Nucleic Acid Biochemistry — MCQs

Nucleic Acid Biochemistry Indian Medical PG Practice Questions and MCQs

Question 251: Which type of RNA is characterized by the presence of a 7-methylguanosine cap at its 5' end?

A. mRNA (Correct Answer)
B. snRNA (small nuclear RNA)
C. tRNA (transfer RNA)
D. rRNA (ribosomal RNA)

Explanation: ***mRNA*** - The **7-methylguanosine cap** at the 5' end of mRNA is crucial for **ribosome binding** during translation initiation in eukaryotes. - This cap also protects the mRNA from degradation by **exonucleases**, thus increasing its stability and half-life in the cytoplasm. *tRNA (transfer RNA)* - tRNA molecules are characterized by a conserved **cloverleaf secondary structure** and an **acceptor stem** for amino acid attachment, not a 7-methylguanosine cap. - Their primary function is to **carry specific amino acids** to the ribosome during protein synthesis. *rRNA (ribosomal RNA)* - rRNA molecules are the main structural and catalytic components of **ribosomes**, where protein synthesis occurs. - They do not possess a 7-methylguanosine cap; instead, they undergo extensive **post-transcriptional modifications** and folding to form functional ribosomal subunits. *snRNA (small nuclear RNA)* - **snRNA** molecules are key components of the **spliceosome**, which catalyzes the removal of introns from pre-mRNA. - Unlike mRNA, snRNA molecules have a **trimethylguanosine cap** (not 7-methylguanosine) and are primarily localized in the nucleus for RNA splicing functions.

Question 252: Simple viruses that can be crystallized, such as tobacco mosaic virus, have been found to be composed of what?

A. Nucleotides
B. Phospholipids
C. Nucleoproteins (Correct Answer)
D. Scleroproteins

Explanation: ***Nucleoproteins*** - Simple viruses like tobacco mosaic virus consist primarily of **nucleic acid** (RNA) encased in a protein coat. - This combination of nucleic acid and protein forms **nucleoproteins**, which are the fundamental structural components of these viruses. *Nucleotides* - **Nucleotides** are the building blocks of nucleic acids (DNA and RNA), but they are not the complete structural component of a virus. - While viruses contain nucleic acids made of nucleotides, the question asks about the overall composition of the virus particle. *Phospholipids* - **Phospholipids** are key components of cell membranes and some enveloped viruses, forming a lipid bilayer. - Simple viruses such as tobacco mosaic virus are **non-enveloped** and therefore lack a phospholipid membrane. *Scleroproteins* - **Scleroproteins** (also known as fibrous proteins) are structural proteins found in animal tissues, like **collagen** and **keratin**. - They are not a primary component of viruses; viral structural proteins are typically globular in nature, forming capsids.

Question 253: Two strands of DNA are joined by:

A. Glycosidic bond
B. Hydrogen bond (Correct Answer)
C. Covalent bond
D. Ionic bond

Explanation: ***Hydrogen bond*** - The two strands of DNA are held together by **hydrogen bonds** formed between complementary nitrogenous bases. - **Adenine (A)** pairs with **thymine (T)** via two hydrogen bonds, and **guanine (G)** pairs with **cytosine (C)** via three hydrogen bonds. *Glycosidic bond* - A **glycosidic bond** links a nitrogenous base to the deoxyribose sugar in a single nucleotide. - It is an intramolecular bond within one strand, not between the two strands. *Covalent bond* - **Covalent bonds**, specifically **phosphodiester bonds**, form the sugar-phosphate backbone of each individual DNA strand. - These bonds are strong and define the primary structure of a single polynucleotide chain. *Ionic bond* - **Ionic bonds** involve the electrostatic attraction between oppositely charged ions and are not the primary forces holding the two DNA strands together. - While ions (like Mg2+) can play roles in DNA structure and stability, they do not directly join the strands.

Question 254: During DNA replication, if the template strand has the sequence 5'-GATTACA-3', what is the sequence of the complementary strand?

