USMLE Step 1 Quiz: Transcription Translation And Regulation

What this quiz covers

This quiz focuses on Transcription Translation And Regulation, giving you a quick way to practice the rules, question types, and explanations that matter most for USMLE Step 1.

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.

Question 1

1. Removal of introns from pre-mRNA (correct answer)
2. Addition of a 7-methylguanosine cap to the 5' end of mRNA
3. Covalent attachment of amino acids to tRNA
4. Degradation of proteins marked with ubiquitin

Explanation: This patient's presentation is classic for Systemic Lupus Erythematosus (SLE). Anti-Smith antibodies, which are highly specific for SLE, target small nuclear ribonucleoproteins (snRNPs). These snRNPs are core components of the spliceosome, the complex responsible for removing non-coding sequences (introns) from pre-mRNA transcripts in the nucleus. The other options describe different, unrelated cellular processes.

Question 2

1. Mannose-6-phosphate (correct answer)
2. Ubiquitin
3. Farnesyl pyrophosphate
4. N-acetylglucosamine

Explanation: This is a classic presentation of I-cell disease (Mucolipidosis II), a lysosomal storage disorder. It is caused by a deficiency in N-acetylglucosaminyl-1-phosphotransferase. This enzyme catalyzes the first step in adding a mannose-6-phosphate (M6P) tag to lysosomal enzymes in the Golgi apparatus. This M6P tag acts as a molecular address label, targeting these enzymes for transport to the lysosome. Without the tag, the enzymes are incorrectly secreted from the cell, leading to their high levels in the serum and deficiency within lysosomes.

Question 3

1. Increased positive charge of histone tails and DNA condensation
2. Decreased affinity of histones for DNA and increased transcription (correct answer)
3. Increased methylation of CpG islands in promoter regions
4. Formation of heterochromatin and gene silencing

Explanation: Histone acetylation removes the positive charge from lysine residues on histone tails, reducing their affinity for negatively charged DNA. This leads to a more relaxed chromatin structure (euchromatin), which is accessible to transcription factors and RNA polymerase, thus promoting gene expression. HDACs remove these acetyl groups, promoting condensation. An HDAC inhibitor would therefore cause an accumulation of acetylated histones, leading to decreased histone-DNA affinity and increased transcription of target genes, such as tumor suppressors.

Question 4

1. Missense
2. Nonsense (correct answer)
3. Silent
4. Frameshift

Explanation: The genetic code includes start and stop codons. The codon UGG codes for the amino acid tryptophan. The codon UGA is one of the three stop codons (along with UAA and UAG). A mutation that changes an amino acid-coding codon into a stop codon is called a nonsense mutation. This results in the premature termination of translation and the production of a truncated, usually nonfunctional, protein.

Question 5

1. Increased transcription of the oncogene
2. Repression of translation of the oncogene mRNA (correct answer)
3. Alternative splicing of the oncogene pre-mRNA
4. Enhanced stability of the oncogene mRNA

Explanation: MicroRNAs (miRNAs) are small, non-coding RNAs that regulate gene expression post-transcriptionally. They bind to complementary sequences, typically in the 3' UTR of target mRNAs. This binding recruits a protein complex (RISC) that either leads to the degradation of the mRNA (if binding is perfectly complementary) or, more commonly in animals, represses translation (if binding is imperfect). Therefore, the normal function of this tumor-suppressive miRNA is to decrease the production of the oncoprotein by repressing translation.

Question 6

1. A frameshift and premature termination of translation (correct answer)
2. A complete failure of transcription of the β-globin gene
3. Synthesis of a structurally normal but unstable β-globin protein
4. Inhibition of the addition of the 5' cap to the mRNA

Explanation: Mutations that create new splice sites (cryptic splice sites) lead to incorrect splicing of pre-mRNA. In this case, including a portion of an intron in the mature mRNA will alter the reading frame because the number of inserted nucleotides is unlikely to be a multiple of three. This frameshift will change the amino acid sequence downstream of the insertion and almost always introduces a premature stop codon, leading to the synthesis of a truncated and nonfunctional β-globin protein. This is a common mechanism in some forms of β-thalassemia.

