Elongation and termination of transcription US Medical PG Practice Questions and MCQs
Practice US Medical PG questions for Elongation and termination of transcription. These multiple choice questions (MCQs) cover important concepts and help you prepare for your exams.
Elongation and termination of transcription US Medical PG Question 1: A 25-year-old female comes to the clinic complaining of fatigue and palpitations. She has been undergoing immense stress from her thesis defense and has been extremely tired. The patient denies any weight loss, diarrhea, cold/heat intolerance. TSH was within normal limits. She reports a family history of "blood disease" and was later confirmed positive for β-thalassemia minor. It is believed that abnormal splicing of the beta globin gene results in β-thalassemia. What is removed during this process that allows RNA to be significantly shorter than DNA?
- A. 3'-poly(A) tail
- B. Exons
- C. Introns (Correct Answer)
- D. microRNAs
- E. snRNPs
Elongation and termination of transcription Explanation: **Introns**
- **Introns** are non-coding regions within a gene that are removed from the pre-mRNA transcript during **splicing**.
- This removal and the subsequent ligation of exons lead to a mature mRNA molecule that is significantly shorter than the initial DNA template.
*3'-poly(A) tail*
- The **3'-poly(A) tail** is an addition to the 3' end of the mRNA molecule, not a removed segment during splicing, and it provides stability and aids in translation.
- While it contributes to mRNA processing, its addition does not involve removing existing sequences to shorten the transcript.
*Exons*
- **Exons** are the coding regions of a gene that are retained and ligated together to form the mature mRNA, which is then translated into protein.
- If exons were removed, the resulting protein would be truncated or non-functional, and the mRNA would not contain the necessary genetic information.
*microRNAs*
- **MicroRNAs (miRNAs)** are small non-coding RNA molecules that regulate gene expression by inhibiting translation or promoting mRNA degradation.
- They are not part of the pre-mRNA transcript that is processed into mRNA; rather, they are distinct regulatory molecules.
*snRNPs*
- **Small nuclear ribonucleoproteins (snRNPs)** are components of the spliceosome, the molecular machine responsible for carrying out splicing.
- They are involved in the process of intron removal but are not themselves removed from the RNA; they are catalytic machinery.
Elongation and termination of transcription US Medical PG Question 2: An investigator is comparing DNA replication in prokaryotes and eukaryotes. He finds that the entire genome of E. coli (4 × 106 base pairs) is replicated in approximately 30 minutes. A mammalian genome (3 × 109 base pairs) is usually replicated within 3 hours. Which of the following characteristics of eukaryotic DNA replication is the most accurate explanation for this finding?
- A. Replication inhibition at checkpoint
- B. Absence of telomerase enzyme activity
- C. DNA compaction in chromatin
- D. Simultaneous replication at multiple origins (Correct Answer)
- E. More efficient DNA polymerase activity
Elongation and termination of transcription Explanation: ***Simultaneous replication at multiple origins***
- Eukaryotic DNA replication initializes at **multiple origins of replication** along each chromosome, allowing synthesis to occur concurrently in many places.
- This strategy compensates for the much larger eukaryotic genome size, enabling its complete replication within a reasonable timeframe despite slower polymerase speed compared to prokaryotes.
*Replication inhibition at checkpoint*
- **Cell cycle checkpoints**, such as those in G1, S, and G2 phases, ensure the integrity of DNA replication and repair.
- While these checkpoints can *pause* or *inhibit* replication if errors occur, they do not fundamentally explain the *speed* or **efficiency** of replication across the entire genome.
*Absence of telomerase enzyme activity*
- **Telomerase** is an enzyme that maintains the ends of eukaryotic chromosomes (telomeres) by adding repetitive DNA sequences.
- Its presence or absence is related to telomere length regulation and cellular aging, not the overall speed of genome replication.
*DNA compaction in chromatin*
- Eukaryotic DNA is compact and organized into **chromatin** within the nucleus, which presents a challenge to replication by limiting access to the DNA.
- While enzymes must overcome this compaction, it is a *hindrance* rather than an enabler of replication speed. If anything, it would slow down replication.
*More efficient DNA polymerase activity*
- In actuality, **prokaryotic DNA polymerases** (e.g., DNA Pol III in *E. coli*) are generally more processive and faster than eukaryotic DNA polymerases.
- Therefore, more efficient polymerase activity is not a characteristic that would explain the relatively fast replication of a larger eukaryotic genome.
