Download Biochemists Break the Code

教
案
~ 2007 学年 第一 学期
2006
学 院
教
名 称
研
生命科学学院
室
课 程
名 称
授 课
对 象
授 课
教 师
陈文利
称
副教授
职
教 材
名 称
2006 年 9 月
生物化学
2005 级生物技术专业
现代生物化学
日
授课题目（教学章、节或主题）
：
教学器材
与工具
第十二章 蛋白质的生物合成与修饰
多媒体设施、黑板与
笔
第 16，17 周一第
授课时间
48-52 节
教学目的、要求（例如识记、理解、简单应用、综合应用等层次）
：
重点介绍原核生物的蛋白质合成过程。
教学内容（包括基本内容、重点、难点）
：
Translation & Post-translational processing, sorting and targeting
Major components
1) Ribosome
2) mRNAs
3) tRNAs
4) aminoacyl-tRNA synthetases (aaRS)
5) IFs, EFs and RFs
General features
Detailed Mechanism
Post-transcriptional Processing
Sorting and Targeting
Functional sites on Ribosome
Three sites for tRNA
1) The A (acceptor) site - where aa-tRNAs come in (except the first one) and where
peptidyl-tRNA is after peptide bond formation and before translocation.
2) The P (peptidyl-tRNA) site - where peptidyl-tRNA is before peptide bond formation.
3) The E (exit) site - where the uncharged tRNA from the P site goes after translocation.
Peptidyl transferase - the active site that catalyzes formation of the peptide bond (23S rRNA in
Prokaryotes)
Polypeptide exit channel
mRNA binding site
The ribosome is a ribozyme?!
“ …. most biologists did not seriously consider the possibility that RNA could be playing more
than a bit part ... In the ribosome it has turned out that most of the intersubunit interface is RNA,
the peptidyl transferase centre is RNA, and the decoding site and most of the A and P sites are
RNA. It appears that the modern ribosome is composed of a somewhat geriatric, but
functionally vital, RNA scaffold that is propped up and doted upon by its protein grandchildren.
The ribosome Is one colossal enzyme.”
James R. Williamson Scripps, Skaggs Inst.
Nature 407: 306
Science 289:878 (T.R. Cech)Science289.878.pdf
tRNAs
Secondary & Tertiary structure
Isoacceptor tRNAs (同工受体tRNA）
Different tRNAs carrying the same aa
tRNA identity- “2nd genetic code”
Special sequence elements on tRNAs that can be recognized by aaRS and then
determine which aa is charged
Positive elements & Negative elements
Some special tRNAs
1) Initiator tRNA: tRNAf Met & tRNAiMet
2) Suppressor tRNA（校正tRNA）
3) tmRNA
Aminoacyl-tRNA Synthetases
Two-step Reaction equation:
1) ATP + amino acid (AA) --> AMP-AA + PPi
2) tRNA + AMP-AA --> tRNA-AA + AMP
Classification
1) Class I aaRSs
2) Class II aaRSs
Proof-reading - quality control at the level of charging
1) The RS is the only place where aa identity is checked
2) The ribosome doesn't care what aa is attached to a tRNA
3) Charged tRNA can be modified and it still works
4) Pre vs post charging editing
5) Double sieve idea
Proof-reading of aaRS
Pre-charging editing
1) Sometimes aaRS will activate the wrong aa
2) Binding of cognate tRNA triggers hydrolysis of the aa-AMP instead of charging
3) aa-AMP is transferred to an editing domain that preferentially hydrolyzes incorrect
aa-AMP
Post-charging editing
1) Sometimes aaRS will activate the wrong aa AND transfer it to the tRNA
2) aa-tRNA is transferred to an editing domain that preferentially hydrolyzes incorrect
aa-AMP before the aa-tRNA can be released from the aaRS.
