Elongation and termination of translation — MCQs

10 questions

An investigator studying protein synthesis in human stem cells isolates tRNA molecules bound to mRNA molecules. The isolated tRNA molecules have inosine in the 5' position of the anticodon; of these, some are bound to adenine, some to cytosine, and some to uracil at the 3' position of the mRNA codon. Which of the following properties of the genetic code is best illustrated by this finding?

A researcher is tracing the fate of C-peptide, a product of preproinsulin cleavage. Which of the following is a true statement regarding the fate of C-peptide?

An investigator is studying the genetic profile of an isolated pathogen that proliferates within macrophages. The pathogen contains sulfatide on the surface of its cell wall to prevent fusion of the phagosome and lysosome. She finds that some of the organisms under investigation have mutations in a gene that encodes the enzyme required for synthesis of RNA from a DNA template. The mutations are most likely to reduce the therapeutic effect of which of the following drugs?

An 18-year old college freshman presents to his university clinic because he has not been feeling well for the past two weeks. He has had a persistent headache, occasional cough, and chills without rigors. The patient’s vital signs are normal and physical exam is unremarkable. His radiograph shows patchy interstitial lung infiltrates and he is diagnosed with atypical pneumonia. The patient is prescribed azithromycin and takes his medication as instructed. Despite adherence to his drug regimen, he returns to the clinic one week later because his symptoms have not improved. The organism responsible for this infection is likely resistant to azithromycin through which mechanism?

An investigator is following a 4-year-old boy with Duchenne muscular dystrophy. Western blot of skeletal muscle cells from this boy shows that the dystrophin protein is significantly smaller compared to the dystrophin protein of a healthy subject. Further evaluation shows that the boy's genetic mutation involves a sequence that normally encodes leucine. The corresponding mRNA codon has the sequence UUG. Which of the following codons is most likely present in this patient at the same position of the mRNA sequence?

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?

You are treating a neonate with meningitis using ampicillin and a second antibiotic, X, that is known to cause ototoxicity. What is the mechanism of antibiotic X?

A 70-year-old man with loose stools over the last 24 hours, accompanied by abdominal pain, cramps, nausea, and anorexia, was hospitalized. Previously, the man was diagnosed with a lung abscess and was treated with clindamycin for 5 days. Past medical history was significant for non-erosive antral gastritis and hypertension. He takes esomeprazole and losartan. Despite the respiratory improvement, fevers and leukocytosis persisted. Which of the following pathogenic mechanisms would you expect to find in this patient?

A 12-year-old male presents to the emergency department following several days of facial edema. A urinalysis confirms proteinuria and hematuria. Once admitted, a kidney biopsy is viewed under an electron microscope to confirm the diagnosis of minimal change disease. In the following electron micrograph, what process occurs in the structure marked with an arrow?

Q10

Given the mRNA sequence shown below, if translation were to start at the first base, what would the tRNA anticodon be for the last amino acid translated in the chain? 5'----GCACCGGCCUGACUAUAA---3'

Elongation and termination of translation US Medical PG Practice Questions and MCQs

Question 1: An investigator studying protein synthesis in human stem cells isolates tRNA molecules bound to mRNA molecules. The isolated tRNA molecules have inosine in the 5' position of the anticodon; of these, some are bound to adenine, some to cytosine, and some to uracil at the 3' position of the mRNA codon. Which of the following properties of the genetic code is best illustrated by this finding?

