1
MB ChB PHASE I
STORING AND USING
GENETIC
INFORMATION
LECTURE 1: DNA Replication
AIMS: To review:
the accuracy of the process
its semi-conservative nature
its bidirectional nature
the enzymes that catalyse the process.
[C&H, Chapter 29]
http://www.abdn.ac.uk/~bch118/index.htm
2
[Refer to Molecules of Life Lecture 2 for
general information about nucleic acid
structure.]
DNA replicates with great, but finite
accuracy.
1 mistake occurs in ~109-1012 bases
incorporated.
Such (small) laxity
variation
required
flexibility.
allows genetic
for
adaptive
The (nevertheless) great accuracy is
achieved because, during replication,
1 specific base complementarity rules
(A pairing with T; G with C),
are generally obeyed;
2 proof-reading (editing) processes
detect and correct errors.
3
Also, damage to non-replicating DNA is
repaired, by mechanisms not discussed
here.
DNA is the only molecule for which such
accurate synthesis and repair processes
exist.
DNA replication follows five ‘Rules’
obeyed by most DNA replicating
systems.
4
I
DNA REPLICATION
CONSERVATIVE
IS
SEMI-
The specific base complementarity part of the
double-helix model suggests a model of DNA
replication,
now confirmed experimentally.
In the model, parental DNA strands separate,
and act as templates, specifying base sequences
of new strands.
In each daughter double-helix is conserved one
-half of the parent: hence ‘semi-conservative’.
5
II DNA REPLICATION BEGINS AT
A PARTICULAR ORIGIN(S) AND
PROCEEDS BIDIRECTIONALLY
Experimental approach:
E. coli grown in medium containing
3
H-thymidine
Radiolabel taken into cells and
incorporated into new DNA
as DNA replicates
DNA extracted
during DNA replication
DNA analysed by autoradiography
(short-range radioactive emissions reduce AgCl
on a photographic plate to Ag atoms, which
aggregate and can be seen microscopically)
‘Theta () structures’ seen
(E. coli DNA is a ds circle)
6
Theta structure:
Ag grains
show position
of radiolabel,
i.e. new DNA
Conclusion:
complete separation of parental strands
is not required before DNA replication
begins.
7
Two possibilities now arise:
1 The origin (of strand separation) is
at one (or other) side of the loop:
origin
strands
are pushed apart and
new DNA made
unidirectionally
2 The origin is in the middle of the
loop:
origin
bidirectional replication
8
A further experiment shows that DNA
replication is bidirectional:
E. coli DNA again radiolabelled
as it is being made
First, precursor of low radioactivity used
Later, precursor of high radioactivity used
DNA again extracted and
analysed by autoradiography
Predicted:
unidirectional
bidirectional
Experimentally,
the bidirectional pattern is observed.
9
Yet another experiment (not shown)
demonstrates that DNA replication
starts at a particular origin(s).
10
On small, circular, prokaryotic DNA
molecules, there is a single, particular
origin.
On long, linear, eukaryotic DNA
molecules, there are multiple, particular
origins:
11
E. coli PROTEINS NEEDED AT THE
ORIGIN FOR INITIATION OF DNA
REPLICATION:
Dna A
recognises origin;
begins DNA unwinding
Dna B
continues DNA unwinding
(i.e. is a ‘helicase’)
(with Dna C)
SSB protein
binds separated DNA strands;
stops them re-forming
double-helix
Primase
synthesises RNA primers
(see later)
Gyrase
removes DNA ‘supercoils’
as DNA unwinds
(e.g. of a ‘topoisomerase’)
12
III DNA SYNTHESIS DURING DNA
REPLICATION IS CATALYSED BY
DNA POLYMERASES
Five have been found in E. coli:
DNA pol I
II
III
IV
V
and a large (and increasing) number in
eukaryotic cells.
All catalyse the same fundamental
reaction.
13
This has been examined using isolated
DNA polymerases.
The enzymes need (in the test-tube):
dATP
dGTP
dCTP
dTTP.
