GenesCodeForProteins
The genetic code is responsible for the construction of proteins,
which may be structural components of cells or metabolism
.ontrollinq enzymes, The various levels of genetic instructions
are illustrated below, together with their 'protein equivalents' ,
Nucleotides are the simplest basic unit of genetic information,
that are read in groups of three (called triplets) , One triplet
provides information to bring in a single amino acid during
protein construction , Series of triplets in a long string allow the
synthesis of polypeptide chains and are called genes, Some
~
~
Thispolypeptide
chain
formsonepartof the
functionalprotein,
Polypeptide chain
I
(0 (0 (0 (0
~
~
!
I
I
Triplet
START
t
-+ -+
I
i
Triplel
Triplet
I,
I
Triplet
I
8 8
Thispolypeptide
chain
formstheotherpartof
thefunctionalprotein,
Functlcnal
protein
Polypeptide chain
I
(0 (0 (0 (0 (0 (0
t
t
I
(0
I
A triplet
­
+ Aminoacids
codesforone
aminoacid
~ Proteinsynthesis:
~
~
~
I
:I
transcription
and
II
I
!
i
,
I
i,
I
I translation
~
1
triplets have a special controlling function in the making of a
polypeptide chain. The equivalent of the triplet on the mRNA
molecule is the codon .Three codons can signify the end point of
polypeptide chain construction in the mRNA : UAG. UAA and
UGA (also called STOP codons) . The triplet ATG is found at the
beginning of every gene (codon AUG on mRNA) and marks the
starting position for reading the gene . Several polypeptide chains
may be needed to form a functional protein, The genes required
to do this are collectively called a transcription unit.
Triplel
t
I
Triplel
i
I
I
I
Triple,
STOP
START
Triplet
I
Triplet
I
,
Triplel
I
Triplet
i
I
I
Triplet
Triplet
STOP
5'
3'
.... DNA
Transcription unit
Three nucleotides
make up a triplet
Note : Thit start code is fer the
codingstrandoftheDNAThe
templateDNAstrandfromwhich
themRNAis madewouldhave
thesequence
: TAC.
Inmodelsof nucleicacids,
nucleotides
aredenoted
bytheirbasetetter.
1, The following exercise is designed to establish an understanding of the terms used in describing protein structure and
the genetic information that determines them. Your task is to consult the diagram above and match the structure in the
level of protein organization with its equivalent genetic information:
(a) Nucleotide
codes for:
(b) TriPlet/eoaon
codes for:
(c) Gene
codes for:
~l~f2C"fH A.sz..
~ (y~c;.:)
(d) Transcription unit
codes for:
a. ~c*i5>na- t pn*~ C~
a
N'\
0. N"Y"-o' 'oA=>
a.. cd
J.
IoeMQAa
c:s£
cd-~ ~)
2, Name the basic building blocks for each of the following levels of genetic information:
(a) Nucleotide
is made up of:
p~bo.1S.LI
s,""'(f\.£o.X"l l
~ 4: n,~
knS.C4
(o-tLu,.y,~R ¥I. ~ I ~~'C4 ~f' ~~l OY"" u..ro..~ l)
...
J
l.o::tCil '''1
(b) Triplet is made up of:
~etbL<
(c) Gene is made up of:
(d) Transcription
3 c..ons.oC.c.url
V't-
m
Ck
ba.~eo
--tbo* a...ce:.
rTead
t&cM...
ttf- 1T\~ld-~ , t±CU"+i
~ wj"",, a. idnK't
O!... tu-~on (s-lnp) e.ccNt
.
a. r.eo,\M>AoI'a
unit is made up of :
themRNAis madewouldhave
•
~-n('h
-two
1
Inmodelsof nucleicacids,
c.e:x4.
P":.
....
GeneExpression
.&.\ RAE)
The process of protein synthesis is fundamental to the
understanding of how a cell can control its activities. Genetic
instructions, in the form of DNA , are used as a blueprint for
designing and manufacturing proteins. Some of these proteins
are the enzymes that control the complex biochemical reactions
in the cell, while others take on a variety of other roles. The
Chromosomal DNA
DNA contains the master copy ot
all the genetic information to
produce proteins for the cell.
Most eukaryotic genes contain
segments of coding sequences
(exons) interrupted by non
coding sequences (introns) .
