IB Biology Quiz: Understand Protein Synthesis
17 questions · exam conditions
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Understand Protein SynthesisQuestion 1 of 17

During translational elongation in a eukaryote, a tRNA molecule carrying the fifth amino acid has just bound to the A site of the ribosome. What is the immediate next step involving this tRNA molecule?

The ribosome translocates, moving the tRNA from the A site to the P site.
The tRNA moves to the E site to be ejected from the ribosome complex.
A peptide bond forms between the fourth amino acid and the fifth amino acid.
The small ribosomal subunit binds to the tRNA to verify the anticodon match.
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IB Biology Quiz

IB Biology Quiz: Understand Protein Synthesis

Practice Understand Protein Synthesis in IB Biology with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

What this quiz covers

This quiz focuses on Understand Protein Synthesis, giving you a quick way to practice the rules, question types, and explanations that matter most for IB Biology.

How to use this quiz

Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.

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Question 1

During translational elongation in a eukaryote, a tRNA molecule carrying the fifth amino acid has just bound to the A site of the ribosome. What is the immediate next step involving this tRNA molecule?

  1. The ribosome translocates, moving the tRNA from the A site to the P site.
  2. The tRNA moves to the E site to be ejected from the ribosome complex.
  3. A peptide bond forms between the fourth amino acid and the fifth amino acid. (correct answer)
  4. The small ribosomal subunit binds to the tRNA to verify the anticodon match.
Explanation: The correct answer is C. After a charged tRNA binds to the A site, the immediate next step is the formation of a peptide bond between the amino acid at the A site (the fifth) and the growing polypeptide chain attached to the tRNA at the P site (containing the fourth amino acid). This reaction is catalyzed by the ribosome's peptidyl transferase activity. Only after the peptide bond is formed does the ribosome translocate (A), which moves the tRNAs and mRNA. Moving to the E site (B) happens after translocation. Verification of the anticodon match (D) occurs upon binding to the A site, before peptide bond formation.

Question 2

The DNA coding strand for a particular gene segment is 5'-AAT GCA GCT-3'. Which sequence represents the anticodon of the tRNA that carries the second amino acid of the resulting peptide?

  1. 3'-UCG-5'
  2. 3'-CGU-5' (correct answer)
  3. 5'-GCA-3'
  4. 5'-CGU-3'
Explanation: The correct answer is B. The DNA coding strand 5'-AAT GCA GCT-3' corresponds to an mRNA sequence 5'-AAU GCA GCU-3'. The second codon is 5'-GCA-3'. The tRNA anticodon must be complementary and antiparallel to this codon. The complementary sequence to GCA is CGU, and since the codon is oriented 5' to 3', the anticodon must be oriented 3' to 5' for proper base pairing. Therefore, the anticodon is 3'-CGU-5'. Choice A is an incorrect complement. Choice C is the codon itself, not the anticodon. Choice D has the correct sequence but incorrect orientation notation.

Question 3

In prokaryotic cells, translation can begin on an mRNA molecule that is still in the process of being transcribed. What feature of prokaryotic cells allows for this coupling of transcription and translation?

  1. The presence of plasmids containing essential genes for protein synthesis.
  2. The relatively simple structure of prokaryotic ribosomes compared to eukaryotic ones.
  3. The use of a single RNA polymerase for transcribing all types of genes.
  4. The absence of a nuclear membrane separating the genetic material from the cytoplasm. (correct answer)
Explanation: The correct answer is D. Prokaryotes lack a nucleus. Their DNA is located in the cytoplasm in a region called the nucleoid. Because there is no physical barrier separating the DNA (where transcription occurs) from the ribosomes (where translation occurs), ribosomes can attach to the 5' end of the growing mRNA and begin translation immediately. In eukaryotes, transcription occurs in the nucleus, and the mRNA must be processed and exported to the cytoplasm before translation can begin. Plasmids (A), ribosome structure (B), and the type of RNA polymerase (C) are features of prokaryotes but do not directly explain the coupling of transcription and translation.

Question 4

Consider the following mRNA sequence: 5'-GCAUGCCGAUAGGCUGA-3'. A mutation inserts a single guanine (G) nucleotide immediately after the start codon. How will this affect the resulting polypeptide chain?