These are precursors for the monomers of
the DNA polymer to be made.
Also needed:
a pre-formed DNA polymer.
The reaction involves:
addition of a monomer to the 3’ end of the
pre-formed polymer.
14
3’ end of the
pre-formed DNA
15
1 of the 4 dNTPs
16
So,
the pre-formed DNA acts as a primer,
onto which, in sequence, monomers are
added.
The DNA made does not have a random
base sequence:
its sequence is directed by
(and is complementary to) the sequence
of the pre-formed DNA,
which, therefore, also acts as a template.
17
The pre-formed DNA in the test-tube
must therefore look something like this:
3’
5’
3’
5’
The diagram emphasises that the
reaction occurs with a
definite directionality:
is 5’
3’
of template reading is 3’
5’.
direction of DNA synthesis
18
In cells,
(as in the test-tube),
DNA polymerases:
use dNTPs,
‘read’ a DNA template,
show this directionality.
They also need a primer, being poor at
starting synthesis on a totally singlestranded template:
3’
5’
but, in cells, the primer is not
pre-formed DNA: it is RNA (Lecture 2).
19
MB ChB PHASE I
STORING AND USING
GENETIC
INFORMATION
LECTURE 2: DNA Replication
AIMS: To review:
DNA polymerases of E. coli
the semi-discontinuous nature of DNA
replication
the involvement of RNA in the process.
[C&H, Chapter 29]
http://www.abdn.ac.uk/~bch118/index.htm
20
DNA POLYMERASES OF E. coli
DNA pol
I
subsidiary role in DNA replication;
DNA repair*
II
DNA repair*
III
major role in DNA replication
IV
DNA repair*
V
DNA repair*
*of non-replicating DNA (Lecture 1)
21
Some E. coli DNA polymerases also act
as nucleases
(i.e. catalyse breakage of 3’, 5’phosphodiester bonds).
In particular,
DNA pol I has
and
5’3’-
exonuclease activities,
and
DNA pol III has
3’exonuclease activity.
These activities are important during
DNA replication.
22
IV DNA SYNTHESIS
DURING DNA REPLICATION
IS SEMI-DISCONTINUOUS
The bidirectional model of DNA
replication suggests that this occurs:
For discussion, we can cut the diagram
in half, and consider directionality of
DNA synthesis in the remaining half.
We might expect:
23
However,
DNA
polymerases
synthesis 5’
3’
(as we saw earlier),
only
catalyse
so how is the lower strand in the
diagram made?
A simple, semi-discontinuous model
seeks to show this.
24
1 DNA synthesis occurs 5’
3’
in the direction of parental strand
separation
for the upper ‘leading’ strand,
and in the opposite direction
for the lower ‘lagging’ strand:
25
2
As parental
continues,
strand
separation
the 3’ end of the ‘leading’ strand is
continuously extended,
but the 3’ end of a ‘lagging’ strand
section can’t add onto the 5’ end of
the preceding section,
so a gap is left:
26
3
As parental
separate,
strands
continue
to
the leading strand is made continuously,
the lagging strand
discontinuously:
27
4
Lagging strand gaps are sealed by
a DNA ligase,
which doesn’t add monomers
(like a polymerase),
but forms phosphodiester bonds
between pre-formed polymers:
Detection of small DNA pieces
(‘Okazaki fragments’)
during DNA replication
suggests that something like this model
actually occurs in cells.
28
V DNA SYNTHESIS DURING DNA
REPLICATION REQUIRES RNA
PRIMERS
The simple semi-discontinuous model
fails to show how DNA synthesis begins.
Presumably, primers (Lecture 1)
are needed for
the leading strand and
each lagging strand section:
29
In cells,
primers of RNA are used (Lecture 1).
They are made by particular
DNA-dependent
RNA
polymerases
(DdRps)
called primases (Lecture 1).
DdRps in general
catalyse RNA synthesis on a DNA
template
in the process called ‘transcription’
(Lecture 3).
