Lecture 1 Flashcards
(1) Which statement best explains why EGFR dimerization increases its tyrosine kinase activity?
A. Dimerization fosters trans‐phosphorylation events in which one kinase domain phosphorylates the other’s activation loop.
B. Dimerization prevents any conformational changes in the extracellular domain.
C. Dimerization blocks ATP binding to the intracellular domain.
D. Dimerization removes the receptor from the membrane surface.
Correct Answer: A
Reasoning: In the active dimeric form, each EGFR kinase domain phosphorylates tyrosine residues on its partner, thereby stabilizing the active conformation.
(2) In the context of membrane proteins, which of the following is least likely to be found in the transmembrane portion of EGFR?
A. Nonpolar amino acid side chains like isoleucine
B. Polar backbone atoms forming α‐helical hydrogen bonds
C. Charged side chains (e.g., arginine) in direct contact with lipid tails
D. Van der Waals interactions among neighboring transmembrane helices
Correct Answer: C
Reasoning: Strongly charged residues are usually unfavorable in the membrane’s nonpolar interior unless specifically stabilized by complementary charges or water channels. Nonpolar side chains and backbone hydrogen bonding are much more common in transmembrane segments.
Which of the following best describes the role of the extracellular domain in EGFR activation?
A. It inhibits ligand binding by blocking the dimerization site.
B. It binds EGF, undergoes a conformational change, and enables dimerization with another EGFR monomer.
C. It has no effect on dimerization but controls ATP hydrolysis in the cytoplasm.
D. It releases phosphate groups to activate downstream signals.
Correct Answer: B
(Rationale: The extracellular domain of EGFR binds EGF, then rearranges to allow two monomers to form an active dimer.)
Once EGFR dimerizes, why do the intracellular tyrosine kinase domains become fully activated?
A. They acquire additional sugars from the Golgi that switch them “on.”
B. They are brought close enough to cross‐phosphorylate each other’s tails, creating phosphotyrosines that recruit downstream proteins.
C. They disassociate from each other, forming separate active kinases.
D. They only function when EGF is hydrolyzed to ADP.
Correct Answer: B
(Rationale: Dimerization physically positions each kinase domain to add phosphate groups to specific tyrosines on the partner’s intracellular tail, turning on the signaling cascade.)
Which statement best describes how ions (e.g., Na⁺, K⁺) move through an ion channel in the cell membrane?
A. They use ATP directly to cross the lipid bilayer.
B. They passively diffuse down their electrochemical gradient through a specialized pore, without directly using ATP.
C. They move against their gradient, requiring no external energy source.
D. They form covalent bonds with the channel protein as they pass.
Correct Answer: B
(Rationale: Ion channels typically allow ions to flow down their concentration/voltage gradient passively. They do not covalently bind ions nor directly consume ATP for this process.)
Question 2
Why are ion channels often highly specific for particular ions like Na⁺ or K⁺?
A. Ion channels form arbitrary pores that let any molecule pass.
B. The selectivity filter in the channel stabilizes certain ions via specific size and coordination interactions.
C. Channels only function if a ligand is hydrolyzed to ADP.
D. Channels have rigid binding pockets that break all hydrogen bonds with water.
Correct Answer: B
(Rationale: The channel’s selectivity filter has precisely arranged amino acid residues that favor certain ionic sizes and charges, enabling selective ion transport.)
A nucleotide is distinguished from a nucleoside primarily by the presence of which group?
A. A nitrogenous base (adenine, thymine, etc.)
B. A sugar (ribose or deoxyribose)
C. One or more phosphate groups attached to the sugar
D. A 3′ hydroxyl group on the sugar ring
Answer: C
Explanation: A nucleoside is sugar + base. When you add a phosphate to the sugar’s hydroxyl group, you get a nucleotide.
When phosphoric acid is esterified to the 5′‐hydroxyl of a nucleoside’s sugar, the resulting product:
A. Is incapable of forming phosphodiester bonds in DNA.
B. Becomes a nucleotide, the basic building block of DNA and RNA.
C. Immediately forms the double‐helix structure.
D. Is called a “peptide bond.”
Answer: B
Explanation: Attaching phosphate (esterification) to the sugar—OH group of a nucleoside converts it into a nucleotide, which can then be polymerized into nucleic acids.
Which functional group defines whether the sugar in a nucleotide is ribose (RNA) or deoxyribose (DNA)?
A. The presence of a 2′‐OH vs. 2′‐H on the sugar ring.
B. Having a purine instead of a pyrimidine base.
C. A methyl group on the phosphate.
D. A disulfide bond linking the sugar to the base.
Correct Answer: A
(Explanation: Ribose has a hydroxyl at the 2′ carbon, deoxyribose has a hydrogen instead.)
Question 2
Why is adenine considered a purine while thymine is considered a pyrimidine?
