BILD 1 Lecture Notes - Lecture 25: Start Codon, Tata Box, Alternative Splicing

45 views4 pages

byzantiumfly635

26 Apr 2016

School

UCSD

Department

Biology/Lower Division

Course

BILD 1

Professor

Stephanie Mel

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

Document Summary

Chapter 14: gene expression from gene to protein. Dna is the genotype, but the physical appearance and function is called the phenotype. How does 4 letters become a whole entire function or appearance. The dna coding strand is completely reflected in the mrna. Same sequence as dna but in rna (u replaces t) Think of it as the top template of the dna being replicated, but it is made out of rna mrna directs the synthesis of polypeptides. The process is similar to bacteria cells, but eukaryotic cells have more complexity. There are 3 nucleotides coding for 20 amino acids and can produce 64 possible combinations. Each amino acid is coded by 3 letter words. 61 of those codons code for amino acids. More than 1 codon can code for the same amino acid. A signal to stop making a certain protein. There is 1 start codons (aug is the point where the codon starts)

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

erinporpoise384

Which type of RNA is unique to eukaryotes?

		mRNA
		rRNA
		tRNA
		snRNA
		Its a trick, all cells have all four types of RNA.

1 points

QUESTION 2

The region of DNA to which RNA polymerase binds to initiate transcription is the:

		origin
		promoter
		enhancer
		none of the above are correct

1 points

QUESTION 3

Holoenzyme:

		is RNA polymerase core enzyme.
		is sigma factor.
		is core plus sigma factor.
		none of the above are correct.

1 points

QUESTION 4

Which of these is formed FIRST when RNA polymerase binds to a promoter?

		The closed promoter complex
		The open promoter complex
		A rho-independent terminator
		none of the above are correct.

1 points

QUESTION 5

A hairpin loop followed by a run of Us:

		is the sequence formed in the sense strand of a gene using a rho-independent terminator.
		is the sequence recognized by the rho protein.
		is the structure formed in RNA that causes rho-independent termination.
		is formed immediately after the open promoter complex is formed.

1 points

QUESTION 6

RNA polymerase:

		reads the template strand of the DNA in the 3' to 5' direction.
		synthesizes the RNA in the 5' to 3' direction.
		can begin transcription without a primer.
		all of the above are correct.

1 points

QUESTION 7

Which of the following statements is FALSE?

		Bacteria like E. coli have only one type of sigma factor.
		RNA polymerase has no proofreading activity.
		The promoter of a gene is upstream of the RNA-coding region.
		Bacterial transcriptional terminators are copied into RNA to cause termination.

1 points

QUESTION 8

If the template strand of a gene has the sequence 5' ATGCGAT 3', then:

		the RNA produced will have the sequence 5' AUGCGAU 3'.
		the RNA produced will have the sequence 5' UAGCGUA 3'.
		the RNA produced will have the sequence 5' AUCGCAU 3'.
		None of the above are correct.

1 points

QUESTION 9

The sense strand:

		is the lagging strand
		is also the template strand.
		always goes left to right.
		none of the above are correct.

1 points

QUESTION 10

Which of these is transcribed but not translated?

		mRNA
		tRNA
		rRNA
		all of the above
		tRNA and rRNA are transcribed but not translated.

please help i am a bit confused

Multiple answers to each question might be possible!

You decide to identify the CFTR mutation by analyzing the genomic DNA of your patients compared to healthy individuals. You specifically are looking to see whether a specific 3' gene truncation has occurred in the patients. You will determine this using hybridization techniques with samples from healthy and CF patients. Which of the following will allow you to accomplish this?

	Using an RNA probe complementary to the region not removed by the truncation.
	Using an RNA probe complementary to the region removed by the truncation.
	Using an DNA probe complementary to the region not removed by the truncation.
	Using an DNA probe complementary to the region removed by the truncation.

To conduct the hybridization experiment, you are trying to decide between using a DNA or RNA probe. Which would be ideal to use and why?

	As both are composed of nucleic acids, using either would result in identical results.
	An RNA probe because RNA has uracil bases.
	An RNA probe because it could also be used in a translation experiment.
	A DNA probe because it is more stable than RNA.
	A DNA probe because RNA cannot bind to DNA.

One step of the Hershey/Chase experiment involved blending the virus/cell mixture before centrifugation and probing the pellet for radioactivity. Why was the blending step necessary?

	To collect the bacteria at the bottom of the tube.
	To break open the bacteria to release the genome.
	To separate the bacteria from the bacteriophages.
	To be able to detect the radioactivity.

Imagine Hershey/Chase had used an RNA virus (genome composed of RNA) instead of a DNA virus in their experiment. Would radioactivity still have been found in the pellet?

	No, because only DNA can be labeled with radioactivity.
	No, because the RNA genome would not enter the bacteria upon infection.
	No, because while DNA and RNA nucleotides are similar, they are not identical.
	Yes, because DNA and RNA nucleotides are similar.
	Yes, because genome in any form (DNA, RNA, protein) would be labeled similarly.

The human genome consists mostly of non-coding DNA. Which of the following are benefits of this?

	Random DNA mutations generally won't affect RNA and protein function.
	It is faster to duplicate the genome when these are present.
	The existence of introns can lead to multiple variations of proteins encoded by a single gene.
	It is unlikely transposons would exist in the genome if there was too much protein coding DNA.

Explain the 5â to 3â directionality of a DNA stand.

	It is due to the fact that the free 5â carbon is on one end and the free 3â carbon is on the other
	It is due to the fact that new nucleotide are added to the 5â carbon of the previous nucleotide
	It is due to the fact that there are 3 phosphate groups attached to the 5â carbon
	It is due to the fact the complementary strand is 3â to 5â
	More than one of the above explain the 5â to 3â directionality

You accidentally add a mutant dNTP (which has an H instead of an OH connected to the 3â carbon) to a reaction where DNA is being replicated. Which of the following is true of this mutant dNTP?

	It can be incorporated into DNA strand but cannot form a phosphodiester bond with an incoming wild type dNTP
	It can be incorporated into a DNA strand but cannot base pair with a complementary nucleotide
	It can be incorporated into a DNA strand and can form a phosphodiester bond with an incoming dNTP, but only if it is another mutant dNTP
	It cannot be incorporated into a DNA strand.

Why does DNA polymerase utilize an RNA primer?

	DNA polymerase is unable to initiate strand synthesis but RNA polymerase can
	DNA polymerase can only add a dNTP to an rNTP
	DNA synthesis proceeds in the 3â to 5â when initiating strand synthesis
	Chromosomal DNA contains interspersed RNA fragments
	The RNA primer increases stability of the newly synthesized strand

rosehippopotamus950

BILD 1 Lecture Notes - Lecture 25: Start Codon, Tata Box, Alternative Splicing

Document Summary

Get access

Related Documents

BILD 1 Chapter Notes - Chapter 17: Neurospora Crassa, Rna Splicing, Alternative Splicing

BILD 1 Chapter Notes - Chapter 14: Gene Expression, Coding Strand, Start Codon

BILD 1 Lecture 19: Lecture 19

Related Questions