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

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

Thispolypeptidechain Functlcnal

Thispolypeptidechain formsonepartof the ~ formstheotherpartof functionalprotein ~ protein thefunctionalprotein

Polypeptide chain I I

(0 (0 (0 (0 8 8 (0 ~ ~ ~ -+ -+ t I I I

tI iI I II 1 I I i I I

START Triplet Triplet Triplel Triplel Triplet Triplel Triple

5

Note Thit start code is fer the codingstrandoftheDNAThe templateDNAstrandfromwhich

A triplet codesforone aminoacid

STOP START

Polypeptide chain I I

Aminoacids (0 (0 (0 (0 (0 (0 +shy

~ ~ ~t t i I

i I I

iII I I I

Triplet Triplet Triplel Triplet Triplet

make up a triplet

~ Proteinsynthesis

I transcriptionand I translation I

Triplet STOP

3

DNA

Three nucleotides Transcription unit

themRNAismadewouldhave InmodelsofnucleicacidsthesequenceTAC nucleotidesaredenoted

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) TriPleteoaon codes for a N 0 NY-ooA=gt a cd J (c) Gene codes for ~l~f2CfH Asz~(y~c) (d) Transcription unit codes for a ~ci5gtna- t pn~ C~ IoeMQAa cspoundcd-~~)

2 Name the basic building blocks for each of the following levels of genetic information

(a) Nucleotide is made up of p~bo1SLI s(fpoundoXl l ~ 4 n~ knSC4

(o-tLuy~R J yenI ~ I ~~C4 ~f ~~l OY uro~ l) lotCil 1 bull

(b) Triplet is made up of 3 Vt- ~-n(h ba~eo rTeadconsoCcurl --tbo ace

~etbLlt m Ck tampcM (c) Gene is made up of a reoMgtAoIa ttf- 1T~ld-~ tplusmnCU+i~ wj a idnKt cex4

(d) Transcription unit is made up of -two O tu-~on (s-lnp) eccNt

themRNAismadewouldhave Inmodelsofnucleicacids1

P shyamp RAE) GeneExpression

The process of protein synthesis is fundamental to the process of transferring the information encoded in a gene to its understanding of how a cell can control its activities Genetic functional gene product is called gene expression It is divided instructions in the form of DNA are used as a blueprint for up into two distinct stages transcription and translation These designing and manufacturing proteins Some of these proteins are summarized below and detailed in the following pages For are the enzymes that control the complex biochemical reactions the sake of simplicity the involvement of introns in gene in the cell while others take on a variety of other roles The expression has been omitted from the following pages

Intron Intron Intron Intron Intron

Most eukaryotic genes contain segments of coding sequences Reverse transcription occurs when retroviruses (eg (exons) interrupted by non shyHIV) invade host cells Their vira l RNA is converted Transcriptioncoding sequences (introns) to DNA and spliced into the hosts genome by an H~ enzyme called reverse transcriptase

Primary RNA Transcript _-----Primary RNA

Both exons and introns are transcribed to produce a long primary RNA transcript

Messenger RNA

The introns are then removed by RNA) are long sequences of codons that have

splicing to form a mature mRNA (as yet) no apparent function They may be the Translation Messenger RNA is an edited remnants of now unused ancient genes It has

copy of the DNA molecule (now been suggested that they might facili tate

axctuding the introns) that codes recomb ination between protein-codinq regions

for the making of a single protein (exons ) of different genes a process known as exon shuffling Th is may accelerate evolution

f Y Y Y y

Structural Regulatory Immunological Transport Catalytic proteins proteins proteins proteins proteins

]

Double stranded molecule of genomic DNA

ExonExonExonExon Exon Exon

DNAChromosomal DNA

DNA contains the master copy ot all the genetic information to produce proteins for the cell

