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Translation Flashcards

1
Q

Learning objectives

A
  • Describe the genetic code
  • Explain the history and experimental design associated with the discovery of the genetic code
  • Explain the redundancy in the genetic code at the mRNA level
  • Explain the redundancy in the recognition of the genetic code by tRNAs using the wobble hypothesis
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2
Q

The Central Dogma

A
  • Translation results in the production of proteins
  • Consist of amino acids joined by peptide bonds
  • It requires
    > mRNA containing the coding region
    > RNA carrying the amino acids
  • Ribosomes
  • A machine consisting of proteins and ribosomal RNA (rRNA)
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2
Q

Overview of gene expression photo

A
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3
Q

The ‘Universal’ Genetic Code

A

Transcription = DNA to RNA
* Similar language, 4 bases matching 4 bases, copy
Translation = RNA Protein
* Different language, 4 bases and 20 amino acids, translate
* DNA/RNA language = 4 bases (GATC/ GAUC).
* Maximum number of codons in groups of three can be calculated:
* 4x4x4 = 64 possible codons
So code is ‘degenerate’ or ‘redundant’

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3
Q

Some codons are unique and others are not recognised by tRNAs

A
  • AUG encodes only for Methionine:
  • This amino acid is found at the start of every open reading frame
  • Initiation signal for translation
    > In prokaryotes, formyl-methionine is the first position in almost all proteins
    > In eukaryotes, it is methionine
  • Three codons are not recognised by tRNAs:
  • These are STOP codons
    > Found at the end of every translated open reading frame
  • Termination signal for translation
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3
Q

Snippet of history: Cracking the Genetic Code, pages 248-247

A

In 1961, Nirenberg and Matthaei produced an artificial RNA composed entirely of uracil, poly(U) and were able to synthesize a protein of poly-phenylalanine. They used a cell-free system:
-E. coli cytosol fraction
-destroyed native mRNA
-added radiolabelled amino acids

Repeated the experiment with poly(A) mRNA to create poly-Lysine
And poly(C) mRNA to create a poly-Proline

1960s - Central Dogma accepted
- 4 bases - codon - 20 aa
- did not know code?
- limited by techniques

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3
Q

Cracking the Genetic Code cont…

A

Next step: Organic chemist Gobind Khorana developed a method of making specific RNA strands of repeating di- tri- and tetra-nucleotide sequences through transcription of DNA into RNA with RNA polymerase.

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4
Q

Cracking the Genetic Code cont… photo

A
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4
Q

Cracking the Genetic Code cont…

A
  • Final step: Nirenberg and Leder realized that they could trap trinucleotides (size of a codon) with the ribosome and this would also trap the corresponding
    charged
    aminoacyl-tRNA
    (radiolabelled). In this way they were able to quickly decipher all the remaining codon sequences i.e what amino acid they code for.

In the final analysis, there are 61 codons for 20 amino acids.
So we need 61 tRNAs for each codon.
But a cell only contains 50 tRNAs!

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4
Q

Degeneracy in tRNA recognition of the genetic code: the Wobble hypothesis photo

A
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5
Q

Wobble hypothesis cont… photo

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6
Q

Summary

A
  • The code is a triplet - each codon consists of a unique combination of 3 nucleotides
  • The code has punctuations only at the end of the coding region in the mRNA:
  • Start codon is AUG
  • Stop codons are UAA, UAG, and UGA
  • Most amino acids are specified by more than one codon, so the code is redundant
    > only methionine and tryptophan are specified by a single codon
  • The code is consistent - each one of the codons specifies a single amino acid
  • The code is universal - viruses, bacteria, plants, and animals all use the same code
  • there are some minor variations in mitochondria and a few fungi
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7
Q

Actual Gene Sequence:
Rat zinc finger protein 96

A

Conventions
You can read your DNA in the 5’ to 3’ direction and using the genetic code indicate what amino acids
should be
encoded
(Short hand without writing it into mRNA first- tedious!)

