Molecular Biology and Genomics Practice Questions

Q: The nucleotide triplet CTC in the sixth position of the beta-globin chain in DNA normally forms a complementary nucleotide on mRNA that codes for glutamic acid. A point mutation in the beta-globin chain resulting in the nucleotide triplet CAC forms a complementary nucleotide on mRNA that codes for valine. In sickle cell anemia, what complementary nucleotide triplet on mRNA, read from 5' to 3', would be expected?

GUG. **Explanation:** The question describes the molecular basis of **Sickle Cell Anemia**, a classic example of a **missense point mutation**. 1. **Why D is correct:** * In the normal $\beta$-globin gene, the DNA template strand triplet is **CTC**, which transcribes into the mRNA codon **GAG** (coding for Glutamic acid). * In Sickle Cell Anemia, a transversion mutation occurs where Adenine replaces Thymine in the DNA template, changing the triplet from CTC to **CAC**. * During transcription, mRNA is synthesized complementary to the DNA template. The DNA base **C** pairs with **G**, and **A** pairs with **U** (in RNA). Therefore, the DNA triplet **5'-CAC-3'** results in the mRNA codon **5'-GUG-3'**. * The codon **GUG** codes for **Valine**, replacing the polar Glutamic acid with a non-polar amino acid at the 6th position. 2. **Analysis of Incorrect Options:** * **A (GAG):** This is the normal mRNA codon for Glutamic acid. * **B (CTC):** This is the DNA sequence, not the mRNA sequence. mRNA contains Uracil (U) instead of Thymine (T). * **C (GTG):** This represents the mutated DNA sequence on the coding (non-template) strand. mRNA would not contain Thymine. **Clinical Pearls for NEET-PG:** * **Mutation Type:** Transversion (Point mutation). Specifically, $A \to T$ on the coding strand ($T \to A$ on the template strand). * **Amino Acid Change:** Glutamate (Acidic/Polar) $\to$ Valine (Neutral/Non-polar). * **Consequence:** Deoxygenated HbS polymerizes, leading to "sickling," hemolysis, and vaso-occlusive crises. * **Electrophoresis:** HbS moves slower than HbA toward the anode because it loses two negative charges (one per $\beta$-chain).

Q: How many high-energy phosphate bonds are required for the formation of one peptide bond?

4. **Explanation:** Protein synthesis (translation) is an energetically expensive process. To form a single peptide bond, **4 high-energy phosphate bonds** are consumed through the following steps: 1. **Amino Acid Activation (2 ATP equivalents):** The enzyme aminoacyl-tRNA synthetase attaches an amino acid to its specific tRNA. This reaction converts **ATP to AMP and inorganic pyrophosphate (PPi)**. Since ATP → AMP is energetically equivalent to breaking two phosphoanhydride bonds (ATP → ADP → AMP), this counts as **2 high-energy bonds**. 2. **Initiation/Elongation (Binding) (1 GTP):** The binding of the incoming aminoacyl-tRNA to the 'A' site of the ribosome requires the hydrolysis of **1 GTP** (mediated by Elongation Factor Tu/EF-1). 3. **Translocation (1 GTP):** After the peptide bond is formed by peptidyl transferase, the ribosome moves one codon along the mRNA. This translocation step requires the hydrolysis of **1 GTP** (mediated by Elongation Factor G/EF-2). **Analysis of Options:** * **A, B, & C are incorrect** because they account for only partial steps of the translation cycle. While the actual formation of the peptide bond by peptidyl transferase itself does not require additional ATP/GTP, the "cost" of making that bond possible includes the activation, binding, and translocation steps mentioned above. **High-Yield Clinical Pearls for NEET-PG:** * **Peptidyl Transferase:** This is a **ribozyme** (23S rRNA in prokaryotes, 28S rRNA in eukaryotes); it does not require ATP/GTP to catalyze the bond formation. * **Diphtheria Toxin & Pseudomonas Exotoxin:** Both inhibit protein synthesis by ADP-ribosylation of **EF-2**, blocking the translocation step (the 4th high-energy bond usage). * **Total Energy Cost:** To synthesize a protein of *'n'* amino acids, the total energy required is **4n - 1** (as the first amino acid does not require translocation).

Q: A change in the second base of the serine codon UCA may result in the codon UAA. This is an example of which type of mutation?

