BIOL 112 Lecture Notes - Lecture 7: Start Codon, Frameshift Mutation, Messenger Rna

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4 Apr 2016

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BIOL 112 Full Course Notes

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In the last lecture we sketched the outline of the central dogma. We showed how information is encoded in the sequence of bases in the dna and how that information determines, via the trna, the sequence of amino acids in the protein encoded by a particular gene. But the information in the dna is vast, 3 billion base pairs worth in humans, and there is no table of contents. In the first part of this lecture we discuss in more detail how the genetic information is extracted. First of all, as we alluded to, the rna polymerase makes an rna polymer that is complementary to one gene at a time. This process, where the dna is transcribed into messenger rna (mrna), is called transcription. The resulting mrna strand starts slightly before the stretch of dna that actually encodes the protein and extends somewhat past the coding region then stops. So the rna polymerase must know where genes start and stop.

Related Questions

During transcription in eukaryotes, a type of RNA polymerase called RNA polymerase II moves along the template strand of the DNA in the 3'â5' direction. However, for any given gene, either strand of the double-stranded DNA may function as the template strand. Which of the following initially determines which DNA strand is the template strand, and therefore in which direction RNA polymerase II moves along the DNA?

A. the location of specific proteins (transcription factors) that bind to the DNA

B. the specific sequence of bases along the DNA strands

C. the location along the chromosome where the double-stranded DNA unwinds

D. the position of the geneâs promoter on the chromosome

E. which of the two strands of DNA carries the RNA primer

When we use DNA to code for a protein, we call the process Transcription and Translation. Below are descriptions of processes that happen when we use DNA to code for a protein. Match the description of each process with the correct term - either Transcription or Translation.

1.A ribosome reads the mRNA and builds a protein from amino acids.
2.A copy is made of a gene in RNA
3.The primary transcript is modified: introns are removed and the transcript gets a cap and a tail
4.tRNA carries amino acids to a ribosome
5.The DNA unwinds, exposing the gene to be copied.

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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BIOL 112 Lecture Notes - Lecture 7: Start Codon, Frameshift Mutation, Messenger Rna

BIOL 112 Full Course Notes

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