Nucleic Acid Biochemistry — MCQs

Nucleic Acid Biochemistry Indian Medical PG Practice Questions and MCQs

Question 211: What is the end product of purine metabolism in humans?

A. Uric acid (Correct Answer)
B. Carbon Dioxide
C. Allantoin
D. None of the options

Explanation: ***Uric acid*** - **Uric acid** is the final breakdown product of **purine metabolism** in humans. - It is formed from the degradation of **adenosine** and **guanosine**, with xanthine oxidase playing a key role in its synthesis. *Allantoin* - **Allantoin** is the end product of **purine metabolism** in most mammals other than primates, as they possess the enzyme **uricase** to further break down uric acid. - Humans lack **uricase**, hence allantoin is not the end product in humans. *Carbon Dioxide* - **Carbon dioxide** is a major end product of **carbohydrate** and **fat metabolism** through cellular respiration. - It is not directly associated with the degradation pathway of purines. *None of the options* - This option is incorrect because **uric acid** is indeed the definitive end product of purine metabolism in humans.

Question 212: Which of the following organs does not primarily utilize the salvage pathway of purine nucleotide synthesis?

A. RBC
B. Leukocytes
C. Liver (Correct Answer)
D. Brain

Explanation: ***Liver*** - The **liver** is capable of both *de novo* synthesis and the salvage pathway of purine nucleotides, but it primarily utilizes the **de novo pathway** due to its high metabolic capacity and central role in biosynthesis for the entire body. - While salvage pathways exist, the liver's robust *de novo* synthesis allows it to readily produce purines from simple precursors, making it less reliant on salvaging pre-formed bases. *Brain* - The **brain** relies heavily on the **salvage pathway** for purine nucleotide synthesis because it has a limited capacity for *de novo* purine synthesis. - This dependency makes the brain particularly vulnerable to deficiencies in salvage enzymes, such as in **Lesch-Nyhan syndrome** where HGPRT deficiency leads to severe neurological dysfunction. *RBC* - **Red blood cells (RBCs)** are anucleated and lack the machinery for *de novo* purine synthesis, making them entirely dependent on the **salvage pathway** to maintain their purine nucleotide pool. - They salvage pre-formed purine bases and nucleosides from the plasma to synthesize necessary adenine and guanine nucleotides. *Leukocytes* - **Leukocytes**, particularly lymphocytes, have a high turn-over rate and metabolic activity, and they primarily rely on the **salvage pathway** for purine nucleotide synthesis. - The **immune system's rapid proliferation** and response demand efficient nucleotide synthesis, and the salvage pathway offers a quick and energy-efficient way to achieve this.

Question 213: Which enzyme polymerises Okazaki fragments?

A. DNA polymerase I
B. DNA polymerase II
C. DNA polymerase III (Correct Answer)
D. RNA polymerase

Explanation: ***DNA polymerase III*** - **DNA polymerase III** is the primary replicative enzyme in **prokaryotes (bacteria)** responsible for synthesizing new DNA strands, including the **polymerization of Okazaki fragments** on the lagging strand. - It possesses high processivity (can add ~500 nucleotides without dissociating), essential for rapid and efficient DNA synthesis during replication, adding nucleotides in a **5' to 3' direction**. - In **eukaryotes**, DNA polymerase δ (delta) performs the analogous function of polymerizing Okazaki fragments. *DNA polymerase I* - **DNA polymerase I** in prokaryotes primarily functions in **removing RNA primers** left by primase and **filling the resulting gaps** with DNA nucleotides. - It has 5' to 3' exonuclease activity for primer removal and polymerase activity for gap filling, but is **not the main enzyme for elongating Okazaki fragments**. - Its role is in **DNA repair and finishing replication**, not the extensive synthesis of Okazaki fragments. *DNA polymerase II* - **DNA polymerase II** in prokaryotes is primarily involved in **DNA repair mechanisms**, particularly in **restarting stalled replication forks** and responding to DNA damage. - It is not the main enzyme responsible for the polymerization of **Okazaki fragments** during normal DNA replication. *RNA polymerase* - **RNA polymerase** (specifically **primase**, a specialized RNA polymerase) synthesizes short **RNA primers** (8-12 nucleotides) during DNA replication, which provide the 3'-OH group necessary to initiate DNA synthesis. - It does not synthesize DNA or polymerize **Okazaki fragments**; its function is to create RNA primers, not extend DNA strands.

Question 214: Rate limiting step in pyrimidine synthesis?