Elongation and termination of transcription US Medical PG Question 3: An investigator isolates bacteria from a patient who presented with dysuria and urinary frequency. These bacteria grow rapidly in pink colonies on MacConkey agar. During replication of these bacteria, the DNA strands are unwound at the origin of replication, forming two Y-shaped replication forks that open in opposite directions. At each replication fork, daughter strands are synthesized from the template strands in a 5′ to 3′ direction. On one strand, the DNA is synthesized continuously; on the other strand, the DNA is synthesized in short segments. The investigator finds that three enzymes are directly involved in elongating the DNA of the lagging strand in these bacteria. One of these enzymes has an additional function that the others do not possess. Which of the following steps in DNA replication is unique to this enzyme?
- A. Elongation of lagging strand in 5'→3' direction
- B. Excision of nucleotides with 5'→3' exonuclease activity (Correct Answer)
- C. Prevention of reannealing of the leading strand and the lagging strand
- D. Creation of ribonucleotide primers
- E. Proofreading for mismatched nucleotides
Elongation and termination of transcription Explanation: ***Excision of nucleotides with 5'→3' exonuclease activity***
- **DNA polymerase I** possesses unique **5'→3' exonuclease activity** that allows it to remove RNA primers synthesized by primase.
- After primer removal, DNA polymerase I synthesizes DNA in the 5'→3' direction to fill the gap.
*Elongation of lagging strand in 5'→3' direction*
- While **DNA polymerase I** elongates the lagging strand, this 5'→3' synthesis function is also shared by **DNA polymerase III**, which is the primary enzyme for DNA synthesis.
- Therefore, this specific function is not unique to the enzyme in question (DNA polymerase I) as DNA polymerase III also performs 5'→3' elongation.
*Prevention of reannealing of the leading strand and the lagging strand*
- This function is carried out by **single-strand binding proteins (SSBs)**, which bind to the separated DNA strands to prevent them from reannealing and protect them from degradation.
- This is not a function of any DNA polymerase.
*Creation of ribonucleotide primers*
- The synthesis of short RNA primers required for initiation of DNA synthesis is performed by **primase**, an RNA polymerase.
- DNA polymerases do not create primers but rather extend them.
*Proofreading for mismatched nucleotides*
- **DNA polymerase I** and **DNA polymerase III** both possess **3'→5' exonuclease activity** for proofreading, allowing them to remove incorrectly incorporated nucleotides.
- Since this function is shared by DNA polymerase III, it is not unique to DNA polymerase I.
Elongation and termination of transcription US Medical PG Question 4: A 67-year-old woman presents to the infectious disease clinic after her PPD was found to be positive. A subsequent chest radiography shows a cavity in the apex of the right upper lobe, along with significant hilar adenopathy. The patient is diagnosed with tuberculosis and is started on the standard four-drug treatment regimen. Four weeks later, she returns for her first follow-up appointment in panic because her eyes have taken on an orange/red hue. Which of the following describes the mechanism of action of the drug most likely responsible for this side effect?
- A. Inhibition of RNA polymerase (Correct Answer)
- B. Inhibition of squalene epoxidase
- C. Inhibition of arabinosyltransferase
- D. Inhibition of mycolic acid synthesis
- E. Inhibition of topoisomerase
Elongation and termination of transcription Explanation: ***Inhibition of RNA polymerase***
- The drug most likely responsible for the **orange/red discoloration of tears, sweat, saliva, and urine is rifampin**.
- **Rifampin exerts its bactericidal effect by inhibiting bacterial DNA-dependent RNA polymerase**, thereby blocking RNA synthesis.
*Inhibition of squalene epoxidase*
- This mechanism of action is characteristic of **terbinafine**, an antifungal drug.
- **Terbinafine is used to treat fungal infections** like dermatophytosis and onychomycosis, not tuberculosis.
*Inhibition of arabinosyltransferase*
- This is the **mechanism of action for ethambutol**, another first-line drug for tuberculosis.
- While ethambutol is part of the standard regimen, its primary side effect is **optic neuritis**, not orange discoloration of bodily fluids.
*Inhibition of mycolic acid synthesis*
- This mechanism is primarily associated with **isoniazid (INH)**, a key drug in tuberculosis treatment.
- **Isoniazid's main toxicities include hepatotoxicity and peripheral neuropathy**, not the red-orange discoloration.
*Inhibition of topoisomerase*
- This mechanism is characteristic of **fluoroquinolone antibiotics**, such as ciprofloxacin and levofloxacin.
- While fluoroquinolones can be used in some tuberculosis regimens, they are typically **second-line agents** and do not cause the orange/red bodily fluid discoloration.
