Biology 1002B Study Guide - Nonsense Mutation, Alternative Splicing, Missense Mutation

82 views5 pages

apricotsheep340

20 Apr 2013

School

Department

Course

Professor

For unlimited access to Study Guides, a Grade+ subscription is required.

Document Summary

Promoter is immediately upstream of where transcription begins and rna polymerase binds and transcribes, the same rna polymerase can transcribe all the various types of genes. Act only after they have been transcribed after which they base pair w/themselves and form a hair- pin loop. Translation initation does not use scanning, but rather the rrna ribosomal subunit finds the region of the start codon directly by base pairing w/ a specific ribosome binding site on the mrna which is upstream of the start codon. Elongation stage of translation is faster in prokaryotes than eukaryotes. Are able to regulate the production of proteins very quickly in response to changing environmental conditions since there is nuclear envelope and so transcription and translation can be coupled very closely. Transcribe and process mrna in the nucleus before exporting it to the cytoplasm for translation. There are different polymerases for transcribing for transcribing different genes (rna polymerase.

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)

1. There are viruses, called bacteriophage, that infectbacterial cells. Following viral entry, the cell begins tosynthesize viral DNA and viral proteins, leading to the productionof mature virus and the death of the bacterial cell. During thecourse of viral infection, transcription of bacterial genes isturned off so that transcription of viral genes predominates. Whichof the following is the most likely explanation for how the virusmanages to shift transcription from bacterial genes to its owngenes?

a. it synthesizes a viral RNA polymerase that recognizes onlyviral promoters

b. it synthesizes a viral DNA polymerase that recognizes onlyviral origins

c. it synthesizes a viral sigma subunit that directs thebacterial RNA polymerase to viral promoters

d. it has special viral ribosomes that only translate viralproteins

2. Unlike bacterial transcripts, eukaryotic RNA transcripts arechemically modified because:

a. Chemical groups help protect the mRNAs from degradation inthe eukaryotic cytoplasm, bacterial mRNAs are not degraded

b. Chemical groups facilitate the transport of eukaryotic RNAsfrom the nucleus to the cytoplasm, bacterial RNAs are synthesizedin the cytoplasm

c. Eukaryotic RNA polymerases cannot transcribe genes unlessthey are chemically modified, bacterial RNA polymerases transcribeunmodified genes

d. Transcripts in eukaryotic cells contain non coding regionsthat must be removed, bacterial transcripts contain only codingsequence

e. b and d

3. Termination of transcription in both E. coli and eukaryoticcells occurs because

a. there is a termination signal in the DNA following the genesequence.

b. there is a hairpin structure followed by a string of U

copperturtle968

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.

Biology 1002B Study Guide - Nonsense Mutation, Alternative Splicing, Missense Mutation

Document Summary

Get access

Related Documents

Biology 1002B Study Guide - Final Guide: Stem-Loop, Start Codon, Elongation Factor

Biology 1002B Chapter Notes - Chapter 13: Ribozyme, Inosine, Peptidyl Transferase

Biology 1002B Lecture Notes - Lecture 6: Permease, Catabolite Activator Protein, Cyclic Adenosine Monophosphate

Related Questions