Question 7

1. Formation of the peptide bond
2. Translocation of the ribosome along the mRNA
3. Binding of aminoacyl-tRNA to the A site (correct answer)
4. Assembly of the 70S initiation complex

Explanation: Tetracycline antibiotics, including doxycycline, work by binding to the 30S ribosomal subunit of prokaryotes. This binding sterically hinders the attachment of aminoacyl-tRNA to the acceptor (A) site of the ribosome, thereby preventing the addition of new amino acids to the growing polypeptide chain and halting protein synthesis. Formation of the peptide bond is inhibited by chloramphenicol. Translocation is inhibited by macrolides and clindamycin.

Question 8

1. A poly(A) tail
2. A mannose-6-phosphate group
3. A polyubiquitin chain (correct answer)
4. A signal peptide sequence

Explanation: The proteasome is a large protein complex responsible for degrading damaged or unneeded proteins. The primary signal that targets a protein for proteasomal degradation is the covalent attachment of a polyubiquitin chain. Proteasome inhibitors block this degradation pathway, causing toxic accumulation of these tagged proteins and triggering apoptosis, an effect that is particularly potent in rapidly dividing cancer cells like myeloma cells.

Question 9

1. The lac repressor protein remains bound to the operator
2. Catabolite activator protein (CAP) does not bind to the promoter region (correct answer)
3. RNA polymerase cannot bind to the operator
4. Allolactose cannot bind to the repressor protein

Explanation: High levels of transcription from the lac operon require two conditions: the presence of lactose (to inactivate the repressor) and the absence of glucose. Glucose presence leads to low cAMP. Cyclic AMP binds to the catabolite activator protein (CAP), and this cAMP-CAP complex binds to the promoter region of the lac operon, greatly enhancing the affinity of RNA polymerase for the promoter. When cAMP is low, CAP is inactive and does not bind, resulting in low levels of transcription even if lactose is present and the repressor is inactive. This mechanism is called catabolite repression.

Question 10

1. mRNA
2. The 60S ribosomal subunit
3. Initiator tRNA (Met-tRNAi) (correct answer)
4. Poly(A)-binding protein

Explanation: During eukaryotic translation initiation, the initiator tRNA (Met-tRNAi) is brought to the 40S ribosomal subunit by eukaryotic initiation factor 2 (eIF2). eIF2 is a G-protein that is active when bound to GTP. The ternary complex of eIF2-GTP-Met-tRNAi binds to the 40S subunit to form the preinitiation complex. The binding of mRNA is mediated by the eIF4F complex. The joining of the 60S subunit occurs later and requires GTP hydrolysis by eIF5B.

Question 11

1. Alternative splicing (correct answer)
2. Gene amplification
3. Promoter activity
4. Post-translational modification

Explanation: Alternative splicing is a regulated process during gene expression that results in a single gene coding for multiple proteins. In this process, particular exons of a gene may be included within or excluded from the final, processed messenger RNA (mRNA). The production of two different mRNA molecules (one with exon 3 and one without) from a single pre-mRNA transcript is a classic example of alternative splicing. In this case, a mutation likely altered a splice site, causing exon 3 to be skipped during processing in a portion of the transcripts.

Question 12

1. Prokaryotic ribosomes (correct answer)
2. Eukaryotic cytoplasmic ribosomes
3. Viral ribosomes
4. Ribosomes of the rough endoplasmic reticulum

Explanation: According to the endosymbiotic theory, mitochondria evolved from aerobic prokaryotes that were engulfed by ancestral eukaryotic cells. As a result, mitochondria retain many prokaryotic features, including their own circular DNA and ribosomes (70S, composed of 50S and 30S subunits) that are structurally similar to bacterial ribosomes. Aminoglycosides target the bacterial 30S subunit. Their toxicity in humans is partly due to their ability to also bind to the mitochondrial 30S subunit, disrupting mitochondrial protein synthesis and leading to cellular dysfunction, especially in energy-intensive tissues like the inner ear and kidney.

Question 13

1. Hypermethylation of the promoter region (correct answer)
2. Creation of a premature stop codon
3. Histone hyperacetylation
4. Impaired polyadenylation of mRNA

Explanation: This patient has Fragile X syndrome, which is caused by a large expansion of CGG repeats in the promoter region of the FMR1 gene. This expansion leads to excessive DNA methylation (hypermethylation) of the promoter's CpG islands. DNA methylation is an epigenetic modification that silences gene expression by recruiting proteins that condense chromatin, making the gene inaccessible for transcription. Histone hyperacetylation is associated with gene activation. The other options describe different mechanisms of mutation.

