MCAT - Bio/Biochem Foundations:: Molecular Biology Flashcards (50 Cards) — Study Online

Curriculum Overview

Comprehensive, high-yield MCAT study deck focusing on Molecular Biology. Features 50 rigorous, curriculum-aligned flashcards designed for advanced-level mastery. Core concepts covered include Molecular Biology, key problem-solving heuristics, foundational formulas, and exam-tested application scenarios. Ideal for active recall review, spaced repetition study, and scoring in the top percentile.

Topics & Key Concepts

Each MCAT Only Seal Unwind Introns Okazaki Ribosomes Synthesize Telomerase

Sample Flashcard Questions & Answers

Showing 8 of 50 cards

Question #1 Active Recall

DNA replication is described as semiconservative, meaning that each newly synthesized double helix consists of:

- **A)** Two entirely newly synthesized strands, with both original parental strands discarded
- **B)** One original (parental) strand and one newly synthesized (daughter) strand, so each of the two resulting double helices retains exactly one strand from the original parent molecule
- **C)** Fragments randomly mixed from multiple different original DNA molecules
- **D)** RNA in place of one of the two DNA strands, permanently

Answer & Explanation:

**Answer: B)**

The semiconservative model (confirmed by the classic Meselson-Stahl experiment using isotope labeling) shows that each daughter DNA molecule retains one intact original parental strand serving as a template, paired with one newly synthesized complementary strand - conserving half of each original molecule in every subsequent generation of replication.

Question #2 Active Recall

DNA helicase, an essential enzyme in the replication process, functions to:

- **A)** Unwind and separate the two parental DNA strands by breaking the hydrogen bonds between complementary base pairs, creating the replication fork and single-stranded template regions needed for synthesis
- **B)** Synthesize the new complementary DNA strand directly
- **C)** Seal nicks in the sugar-phosphate backbone of a completed strand
- **D)** Remove RNA primers after replication

Answer & Explanation:

**Answer: A)**

Helicase uses ATP hydrolysis to unwind the DNA double helix ahead of the replication machinery, breaking the hydrogen bonds holding the two parental strands together and creating the Y-shaped replication fork with exposed single-stranded template DNA that other replication enzymes can then act upon.

Question #3 Active Recall

DNA polymerase, the central enzyme actually synthesizing new DNA strands, has an absolute requirement for:

- **A)** A separate DNA template but no primer at all
- **B)** RNA nucleotides exclusively as its substrate, never DNA nucleotides
- **C)** A pre-existing primer (a short segment with a free 3'-OH group) to which it can add new nucleotides - DNA polymerase cannot initiate synthesis of a completely new strand from scratch, but can only extend an existing primer strand, and it synthesizes new DNA exclusively in the 5' to 3' direction
- **D)** A completely single-stranded, primer-free template, with synthesis proceeding equally well in either the 5' to 3' or 3' to 5' direction

Answer & Explanation:

**Answer: C)**

DNA polymerase can only add new nucleotides onto an existing free 3'-OH end (extending a primer), never initiating synthesis de novo on a bare template - this is why RNA primase must first lay down a short RNA primer, and why DNA polymerase synthesizes exclusively in the 5' to 3' direction (adding nucleotides to the growing 3' end).

Question #4 Active Recall

Because DNA polymerase synthesizes only in the 5' to 3' direction, and the two parental template strands are antiparallel, DNA replication at a single replication fork proceeds:

- **A)** Identically and continuously on both strands, with no structural difference between them
- **B)** In the 3' to 5' direction on both strands simultaneously
- **C)** Only on one of the two strands, with the other strand never being replicated at all
- **D)** Continuously on the leading strand (synthesized smoothly in the same direction as the replication fork is moving) but discontinuously on the lagging strand (synthesized in short Okazaki fragments, each still individually made 5' to 3', but overall moving away from/opposite to the replication fork's direction of movement, requiring the fragments to later be joined together)

Answer & Explanation:

**Answer: D)**

Because the two template strands run antiparallel but DNA polymerase only synthesizes 5' to 3', one new strand (the leading strand) can be synthesized continuously in the same direction the replication fork opens, while the other new strand (the lagging strand) must be synthesized discontinuously, in short Okazaki fragments, each primed separately and later joined together - an elegant solution to the directionality constraint.