A. 5'-GATTACA-3'
B. 3'-GATTACA-5'
C. 5'-ACATTAG-3'
D. 5'-TGTAATC-3' (Correct Answer)

Explanation: **5'-TGTAATC-3'** - DNA replication involves **base pairing rules**: **adenine (A)** pairs with **thymine (T)**, and **guanine (G)** pairs with **cytosine (C)**. - The complementary strand is synthesized in an **antiparallel direction**: if the template is 5'-GATTACA-3', the new strand will be 3'-CTAATGT-5'. When written in the conventional 5' to 3' direction, this becomes 5'-TGTAATC-3'. *5'-GATTACA-3'* - This sequence is identical to the template strand, which would only occur if the DNA were to replicate in a **non-complementary manner**, violating base pairing rules. - Direct duplication of the template sequence does not produce a complementary strand. *3'-GATTACA-5'* - This sequence is the **template sequence written in the antiparallel direction** but is not the complementary strand. - It fails to apply the correct base pairing rules (A with T, G with C). *5'-ACATTAG-3'* - This sequence incorrectly pairs the bases and does not maintain the **antiparallel orientation** correctly. - For example, the first base G in the template would pair with C, not A.

Question 255: Which of the following statements is true regarding DNA polymerases?

A. Leading strand is synthesized by DNA polymerase I.
B. DNA polymerase I has significant repair activity. (Correct Answer)
C. Okazaki fragments are synthesized by DNA polymerase I.
D. Only DNA polymerase III has proofreading activity.

Explanation: ***DNA polymerase I has significant repair activity*** - DNA polymerase I plays a crucial role in **DNA repair**, including **excising RNA primers** and filling in the resulting gaps during replication. - Its **5' to 3' exonuclease activity** allows it to remove nucleotides ahead of its synthetic work, which is essential for correcting damage and removing primers. - This unique combination of activities makes DNA Pol I particularly important in **DNA repair mechanisms**. *Leading strand is synthesized by DNA polymerase I* - The **leading strand** in prokaryotes is primarily synthesized by **DNA polymerase III**, not DNA polymerase I. - DNA polymerase I is mainly involved in **primer removal** and filling gaps, not the bulk synthesis of new DNA strands. *Okazaki fragments are synthesized by DNA polymerase I* - **Okazaki fragments** are synthesized by **DNA polymerase III** in prokaryotes. - DNA polymerase I then replaces the RNA primers with DNA nucleotides within these fragments. *Only DNA polymerase III has proofreading activity* - This is **incorrect** - **DNA polymerases I, II, and III** all have **3' to 5' exonuclease proofreading activity**. - This proofreading function allows these enzymes to correct errors during DNA synthesis by removing incorrectly paired nucleotides.

Question 256: If a sequence of 4 nucleotides codes for 1 amino acid, how many amino acids can be theoretically formed?

A. 4
B. 64
C. 16
D. 256 (Correct Answer)

Explanation: ***256*** - With **4 distinct nucleotides** and a code sequence of **4 nucleotides** per amino acid, the number of possible unique combinations is calculated as 4^4. - This results in 4 × 4 × 4 × 4 = **256 theoretically possible amino acids**. - This is a mathematical combinatorics calculation: with 4 choices at each of 4 positions, total combinations = 4^4 = 256. *64* - This number represents the combinations if **3 nucleotides** coded for one amino acid (4^3 = 64), which is the actual case in the **standard genetic code** (triplet codons). - However, the question specifies a hypothetical sequence of **4 nucleotides** per amino acid, making this option incorrect. *16* - This number would be correct if **2 nucleotides** coded for one amino acid (4^2 = 16). - The problem explicitly states that **4 nucleotides** code for each amino acid in this theoretical scenario. *4* - This would only be the case if each **single nucleotide** coded for one amino acid (4^1 = 4). - Given **4 distinct nucleotides** and a sequence length of 4, the potential for combinations is much higher.

Question 257: What is meant by the melting of double-stranded DNA?