Elongation and termination of transcription US Medical PG Question 5: A 25-year-old male is brought to the emergency department by his friends after a camping trip. He and his friends were in the woods camping when the patient started experiencing severe right upper quadrant abdominal pain after foraging and ingesting some wild mushrooms about 3 hours earlier. The patient is lethargic on exam and appears jaundiced. He has scleral icterus and is severely tender to palpation in the right upper quadrant. He has scattered petechiae on his extremities. Liver function tests are:
Serum:
Na+: 134 mEq/L
Cl-: 100 mEq/L
K+: 4.2 mEq/L
HCO3-: 24 mEq/L
Urea nitrogen: 50 mg/dL
Glucose: 100 mg/dL
Creatinine: 1.4 mg/dL
Alkaline phosphatase: 400 U/L
Aspartate aminotransferase (AST, GOT): 3278 U/L
Alanine aminotransferase (ALT, GPT): 3045 U/L
gamma-Glutamyltransferase (GGT): 100 U/L
The most likely cause of this patient’s clinical presentation acts by inhibiting which of the following molecules?
- A. RNA polymerase II (Correct Answer)
- B. RNA polymerase III
- C. Topoisomerase
- D. RNA polymerase I
- E. Prokaryote RNA polymerase
Elongation and termination of transcription Explanation: ***RNA polymerase II***
- The clinical presentation with severe hepatotoxicity (jaundice, elevated AST/ALT, RUQ pain, petechiae, lethargy) following wild mushroom ingestion is highly suggestive of poisoning by **Amanita phalloides** (death cap mushroom).
- The primary toxin in *Amanita phalloides* is **alpha-amanitin**, which specifically inhibits **RNA polymerase II**, thereby halting mRNA synthesis and leading to cellular death, particularly in rapidly dividing cells and hepatocytes.
*RNA polymerase III*
- **RNA polymerase III** is responsible for synthesizing **tRNA** and **5S ribosomal RNA**.
- While essential for cell function, it is not the primary target of amanitin toxins, and its inhibition would not directly cause the severe hepatotoxicity observed.
*Topoisomerase*
- **Topoisomerases** are enzymes that regulate the supercoiling of **DNA** during replication, transcription, and repair.
- While critical for cell survival, they are not the target of the toxins found in *Amanita phalloides* mushrooms.
*RNA polymerase I*
- **RNA polymerase I** is responsible for synthesizing most **ribosomal RNA (rRNA)**.
- While also essential, it is less sensitive to **alpha-amanitin** than RNA polymerase II, requiring much higher concentrations for inhibition.
*Prokaryote RNA polymerase*
- **Prokaryote RNA polymerase** is fundamentally different in structure and function from eukaryotic RNA polymerases.
- **Alpha-amanitin** specifically targets eukaryotic RNA polymerases and has no significant inhibitory effect on prokaryotic RNA polymerase.
Elongation and termination of transcription US Medical PG Question 6: A group of researchers is studying molecules and DNA segments that are critical for important cellular processes in eukaryotic cells. They have identified a region that is located about 28 bases upstream of the 5’ coding region. This region promotes the initiation of transcription by binding with transcription factors. Which of the following regions have these researchers most likely identified?
- A. TATA Box (Correct Answer)
- B. RNA polymerase II
- C. Small nuclear ribonucleoprotein (SnRNPs)
- D. DNA methyltransferase
- E. CAAT Box
Elongation and termination of transcription Explanation: ***TATA Box***
- The **TATA box** is a core promoter element found in eukaryotic genes, typically located **25-35 base pairs upstream** of the transcription start site.
- It plays a crucial role in initiating transcription by serving as a binding site for **transcription factors**, which in turn recruit **RNA polymerase II**.
*RNA polymerase II*
- **RNA polymerase II** is the enzyme responsible for transcribing protein-coding genes into mRNA.
- While essential for transcription, it is an enzyme that binds to the promoter region (which includes the TATA box), rather than a regulatory DNA sequence itself.
*Small nuclear ribonucleoprotein (SnRNPs)*
- **SnRNPs** are components of the spliceosome, involved in the **splicing of pre-mRNA** to remove introns.
- They are involved in post-transcriptional modification, not in the initiation of transcription.
*DNA methyltransferase*
- **DNA methyltransferase** is an enzyme involved in **DNA methylation**, a process that typically represses gene expression.
- This enzyme modifies DNA, but it is not a DNA region that promotes transcription initiation.
*CAAT Box*
- The **CAAT box** is another common promoter element in eukaryotes, usually located further **upstream (70-80 base pairs)** from the transcription start site.
- While it also binds transcription factors and influences transcription initiation, its location is generally *more distant* than the 28 bases upstream described, making the TATA box a more accurate fit for the given distance.