Nonsense. ### Explanation **1. Why Nonsense is Correct:** A **nonsense mutation** occurs when a single nucleotide substitution results in a premature **stop codon** (UAA, UAG, or UGA). In this scenario, the codon **UCA** (which codes for the amino acid Serine) undergoes a point mutation at the second base (C → A), transforming it into **UAA**. Since UAA is a termination signal, translation stops prematurely, leading to a truncated, usually non-functional protein. **2. Why Other Options are Incorrect:** * **Missense:** This occurs when a point mutation changes one codon to another that codes for a *different* amino acid (e.g., UCA to UUA/Leucine). It does not stop translation. * **Silent:** This is a "synonymous" mutation where the nucleotide change results in a codon that codes for the *same* amino acid (e.g., UCA to UCG, both coding for Serine), usually due to the degeneracy of the genetic code. * **Frameshift:** This occurs due to the **insertion or deletion** of nucleotides (not divisible by three), which shifts the entire reading frame downstream. The question describes a single base substitution, not an indel. **3. NEET-PG Clinical Pearls:** * **Stop Codons:** Remember them with the mnemonic: **U** **A**re **A**way (**UAA**), **U** **A**re **G**one (**UAG**), **U** **G**o **A**way (**UGA**). * **Transition vs. Transversion:** Changing C (pyrimidine) to A (purine) as seen here is a **transversion**. * **Clinical Example:** Nonsense mutations are a common cause of **β-thalassemia major** (e.g., CAG → UAG at codon 39). * **Read-through therapy:** Drugs like **Aminoglycosides** (Gentamicin) or **Ataluren** can sometimes induce "ribosomal oversight" to bypass premature stop codons in genetic diseases.

Q: Random inactivation of the X chromosome is known as:

Lyonization. ### Explanation **1. Why Lyonization is Correct:** Lyonization (named after Mary Lyon) refers to the process in female mammals where one of the two X chromosomes is randomly inactivated during early embryonic development. This occurs to achieve **dosage compensation**, ensuring that females (XX) do not produce double the amount of X-linked gene products compared to males (XY). The inactivated X chromosome condenses into a heterochromatic structure known as a **Barr body**, visible at the periphery of the nucleus. This process is mediated by the **XIST gene** (X-inactive specific transcript), which produces a non-coding RNA that "coats" the chromosome to silence it. **2. Analysis of Incorrect Options:** * **B. Allelic exclusion:** This is a process where only one allele of a gene is expressed while the other is silenced (commonly seen in B-lymphocytes for immunoglobulin gene expression), but it does not involve an entire chromosome. * **C. Randomization:** This is a general statistical term and not a specific biological mechanism for gene silencing. * **D. Genomic imprinting:** This involves the epigenetic silencing of specific genes based on their **parent of origin** (e.g., Prader-Willi and Angelman syndromes). Unlike Lyonization, which is random, imprinting is predictable and specific to certain loci. **3. High-Yield Clinical Pearls for NEET-PG:** * **Mosaicism:** Because Lyonization is random, females are "genetic mosaics." This explains why female carriers of X-linked recessive disorders (like Hemophilia or G6PD deficiency) may show varying degrees of clinical symptoms (skewed lyonization). * **Barr Body Formula:** Number of Barr bodies = (Total number of X chromosomes – 1). * Turner Syndrome (45, XO): 0 Barr bodies. * Klinefelter Syndrome (47, XXY): 1 Barr body. * **XIST Gene:** Located in the X-inactivation center (Xic) at **Xq13**. It is essential for the initiation of inactivation.

Q: Which enzyme forms the RNA primer in eukaryotes?

RNA polymerase. ### Explanation **Correct Answer: B. RNA polymerase** In eukaryotic DNA replication, DNA polymerase cannot initiate the synthesis of a new strand *de novo*; it requires a free 3'-OH group. This is provided by a short sequence of RNA known as a **primer**. The enzyme responsible for synthesizing this primer is **DNA-dependent RNA polymerase**, specifically a specialized form called **DNA Primase**. In eukaryotes, Primase exists in a complex with **DNA Polymerase α**. The Primase subunit synthesizes a short RNA stretch (approx. 10 nucleotides), which is then extended by Polymerase α with a short segment of DNA before the "polymerase switching" occurs to Polymerase δ or ε. **Why other options are incorrect:** * **A. Ligase:** Its role is to join DNA fragments (like Okazaki fragments) by catalyzing the formation of phosphodiester bonds. It does not synthesize new strands. * **C. Topoisomerase:** These enzymes (Type I and II) relieve torsional strain and supercoiling ahead of the replication fork by creating transient breaks in the DNA backbone. * **D. Helicase:** This enzyme (MCM complex in eukaryotes) uses ATP to unwind the DNA double helix into single strands; it has no synthetic activity. --- ### High-Yield Clinical Pearls for NEET-PG * **Polymerase Switching:** In eukaryotes, **Pol α** initiates (Primase activity), **Pol δ** synthesizes the lagging strand, and **Pol ε** synthesizes the leading strand. * **Prokaryotic Equivalent:** In *E. coli*, the primase is the **DnaG** protein. * **Directionality:** Primers are always synthesized in the **5' to 3' direction**. * **Drug Link:** Many antiviral and anticancer drugs (like Cytarabine) target DNA polymerases, inhibiting the elongation phase after the primer is formed.