A. Unambiguity
B. Non-overlapping
C. Degeneracy (Correct Answer)
D. Specificity of the start codon
E. Specificity of stop codons

Explanation: ***Degeneracy*** - The finding that a single tRNA anticodon (with **inosine** at the 5' position) can bind to multiple different mRNA codons (ending in **adenine, cytosine, or uracil**) illustrates the concept of **degeneracy** in the genetic code. - This **wobble hypothesis** allows fewer tRNAs to recognize more than one codon for a given amino acid, meaning multiple codons can code for the same amino acid. *Unambiguity* - The genetic code is unambiguous, meaning that each codon specifies **only one specific amino acid** (or a stop signal) and never two different amino acids. - This finding, however, shows one tRNA recognizing multiple codons, not one codon coding for multiple amino acids. *Non-overlapping* - The **non-overlapping** nature of the genetic code means that each nucleotide in an mRNA sequence is read only once as part of a single codon, without sharing nucleotides between adjacent codons. - This concept describes how codons are read sequentially, not the flexibility of codon-anticodon pairing. *Specificity of the start codon* - The **start codon (AUG)** specifically initiates translation, coding for methionine, and signals the beginning of a polypeptide chain. - This finding relates to the wobble pairing at the 3' end of the codon, not the initiation of translation. *Specificity of stop codons* - **Stop codons (UAA, UAG, UGA)** specifically signal the termination of translation without coding for any amino acid. - This finding describes the flexibility of codon-anticodon pairing, not the distinct function of termination codons.

Question 2: A researcher is tracing the fate of C-peptide, a product of preproinsulin cleavage. Which of the following is a true statement regarding the fate of C-peptide?

A. C-peptide exits the cells via a protein channel
B. C-peptide is further cleaved into insulin
C. C-peptide is packaged with insulin in secretory vesicles (Correct Answer)
D. C-peptide is immediately degraded by the proteasome
E. C-peptide activates an intracellular signaling cascade

Explanation: ***C-peptide is packaged with insulin in secretory vesicles*** - Preproinsulin is cleaved in the **endoplasmic reticulum** to proinsulin (signal peptide removal), which is then transported to the **Golgi apparatus**. - In the Golgi, proinsulin is cleaved by **prohormone convertases** into **insulin** and **C-peptide**, and both are stored together in **secretory vesicles** within the pancreatic beta cells. - Upon stimulation, both insulin and C-peptide are **co-secreted** via exocytosis in equimolar amounts, making C-peptide a useful marker of endogenous insulin secretion. *C-peptide exits the cells via a protein channel* - C-peptide exits the beta cells via **exocytosis** of secretory granules, not through specific protein channels. - It is **co-secreted with insulin** when secretory vesicles fuse with the plasma membrane. - Its presence in the bloodstream in equimolar amounts with insulin makes it an indirect measure of **insulin secretion**. *C-peptide is further cleaved into insulin* - **C-peptide** is a product of proinsulin cleavage, alongside insulin; it is not further processed into insulin. - Insulin itself is composed of two **peptide chains (A and B)** linked by disulfide bonds, formed after C-peptide is removed from proinsulin. *C-peptide is immediately degraded by the proteasome* - C-peptide is not immediately degraded by the **proteasome** upon synthesis. - After secretion, it circulates in the blood with a **longer half-life** than insulin (approximately 30 minutes versus 4-6 minutes), allowing it to be a useful marker of endogenous insulin production. - Its degradation occurs primarily in the **kidney**. *C-peptide activates an intracellular signaling cascade* - While there is some research suggesting C-peptide may have independent **biological activity** and activate certain signaling pathways extracellularly, its primary role in the context of the insulin synthesis pathway is as a **byproduct** of proinsulin processing. - Its clinical utility is primarily as a **biomarker** of endogenous insulin secretion, particularly useful in distinguishing between endogenous and exogenous insulin in diabetic patients.

Question 3: An investigator is studying the genetic profile of an isolated pathogen that proliferates within macrophages. The pathogen contains sulfatide on the surface of its cell wall to prevent fusion of the phagosome and lysosome. She finds that some of the organisms under investigation have mutations in a gene that encodes the enzyme required for synthesis of RNA from a DNA template. The mutations are most likely to reduce the therapeutic effect of which of the following drugs?