-~II-I---

Exons are spliced together

- rII Introns

I~

Introns in the DNA (also copied to the primary

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

~ CJpoundIPreoo ~ -tbt f20CCOO ~ whtch fut ccll (or c~~i Cl

rfONAtto~ rtcMo ~ ~ ~ flndMc~ ~A ~

-+btlnftx~Sgt00J2 c0c4 or ostltudiori foe ~Pc1d~ a ~ Thnk nb=nanoo I s -h-oJ($m~ ~~crfi~o IcrJA ~

-tcoms~S)vU~ maJk~ 0 ~K 1h ~ crl ~ ~ s mflb~~ 2 Explain the significance of introns and exons found in DNA and primary RNA

k-CJD--_-i+cxt-=-L-~--L__ LUo~---LL~

(a) Intron sltLtpound nt~ ~~qQ

W~I oeca-tr()NSlo~ 1hs s CO l f 9 d

l~ dA NoT ~ -=-ccv1L ceri~uvUmiddot _

(b) Exon rCJrrA ~ n -tbL mt(NA o~ ~ -fur a

- bullbull _ bull bullbullbullbullbull _ bull _~_ ~ bullbullbull - - -- ---~ _ _ _ ~ - -r r ---_ _~

P shyamp RAE) GeneExpression

The process of protein synthesis is fundamental to the process of transferring the information encoded in a gene to its understanding of how a cell can control its activities Genetic functional gene product is called gene expression It is divided instructions in the form of DNA are used as a blueprint for up into two distinct stages transcription and translation These designing and manufacturing proteins Some of these proteins are summarized below and detailed in the following pages For are the enzymes that control the complex biochemical reactions the sake of simplicity the involvement of introns in gene in the cell while others take on a variety of other roles The expression has been omitted from the following pages

Intron Intron Intron Intron Intron

Most eukaryotic genes contain segments of coding sequences Reverse transcription occurs when retroviruses (eg (exons) interrupted by non shyHIV) invade host cells Their vira l RNA is converted Transcriptioncoding sequences (introns) to DNA and spliced into the hosts genome by an H~ enzyme called reverse transcriptase

Primary RNA Transcript _-----Primary RNA

Both exons and introns are transcribed to produce a long primary RNA transcript

Messenger RNA

The introns are then removed by RNA) are long sequences of codons that have

splicing to form a mature mRNA (as yet) no apparent function They may be the Translation Messenger RNA is an edited remnants of now unused ancient genes It has

copy of the DNA molecule (now been suggested that they might facili tate

axctuding the introns) that codes recomb ination between protein-codinq regions

for the making of a single protein (exons ) of different genes a process known as exon shuffling Th is may accelerate evolution

f Y Y Y y

Structural Regulatory Immunological Transport Catalytic proteins proteins proteins proteins proteins

]

Double stranded molecule of genomic DNA

ExonExonExonExon Exon Exon

DNAChromosomal DNA

DNA contains the master copy ot all the genetic information to produce proteins for the cell

-~II-I---

Exons are spliced together

- rII Introns

I~

Introns in the DNA (also copied to the primary

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

~ CJpoundIPreoo ~ -tbt f20CCOO ~ whtch fut ccll (or c~~i Cl

rfONAtto~ rtcMo ~ ~ ~ flndMc~ ~A ~

-+btlnftx~Sgt00J2 c0c4 or ostltudiori foe ~Pc1d~ a ~ Thnk nb=nanoo I s -h-oJ($m~ ~~crfi~o IcrJA ~

-tcoms~S)vU~ maJk~ 0 ~K 1h ~ crl ~ ~ s mflb~~ 2 Explain the significance of introns and exons found in DNA and primary RNA

k-CJD--_-i+cxt-=-L-~--L__ LUo~---LL~

(a) Intron sltLtpound nt~ ~~qQ

W~I oeca-tr()NSlo~ 1hs s CO l f 9 d

l~ dA NoT ~ -=-ccv1L ceri~uvUmiddot _

(b) Exon rCJrrA ~ n -tbL mt(NA o~ ~ -fur a

- bullbull _ bull bullbullbullbullbull _ bull _~_ ~ bullbullbull - - -- ---~ _ _ _ ~ - -r r ---_ _~

