BISC207 Lecture Notes - Lecture 19: Pyrimidine, Polyadenylation, Phosphodiester Bond

47 views6 pages

violetgrasshopper2

3 Feb 2017

School

University of Delaware

Department

Biological Sciences

Course

BISC207

Professor

Nauen Jennifer

For unlimited access to Class Notes, a Class+ subscription is required.

BISC207 Full Course Notes

Verified Note

25 documents

Document Summary

Chapter 17: genes and making gene products: gene structure, transcription, eukaryotic rna processing, translation, a gene: a sequence of dna that codes for a product, and its associated control regions. Product- sequence that is transcribed prokaryotic gene structure (represents dna double helix) Transcription factors- proteins you need to run transcription. Apoptosis- programmed cell death: transcription: making an rna product from a gene. Horizontal lines represent the bases of the nucleotides. Non variable part- vertical line represents alternating phosphate sugar phosphate sugar. Rna: u instead of t; 2"c of sugar has oh. Initiation: specific proteins associate with the control region(s) of a gene. In prokaryotes, sigma and rna polymerase form a holoenzyme (fig. Rna polymerase separates dna strands to form the transcription bubble. Ase = enzyme- often a protein that catalyzes (takes part in but is not chemically consumed) a chemical reaction. Once the first few rntps are incorporated rna polymerase dissociates from the promoter and sigma.

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

QUESTION 10

If you can drink milk as an adult, it means that you have inherited a mutation in the promoter of your lactase gene (the gene that encodes the enzyme you need to break down lactose). Predict the effect of this mutation:

		The mutation changes the number of domains in the enzyme, which makes it work more efficiently
		The mutation changes the amino acid sequence of the lactase protein
		The mutation increases the number of copies of the lactase gene that will be found in your genome
		The mutation changes whether the lactase sequence is found in an intron or exon
		The mutation affects the expression of the lactase gene

1.2 points

QUESTION 11

A competitive inhibitor is decreasing the activity of an enzyme. Predict the effect of adding more substrate to the reaction.

		The substrate will increase the reaction rate by binding to the allosteric site
		The substrate will increase the reaction rate by competing with the inhibitor for the active site
		The reaction rate will not change unless the inhibitor can be removed
		The enzyme adjusts its shape so that the substrate, but NOT the competitive inhibitor, can bind
		The substrate will bind to the competitive inhibitor and block its ability to bind to the enzyme

1.2 points

QUESTION 12

What determines where in the genome a transcription regulator will bind?

		Transcription regulators bind to the 5' UTR region of a gene
		Regulators bind via complementary base-pairing to certain DNA molecules
		Covalent bonds form between the transcription regulator and the atoms of the DNA backbone
		Every eukaryotic gene has a different transcription regulator that will bind to the 5' end of the gene
		Transcription regulators bind to specific DNA sequences via multiple weak non-covalent interactions

1.2 points

QUESTION 13

What is the basic premise of cell theory?

		DNA -> RNA -> protein
		All cells arise from pre-existing cells
		DNA provides the complete instructions to create a cell
		The identity of a cell is determined through gene expression patterns
		All cells contain the same four basic macromolecules

1.2 points

QUESTION 14

What is the benefit of using BOTH the lac activator and the lac repressor to control gene expression?

		Using both an activator and repressor enables cells to more accurately determine the amount of lactose available in the environment
		Enzymes to digest lactose are only made when energy is low and lactose is available
		The activator can override the inhibition of the lac operon by the repressor
		The repressor can control the enhancer, while the activator can control the promoter
		When neither the lac activator or repressor is present, expression of the lac operon is too high

1.2 points

QUESTION 15

What is the histone code used for?

		Phosphorylation and acetylation of DNA affect its ability to be compacted
		Changes to the sequence of DNA change whether DNA will wrap around histone proteins
		Covalent modifications of histones affect the ability of the transcription initiation complex to form
		Histones provide the codon sequences needed for translation to occur
		The histone code affects which amino acids will get added to a polypeptide

BISC207 Lecture Notes - Lecture 19: Pyrimidine, Polyadenylation, Phosphodiester Bond

BISC207 Full Course Notes

Document Summary

Get access

Related Documents

BISC207 Lecture Notes - Lecture 2: Purine, Covalent Bond, Thymine

BISC207 Chapter Notes - Chapter 17: Uracil, Adenine, Pyrimidine

BISC207 Lecture Notes - Lecture 12: Knockout Mouse, Srb, Metabolic Pathway

Related Questions