Lecture 7 - From DNA to Protein: Gene Expression Flashcards
Central Dogma
DNA (transcription) -> RNA (translation) -> Polypeptide
Some viruses have ___ as their main basis of information storage.
RNA
RNA (transcription) -> RNA (translation) -> Polypeptide
One gene-one enzyme hypothesis was validated by the study of _________.
Alkaptonuria
- mutation that causes accumulation of homogentistic acid.
- when the allele is mutated, the enzyme to break down homogentistic acid is inactive.
Testing the one gene-one enzyme hypothesis required model organisms which have these four characteristics (4)
- easy to grow in lab
- short generation times
- easy to manipulate genetically
- produce large numbers of progeny
ex. pea plant, mouse, drosophila, bread mold
Testing the one gene-one enzyme hypothesis by comparing ______ strains with the wild type (3)
mutant
- mutations were induced with X-rays
- the mutant strains needed additional nutrients
- they were able to determine the metabolic pathway of arginine.
Two main players of transcription (2)
- RNA: like DNA, is formed from nucleotides (has ribose sugar, uracil instead of thymine)
- RNA polymerase: catalyzes transcription
There are ___ types of RNA polymerases in Eukaryotes:
three
- RNA polymerase I transcribes genes encoding ribosomal RNAs.
- RNA polymerase II transcribes genes encoding proteins.
- RNA polymerase III transcribes genes encoding transfer and ribosomal RNAS (tRNA & rRNA).
RNA polymerase catalysis reaction (4 main points)
- 5’-3’ direction
- copies only one strand at a time
- processive - one enzyme-template binding results in polymerization of hundreds of RNA bases
- does not need a primer.
Transcription Initiation (2)
- RNA polymerase will bind to the promoter (promoter is a sequence of DNA in front of the gene of interest)
- This binding will determine directionality and which of the two DNA strands is going to be transcribed.
Transcription (2)
- template is the sequence of DNA being copied.
- non-template is the sequence of DNA that codes for the mRNA being produced. also, non-template has promoter and has identical base sequence to RNA (except for T for U).
Transcription Elongation (2)
- RNA polymerase unwinds DNA about 10 base pairs at a time; reads template in 3’ to 5’ direction.
- RNA polymerases do not proofread and correct mistakes.
Transcription termination (2)
- specified by a specific DNA sequence.
- In prokaryotes, helper protein binds to the transcript and causes it to detach from the DNA. The transcript forms a loop (hairpin) and falls away from the DNA.
Transcription of eukaryotes require RNA processing (3 steps)
- splicing
- capping
- polyadenylation
Eukaryotic mRNA contains ____.
introns (INTervening regiONS)
Splicing (2)
- splicing is the process by which introns are removed to produce a mature mRNA.
- process is performed by the spliceosome complex which is a mixture of RNA and protein.
Spliceosome assembly (4)
- specific sequences on the pre-mRNA identify the presence of an intron.
- snRNPS (small nuclear ribonucleoproteins) bind to these sequences and recruit components of the spliceosome.
- the hydroxyl (-OH) group of the new 3’ end and the phosphate (PO4) group of the new 5’ end react to connect exons, relieving the intron and spliceosome complex to create mature mRNA.
- the excised intron and spliceosome are degraded.
mRNA capping (3)
- a 5’-cap is added at the 5’ end
- the cap is a modified GTP which facilitates mRNA binding to a ribosome and helps translation initiation.
- also protects mRNA from being digested by ribonucleases.
Polyadenylation (2)
- a poly A tail is added at the 3’ end. This is an addition of ATP at the end of the mRNA.
- The tail assists in export from the nucleus and is important for stability of the mRNA.
Translation has __ players.
- mRNA: codes for the protein that is being produced. also note that mRNA is the product of transcription.
- tRNA: brings the amino acids and reads the code presented by mRNA.
