BIO 351 Lecture Notes - Lecture 11: Rna Polymerase Ii, Consensus Sequence, Speed Bump

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13 May 2018

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Transcription, RNA Processing, and Translation Objectives

1. Be ale to da DNA ad RNA uleotides ad distiguish the fo eah othe ou do’t eed to da the

seuees of the itoge ases… just the deoiose ad the phosphate

RNA DNA

2. Not all ateial pootes hae the eat sae DNA uleotides ithi the osesus seuee at -10 and -35.

Explain the significance of these sequence differences.

− Need consensus to have a generally agreed upon, widely recognized promoter sequence

− If there were all the same sequences then they would all have the exact same level of expression

− B hagig the seuees, ou affet the affiit of the “iga Fato o TF II D i eukaotes ad that’s

going to allow for different levels of expression

− Expression level is coded by -35 to -10 (in prokaryotes)

3. Explain the role of RNA secondary structure in termination of transcription (hairpin)

− Termination sequences include inverted repeats

o CpG rich region of mRNA will bend back on itself and make base pairs

o This secondary structure, called a hairpin, acts like a speed bump and makes the RNA Polymerase slow

down to a halt (through a poorly understood mechanism)

o With RNA Polymerase already halted, it just needs to dissociate from the DNA strand

o These base pairs are broken either in a Rho independent or Rho dependent method

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4. Initiation of transcription in eukaryotes is more complex than in prokaryotes, involving many proteins and DNA

sequences. Why do multicellular organisms require more complex regulation of gene expression?

− Default epessio i pokaotes is o

− Default epessio i eukaotes is off and must go to great lengths to turn them on

o This is likely because eukaryotes are more complicated, have different cell types, have different cell

regions, and need a lot more coordination

o Having all of them off by default, allows the cell to them go through and turn on the ones that only need

to be turned on (or increased) for specific cells in specific regions

o In patiula, the uleus is seletie o hat it allos to tael i ad out, heeas pokaotes do’t

have a nucleus or specialized organelles

5. What are the roles of TFIID in eukaryotic transcription? Distinguish between transcription factors and

transcriptional activators.

− Transcription Factors (TFs)

o Help RNA Polymerase II position near the promoter in Eukaryotes

o We did’t ae a lot of the, ut e did ae oe

o TF II D is the primary one that interacts with RNA Polymerase

o It anchors itself around the -25 region and serves as a landing platform for RNA Polymerase to land on

o TF II D and many other TFs are necessary for the proper affinity and binding of RNA Polymerase

− Transcriptional Activator Proteins (TAPs)

o Proteins that bind to enhancer regions

o These regions are quite upstream from the start of transcription

o An TAP bound to an enhancer region will bend around and enzymatically interact with RNA

Polymerase, tapping it and giving it the green light to start transcription

6. We have discussed gain/loss of function mutations. How could either of these be explained by changes to

promoter sequences?

− Mutations in the promoter or enhancer region could increase or decrease affinity of their corresponding

proteins

− Mutation in promoter that makes it unrecognizable → LOF

− Mutation in enhancer that makes it unrecognizable → LOF

− Mutation in enhancer that makes for stronger affinity (or perhaps a duplication and the additive effect of

having more enhancer regions) → GOF

7. Distinguish between the roles of promoter sequences and enhancer sequences in eukaryotic gene expression. How

do promoters and enhancers differ in terms of their proximity to the coding region of a gene? What proteins bind