A. Adenine has a single 6‐membered ring, while thymine has two fused rings.
B. Adenine is found only in RNA, thymine only in DNA.
C. Purines (like adenine) have two rings, whereas pyrimidines (like thymine) have one ring.
D. Purines lack nitrogen, whereas pyrimidines contain nitrogen.
Correct Answer: C
(Explanation: Purines are two‐ring nitrogenous bases; pyrimidines have a single ring.)
Which functional group on the sugar of one nucleotide reacts with the phosphate on the next nucleotide to form the phosphodiester bond?
A. 3′‐OH group
B. 2′‐OH group
C. 1′ carbon
D. Base’s amine group
Answer: A
(Explanation: The 3′ hydroxyl is the key site that forms an ester linkage with the 5′ phosphate, resulting in the phosphodiester bond.)
Why is DNA referred to as a polymer with an alternating sugar‐phosphate backbone?
A. Each sugar is randomly attached to different functional groups with no pattern.
B. Phosphodiester bonds link a phosphate to the 3′ carbon of one sugar and the 5′ carbon of the next, creating a repeating sugar‐phosphate sequence.
C. The bases alternate instead, forming the backbone.
D. DNA lacks phosphate and is formed exclusively by hydrogen bonding between sugars.
Answer: B
(Explanation: The polymer structure arises from repeated sugar‐phosphate links along the chain, with the bases branching off the sugar.)
Which statement best explains why A pairs with T and G pairs with C in the DNA double helix?
A. A–A and C–T pairs form stronger bonds but are less common in the cell.
B. A–T and G–C pairs each combine a purine with a pyrimidine, ensuring the correct helix diameter and optimal hydrogen bonding.
C. Base pairing is random; A–T is just more abundant by chance.
D. Because the sugar backbone enforces covalent links between opposite bases.
Answer: B
(Rationale: A pairs with T and G pairs with C to match a purine with a pyrimidine, giving consistent base‐pair geometry and strong hydrogen bonding.)
Why are the two DNA strands said to be “complementary”?
A. They contain identical sequences going in the same direction.
B. They have matching base pairs: each position on one strand is paired with the corresponding complementary base on the other.
C. The sugar‐phosphate backbones have no negative charges.
D. They only pair up at the 5′ ends.
Answer: B
(Rationale: “Complementary” means if one strand has A, the other has T at the same rung; if one has G, the other has C, etc. This is the basis for the double helix structure.)
Which form of DNA is generally most common under normal physiological conditions?
A. A‐DNA
B. B‐DNA
C. Z‐DNA
D. All forms are equally common in the cell
Answer: B
(Rationale: B‐DNA is the standard “Watson–Crick” form found under typical cellular conditions.)
How does Z‐DNA differ from B‐DNA?
A. Z‐DNA is right‐handed, with 10 base pairs per turn, whereas B‐DNA is left‐handed.
B. Z‐DNA is left‐handed and has a thinner diameter (∼1.8 nm) plus ~12 base pairs/turn, whereas B‐DNA is right‐handed, ~2.0 nm in diameter, with ~10.5 base pairs/turn.
C. B‐DNA has no major groove, whereas Z‐DNA does.
D. They differ only in having different base sequences but the same helical geometry.
Answer: B
(Rationale: Z‐DNA is left‐handed, typically has 12 bp/turn, ~1.8 nm diameter, in contrast to B‐DNA’s right‐handed ~10.5 bp/turn, ~2.0 nm diameter.)
Which type of RNA is primarily responsible for carrying the genetic message from DNA to the ribosome?
A. mRNA
B. tRNA
C. rRNA
D. siRNA
Correct Answer: A
(Explanation: mRNA or “messenger” RNA is the copy of a gene from DNA that directs protein synthesis at the ribosome.)
Which RNA can catalyze biochemical reactions, acting as an enzyme?
A. tRNA
B. mRNA
C. Ribozyme
D. rRNA
Correct Answer: C
(Explanation: Ribozyme refers to RNA with enzymatic (catalytic) activity, like self‐splicing introns or peptidyl transferase activity in the ribosome.)
Which statement best describes why RNA often has loops and bulges in its structure?
A. Because it is single‐stranded, it can form partial double‐helical stems plus regions of unpaired or mismatched bases, creating loops and bulges.
B. Because it never hydrogen‐bonds, it stays linear.
C. Because all bases are identical in RNA, it cannot fold.
D. Because the sugar is deoxyribose, it forms a stable double helix with no loops.
Correct Answer: A
(Rationale: RNA typically folds back on itself, forming base‐paired stems interspersed with mismatches/unpaired bases, leading to loops/bulges.)
Why can RNA adopt complex tertiary structures much like proteins?
A. RNA is entirely double‐stranded, so it forms only a simple helix.
B. The sugar‐phosphate backbone has fewer negative charges, so it folds easily.
C. Internal base pairing (even imperfect) plus unpaired regions create intricate shapes, analogous to protein folding.
D. RNA is always a straight chain, so it has no 3D shape.
Correct Answer: C
(Rationale: Base‐paired stems and unpaired loops give rise to distinct 3D folds in RNA, akin to how proteins fold into complex tertiary shapes.)