The double sieve idea
1) Hard to build the perfect active site
2) activation and editing sites each act as sieves to reject different kinds of errors
An Example - IleRS
Isoleucine is a large hydrophobic aa.
hydrophobic aas larger than Ile can't fit into the activation pocket - the first sieve
Valine looks like Ile minus a terminal CH3 group - valine fits the activation pocket
Valine leaves a hole in the complex, so Ile still binds better
Valine binds well enough to give significant misactivation
The editing domain has an active site that fits valine perfectly
Ile is too large to fit into the editing site, so it does not get rejected - the second sieve
Valine and smaller aas - e.g. Alanine - are hydrolyzed
IFs, EFs and RFs
Initiator factors (IF)
Prokaryotes: IF1, IF2 and IF3
Eukaryotes: eIFs
Elongation factors (EF)
Prokaryotes: EF-Tu, EF-Ts and EF-G
Eukaryotes: eEF-1 and eEF-2
Release Factors (RF)
Prokaryotes: RF-1,RF-2 and RF-3
Eukaryotes: eRF
General features
With mRNA as template, tRNA as carrier for aa, ribosome as assembly site
Translational polarity
1) Extends in N-end → C-end
how to prove?
2) Reads mRNAs in 5’ → 3’
Triplet codon
1) How many bases determine one aa?
“three determine one”
2) Cracking of the genetic code
3) Features of the genetic code
Ribosomes recognize aa-tRNA just by virtue of the base-pairing interaction between codons
and anticodons
Wobble hypothesis
Biochemists Break the Code
Assignment of "codons" to their respective amino acids
Marshall Nirenberg and Heinrich Matthaei
Experiment:
They set up 20 different test tube reactions
Each one was spiked with a different radioactive aa
They programmed each with the poly-U RNA
Then recovered the proteins by acid precipitation
Under these conditions the proteins precipitate but the free amino acids do not
Then they asked which reaction (out of the 20) has radioactivity in the protein pellet?
Biochemists Break the Code
Results
Marshall Nirenberg and Heinrich Matthaei showed that poly-U produced polyphenylalanine in a
cell-free solution from E. coli. In other words, only the test tube reaction spiked with radioactive
Phe generated a radioactive pellet
They repeated the experiment with other synthetic homopolymer RNAs
Poly-A gave polylysine (AAA = Lys)
Poly-C gave polyproline (CCC = Pro)
Poly-G gave polyglycine (GGG = Gly)
Getting at the Rest of the Code
Work with nucleotide copolymers (poly (A,C), etc.), revealed some of the codes
Gobind Khorana (organic chemist)
-synthesized DNA composed of alternating copolymers eg: ACACACACACAC…..
Then used RNAP to make RNA from the DNA template eg: UGUGUGUGUGUGU……
This RNA transcript has two possible alternating codons: UGU GUG UGU GUG
In a translation extract you should get a protein with 2 alternating amino acids
Ribosome binding assay devised by Nirenberg and Leder
Took a cell-free translation extract (ribosomes and tRNAs charged with their specific amino
acid)
Added a synthetic triplet RNA (a codon) eg UUU
They found that addition of that simple triplet RNA to the cell-free extract could stimulate the
binding of the tRNA that recognized that codon to a ribosome
Since the tRNA is covalently linked to the amino acid that is coded for by the codon, therefore
that amino acid gets localized to the ribosome
If they collect the ribosomes from the experiment they can identify which amino acid was
brought to the ribosome by that triplet codon
Getting at the Rest of the Code
Finally Marshall Nirenberg and Philip Leder cracked the entire code in 1964
They showed that trinucleotides bound to ribosomes could direct the binding of specific
aminoacyl-tRNAs By using C-14 labeled amino acids with all the possible trinucleotide codes,
they elucidated all 64 correspondences in the code
Found that all the codons (except the 3 stop codons) specified an amino acid
There are 64 codons and 20 amino acids
Therefore amino acids can be encoded by >1 codon
The Nature of the Genetic Code
All the codons have meaning: 61 specify amino acids, and the other 3 are "nonsense" or "stop"
codons
The code is unambiguous - only one amino acid is indicated by each of the 61 codons
The code is degenerate - except for Trp and Met, each amino acid is coded by two or more
codons
Codons representing the same or similar amino acids are similar in sequence
2nd base pyrimidine: usually nonpolar amino acid
2nd base purine: usually polar or charged aa
The code is not overlapping
The base sequence is read from a fixed starting point, with no punctuation
Universal & Unusual
Third-Base Degeneracy and the Wobble Hypothesis
Codon-anticodon pairing is the crucial feature of the "reading of the code"
But what accounts for "degeneracy": are there 61 different anticodons, or can you get by with
fewer than 61, due to lack of specificity at the third position?
