Chap 13 Problem Solutions

What are the three essential functional components of a standard bacterial transcription unit?

A) Origin, replication fork, and termination sequence

B) Promoter, RNA-coding region, and terminator

C) Core promoter, enhancer, and poly-A tail

D) Operator, Shine-Dalgarno sequence, and stop codon

Promoter, RNA-coding region, and terminator. A transcription unit consists of a promoter (specifies where transcription starts and which strand is read), an RNA-coding region (the transcribed sequence), and a terminator (signals the end of transcription).

How are ribonucleotides added to a growing RNA transcript during elongation?

A) Added to the 5'-phosphate group using ATP hydrolysis

B) Added to the 3'-OH group by cleaving two phosphates from incoming rNTPs

C) Added at both 5' and 3' ends simultaneously

D) Linked directly to DNA bases via peptide bonds

Added to the 3'-OH group by cleaving two phosphates from incoming rNTPs. RNA polymerase attaches the 5'-phosphate of an incoming ribonucleoside triphosphate (rNTP) to the free 3'-OH of the growing chain, releasing pyrophosphate (PPi) to drive phosphodiester bond formation.

Which protein subunit distinguishes the bacterial RNA polymerase holoenzyme from the core enzyme?

A) Alpha subunit

B) Beta prime subunit

C) Sigma factor

D) Rho protein

Sigma factor. Bacterial core enzyme consists of five subunits (two alpha, one beta, one beta-prime, and one omega). The addition of the sigma factor converts the core enzyme into the holoenzyme.

What distinct functional roles do the bacterial RNA polymerase holoenzyme and core enzyme play?

A) Core enzyme binds the promoter; holoenzyme completes termination

B) Holoenzyme recognizes the promoter and initiates transcription; core enzyme carries out elongation after sigma release

C) Holoenzyme synthesizes RNA primers; core enzyme synthesizes DNA

D) Holoenzyme terminates transcription; core enzyme initiates it

Holoenzyme recognizes the promoter and initiates transcription; core enzyme carries out elongation after sigma release. Sigma factor within the holoenzyme confers promoter specificity. Once initiation occurs and RNA synthesis begins, sigma is released and the core enzyme carries out elongation.

Which sequence correctly outlines the three stages of bacterial transcription?

A) Origin unwinding, leading strand synthesis, termination

B) Holoenzyme assembly, sigma binding, promoter degradation

C) Initiation (promoter binding and start of synthesis), Elongation (5' to 3' synthesis as DNA unwinds), Termination (separation of RNA and polymerase)

D) Primer annealing, strand extension, nick translation

Initiation (promoter binding and start of synthesis), Elongation (5' to 3' synthesis as DNA unwinds), Termination (separation of RNA and polymerase). Transcription proceeds through Initiation (promoter recognition/bubble formation), Elongation (reading template 3' to 5' while making RNA 5' to 3'), and Termination (release of RNA and enzyme).

Where are the two critical consensus sequences located in a bacterial promoter, and why are they essential?

A) Centered at +1 and +10; required for ribosome binding

B) Centered at -10 and -35; required for RNA polymerase holoenzyme binding and initiation

C) Centered at -50 and -100; required for DNA gyrase binding

D) Centered at the termination site; required for hairpin formation

Centered at -10 and -35; required for RNA polymerase holoenzyme binding and initiation. The -10 (Pribnow box) and -35 consensus sequences are recognized by the sigma subunit of the holoenzyme to properly position RNA polymerase and initiate unwinding.

How does rho-independent transcription termination operate in bacterial cells?

A) Rho protein binds a Rut site and catches up to paused polymerase

B) An inverted repeat forms an RNA hairpin, followed by a string of uracils that breaks weak A-U hybrid base pairs

C) A stop codon stalls the polymerase

D) DNA ligase seals the termination loop to eject the transcript

An inverted repeat forms an RNA hairpin, followed by a string of uracils that breaks weak A-U hybrid base pairs. Transcribing the inverted repeats forms a stable hairpin structure that causes RNA polymerase to pause, destabilizing the weak A-U bonds between the DNA poly-A run and the RNA poly-U tract, releasing the transcript.

How does rho-dependent transcription termination function in bacterial cells?

A) Rho binds directly to the promoter to halt initiation

B) Rho binds to the Rut site on nascent RNA, tracks toward paused polymerase at a hairpin, and unwinds the RNA-DNA hybrid

C) Rho cleaves the nascent RNA at the 5' cap

D) Rho replaces the sigma factor during elongation

Rho binds to the Rut site on nascent RNA, tracks toward paused polymerase at a hairpin, and unwinds the RNA-DNA hybrid. Rho binds the RNA at an upstream Rut (rho utilization) site and migrates toward the 3' end. When RNA polymerase pauses at a hairpin, rho catches up and uses its helicase activity to unwind the RNA-DNA duplex.

Which of the following characteristics is UNIQUE to transcription compared to DNA replication?

A) Synthesizes nucleic acid chains in the 5' to 3' direction

B) Utilizes a DNA template strand

C) Initiation does not require a pre-existing primer

D) Uses complementary base-pairing rules

Initiation does not require a pre-existing primer. Unlike DNA polymerases, RNA polymerases can initiate synthesis de novo without a primer; transcription is also unidirectional for a single gene rather than replicating both strands of the entire genome.

In eukaryotic transcription, why does RNA polymerase II bind specifically to the core promoter rather than the regulatory promoter?

A) The core promoter contains enhancers that activate transcription

B) The core promoter is immediately adjacent to the +1 transcription start site

C) The regulatory promoter degrades RNA polymerase

D) Eukaryotic RNA polymerase binds only to the poly-A tail

The core promoter is immediately adjacent to the +1 transcription start site. The core promoter contains elements (such as the TATA box) immediately upstream of +1 that correctly position RNA polymerase II to start transcribing at the proper nucleotide.