Transcription 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 cells 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

Template strand of DNA containstheinformationfor the constructionof a protein

Formationof a singlestrandof mRNAthatis complementarytemplatestrand(thereforethesame message as the codingstrand)

RNApolymerase enzyme

Freenucleolides usedto construct the mRNAstrand

5

5

The two strands of DNAcoil up into a helix

Single-armed enrornosomeas found in non-diVidingcell

shy Coding strand of ~ nucleotidesequence - complementary to the templatestrand

DNAhasa

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 nucleusand thechromosomes are in directcontact with the cytoplasmThis means that the next stage (translation ) can begin immediately with the mRNA still being synthesized by enzymes on the DNA molecule

Pore(hole) in the nuclear mpmhrlnethrouah which the mRNApassesto enter ~ the cytoplasm - ~~j

-IX

Oncein thecytoplasm to the the mRNAwill engage

ribosomesto beginthe nex1stagein protein synthesis translation

Nuclear membranethat enclosesthenucieus

Nucleus Cytoplasm

1 Explain the role of messenger RNA (m R NA) in protein synthesis m ~NA COKrieo 0 ~ of -tht ~c ~~ fnm -l-bDDNA W fun ncJ~ ~ rmiddotW~

0 +Vwo~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 --LAul-lcr04- _ (b) Stop (termination ) codons (AM lAAGUGrA

For the following triplets on the DNA determine the~ sequence for the mRNA that would be synthes ized

(a ) Triplets on the DNA T A C ~A G C C G C G A T T T

Codons on the mRNA AlA cr iluc GGcC GrCLl AAJA (b) Triplets on the DNA T A C A A G C C T A T A A A A

Codons on the mRN A ALJG lAUe GMA lAAL tAlAlI

Coding strand of n NA h~~ ~ ~~L ~~L_ ~~~

Translation The diagram below shows the translation phase of protein more of the mRNA than the ribosome to the left The antishysynthesis 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 unlOBtThrmiddotIRNA

Polypeptide chain

Ii This chain is in an advanced stage of synthesis

unloa~ Polypeptide chain in an early stage ~~

of synthesis

unloadtedThrmiddotIRNA

---- Ii

5 JIJdJjbtDjjoiIilllUlnllillUlJlllJdlltlJmRNA_~ =-__ ---gt

tRNA molecules move into the ribosome bringing in amino acids 10add to the polypeptide cham under construction

Ribosome t i

The anticodon is the site of the 3middotbase sequence that recognizes and L=--J

Large Small matches up with the codon Anticodonsubunit subuni t on the mRNA molecule

(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 an1lcodons for each tRNA that would deliver the amino acids

Codons on the mRNA U A C U A G C C G C G A 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 J2J _ Ib) Give a reason for your answer in ra) above fbAltecvye ltellfb~ bls~ -mrWlIlNA

bu 3 C1Jlaquo-s-kgtf~ Col vxtot$ ~r m tNA rrqupoundG=21~NAs

etALhWeamp0 ea_ ltoctonsmiddot bringing in amino acids 10add to the attachment site - n

Ribosomes rnovmq In this direction

Amino acid attachment site

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

TheGeneticCode The genetic information that cooes for the assembly of amino each amino acid there may be more than one codon Most of acias is stored as three-ietter codes calied codons Each codon this aegenerac y involves the third nucleotide of a codon The ioresents one of 20 amino acids used In the construction of genetic code is universal all liVing organisms on Earth from olyoeptide chains The mANA amino acid table (bottom 01 viruses and bacteria to plants and humans share the same

page) can be used to identify the amino acid encoded by each of genetic code book (with a few minor exceotions representing the mRNA cocons NOTethat the code is degenerate in that for mutations tnat have occurred over the iong history of evolution)