They catalyse a reaction very similar to
that of DNA polymerases,
except that
ATP, CTP, GTP, UTP are used
and RNA is synthesised.
30
Like DNA polymerases,
pre-formed DNA is needed,
and acts as a template,
but,
unlike the situation for DNA polymerases,
DdRps don’t use the pre-formed DNA as a
primer.
This is because,
unlike DNA polymerases,
DdRps can start synthesis on a totally
single-stranded template:
DNA pol
DdRp
3’
3’
5’
3’
5’
5’
31
The simple semi-discontinuous model
can be modified to show
production and removal of RNA primers.
It is also possible to include particular
E. coli DNA polymerases
in the model.
32
1 At the origin, primase makes an RNA
primer to start the (upper) leading
strand.
Then DNA pol III extends the primer
with DNA.
At the fork, primase makes an RNA
primer for the first section of the
(lower) lagging strand.
Then DNA pol III extends the primer
with DNA.
33
2 DNA pol III extends the leading strand
continuously.
Each successive section of the lagging
strand starts with an RNA primer, and
is extended with DNA by DNA pol III.
34
3 Focusing on the lagging strand:
RNA primers are excised by 5’exonuclease of DNA pol I.
4 DNA pol I extends 3’ ends of the
lagging strand sections to replace
excised parts with DNA.
35
5 DNA ligase joins the lagging strand
sections.
36
PROOF-READING (EDITING) IN DNA
REPLICATION
(Lecture 1)
As we have seen,
DNA pol I 5’-exonuclease
removes RNA primers.
DNA pols I and III both have
3’- exonuclease activity.
This is used to proof-read DNA as it is
being made by the enzymes.
If either enzyme inserts an incorrect
(i.e. non-complementary) base,
the 3’-exonuclease of the pol
excises that monomer
and
the polymerase activity of the pol
replaces it with the correct monomer.
37
MB ChB PHASE I
STORING AND USING
GENETIC
INFORMATION
LECTURE 3: Transcription
AIMS: To review:
RNA structure and functions
RNA synthesis (‘transcription’)
Transcriptional control of gene
expression
Post-transcription modification of RNA
[C&H, Chapter 30]
http://www.abdn.ac.uk/~bch118/index.htm
38
Genetic information of a cell or organism
is stored in the base sequences of its DNA.
The stored information is passed on,
accurately,
to daughter cells,
when DNA replicates before a cell
divides;
to an embryo,
which inherits DNA-containing
chromosomes from
parent-derived gametes.
The stored information is used
(or ‘expressed’)
through synthesis of RNA.
39
THE ‘CENTRAL DOGMA
OF MOLECULAR BIOLOGY’
Parts of the DNA base sequence
(‘genes’ aka ‘cistrons’)
are used as templates
to make RNA molecules
with complementary base sequences.
Some of these molecules (called messenger RNA)
are then used as templates
to make polypeptides
with defined amino-acid sequences.
The polypeptides made
determine the phenotype of the cell
or organism (Proteins Lectures 1, 2).
Direction of flow of genetic information:
DNA
messenger (m)RNA
protein
ribosomal (r)RNA
transfer
transcription
(t)RNA
translation
40
RNA STRUCTURE
RNA has the same polymeric structure
as DNA,
(mononucleotides linked by
3’, 5’-phosphodiester bonds;
5’ and 3’ ends)
but with D-ribose rather than
2-deoxy D-ribose,
and U rather than T.
Most RNA molecules are single-stranded,
but many have
intra-strand double-helicity
involving specific base complementarity
(the single strand folds back on itself);
e.g. the ‘clover-leaf’ structure of tRNA
(Lecture 5).
41
RNA FUNCTIONS
mRNA
base-sequence acts as template for
polypeptide synthesis in translation
tRNA
‘adapter’ molecule between a coded
amino-acid and the mRNA codeword that specifies the amino-acid
rRNA
component of ribosomes, the RNAprotein complexes on which
translation occurs
In eukaryotic cells,
all three are transcribed from DNA
in the nucleus (rRNA in the nucleolus)
and
move to the cytoplasm to function.
