Week 28 / Transcription

Question 1

Q: What is transcription?
[what is it ?
what does it form?
]

Answer 1

A: Transcription is the enzymatic synthesis of RNA from a DNA template and

forms the first step in gene expression.

Question 2

Q: What is the product of transcription?

Answer 2

A: The product of transcription is messenger RNA (mRNA).

Question 3

Q: What is translation?

Answer 3

A: Translation is the enzymatic synthesis of protein from a transcribed gene sequence into a functional RNA molecule (mRNA).

Question 4

Q: What enzyme catalyzes transcription?

Answer 4

A: Transcription is catalyzed by an RNA polymerase enzyme complex.

Question 5

Q: What are the requirements for transcription? [2]

Answer 5

A: Transcription requires
a dsDNA template and ribonucleotides (ATP, GTP, CTP, and UTP).

Question 6

Q: In which direction does RNA synthesis occur?

Answer 6

A: RNA synthesis occurs in the 5’ to 3’ direction.

Question 7

Q: What are the two DNA strands used in transcription?

Answer 7

A:

Sense strand: Carries the gene sequence that will be copied into an RNA molecule for protein translation.

Antisense strand: Used as a template to generate a complementary RNA copy and also carries sequences for non-coding RNA molecules with structural or regulatory functions.

Question 8

Q: How many RNA polymerases are present in prokaryotes?

Answer 8

A: Prokaryotes have a single RNA polymerase.

Question 9

Q: How many RNA polymerases are present in eukaryotes?

Answer 9

A: Eukaryotes have three RNA polymerases: I, II, and III.

Question 10

Q: What is the function of RNA Polymerase II in eukaryotes?

Answer 10

A: RNA Polymerase II transcribes all protein-coding genes.

Question 11

Q: What is the function of RNA Polymerase I in eukaryotes?

Answer 11

A: RNA Polymerase I transcribes most ribosomal RNAs (rRNAs).

Question 12

Q: What is the function of RNA Polymerase III in eukaryotes?

Answer 12

A: RNA Polymerase III transcribes transfer RNAs (tRNAs).

Question 13

Q: What are the steps involved in transcription initiation? [5]

Answer 13

Binding – RNA polymerase binds to DNA.

Promoter Find – It recognizes the promoter (a special starting point).

Promoter Power – Promoters vary in strength (strong/weak) and can be regulated.

Unwind – DNA is unwound locally to expose the template.

Start – RNA polymerase starts transcription at the +1 site.

A:
RNA Polymerase Binding – RNA polymerase binds to the dsDNA.

Promoter Recognition – The polymerase binds to a promoter sequence upstream of the gene.

Promoter Strength & Regulation – Promoters can be weak or strong depending on sequence elements and protein stimulation.

DNA Unwinding – The dsDNA is locally unwound so RNA polymerase can access the template strand.

Start Site Selection – RNA polymerase begins RNA synthesis at the start site (position +1) of the gene.

Question 14

Q: What are the steps involved in transcription elongation? [4]

Answer 14

Add Nucleotides – RNA polymerase adds bases to the 3’ end of mRNA (grows 5’ → 3’).

Read Template – It reads the DNA template strand in the 3’ → 5’ direction.

Unwind & Rewind – DNA is unwound ahead and rewound behind the polymerase.

Speed – In E. coli, about 40 bases/second at 37°C.

A:

Nucleotide Addition – RNA polymerase covalently adds ribonucleotides to the 3’-end of the growing mRNA molecule in the 5’ to 3’ direction.

Template Strand Reading – The polymerase moves along the antisense/template DNA strand in the 3’ to 5’ direction.

DNA Unwinding & Rewinding – The polymerase locally unwinds the DNA ahead and reforms the helix behind it.

Elongation Speed – In E. coli, RNA polymerase synthesizes around 40 bases per second at 37°C.

Question 15

What is the termination step in transcription?

Answer 15

The polymerase dissociates from the DNA upon reaching a termination sequence.

The common stop signal is an RNA hairpin formed by self-complementary sequence areas in the mRNA.

