BIOL10005 Lecture Notes - Lecture 9: Stop Codon, Start Codon, Release Factor

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jadebutterfly872

29 Jun 2018

School

University of Melbourne

Department

Biology

Course

BIOL10005

Professor

All

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Lecture 9: Gene expression: transcription, translation

and the genetic code

GENE EXPRESSION:

Genetic information is decoded into functional protein

DNA --> RNA --> PROTEIN

oIn mutant gene product: mRNA is still produced but the produced protein is

non-functional

oDNA --> RNA: transcription

oRNA --> PROTEIN: translation

GENES

Gene is sequence of DNA that encodes a functional product

oCoding sequence: found on one strand of DNA

Same as the RNA

Non-coding = complementary to RNA = template strand

Used to produce mRNA

oRegulatory sequence: allow for the gene to be expressed

oAlways oriented from 5' --> 3'

TRANSCRIPTION:

Initiation

oRNA pol binds to promoter

oUnwinds DNA

oUses template strand to produce RNA

Elongation

oRNA bases complementary to DNA template added

Termination

oRNA pol reaches a chain termination sequence

RNA PROCESSING IN EUKARYOTES (changing pre-mRNA)

Modification to increase mRNA stability

o5' methylated cap

o3' poly(A) tail

Splicing to remove introns (non-coding RNA regions)

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Document Summary

Lecture 9: gene expression: transcription, translation and the genetic code. Genetic information is decoded into functional protein. In mutant gene product: mrna is still produced but the produced protein is non-functional: dna --> rna: transcription, rna --> protein: translation. Gene is sequence of dna that encodes a functional product: coding sequence: found on one strand of dna. Non-coding = complementary to rna = template strand. Used to produce mrna: regulatory sequence: allow for the gene to be expressed, always oriented from 5" --> 3" Initiation: rna pol binds to promoter, unwinds dna, uses template strand to produce rna. Elongation: rna bases complementary to dna template added. Termination: rna pol reaches a chain termination sequence. Modification to increase mrna stability: 5" methylated cap, 3" poly(a) tail. Splicing to remove introns (non-coding rna regions) Ribosomes read codons on mrna o t-rna with complementary anticodons bring specific aa: ribosomes move along mrna & aa brought by trna forms peptide bonds o.

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

1. _________________________ is the name of the process that produces proteins using RNA molecules.

a.Transcription

b Translation

c. Replication

d. Mutagenesis

2. The peptide bonding site on the ribosome:

a. A site

b. P site

c. E site

d. D site

3. The name given to the bond formed between two amino acids.

a. Hydrogen bond

b. Double bond

c. Peptide bond

d. Negative bond

4. __________________________ is the movement of the mRNA molecule on the ribosome to open up the A-site for the new incoming amino acid.

a. Transformation

b. Mutation

c. Translocation

d. Initiation

5. What does the term semi-conservative mean.

a. Replication of RNA is semi-conservative, each new strand consists of one old strand of RNA and one new strand of RNA.

b. Replication of DNA is semi-conservative, each new strand consists of one old strand of DNA and one new strand of DNA.

c. Replication of DNA is semi-conservative, each new strand consists of new strands of nucleotides.

d. Transcription of RNA is semi conservative.

6. This molecule functions during protein synthesis to recognize the codon and bring the proper amino acid back to the ribosome for translation.

a. mRNA

b. tRNA

c. rRNA

d. DNA polymerase

7. What is meant by the base-pairing rule.

a. Bases on a single strand of DNA are always found as pairs

b. Adenine always base pairs with guanine; cytosine with thymine

c. Adenine always base pairs with thymine; cytosine with guanine

d. Bases on a single strand of RNA are found in pair.

8. In this step of protein synthesis, amino acids are added to a growing polypeptide chain.

a. Initiation.

b. Elongation.

c. Termination.

d. Transpiration.

9. Three complementary bases found in a mRNA are called ____________________.

a. Anti-codon.

b. Triplet.

c. Doublet.

d. Codons.

10. Which statement best describes the job or function of the messenger RNA.

a. Carries the map of the gene sequence in the form of codons to the cytoplasm

for translation of the protein.

b. It transfers amino acids to the ribosome for protein synthesis.

c. It forms the ribosomes.

d. It helps replicate DNA.

ceruleangnat663

22. What is probably the single most important distinction between bacteria and eukaryotes?

a. transcription

b. the separation of the cell's genetic material from the surrounding cytoplasm

c. the existence of ribosomes

d. RNA processing in bacteria

e. antibiotics

23. Which one does NOT characterize an open reading frame (ORF)

a. Uses six reading frames during translation

b. Contains a start codon

c. Contains a stop codon

d. Its length is in multiples of 3 nucleotides

e. It should have the number of codons for a protein or on average ~250 codons in length.

24. Nested intronic genes lie within the non-coding intronic region of a larger gene, and occur in the human genome.

a. True

b. False

25. How would you characterize this statement: EF-Tu will hydrolyze GTP only if the codon:anticodon & tRNA:amino acid matchings are correct

a. Transcription; initiation steps

b. Transcription; termination steps

c. Translation: initiation steps

d. Tanslation: elongation steps

e. Spliceosome

26. Regarding the transcription unit architecture of the Escherichia coli genome, which one is FALSE?

a. Each of the 4,661 transcription units comprised an average of 1.1 modular units.

b. 3,010 transcription units (~65%) are monocistronic.

c. 1,652 transcription units (35%) contain more than one ORF (polycistronic)

d. The largest transcription unit is equivalent to 16 ORFs with numerous introns.

e. 398 transcription units (~9%) are nested within each other in opposite transcriptional orientation

27. Regarding RNA interference (RNAi), which one is FALSE?

a. It is a common and conserved mechanism in eukaryotic cells.

b. Originally discovered in bacteria and plant viruses.

c. A form of primitive immunity to protect the genome.

d. Pre microRNA (miRNA) genes have been found in introns.

e. Recent research shows that exogenous plant miRNAs in food can regulate the expression of target genes in mammals.

28. Aminoacyl-tRNA synthetases

a. Recognize the 3âUTR

b. The genes for these enzymes fall into the family of inducible genes

c. The enzymes are imported into the nucleus

d. Covalently attaches an amino acid to the appropriate tRNA. This takes place in the cytoplasm.

e. Forms the peptide bond during protein synthesis

29. What is the process by which complementary RNAs are made from a DNA template?

a. translation

b. transcription

c. protein synthesis

d. transduction

e. transliteration

cyanbee225

BIOL10005 Lecture Notes - Lecture 9: Stop Codon, Start Codon, Release Factor

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