Ala Alanine

Arg Arginine

I GCU GCC GCA GCG 4 Leu

Lys

Leucine

Lysine

IUAA UUGr CW-- Irs u rdl cuLr

AAAA~

Asn Asparagine I AAltA AAC 12 Met Methionine

Asp Asparticacid Phe Phenylalanine UtAU llAC 12 Cys Cysteine ItAGfU AGrC 12 Pro Proline

Gin Glutamine

Glu Glutamicacid IGrAA GtAGr Z

Ser

Thr

Serine

Threonine

lAGlA uccI vLA I I(Jf rA(ret A~c b

I ~ Ace kA AcampIlf Gly Glycine

His Histidme

r so Isoleucine

CAuCAc

IAUU NACAUA

Try

Tyr

Val

Tryptophan

Tyrosine

Valine

I

I I tAACAI lAAc 1 2

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 --lb _

(b) Explain why this number of bases is inadequate to code for the 20 amino acids required to make proteins

~-~lt ~ Ngtt pngtvido enrwt -~cla Glt)tnbiOQh~

qxc 2-0 cAmiddotIfpound~ 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

~kVamp- p(m ~~nQ ood ~-t ~ -u ~ acol-eamppC-ct~ rf

fha 0wM2trv6tgt1) (~l In +tu b)plusmn bat( 6r ~ CecA opt - ~tz~ ~pRj if --~ cOD6NCHART rgtkf ) ~ ~r

mRNA-Amino Acid Table Read secon d

lener nere Second Letter Read lhrrdIe)here A ITd

How to read the table The table on the right

Head firs letter nere ~ u c A G ~f

IS used to oecods the genetic code as a UUU Pne UCU se- UAU Ty UGU e VE

sequence of amino acids in a polypentide chain from a given mRNA sequence To work

UUC Pn UUA Leu UUG Leu

UCC UCA UCG

Ser Se t Ser

UAC i yr

UAA STOP

UAG STOP

UGC cs UGA STOo

UGG Try out whic h amino acid is code d tor by a codon (trip let of oases ) look for the first tette r of the CUU Leu CCU Pro CAU -1 1 ~ CGU Ar

codon in tne row label on the iei hand side cue L et

CUA Lel

CCC CCA

Fc Pre

CAe HI S

CAf Gl CGC CGA

A9 A r~

]ben lo ok for the column that Intersects the CUG Le u CCG rc CAG Gte CGG Ag

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

AUU rsc AUC ISO

AUA Iso

ACU ACC ACA

Tnr in -I n

AAU ASr

AAC psn

AAA LYS

AGU AGC AGA

Se r Ser

Arg

rrgnt nand end tha t matcnes you r codon AUG Me ACG ~ AAG LYS AGG Arg

Example Determine CAG GUU va GCU Al e GAU ASL GGU Gry

C on the left row flon the too colu mn G GUC Va

GUA Va

GCe GCA

AlE

Ali

GAC GAt

Aso

Glu

GGC Gl

GGA Gy on the right row GUG jo GCG Ali GAG GI GGG Gr

CAG is Gin Iglutamine )

and the number of different codons that can code tor each amino acid (the first amino acid has been done for vou)

ProteinSynthesisSummary ~

middot0middotmiddotmiddotmiddot ~ 6

~ b shytJ 0- 4-

8 ee 8eO08~aQ

I~ V

Cytoplasm

~

ltlr J

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 ~Ihileuro structures r9 labeled with letters

1 Write a brief description of each numbered process in the diagram above

(a) Process 1 AKwi ~ DNA

(b) Process 2 -traoWript6gtn-mitNA s~w -rAlcJdic44 QQ(dpt 10 ~3 mleNf

(c) Process 3 DNA febgt rdo

~ ~=~r~ai(f) Process 6 ~~t ~h 4eJlWc~crxAAn f QgtclAnknL -pMLf arn(1() 1laquo~

(g) Process 7 l(NA leavCoihQ ~ny o~~ L bull II -J ~ I~- Lt - bull d

(h) Process 8 ~ NA u rcc~ Wl=rn ~~ 0(4

2 Identify each of the structures marked with a letter and write their names below in the spaces provided

(a) Structure A ])NL1Lf~ _ (f) Structure F NACg De fDr-

(b) Structure B FCce WM(ld ertickoe (g) Structure G_=-eltLNoA-=- _

(c ) Structure c RNA Pl)l~ (h) Structure H ~Q ocAcLo m~RN3IJPL- (i)(d) Structure D _ Structure I ~ 1rc~rYA-