An RNA molecule has 23% A, 42% U, 21% C, and 14% G. Is this RNA single- or double-stranded?

A) Double-stranded because it contains all four bases

B) Double-stranded because %A + %U equals %G + %C

C) Single-stranded because %A does not equal %U and %G does not equal %C

D) Single-stranded because it lacks thymine

Single-stranded because %A does not equal %U and %G does not equal %C. A complementary double-stranded duplex must have equal percentages of pairing bases (%A = %U and %G = %C). The unequal proportions demonstrate it is single-stranded.

An RNA molecule has 23% A, 42% U, 21% C, and 14% G. What are the base percentages in the template DNA strand?

A) A = 23%, T = 42%, C = 21%, G = 14%

B) A = 42%, T = 23%, C = 14%, G = 21%

C) A = 14%, T = 21%, C = 42%, G = 23%

D) A = 21%, T = 14%, C = 23%, G = 42%

A = 42%, T = 23%, C = 14%, G = 21%. Complementary base pairing dictates that RNA U (42%) pairs with DNA A (42%), RNA A (23%) pairs with DNA T (23%), RNA G (14%) pairs with DNA C (14%), and RNA C (21%) pairs with DNA G (21%).

A template DNA strand has the sequence 3'-TATCCGCTACGGT-5'. What is the transcribed RNA sequence?

A) 5'-TATCCGCTACGGT-3'

B) 3'-AUAGGCGAUGCCA-5'

C) 5'-AUAGGCGAUGCCA-3'

D) 5'-TGCCAUCGGCUAU-3'

5'-AUAGGCGAUGCCA-3'. RNA is antiparallel and complementary to the template strand, substituting uracil for thymine: 3'-T-A-T-C-C-G-C-T-A-C-G-G-T-5' transcribes into 5'-A-U-A-G-G-C-G-A-U-G-C-C-A-3'.

If RNA polymerase moves from left to right along a template DNA strand, which end of the template is at the left?

A) 5' end

B) 3' end

C) The amino terminus

D) Either end depending on the promoter

3' end. RNA polymerase synthesizes RNA in the 5' to 3' direction and reads the template strand in the 3' to 5' direction. Moving left to right means the template must start at the 3' end on the left and end at the 5' end on the right.

If RNA polymerase moves left to right along the template 3'-ATTGCCAGATCATCCCAATAGAT-5', what RNA sequence is produced?

A) 5'-TAACGGTCTAGTAGGGTTATCTA-3'

B) 5'-UAACGGUCUAGUAGGGUUAUCUA-3'

C) 3'-UAACGGUCUAGUAGGGUUAUCUA-5'

D) 5'-AUUGCCAGAUCAUCCCAAUAGAU-3'

5'-UAACGGUCUAGUAGGGUUAUCUA-3'. Reading 3' to 5' from left to right produces a 5' to 3' complementary RNA strand where A pairs with U, T pairs with A, G pairs with C, and C pairs with G.

Why do DNA polymerases synthesize nucleic acids at much faster rates than RNA polymerases?

A) RNA nucleotides are heavier than DNA nucleotides

B) DNA polymerases replicate entire large genomes, whereas RNA polymerases transcribe individual genes

C) RNA polymerase must synthesize both strands simultaneously

D) RNA polymerase lacks catalytic active sites

DNA polymerases replicate entire large genomes, whereas RNA polymerases transcribe individual genes. Cells must replicate millions or billions of base pairs across the whole genome within a strict cell-cycle window, requiring extreme speed, whereas transcription targets small, discrete gene intervals as needed.

What is the most probable consequence of a single-base substitution in the -10 consensus sequence of a bacterial promoter?

A) Premature transcription termination

B) Inability of DNA ligase to seal nicks

C) Decreased transcription rate due to reduced RNA polymerase holoenzyme binding affinity

D) Alteration of the downstream amino acid sequence without affecting transcription rate

Decreased transcription rate due to reduced RNA polymerase holoenzyme binding affinity. The -10 region (Pribnow box) is a core recognition site for the sigma subunit; mutating it impairs promoter recognition and unwinding, lowering the frequency of transcription initiation.

What is the most likely outcome of a point mutation at the -35 consensus sequence of a bacterial promoter?

A) Termination of transcription fails

B) A reduced rate of transcription due to impaired promoter recognition by holoenzyme

C) A shift in the reading frame of the protein

D) Increased binding of rho factor

A reduced rate of transcription due to impaired promoter recognition by holoenzyme. The -35 sequence is critical for initial recognition by the sigma subunit of RNA polymerase holoenzyme; mutations here reduce binding efficiency and lower the initiation rate.

What is the most likely effect of a single base mutation at position -20 in a bacterial promoter?

A) Complete loss of transcription initiation

B) No significant effect on transcription

C) A change in the first nucleotide of the RNA transcript

D) Immediate rho-dependent termination

No significant effect on transcription. Position -20 lies in the spacer region between the -35 and -10 consensus sequences. While the distance between the two consensus sequences is critical, the specific sequence of the spacer is generally not important for polymerase binding.

What is the most likely consequence of a mutation at the +1 transcription start site of a bacterial gene?

A) RNA polymerase cannot bind to the promoter

B) Transcription rate drops to zero

C) Transcription rate is unchanged, but the first base of the RNA transcript is altered

D) Rho factor can no longer terminate the transcript

Transcription rate is unchanged, but the first base of the RNA transcript is altered. The +1 position is determined by spacing relative to the -10 and -35 consensus sequences. A change at +1 does not prevent polymerase binding or initiation, but it alters the initial nucleotide incorporated into the RNA molecule.