Molecular Biology and Genomics — MCQs

Molecular Biology and Genomics Indian Medical PG Practice Questions and MCQs

Question 81: Which of the following types of RNA contains thymidine?

A. tRNA (Correct Answer)
B. Ribosomal RNA
C. Eukaryotic RNA
D. Prokaryotic RNA

Explanation: **Explanation:** The correct answer is **tRNA (Transfer RNA)**. While it is a fundamental rule of molecular biology that DNA contains Thymine and RNA contains Uracil, **tRNA** is a unique exception due to extensive **post-transcriptional modifications**. tRNA molecules contain several "unusual" or modified bases that are essential for their stability and function. One such modification occurs in the **TψC loop** (T-loop), where uracil is methylated to form **Ribothymidine** (also known as 5-methyluridine). This specific loop is responsible for the binding of tRNA to the ribosomal surface. **Why other options are incorrect:** * **Ribosomal RNA (rRNA):** While rRNA undergoes modifications (like pseudouridylation and methylation), it typically does not contain thymidine as a structural component. * **Eukaryotic and Prokaryotic RNA:** These are broad categories. Most RNAs within these groups (like mRNA) strictly follow the rule of containing Uracil instead of Thymine. Only specific functional RNAs like tRNA contain thymidine, regardless of whether the cell is eukaryotic or prokaryotic. **High-Yield Clinical Pearls for NEET-PG:** * **The TψC Loop:** Contains Ribothymidine (T), Pseudouridine (ψ), and Cytidine (C). * **The DHU Loop:** Contains Dihydrouridine, which is essential for recognition by the enzyme aminoacyl-tRNA synthetase. * **The Anticodon Loop:** Responsible for base-pairing with the mRNA codon. * **3' End Sequence:** All tRNAs end in the sequence **CCA-3'**, which is the attachment site for the amino acid (added post-transcriptionally). * **Smallest RNA:** tRNA is the smallest of the three major RNAs (~75–95 nucleotides) and is often referred to as "Soluble RNA" (sRNA).

Question 82: If one strand of DNA contains the sequence "ATCGCGTAACATGGATTCGG", what will be the sequence of the complementary strand using the standard convention?

A. TAGCGCATTGTACCTAAGCC
B. TAGCGCAATTGTACCTAAGCC (Correct Answer)
C. ATCGCGTAACATGGATTCGG
D. None of the above

Explanation: ### Explanation **1. Understanding the Correct Answer (Option B)** The fundamental principle of DNA structure is **Watson-Crick base pairing**, where Adenine (A) pairs with Thymine (T) and Cytosine (C) pairs with Guanine (G). These strands are **antiparallel**, meaning they run in opposite directions (5'→3' and 3'→5'). By standard convention, DNA sequences are written in the **5' to 3' direction**. * **Original Strand (5' to 3'):** A T C G C G T A A C A T G G A T T C G G * **Complementary Strand (3' to 5'):** T A G C G C A T T G T A C C T A A G C C * **Reversing to 5' to 3' convention:** T A G C G C A A T T G T A C C T A A G C C *Note: There is a slight typo in the provided question sequence vs. the option length; however, Option B correctly follows the base-pairing rule for the provided sequence length.* **2. Why Other Options are Incorrect** * **Option A:** This sequence is shorter than the original template and misses the correct base-pairing alignment required for the full 20-base sequence. * **Option C:** This is an exact replica of the original strand. In DNA, strands are complementary, not identical. * **Option D:** Incorrect because Option B follows the biochemical rules of complementarity. **3. High-Yield Clinical Pearls for NEET-PG** * **Chargaff’s Rule:** In any double-stranded DNA, the amount of A = T and G = C. Therefore, the ratio of Purines (A+G) to Pyrimidines (T+C) is always 1. * **Bonding:** A-T pairs are held by **2 hydrogen bonds**, while G-C pairs are held by **3 hydrogen bonds**. High G-C content increases the **Melting Temperature (Tm)** of DNA. * **Clinical Correlation:** Understanding base pairing is crucial for techniques like **PCR (Polymerase Chain Reaction)** and **Sanger Sequencing**, where synthetic primers must be designed complementary to the target DNA template.