42
RNA SYNTHESIS ON A DNA TEMPLATE
(‘TRANSCRIPTION’)
Transcription is
catalysed by
(DNA-dependent) RNA polymerases (DdRps)
[‘primase’ (Lecture 2) is a specialised DdRp],
in a reaction having the same mechanism as that
catalysed by DNA polymerases (Lecture 1)
i.e. addition of monomers to the 3’ end
of growing chain,
to give a 5’ to 3’ direction of synthesis,
The base-sequence of the synthesised polymer is
determined by the base sequence of the DNA
template,
which is ‘read’ in a 3’ to 5’ direction.
43
However, unlike DNA polymerases,
DdRps
use ATP, GTP, CTP and UTP as the
monomer precursors;
do not need a primer (with a 3’ end)
in order to begin synthesis
(hence the role of primases in DNA replication).
DNA polymerase
DdRp
primer
DNA
template
DNA
new DNA
template
new RNA
44
OTHER PATTERNS OF
NUCLEIC ACID SYNTHESIS
template product enzyme
DNA replication DNA
DNA
DNA polymerase
transcription
RNA
(DNA-dependent)
RNA polymerase
DNA
In RNA-containing retroviruses (e.g. HIV)
that integrate, as DNA, into the DNA of the infected cell:
reverse
transcription
RNA
DNA
RNA-dependent
DNA polymerase
(‘reverse transcriptase’)
In other RNA-containing viruses
(e.g. influenza, mumps, measles, polio):
RNA replication
RNA
RNA
(RNA-dependent)
RNA polymerase
45
(DNA-DEPENDENT)
RNA POLYMERASES
(DdRps)
Prokaryotes: a single, multimeric protein.
Eukaryotes:
different, multimeric proteins
catalyse
mRNA,
rRNA
and tRNA synthesis.
46
INITIATION OF TRANSCRIPTION
Particular sequences of DNA
(‘genes’ aka ‘cistrons’)
are transcribed into RNA.
A DdRp must recognise and bind to DNA
at the start of these sequences.
In prokaryotes,
DdRp binds to sites called ‘promoters’,
all of which have a similar structure:
47
STRUCTURE OF A PROMOTER
TTGACA
~19 bases
TATAAT ~7 bases
template strand
transcription
starts
(i.e. start of gene)
48
TERMINATION OF TRANSCRIPTION
A DdRp must recognise a stop signal
at the end of a gene.
A DdRp progresses along the DNA
template
in a 3’ to 5’ direction
(or the template is pulled past a stationary DdRp)
catalysing RNA polymerisation
in a 5’ to 3’ direction
until a particular stop sequence is reached.
DdRp and the new RNA
then dissociate from the template.
In some cases, in prokaryotes,
this involves another protein,
the rho () factor.
49
THE BACTERIAL OPERON:
A CONTROL OF GENE EXPRESSION
AT THE TRANSCRIPTIONAL LEVEL
The E. coli lac operon consists of 3 genes
encoding enzymes that catabolise lactose,
(a potential food for the bacterium),
together with sequences
(‘promoter’ and ‘operator’)
that control transcription of the genes.
Under most circumstances,
when E. coli is not in contact with lactose,
transcription is switched off,
because a ‘repressor’ protein
[Proteins Lecture 1 Function 10]
binds to the operator and
prevents progress of DdRp
from the promoter, through the genes.
Thus, in the absence of lactose,
enzymes for its catabolism are not made.
50
The lac operon
(E. coli dsDNA shown as a single line)
In absence of lactose
promoter operator
DdRp
repressor
bound
gene A
gene B
gene C
no transcription
lactose-catabolising
enzymes A, B, C not made
51
When E. coli encounters lactose,
it enters the cell
and (indirectly) inactivates the repressor,
which no longer binds to the operator.