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7
Q

The ‘Universal’ Genetic Code

A

Transcription = DNA → RNA
-similar language, 4 bases matching 4 bases, copy
Translation = RNA → Protein
-different language, 4 bases and 20 amino acids, translate
DNA/RNA language = 4 bases (GATC/ GAUC).
Maximum number of codons in groups of three can
be calculated:
4x4x4 = 64 possible codons
So code is ‘degenerate’ or ‘redundant’

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8
Q

Exercise 1: Translating Competition
RNA to Protein photo

A
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8
Q

Exercise 2 photo

8
Q

Lecture Outcomes

A
  • Describe the three stages of protein synthesis
  • Understand the term translational reading frames
  • Describe and know the differences in initiation of translation in eukaryotes and prokaryotes
  • Describe the initiation signals in eukaryotes and prokaryotes
  • Describe the difference between the initiation methionine in eukaryotes and prokaryotes
  • Describe the processes of elongation and termination of translation
  • Describe why certain antibiotics can specifically inhibit bacterial translation and not eukaryotic translation
9
Q

The Central Dogma

A
  • Translation results in the production of proteins
  • Consist of amino acids joined by peptide bonds
    It requires
    > mRNA containing the coding region
    > IRNA carrying the amino acids
  • Ribosomes
  • A machine consisting of proteins and ribosomal RNA (rRNA)
10
Q

Protein Synthesis

A

Divided into 3 stages:
50S
1. Initiation - mRNA + ribosomes + initiating tRNA
2. Elongation - peptide bonds + movement
3. Termination - dissociation of ribosomes & peptides

10
Q

A genetic code based on groups of three nucleotides results in three potential ways of reading a sequence!

A
  • Since there are three nucleotides per codon,
    potential
    there are three potential translational reading frames per strand of mRNA
  • Frame 1
  • Frame 2
  • Frame 3
  • But only one frame is used for translation and the other frames are ignored. How?
11
Q

A genetic code based on groups of three nucleotides results in three potential ways of reading a sequence! photo

12
Q

Revision: Ribosomes and their binding sites photo

12
Q

Revision: Ribosomes and their binding sites

A
  • Each ribosome has one binding site for mRNA and three binding sites for tRNA
  • tRNA binding sites:
  • A = aminoacyl-tRNA : new tRNA enters ribosomal complex
  • P = peptidyl-tRNA : tRNA attached to the polypeptide chain
  • E = exit: de-acylated tRNA exits ribosomal complex
  • mRNA is translated in the 5’-to-3’ direction, and the N-terminal end of a protein is made first, with each cycle adding one amino acid to the C-terminus of the polypeptide chain
13
Q