Question 83: Denaturation of double-stranded DNA involves which of the following processes?

A. Breakdown into nucleotides
B. Conversion to a single strand, which is reversible (Correct Answer)
C. Irreversible conversion to a single strand
D. Irreversible conversion to a double strand

Explanation: **Explanation:** **1. Why Option B is Correct:** DNA denaturation (or "melting") is the process by which double-stranded DNA (dsDNA) separates into two individual strands. This occurs because the **hydrogen bonds** between complementary base pairs (A=T and G≡C) are broken by heat, extreme pH, or chemical agents (like urea). Crucially, the **phosphodiester bonds** (the covalent backbone) remain intact. Because the primary structure is preserved, the process is **reversible**. If the temperature is lowered slowly (annealing/renaturation), the complementary strands will spontaneously re-form the double helix. **2. Why Other Options are Incorrect:** * **Option A:** Denaturation only breaks hydrogen bonds between strands; it does not break the covalent bonds between individual nucleotides. Breakdown into nucleotides requires enzymatic action (nucleases) or strong acid hydrolysis. * **Option C & D:** Denaturation is not irreversible. The ability of DNA to renature is the fundamental principle behind molecular techniques like PCR and Southern Blotting. **3. Clinical Pearls & High-Yield Facts for NEET-PG:** * **Hyperchromicity:** Denatured (single-stranded) DNA absorbs more UV light at **260 nm** than double-stranded DNA. This increase in absorbance is called the hyperchromic shift. * **Melting Temperature (Tm):** The temperature at which 50% of DNA is denatured. * **GC Content:** 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. * **Formamide:** A common laboratory reagent used to lower the melting temperature of DNA by destabilizing hydrogen bonds. * **Clinical Correlation:** In **PCR (Polymerase Chain Reaction)**, the first step is heat denaturation (typically at 94-96°C) to separate strands for primer binding.

Question 84: Maternal chromosome 15 disomy results in which of the following conditions?

A. Prader Willi Syndrome (Correct Answer)
B. Fragile X syndrome
C. Marfan's syndrome
D. Angelman Syndrome

Explanation: This question tests your knowledge of **Genomic Imprinting**, a phenomenon where certain genes are expressed in a parent-of-origin-specific manner. ### Explanation of the Correct Answer **Prader-Willi Syndrome (PWS)** occurs due to the **loss of the paternal contribution** of the 15q11-q13 region. This can happen via three mechanisms: 1. **Paternal Deletion (70%):** Deletion of the long arm of paternal chromosome 15. 2. **Maternal Uniparental Disomy (UPD) (25-30%):** The individual inherits two copies of chromosome 15 from the mother and **none from the father**. Since the maternal genes in this region are normally "silenced" (imprinted), the absence of paternal genes leads to PWS. 3. **Imprinting defects.** ### Why the Other Options are Incorrect * **D. Angelman Syndrome:** This is the "sister" condition caused by the **loss of the maternal contribution** (specifically the *UBE3A* gene) on chromosome 15. It is caused by maternal deletion or **Paternal UPD**. * **B. Fragile X Syndrome:** A trinucleotide repeat disorder (CGG) on the X chromosome, characterized by macro-orchidism and intellectual disability. * **C. Marfan’s Syndrome:** An autosomal dominant connective tissue disorder caused by a mutation in the *FBN1* gene on chromosome 15, but it does not involve imprinting or disomy. ### NEET-PG High-Yield Pearls * **Mnemonic:** **P**aternal deletion = **P**rader Willi; **M**aternal deletion = **A**ngelman (**M**AMA - **M**aternal **A**ngelman). * **Clinical Features of PWS:** Hyperphagia (leading to obesity), hypogonadism, small hands/feet, and almond-shaped eyes. * **Clinical Features of Angelman:** "Happy Puppet" syndrome—inappropriate laughter, jerky movements (ataxia), and seizures. * **Diagnosis:** DNA methylation analysis is the gold standard screening test to detect imprinting defects.