DdRp is now able to move
from the promoter, through the genes,
transcribing them all to form a
‘polycistronic mRNA’.
This is then translated, producing the
enzymes that catabolise lactose.
Thus, in the presence of lactose,
synthesis of enzymes for its catabolism
is switched on. In the presence of lactose
52
In presence of lactose
promoter operator
gene A
gene B
gene C
DdRp
transcription
lactose inactivates repressor
polycistronic mRNA
translation
lactose-catabolising
enzymes A, B, C made
53
Notice that:
clustering of genes the products of which
have related functions
(a common feature of prokaryote DNA)
means that the genes can be expressed
in a co-ordinated way;
a similar transcriptional contol of gene
expression occurs in eukaryotes,
where histones bind to DNA
and silence genes in differentiated cells
(Proteins Lecture 1 Function 10).
54
POST-TRANSCRIPTIONAL
MODIFICATION OF EUKARYOTIC
RNA
In some eukaryotic RNAs
some bases (A, U, G, C) are converted into
other, ‘minor’ bases
e.g. in tRNA (Lecture 5)).
Function:
prevents intra-strand double-helical structures
forming?
In most eukaryotic mRNAs,
a 5’ cap made of a complex modified G and
a 3’ ‘poly A tail’
are added.
Function:
protects ends;
tail helps transport to cytoplasm;
tail determines longevity of mRNA;
cap helps translation?
55
Many eukaryotic genes encoding proteins
consist of coding sequences
(‘exons’)
interspersed with large, non-coding regions
(‘introns’).
gene
e
i
e
i
e
The entire gene is transcribed.
Intron transcripts are removed
by an endonuclease.
Exon transcripts are joined by a ligase
to form the functional, coding mRNA
(the process is called ‘splicing’).
Function of introns:
unknown.
56
THE CODING FUNCTION
OF HUMAN DNA
~30% consists of genes
transcribed into the precursors
of mRNA, rRNA, tRNA.
The rest is ‘non-coding’,
in some cases consisting of
highly repetitive sequences.
Of the 30%,
much consists of non-coding introns
in genes encoding mRNA.
So, in fact, ~ only 1.5%
of the total DNA
is ‘coding’ DNA.
57
MB ChB PHASE I
STORING AND USING
GENETIC
INFORMATION
LECTURE 4: The Genetic Code
AIMS: To review:
Features of the Code
[C&H, Chapter 31]
http://www.abdn.ac.uk/~bch118/index.htm
58
THE CENTRAL DOGMA
OF MOLECULAR BIOLOGY:
A RECAP
DNA
DNA
replication
RNA
transcription
protein
translation
Sequences of coding DNA are arranged
into 3-base code-words (‘triplets’).
These are transcribed into
complementary mRNA 3-base ‘codons’.
Particular triplets/codons specify
particular amino-acids.
mRNA codon sequences are translated
into sequences of amino-acids
in polypeptides.
59
FEATURES OF THE CODE
(a) Degeneracy.
Most amino-acids
assigned to them.
20
43 = 64
61
3
have
>1
codon
coded amino-acids
possible code-words
specify particular aminoacids
(UAG, UGA, UAA, read 5’ to 3’)
are translation stop signals
(see later).
(b) The code is ‘non-overlapping’ and
‘comma-less’.
A sequence of bases is read
… ABC DEF GHI JKL …
aa1 aa2 aa3 aa4
60
and not
… ABC
CDE
EFG
aa1
aa2
aa3
or
… ABC D EFG H IJK …
aa1
aa2
aa3
comma
comma
etc
61
(c) The triplet nature of the code
explains the effects of
frame-shift mutations.
These arise from insertion or deletion of
a piece of DNA.
Base-sequence
Of ‘normal’ DNA
… ABC DEF GHI JKL MNO …
aa1 aa2 aa3 aa4 aa5
A piece of
DNA inserted
… ABC DEF XXX XXX XGH IJK …
aa1 aa2 ?
?
?
?
From this point, succeeding
code-words are out-of-frame
(even after the inserted
DNA), so the protein may
well be defective.