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Document Summary

Transcription, rna processing, and translation objectives: be a(cid:271)le to d(cid:396)a(cid:449) dna a(cid:374)d rna (cid:374)u(cid:272)leotides a(cid:374)d disti(cid:374)guish the(cid:373) f(cid:396)o(cid:373) ea(cid:272)h othe(cid:396) (cid:894)(cid:455)ou do(cid:374)"t (cid:374)eed to d(cid:396)a(cid:449) the se(cid:395)ue(cid:374)(cid:272)es of the (cid:374)it(cid:396)oge(cid:374) (cid:271)ases just the (cid:894)deo(cid:454)(cid:455)(cid:895)(cid:396)i(cid:271)ose a(cid:374)d the phosphate(cid:895) Dna: not all (cid:271)a(cid:272)te(cid:396)ial p(cid:396)o(cid:373)ote(cid:396)s ha(cid:448)e the e(cid:454)a(cid:272)t sa(cid:373)e dna (cid:374)u(cid:272)leotides (cid:449)ithi(cid:374) the (cid:862)(cid:272)o(cid:374)se(cid:374)sus se(cid:395)ue(cid:374)(cid:272)e(cid:863) at -10 and -35. Need consensus to have a generally agreed upon, widely recognized promoter sequence. B(cid:455) (cid:272)ha(cid:374)gi(cid:374)g the se(cid:395)ue(cid:374)(cid:272)es, (cid:455)ou affe(cid:272)t the affi(cid:374)it(cid:455) of the ig(cid:373)a fa(cid:272)to(cid:396) (cid:894)o(cid:396) tf ii d i(cid:374) euka(cid:396)(cid:455)otes(cid:895) a(cid:374)d that"s. If there were all the same sequences then they would all have the exact same level of expression going to allow for different levels of expression. Expression level is coded by -35 to -10 (in prokaryotes: explain the role of rna secondary structure in termination of transcription (hairpin) Initiation of transcription in eukaryotes is more complex than in prokaryotes, involving many proteins and dna sequences.

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.

5. Alpha amanitin is a drug that binds RNA polymerase andprevents it from functioning in the cell (this is the compound thatis found in poison mushrooms). The reason people die when theyconsume alpha amanitin is:

a) Alpha amanitin prevents gene expression from occurring in thecell

b) DNA replication is directly inhibited by alpha amantin

c) Alpha amanitin binds tightly to the messenger RNAs that aremade in the cell, preventing translation

d) Alpha amanitin binds to nuclear pores preventing thetransport of messenger RNA into the cytoplasm

6. In order for transcriptional initiation to begin in E. colithe following elements are needed:

a) an origin and RNA polymerase

b) initiator protein and a promoter

c) the sigma subunit of RNA polymerase and an RNA template

d) RNA polymerase and a promoter

e) helicase and primase

Questions 7 and 8. You are studying two species of closelyrelated snakes, one species lives in sandy desert area and has paletan skin and the species lives in a volcanic rock area and has darkbrown skin. You a sequence the DNA from each type of snake anddiscover a gene encoding a protein that transports pigment into theskin cells. The pigment transporter protein is responsible for skincolor. When you analyze the DNA coding sequence of the pigmenttransporter from each snake you find these two DNA sequences (hashmarks have been added to indicate the reading frame):

pale tan snake: GCG/TTG/GAC/TGC/ACC (DNA template strand)CGC/AAC/CTG/ACG/TGG (complementary strand) dark brown snake:ACG/TTG/GAC/TGC/ACC (DNA template strand) TGC/AAC/CTG/ACG/TGG(complementary strand)

7. From these sequences you conclude:

a) that the amino acid sequence of the pigment transporters forthe two proteins are almost identical

b) that the amino acid sequence of the pigment transporters forthe two proteins are very dissimilar

c) that the amino acid sequence of the pigment transporters forthe two proteins are identical

8. Since the skin color is different in each snake, you concludefrom the sequence data above that:

a) the genetic code in the skin cells in each of these snakes isdifferent

b) the DNA sequence encoding the entire gene for pigmenttransporters in these two snakes is different

c) that one of the codons in the gene for the pigmenttransporter in these two snakes is different

