BSC 315 Chapter Notes - Chapter 8: Sigma Factor, Methionine, Stem-Loop

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byzantiumgiraffe268

16 Oct 2016

School

University of Alabama

Department

Biological Sciences

Course

BSC 315

Professor

Dr.Jeffrey Lozier

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Document Summary

11:00 pm: triplet codons of nucleotides represent individual amino acids, each nucleotide triplet is termed a codon, a gene"s nucleotide sequence is co-linear with the amino acid sequence of the. Encoded polypeptide: speaks for itself really , evidence that a codon is composed of more than one nucleotide, not all point mutations result in a missense mutation. Intrinsic - do not require addition loops (ex. Hairpin loops): primary transcript is the result of transcription, only in eukaryotes must the primary transcript undergo posttranscriptional modification, methylated 5" cap (g) Poly-a 3" tail: exons and introns, exons. Sequences found within a gene"s dna and mature mrna: not spliced out by spliceosome, small nuclear rna w proteins assemble into these spliceosomes, introns. Sequences found within a gene"s dna but not in mrna. Spliced out by spliceosome: hence a primary transcript contains both introns and exons, short sequences in the primary transcript determines where the splicing occurs.

Related Questions

Part B

Think about the DNA coding sequence of a gene. If an A were swapped for a T, what kind of mutation could it cause and why?

a) It could cause a frameshift nonsense or frameshift missense mutation because it would change the reading frame of the codon triplet.

b) It could cause a silent, missense, or nonsense mutation because those are the types that can be caused by a nucleotide-pair substitution like this one.

c) It could cause a nonsense mutation because the sequence would no longer be the same, so the protein would be shorter and non-functional.

d) It could cause a silent mutation because A and T are complementary to each other so it is not really a substitution mutation.

Part C

Why is a frameshift missense mutation more likely to have a severe effect on phenotype than a nucleotide-pair substitution missense mutation in the same protein?

a) A substitution missense mutation causes the protein to be shorter and thus non-functional.

b) A frameshift missense will cause the codons to be out of order, but a substitution missense does not change the order of the codons.

c) A frameshift missense mutation will cause an early Stop codon, but a substitution missense might be silent.

d) A substitution missense affects only one codon, but a frameshift missense affects all codons downstream of the frameshift.

Part D

Which of the following sequences shows a frameshift mutation compared to the wild-type mRNA sequence?

a) wild-type	5â-AUGCAUACAUUGGAGUGA-3â
mutant	5â-AUGCAUACAGAGUGA-3â

b) wild-type	5â-AUGCAUACAUUGGAGUGA-3â
mutant	5â-AUGCAUACGUUGGAGUGA-3â

c) wild-type	5â-AUGCAUACAUUGGAGUGA-3â
mutant	5â-AUGCAUACAUCUGGAGUGA-3â

d) wild-type	5â-AUGCAUACAUUGGAGUGA-3â
mutant	5â-AUGCAUGUGACAUUGGAGUGA-3â

Part F

Suppose that the triplet of nucleotides indicated in bold (AGC) spans two codons, that is, CTA and GCC. If the triplet AGC were deleted from this DNA coding sequence, what effect would it have on the resulting protein?

5â-ATGCTAGCCTATCGTAAC-3â

a) The two flanking codons would be altered, but the rest of the amino acid sequence would be the same because there would be no frameshift.

b) The entire amino acid sequence would be altered due to the frameshift.

c) All of the amino acids after the deletion would be altered due to the frameshift.

d) All of the amino acids up to the deletion would be altered due to the frameshift.

mauvetrout554

Fill in the blank. Elongation during translation does NOT involve ____________.

Question 16 options:

	the translation of codons according to the genetic code
	the formation of bonds catalyzed by the ribosome
	complementary base pairing between RNA molecules
	amino acids being linked together in a polypeptide
	reading the DNA template 3' to 5'

For a given gene, what establishes the reading frame for translation?

Question 17 options:

	the location of the enhancer relative to the gene
	the first three nucleotides at the 5' end of the mRNA
	the first three nucleotides at the 3' end of the mRNA
	the start codon in the mRNA
	the location of the promoter relative to the gene

Which of the following is the LEAST likely direct consequence of a substitution mutation?

Question 18 options:

	changing the length of a protein coded for by a gene
	changing one amino acid in a protein
	creating a stop codon
	eliminating a start codon
	changing the length of the DNA molecule containing a gene

Suppose that the pre-mRNA transcript from a eukaryotic gene is 30,000 nucleotides long, and the gene codes for a sequence of 300 amino acids. What is the best explanation for the relationship between these numbers?

Question 19 options:

	only the first 900 nucleotides of the pre-mRNA transcript are translated
	it takes 100 nucleotides to specify a single amino acid
	300 of the nucleotides in the transcript are important, and the rest are "junk"
	only the last 900 nucleotides of the pre-mRNA transcript are translated
	large portions of pre-mRNA transcripts are cut out during RNA processing

Suppose an individual is born into a population with a novel mutation. Is the new mutation an evolutionary change, and why?

Question 20 options:

	no, because it is not a big enough change to count
	yes, because new mutations are always adaptive
	yes, because the appearance of a new genetic variant is a genetic change in a population
	no, because not enough individuals have the mutation for it to matter
	no, because most mutations are not adaptive

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

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BSC 315 Chapter Notes - Chapter 8: Sigma Factor, Methionine, Stem-Loop

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