A. Aspartate transcarbamoylase (ATCase)
B. Dihydroorotate dehydrogenase
C. Dihydro-orotase
D. Carbamoyl phosphate synthase-II (Correct Answer)

Explanation: ***Carbamoyl phosphate synthetase II (CPS-II)*** - **CPS-II** is the **committed and rate-limiting enzyme** in **de novo pyrimidine synthesis** in **mammals (including humans)** - It catalyzes the formation of **carbamoyl phosphate** from glutamine, CO₂, and 2 ATP in the **cytoplasm** - This is the **first committed step** and the main **regulatory checkpoint**, inhibited by UTP (feedback inhibition) and activated by PRPP and ATP - CPS-II is part of the **CAD complex** (carbamoyl phosphate synthetase, aspartate transcarbamoylase, dihydroorotase) in mammals *Aspartate transcarbamoylase (ATCase)* - ATCase catalyzes the **second step**: condensation of carbamoyl phosphate with aspartate to form carbamoyl aspartate - While ATCase is the **rate-limiting step in bacteria** (E. coli), in **mammals** it is part of the CAD complex and **not the primary regulatory step** - This option is incorrect for human/mammalian biochemistry tested in NEET PG *Dihydro-orotase* - The **third enzyme** in the pathway, converting carbamoyl aspartate to dihydroorotate - Part of the CAD complex in mammals but **not the rate-limiting step** *Dihydroorotate dehydrogenase* - Catalyzes the **fourth step**: oxidation of dihydroorotate to orotate - Located on the **outer surface of the inner mitochondrial membrane** (only mitochondrial enzyme in the pathway) - Important enzyme but **not rate-limiting**

Question 215: Which of the following molecular interactions are found in the structure of DNA?

A. Hydrogen bond
B. Glycosidic bond
C. Covalent interactions
D. All of the options (Correct Answer)

Explanation: ***All of the options*** - All three types of molecular interactions listed are present in DNA structure, making this the correct answer. - **Hydrogen bonds** hold together the two strands of the DNA double helix, forming between complementary base pairs (A-T with 2 hydrogen bonds, G-C with 3 hydrogen bonds). - **Glycosidic bonds** (N-glycosidic bonds) link the nitrogenous bases to the C1' carbon of the deoxyribose sugar in each nucleotide. - **Covalent interactions** (phosphodiester bonds) form the strong, stable sugar-phosphate backbone by linking the 3' hydroxyl group of one sugar to the 5' phosphate group of the next. *Hydrogen bond* - This is a **true statement** - hydrogen bonds are essential structural components of DNA. - However, this option alone is **incomplete** as DNA structure also contains glycosidic bonds and covalent phosphodiester bonds. - If only hydrogen bonds were present, there would be no nucleotides or backbone structure. *Glycosidic bond* - This is a **true statement** - glycosidic bonds are present in every nucleotide of DNA. - However, this option alone is **incomplete** as DNA also requires hydrogen bonds for base pairing and phosphodiester bonds for the backbone. - Without other bonds, individual nucleotides could not form a functional double helix. *Covalent interactions* - This is a **true statement** - covalent phosphodiester bonds form the DNA backbone within each strand. - However, this option alone is **incomplete** as it doesn't account for glycosidic bonds (nucleotide formation) or hydrogen bonds (strand pairing). - While the strongest bonds in DNA, they alone cannot create the complete double helix structure.

Question 216: In eukaryotic cells, where does the majority of functional RNA activity occur?

A. Nucleus
B. Ribosome
C. Cytoplasm (Correct Answer)
D. None of the options

Explanation: ***Cytoplasm*** - The **cytoplasm** is the cellular compartment where the **majority of functional RNA activity** occurs, including **translation** (protein synthesis) involving mRNA, tRNA, and rRNA. - **Ribosomes** (the sites of translation) are located in the cytoplasm, either free-floating or bound to the endoplasmic reticulum. - Many types of **regulatory RNAs** such as microRNAs (miRNAs) and small interfering RNAs (siRNAs) exert their functions in the cytoplasm by targeting mRNAs for degradation or translational repression. - **mRNA degradation** and **RNA interference pathways** primarily operate in the cytoplasm. - The question asks for the broader location rather than the specific molecular machinery, making cytoplasm the most comprehensive answer. *Nucleus* - While RNA is **transcribed** from DNA and **processed** (capping, polyadenylation, splicing) in the nucleus, these are preparatory steps. - The nucleus is primarily the site of **RNA synthesis**, not where most RNA performs its functional roles. - Only a small fraction of functional RNA activity (like rRNA processing in the nucleolus) occurs here compared to the cytoplasm. *Ribosome* - While **ribosomes are the specific sites of translation** and are composed of rRNA and proteins, they represent molecular machinery rather than a cellular location. - Ribosomes themselves are located **within the cytoplasm**, making cytoplasm the more inclusive answer for where RNA activity occurs. - The question asks "where" in terms of cellular compartment, not which molecular complex. *None of the options* - This is incorrect as the cytoplasm is indeed the primary site where the majority of functional RNA activities occur in eukaryotic cells.