Q: Which enzyme synthesizes the RNA primer?

DNA primase. **Explanation:** **1. Why DNA Primase is Correct:** DNA replication cannot be initiated *de novo* because DNA polymerases require a free **3'-OH group** to begin adding nucleotides. To overcome this, **DNA primase** (a specialized RNA polymerase) synthesizes a short sequence of RNA (approximately 10 nucleotides long) known as the **RNA primer**. This primer provides the necessary 3'-OH terminus that DNA Polymerase α (in eukaryotes) or DNA Polymerase III (in prokaryotes) uses to start DNA synthesis. **2. Why the Other Options are Incorrect:** * **Topoisomerase:** These enzymes (e.g., DNA Gyrase) relieve the **torsional strain** and supercoiling caused by the unwinding of the DNA double helix. * **Helicase:** This enzyme uses ATP to break hydrogen bonds between nitrogenous bases, effectively **unzipping** the double-stranded DNA at the replication fork. * **DNA Ligase:** This enzyme acts as "molecular glue." It catalyzes the formation of phosphodiester bonds to join **Okazaki fragments** on the lagging strand or to seal nicks during DNA repair. **3. High-Yield Clinical Pearls for NEET-PG:** * **Primosome:** This is a functional complex consisting of **DNA Helicase + DNA Primase**. * **Directionality:** RNA primers are synthesized in the **5' to 3' direction**. * **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**. * **Drug Target:** Fluoroquinolones (e.g., Ciprofloxacin) inhibit bacterial DNA Gyrase (Topoisomerase II) and Topoisomerase IV, preventing replication.

Q: Linker DNA is bound to which of the following histone?

H1. ### Explanation The fundamental unit of chromatin is the **nucleosome**, which consists of a protein core wrapped by DNA. **1. Why H1 is the Correct Answer:** Histone **H1** is known as the **linker histone**. Unlike the other histones, it is not part of the nucleosome core particle. Instead, it binds to the "linker DNA" (the segment of DNA between adjacent nucleosomes) 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" chromatin into more complex, higher-order structures like the **30-nm fiber**. **2. Why the Other Options are Incorrect:** * **H2A, H2B, H3, and H4** are known as **Core Histones**. * Two molecules of each (H2A, H2B, H3, and H4) combine to form an **octamer**. * Approximately 146 base pairs of DNA wrap around this octamer core to form the nucleosome core particle. These histones are involved in the internal structural integrity of the core, not the binding of linker DNA. **3. Clinical Pearls & High-Yield Facts for NEET-PG:** * **Charge:** Histones are highly **basic** proteins, rich in **Lysine and Arginine**, which allows them to bind strongly to the negatively charged phosphate backbone of DNA. * **Epigenetics:** Histone modification (Acetylation/Methylation) occurs primarily at the N-terminal tails. **Acetylation** (by HATs) neutralizes the positive charge, relaxing chromatin and increasing transcription (**Euchromatin**). * **Drug Connection:** **Hydralazine** and **Procainamide** can induce Systemic Lupus Erythematosus (Drug-induced SLE), where **Anti-histone antibodies** are a characteristic diagnostic marker. * **Evolutionary Conservation:** H3 and H4 are among the most highly conserved proteins across species.

Q: Which of the following diseases is NOT characterized by a defect in DNA repair mechanisms?

Huntington's disease. **Explanation:** The correct answer is **Huntington’s disease**. Unlike the other options, Huntington’s disease is a **Trinucleotide Repeat Expansion disorder** (specifically CAG repeats on chromosome 4). It is characterized by protein misfolding and aggregation (Huntingtin protein) rather than a primary defect in DNA repair. **Analysis of Options:** * **Xeroderma Pigmentosa (XP):** A classic DNA repair disorder caused by a defect in **Nucleotide Excision Repair (NER)**. Patients cannot repair pyrimidine dimers formed by UV light, leading to extreme photosensitivity and early skin cancers. * **Fanconi Anemia (Note: Option says Fanconi syndrome, but in the context of DNA repair, Fanconi Anemia is implied):** This is caused by defects in the **interstrand cross-link repair** mechanism. It presents with bone marrow failure, physical abnormalities, and high cancer risk. (Note: *Fanconi Syndrome* is a renal proximal tubule defect; *Fanconi Anemia* is the DNA repair defect). * **Hereditary Nonpolyposis Colon Cancer (HNPCC/Lynch Syndrome):** Caused by mutations in **Mismatch Repair (MMR)** genes (e.g., MSH2, MLH1). This leads to microsatellite instability and increased risk of colorectal and endometrial cancers. **High-Yield Clinical Pearls for NEET-PG:** * **Ataxia Telangiectasia:** Defect in **Non-homologous end joining (NHEJ)** or ATM gene (double-strand break repair). * **Bloom Syndrome / Cockayne Syndrome:** Also categorized as DNA repair/stability defects. * **Huntington’s Disease Hallmark:** Shows **"Anticipation"** (earlier onset in successive generations) and is inherited in an Autosomal Dominant pattern.