U) Structure J p~

3 Explain the purpose of protein synthesis (gene expression) 7~k+An cwn (JI) fStlJj bla -ftir

So ~ fUMcA-iC)n1witWn CA Wl -EN~J ~pgtrt Prb~J

~iNe-hL(ai ~ ~LJ -KnCdat(f ~ GoDec ~ ~

(e) Structure E udDl1 ~~ Po~r~h ckaO

d hey -tho nnC4-C-uf ~ ceM - PAMt9V vsmiddot rCEAblotd ~tl l)n ~~ bull J

110

Translation The diagram below shows the translation phase of protein more of the mRNA than the ribosome to the left The antishysynthesis 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 unlOBtThrmiddotIRNA

Polypeptide chain

Ii This chain is in an advanced stage of synthesis

unloa~ Polypeptide chain in an early stage ~~

of synthesis

unloadtedThrmiddotIRNA

---- Ii

5 JIJdJjbtDjjoiIilllUlnllillUlJlllJdlltlJmRNA_~ =-__ ---gt

tRNA molecules move into the ribosome bringing in amino acids 10add to the polypeptide cham under construction

Ribosome t i

The anticodon is the site of the 3middotbase sequence that recognizes and L=--J

Large Small matches up with the codon Anticodonsubunit subuni t on the mRNA molecule

(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 an1lcodons for each tRNA that would deliver the amino acids

Codons on the mRNA U A C U A G C C G C G A 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 J2J _ Ib) Give a reason for your answer in ra) above fbAltecvye ltellfb~ bls~ -mrWlIlNA

bu 3 C1Jlaquo-s-kgtf~ Col vxtot$ ~r m tNA rrqupoundG=21~NAs

etALhWeamp0 ea_ ltoctonsmiddot bringing in amino acids 10add to the attachment site - n

Ribosomes rnovmq In this direction

Amino acid attachment site

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

TheGeneticCode The genetic information that cooes for the assembly of amino each amino acid there may be more than one codon Most of acias is stored as three-ietter codes calied codons Each codon this aegenerac y involves the third nucleotide of a codon The ioresents one of 20 amino acids used In the construction of genetic code is universal all liVing organisms on Earth from olyoeptide chains The mANA amino acid table (bottom 01 viruses and bacteria to plants and humans share the same

page) can be used to identify the amino acid encoded by each of genetic code book (with a few minor exceotions representing the mRNA cocons NOTethat the code is degenerate in that for mutations tnat have occurred over the iong history of evolution)

Ala Alanine

Arg Arginine

I GCU GCC GCA GCG 4 Leu

Lys

Leucine

Lysine

IUAA UUGr CW-- Irs u rdl cuLr

AAAA~

Asn Asparagine I AAltA AAC 12 Met Methionine

Asp Asparticacid Phe Phenylalanine UtAU llAC 12 Cys Cysteine ItAGfU AGrC 12 Pro Proline

Gin Glutamine

Glu Glutamicacid IGrAA GtAGr Z

Ser

Thr

Serine

Threonine

lAGlA uccI vLA I I(Jf rA(ret A~c b

I ~ Ace kA AcampIlf Gly Glycine

His Histidme

r so Isoleucine

CAuCAc

IAUU NACAUA

Try

Tyr

Val

Tryptophan

Tyrosine

Valine

I

I I tAACAI lAAc 1 2

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 --lb _

(b) Explain why this number of bases is inadequate to code for the 20 amino acids required to make proteins