A. Pyrazinamide
B. Ethambutol
C. Rifampin (Correct Answer)
D. Streptomycin
E. Levofloxacin

Explanation: ***Rifampin*** - **Rifampin** specifically targets bacterial **DNA-dependent RNA polymerase**, inhibiting **RNA synthesis**. Mutations in the gene encoding this enzyme would directly reduce rifampin's binding and effectiveness. - The description of the pathogen thriving within macrophages and using **sulfatide to evade lysosomal fusion** strongly suggests **Mycobacterium tuberculosis**, a bacterium for which rifampin is a cornerstone treatment. *Pyrazinamide* - **Pyrazinamide** is a prodrug that, once converted to **pyrazinoid acid**, disrupts **mycobacterial membrane potential** and metabolism. Its primary target is not RNA synthesis. - Its efficacy is pH-dependent and it acts optimally in acidic environments, such as within macrophages, but mutations affecting RNA synthesis would not directly compromise its action. *Ethambutol* - **Ethambutol** inhibits **arabinosyl transferase**, an enzyme essential for the synthesis of the **mycobacterial cell wall component arabinogalactan**. - Its mechanism of action is distinct from RNA synthesis, thus mutations affecting RNA polymerase would not impact its efficacy. *Streptomycin* - **Streptomycin** is an **aminoglycoside antibiotic** that binds to the **30S ribosomal subunit**, inhibiting bacterial **protein synthesis**. - This mechanism is unrelated to DNA-dependent RNA polymerase, so mutations in RNA synthesis enzymes would not affect streptomycin's action. *Levofloxacin* - **Levofloxacin** is a **fluoroquinolone antibiotic** that inhibits **bacterial DNA gyrase (topoisomerase II)** and **topoisomerase IV**, thereby blocking DNA replication and transcription. - While it affects processes related to DNA, its direct target is not the DNA-dependent RNA polymerase enzyme itself, distinguishing it from rifampin's specific mechanism.

Question 4: An 18-year old college freshman presents to his university clinic because he has not been feeling well for the past two weeks. He has had a persistent headache, occasional cough, and chills without rigors. The patient’s vital signs are normal and physical exam is unremarkable. His radiograph shows patchy interstitial lung infiltrates and he is diagnosed with atypical pneumonia. The patient is prescribed azithromycin and takes his medication as instructed. Despite adherence to his drug regimen, he returns to the clinic one week later because his symptoms have not improved. The organism responsible for this infection is likely resistant to azithromycin through which mechanism?

A. Mutation in topoisomerase II
B. Methylation of ribosomal binding site
C. Presence of a beta-lactamase
D. Decreased binding to RNA polymerase
E. Insertion of drug efflux pumps (Correct Answer)

Explanation: ***Insertion of drug efflux pumps*** - **Azithromycin** is a macrolide antibiotic that inhibits bacterial protein synthesis by binding to the **50S ribosomal subunit**. - In **Mycoplasma pneumoniae** (the most common cause of atypical pneumonia in young adults), the **most common** mechanism of macrolide resistance is through **efflux pumps**, particularly the **mef genes**. - These efflux pumps actively transport macrolides out of the bacterial cell, reducing intracellular drug concentration and conferring resistance. - This mechanism is responsible for the majority of macrolide-resistant *M. pneumoniae* isolates worldwide. *Methylation of ribosomal binding site* - **Methylation** of the ribosomal binding site (specifically the **23S rRNA** via erm genes) does prevent azithromycin from binding effectively. - While this is a valid macrolide resistance mechanism seen in organisms like *Streptococcus pneumoniae* and *Streptococcus pyogenes*, it is **less common** in *Mycoplasma pneumoniae*. - Efflux pumps (mef) are the predominant mechanism in *M. pneumoniae* resistant strains. *Mutation in topoisomerase II* - **Topoisomerase II** (DNA gyrase) is the target of **fluoroquinolone antibiotics**, not macrolides. - Mutations in this enzyme lead to resistance against fluoroquinolones, such as **ciprofloxacin**. *Presence of a beta-lactamase* - **Beta-lactamase enzymes** inactivate **beta-lactam antibiotics** (e.g., penicillin, cephalosporins) by hydrolyzing their beta-lactam ring. - Additionally, *Mycoplasma pneumoniae* **lacks a cell wall**, making it inherently resistant to all beta-lactam antibiotics regardless of beta-lactamase production. *Decreased binding to RNA polymerase* - **RNA polymerase** is the target for antibiotics like **rifampin**, which inhibits bacterial transcription. - Decreased binding to RNA polymerase would lead to rifampin resistance, not azithromycin resistance.