A. Splitting of double strands into single strands (Correct Answer)
B. Splitting of DNA into fragments
C. Formation of triple helix
D. Separation of double-stranded bases

Explanation: ***Splitting of double strands into single strands*** * **DNA melting**, also known as **DNA denaturation**, refers to the process where the two complementary strands of a **double-stranded DNA** molecule separate to form two individual single strands. * This process involves the breaking of the **hydrogen bonds** between the paired bases (**A-T and G-C**) due to increased temperature or changes in pH. * The temperature at which 50% of the DNA is denatured is called the **melting temperature (Tm)**, which depends on GC content (higher GC = higher Tm due to three hydrogen bonds vs. two in AT pairs). *Splitting of DNA into fragments* * The splitting of DNA into fragments is referred to as **DNA fragmentation**, which typically occurs due to processes like **restriction enzyme digestion**, mechanical shearing, or programmed cell death (apoptosis). * This process involves the breaking of the **phosphodiester bonds** within the DNA backbone, not just the hydrogen bonds between strands. *Formation of triple helix* * The formation of a **triple helix** (triplex DNA) is a less common DNA structure where a third oligonucleotide strand binds into the major groove of a **B-form DNA duplex**. * This process is distinct from DNA melting, which involves the *separation* of existing double strands rather than the *addition* of a third strand. *Separation of double-stranded bases* * The term "double-stranded bases" is imprecise terminology; bases are paired (e.g., A with T, G with C) within the double helix structure. * While the separation of base pairs does occur during melting, the more accurate description is the **separation of the entire double helix into two single strands**, not just the individual bases.

Question 258: Which of the following is not an example of a point mutation?

A. Nonsense mutation
B. Silent mutation
C. Frame-shift mutation (Correct Answer)
D. Missense mutation

Explanation: ***Frame-shift mutation*** - A **frame-shift mutation** involves the insertion or deletion of nucleotides not in multiples of three, altering the reading frame and typically leading to a completely different protein sequence or a premature stop codon. - While it can result from an insertion or deletion of a *single nucleotide*, its impact on the reading frame goes beyond a simple point alteration. *Silent mutation* - A **silent mutation** is a type of point mutation where a single nucleotide change in the DNA sequence does not change the amino acid sequence of the protein. - This is due to the **degeneracy of the genetic code**, where multiple codons can code for the same amino acid. *Nonsense mutation* - A **nonsense mutation** is a type of point mutation where a single nucleotide change results in a premature stop codon, leading to a truncated and often non-functional protein. - This significantly impacts protein synthesis by causing early termination of translation. *Missense mutation* - A **missense mutation** is a type of point mutation where a single nucleotide change results in a codon that codes for a different amino acid. - This can alter the protein's function, depending on the biochemical properties of the new amino acid and its location within the protein.

Question 259: Which component is part of the 50S ribosomal subunit?

A. 5.8S
B. 28S
C. 23S (Correct Answer)
D. 25S

Explanation: ***23S*** - The **23S ribosomal RNA** is a key structural and catalytic component of the **50S ribosomal subunit** in prokaryotes. - It forms the **peptidyl transferase center**, responsible for catalyzing the formation of peptide bonds during protein synthesis. *28S* - The **28S ribosomal RNA** is a component of the **large ribosomal subunit (60S)** in eukaryotes. - It is crucial for the structural integrity and catalytic activity of the eukaryotic ribosome. *5.8S* - The **5.8S ribosomal RNA** is another component of the **large ribosomal subunit (60S)** in eukaryotes. - It helps in the **assembly and stability** of the eukaryotic ribosomal complex. *25S* - The **25S ribosomal RNA** is found in the **large subunit (60S)** of ribosomes in **plants and some lower eukaryotes**. - It is functionally analogous to the 28S rRNA found in other eukaryotes.

Question 260: What is the function of the UGA codon?

A. Initiates transcription
B. Translates
C. Terminates protein synthesis (Correct Answer)
D. None of the options

Explanation: ***Terminates protein synthesis*** - The **UGA codon** is one of the three **stop codons** (UAA, UAG, UGA) that signal the termination of translation. - When a ribosome encounters a UGA codon, there is no corresponding **tRNA** with an anticodon, leading to the binding of release factors and dissociation of the ribosomal complex. *Initiates transcription* - **Transcription initiation** involves RNA polymerase binding to a promoter region, which is a DNA sequence, not a specific mRNA codon. - The UGA codon is part of an mRNA sequence and functions during translation. *Translates* - While translation is the process of synthesizing protein from an mRNA template, the **UGA codon** specifically acts as a signal to **stop** this process. - It does not directly code for an amino acid, unlike other codons that are "translated" into specific amino acids. *None of the options* - This option is incorrect because **UGA** has a very specific and critical function in **terminating protein synthesis**.

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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