~-~lt ~ Ngtt pngtvido enrwt -~cla Glt)tnbiOQh~

qxc 2-0 cAmiddotIfpound~ 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

~kVamp- p(m ~~nQ ood ~-t ~ -u ~ acol-eamppC-ct~ rf

fha 0wM2trv6tgt1) (~l In +tu b)plusmn bat( 6r ~ CecA opt - ~tz~ ~pRj if --~ cOD6NCHART rgtkf ) ~ ~r

mRNA-Amino Acid Table Read secon d

lener nere Second Letter Read lhrrdIe)here A ITd

How to read the table The table on the right

Head firs letter nere ~ u c A G ~f

IS used to oecods the genetic code as a UUU Pne UCU se- UAU Ty UGU e VE

sequence of amino acids in a polypentide chain from a given mRNA sequence To work

UUC Pn UUA Leu UUG Leu

UCC UCA UCG

Ser Se t Ser

UAC i yr

UAA STOP

UAG STOP

UGC cs UGA STOo

UGG Try out whic h amino acid is code d tor by a codon (trip let of oases ) look for the first tette r of the CUU Leu CCU Pro CAU -1 1 ~ CGU Ar

codon in tne row label on the iei hand side cue L et

CUA Lel

CCC CCA

Fc Pre

CAe HI S

CAf Gl CGC CGA

A9 A r~

]ben lo ok for the column that Intersects the CUG Le u CCG rc CAG Gte CGG Ag

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

AUU rsc AUC ISO

AUA Iso

ACU ACC ACA

Tnr in -I n

AAU ASr

AAC psn

AAA LYS

AGU AGC AGA

Se r Ser

Arg

rrgnt nand end tha t matcnes you r codon AUG Me ACG ~ AAG LYS AGG Arg

Example Determine CAG GUU va GCU Al e GAU ASL GGU Gry

C on the left row flon the too colu mn G GUC Va

GUA Va

GCe GCA

AlE

Ali

GAC GAt

Aso

Glu

GGC Gl

GGA Gy on the right row GUG jo GCG Ali GAG GI GGG Gr

CAG is Gin Iglutamine )

and the number of different codons that can code tor each amino acid (the first amino acid has been done for vou)

ProteinSynthesisSummary ~

middot0middotmiddotmiddotmiddot ~ 6

~ b shytJ 0- 4-

8 ee 8eO08~aQ

I~ V

Cytoplasm

~

ltlr J

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 ~Ihileuro structures r9 labeled with letters

1 Write a brief description of each numbered process in the diagram above

(a) Process 1 AKwi ~ DNA

(b) Process 2 -traoWript6gtn-mitNA s~w -rAlcJdic44 QQ(dpt 10 ~3 mleNf

(c) Process 3 DNA febgt rdo

~ ~=~r~ai(f) Process 6 ~~t ~h 4eJlWc~crxAAn f QgtclAnknL -pMLf arn(1() 1laquo~

(g) Process 7 l(NA leavCoihQ ~ny o~~ L bull II -J ~ I~- Lt - bull d

(h) Process 8 ~ NA u rcc~ Wl=rn ~~ 0(4

2 Identify each of the structures marked with a letter and write their names below in the spaces provided

(a) Structure A ])NL1Lf~ _ (f) Structure F NACg De fDr-

(b) Structure B FCce WM(ld ertickoe (g) Structure G_=-eltLNoA-=- _

(c ) Structure c RNA Pl)l~ (h) Structure H ~Q ocAcLo m~RN3IJPL- (i)(d) Structure D _ Structure I ~ 1rc~rYA-