Crick's Wobble Hypothesis argues for the second possibility - the first base of the anticodon
(which matches the 3rd base of the codon) is referred to as the "wobble position"
The Wobble Hypothesis
The first two bases of the codon make normal (canonical) H-bond pairs with the 2nd and 3rd
bases of the anticodon
At the remaining position, less stringent rules apply and non-canonical pairing may occur
The rules: first base U can recognize A or G, first base G can recognize U or C, and first base I
can recognize U, C or A (I comes from deamination of A)
Advantage of wobble: dissociation of tRNA from mRNA is faster and protein synthesis too
Translation in Prokaryotes
Activation of AAs
1) Formation of fMet-tRNA fMet
2) Formation of Sec-tRNASec
3) Formation of some other aa-tRNAs
Initiation
Elongation
Termination
Initiation of translation
in Prokaryotes
Recognition of the start codon
AUG(90%),GUG(8%) and UUG(1%)
1) SD sequence and S'D' sequence
How to prove?
Colicin E3(大肠杆菌素） and mutation experiments
2) Secondary structure of mRNAs
Formation of initiation complex
1) Binding of the ribosome 30S subunit with Initiation Factors
2) Binding of the mRNA and the fMet-tRNAfMet
3) Binding of the ribosome 50S subunit and release of Initiation Factors
Initiation
Binding of the ribosome 30S subunit with IFs
1) IF3 promotes the dissociation of the ribosome into its two component subunits. The
presence of IF3 permits the assembly of the initiation complex and prevents binding of the 50S
subunit prematurely.
2) IF1 assists IF3 in some way.
Binding of the mRNA and the fMet-tRNAfMet
1) IF3 assists the mRNA to bind with the 30S subunit of the ribosome so that the start
codon is correctly positioned at the P site of the ribosome. The mRNA is positioned by means
of base-pairing between the S'D' of the 16S rRNA with the SD sequence immediately upstream
of the start codon.
Elongation of translation
in Prokaryotes
3 distinct steps to add one amino acid to the growing polypeptide chain.
Occurs many times per polypeptide, the number of which depends upon the number of mRNA
codons or amino acids in the protein
The Elongation Cycle is similar in prokaryotes and eukaryotes.
Fast: 15-20 amino acids added per second
Accurate: 1 mistake every ~10,000 amino acids
Elongation of translation
in Prokaryotes
Binding of a new aa-tRNA at the A site -EF-Tu (GTP), EF-Ts
Formation of the new peptide bond (Transpeptidation)-23S rRNA
Translocation of the Ribosome -EF-G(GTP)
Repeat and Repeat until the stop codon enters the A site
Binding of a new aa-tRNA at the A site
At the start of each cycle, the A site on the ribosome is empty, the P site contains a
peptidyl-tRNA, and the E site contains an uncharged tRNA.
EF-Tu (GTP) binds with an aa-tRNA and brings it to the ribosome. Once the correct aa-tRNA is
positioned in the ribosome, GTP is hydrolyzed and EF-Tu (GDP) dissociates away from the
ribosome.
There are two ways that EF-Tu functions to ensure that the correct aa-tRNA is in place:
1) EF-Tu prevents the aa end of the charged tRNA from entering the A site on the ribosome.