  1. It will cause a single amino acid substitution but the rest of the chain will be normal.
  2. It will have no effect on the polypeptide due to the degeneracy of the code.
  3. It will cause a frameshift, leading to a completely different downstream amino acid sequence and likely early termination. (correct answer)
  4. It will prevent the initiation of translation, so no polypeptide will be formed.
Explanation: The correct answer is C. The start codon is AUG. The original reading frame is AUG-CCG-AUA-GGC-UGA. The insertion of a G after AUG changes the sequence to AUG-GCC-GAU-AGG-CUG-A. This shifts the reading frame for all subsequent codons. The new codons (GCC, GAU, AGG, etc.) are completely different from the original ones (CCG, AUA, GGC). This is a frameshift mutation and almost always results in a non-functional protein, often because a premature stop codon is created. An insertion causes a frameshift, not a single substitution (A). Degeneracy (B) cannot compensate for a frameshift. The mutation occurs after the start codon, so initiation of translation (D) will still occur.

Question 5

Which molecule is directly responsible for bringing a specific amino acid to the ribosome and matching it to the correct codon on the mRNA?

  1. Ribosomal RNA (rRNA)
  2. Aminoacyl-tRNA synthetase
  3. Transfer RNA (tRNA) (correct answer)
  4. Small nuclear RNA (snRNA)
Explanation: The correct answer is C. Transfer RNA (tRNA) has a dual function: it carries a specific amino acid at one end, and at the other end, it has an anticodon that is complementary to an mRNA codon. This structure allows tRNA to act as the adapter molecule that physically links the codon sequence to the amino acid sequence. Ribosomal RNA (rRNA) (A) is a structural and catalytic component of the ribosome itself. Aminoacyl-tRNA synthetase (B) is the enzyme that attaches the correct amino acid to its corresponding tRNA, a process called tRNA charging, but it does not participate directly at the ribosome. Small nuclear RNA (snRNA) (D) is involved in splicing pre-mRNA in eukaryotes.

Question 6

The process of charging a tRNA molecule, where an amino acid is attached to it, is essential for translation. Which molecule provides the energy for this process and what is the name of the enzyme that catalyzes it?

  1. ATP; Aminoacyl-tRNA synthetase (correct answer)
  2. ATP; RNA polymerase
  3. GTP; Peptidyl transferase
  4. GTP; Aminoacyl-tRNA synthetase
Explanation: The correct answer is C. The attachment of an amino acid to its specific tRNA is catalyzed by a family of enzymes called aminoacyl-tRNA synthetases. There is a specific synthetase for each amino acid. This process requires energy, which is supplied by the hydrolysis of ATP to AMP and pyrophosphate. This creates a high-energy bond between the amino acid and the tRNA, and this stored energy is later used to form the peptide bond. GTP (A, D) is primarily used for energy during the initiation and elongation steps of translation on the ribosome. Peptidyl transferase (A) is the ribosomal activity that forms peptide bonds. RNA polymerase (B) is the enzyme for transcription.

Question 7

What is the primary advantage of forming polysomes during protein synthesis?

  1. It ensures that only one type of protein is made at any given time, preventing errors.
  2. It allows for the rapid synthesis of many copies of a polypeptide from a single mRNA molecule. (correct answer)
  3. It facilitates the post-transcriptional modification of mRNA by slowing down translation.
  4. It helps to stabilize the mRNA molecule, increasing its lifespan within the cytoplasm.
Explanation: The correct answer is B. A polysome (or polyribosome) is a complex of one mRNA molecule with several ribosomes actively translating it simultaneously. As soon as the first ribosome moves past the start codon, a second ribosome can bind and begin translation. This assembly line-like process allows for the production of a large number of polypeptide chains from a single mRNA transcript in a relatively short amount of time, amplifying the expression of the gene. It does not limit protein synthesis to one type (A), slow down translation (C), or primarily function to stabilize mRNA (D), although the presence of ribosomes might offer some protection from nucleases.

Question 8

A point mutation alters a DNA triplet on the template strand from 3'-GTC-5' to 3'-GTT-5'. Given that the codon CAG codes for glutamine (Gln) and the codon CAA also codes for glutamine (Gln), what is the most likely consequence of this mutation?

  1. A nonsense mutation, causing premature termination of the polypeptide chain.
  2. A frameshift mutation, altering the entire downstream amino acid sequence.
  3. A silent mutation, resulting in no change to the polypeptide's primary structure. (correct answer)
  4. A missense mutation, causing the substitution of glutamine with a different amino acid.
Explanation: The correct answer is C. The original DNA template strand is 3'-GTC-5', which is transcribed into the mRNA codon 5'-CAG-3'. This codon codes for glutamine. The mutated DNA template strand is 3'-GTT-5', which is transcribed into the mRNA codon 5'-CAA-3'. Since the problem states that CAA also codes for glutamine, the amino acid sequence is unchanged. This is a silent mutation due to the degeneracy of the genetic code. A nonsense mutation (A) would result from a change to a stop codon. A frameshift mutation (B) is caused by an insertion or deletion, not a substitution. A missense mutation (D) would occur if the new codon specified a different amino acid.