Intron
DNA
Intron
Intron
Intron
]
Exon
Exon Exon
Transcription
H~
Exon
Exon
Double stranded
molecule of
genomic DNA
Exon
Reverse transcription occurs when retroviruses (e.g.
HIV) invade host cells. Their vira l RNA is converted
to DNA and spliced into the host's genome by an
enzyme called reverse transcriptase .
._----- -~II-I--- .........
Primary RNA Transcript
Both exons and introns are
transcribed to produce a long
primary RNA transcript.
Intron
process of transferring the information encoded in a gene to its
functional gene product is called gene expression. It is divided
up into two distinct stages: transcription and translation .These
are summarized below and detailed in the following pages. For
the sake of simplicity, the involvement of introns in gene
expression has been omitted from the following pages.
Primary RNA
II
-,
r
Introns
I::::~.,
Exons are
spliced together
Messenger RNA
The introns are then removed by
splicing to form a mature mRNA.
Messenger RNA is an edited
copy of the DNA molecule (now
axctuding the introns) that codes
for the making of a single protein.
Introns in the DNA (also copied to the primary
RNA) are long sequences of codons that have
(as yet) no apparent function . They may be the
remnants of now unused ancient genes . It has
been suggested that they might facili tate
recomb ination between protein-codinq regions
(exons ) of different genes; a process known as
exon shuffling . Th is may accelerate evolution .
Translation
Structural
proteins
Y
Y
Y
'f
Immunological
proteins
Regulatory
proteins
y
Transport
proteins
Catalytic
proteins
1. The hypothesis known as the central dogma of biology states that: "genetic information can only flow in the direction of
DNA to proteins and not in the opposite directiort'. Accounting for the ideas in the diagram above , form a discussion
group with 2-3 of your classmates and discuss the merits of this statement. Summarize your group 's response below :
~ CJ£.IPreoo
i C\l"\ ',~ -tb.t. f20CC;OO
~ wh't c..h ..fut. c..cll (or c~ .~
rf'O,N\A.tto..~ rtcM,o
~ ~ ~ fln:d..Mc~. ~A ~
-+bt.. "lnftx~S>00J2 c0c4" or "ost<udi.ori foe ~'Pc1d~ a.. ~.
'Thnk \nb=n"'\a.noo o\" I,crJ...A"s. -h-o.J(\$m"~ ~~c.r'f'\i~ ~
I,
-tco.m.s.~S*'\)v..U\~ maJk'~
0.
~K'\. 1'h;.. ~ crl ~ ~ \s ',mflb~~
.
2. Explain the significance of introns and exons found in DNA and primary RNA:
(a) Intron:
.s <Lt£.
W'\~I\
l~
(b) Exon:
"..
,nt~ ~~q.....Q....k\....-CJ:D"""""",--_-i+cx;t""""'-=-:.L.-~""""--L__ LUo""""''''''''~'''''''''''''''''--''-....L..L~
\oe.!ca.-tr()N\.S.lo~, '1h\s \s
dA NoT
ccv1L
rCJrrA ~ \n -tb.L
."'
""c.......
....
·...........
~ eri~u..vU -=-.
mt(.NA.
- " •• _ • •••••
CO l f 9 d
_ .•
o.~
~
_~:_ : : : : ~
•••
_
-fur
- - "
'--
a..
"
---~. _ ..
_ .. _ ......
~.
. .. .- ... -r r
---_ ._~
Transcripti
on
Transcription is the process by which the code contained in the
DNA molecule is transcribed (rewritten) into a mRNA molecule .
Transcription
is under the control of the cell's metabolic
processes which must activate a gene before this process can
begin . The enz yme that directly controls the process is RNA
polymerase, which makes a strand of mRNA using the single
antisense (template ) strand of DNA as a template . The enzyme
transcribes onl y a gene length of DNA at a time and therefore
recognizes star t and stop signals (codes ) at the beginning and
end of the gene . Only RNA polymerase is involved in mRNA
synthesis ; it causes the unwinding of the DNA as well. It is
common to find several RNA polymerase enzyme mo lecules on
the same gene at anyone time, allowing a high rate of mRNA
synthesis to occur.