62
Note:
1 the 3-base nature of the code means
that there is a 1 in 3 chance of
information getting back in frame
after the insertion;
2 a deletion, in this respect, produces
the same effect as an insertion.
63
Much more common than frame-shifts
are ‘point’ mutations, produced by
single-base changes:
... ABC DEF GHX JKL MNO …
E.g. in human haemoglobin  chain
position 6, normal protein has
glutamate,
but sickle-cell haemoglobin has
valine.
This is caused by a T to A change on the
DNA
… CTC …
(codes for glutamate)
… CAC …
(codes for valine).
64
(d) Degeneracy of the code minimises
effects of mutations.
1 Why spread 61 triplets over 20
amino-acids? Why not have just 20
code-words, and 40-or-so triplets
meaning nothing?
With a degenerate code, a point
mutation is likely to change a triplet to
another specifying an amino-acid, (so a
protein may still be made), rather than
to one meaning nothing, (in which case,
no protein would be made).
2 Because
of
the
non-random
assignment of triplets to amino-acids
(see C&H p.432),
it is quite possible that a point mutation
will change a triplet to another assigned
to the same amino-acid.
This is a ‘silent’ (aka ‘synonymous’)
mutation.
65
Look at the genetic code-word
dictionary (C&H, p.432) and pick out
code-word assignments for individual
amino-acids, to illustrate the point made
in Section 2 above.
66
3 Because the non-random assignment
of triplets to amino-acids extends to
groups of amino-acids of similar type,
even if a point mutation changes a
triplet to one meaning another aminoacid, it is likely to be similar to the
originally-coded amino-acid.
This is a ‘conservative’ mutation.
The mutant protein may well function,
possibly even better than the original.
Much evolution works at the molecular
level in this way.
67
Look at the genetic code-word
dictionary (C&H, p.432) and pick out
code-word assignments for sets of
amino-acids with similar R groups, to
illustrate the point made in Section 3
above.
68
(e) The code is nearly universal.
The same assignment of code-words to
amino-acids is used by all living
organisms (and viruses).
There are minor exceptions:
in
some bacteria,
mitochondria,
some unicellular eukaryotes.
Often the exceptions involve one of the 3
translation stop signals (particularly
UGA) being used to specify an aminoacid.
69
OPEN READING FRAMES
A sequence of bases has three possible
reading frames:
… ABC DEF GHI JKL …
… . . A BCD EFG HIJ KL . …
… . AB CDE FGH IJK L .. …
Which frame is used depends on
recognition by the translation apparatus
of a translation start signal. This is the
sequence AUG.
(We will see later how this acts to start the
process.)
The bases are then read in groups of
three until, in frame, a translation stop
codon (UAA, UGA or UAG) is reached.
70
So, an ‘open reading frame’
on a mRNA looks like this:
5’ … XXX AUG XXX XXX XXX … XXX UAA XXX …3’
or UAG
or UGA
untranslated start
translated
stop untranslated
71
MB ChB PHASE I
STORING AND USING
GENETIC
INFORMATION
LECTURE 5: Translation
AIMS: To review:
Properties and functions of tRNA
Aminoacyl tRNA synthetases
Wobble
[C&H, Chapter 31]
http://www.abdn.ac.uk/~bch118/index.htm
72
TRANSFER RNA (tRNA)
ACTS AS AN ADAPTER
There is no direct, specific interaction
between an amino-acid and the
sequence of three bases encoding it.
This is not surprising:
the two have very different structures.
For the genetic code to work,
an adapter molecule is needed,
to bring these different structures
together
(rather like a 2-/3-pin electrical plug
adapter).
The adapter is transfer (t) RNA.
73
PROPERTIES OF tRNA
1 Acts as adapter between
mRNA codon and amino-acid.
2 Small for a nucleic acid
(73-93 nucleotides).
3 Like other RNA,
is transcribed from DNA.
4 Comprises about 15%
of total cell RNA.