Question 9

The gene for human insulin can be inserted into a bacterium such as E. coli, which can then produce human insulin. Which feature of the genetic code makes this possible?

  1. The code is degenerate, meaning multiple codons can specify the same amino acid.
  2. The code is non-overlapping, meaning each base is part of only one codon.
  3. The code includes specific start and stop signals for translation.
  4. The code is nearly universal, meaning a given codon specifies the same amino acid in most organisms. (correct answer)
Explanation: The correct answer is D. The fact that a gene from one species (human) can be correctly translated into a functional protein in a completely different species (bacterium) demonstrates the universality of the genetic code. The codon CUU, for example, codes for leucine in humans, bacteria, and almost all other life forms. Degeneracy (A) explains silent mutations but not interspecies gene expression. The non-overlapping nature (B) and start/stop signals (C) are crucial features for translation within any single organism, but it is the universality that allows for the transfer of genes between them.

Question 10

Crick and Brenner's experiments with frameshift mutations in bacteriophage T4 were crucial in determining a key feature of the genetic code. What did their results strongly suggest?

  1. That the genetic code is universal across all organisms.
  2. That the code is read in non-overlapping triplets of bases. (correct answer)
  3. That some amino acids are coded for by more than one codon.
  4. That DNA, not protein, is the hereditary material.
Explanation: The correct answer is B. The experiments by Crick and Brenner showed that inserting or deleting one or two nucleotides caused a frameshift mutation that scrambled the entire downstream message, resulting in a non-functional protein. However, inserting or deleting three nucleotides often resulted in a functional or near-functional protein, with just one amino acid added or missing. This provided powerful evidence that the genetic code is read in sequential, non-overlapping groups of three bases (triplets or codons). Their work did not directly address universality (A), degeneracy (C), or the identity of the hereditary material (D), which was shown by Hershey and Chase.

Question 11

Which modification to eukaryotic pre-mRNA is crucial for the initiation of translation and for protecting the transcript from degradation by exonucleases?

  1. The addition of a methylated guanine cap to the 5' end. (correct answer)
  2. The addition of a poly-adenine (poly-A) tail to the 3' end.
  3. The removal of exons and splicing of introns.
  4. The attachment of small ubiquitin-related modifier (SUMO) proteins.
Explanation: The correct answer is C. The 5' cap, which is a modified guanine nucleotide added to the 5' end of the pre-mRNA, serves two main functions. It is recognized by the ribosome initiation complex, making it essential for starting translation. It also protects the mRNA from being broken down by 5' exonucleases, increasing its stability. The poly-A tail (B) also contributes to stability and helps with export from the nucleus, but is less directly involved in the initiation of translation. Splicing (A) removes non-coding introns but is not primarily for stability or translation initiation. SUMOylation (D) is a type of post-translational modification of proteins, not mRNA.

Question 12

During transcription, the enzyme RNA polymerase performs several functions. Which of the following is NOT a function of RNA polymerase?

  1. Unwinding the DNA double helix at the start of a gene.
  2. Synthesizing an RNA primer to initiate transcription. (correct answer)
  3. Reading the DNA template strand in the 3' to 5' direction.
  4. Joining ribonucleotides together to form a pre-mRNA strand.
Explanation: The correct answer is B. Unlike DNA polymerase, RNA polymerase does not require a primer to initiate synthesis. It can start a new RNA chain de novo at a specific promoter sequence on the DNA. RNA polymerase does unwind the DNA helix to access the template strand (A), read that template in the 3' to 5' direction (C), and polymerize ribonucleotides in the 5' to 3' direction to create the transcript (D). The need for a primer is a key feature of DNA replication, not transcription.

Question 13

If a segment of a DNA template strand is 3'-GGC AAT CAT-5', what is the amino acid sequence that would be produced? (Use the genetic code: CCG=Pro, UUA=Leu, GUA=Val)

  1. Pro - Leu - Val (correct answer)
  2. Pro - Val - Leu
  3. Val - Leu - Pro
  4. Gly - Asn - His
Explanation: The correct answer is A. First, transcribe the DNA template strand 3'-GGC AAT CAT-5' into mRNA. The mRNA is complementary and antiparallel: 5'-CCG UUA GUA-3'. Then translate using the genetic code: CCG codes for Proline (Pro), UUA codes for Leucine (Leu), and GUA codes for Valine (Val). Therefore, the amino acid sequence is Pro - Leu - Val. Option D represents the amino acids that would correspond to the template strand if read directly as codons (incorrect process).