Elongation and termination of transcription US Medical PG Question 7: An investigator is studying the genotypes of wingless fruit flies using full exome sequencing. Compared to wild-type winged fruit flies, the wingless fruit flies are found to have a point mutation in the gene encoding wing bud formation during embryogenesis. The point mutation in the gene causes the mRNA transcript to have a 'UUG' segment instead of an 'AUG' segment. Which of the following processes is most likely affected by this mutation?
- A. Cleavage of 5' intron
- B. Binding of met-tRNA to 40S complex (Correct Answer)
- C. Catalyzation of peptide bond formation
- D. Dissociation of mRNA from ribosome complex
- E. Shift of peptidyl-tRNA from A to P site
Elongation and termination of transcription Explanation: ***Binding of met-tRNA to 40S complex***
- The **start codon AUG** is essential for the initiation of translation, as it signals where the ribosome should begin synthesizing the polypeptide chain and recruits the initiator tRNA carrying **methionine (met-tRNA)** to the 40S ribosomal subunit.
- A mutation from **AUG to UUG** means the ribosome will not recognize the correct start site, preventing the initial binding of met-tRNA and the formation of the **initiation complex**.
*Cleavage of 5' intron*
- This process is part of **RNA splicing**, which occurs after transcription in the nucleus, where introns are removed from the **pre-mRNA**.
- The described mutation affects a **codon sequence** in the mRNA, which is a post-splicing event related to translation, not intron cleavage.
*Catalyzation of peptide bond formation*
- This occurs during the **elongation phase of translation**, where the peptidyl transferase activity of the ribosome forms peptide bonds between amino acids.
- The mutation prevents the **initiation of translation** altogether, meaning elongation and peptide bond formation will not even begin.
*Dissociation of mRNA from ribosome complex*
- This event happens at the **termination phase of translation**, when a stop codon is reached, and release factors cause the ribosome to dissociate from the mRNA and the newly synthesized polypeptide.
- The mutation prevents the **start of translation**, so the ribosome will not reach the stage where it would dissociate from the mRNA.
*Shift of peptidyl-tRNA from A to P site*
- This is a step in the **elongation phase of translation**, specifically the **translocation process**, where the ribosome moves along the mRNA, shifting the peptidyl-tRNA from the A (aminoacyl) site to the P (peptidyl) site.
- Since the **initiation of translation** is blocked by the mutated start codon, the ribosome cannot begin polypeptide synthesis, and thus, elongation steps like translocation cannot occur.
Elongation and termination of transcription US Medical PG Question 8: A 55-year-old man, who was recently diagnosed with tuberculosis, presents to his primary care provider as part of his routine follow-up visit every month. He is currently in the initial phase of anti-tubercular therapy. His personal and medical histories are relevant for multiple trips to Southeast Asia as part of volunteer activities and diabetes of 5 years duration, respectively. A physical examination is unremarkable except for a visual abnormality on a color chart; he is unable to differentiate red from green. The physician suspects the visual irregularity as a sign of toxicity due to one of the drugs in the treatment regimen. Which of the following is the mechanism by which this medication acts in the treatment of Mycobacterium tuberculosis?
- A. Inhibition of DNA-dependent RNA polymerase
- B. Inhibition of arabinosyltransferase (Correct Answer)
- C. Inhibition of protein synthesis by binding to the 30S ribosomal subunit
- D. Inhibition of mycolic acid synthesis
- E. Induction of free radical metabolites
Elongation and termination of transcription Explanation: ***Inhibition of arabinosyltransferase***
- The patient's inability to differentiate red from green is a classic symptom of **optic neuritis** (specifically retrobulbar neuritis), a known side effect of **ethambutol**.
- **Ethambutol** works by inhibiting **arabinosyltransferase**, an enzyme essential for the synthesis of the mycobacterial cell wall component **arabinogalactan**.
*Inhibition of DNA-dependent RNA polymerase*
- This is the mechanism of action for **rifampin**, another first-line anti-TB drug.
- While rifampin has various side effects (e.g., **hepatotoxicity**, **red-orange discoloration of bodily fluids**), **optic neuritis** is not its primary or common adverse effect.
*Inhibition of protein synthesis by binding to the 30S ribosomal subunit*
- This mechanism is characteristic of **aminoglycosides** (e.g., streptomycin, amikacin) and **tetracyclines**, which are used in certain TB regimens, especially for **drug-resistant cases**.
- Common side effects include **ototoxicity** and **nephrotoxicity**, not optic neuritis.
*Inhibition of mycolic acid synthesis*
- This describes the mechanism of action of **isoniazid**, a cornerstone anti-TB drug.
