BIO130H1 Lecture 10: lecture 10 - midterm notes

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3 Mar 2017

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Everything 5" of the +1 on the template strand is downstream, and 3" is upstream and the: coli promoters are upstream of the +1 site. 2 major promoters: -35 upstream of the +1 site. 10 upstream of the +1 site (pribno box) consensus sequences = conserved and similar in function and structure across organisms: different sigma factors recognize different sequences. We have an operator and oppressor proteins regulate transcription. Eukaryotic transcription: primary mrna transcript, 5" cap on the rna transcript, and as you are transcribing, eventually will cut the transcript and more processing of the rna will occur splicing. You go from primary rna to mature rna that will be exported out of the nucleus into the plasma for processing of the rna: not polycistronic. Telomerase has its own template, and a noncoding rna. Transcription part 2: the polycystronic mrna, generates multiple proteins.

Related Questions

Please help on these. Thank you!

Question 1

Which of the following is NOT a channel in the RNA polymeraseenzyme?

		RNA exit channel
		DNA entry channel
		DNA exit channel
		Nucleotide exit channel

Question 2

A promoter region upstream of a bacterial gene typically has thesequence TTGACA in the -35 position and TATAAT in the -10 position.If there was a 5 bp insertion at the -40 position, such thateverything downstream was shifted (ie: the "-35 element" is now atthe -30 from the transcription start site), which of the followingwill occur?

		Sigma protein will not bind and so no transcription willoccur.
		RNA polymerase will not bind and so no transcription willoccur.
		Sigma protein will bind to its target sequences and sotranscription will start 5bp downstream from the originaltranscription start site.
		Sigma protein will bind as usual and transcription will startappropriately at the transcription start site.

Question 3

A person has a mutation that prevents TFIIH from being produced.Which of the following outcomes will occur?

		No transcription because TFIID will not be able to bind.
		No transcription because the DNA strand will not be opened.
		Transcription will occur, but it will not be in the correctposition.
		Transcription will occur as usual.

Question 4

There are 4 rRNAs produced to make both subunits of theribosome. How many transcripts are required to produce these 4rRNAs?

		1
		2
		3
		4

Question 5

True or False: the 5' 7m-Gppp cap is attached to the start codonof the mRNA.

		True.
		False, the cap is attached to the 5'-untranslated region.
		False - the cap is attached to the 3'-untranslated region.
		False, the cap is attached to the polyA tail.

Question 6

During the splicing mechanism, which of the following joins twoexons together?

		A 2'OH attacks a 5' phosphate
		A 5' phosphate attacks a 2'OH
		A 3'OH attacks a 5' phosphate
		A 5' phosphate attacks a 3'OH Question 7 After their experiment in Neurospora, Beadle and Tatum concludedthat each single genes code for single enzymes. Do you agree withtheir statement, why or why not?

burgundyhedgehog579

1 DNA sequences found in introns provide [A] amino acid sequence information.

[B] regulatory information.
[C] no known useful information.

[D] structure for the gene.
[E] alternative DNA splicing possibilities.

3. The TATA box is a(n)
[A] sequence close to the promoter region of many genes.
[B] square-shaped sequence.
[C] enhancer consensus sequence.
[D] activator sequence necessary for proper translation.
[E] None of the above

5. Regarding genes and gene expression, which of the following is not a difference between prokaryotes and eukaryotes
[A] Eukaryotic DNA does not contain a TATA box close to the initiation site as does prokaryotic DNA.

[B] In eukaryotes, RNA polymerase requires

'transcription factors'or initiation.

[C] Prokaryotic genes are often grouped together in

operons.

[D] In eukaryotes, the regulation of several genes at once requires common control elements in each of the genes.
[E] There is a diversity of eukaryotic polymerases that is reflected in the diversity of eukaryotic promoters.

10) In eukaryotic cells, a repressor

[A] is made of DNA.
[B] binds to the enhancer region to block transcription. [

C] is located both upstream and downstream from the promoter.
[D] binds to the operator to block RNA polymerase.

[E] binds to a silencer to reduce transcription rates.

Coordinated regulation of genes in eukaryotic cells [A] is the result of positioning the same regulatory sequence in front of each gene.
[B] results from all of those genes being under the control of one promoter.

[C] occurs because related genes all have the same

operons.
[D] occurs because enhancers cause DNA to bend.

Match the following:

1	Region which lies upstream of thetranslation start site contains the Shine Dalgarno sequence
2	Sequence added to the 3' end of thepre-mRNA which affects mRNA stability
3	Establishes the reading frame andspecifies the first amino acid of the protein chain
4	DNA sequence that the transcriptionapparatus recognizes and binds to initiate transcription
5	Expressed sequences of DNA that arenot removed during alternative splicing
6	Signals the end of translation andthe protein molecule
7	Group of three nucleotides that codefor an amino acid
8	Site begins the coding region of thegene and is located 25-30 nucleotides downstream of the TATAbox
9	Non-coding sequences of DNA thatintervene within coding regions of a gene
	10 Ribosome binding site ineukaryotes; functions in the initiation of translation and mRNAstability
11	A sequence which lies downstream ofthe coding region of the gene; determines the location of the 3'cleavage site in the pre-mRNA molecule
12	Sequence of nucleotides at the 3' endof the mRNA that is not translated into protein; affects thetranslation and stability of the mRNA molecule

A.	5' untranslated region
B.	AAUAAA consensus sequence
C.	Promoter
D.	Transcription start site
E.	3' untranslated region
F.	Introns
G.	Initiation codon (AUG)
H.	Poly(A) tail
I.	5' cap
J.	Termination codon (UAA, UGA, UAG)
K.	Exons
L.	Sense codon

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BIO130H1 Lecture 10: lecture 10 - midterm notes

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