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BIO 1140 Lecture Notes - Lecture 16: Sigma Factor, Polyadenylation, Eukaryote

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plumferret19
14 Mar 2018
School
University of Ottawa
Department
Biology
Course
BIO 1140
Professor
Caroline Petit- Turcotte

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Document Summary

Cells can derive genes from any given protein. Which of these statements corresponds to a difference in transcription between prokaryotes and eukaryotes: Always proceeds in 5" to 3" direction. What is the role of the sigma alpha factor in transcription. Provide 1 key difference between prokaryotes and eukaryotes regarding termination of transcription. Mrna sequences that recruits an enzymes (poly a tail) Prokaryotes: role dependent/ intrinsic, proteins that recognizes sequences on the dna and acts as a stopper to prevent transcription from continuing. Our genome is organized by nucleotides who assemble together. Genes are separated by intergenic regions that don"t code for specific genes. The flow of genetic information: central dogma. Prokaryotes have one rna polymerase that transcribes dna to rna. Eukaryotes have distinct enzymes that transcribes dna to rna. Ribosomal rna is used to build protein ( nucleolus) Bond at 3rd sugar and 5th carbon nucleotide. Dna is directly transcribed to mrna which can occur anytime.

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Related Questions

Eukaryotic transcription normally takes place on a chromatin template at nuclease-accessible sites.(True/False)

*After promoter clearance by RNA polymerase II:

A) all associated factors remain bound to the polymerase for elongation.

B) a subset of factors remain at the promoter region.

*Which of the following statements is FALSE?

A) RNA polymerase II is responsible for the transcription of genes encoding protein-coding sequences only.

B) Many promoters recognized by RNA polymerase II contain regulatory elements several kbp upstream of the initiation site.

C) TFIIB has a role in converting a closed-promoter into an open-promoter complex.

D) The C-terminal domain of the largest subunit of RNA pol II undergoes extensive phosphorylation prior to transcription elongation.

E) A multiprotein complex is involved in communication between upstream control elements and proteins bound at the promoter.

C) no further elongation factors are acquired by the polymerase.

D) no further chromatin remodeling is required.

E) transcription continues until the polymerase reaches the poly (A) tail

*In eukaryotes RNA polymerase I transcribes all of the ribosomal RNA components.(TRUE/FALSE)

Molecular Biology Help!

Histone chaperones play a role in replication by:

Histone chaperones play a role in replication by:

preventing specific interactions between the positively charged histones and negatively charged DNA.
preventing nonspecific interactions between the positively charged histones and negatively charged DNA.
facilitating specific interactions between the positively charged histones and negatively charged DNA.
facilitating nonspecific interactions between the positively charged histones and negatively charged DNA.
none of the above.

Histone chaperones play a role in replication by:

Histone chaperones play a role in replication by:

preventing specific interactions between the positively charged histones and negatively charged DNA.
preventing nonspecific interactions between the positively charged histones and negatively charged DNA.
facilitating specific interactions between the positively charged histones and negatively charged DNA.
facilitating nonspecific interactions between the positively charged histones and negatively charged DNA.
none of the above.

Histone chaperones play a role in replication by:

Histone chaperones play a role in replication by:

preventing specific interactions between the positively charged histones and negatively charged DNA.
preventing nonspecific interactions between the positively charged histones and negatively charged DNA.
facilitating specific interactions between the positively charged histones and negatively charged DNA.
facilitating nonspecific interactions between the positively charged histones and negatively charged DNA.
none of the above.

Histone chaperones play a role in replication by:

Histone chaperones play a role in replication by:

preventing specific interactions between the positively charged histones and negatively charged DNA.
preventing nonspecific interactions between the positively charged histones and negatively charged DNA.
facilitating specific interactions between the positively charged histones and negatively charged DNA.
facilitating nonspecific interactions between the positively charged histones and negatively charged DNA.
none of the above.

Histone chaperones play a role in replication by:

Histone chaperones play a role in replication by:

preventing specific interactions between the positively charged histones and negatively charged DNA.
preventing nonspecific interactions between the positively charged histones and negatively charged DNA.
facilitating specific interactions between the positively charged histones and negatively charged DNA.
facilitating nonspecific interactions between the positively charged histones and negatively charged DNA.
none of the above.

________________ may play a role in aging and cancer.

________________ may play a role in aging and cancer.

telomerase
telomerase RNA
DNA polymerase III
restriction enzymes

none of the above

Telomeres consist of :

Telomeres consist of :

long random sequences at the ends of the linear chromosomes
short random sequences at the ends of the linear chromosomes
long TG-rich repeats at the ends of the linear chromosomes
long AT-rich repeats at the ends of the linear chromosomes

long AC-rich repeats at the ends of the linear chromosomes

Telomerase is a _____________________.

