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BIOA01H3 Lecture Notes - Lecture 33: Dna Replication, Polynucleotide, Uracil

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maroonalpaca618
13 May 2018
School
UTSC
Department
Biological Sciences
Course
BIOA01H3
Professor
Maydianne Andrade

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BIOA01
LEC 33 Transcription Chapters 13 & 14
Nov 24, 2015
→
- change in gene expression → turned on or off
→ What determines when a gene is to be transcribed?
Transcriptional regulation
- regulation of transcription initiation
Posttranscriptional regulation
-variations in pre-mRNA processing
- we can make a protein stay around for a long time
- you want signal proteins to be there for short time → make them unstable
*** all this to fine-tune final product
→ DNA-Directed RNA Synthesis
transcription→ dna to rna
- rna is READ in 3’ to 5’
- rna is SYNTHESIZED in 5’ to 3’
→ Differences bw DNA replication and transcription
- in RNA→ only one of two DNA strands acts as template to make a
complementary copy
- only the sequence encoding one gene is copied
- RNA polymerase puts together ribonucleotides into single RNA strand
*** does not use primers for help
- RNA molecules are single polynucleotide chains
- Uracil pairs with adenine
→ Gene Structure
two main parts:
- promoter(control sequence for transcription)
- transcription unit )section of the gene that is copied into RNA molecule by
RNA polymerase until it reaches the stop point)
→ Three stages of transcription
1) initiation→ molecules assemble at promoter and synthesis of an RNA copy of
gene begins
2) elongation→ RNA polymerase moves along gene extending RNA chain
- RNA polymerase unwinds DNA 5’ to 3’
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Document Summary

Nov 24, 2015 change in gene expression turned on or off. We can make a protein stay around for a long time. You want signal proteins to be there for short time make them unstable. Dna-directed rna synthesis transcription dna to rna rna is read in 3" to 5" rna is synthesized in 5" to 3". Differences bw dna replication and transcription in rna only one of two dna strands acts as template to make a complementary copy. Only the sequence encoding one gene is copied. Rna polymerase puts together ribonucleotides into single rna strand. Promoter(control sequence for transcription) transcription unit )section of the gene that is copied into rna molecule by. Rna polymerase until it reaches the stop point) Three stages of transcription: initiation molecules assemble at promoter and synthesis of an rna copy of, elongation rna polymerase moves along gene extending rna chain gene begins.

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Related Questions

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

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