Question 14

A scientist identifies a protein that is destined to be secreted from a eukaryotic cell. Where in the cell was this protein most likely synthesized?

  1. On free ribosomes in the cytoplasm, followed by transport into the nucleus.
  2. On ribosomes attached to the rough endoplasmic reticulum. (correct answer)
  3. Inside the Golgi apparatus, where ribosomes are temporarily located.
  4. On free ribosomes in the cytoplasm, from where it diffuses out of the cell.
Explanation: The correct answer is B. Proteins destined for secretion, insertion into membranes, or delivery to certain organelles (like lysosomes) are synthesized on ribosomes attached to the rough endoplasmic reticulum (RER). A signal peptide at the beginning of the polypeptide directs the ribosome to the RER. The protein is then synthesized directly into the RER lumen, from which it can be packaged and transported out of the cell via the Golgi apparatus and vesicles. Free ribosomes in the cytoplasm (A, D) synthesize proteins that will function within the cytoplasm, nucleus, mitochondria, or chloroplasts. Ribosomes are not found inside the Golgi apparatus (C).

Question 15

Alternative splicing is a process in eukaryotes that can generate multiple different proteins from a single gene. What is the mechanism by which alternative splicing achieves this?

  1. By using different start codons on the same mRNA molecule to produce truncated proteins.
  2. By treating certain exons as introns, leading to their removal from the final mRNA. (correct answer)
  3. By altering the sequence of nucleotides within exons during mRNA processing.
  4. By attaching different amino acids to the same tRNA molecules.
Explanation: The correct answer is B. Alternative splicing allows a single pre-mRNA transcript to be processed in different ways. By selectively including or excluding certain exons (or parts of exons), different combinations of exons can be joined together to form different mature mRNA molecules. These different mRNAs are then translated into different protein isoforms, which may have different functions. This is achieved by the splicing machinery (spliceosome) recognizing different splice sites and treating what would normally be an exon as part of an intron, which is then removed. It does not involve changing start codons (A), altering exon sequences (C), or changing tRNA charging (D).

Question 16

During the initiation of translation in eukaryotes, what is the sequence of binding events?

  1. The small ribosomal subunit, with the initiator tRNA already bound, binds to the 5' cap of the mRNA and scans for the start codon. (correct answer)
  2. The initiator tRNA binds to the start codon, followed by the binding of the small and then the large ribosomal subunits.
  3. The large ribosomal subunit binds to the mRNA, followed by the small subunit and then the initiator tRNA.
  4. The small and large ribosomal subunits assemble on the mRNA first, creating a vacant A site for the initiator tRNA.
Explanation: The correct answer is C. In eukaryotic translation initiation, the initiator tRNA (carrying methionine) first binds to the small ribosomal subunit (40S). This complex then recognizes and binds to the 5' cap of the mRNA molecule. The complex then scans along the mRNA in the 5' to 3' direction until it encounters the first AUG start codon. Once the initiator tRNA's anticodon pairs with the start codon, the large ribosomal subunit (60S) joins the complex, completing the initiation phase. The other options describe incorrect sequences of these events.

Question 17

Which of the following comparisons between eukaryotic transcription and DNA replication is correct?

  1. Both processes synthesize a new strand in the 3' to 5' direction.
  2. Both processes involve the unwinding of the entire DNA molecule.
  3. RNA polymerase is used in transcription, while DNA polymerase is used in replication. (correct answer)
  4. Both processes result in a semi-conservative product containing one old and one new strand.
Explanation: The correct answer is C. Transcription is catalyzed by RNA polymerase, which synthesizes an RNA strand, while DNA replication is catalyzed by DNA polymerase, which synthesizes a new DNA strand. Choice A is incorrect; both processes synthesize the new strand in the 5' to 3' direction. Choice B is incorrect; only a specific gene region of the DNA unwinds during transcription, whereas the entire chromosome is typically unwound (in sections) during replication. Choice D is incorrect; only DNA replication is semi-conservative. The product of transcription is a single-stranded RNA molecule, and the original DNA double helix reforms after the polymerase passes.