5'
A copy of the genetic information for making a
protein is made in the form of messenger RNA
(mRNA). Many mRNA copies may be made from
a single gene on the DNA molecule. Once the
mRNA is complete and has been released from
the chromosome, it travelsto the edgeof the nucleus
where it gains access to the cytoplasm through a
tiny hole called a nuclear pore. In prokaryoticcells
(bacteria) there is no nucleus, and thechromosomes
are in direct contact with the cytoplasm.This means
that the next stage (translation ) can begin
immediately, with the mRNA still being synthesized
by enzymes on the DNA molecule.
Freenucleolides
usedto construct
the mRNAstrand.
Single-armed
enrornosomeas found
in non-diVidingcell.
RNApolymerase
enzyme
Template strand of DNA
containstheinformationfor
the constructionof a protein.
Pore(hole) in the nuclear
mp.mhr;:l:ne
throuah which
"",
"
Coding strand of
DNAhas a
nucleotidesequence
complementary to
the templatestrand.
the mRNApassesto enter
the cytoplasm.
-.
.
. ,~
:'?~~j
"'-I'"
X '
~
'.
-,
Oncein thecytoplasm,
the mRNAwill engage
ribosomesto beginthe
nex1stagein protein
synthesis:translation
Formationof a singlestrandof
mRNAthatis complementaryto the
templatestrand(therefore thesame
"message" as the codingstrand).
The two strands
of DNAcoil up
into a helix.
Nuclear membranethat
enclosesthe nucieus.
Nucleus
5'
1 . Explain the role of messenger
m ~NA COKrieo
.fn:m. -l-bDDNA 'W\. fun n"cJ~
RNA (m R N A) in protein synthesis :
~c. \~~
',0.
Cytoplasm
0..
~
of -tht..
~ r·'W;;;.~
+Vw.o.~p~m.
2 . The genetic code contains punctuation codons to mark the starting and finishing points of the code for synthesis of
polypeptide chains and proteins . Consult the mRNA-amino
acid table earlier in this manua l and state the codes for :
(a) Start codon:
--LA....u,l""'-l""c:.r04- _
For the following
triplets
on the DNA , determine
(a ) Triplets on the DNA:
Codons
on the mRNA:
(b ) Triplets on the DNA :
Codons
on the mRN A :
Coding strand of
n NA
h~~ ~
(b) Stop (termination ) codons :
~~L
~A G
for the mRNA that would be synthes ized :
G A
C C G
C
c:r iluc..
GGcC
GrCL.l
AAJA...
G
C C T
A T A
A A A
T A C
AlA
the~ sequence
(A.M, lAAG",UGrA
T A C
A A
ALJ.G
lAUe
GMA lAAL\.
" ' ~~L_
T T T
tAlA.lI\.
~~~..
Translati
on
The diagram below shows the translation phase of protein
­
more of the mRNA than the ribosome to the left. The anti
synthesis . The scene shows how a single mRNA molecule
codon at the base of each tRNA must make a perfect
can be 'serviced' by many ribosomes at the same time. The
complementary match with the codon on the mRNA before
ribosome on the right is in a more advanced stage of
the amino acid is released. Once released, the amino acid is
constructing a polypeptide chain because it has 'translated'
added to the growing polypeptide chain by enzymes.
dBd
unlOB:t:
Thr·IRNA
Polypeptide chain
Ii
This chain is in an advanced
stage of synthesis.
unloa~
""~~
Polypeptide chain
in an early stage
of synthesis
unload:t;ed
Thr·IRNA
-'---" ' "
Ii
JIJdJjbtDjjoiIilllUl.nllillUlJlllJdlltlJ
=-__
5'
mRNA_
.~
---:>
......
Ribosomes rnovmq In this direction
Amino acid
attachment site
tRNA molecules move into the ribosome .
bringing in amino acids 10add to the
polypeptide cham unde r construction.
t
Ribosome
i
Large
subunit
The anticodon is the site
of the 3·base sequence
that 'recognizes' and
matches up with the codon
on the mRNA molecule.
L=--.J
Small
subuni t
Anticodon
Ribosomes are maoe up of a complex of ribosomal RNA (rRNA)
and proteins. They exist as two seoarate sub-units until they
are attracted to a binding site on the mRNA molecule . when
they join together . Ribosomes have binding sites that attract
tRNA molecules loaded with amino acids. The transfer RNA
(tRNA) molecules are about 80 nucleotides in length and are
made under the direction ot genes in the chromosomes. There
is a different IRNA molecu le for each of the differenl possible
anticodons (there may be up to six different tRNAs carrying the
same amino acid).