Question #5 Active Recall

DNA ligase, an essential enzyme completing the lagging strand synthesis process, functions to:

- **A)** Unwind the DNA double helix ahead of the replication fork, identical to helicase
- **B)** Seal the remaining nicks (gaps in the sugar-phosphate backbone) between adjacent Okazaki fragments after their RNA primers have been removed and replaced with DNA, joining the fragments into one continuous, covalently intact lagging strand
- **C)** Synthesize the RNA primers needed to initiate each Okazaki fragment
- **D)** Proofread and remove misincorporated nucleotides

Answer & Explanation:

**Answer: B)**

After RNA primers are removed and the resulting gaps filled with DNA (typically by a specialized DNA polymerase), a nick remains in the sugar-phosphate backbone between adjacent Okazaki fragments; DNA ligase catalyzes formation of the final phosphodiester bond sealing this nick, producing one continuous, intact lagging strand.

Question #6 Active Recall

Telomerase, an enzyme with reverse transcriptase activity, addresses which specific structural problem inherent to linear chromosome replication?

- **A)** The need to unwind the DNA double helix
- **B)** The need to correct base-pairing errors during replication
- **C)** The 'end replication problem' - because DNA polymerase requires a primer and synthesizes only 5' to 3', the very end of the lagging strand template cannot be fully replicated after the final RNA primer is removed, causing chromosomes to progressively shorten with each round of replication; telomerase extends the repetitive telomeric DNA sequences at chromosome ends, helping counteract (though not always completely preventing) this progressive shortening
- **D)** Preventing chromosomal translocations between nonhomologous chromosomes

Answer & Explanation:

**Answer: C)**

Linear chromosomes face an inherent 'end replication problem': after the terminal RNA primer on the lagging strand is removed, there's no way for DNA polymerase to fill that final gap (no upstream primer to extend from), causing progressive chromosome shortening each replication cycle. Telomerase (using an internal RNA template and reverse transcriptase activity) extends repetitive, non-coding telomeric sequences at chromosome ends, helping offset (though not fully eliminating, in most somatic cells) this shortening - telomerase activity is notably high in germ cells and many cancer cells, but low/absent in most normal somatic cells, contributing to cellular replicative aging (the Hayflick limit).

Question #7 Active Recall

Transcription, the synthesis of an RNA molecule from a DNA template, is catalyzed primarily by:

- **A)** RNA polymerase, which synthesizes a complementary RNA strand using one of the two DNA strands (the template/antisense strand) as its guide, without requiring a separate primer (unlike DNA polymerase)
- **B)** DNA polymerase, identical to DNA replication
- **C)** Ribosomes directly
- **D)** Telomerase

Answer & Explanation:

**Answer: A)**

RNA polymerase synthesizes RNA complementary to the DNA template strand, reading 3' to 5' along that template and synthesizing the new RNA 5' to 3' - notably, unlike DNA polymerase, RNA polymerase can initiate synthesis de novo without requiring a pre-existing primer.

Question #8 Active Recall

The promoter region of a gene functions to:

- **A)** Encode the actual amino acid sequence of the resulting protein
- **B)** Terminate transcription at the end of a gene
- **C)** Serve as the site of intron removal during RNA processing
- **D)** Serve as the specific DNA sequence where RNA polymerase (often assisted by additional transcription factor proteins) binds to initiate transcription, positioned just upstream of a gene's coding sequence and helping determine both where and how efficiently that gene is transcribed

Answer & Explanation:

**Answer: D)**

The promoter is a regulatory DNA sequence upstream of a gene's transcribed region, providing the binding site for RNA polymerase (and, in eukaryotes, an array of general and gene-specific transcription factors) to properly position and initiate transcription - promoter sequence/strength substantially influences how actively a given gene is transcribed.

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