U) Structure J p~

3 Explain the purpose of protein synthesis (gene expression) 7~k+An cwn (JI) fStlJj bla -ftir

So ~ fUMcA-iC)n1witWn CA Wl -EN~J ~pgtrt Prb~J

~iNe-hL(ai ~ ~LJ -KnCdat(f ~ GoDec ~ ~

(e) Structure E udDl1 ~~ Po~r~h ckaO

d hey -tho nnC4-C-uf ~ ceM - PAMt9V vsmiddot rCEAblotd ~tl l)n ~~ bull J

110

TheGeneticCode The genetic information that cooes for the assembly of amino each amino acid there may be more than one codon Most of acias is stored as three-ietter codes calied codons Each codon this aegenerac y involves the third nucleotide of a codon The ioresents one of 20 amino acids used In the construction of genetic code is universal all liVing organisms on Earth from olyoeptide chains The mANA amino acid table (bottom 01 viruses and bacteria to plants and humans share the same

page) can be used to identify the amino acid encoded by each of genetic code book (with a few minor exceotions representing the mRNA cocons NOTethat the code is degenerate in that for mutations tnat have occurred over the iong history of evolution)

Ala Alanine

Arg Arginine

I GCU GCC GCA GCG 4 Leu

Lys

Leucine

Lysine

IUAA UUGr CW-- Irs u rdl cuLr

AAAA~

Asn Asparagine I AAltA AAC 12 Met Methionine

Asp Asparticacid Phe Phenylalanine UtAU llAC 12 Cys Cysteine ItAGfU AGrC 12 Pro Proline

Gin Glutamine

Glu Glutamicacid IGrAA GtAGr Z

Ser

Thr

Serine

Threonine

lAGlA uccI vLA I I(Jf rA(ret A~c b

I ~ Ace kA AcampIlf Gly Glycine

His Histidme

r so Isoleucine

CAuCAc

IAUU NACAUA

Try

Tyr

Val

Tryptophan

Tyrosine

Valine

I

I I tAACAI lAAc 1 2

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 --lb _

(b) Explain why this number of bases is inadequate to code for the 20 amino acids required to make proteins

~-~lt ~ Ngtt pngtvido enrwt -~cla Glt)tnbiOQh~

qxc 2-0 cAmiddotIfpound~ 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

~kVamp- p(m ~~nQ ood ~-t ~ -u ~ acol-eamppC-ct~ rf

fha 0wM2trv6tgt1) (~l In +tu b)plusmn bat( 6r ~ CecA opt - ~tz~ ~pRj if --~ cOD6NCHART rgtkf ) ~ ~r

mRNA-Amino Acid Table Read secon d

lener nere Second Letter Read lhrrdIe)here A ITd

How to read the table The table on the right

Head firs letter nere ~ u c A G ~f

IS used to oecods the genetic code as a UUU Pne UCU se- UAU Ty UGU e VE

sequence of amino acids in a polypentide chain from a given mRNA sequence To work

UUC Pn UUA Leu UUG Leu

UCC UCA UCG

Ser Se t Ser

UAC i yr

UAA STOP

UAG STOP

UGC cs UGA STOo

UGG Try out whic h amino acid is code d tor by a codon (trip let of oases ) look for the first tette r of the CUU Leu CCU Pro CAU -1 1 ~ CGU Ar

codon in tne row label on the iei hand side cue L et

CUA Lel

CCC CCA

Fc Pre

CAe HI S

CAf Gl CGC CGA

A9 A r~

]ben lo ok for the column that Intersects the CUG Le u CCG rc CAG Gte CGG Ag

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

AUU rsc AUC ISO

AUA Iso

ACU ACC ACA

Tnr in -I n

AAU ASr

AAC psn

AAA LYS

AGU AGC AGA

Se r Ser

Arg

rrgnt nand end tha t matcnes you r codon AUG Me ACG ~ AAG LYS AGG Arg

Example Determine CAG GUU va GCU Al e GAU ASL GGU Gry

C on the left row flon the too colu mn G GUC Va

GUA Va

GCe GCA

AlE

Ali

GAC GAt

Aso

Glu

GGC Gl

GGA Gy on the right row GUG jo GCG Ali GAG GI GGG Gr

CAG is Gin Iglutamine )

and the number of different codons that can code tor each amino acid (the first amino acid has been done for vou)

