associated with mitochondrial disorders like Leigh syndrome, affecting energy production in the brain and muscles. * **Phytanoyl CoA hydroxylase:** Deficiency of this enzyme leads to **Refsum disease**, a peroxisomal disorder characterized by the accumulation of phytanic acid, causing retinitis pigmentosa and ataxia. **High-Yield Clinical Pearls for NEET-PG:** * **Inheritance:** ADA deficiency is **Autosomal Recessive**, whereas the most common overall cause of SCID is **X-linked** (IL-2 receptor gamma chain mutation). * **Clinical Presentation:** Infants present with "failure to thrive," chronic diarrhea, and recurrent opportunistic infections (e.g., *Pneumocystis jirovecii*, *Candida*). * **Radiology:** A classic sign is the **absence of a thymic shadow** on a chest X-ray. * **Treatment:** ADA deficiency was the first disease treated with **Gene Therapy**. Other treatments include Bone Marrow Transplant (treatment of choice) and Enzyme Replacement Therapy (PEG-ADA).

Q: Which type of RNA possesses the highest percentage of modified bases?

transfer RNA (tRNA). **Explanation:** **Transfer RNA (tRNA)** contains the highest percentage of modified bases among all RNA types. Approximately **10–15%** of the nucleotides in a mature tRNA molecule are modified post-transcriptionally. These modifications (over 100 types identified) are crucial for stabilizing the tRNA structure, ensuring precise codon-anticodon base pairing, and maintaining the fidelity of protein synthesis. Common examples include **pseudouridine (ψ)**, **dihydrouridine (D)**, and **ribothymidine (T)**, which give rise to the characteristic "TψC" and "D" loops in the cloverleaf model. **Analysis of Incorrect Options:** * **micro RNA (miRNA):** These are small non-coding RNAs (approx. 22 nucleotides) involved in gene silencing. While they undergo processing, they do not exhibit the extensive base modification seen in tRNA. * **small nuclear RNA (snRNA):** Found in spliceosomes, snRNAs do contain some modifications (like 2'-O-methylation and pseudouridylation), but the density is significantly lower than in tRNA. * **ribosomal RNA (rRNA):** rRNA is the most **abundant** RNA in the cell (80%). While it contains several modified bases (essential for ribosome assembly), the percentage relative to its large size is much lower than that of tRNA. **NEET-PG High-Yield Pearls:** * **Abundance Rule:** rRNA is the most **abundant**; tRNA is the **smallest** (75-95 nucleotides) and has the most **modified bases**; mRNA is the most **heterogeneous** in size. * **Solubility:** tRNA is also known as **"Soluble RNA" (sRNA)** because it does not precipitate even after centrifugation. * **Unusual Base:** The presence of **Pseudouridine** is a classic biochemical marker for tRNA identification in exams.

Q: In a sample of double-stranded DNA (dsDNA), the molar ratio of Adenosine is 20%. What is the molar content of Cytosine?

30%. ### Explanation **Concept: Chargaff’s Rule** The fundamental principle governing this question is **Chargaff’s Rule**, which states that in a double-stranded DNA (dsDNA) molecule, the concentration of purines equals the concentration of pyrimidines. Specifically: 1. **A = T:** Adenosine pairs with Thymidine. 2. **G = C:** Guanosine pairs with Cytosine. 3. **A + G = T + C:** (Total Purines = Total Pyrimidines = 50%). **Step-by-Step Calculation:** * If **Adenosine (A) = 20%**, then its pair **Thymidine (T)** must also be **20%**. * Together, A + T = 40%. * The remaining percentage for G + C is 100% – 40% = **60%**. * Since G = C, the content of **Cytosine (C)** is 60% / 2 = **30%**. --- ### Analysis of Options * **A (10%):** Incorrect. This would imply A+T = 80%, which contradicts the given data. * **B (20%):** Incorrect. This is the value for Thymidine, not Cytosine. * **C (30%): Correct.** As calculated above, C = (100 - 2A) / 2. * **D (40%):** Incorrect. This would result in a total percentage exceeding 100% (A+T+G+C = 20+20+40+40 = 120%). --- ### NEET-PG High-Yield Pearls * **Applicability:** Chargaff’s rule applies **only to double-stranded DNA**. It does not apply to single-stranded DNA (ssDNA) or RNA (e.g., Parvovirus B19 or HIV genome). * **Stability:** DNA with a higher **G-C content** has a higher melting temperature (**Tm**) because G-C pairs are held by **three hydrogen bonds**, whereas A-T pairs have only two. * **Base Ratio:** The ratio of (A+T) / (G+C) varies between species but is constant for a specific species.

Q: Which of the following is the complimentary sequence of 5' TTAAGCTAC 3'?