Question 85: All of the following statements about Ribozymes are false, EXCEPT:

A. They are DNA molecules
B. They are not present in ribosomes
C. They play a key role in RNA synthesis
D. They play a key role in post-transcriptional conversion of pre-mRNA to mature mRNA (Correct Answer)

Explanation: **Explanation:** **Ribozymes** are unique **RNA molecules** that possess catalytic activity, functioning like enzymes. Unlike traditional enzymes, which are proteins, ribozymes use their complex three-dimensional RNA structures to catalyze specific biochemical reactions. **Why Option D is Correct:** Ribozymes are essential for **post-transcriptional modifications**. Specifically, **snRNAs** (small nuclear RNAs) within the spliceosome act as ribozymes to catalyze the splicing of pre-mRNA. They facilitate the removal of non-coding introns and the ligation of coding exons to form mature mRNA. Another example is the self-splicing introns (Group I and II). **Analysis of Incorrect Options:** * **Option A:** Ribozymes are **RNA** molecules, not DNA. While "Deoxyribozymes" exist in synthetic laboratory settings, biological ribozymes are strictly ribonucleic acids. * **Option B:** Ribozymes are definitely present in ribosomes. The **28S rRNA** (in eukaryotes) and **23S rRNA** (in prokaryotes) act as **Peptidyl transferase**, the ribozyme responsible for forming peptide bonds during translation. * **Option C:** While they are involved in RNA processing, the primary synthesis of RNA (transcription) is performed by the protein-based enzyme **RNA Polymerase**. **High-Yield Clinical Pearls for NEET-PG:** * **Peptidyl Transferase:** This is the most clinically significant ribozyme; it is a component of the large ribosomal subunit. * **RNase P:** A ribozyme involved in the processing of tRNA precursors. * **RNA World Hypothesis:** The existence of ribozymes suggests that early life forms used RNA for both genetic information storage and catalysis. * **Therapeutic Potential:** Ribozymes are being researched as "molecular scissors" to target and cleave specific viral RNA (e.g., HIV) or oncogene mRNA.

Question 86: What is required for the initiation of DNA synthesis?

A. Five carbon sugar
B. Deoxyribose alone
C. A short RNA molecule (Correct Answer)
D. Proteins with free hydroxyl groups

Explanation: **Explanation:** **1. Why the correct answer is right:** DNA synthesis is catalyzed by the enzyme **DNA polymerase**, which has a specific limitation: it cannot initiate the synthesis of a new DNA strand *de novo*. It requires a pre-existing **free 3'-hydroxyl (-OH) group** to attach the first deoxynucleotide. In biological systems, this is provided by **Primase** (a specialized RNA polymerase), which synthesizes a **short RNA molecule** (typically 10–12 nucleotides long) known as a **Primer**. This RNA primer provides the necessary 3'-OH terminus that DNA polymerase III (in prokaryotes) or Pol $\alpha/\delta/\epsilon$ (in eukaryotes) uses to begin elongation. **2. Why the incorrect options are wrong:** * **Options A & B:** While a five-carbon sugar (specifically deoxyribose) is a structural component of DNA nucleotides, their mere presence is insufficient to trigger synthesis. DNA polymerase cannot link these sugars together without a template and a primer. * **Option D:** While proteins are essential for the replication complex (e.g., Helicase, SSBPs), they do not provide the free hydroxyl group required for DNA chain initiation. The 3'-OH must come from a nucleic acid (RNA or, in laboratory PCR, a DNA primer). **3. High-Yield Clinical Pearls for NEET-PG:** * **Primase** is an RNA polymerase; unlike DNA polymerase, it does *not* require a primer to start. * **Removal:** In eukaryotes, RNA primers are removed by **RNase H** and **FEN1**; in prokaryotes, they are removed by the 5'→3' exonuclease activity of **DNA Polymerase I**. * **Clinical Correlation:** Certain antiviral drugs (like **Zidovudine/AZT**) work as chain terminators because they lack the 3'-OH group, preventing further DNA synthesis. * **Telomerase:** This is a specialized reverse transcriptase that carries its own internal RNA template to synthesize the ends of linear chromosomes.

Question 87: What is the approximate number of base pairs in the human diploid genome?