Question 5: An investigator is following a 4-year-old boy with Duchenne muscular dystrophy. Western blot of skeletal muscle cells from this boy shows that the dystrophin protein is significantly smaller compared to the dystrophin protein of a healthy subject. Further evaluation shows that the boy's genetic mutation involves a sequence that normally encodes leucine. The corresponding mRNA codon has the sequence UUG. Which of the following codons is most likely present in this patient at the same position of the mRNA sequence?

A. AUG
B. UCG
C. UAG (Correct Answer)
D. UUU
E. GUG

Explanation: ***UAG*** - A premature **stop codon** (UAG, UAA, UGA) leads to a truncated protein, which explains the significantly smaller dystrophin protein observed in the Western blot. - The mutation converts a leucine codon (UUG) into a stop codon, resulting in a **nonsense mutation**. *AUG* - This codon codes for **methionine** and serves as the **start codon** for protein synthesis. - A mutation from UUG to AUG would change one amino acid to another, but it would not result in a significantly smaller protein. *UCG* - This codon codes for **serine**. - A mutation from UUG to UCG would result in a **missense mutation**, substituting leucine with serine, and would not lead to a significantly shorter protein. *UUU* - This codon codes for **phenylalanine**. - A mutation from UUG to UUU would be a **missense mutation**, substituting leucine with phenylalanine, and would not cause protein truncation. *GUG* - This codon codes for **valine**. - A mutation from UUG to GUG would result in a **missense mutation**, substituting leucine with valine, and would not lead to an abnormally short protein.

Question 6: 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

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.

Question 7: You are treating a neonate with meningitis using ampicillin and a second antibiotic, X, that is known to cause ototoxicity. What is the mechanism of antibiotic X?

A. It binds the 50S ribosomal subunit and inhibits formation of the initiation complex
B. It binds the 30S ribosomal subunit and inhibits formation of the initiation complex (Correct Answer)
C. It binds the 30S ribosomal subunit and reversibly inhibits translocation
D. It binds the 50S ribosomal subunit and inhibits peptidyltransferase
E. It binds the 50S ribosomal subunit and reversibly inhibits translocation

Explanation: ***It binds the 30s ribosomal subunit and inhibits formation of the initiation complex*** - The second antibiotic, X, is likely an **aminoglycoside**, such as **gentamicin** or **amikacin**, which are commonly used in combination with ampicillin for neonatal meningitis and are known to cause ototoxicity. - Aminoglycosides exert their bactericidal effect by **irreversibly binding to the 30S ribosomal subunit**, thereby **inhibiting the formation of the initiation complex** and leading to misreading of mRNA. *It binds the 50S ribosomal subunit and inhibits formation of the initiation complex* - This mechanism is characteristic of **linezolid**, which targets the 50S ribosomal subunit to prevent the formation of the initiation complex. - While linezolid can cause side effects, **ototoxicity** is less commonly associated with it compared to aminoglycosides, and it is not a primary drug for neonatal meningitis alongside ampicillin. *It binds the 50S ribosomal subunit and inhibits peptidyltransferase* - This is the mechanism of action for **chloramphenicol**, which inhibits **peptidyltransferase** activity on the 50S ribosomal subunit, preventing peptide bond formation. - Although chloramphenicol can cause **ototoxicity** and **aplastic anemia**, its use in neonates is limited due to the risk of **Gray Baby Syndrome**. *It binds the 30s ribosomal subunit and reversibly inhibits translocation* - This describes the mechanism of action of **tetracyclines**, which reversibly bind to the 30S ribosomal subunit and prevent the attachment of aminoacyl-tRNA, thereby inhibiting protein synthesis. - Tetracyclines are **contraindicated in neonates** due to their potential to cause **tooth discoloration** and **bone growth inhibition**, and ototoxicity is not their primary adverse effect. *It binds the 50s ribosomal subunit and reversibly inhibits translocation* - This mechanism of reversibly inhibiting translocation by binding to the 50S ribosomal subunit is characteristic of **macrolides** (e.g., erythromycin, azithromycin) and **clindamycin**. - While some macrolides can cause **transient ototoxicity**, they are not typically the second antibiotic of choice for neonatal meningitis in combination with ampicillin, and clindamycin's side effect profile is different.

