Unit 1: Molecular and Cellular Biology

Answer: B — The correct answer is A. uORFs (upstream open reading frames) contain start codons (AUG) upstream of the actual protein-coding sequence. When ribosomes scan from the 5' cap and encounter the first AUG, they initiate translation of the uORF. After translating and terminating at the uORF's stop codon, the ribosome must reinitiate translation at a downstream AUG to translate the target protein. Not all ribosomes successfully reinitiate; many dissociate from the mRNA after uORF termination, resulting in reduced translation of the main protein coding sequence. This is a legitimate regulatory mechanism found in nature. Why the other options are incorrect: B) This is incorrect because uORFs do not prevent ribosome binding. The ribosomal scanning mechanism can still proceed through the 5' UTR, and while secondary structures can impede scanning, the question specifies that mRNA levels are normal and doesn't indicate structural obstruction. C) This is incorrect because uORFs do not sequester tRNAs in any meaningful way that would deplete the cellular tRNA pool. Translation of short uORFs uses relatively few tRNAs, and the cell maintains large pools of aminoacyl-tRNAs for protein synthesis. D) This is incorrect because uORFs do not inherently trigger nonsense-mediated decay (NMD). NMD is triggered by premature termination codons typically located more than 50-55 nucleotides upstream of an exon junction complex, and normal uORFs followed by normal reinitiation do not meet these criteria for NMD activation.

Answer: D — The correct answer is A. The original codon UAC codes for tyrosine (a sense codon), while UAA is a stop codon (nonsense codon). This mutation converts a sense codon to a stop codon, which causes premature termination of translation. The ribosome will cease protein synthesis when it encounters UAA, resulting in a truncated (shorter) protein that is likely to be nonfunctional. Option B is incorrect because UAA is not a codon that codes for an amino acid—it is a stop signal. Option C is backwards; the mutation converts a sense codon TO a stop codon, not from a stop codon to a sense codon, so the protein would be shorter, not longer. Option D is incorrect because a single nucleotide change in a codon would not significantly alter mRNA secondary structure in a way that would be the primary consequence of this mutation; the primary effect is the change in the genetic code itself, not mRNA folding.

Answer: C — Passive transport is the correct answer because it is the type of transport that occurs without the use of energy or transport proteins. The molecule is likely crossing the membrane through diffusion, which is a type of passive transport. Active transport (A) requires energy and transport proteins, endocytosis (B) involves the uptake of molecules into the cell through vesicles, and exocytosis (C) involves the release of molecules from the cell through vesicles. Both endocytosis and exocytosis require energy and are not types of passive transport.

Answer: D — Exocytosis is the correct answer because it is the process by which large, polar molecules such as proteins are transported from the cytosol to the outside of the cell. This process involves the fusion of vesicles containing the molecule with the cell membrane, releasing the molecule outside the cell. Simple diffusion (A) is incorrect because it is the passive movement of non-polar molecules across the cell membrane, which does not apply to large polar proteins. Facilitated diffusion (B) is also incorrect because it involves the movement of molecules down their concentration gradient with the help of transport proteins, but it does not allow for the transport of large polar molecules. Osmosis (D) is incorrect because it is the movement of water molecules across the cell membrane, not the transport of large polar proteins. The correct answer requires an understanding of the different transport mechanisms across the cell membrane and their specific functions.

Answer: C — The correct answer is A. Enzymes function by stabilizing the transition state and providing an alternative reaction pathway with a lower activation energy barrier. This means reactants require less energy to reach the transition state, allowing more molecules to have sufficient energy to react at a given temperature, which increases reaction rate. Option B is incorrect because enzymes do not increase the kinetic energy or temperature of substrate molecules; the reaction occurs at the same temperature. Option C is incorrect because enzymes are not permanently altered and do not permanently change substrates—they form temporary enzyme-substrate complexes. Option D is incorrect because enzymes do not change the equilibrium position (ΔG) of a reaction; they only affect the rate at which equilibrium is reached. This distinction between activation energy and free energy change is fundamental to understanding enzyme catalysis.