Translation initiation signals

A
  • AUG is a translation initiation signal.
  • But methionine can occur anywhere in a protein
  • How does the translation machinery pick the right AUG?
  • Must have other initiation signals such as a ribosome binding site motif in the mRNA
  • In eukaryotes, the ribosomes have a scanning ability to find the right site
  • In bacteria, the small subunit of the ribosome binds to the Shine Dalgarno sequence (5’AGGAGGU-3’) located upstream of the AUG codon
13
Initiation in eukaryotes
* Initiation site is crucial as it will define the correct open reading frame. * First step: Initiation Met-tRNAi first loaded into the "p" site of small ribosomal subunit with initiation factors (elF2-GTP, eukaryotic Initiation Factor) * Second step: The loaded small ribosomal subunit attaches to the 5 end of the mRNA (assisted by 5'end cap) * Third step: It then scans along the mRNA (5' to 3' direction) until it identifies the first AUG codon, surrounded by a long consensus sequence (Kozac sequence) * If the Kozac sequence is degenerate, the ribosome will initiate translation from multiple start codons producing truncated polypeptides * Fourth step: The initiation factors detach and the large subunit can bind and complete the ribosomal complex * The first amino acid is in the 'P' site ready for chain elongation
14
Initiation in eukaryotes photo
15
Initiation of translation in bacteria
* First step: 30S interacts with initiation factor (IF) and this complex binds the Shine Dalgarno sequence upstream of the AUG start codon * Second step: Initiator tRNA binds (fMet-tRNAi) aligns with start codon (AUG) in mRNA * Third step: 50S associates with 30S, releasing IF and forming 70S complex * ERNA occupies peptidyl (P) site of 50S subunit (aminoacyl (A) site is empty) (note the exit site present but not shown in this diagram)
15
Initiation of translation in bacteria photo
16
Elongation and Initiation tRNA-Met is different between bacteria and eukaryotes
* Initiation - bacteria * The initiator tRNA in bacteria has a formylated methionine (N-formyl-methionine RNA) or Met-tRNAi * Initiation - eukaryotes > The initiator tRNA in eukaryotes is Met-tRNAi (not formylated) * Elongation in both use * When a methionine codon appears inside a coding region, an elongation Met-tRNA is used. > Met-tRNA which has a different stem loop structure that preferentially binds elongation co-factors
17
Elongation
* Elongation is a three-step cycle which is repeated over and over during the synthesis of a protein chain * Step 1: an aminoacyl-tRNA molecule binds to a vacant A-site on the ribosome * Step 2: a new peptide bond is formed between the amino acids attached to the tRNAs in the P and A sites, to create tRNA-peptide in the A-site > Step 3: the mRNA moves a distance of three nucleotides through the small subunit (uses energy from ATP to ADP+PPi) * the de-acylated tRNA molecule is ejected from the E site * The A-site-RNA-peptide moves into the P site * The A-site is now empty and located over the downstream codon * The A-site is ready to receive another aminoacyl-tRNA molecule which base pairs with this codon * The position at which the growing peptide chain is attached to a tRNA does not change during the elongation cycle: it is always linked to the tRNA present at the P-site of the large subunit
18
Elongation continued photo
19
Elongation cont...
* Ribosomes are made of RNA and polypeptides * Polypeptides are structural > IRNA forms a highly structured pocket of hydrogen networks and has catalytic properties * Therefore ribosomes are ribozymes- that is the RNA performs the condensation of the carboxy-terminus of the amino acid on the tRNA in the P site with the amino-terminus of the amino acid in the tRNA in the A-site * The peptide-tRINA in the A-site moves to the P-site as the ribosome moves down the mRNA in the 3' direction.
20
Elongation cont... photo
21
Termination of translation
* Termination is initiated by stop codons: - UAA, UAG, UGA - do not match a tRNA * Releasing factors bind to a stop codons that reaches the 'A' on the ribosome - this alters the activity of the peptidyl transferase **adds a water molecule to the peptide and releases it** The ribosome releases the mRNA and then the 2 subunits dissociate and can bind to the same or another mRNA at the 5'end
22
Overview of Translation in Bacteria
22
A question about antibiotics! photo
23
Comparison/contrast between Prokaryotic and Eukaryotic Translation machinery
The subtle differences between prokaryotic (bacterial) and eukaryotic ribosomes is enough to create high affinity interactions between antibiotics and the prokaryotic ribosome. Eukaryotic ribosomes are not usually inhibited (except for ribosomes in the mitochondria which are more like those of prokaryotes eg chloramphenicol)
24
Comparison/contrast between Prokaryotic and Eukaryotic Translation machinery photo
24
Summary
* There are three potential translational frames in any mRNA- however, only one is used. * Translation occurs in three steps: initiation, elongation and termination * Initiation of translation requires initiator Met-tRNAi in eukaryotes and fMet-tRNAi in prokaryotes. * When Met appears inside an open reading frame, Met-tRNA is used and this binds elongation co-factors. * Elongation is a process using ATP to move the ribosome in the 5' to 3' direction, inserting tRNAs above codons and catalysing peptide bond formation between peptides. * Termination occurs when the ribosome encounters a stop codon in the A-site, leading to a hydrolysis reaction, releasing the peptide from the final tRNA and causing dissociation of the ribosome from the mRNA.

Molecular Biology Flashcards (28 decks)

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