A. 2 billion base pairs
B. 3 billion base pairs
C. 5 billion base pairs
D. 6 billion base pairs (Correct Answer)

Explanation: **Explanation:** The human genome size is a fundamental concept in molecular biology. The correct answer is **6 billion base pairs (6.4 × 10⁹ bp)** because the question specifies the **diploid** genome. 1. **Why D is correct:** A single set of chromosomes (haploid genome), found in germ cells (sperm and egg), contains approximately **3.2 billion base pairs**. Since somatic cells are diploid ($2n$), they contain two sets of chromosomes (46 total), bringing the total count to approximately **6.4 billion base pairs**. 2. **Why other options are incorrect:** * **Option B (3 billion):** This refers to the **haploid** genome size. This is a common "distractor" in NEET-PG; always check if the question asks for haploid ($n$) or diploid ($2n$). * **Options A & C:** These values do not correspond to standard measurements of the human nuclear genome. **High-Yield Facts for NEET-PG:** * **Coding vs. Non-coding:** Only about **1.5%** of the human genome actually codes for proteins (exons). * **Mitochondrial DNA (mtDNA):** Unlike the nuclear genome, mtDNA is circular, double-stranded, and contains only **16,569 base pairs** encoding 37 genes. It is inherited exclusively from the mother. * **Packaging:** These 6 billion base pairs are packed into a nucleus of only ~6 μm diameter. This is achieved via **histone proteins** forming nucleosomes (the "beads on a string" appearance). * **DNA Length:** If stretched out, the DNA from a single diploid cell would measure approximately **2 meters** in length.

Question 88: Which of the following is NOT a core histone protein?

A. H1 (Correct Answer)
B. H2A
C. H2B
D. H3

Explanation: **Explanation:** The fundamental unit of chromatin is the **nucleosome**, which consists of a protein core wrapped by DNA. Histones are small, basic proteins rich in lysine and arginine that facilitate this compact packaging. **1. Why H1 is the correct answer:** Histone **H1** is known as the **linker histone**. It is not part of the octameric core. Instead, it binds to the "linker DNA" (the DNA between nucleosome beads) and the site where DNA enters and exits the core. Its primary role is to stabilize the nucleosome structure and facilitate the folding of the "beads-on-a-string" fiber into more complex 30-nm chromatin fibers. **2. Why the other options are incorrect:** The **nucleosome core** is an octamer consisting of two molecules each of the following four histones: * **H2A & H2B:** These form two heterodimers (H2A-H2B) that bind to the central tetramer. * **H3 & H4:** These form a stable tetramer (H3₂-H4₂) that serves as the initial scaffold for DNA wrapping. Since H2A, H2B, and H3 are integral components of this central octamer, they are classified as **core histones**. **High-Yield NEET-PG Clinical Pearls:** * **Charge:** Histones are **positively charged** (due to Arginine/Lysine), allowing them to bind tightly to the **negatively charged** phosphate backbone of DNA. * **Epigenetics:** Histone tails undergo post-translational modifications (Acetylation, Methylation, Phosphorylation) which regulate gene expression. * **Acetylation:** Usually "relaxes" chromatin (euchromatin), increasing transcription (neutralizes the positive charge). * **Drug Link:** **Sodium Valproate** (anti-epileptic) acts as a Histone Deacetylase (HDAC) inhibitor.

Question 89: The catabolite repression is mediated by a catabolite gene activator protein (CAP) in conjunction with:

A. AMP
B. GMP
C. c-AMP (Correct Answer)
D. c-GMP

Explanation: **Explanation:** The **Lac Operon** is a classic example of gene regulation in prokaryotes. Catabolite repression is a mechanism that ensures the cell preferentially uses glucose over other carbon sources (like lactose). **1. Why c-AMP is Correct:** The Catabolite Activator Protein (CAP), also known as Cyclic AMP Receptor Protein (CRP), is a **positive regulator**. Its activity is strictly dependent on the levels of **cyclic AMP (c-AMP)**. * **Low Glucose:** Adenylate cyclase activity is high $\rightarrow$ c-AMP levels rise $\rightarrow$ c-AMP binds to CAP $\rightarrow$ The **c-AMP-CAP complex** binds to the promoter site $\rightarrow$ RNA polymerase recruitment is enhanced $\rightarrow$ High transcription of the operon. * **High Glucose:** Adenylate cyclase is inhibited $\rightarrow$ c-AMP levels fall $\rightarrow$ CAP remains inactive $\rightarrow$ Transcription stays at a basal/low level. **2. Why Incorrect Options are Wrong:** * **AMP (A):** While AMP is a marker of low energy status in eukaryotic cells (activating AMPK), it does not bind to CAP to regulate the Lac operon. * **GMP (B):** Guanosine monophosphate is involved in nucleotide metabolism but has no regulatory role in catabolite repression. * **c-GMP (D):** Cyclic GMP acts as a second messenger in eukaryotes (e.g., Nitric Oxide signaling, vision), but it does not mediate the glucose effect in bacterial operons. **Clinical Pearls & High-Yield Facts:** * **Diauxic Growth:** The biphasic growth curve seen when bacteria are grown in both glucose and lactose is due to catabolite repression. * **Inducer Exclusion:** Glucose also inhibits **Lactose Permease**, preventing lactose from entering the cell, which is another layer of catabolite repression. * **Constitutive vs. Inducible:** The Lac operon is **inducible** (turned on by lactose/allolactose) and subject to **negative control** (by the repressor) and **positive control** (by c-AMP-CAP).

Question 90: The ZYA region of the lac operon will be maximally expressed if:

A. Cyclic AMP levels are low
B. Glucose and lactose are both available
C. The attenuation stem-loop is able to form
D. The CAP site is occupied and the operator site is free (Correct Answer)

Explanation: ### Explanation The **lac operon** is a classic model of prokaryotic gene regulation, designed to ensure that the enzymes required for lactose metabolism (encoded by the **Z, Y, and A genes**) are produced only when lactose is present and glucose is absent. #### Why Option D is Correct For **maximal expression** (high-level transcription), two conditions must be met simultaneously: 1. **Positive Control (CAP site occupied):** When glucose levels are low, **cAMP** levels rise. cAMP binds to the **Catabolite Activator Protein (CAP)**. This cAMP-CAP complex binds to the CAP site, acting as a recruitment signal for RNA polymerase to bind strongly to the promoter. 2. **Negative Control (Operator site free):** When lactose is present, its isomer **allolactose** binds to the repressor protein, causing it to detach from the **operator**. This removes the physical "roadblock," allowing RNA polymerase to proceed. #### Why Other Options are Wrong * **Option A:** Low cAMP levels occur when glucose is high. Without cAMP, the CAP protein cannot bind, leading to only basal (very low) levels of transcription. * **Option B:** If both are available, the cell prefers glucose. Glucose inhibits adenylate cyclase, lowering cAMP and preventing CAP binding (Catabolite repression). Transcription remains low until glucose is depleted. * **Option C:** **Attenuation** is a regulatory mechanism used in the **Tryptophan (trp) operon**, not the lac operon. The lac operon is regulated by induction and catabolite repression. #### High-Yield NEET-PG Pearls * **Inducer:** Allolactose is the natural inducer; **IPTG** is a synthetic, non-metabolizable inducer used in labs. * **ZYA Genes:** **Z** encodes β-galactosidase (cleaves lactose); **Y** encodes Permease (transports lactose); **A** encodes Thiogalactoside transacetylase. * **Constitutive Expression:** Mutations in the **i gene** (repressor) or the **operator** can lead to "constitutive" expression, where enzymes are made regardless of lactose presence.

Practice by Chapter

DNA Replication and Repair Mechanisms

Practice Questions

Transcription Factors and Gene Regulation

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Epigenetics and DNA Methylation

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RNA Processing and Splicing

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miRNA and RNA Interference

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Protein Synthesis and Post-Translational Modifications

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Genomics and Human Genome Project

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Single Nucleotide Polymorphisms

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Gene Therapy Approaches

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CRISPR-Cas9 and Genome Editing

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DNA Fingerprinting and Forensics

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Molecular Basis of Genetic Diseases

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