Telomerase is a _____________________.

restriction enzyme
protease
reverse transcriptase
exonuclease

none of the above

Replication of the eukaryotic nuclear genome requires which of the following?

Replication of the eukaryotic nuclear genome requires which of the following?

DNA polymerase α
DNA polymerase γ
DNA polymerase δ
DNA polymerase ε
all but one of the above

Which of the following is the eukaryotic DNA replication initiator?

Which of the following is the eukaryotic DNA replication initiator?

DnaA
replication protein A
replication protein C
ORC
MCM

The yeast ARS element has which of the following functions and properties:

The yeast ARS element has which of the following functions and properties:

they are AT-rich and serve as replication termination sites
they are GC-rich and serve as replication termination sites
they have random sequence and serve as replication termination sites
they are AT-rich and serve as replication initiation sites
they are GC-rich and serve as replication initiation sites

Replication factor C plays which of the following roles in eukaryotic DNA replication?

Replication factor C plays which of the following roles in eukaryotic DNA replication?

it is the sliding clamp
it is the clamp loader
it is the DNA ligase
it is a single-stranded DNA binding protein

it is a cofactor required for activating the DNA polymerase

Which of the following is the replicative DNA polymerase for both the leading and lagging strands of SV40?

Which of the following is the replicative DNA polymerase for both the leading and lagging strands of SV40?

DNA polymerase α
DNA polymerase β
DNA polymerase γ
DNA polymerase δ

DNA polymerase ε

SV40 T antigen recruits which of the following to the proto-replication bubble?

SV40 T antigen recruits which of the following to the proto-replication bubble?

DNA polymerase α
DNA polymerase β
DNA polymerase γ
DNA polymerase δ

DNA polymerase ε

The SV40 large tumor (T) antigen has which of the following functions:

The SV40 large tumor (T) antigen has which of the following functions:

helicase activity only
DNA primase and DNA helicase activity
origin binding and DNA primase activity
origin binding and DNA helicase activity
origin binding and DNA helicase activity

Which of the following did Erwin Chargaff observe?

A)

Base composition changes as the organism ages.

B)

Base composition of DNA does not vary from one species toanother.

C)

In all species the number of adenines equals the number ofguanines.

D)

In all species A + T = G + C.

E)

DNAs isolated from different tissues of the same species havethe same base composition.

The melting temperature (Tm) of a DNA duplex is:

A)

the temperature at which double-stranded nucleic acids degradeinto free nucleotides.

B)

the temperature at which double-stranded DNA is toosingle-stranded to ever renature back to double strands.

C)

the temperature at which double-stranded DNA is completelymelted.

D)

the temperature at which double-stranded DNA has become 50%single-stranded.

E)

the temperature at which double-stranded DNA begins to melt.

Given your knowledge and the descriptions of each of thesemethods, which of the following does NOT rely on the ability ofnucleic acids to hybridize with each other?

A)

Southern blotting for the detection of specific DNA fragmentsthat have been separated by gel electrophoresis.

B)

Colony blot hybridization to identify a specific sequence from acollection of sequences that have been inserted into bacteria.

C)

Northern blotting for the detection of specific RNA fragmentsthat have been separated by gel electrophoresis.

D)

A DNA oligonucleotide microarray incubated with probes made fromcDNA.

E)

Western blotting for the detection of specific proteins thathave been separated by gel electrophoresis.

Which of the following is a possible reason why DNA uses thymineinstead of uracil?

A)

Cytidine is deaminated to uracil so it would be difficult forany repair mechanism to differentiate between uracil that belongsin the sequence and uracil that resulted from deamination.

B)

Adenine is deaminated to uracil so it would be difficult for anyrepair mechanism to differentiate between uracil that belongs inthe sequence and uracil that resulted from deamination.

C)

When uracil is bound to a deoxyribose it will become morereactive, forming covalent bonds with other bases.

D)

None of the above.

Which of the following is an enzyme used in cloning to breakcovalent bonds?

A)

DNA ligase

B)

Restriction endonuclease

C)

Reverse transcriptase

D)

DNA polymerase I

E)

Alkaline phosphatase

Which of the following could you use to remove the 5' phosphatesafter cleavage of a plasmid with a single type of restrictionenzyme to ensure that the plasmid would not simply fuse backtogether upon addition of ligase?