Q: What are the components of the 50S ribosomal subunit?

23S RNA & 5S RNA. ### Explanation The ribosome is the cellular machinery responsible for protein synthesis (translation). In prokaryotes (like bacteria), the ribosome is **70S**, consisting of a small **30S** subunit and a large **50S** subunit. **1. Why Option D is Correct:** The **50S (large) subunit** of the prokaryotic ribosome is composed of two types of ribosomal RNA (rRNA) and approximately 34 proteins (L1–L34). The specific rRNA components are: * **23S rRNA:** Acts as a ribozyme (peptidyl transferase) to catalyze peptide bond formation. * **5S rRNA:** Provides structural stability. **2. Analysis of Incorrect Options:** * **Option A (5S, 5.8S, & 28S):** These are the components of the **60S (large) subunit of Eukaryotic ribosomes**. Eukaryotes have an 80S ribosome (40S + 60S). * **Option B (16S & 23S):** 16S rRNA is the signature component of the **30S (small) subunit** in prokaryotes, not the 50S subunit. * **Option C (23S & 5.8S):** This is a mismatch. 5.8S is exclusively eukaryotic, while 23S is exclusively prokaryotic. **3. High-Yield Clinical Pearls for NEET-PG:** * **Antibiotic Targets:** Many antibiotics selectively target the 50S subunit (e.g., **Macrolides, Chloramphenicol, Clindamycin, and Linezolid**) or the 30S subunit (e.g., **Aminoglycosides, Tetracyclines**). This selectivity is based on the structural differences between prokaryotic (70S) and eukaryotic (80S) ribosomes. * **Shine-Dalgarno Sequence:** The **16S rRNA** (of the 30S subunit) contains a sequence complementary to the mRNA's Shine-Dalgarno sequence, which is essential for initiation of translation in bacteria. * **Svedberg Unit (S):** It measures the sedimentation rate, which depends on both mass and shape (which is why 50S + 30S = 70S, not 80S).

Question 1

The nucleotide triplet CTC in the sixth position of the beta-globin chain in DNA normally forms a complementary nucleotide on mRNA that codes for glutamic acid. A point mutation in the beta-globin chain resulting in the nucleotide triplet CAC forms a complementary nucleotide on mRNA that codes for valine. In sickle cell anemia, what complementary nucleotide triplet on mRNA, read from 5' to 3', would be expected?

Accepted Answer

GUG

Answer

GAG

Answer

CTC

Answer

GTG

Question 2

How many high-energy phosphate bonds are required for the formation of one peptide bond?

Accepted Answer

4

Answer

0

Answer

1

Answer

2

Question 3

A change in the second base of the serine codon UCA may result in the codon UAA. This is an example of which type of mutation?

Accepted Answer

Nonsense

Answer

Missense

Answer

Silent

Answer

Frameshift

Question 4

Random inactivation of the X chromosome is known as:

Accepted Answer

Lyonization

Answer

Allelic exclusion

Answer

Randomization

Answer

Genomic imprinting

Question 5

What is essential for the Polymerase Chain Reaction (PCR)?

Accepted Answer

A thermostable enzyme is needed.

Answer

Multiple primer annealing steps are required.

Answer

Non-specific primer binding occurs.

Answer

A thermolabile enzyme is used.

Question 6

Which enzyme forms the RNA primer in eukaryotes?

Accepted Answer

RNA polymerase

Answer

Ligase

Answer

Topoisomerase

Answer

Helicase

Question 7

Which enzyme synthesizes the RNA primer?

Accepted Answer

DNA primase

Answer

Topoisomerase

Answer

Helicase

Answer

DNA ligase

Question 8

Linker DNA is bound to which of the following histone?

Accepted Answer

H1

Answer

H2A

Answer

H2B

Answer

H3

Question 9

Which of the following diseases is NOT characterized by a defect in DNA repair mechanisms?

Accepted Answer

Huntington's disease

Answer

Xeroderma Pigmentosa

Answer

Fanconi syndrome

Answer

Hereditary nonpolyposis colon cancer

Question 10

What are the components of the 50S ribosomal subunit?

Accepted Answer

23S RNA & 5S RNA

Answer

5S RNA, 5.8S & 28S RNA

Answer

16S RNA & 23S RNA

Answer

23S RNA & 5.8S RNA

Molecular Biology and Genomics — MCQs

Molecular Biology and Genomics — MCQs

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