Question 8: A 70-year-old man with loose stools over the last 24 hours, accompanied by abdominal pain, cramps, nausea, and anorexia, was hospitalized. Previously, the man was diagnosed with a lung abscess and was treated with clindamycin for 5 days. Past medical history was significant for non-erosive antral gastritis and hypertension. He takes esomeprazole and losartan. Despite the respiratory improvement, fevers and leukocytosis persisted. Which of the following pathogenic mechanisms would you expect to find in this patient?

A. Glucosylation of Rho family GTPases (Correct Answer)
B. Inactivation of elongation factor EF-2
C. Inactivation of the 60S ribosome subunit
D. Cell membrane degradation by lecithinase
E. ADP-ribosylation of Gs-alpha subunit of G-protein coupled receptors

Explanation: ***Glucosylation of Rho family GTPases*** - The clinical presentation (clindamycin use, loose stools, abdominal pain, fever, leukocytosis) strongly suggests **_Clostridioides difficile_ infection (CDI)**. The **_C. difficile_ toxins A and B** are glucosyltransferases that modify and inactivate Rho family GTPases. - **Inactivation of Rho GTPases** leads to disruption of the **cytoskeleton**, loss of tight junctions between enterocytes, and ultimately causes **cell death and colonic inflammation**, resulting in pseudomembranous colitis. *Inactivation of elongation factor EF-2* - This is the mechanism of action of **diphtheria toxin** (produced by **_Corynebacterium diphtheriae_**) and **_Pseudomonas aeruginosa_ exotoxin A**. - These toxins **ADP-ribosylate elongation factor 2 (EF-2)**, inhibiting protein synthesis and leading to cell death. This does not align with the patient's symptoms or antibiotic history. *Inactivation of the 60S ribosome subunit* - This mechanism is associated with **Shiga toxin** (produced by **_Shigella dysenteriae_** and **enterohemorrhagic _E. coli_ (EHEC)**) and **ricin toxin**. - These toxins enzymatically remove an adenine residue from the 28S rRNA of the 60S ribosomal subunit, thereby **halting protein synthesis** and causing cell damage. *Cell membrane degradation by lecithinase* - **Lecithinase (alpha-toxin)** produced by **_Clostridium perfringens_** is a phospholipase that degrades **lecithin** in cell membranes. - This leads to **hemolysis, myonecrosis, and tissue destruction** characteristic of gas gangrene, which is not consistent with the patient's diarrheal illness. *ADP-ribosylation of Gs-alpha subunit of G-protein coupled receptors* - This is the mechanism of action of **cholera toxin** (produced by **_Vibrio cholerae_**) and **heat-labile enterotoxin (LT)** of **enterotoxigenic _E. coli_ (ETEC)**. - **ADP-ribosylation of Gs-alpha subunit** permanently activates adenylate cyclase, leading to increased intracellular **cAMP**, which causes excessive **secretion of water and electrolytes** into the gut lumen, resulting in watery diarrhea, but without significant inflammation as seen in the patient.