5 Rich in unusual bases
(made by modification of A,U,G,C).
6 Much intra-strand double-helicity.
7 All tRNAs have a similar
and
2-D structure
3-D structure
(‘clover-leaf’)
(upper-case L)
stabilised by intra-strand H-bonding.
74
Look at C&H Figure 30.3, which shows
2- and 3-dimensional structures of
tRNA.
75
HOW tRNA ACTS AS AN ADAPTER
1 Activation of amino-acids.
NH2
adenine
R - C - C - OH
+
P-P-P-ribose
H O
amino-acid
ATP
PPi
NH2
O
adenine
R – C – C – O – P – O – ribose
H O
O-
aminoacyl AMP
(‘activated amino-acid’)
76
2 Activated amino-acid is loaded onto
3’ end of a tRNA.
O
O P OO
H2C O Base
adenine
OH OH
3’ end of a tRNA
amino-acid – P - ribose
aminoacyl AMP
AMP
77
O
O P OO
H2C O Base
O
OH
C O
H2N – C – H
R
aminoacyl tRNA
78
The same set of enzymes catalyses both
steps, i.e.
1 activation;
2 loading.
They are aminoacyl tRNA synthetases
(i.e. they are named for step 2).
Each of the 20 coded amino-acids is
assigned
at least one, specific
aminoacyl tRNA synthetase
and
at least one, specific tRNA.
79
Aminoacyl tRNA synthetases
are very specific in their action.
Each catalyses activation of
one, particular amino-acid,
and then loads that amino-acid
only onto the tRNA assigned to it.
This specificity is crucial,
because one of the tRNA loops
has a 3-base sequence,
called the anticodon,
that is complementary
to the codon of the amino-acid
assigned to that tRNA.
80
We can now visualise the adapter role of
tRNA between codon and amino-acid.
activated, loaded
amino-acid
anticodon
codon on mRNA
81
So, recognition of the mRNA codon
is through its interaction,
by specific base complementarity,
with a tRNA anticodon
(tRNA acting as the ‘adapter’),
and NOT by interaction with the aminoacid itself,
which, once loaded onto its tRNA,
isn’t recognised by anything
(i.e. it’s anonymous).
For this adapter system to work,
the correct amino-acid MUST be loaded
onto the correct tRNA.
So, the specificity of the aminoacyl
tRNA synthetases is crucial.
82
WOBBLE
With the adapter system just described,
61 tRNAs would be expected,
each with a different anticodon,
complementary to the 61 coding
mRNA codons.
In fact, there are <61 tRNAs.
This is because several tRNAs recognise
>1 codon.
This is possible
Because, in some anticodons,
the base in position 3
pairs with >1 codon base.
83
3’
5’
anticodon
--- X Y Z ---
codon
- - - X’ Y’
5’
specific base
complementarity
--3’
less specific
pairing (‘wobble’)
Codon redundancy occurs mainly in this third
position (Lecture 4),
and all codons pairing with XYZ in the
diagram above encode the same amino-acid.
What evolutionary advantage might
‘wobble’ confer?
It means that the codon-anticodon
interaction,
although strong and specific enough
to allow the adapter function of tRNA,
is not so strong
that need for its disruption
slows the process of translation.
84
PROOF-READING BY AMINOACYL
tRNA SYNTHETASES
Some of these enzymes ‘check’
that they have activated/loaded
the correct amino-acid.
E.g. the E.coli enzyme
for isoleucine
sometimes activates and loads
valine (it has a similar structure).
The enzyme has an additional site
at which hydrolysis of the aminoacyl
tRNA occurs,
releasing free amino-acid.
Valine fits this site better than
isoleucine.
The overall fidelity of translation
is about 1 mistake / 104 amino-acids
polymerised.
So, although accurate,
it’s less so than replication of DNA.