1, For the following cedens on the mRNA, determine the an1l;codons for each tRNA that would deliver the amino acids:
Codons on the mRNA:
U A C
U A G
C C G
C GA
U U U
Ant i-codons on the tRNAs:
2. There are many different types of tRNA molecules , each with a different anti-codon (HINT: see the mRNAtable ).
(a) State how many different tRNA types there are , each with a unique anticodon :
bu* 3
C1J«.. -s-k>f
~.
etALh.We&.0..
bringing in amino acids 10add to the
_
:fb.A..<e
c::vye. <ell'f'b~; bls..~ -mr:
WlIl..NA}
Col vxt.ot'\$ ~r m tNA rr:qu:£G'=21~NAs
ea_
Ib) Give a reason for your answer in ra) above:
.J2J
<;.oc;ton.s·
attachment site -
n
TheGeneticCode
The genetic information that cooes for the assembly of amino
acias is stored as three-ietter codes. calied codons. Each codon
ioresents one of 20 amino acids used In the constructio n of
olyoeptide chains, The mANA amino acid table (bottom 01
page) can be used to identify the amino acid encoded by each of
the mRNA cocons. NOTethat the code is degenerate in that for
I
eac h amino acid. there may be more than one codon. Most of
this aegenerac y involves the third nucleotide of a codon. The
genetic code is universal ; all liVing organisms on Earth, from
viruses and bacteria, to plants and humans, share the same
genetic code book (with a few minor exceotions representing
mutations tnat have occurred over the iong history of evolution),
IUAA, UUGr, CW--\..
Ir's u: r'dl
c:.uL::r
Leu
Leucine
Lys
Lysine
Met
Methionine
Phe
Phenylalanine
Pro
Proline
Glutamine
Ser
Serine
Glu
Glutamicacid
Thr
Threonine
Gly
Glycine
Try
Tryptophan
His
Histidme
CAu,C,Ac..
Tyr
Tyrosine
Isoleucine
IAUU., N..AC,AUA
Val
Ala
Alanine
Arg
Arginine
Asn
Asparagine
Asp
Asparticacid
Cys
Cysteine
Gin
r' s o
4
GCU, GCC, GCA, GCG
I AA.<A,
12
AAC
ItAGfU, \AGrC
12.
IGrAA ..Gt'AGr
Z.
AAA,A~
UtA.U..,
l,AGlA,
I(J.f
12.
\..llAC,
I
uc.c. v.LA,
r....A
I
(;ret
I ~ , Ace,
A~c..
k..A. Ac&
b
Ilf
I
II
tA.ACAI lAAc
1
2
Valine
Use tne mRNA amino acid table (below ) to us: In the taole above all tne codons that code for each of the amino acids
and the number of different codons that can code tor each amino acid (the first amino acid has been done for you).
2. (a) State how many amino acids could be coded for if a codon consisted of just TWO bases:
--l/....
b....,.
_
(b) Explain wh y this number of bases is inadequate to code for the 20 amino acids required to make proteins :
~-~<. ~ c:la N::>t pn:>vido
q,xc:
enr.::.wt'
G<:)tnbiOQh'~
- ~
2-0 c:A.·If£~ a~D o.~-elA
3, There are multiple codons for a single amino acid, Comment on the significance of this with respect to point mutations:
~kV'&-
p:(m
~~n,Q ""o"od
..fha. 0"wM2trv6t>1')
(~'l\
'In
+tu
mRNA-Amino Acid Table
~-t ~ -#u. ~ ac:'ol-e&pC-c.t~ 'rf
b)± ba.t(., 6.,r ~
~tz.~ ..~p,R"j if *'--~'.
c..OD6'NCHART \r>,kf " \ ) ~ ~r'""
Rea d secon d
lener nere
Example : Determine CAG
C on the left row, fl. on the too colu mn, G
on the right row
CAG is Gin I,glutamine )
Ie)
Read lhrrd
Second Letter
Head firs'
letter nere ~
How to read the table: The table on the right
IS used to 'oecods ' the genetic code as a
sequence of amino acids in a polypentide
chain . from a given mRNA sequence To work
out whic h amino acid is code d tor by a codon
(trip let of oases ) look for the first tette r of the
codon in tne row label on the iei! hand side.