- Isoniazid's main side effects are **hepatotoxicity** and **peripheral neuropathy**, which is often prevented by **pyridoxine (vitamin B6) supplementation**.
*Induction of free radical metabolites*
- This is the mechanism by which **pyrazinamide** is thought to act, although its precise mechanism is not fully understood.
- Pyrazinamide is primarily associated with **hepatotoxicity** and **hyperuricemia** leading to **gout**, not optic neuritis.
Elongation and termination of transcription US Medical PG Question 9: A group of microbiological investigators is studying bacterial DNA replication in E. coli colonies. While the cells are actively proliferating, the investigators stop the bacterial cell cycle during S phase and isolate an enzyme involved in DNA replication. An assay of the enzyme's exonuclease activity determines that it is active on both intact and demethylated thymine nucleotides. Which of the following enzymes have the investigators most likely isolated?
- A. DNA ligase
- B. Telomerase
- C. Primase
- D. DNA topoisomerase
- E. DNA polymerase I (Correct Answer)
Elongation and termination of transcription Explanation: ***DNA polymerase I***
- **DNA polymerase I** possesses **5' to 3' exonuclease activity**, which is crucial for removing **RNA primers** (intact nucleotides) laid down by primase during DNA replication.
- This 5' to 3' exonuclease activity also allows it to excise damaged DNA, including DNA containing **demethylated thymine nucleotides**.
- It also has 3' to 5' exonuclease activity for proofreading.
- **Key distinction:** While DNA polymerase III (the main replicative enzyme) only has 3' to 5' exonuclease activity, DNA polymerase I has **both** 3' to 5' and 5' to 3' exonuclease activities, making it essential for primer removal and DNA repair.
*DNA ligase*
- **DNA ligase** functions to form a **phosphodiester bond** between adjacent nucleotides to seal nicks in the DNA backbone, but it does not have exonuclease activity.
- Its primary role is in joining Okazaki fragments and repairing single-strand breaks.
*Telomerase*
- **Telomerase** is a specialized reverse transcriptase that extends the telomeres at the ends of eukaryotic chromosomes, but is not present in prokaryotes like *E. coli*.
- It uses an RNA template to synthesize DNA, and it lacks exonuclease activity.
*Primase*
- **Primase** is an RNA polymerase that synthesizes short **RNA primers** on the DNA template, providing a starting point for DNA synthesis.
- It is involved in synthesizing primers, not in removing or excising nucleotides, and has no exonuclease activity.
*DNA topoisomerase*
- **DNA topoisomerases** relieve supercoiling in DNA during replication and transcription by cutting and rejoining DNA strands.
- While they act on DNA, their function is to manage topological stress, and they do not exhibit exonuclease activity on nucleotides.
Elongation and termination of transcription US Medical PG Question 10: E. coli has the ability to regulate its enzymes to break down various sources of energy when available. It prevents waste by the use of the lac operon, which encodes a polycistronic transcript. At a low concentration of glucose and absence of lactose, which of the following occurs?
- A. Decreased cAMP levels result in poor binding to the catabolite activator protein
- B. Increased cAMP levels result in binding to the catabolite activator protein (Correct Answer)
- C. Increased allolactose levels bind to the repressor
- D. Repressor releases from lac operator
- E. Transcription of the lac Z, Y, and A genes increase
Elongation and termination of transcription Explanation: ***Increased cAMP levels result in binding to the catabolite activator protein***
- In the absence of glucose, **adenylate cyclase** activity increases, leading to higher levels of **cAMP**.
- **cAMP** then binds to the **catabolite activator protein (CAP)**, forming the **cAMP-CAP complex**, which is crucial for activating lac operon transcription in the absence of glucose.
*Decreased cAMP levels result in poor binding to the catabolite activator protein*
- **Decreased glucose levels** actually lead to **increased cAMP** synthesis, not decreased.
- High **cAMP** levels enhance, not hinder, its binding to **CAP**.
*Increased allolactose levels bind to the repressor*
- **Allolactose** is an inducer that forms in the presence of **lactose**, which is stated to be absent in this scenario.
- Therefore, **allolactose levels** would be low, and it would not bind to the **repressor**.
*Repressor releases from lac operator*
- The **repressor protein** is bound to the **lac operator** in the absence of lactose.
- For the **repressor to be released**, **allolactose** (formed from lactose) must be present to bind to it.
*Transcription of the lac Z, Y, and A genes increase*
- While **cAMP-CAP binding** would promote transcription, the **absence of lactose** means the **repressor remains bound** to the operator.
- This binding effectively blocks RNA polymerase, preventing significant transcription of the **lac Z, Y, and A genes**, regardless of high **cAMP** levels.
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