85
MB ChB PHASE I
STORING AND USING
GENETIC
INFORMATION
LECTURE 6: Translation
AIMS: To review:
The initiation, elongation, and
termination steps of polypeptide
synthesis
[C&H, Chapter 31]
http://www.abdn.ac.uk/~bch118/index.htm
86
RIBOSOMES
These organelles are the site of
codon-anticodon recognition,
and of polypeptide synthesis.
Ribosomes have a generally similar
structure in all cells.
In E. coli
30S subunit
(16S RNA +
21 proteins)
50S subunit
(23S RNA,
5S RNA +
36 proteins)
S= Svedborg unit
(to do with how fast particles sediment in an
ultracentrifuge)
87
DIRECTIONALITY IN TRANSLATION
DNA replication (recap)
DNA polymerised:
DNA template ‘read’:
5’
3’
3’
5’
5’
3’
3’
5’
N
5’
C
3’
Transcription (recap)
RNA polymerised:
DNA template ‘read’:
Translation
Polypeptide polymerised
mRNA template ‘read’
88
AUG ACTS AS A TRANSLATION
START SIGNAL (Lecture 4)
AUG is (also) the codon for the aminoacid methionine (Met).
In E.coli,
There are two tRNAs for Met.
Both are loaded with activated Met.
Both have an anticodon complementary
to AUG.
In only one,
the activated, loaded Met is converted to
N-formyl Met (fMet)
C O
H2N CH
(CH2)2
S
CH3
3’ end of tRNA,
loaded with Met
C O
OHC N CH
H (CH2)2
S
CH3
3’ end of tRNA,
loaded with fMet
89
It is this tRNA, loaded with fMet, that
plays a crucial role at the start of
polypeptide synthesis.
The other tRNA (loaded with Met)
recognises AUG in the middle of a
message, and inserts Met into the
middle of a polypeptide.
What is it about the tRNA loaded with
fMet that makes it active in polypeptide
initiation?
It is the only tRNA that can bind directly
to the peptidyl site of the 50S ribosome
subunit (which we see later).
90
THE START OF TRANSLATION
(a) mRNA binds to the 30S subunit.
(At this stage, the two subunits are
separate.)
The initial interaction is between 4-9
bases of the 16S RNA, and
complementary bases, just to the 5’
side of an AUG codon.
It is these 4-9 bases that differentiate
the AUG from other AUGs within
the message.
(b) tRNA, loaded with fMet, binds to the
30S subunit/mRNA complex,
interacting, through its anticodon,
with the AUG.
Processes (a) and (b) need three proteins
(IF-1, IF-2, IF-3) and GTP.
91
(c) The 50S subunit binds to the 30S
subunit.
The fMet – loaded tRNA is now on
the peptidyl site of the 50S subunit.
Process (c) needs GTP dephosphorylation.
92
A diagram showing steps (a) –(c)
is shown in the Lecture.
93
94
95
FORMATION OF THE PEPTIDE BOND
Three, repeated steps occur:
Step 1
tRNA, loaded with amino-acid, binds to
the aminoacyl site.
This requires 2 proteins (EF-Tu, EF-Ts)
and
GTP dephosphorylation.
Step 2
The peptide bond is formed.
This requires ‘peptidyl transferase’,
a ribozyme activity of the 23S RNA.
Step 3
The ribosome moves along the mRNA
in a 5’ to 3’ direction.
This requires a protein (EF-G)
and
GTP dephosphorylation.
96
A diagram showing the three steps is in
the Learning Guide.
97
Notice that all the loaded tRNAs bind
first to the aminoacyl site, and then
move across to the peptidyl site.
To start the process, ONE tRNA must
bind directly to the peptidyl site.
The only one that can do this is the
tRNA loaded with fMet (as we saw
earlier).
98
TRANSLATION TERMINATION
When UAA, UAG or UGA occur,
in frame, in a message,
one or other of three proteins
(RF1, RF2, RF3) bind to the ribosome.
This causes hydrolysis of the bond
between the last amino-acid of the
polypeptide and the last tRNA.
Free polypeptide and free tRNA
are released,
and the 30S and 50S subunits dissociate.