]ben lo ok for the column that Intersects the
I ----e
row from above tnat matches tne
___
.ronc base. Finally, locate the third base in
the codon by looking along the row from the
rrgnt nand end tha t matcnes you r codon,
CecA opt -
u
c
Pne
A
UCU seUCC Ser
UCA Se t
UCG Ser
UAU
UAC
UAA
UAG
Ty'
i yr
STOP
CCU
CCC
CCA
CCG
Pro
CAU
CAe
!-1 1 ~
AUA Iso
AUG Me:
ACU
ACC
ACA
ACG
Tnr
in I n:
GUU va
GUC Va
GUA Va
GUG ",jo
GCU
GCe
GCA
GCG
Al e
UUU
UUC
UUA
UUG
Pn"
Leu
Leu
CUU Leu
cue
L et,
CUA Lel.
CUG Le u
AUU rsc
AUC
ISO
F :,c
Pre
" rc
.,...,
~
AlE:
Ali
Ali
here
STOP
G
UGU e VE
UGC c."s
UGA STOo
UGG Try
CAf... Gl....
CAG Gte
CGU Ar"
CGC A'9
CGA A r~
CGG A,g
AAU .A.Sr.
AAC p,sn
AAA LYS
AAG LYS
AGU Se r
AGC Ser
AGA Arg
AGG Arg
GAU ASL
GAC Aso
GAt. Glu
GAG GI...
GGU Gry
GGC G l,!
GGA G'y
HI S
GGG
Gr.'
and the number of different codons that can code tor each amino acid (the first amino acid has been done for vou),
A.
ITd'
~f\.
ProteinSynthesis
Summary
~
8
.
ee
8eO
08
. :·..0.····
~aQ
I~ V
: ~ 6
,' ~ b ­.'.tJ '0- .
..
~....
' 4-..
ltlr J
Cytoplasm
The diagram above shows an overview of the process of protein synthesis . It is a combination of the diagrams from the
previous two pages. Each of the major steps in the process are numbered , \~Ihil€' structures :r9 labeled with letters.
1. Write a brief description of each numbered process in the diagram above:
(a) Process 1: \AK\wi ~
DNA
(b) Process 2 :
-tra.,oWr'ipt6>n-mitNA s..~w -r\A.lcJ,dic44
(c) Process 3 :
DNA
feb>"
Q"Q(dp.,t
10 ~"3 mleNf\
rd.o
:::
::::::
:::::a:::;i:::!'::::::::
~ :::~::=~r~
~ ~t ... ~h 4eJlW*c~crxAAn f Q'>clAnkn'L
(g) Process 7 : l:,(?NA lea.v'Co
i'h.Q ";~ny"
(f) Process 6 :
.L'"
(h) Process 8: ~ ....NA
•
u
II
.-J ~
rc.c:~""
'I~-
. ..Lt.
.-
•
~ ~ 0.(.4
Wl=rn
-pMLf a.rn;(1() "1«~
o;:.~~
'd
2 . Identify each of the structures marked with a letter and write their names below in the spaces provided :
]).....
NL.1Lf\~
(a) Structure A: .....
_
(b) Structure B: "FCc.e. W'\M("ld
(c ) Structure
c RNA
ertickoe
3
_
"'udDl1#' ~~
fU.McA-iC)n1
wi:tWn
CA. W.l
~Q
(i) Structure I:
~ 1r::c~r'Y'A-
U) Structure J :
Po\~r~h c.k\a.,,:O
/p~
7~k+,An c:.wn:.
o.cAc::Lo
(J'!I)
fS'tlJ.j bla. -ftir:
-:E'N\..~J ~p>rt Prb~J
~iN.e-hL(a..i ~} ~'!LJ """"-KnCdat(f
d .~
-tho. n"nC4-\C*"- uf ~ c..eM - PAMt'9V\ vs· rCEA,blotd
........ 11'0.
_
(h) Structure H:
Explain the purpose of protein synthesis (gene expression):
So ~
N. ACg De,., fDr"-'
(g) Structure G:_"'=-""""""e..!!!<L...N::..oA-=-
Pl)l~
(d) Structure D: ......
m~R"""""N.3I.J'P.::L(e) Structure E
(f) Structure F:
Go.
Dec
h..ey ~
~tl.
l')n
~
~,~
•
.J