You are carrying out experiments to study transcription...

You are carrying out experiments to study transcription termination in an Ecoli gene. You sequence the $3^{\prime}$-end of the gene and get the following results: $5^{\prime}$ - CGAAGCGCCGATTGCCGGCGCTTTTTTTT - $3^{\prime}$ $3^{\prime}$ - GCTTCGCGGCTAACGGCCGCGAAAAAAAAA - ' $^{\prime}$ You then create mutant genes with this sequence changed to the following (top, or nontemplate strand, $5^{\prime} \rightarrow 3^{\prime}$ ): Mutant A: CGAAACTAAGATTGCAGCAGTTITTTTT Mutant B: CGAAGCGCCGTAGCACGGCGCTITTTTTT Mutant C: CGAAGCGCCGATTGCCGGCGCTTACGGCCC You put each of the mutant genes into an assay that measures termination and get the following results: (TABLE CAN'T COPY) a. Draw the structure of the RNA molecule that results from transcription of the wild-type sequence above. b. Explain these experimental results as completely as possible.

You are carrying out experiments to study transcription termination in an Ecoli gene. You sequence the $3^{\prime}$-end of the gene and get the following results:
$5^{\prime}$ - CGAAGCGCCGATTGCCGGCGCTTTTTTTT - $3^{\prime}$
$3^{\prime}$ - GCTTCGCGGCTAACGGCCGCGAAAAAAAAA - ' $^{\prime}$
You then create mutant genes with this sequence changed to the following (top, or nontemplate strand, $5^{\prime} \rightarrow 3^{\prime}$ ):
Mutant A: CGAAACTAAGATTGCAGCAGTTITTTTT
Mutant B: CGAAGCGCCGTAGCACGGCGCTITTTTTT
Mutant C: CGAAGCGCCGATTGCCGGCGCTTACGGCCC
You put each of the mutant genes into an assay that measures termination and get the following results:
(TABLE CAN'T COPY)
a. Draw the structure of the RNA molecule that results from transcription of the wild-type sequence above.
b. Explain these experimental results as completely as possible.

Key Concepts

U-rich Tract

Following the hairpin structure, a sequence rich in uracil residues (poly-U tract) is typically present. This sequence forms a weak RNA-DNA hybrid due to the lower stability of A-U pairs compared to G-C pairs. The weak hybrid further facilitates the detachment of the RNA transcript from the polymerase, ensuring effective termination.

Impact of Mutations on Termination Efficiency

Mutational analysis helps determine the importance of specific RNA sequences and structures in transcription termination. Changes in either the hairpin-forming inverted repeats or the U-rich region can alter the stability or formation of the hairpin, directly affecting the efficiency of RNA polymerase release. By examining the effects of these mutations through experiments, one can infer the structural requirements necessary for effective intrinsic termination.

Intrinsic Transcription Termination

Intrinsic termination in bacteria is a mechanism where the RNA molecule itself, without the need for additional factors, signals the end of transcription. This is achieved when a specific sequence in the DNA coding region is transcribed into RNA, allowing the formation of a secondary structure that causes RNA polymerase to dissociate from the DNA template, thereby terminating transcription.

RNA Hairpin Structure

An RNA hairpin is a secondary structure formed when sequences that are complementary to each other within the nascent RNA strand base pair to create a stem, with a loop at the top. This structure is key in intrinsic termination as it disrupts the transcription complex by causing physical stress and promoting the release of the RNA transcript from the DNA template.

Transcript

00:01 Okay, so we are given the segment of dna and asked to find the termination site for transcription, figure out if our termination is row dependent or independent and transcribe our rna from our dna segment.

00:13 So to figure out where our transcription site is, we first have to determine if our termination is row dependent or independent.

00:22 So if our termination is row dependent, we'll find long stretches of cysteen in our dna.

00:27 The long stretches of cysteine is where the row factor will bind to.

00:32 And if our termination is row independent, we'll find long stretches of adenine and inverted repeats.

00:39 So just by looking at our dna segment, we do see this long sequence of adenines.

00:46 I don't see any long stretches of cisterines.

00:49 Now we also need to look for inverted repeats.

00:52 Inverted repeats are nucleotides that are inverted and complementary.

00:55 And we'll bind in a way that will form a loop.

00:59 I'll show you what that looks like in a little bit.

01:01 But just to show you where our inverted repeats are on here.

01:06 So our first one is over here.

01:09 So c -a -g -a -c -c.

01:13 And our second one is over here.

01:16 G -g -t -t -t -g.

01:20 So we can see where our nucleotides are, well, base pair.

01:24 So our first nucleotide is c, which will pair with our last nucleotide g.

01:31 Then our second and the last t over here will pair with our a and so on and so on.

01:38 So since our segment of dna has this long stretches of adenine and inverted repeats, we can determine that this termination is row independent.

01:53 All right...

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