Question 9: A 12-year-old male presents to the emergency department following several days of facial edema. A urinalysis confirms proteinuria and hematuria. Once admitted, a kidney biopsy is viewed under an electron microscope to confirm the diagnosis of minimal change disease. In the following electron micrograph, what process occurs in the structure marked with an arrow?

A. Podocyte foot process effacement (Correct Answer)
B. Normal podocyte foot process interdigitation
C. Mesangial cell proliferation
D. Endothelial cell fenestration
E. Glomerular basement membrane thickening

Explanation: **Podocyte foot process effacement** - Minimal change disease is characterized by the **effacement (flattening and fusion)** of podocyte foot processes, which are the structures indicated by the arrow in the image. - This effacement leads to the loss of the **slit diaphragm barrier**, causing massive proteinuria. *Normal podocyte foot process interdigitation* - Normal podocyte foot processes exhibit a distinct, highly organized **interdigitating pattern**, which is clearly not observed in the image due to the flattening. - This normal interdigitation is crucial for maintaining the **glomerular filtration barrier** and preventing protein leakage. *Mesangial cell proliferation* - **Mesangial cell proliferation** is characteristic of conditions like IgA nephropathy or mesangioproliferative glomerulonephritis, and is not the primary feature of minimal change disease. - The image illustrates changes in the podocytes, not the mesangial cells (labeled 'M'). *Endothelial cell fenestration* - **Endothelial cell fenestrations** are normal pores in the glomerular endothelial cells that allow for filtration, and are not directly affected or effaced in minimal change disease. - The arrow in the image points to the podocyte layer, not the endothelial cells. *Glomerular basement membrane thickening* - **Glomerular basement membrane (GBM) thickening** is seen in conditions like diabetic nephropathy or membranous nephropathy, but not typically in minimal change disease. - In minimal change disease, the GBM typically appears normal under electron microscopy.

Question 10: Given the mRNA sequence shown below, if translation were to start at the first base, what would the tRNA anticodon be for the last amino acid translated in the chain? 5'----GCACCGGCCUGACUAUAA---3'

A. 3' GCG 5'
B. 3' CGC 5'
C. 5' CGG 3'
D. 3' CGG 5' (Correct Answer)
E. 3' GAU 5'

Explanation: ***3' CGG 5'*** - The mRNA sequence is 5'-GCACCGGCCUGACUAUAA-3'. We need to identify the **open reading frame** starting from the first base and translate codons until a stop codon is reached. - The codons are **GCA** (Ala), **CCG** (Pro), **GCC** (Ala), **UGA** (Stop). The **last amino acid** translated is Alanine, corresponding to the mRNA codon **GCC**. The tRNA anticodon for GCC is **3'-CGG-5'** because base pairing rules dictate C pairs with G, and G pairs with C, in an antiparallel orientation. *3' GCG 5'* - This anticodon would pair with an mRNA codon of 5'-CGC-3', which codes for Arginine, not the alanine derived from the last amino acid in the given sequence. - It does not correctly reflect the antiparallel binding and base pairing required for the mRNA codon GCC. *5' CGG 3'* - While it contains the correct bases for pairing with GCC, the **orientation is incorrect**. tRNA anticodons are written 3' to 5'. - A 5'-CGG-3' anticodon would pair with an mRNA codon of 3'-GCC-5', which is not consistent with the standard 5' to 3' mRNA codon reading. *3' GAU 5'* - This anticodon would pair with an mRNA codon of 5'-CUA-3', which codes for Leucine. - Leucine is not the last amino acid translated from the given mRNA sequence before a stop codon. *3' CGC 5'* - This anticodon would pair with an mRNA codon of 5'-GCG-3', which codes for Alanine. - However, the last amino acid translated is encoded by 5'-GCC-3', not 5'-GCG-3'.

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