This ensures that codon-anticodon pairing is checked first before the charged tRNA is
irreversibly bound in the A site and a new, potentially incorrect, peptide bond is made.
2) GTP hydrolysis is SLOW and EF-Tu cannot dissociate from the ribosome until it occurs. The
amount of time prior to GTP hydrolysis allows the final fidelity check to take place. If the
anticodon-codon interaction is incorrect, the aa-tRNA simply dissociates and a new one is
brought in.
Experiments using GTP analogues have been used to establish these results:
1)If a GTP analogue such as GTP-ã-S, which is hydrolyzed very slowly, is used then protein
synthesis slows down because of the slow rate of hydrolysis but it also becomes more
accurate because there is more time to check that the correct aa-tRNA is in place.
2)If a GTP analogue such as GMP-PCP, is used then protein synthesis stops because
EF-Tu cannot dissociate from the ribosome.
EF-Tu is the most abundant protein in the E. coli cell. There are approximately ~100,000
molecules/cell which is 5% of the total cell protein. There are also approximately ~100,000
tRNA molecules/cell. Nearly all of the aa-tRNA in the cell is bound by EF-Tu.
EF-Tu cannot bind with tRNAfMet.
EF-Tu (GDP) is inactive and cannot function to bind aa tRNAs. In order to recycle EF-Tu,
EF-Ts binds to the EF-Tu (GDP) complex to displace the GDP. GTP then, in turn, displaces
EF-Ts.
Formation of the new peptide bond (Transpeptidation)
Peptide bond formation is simple. It is just a kind of nucleophilic reaction
The peptidyltransferase activity of the ribosome which catalyzes this reaction is located on the
23S rRNA though it will be assisted by some of the ribosomal protein subunits. In other words,
peptidyl transferase is a ribozyme - another example of a catalytic RNA.
Termination
A stop codon enters the A site.
There are no tRNAs that recognize the stop codons. Rather they are recognized by release
factor RF1 (which recognizes the UAA and UAG stop codons) or RF2 (which recognizes the
UAA and UGA stop codons). These RFs act at the A site. A third release factor, RF3 (GTP),
stimulates the binding of RF1 and RF2.
Binding of the release factors alters the peptidyltransferase activity so that water is now the
nucleophilic attack agent. The result is hydrolysis of the peptidyl-tRNA and release of the
completed polypeptide chain. The uncharged tRNA then dissociates as do the release factors.
GTP is hydrolyzed.
Finally, the ribosome dissociates into its 30S and 50S subunits and the mRNA is released. IF3
may help this process.
Translation In Eukaryotes
Differences between the two systems
1) Translation & transcription are not coupled
2) Ribosomes are larger.
3) The initiating amino acid is still Met, but it is not formylated.
4) Eukaryotic mRNA is capped. This is used as the recognition feature for ribosome binding
-- not the 18S rRNA.
5)The initiation phase of protein synthesis requires over 10 eIFs, one of which is the cap
binding protein.
6) The elongation phase requires two eEFs
7) The termination phase require just a single release factor, eRF.
Detailed mechanism
1) Activation of AAs
2) Initiation
How to find the start codon?
-Scanning model & Internal entry model
How to form the 80S initiation complex?
3) Elongation
Similar to that in Prokaryotes
eEF-1 is equivalent to EF-Tu and EF-Ts
eEF-2 is equivalent to EF-G
4) Termination
Similar to that in Prokaryotes
eRF
重点：重点介绍原核生物的蛋白质合成过程。
难点：蛋白质合成抑制剂的作用原理
教学过程设计（要求阐明对教学基本内容的展开及教学方法与手段的应用、讨论、作业布置）：
利用课件结合板书介绍蛋白质合成的基础知识，了解蛋白质合成的研究新进展，重点掌握原
核生物蛋白质合成的过程及简单的能量供给计算，在理解的基础上布置作业，让学生在作业中发
现问题提出问题，对于比较难理解的老师在课堂上再次强调。在教学过程中给学生介绍学习方法
及鼓励学生拓展知识。