Question 217: Which type of RNA is primarily involved in gene silencing?

A. rRNA
B. tRNA
C. miRNA (Correct Answer)
D. mRNA

Explanation: ***miRNA*** - **miRNA** (microRNA) is a small non-coding RNA molecule that plays a crucial role in **post-transcriptional regulation of gene expression**. - It functions by binding to complementary messenger RNA (mRNA) molecules, leading to **mRNA degradation** or **inhibition of translation**, thereby silencing genes. - miRNA is the primary RNA type involved in **gene silencing** through the RNA interference (RNAi) pathway. *rRNA* - **rRNA** (ribosomal RNA) is a primary component of **ribosomes**, the cellular machinery responsible for protein synthesis. - Its main function is to **catalyze peptide bond formation** and provide structural integrity to the ribosome, not gene silencing. *tRNA* - **tRNA** (transfer RNA) is responsible for carrying specific **amino acids** to the ribosome during protein synthesis. - It acts as an adapter molecule, translating the **genetic code** in mRNA into an amino acid sequence. *mRNA* - **mRNA** (messenger RNA) carries genetic information from **DNA to ribosomes** for protein synthesis. - While mRNA can be targeted by gene silencing mechanisms (like miRNA), it is not the RNA type that performs the silencing function itself.

Question 218: A frameshift mutation does not affect the complete amino acid sequence if it occurs in multiples of what number?

A. 1
B. 2
C. 3 (Correct Answer)
D. None of the options

Explanation: ***3*** - A **frameshift mutation** occurs when nucleotides are inserted or deleted in a number not divisible by three, altering the **reading frame** of the codons. - If insertions or deletions occur in multiples of **three**, the reading frame is restored after the mutation, largely preserving the downstream amino acid sequence. *1* - An insertion or deletion of a single nucleotide (1) definitively causes a **frameshift mutation**. - This alters all subsequent **codons**, leading to a completely different amino acid sequence downstream from the mutation. *2* - An insertion or deletion of two nucleotides (2) also results in a **frameshift mutation**. - This change shifts the **reading frame**, leading to the production of an altered protein or a premature stop codon. *None of the options* - This option is incorrect because a specific number, **three**, can allow for a frameshift mutation to not affect the complete amino acid sequence. - Multiples of three maintain the original **reading frame** (although potentially adding or removing a specific amino acid), whereas other numbers guarantee a frameshift.

Question 219: What is attached to the 3' end of mRNA after transcription?

A. CCA
B. Intron
C. 7-methylguanosine
D. Poly-A tail (Correct Answer)

Explanation: ***Poly-A tail*** - A **poly-A tail**, consisting of multiple adenosine monophosphates, is added to the **3' end of mRNA** after transcription to protect it from degradation. - This modification aids in the **transport of mRNA from the nucleus to the cytoplasm** and in its translation. *CCA* - The **CCA sequence** is found at the **3' end of tRNA**, not mRNA, and is critical for amino acid attachment. - It is added post-transcriptionally to tRNA molecules by the enzyme **tRNA nucleotidyltransferase**. *Intron* - **Introns** are non-coding regions within a gene that are transcribed into mRNA but are subsequently removed during **RNA splicing**, not added to the 3' end. - Their removal ensures that only the **coding regions (exons)** are translated into protein. *7-methylguanosine* - **7-methylguanosine** forms the **5' cap** of mRNA, which is added to the 5' end, not the 3' end. - This cap is important for **mRNA stability**, ribosome binding, and protection against degradation.

Question 220: What does Chargaff's rule state regarding the base pairing in DNA?

A. A=T, G=C (Correct Answer)
B. A=G, T=C
C. A=C, G=T
D. Any combination possible

Explanation: ***A=T, G=C*** - **Chargaff's rules** state that in any double-stranded DNA, 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)**. - This equivalency reflects the specific **base pairing** in the DNA double helix, where A always pairs with T, and G always pairs with C. *A=G, T=C* - This statement is incorrect as it proposes an atypical and biologically inaccurate pairing between a **purine (A)** and another **purine (G)**, and a **pyrimidine (T)** with a **pyrimidine (C)**. - This combination would disrupt the uniform diameter of the DNA double helix required for its structural stability. *A=C, G=T* - This option is incorrect because it suggests pairing a purine (A) with a pyrimidine (C) and a purine (G) with a pyrimidine (T) in a way that is not observed in natural DNA. - Such pairings would also lead to an irregular width of the DNA molecule, destabilizing its structure. *Any combination possible* - This statement is false; base pairing in DNA is **highly specific** and not random due to chemical and structural constraints. - The specific pairing rules (**A with T, G with C**) are crucial for maintaining the consistent structure of the DNA double helix and for accurate DNA replication and transcription.

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

Want unlimited practice?

Get full access to all questions, explanations, and performance tracking.

Start For Free