- rRNA: ribosomal RNA that makes up the ribosome which catalyzes the translation reaction.
tRNAs (4)
- each tRNA will bind to a specific enzyme (aminoacyl tRNA synthetase) that will catalyze the attachment of a specific amino acid.
- the tRNA recognizes the codon of the mRNA
- binds to the ribosome
- has attached specific amino acid (the amino acid attachment site is always CCA)
tRNA charging (2)
- charging requires ATP; a high-energy bond forms between the amino acid and the tRNA.
- this energy is later used to form the peptide bond.
Decoding the code (2)
- Nirenberg and Matthaei used simple artificial mRNAs of known composition to identify the polypeptide that resulted.
- this led to the identification of the first three codons.
The code (3)
- AUG is the start codon: initiation signal for translation
- Three different stop codons: termination signals, including UAA, UAG, and UGA
- For most amino acids, there is more than one codon: the genetic code is redundant.
Wobble - a redundant code (3)
- Wobble: Specificity for the base at the 3’ end of the codon is not always observed.
- Example: Codons for alanine - GCA, GCC, and GCU - are recognized by the same tRNA.
- As a result of wobble there are fewer tRNA species, but the genetic code is not ambiguous.
The ribosome (3)
- catalyzes translation
- complex composed of rRNA and proteins (catalysis is performed by the rRNA)
- ribosome reads the mRNA 5’-3’ direction and catalyzes the reaction from the N- to the C-terminal.
There are ___ subunits of the ribosome.
two: large and small
There are ___ binding sites on the large subunit of the ribosome.
three
- A (aminoacyl tRNA) - binds with anticodon of charged tRNA
- P (peptidyl tRNA) - where tRNA adds its amino acid to the growing chain
- E (exit site) - where tRNA sits before being released from the ribosome.
Translation initiation (2)
Prokaryotes:
- rRNA binds to the Shine-Dalgarno sequence of the mRNA
- An initiation complex forms - a charged tRNA and small ribosomal subunit, both bound to mRNA
Eukaryotes:
- ribosome binds to the 5’ cap to initiate translation
Translation elongation (3 steps)
- Incoming charged tRNA pairs with the codon at the A-site
- The free end of the amino acid on the A site reacts with the polypeptide chain, releasing the tRNA on the P site; ribosome shifts
- Shifting of ribosome opens a new A site.
Translation termination (2)
- translation ends when a stop codon enters the A site
- Stop codons bind a protein release factor which hydrolyzes bond between the polypeptide and the tRNA in the P site.
Proofreading mechanism (3)
- ribosome checks for correct codon-anti codon pairing.
- charged tRNA binds to EF-Tu (elongation factor)
- If correct codon-anti codon match occurs, EF-Tu releases the charged tRNA and reaction proceeds.
Protein processing: Post-translation modifications (3)
Most polypeptides are modified after translation: not all proteins have methionine (AUG) as the first AA. This is due to post-translation modification.
- Phosphorylation: addition of phosphate groups catalyzed by protein kinases (the charged phosphate groups change the conformation and may expose active sites or binding sites)
- Glycosylation: addition of sugars to form glycoproteins.
- Proteolysis: polypeptide is cut by proteases (e.g., signal sequence is removed)
- Pro-Opiomelancortin (POMC) is a pro-hormone.
Differences between prokaryotes and eukaryotes (2)
- eukaryotes have a nucleus so transcription occurs inside the nucleus; translation occurs in the cytoplasm. prokaryotes’ transcription and translation occur in cytoplasm.
- eukaryotes mRNA gets capped, spliced, and polyadenylated.
Mature mRNA leaves the nucleus through ______ ___.
nuclear pores.
Targeting (3)
- protein sequences (targeting sequences) can be recognized by targeting proteins to bring the newly synthesized protein to the right place (e.g. nuclear localization sequence, export sequence)
- ER sequence will target the protein to be synthesized in the ER
- In the ER, the protein can be targeted to a specific region (plasma membrane, vesicles, lysosome)