The hairpin is GC-rich and stable, stalling the polymerase.

After the hairpin, a sequence of 4 or more uridine (U) residues weakly binds to the adenine (A) residues on the template strand, causing dissociation of the core enzyme.

Question 16

Q: How many subunits are there in E. coli RNA polymerase?

Answer 16

A: E. coli RNA polymerase consists of at least five known subunits.

Question 17

Q: What is the role of the α-subunits in E. coli RNA polymerase?

Answer 17

A: The two α-subunits are required for holoenzyme assembly.

Question 18

Q: What does the β-subunit do in E. coli RNA polymerase? [2]

Answer 18

A: The β-subunit is the catalytic center of the enzyme and is key for initiation and elongation.

Question 19

Q: What is the function of the β’-subunit in E. coli RNA polymerase?

Answer 19

A: The β’-subunit binds two Zn²⁺ ions to help catalyze the joining of ribonucleotides.

Question 20

Q: What is the function of the σ-subunit in E. coli RNA polymerase?

Answer 20

A: The σ-subunit is responsible for promoter recognition.

Question 21

Q: What is the role of the ω-subunit in E. coli RNA polymerase?

Answer 21

A: The ω-subunit stabilizes the assembled holoenzyme.

Question 22

Q: What determines which promoter is recognized by E. coli RNA polymerase?

Answer 22

A: Different σ factors/subunits recognize different promoters. The most common σ factor in E. coli is the σ⁷⁰ factor.

Question 23

Q: What does the E. coli RNA polymerase bind to in the promoter region?

Answer 23

A: The E. coli RNA polymerase binds to sequences within the promoter region upstream of the initiation site.

Question 24

Q: How is the transcription start site and promoter sequence numbered in E. coli?

Answer 24

A: The transcription start site is denoted by position +1, while the promoter sequence is denoted by a negative number since it is upstream of the initiation site.

Question 25

Q: What is the length of the σ⁷⁰ promoter in E. coli?

Answer 25

A: The σ⁷⁰ promoter is between 40 and 60 base pairs long.

Question 26

Q: Which regions of the σ⁷⁰ promoter are associated with RNA polymerase holoenzyme binding?

Answer 26

A: The region from -55 to +20 binds the RNA polymerase holoenzyme, and the region from -20 to +20 is very strongly associated with the holoenzyme.

Question 27

Q: What is required for efficient transcription in E. coli?

Answer 27

A: Sequences up to -40 are required for efficient transcription.

Question 28

Q: What happens during Step 1 (Promoter Binding) of E. coli transcription?

Answer 28

Core Polymerase Alone: Has loose, non-specific DNA binding. σ Factor Joins → Holoenzyme: Forms the holoenzyme. Gains 100x stronger affinity for specific promoter regions. Promoter Search: Holoenzyme slides along DNA, looking for -35 and -10 promoter sites. Closed Complex Formation: Binds to promoter but DNA is still closed. Waits for signals at important genes to begin transcription. A: The core polymerase enzyme has a loose, non-specific affinity for DNA. Association of the σ factor forms the holoenzyme, which binds to specific promoter sequences with a 100-fold increase in affinity. The holoenzyme is thought to slide along the DNA, searching for the promoter’s -35 and -10 sites. This closed complex awaits stimulation at highly expressed or important genes.

Question 29

Q: What happens during Step 2 (DNA Unwinding) of E. coli transcription?

Answer 29

Polymerase Unwinds DNA: To access the antisense strand (the template for transcription). DNA Topology Helps: Negative supercoiling at active promoters makes DNA easier to unwind. Promoter Site Positioning: The -35 and -10 sites are spaced to support DNA bending/unwinding. Open Complex Forms: Unwinding leads to an open complex with the holoenzyme bound to single-stranded DNA. A: The polymerase unwinds the DNA to access the antisense strand. The DNA topology plays a key role, with negative supercoiling used at active gene promoters to help melt the duplex. The -35 and -10 sites' relative positions facilitate DNA conformation changes. The initial unwinding forms an open complex with the holoenzyme.