ProteinSynthesisSummary ~

middot0middotmiddotmiddotmiddot ~ 6

~ b shytJ 0- 4-

8 ee 8eO08~aQ

I~ V

Cytoplasm

~

ltlr J

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 ~Ihileuro structures r9 labeled with letters

1 Write a brief description of each numbered process in the diagram above

(a) Process 1 AKwi ~ DNA

(b) Process 2 -traoWript6gtn-mitNA s~w -rAlcJdic44 QQ(dpt 10 ~3 mleNf

(c) Process 3 DNA febgt rdo

~ ~=~r~ai(f) Process 6 ~~t ~h 4eJlWc~crxAAn f QgtclAnknL -pMLf arn(1() 1laquo~

(g) Process 7 l(NA leavCoihQ ~ny o~~ L bull II -J ~ I~- Lt - bull d

(h) Process 8 ~ NA u rcc~ Wl=rn ~~ 0(4

2 Identify each of the structures marked with a letter and write their names below in the spaces provided

(a) Structure A ])NL1Lf~ _ (f) Structure F NACg De fDr-

(b) Structure B FCce WM(ld ertickoe (g) Structure G_=-eltLNoA-=- _

(c ) Structure c RNA Pl)l~ (h) Structure H ~Q ocAcLo m~RN3IJPL- (i)(d) Structure D _ Structure I ~ 1rc~rYA-

U) Structure J p~

3 Explain the purpose of protein synthesis (gene expression) 7~k+An cwn (JI) fStlJj bla -ftir

So ~ fUMcA-iC)n1witWn CA Wl -EN~J ~pgtrt Prb~J

~iNe-hL(ai ~ ~LJ -KnCdat(f ~ GoDec ~ ~

(e) Structure E udDl1 ~~ Po~r~h ckaO

d hey -tho nnC4-C-uf ~ ceM - PAMt9V vsmiddot rCEAblotd ~tl l)n ~~ bull J

110

ProteinSynthesisSummary ~

middot0middotmiddotmiddotmiddot ~ 6

~ b shytJ 0- 4-

8 ee 8eO08~aQ

I~ V

Cytoplasm

~

ltlr J

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 ~Ihileuro structures r9 labeled with letters

1 Write a brief description of each numbered process in the diagram above

(a) Process 1 AKwi ~ DNA

(b) Process 2 -traoWript6gtn-mitNA s~w -rAlcJdic44 QQ(dpt 10 ~3 mleNf

(c) Process 3 DNA febgt rdo

~ ~=~r~ai(f) Process 6 ~~t ~h 4eJlWc~crxAAn f QgtclAnknL -pMLf arn(1() 1laquo~

(g) Process 7 l(NA leavCoihQ ~ny o~~ L bull II -J ~ I~- Lt - bull d

(h) Process 8 ~ NA u rcc~ Wl=rn ~~ 0(4

2 Identify each of the structures marked with a letter and write their names below in the spaces provided

(a) Structure A ])NL1Lf~ _ (f) Structure F NACg De fDr-

(b) Structure B FCce WM(ld ertickoe (g) Structure G_=-eltLNoA-=- _

(c ) Structure c RNA Pl)l~ (h) Structure H ~Q ocAcLo m~RN3IJPL- (i)(d) Structure D _ Structure I ~ 1rc~rYA-

U) Structure J p~

3 Explain the purpose of protein synthesis (gene expression) 7~k+An cwn (JI) fStlJj bla -ftir

So ~ fUMcA-iC)n1witWn CA Wl -EN~J ~pgtrt Prb~J

~iNe-hL(ai ~ ~LJ -KnCdat(f ~ GoDec ~ ~

(e) Structure E udDl1 ~~ Po~r~h ckaO

d hey -tho nnC4-C-uf ~ ceM - PAMt9V vsmiddot rCEAblotd ~tl l)n ~~ bull J

110