A)

Reverse transcriptase

B)

Alkaline phosphatase

C)

DNA ligase

D)

DNA polymerase I

E)

Restriction endonuclease

Which of the following would you use to create a cDNA librarythat you would not use for a genomic library?

A)

Alkaline phosphatase

B)

Reverse transcriptase

C)

Restriction endonuclease

D)

DNA ligase

E)

DNA polymerase I

A plasmid vector typically has which of the followingfeatures?

A)

An origin of replication

B)

A selectable marker to ensure that the only bacteria growingcontain the intact or recombinant vector.

C)

Unique restriction sites for insertion of DNA.

D)

Small size to facilitate entry into the cell and biochemicalmanipulation of the DNA.

E)

All of the above are features of plasmid vectors.

Which of the following is FALSE of genomic or cDNAlibraries?

A)

Genes represented in cDNA libraries include control sequencesthat direct transcription.

B)

cDNA libraries contain cDNAs made from mRNA.

C)

Genomic libraries contain fragments of genomic DNA that aregenerated by partial digests with restriction enzymes.

D)

Genomic libraries usually contain large fragments and are likelyto be constructed in YAC or BAC vectors.

E)

cDNA libraries contain sequences from expressed genes only.

Which of the following is NOT a typical component of apolymerase chain reaction (PCR)?

A)

dNTPs

B)

DNA containing the sequences to be amplified

C)

DNA ligase to connect the fragments together

D)

Primers complementary to each end of the sequence to beamplified

E)

Heat stable Taq polymerase

How can a fusion protein be formed?

A)

Site-directed mutagenesis where a fragment is replaced with asynthetic sequence containing the altered DNA sequence.

B)

Two proteins can recombine in the bacterial cell to form a newhybrid protein.

C)

Portions of two different genes are ligated together to create anew hybrid gene. When expressed the gene product is a fusion of thetwo gene products.

D)

Using oligonucleotide-directed mutagenesis to introduce specificmutations that cause the expressed protein to fuse to otherproteins.

E)

A portion of the gene can be removed to form a shorter gene andtherefore a shorter protein.

Why is green fluorescent protein (GFP) so useful in visualizingfusion proteins in eukaryotes?

A)

The GFP protein recognizes and binds to fusion proteins allowingthe location of the fusion protein to be determined.

B)

The reaction is only dependent on the presence of GFP and oneother cofactor.

C)

It is easily cloned and expressed in bacterial cells.

D)

The presence of just a few molecules of GFP fusion proteins canbe sufficient for observing the proteins microscopically allowingthe study of its location and movements in the cell.

E)

It is derived from fire flies, which are easy to cultivate inthe lab.

You wish to determine the cellular location of a protein butunfortunately it is not particularly antigenic (can t develop astrong antibody response). How could you solve this problem?

A)

Create a fusion protein with an epitope tag that can bevisualized by incubation with fluorescently tagged antibody to theepitope.

B)

Create a fusion with green fluorescent protein (GFP).

C)

Express the protein as a fusion with biotin. The addition of GFPwill cause fluorescence.

D)

A and B might work.

E)

A, B, and C might work.

F)

None of the above will work.

A positive result for the yeast two-hybrid analysis means thefollowing:

A)

The proteins that you are studying can be made into fusionproteins, but do NOT interact with each other within the nucleus ofyeast.

B)

A fusion protein containing the Gal4p activation domain, hasbound to RNA polymerase resulting increase transcription ofnumerous genes, including the reporter gene, and formation ofcolonies.

C)

Two fusion proteins, one containing the Gal4p binding sitedomain and one containing the Gal4p activation domain, haveinteracted through their Gal4p domains resulting in transcriptionof the reporter gene.

D)

Two fusion proteins interact with each other and because of theGal4p binding site domain and the Gal4p activation domain containedin each fusion protein, the reporter genes are activated and thecells die resulting in no colonies.

E)

The colonies growing are expressing the two fusion proteins, butthe two proteins have not interacted with each other.

Which of the following is the approximate size of the humangenome?

A)

3 Gm

B)

300 Kb

C)

3 Mb

D)

3 Gb

E)

3 Mm

Which of the following is correct about the structure orlocation of genes?

Question 22 options:

A)

Prokaryotic genes typically possess their own regulatorysequences that only control expression of a single gene.

B)

Eukaryotic genes often encode introns that are spliced outbefore translation.

C)

Eukaryotic genes are mostly located at the telomeres and thecentromeres of chromosomes.

D)

Eukaryotic genes are typically arranged in operons.

E)

Prokaryotic genes often encode exons that are spliced out beforetranslation