Question 14

1. Formation of a stable triple helix (correct answer)
2. Secretion of procollagen from the cell
3. Cleavage of N- and C-terminal propeptides
4. Cross-linking of tropocollagen molecules

Explanation: The hydroxylation of specific proline and lysine residues occurs within the rough endoplasmic reticulum. The resulting hydroxyproline and hydroxylysine residues are essential for the formation of hydrogen bonds that stabilize the triple-helical structure of procollagen. Without this modification (e.g., in vitamin C deficiency or certain genetic defects), the triple helix is unstable and is degraded within the cell, leading to impaired collagen synthesis. Cross-linking occurs later in the extracellular space and is mediated by lysyl oxidase.

Question 15

1. Increased recruitment of RNA polymerase II to target gene promoters (correct answer)
2. Inhibition of ribosome assembly on target mRNAs
3. Methylation of CpG islands in target genes
4. Decreased stability of target gene mRNA transcripts

Explanation: p53 is a transcription factor, a protein that controls the rate of transcription of genetic information from DNA to messenger RNA. When p53 binds to its specific DNA response elements in the regulatory regions of its target genes (like p21), it recruits co-activator proteins and general transcription factors. This complex then facilitates the binding and activity of RNA polymerase II at the promoter, leading to increased transcription of the target gene. The other options describe post-transcriptional or epigenetic silencing mechanisms.

Question 16

1. RNA polymerase I
2. RNA polymerase II (correct answer)
3. RNA polymerase III
4. DNA polymerase III

Explanation: The toxin described is α-amanitin, found in Amanita phalloides mushrooms. It is a potent inhibitor of eukaryotic RNA polymerase II, the enzyme responsible for synthesizing mRNA. Inhibition of mRNA synthesis halts the production of new proteins, leading to cell death, particularly in metabolically active cells like hepatocytes. RNA polymerase I (synthesizes most rRNA) and RNA polymerase III (synthesizes tRNA and 5S rRNA) are much less sensitive to α-amanitin. DNA polymerase III is a prokaryotic enzyme involved in DNA replication.

Question 17

1. Promoters
2. Enhancers (correct answer)
3. Operators
4. Shine-Dalgarno sequences

Explanation: Steroid hormone receptors, such as the androgen receptor, are ligand-activated transcription factors. Upon binding their hormone, they translocate to the nucleus and bind to specific DNA sequences called hormone response elements (HREs). These HREs function as enhancers or silencers to modulate the transcription of target genes. Promoters are regions where RNA polymerase binds. Operators and Shine-Dalgarno sequences are features of prokaryotic gene regulation and translation, respectively.

Question 18

1. Elongation factor 2 (eEF-2) (correct answer)
2. Initiation factor 4E (eIF-4E)
3. The 60S ribosomal subunit
4. Aminoacyl-tRNA synthetase

Explanation: Diphtheria toxin catalyzes the ADP-ribosylation of eukaryotic elongation factor 2 (eEF-2). This modification inactivates eEF-2, which is required for the translocation step of polypeptide chain elongation. By blocking this essential step, the toxin halts protein synthesis, leading to cell death and the characteristic necrosis seen in diphtheria.

Question 19

1. Translation will not initiate, and no protein will be made.
2. Transcription will be significantly reduced.
3. The mRNA will be rapidly degraded in the cytoplasm.
4. Translation will proceed normally. (correct answer)

Explanation: The Shine-Dalgarno sequence is a ribosomal binding site in prokaryotic mRNA, generally located around 8 bases upstream of the start codon AUG. It is essential for the initiation of translation in bacteria. Eukaryotic cells use a different mechanism for translation initiation, where the 40S ribosomal subunit typically binds to the 5' cap of the mRNA and scans downstream until it encounters the first AUG codon (Kozak sequence context). Therefore, the absence of a Shine-Dalgarno sequence will have no effect on translation in a human (eukaryotic) cell line.

Question 20

1. Splicing of the final intron
2. Binding of the ribosome for translation initiation
3. Addition of the poly(A) tail (correct answer)
4. Termination of transcription by RNA polymerase II

Explanation: The sequence AAUAAA is the consensus polyadenylation signal sequence in eukaryotes. It is recognized by an endonuclease and polyadenylate polymerase, which cleave the pre-mRNA and add a long chain of adenine nucleotides (the poly(A) tail) to the 3' end. This poly(A) tail is crucial for protecting the mRNA from degradation and facilitating its export from the nucleus and translation in the cytoplasm. Disruption of this signal impairs polyadenylation, leading to mRNA instability and lower cytoplasmic levels.