Question 30

Q: What occurs in Step 3 (Transcription Initiation) of E. coli transcription?

Answer 30

No Primer Needed: RNA synthesis starts without a primer (unlike DNA). First Bond Formation: Begins with GTP or ATP, forming a phosphodiester bond between the first 2 ribonucleotides. Abortive Initiation: First 9 bases are added while the enzyme is stationary—may abort and retry multiple times. Speed Matters: Especially during rapid growth, transcription decisions must be quick and efficient. A: RNA synthesis begins without a primer (unlike DNA synthesis). The first two ribonucleotides are incorporated to form a phosphodiester bond, starting with GTP or ATP. The first 9 bases are joined without the enzyme moving; transcription can be aborted during this phase. The process is fast, especially during rapid growth, requiring quick decision-making for transcription initiation.

Question 31

Q: What happens during Step 4 (mRNA Elongation) of E. coli transcription?

Answer 31

Ternary Complex Forms: σ-factor + Polymerase + DNA + nascent RNA → forms a ternary complex. This triggers polymerase movement and promoter clearance. Elongation Begins: Polymerase moves at ~40 bases/sec. Keeps 17 base pairs unwound and maintains a 12 bp RNA-DNA hybrid. σ-factor Release: Once the polymerase starts elongation, the σ-factor is released. It’s free to join another core polymerase and repeat the process. A: The σ-factor forms a ternary complex with polymerase, DNA, and nascent RNA, causing the polymerase to move forward and clear the promoter. The polymerase moves at 40 bases per second, keeping 17 base pairs unwound and a 12 base pair RNA-DNA hybrid constant. The σ-factor is released once the polymerase is moving, and it is free to form a new holoenzyme.

Question 32

Q: What occurs during Step 5 (Termination) of E. coli transcription?

Answer 32

Termination Signal Reached: RNA polymerase stops when it hits a termination sequence. Hairpin Structure Forms: RNA folds into a GC-rich hairpin using self-complementary regions. This hairpin is very stable and stalls the polymerase. U-Rich Sequence Follows: After the hairpin, RNA has 4+ uridine (U) bases. These weakly bind to the A bases on the DNA template. Dissociation: The weak U-A bonds + stalled polymerase cause the enzyme to fall off the DNA. A: The polymerase dissociates from the DNA upon reaching a termination sequence. The common stop signal is an RNA hairpin formed by self-complementary sequence areas in the mRNA. The hairpin is GC-rich and stable, stalling the polymerase. After the hairpin, a sequence of 4 or more uridine (U) residues weakly binds to the adenine (A) residues on the template strand, causing dissociation of the core enzyme.

Question 33

Q: How is transcription controlled in most genes?

Answer 33

A: Transcription of a particular gene is controlled by a regulatory region of DNA near the site of transcription. This is common in prokaryotes.

Question 34

Q: What are the two types of regulatory regions in transcription control?

Answer 34

A: Some regulatory regions act like simple switches. Others act like a microprocessor, responding to a variety of signal inputs, which is more common in eukaryotes.

Question 35

Q: How do gene regulatory proteins control gene transcription?

Answer 35

A: Gene regulatory proteins bind to the regulatory region of the DNA to turn genes on or off.

Question 36

Q: How do gene regulatory proteins interact with DNA?

Answer 36

A: The proteins bind to the outside of the DNA by recognizing specific sequences, interacting with the exposed edges of base pairs via hydrogen bonding and hydrophobic interactions in both the minor and major grooves.

Question 37

Q: Where do most gene regulatory proteins bind on the DNA?

Answer 37

A: Most regulatory proteins bind to the major groove of the DNA because the patterns for each of the 4 base pairs are unique there.

Question 38

Q: How long are the regulatory regions of DNA that gene regulatory proteins bind to?

Answer 38

A: These regulatory regions typically contain less than 20 nucleotide base pairs.

Question 39

Q: What is the role of Zn²⁺ ions in some DNA-binding proteins?

Answer 39

A: Some DNA-binding proteins use Zn²⁺ ions to help maintain the correct protein folding.