9. You are analyzing eukaryotic cells that have been exposed tothe radioactive uracil. In most cells the radioactive uracil isinitially localized in the nucleus and then moves from the nucleusto the cytoplasm. However, you observe that in some cells theradioactive molecules stay in the nucleus and never arrive in thecytoplasm. This is most likely a defect in:

a) the transcription of the DNA

b) the processing of the mRNA

c) the translation of the RNA

d) the labeling of the RNA

e) the replication of the DNA

10. Which of the following is true of both eukaryotic andbacterial transcription? (select all that are true)
a) Transcription factors must bind the promoter before RNApolymerase initiates transcription

b) Messenger RNAs must be spliced before translation begins inorder to produce the appropriate gene product

c) During transcription only one DNA strand is used as atemplate to make the messenger RNA

d) Termination of transcription occurs when the RNA polymerasetranscribes a hairpin structure in the RNA followed by a string ofU

erinporpoise384

1. Which of the following is the correct sequence of events (from start to finish) in cellular fatty acid oxidation?

A. Fatty acid activationÃ carnitine shuttleÃ beta-oxidation

B. Beta-oxidationÃ carnitine shuttleÃ Fatty acid activation

C. Fatty acid activationÃ carnitine shuttleÃ beta-oxidation

D. Carnitine shuttleÃ Fatty acid activationÃ beta-oxidation

2. What reaction is catalyzed by lipoprotein lipase, or lipase?

A. Hydrolysis of TAGs into fatty acids and glycerol

B. Hydrolysis of TAGs into amino acids and glycerol

C. Hydrolysis of cholesterol in to fatty acid and water

D. Hydrolysis of cholesterol into acetyl-CoA

3. How do fatty acids of different carbon lengths cross into the mitochondrial matrix for oxidation?

A. Small FAs of 10 carbons or fewer can freely diffuse through the membrane

B. Large FAs are cleaved into two smaller units that are actively transported across the membrane

C. A membrane-bound transferase catalyzes the conversion of long-chain FAs and FA-carnitine esters, which are transported across the membrane via a specific carrier

D. The breakdown of large FAs starts outside the matrix, allowing for the smaller particles to pass easily through the membrane

4. How can any RNA nucleotide be easily distinguished from a DNA nucleotide?

A. RNA nucleotides have a unique phosphodiester bond linkage

B. RNA nucleotides have uracil nitrogenous bases

C. RNA nucleotides have an -OH on the 2â carbon of the ribose

D. RNA nucleotides cannot be easily distinguished from DNA nucleotides

5. Which of the following statement is true concerning the sequence pTACGTAACGTA?

A. This is an RNA sequence

B. In order to have a more rigid structure in a double-stranded molecule, the sequence given would need more adenosine

C. If the complementary strand were present, the number of purines would be equal to the number of pyrimidines

D. The amount of purines is equal to the amount to pyrimidines

6. Complementary base pairing is considered which level of structure?

A. Secondary

B. Tertiary

C. Quaternary

D. Primary

7. Which of the following is/are required for the polymerization reaction catalyzed by DNA polymerase III?

I. RNA primer

II. DNA primer

III. RNA template

IV. DNA template

V. NTPs

VI. dNTPs

VII. Magnesium

VIII. Calcium

A. II, IV, V, and VII

B. II, IV, VI, and VII

C. I, IV, V, and VIII

D. I, IV, VI, and VII

8. DNA polymerase I, from E. coli, has three different catalytic sites. Which of the following is NOT one of them?

A. RNA-dependent RNA polymerase

B. RNA-dependent DNA polymerase

C. DNA-dependent RNA polymerase

D. DNA-dependent DNA polymerase

9. Eukaryotes solve the problem of replicating the ends of their linear chromosomes by:

A. Using a protein as primer

B. Enzyme activity that uses an RNA template and reverse transcription in a series of repeat cycles

C. All of the listed solution

D. Terminal redundancy whereby each end of the chromosome is duplicated allowing recombination

BIO 351 Lecture Notes - Lecture 11: Rna Polymerase Ii, Consensus Sequence, Speed Bump

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