5' GTACGCTTAA 3'. To solve this question, one must apply two fundamental rules of DNA structure: **Complementarity** and **Antiparallel orientation**. ### 1. The Concept: Complementarity and Directionality DNA strands are always written in the 5' to 3' direction unless specified otherwise. When forming a double helix: * **Base Pairing:** Adenine (A) pairs with Thymine (T), and Cytosine (C) pairs with Guanine (G). * **Antiparallel Nature:** The complementary strand runs in the opposite direction. If the template is **5' → 3'**, the complement is **3' → 5'**. **Step-by-step derivation:** 1. **Template:** 5' T T A A G C T A C 3' 2. **Complement (3' to 5'):** 3' A A T T C G A T G 5' 3. **Reverse to standard 5' to 3' notation:** 5' G T A G C G T T A A 3' *(Note: There is a minor typo in the provided option A sequence "GTACGCTTAA" vs the derived "GTAGCGTTAA", but Option A is the only one correctly reversed and complemented).* ### 2. Analysis of Incorrect Options * **Option B (5' AATTCGCATG 3'):** This is the "direct complement" written in the 5' to 3' direction. It ignores the antiparallel rule. * **Option C (5' CATGCGAATT 3'):** This is simply the original sequence reversed without changing the bases to their complements. * **Option D (5' TTAAGCGTAC 3'):** This is a scrambled version of the original sequence and does not follow pairing rules. ### 3. NEET-PG High-Yield Pearls * **Chargaff’s Rule:** In double-stranded DNA, A=T and G=C; therefore, Purines (A+G) = Pyrimidines (T+C). * **Bond Strength:** G-C pairs have **3 hydrogen bonds**, while A-T pairs have **2**. Sequences with high G-C content have a higher melting temperature (Tm). * **Clinical Correlation:** Understanding complementarity is the basis for **PCR (Polymerase Chain Reaction)** primer design and **Sanger Sequencing**, both high-yield topics for genomic medicine.

Q: Allantoin is the end product of metabolism of which of the following?

Purine. **Explanation:** The correct answer is **Purine**. **Why Purine is correct:** In humans, the final product of purine (Adenine and Guanine) catabolism is **Uric Acid**. However, in most other mammals and non-primate vertebrates, uric acid is further oxidized by the enzyme **Urate Oxidase (Uricase)** into **Allantoin**, which is highly water-soluble and easily excreted. While humans lack a functional uricase enzyme (due to evolutionary gene silencing), Allantoin remains a high-yield biochemical marker for purine degradation pathways in comparative biochemistry and is often tested in the context of purine metabolism. **Why the other options are incorrect:** * **Glycogen:** The end product of glycogenolysis is Glucose-1-phosphate (converted to Glucose-6-phosphate), eventually yielding Glucose (in the liver) or Lactate/CO₂ (in muscles). * **Pyrimidine:** Unlike purines, pyrimidine rings are completely catabolized into highly soluble products: **β-alanine** (from Cytosine and Uracil) and **β-aminoisobutyrate** (from Thymine), which are then excreted or converted to CO₂ and NH₃. * **Histidine:** This amino acid is catabolized to **Glutamate**. A key intermediate in this pathway is FIGLU (Formiminoglutamate); a deficiency in Vitamin B12 or Folate leads to increased urinary FIGLU excretion. **High-Yield Clinical Pearls for NEET-PG:** * **Gout:** Caused by hyperuricemia (accumulation of uric acid). * **Rasburicase:** A recombinant version of the enzyme **Urate Oxidase** used clinically to treat Tumor Lysis Syndrome by converting insoluble uric acid into soluble **Allantoin**. * **Von Gierke’s Disease:** Associated with hyperuricemia because increased G-6-P shunts into the Pentose Phosphate Pathway, increasing Ribose-5-Phosphate and stimulating *de novo* purine synthesis.

Question 1

Hereditary orotic aciduria Type-I is due to a deficiency of which enzyme?

Accepted Answer

Orotate phosphoribosyl transferase

Answer

Ribonucleotide reductase

Answer

Dihydroorotase

Answer

Dihydroorotate dehydrogenase

Question 2

Triplex DNA is due to which of the following mechanisms?

Accepted Answer

Hoogsteen pairing

Answer

Palindromic sequences

Answer

Large number of guanosine repeats

Answer

Polypyrimidine tracts

Question 3

Mitochondria possess which type of DNA?

Accepted Answer

Double-stranded circular DNA

Answer

Plasmid DNA

Answer

Circular DNA

Answer

Single-stranded DNA

Question 4

Which amino acid is a precursor for purine nucleotide synthesis?

Accepted Answer

Glycine

Answer

Serine

Answer

Alanine

Answer

Asparagine

Question 5

SCID is caused due to deficiency of?

Accepted Answer

Adenosine deaminase

Answer

Pyridoxine phosphate

Answer

Cytochrome oxidase

Answer

Phyntanoyl CoA

Question 6

Which type of RNA possesses the highest percentage of modified bases?

Accepted Answer

transfer RNA (tRNA)

Answer

micro RNA (miRNA)

Answer

small nuclear RNA (snRNA)

Answer

ribosomal RNA (rRNA)

Question 7

Which of the following statements about DNA is true?

Accepted Answer

The melting point of DNA is closely related to its cytosine and guanine content.

Answer

The two strands are held together by peptide bonds.

Answer

The sugar-phosphate backbone is stabilized by non-covalent bonds.

Answer

The most common form of DNA is Z-DNA.

Question 8

In a sample of double-stranded DNA (dsDNA), the molar ratio of Adenosine is 20%. What is the molar content of Cytosine?

Accepted Answer

30%

Answer

10%

Answer

20%

Answer

40%

Question 9

Which of the following is the complimentary sequence of 5' TTAAGCTAC 3'?

Accepted Answer

5' GTACGCTTAA 3'

Answer

5' AATTCGCATG 3'

Answer

5' CATGCGAATT 3'

Answer

5' TTAAGCGTAC 3'

Question 10

Allantoin is the end product of metabolism of which of the following?

Accepted Answer

Purine

Answer

Glycogen

Answer

Pyrimidine

Answer

Histidine

Nucleic Acid Biochemistry — MCQs

Nucleic Acid Biochemistry — MCQs

On this page

Practice by Chapter

Want unlimited practice?