BIOL 401 Lecture Notes - Lecture 9: Cyclic Adenosine Monophosphate, Galactose, Permease

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greenfly749

1 Mar 2017

School

Binghamton University

Department

Biology

Course

BIOL 401

Professor

Steven P Tammariello

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Document Summary

This review sheet will focus on how gene expression is controlled in prokaryotes. Genes only need to be turned on at certain times, so they must be controlled somehow! Prokaryotic transcription, a process of reading a template and synthesizing rna, requires two types of dna/protein interactions: the first interaction is the determination of where transcription will begin, which is determined by the promoter. Transcription begins at +1, and anything to the right of it is considered downstream, while anything to the left (-1) is considered upstream. It"s important to recognize that upstream factors will not be transcribed (only downstream), but they still play a role in gene regulation. Another very important note is that the promoter is found upstream. The rna polymerases, which are dna directed, will bind and begin to function. Must have a promoter: the second interaction is located at operator sites, which regulate promoter- driver transcription via activators, which enhance transcription, and repressors, which prevent transcription.

Related Questions

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.

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

This is a longer question i know, but if you could help me out with this, it would be much apreciated.

READ ALL THESE INSTRUCTIONS BEFORE BEGINNING! There is no time limit, but there is a due date!

You will be provided with a prokaryotic DNA sequence. It is formatted in a way that provides an easier way to keep track of which base you are looking at. That is all the numbers are there for.

You need to locate the CORRECT protein coding region (the upstream sites, start codon, etc.) and submit the corresponding amino acid sequence using the single-letter abbreviations without spaces or hyphens (ie, KDELHDN....) in the submission box provided.

As this is computer-graded, you MUST enter the information correctly as instructed above!!

Some reminders:

Never assume that the only place ATG appears is at the beginning of a coding sequence. Many proteins include methionine as part of their internal sequence.

Finding an ORF (open-reading frame) does not mean that it is a functional gene. Those must have the upstream elements- promoter, ribosome binding site, etc.

For the purposes of this exercise, you may assume that the start codon is 10 bases downstream of the +1 site.

You may also assume that other upstream elements are centered at exactly -10 and -35, but may not be exactly the same as the consensus sequences.

You may use any resources to complete this assignment.

You may attempt this twice.

QUESTION (The teacher said that when it is finishe it should spell out and write some type of sentence

1 ATGAAACGAA AGCGATGCGC GCGAAAGAAC

31 TCATGCATCG CGATAATCTG ATGGAAAGCA

61 ACGCGGGCGA AAGCGCTGAC GAAGTGGAAC

91 ACTGAACCGT CGAAAGCAAA ACACCGCGAT

121 AAGCGGTTGA CAGACGGGCG AGACGACAAA

151 ATATAATCGT GTCTAAGGAG GTCCATGGCG

181 AACTATATTA ACACCGAACG CAACGAAACC

211 AGCATTACCG AAAGCGCGCG CGAAATGATT

241 AACGATCTGG AAAGCAGCTT TCGCATTCCG

271 CCGGAACGCT ATTGATAACA AGTATCCGTA

301 CTGGAAGAGC CCCCGCTATA GATGCCACAC

331 TGTTCAATTT TGGACTAATC TGAATACTTT

361 TCAGCTTGAC AGTATCAAAC CGGCAACGTG

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BIOL 401 Lecture Notes - Lecture 9: Cyclic Adenosine Monophosphate, Galactose, Permease

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