QuantitativeCentralDogmaI:Reference:http://book.bionumbers.orgBasicSizeandGeometryhttp://book.bionumbers.org/what-is-the-range-of-cell-sizes-and-shapes/Volume:onelitre=1L=10−3m3 ;E.Colisizeisabout1µm ,andthevolumeisestimatedtobe VEcoli !1µm

3 =10−18m3 ( 1µm3 × 10−6miµm−1( )3 =10−18m3 ).

Concentration: 1M =1mol iL−1 ,but1mol =6.023×1023 ,so1M =6.023×1026m−3 ,Inassignment4,youwillshowthat 1nM ∼onemoleculeinavolumeof1µm

3 .Mass:OneDalton=MassofHydrogen=1Da=1.66×10

−27kg=1.66×10−24 g ,andonemole(Avogadro’snumberNA =6.023×1023 )ofhydrogenweighs1gram=1g

( 1Da( )−1 = 1.66×10−24 g( )−1 =6.022×1023 g−1 ):1g=6.023×1023Da .

Oneaminoacid(aa)is ∼100Da=1.66×10−25kg : 1aa∼100Da .

Anaverageprotein(gene)hasabout300aminoacids: 1protein ∼300aa .Onebasepair(bp)is ∼650Da : 1bp∼650Da ProteinConcentration:http://book.bionumbers.org/how-many-proteins-are-in-a-cell/

Datashowsthattheproteinmassperunitvolumeisabout0.2 g

mL,but

1mL=10−6m3 =10−6m3 × 106 µm

m⎛⎝⎜

⎞⎠⎟

3

=1012µm3 ,soAverageMassconcentrationof

proteinsincellisMV

⎛⎝⎜

⎞⎠⎟=0.20 g

ml×6.023×1023 Da

g×10−12 ml

µm3 =1.2×1011Daµm3

AverageNumberconcentrationofproteinsincellisfoundby:

NV

⎛⎝⎜

⎞⎠⎟= M

V⎛⎝⎜

⎞⎠⎟

1100Da iaa−1 ×300aa iprotein−1 = 4×106

proteinµm3 .Notethattheanswerin

thewebpageiswrong!Therewillbeasimilarquestionontheassignment.

InitiationTranscription:RNApolymeraseproteinandTranscriptionFactor(TF)proteinsbindstopromoterregionofDNA

Transcription:MessengerRNA,mRNA,isproducedandtransportedtoRibosomes

Translation:mRNAreadbyRibosomestoproduceproteins

mRNAinCells:http://book.bionumbers.org/how-many-mrnas-are-in-a-cell/Experimentally,itisknownthatabacteriumhasaboutNmRna

bacteria ≈103 mRNA,andamammaliancellhasaboutNmRna

mamalian ≈105 .ItisknownthatthenumberofmRNAinacellremainsrelativelyconstant.ItseemsthatmRNAconcentrationscaleswithsize.

Bacteria:Let’sassumethattheproteinconcentrationinacellis ∼4×106 proteins

µm3 ,

whichmeansthatabacteriumofvolume1µm3 hasabout4×106 proteins.Itis

knownthat,onaverage,bacteriadivideeveryhour.Thismeansthatthenumberofproteinsmustdoubleeveryhour,soacellmustproduce4×106newproteinsevery

hour,givingarate dNprotein

bacteria

dt= 4×10

6proteins3600s ∼103 proteins

s.Assumenowthatone

mRNAcanbetranslatedtoabout1proteinpersecond,or rmRNA→protein ∼1

proteinsmRNA is ,

whichgivesnumberofmRNA NmRNAbacteria ∼ dNprotein

bacteria /dt( )/ rmRNA→proteins( ) =103 .MammalianCell:Assumeavolumeof3000µm

3 or1.2×1010 proteins.Mammalian

cellsdividesevery24hr, dNprotein

mammalian

dt= 1.2×1010proteins24hr ×3600s ihr −1 ∼1.4×10

5 proteinss

,and

with rmRNA→protein ∼1

proteinsmRNA is , NmRNA

mammalian ∼ dNproteinmammalian /dt( )/ rmRNA→proteins( ) =1.4×105 .

ProteinandmRNAdegradation:http://book.bionumbers.org/how-fast-do-rnas-and-proteins-degrade/Preamble:AsmentionedmRNAnumberremains,moreorless,constant,and

describedbytheequation,

dNmRNA

dt= rmRNA −kmRNA

decayNmRNA ,whereNmRNA isthenumberof

mRNA(inacell)

dNmRNA

dtistherateofincrease(ordecrease)ofmRNAinunitsof

mRNA i s−1 ,rmRNA istherateofmRNAproduction,in mRNA i s−1 ,andkmRNAdecay istherate

ofmRNAdecayinunitofs−1 .HowaremRNAdegraded?Inlivingcells,mRNAsaredegradedbyenzymes(proteins),suchasRibonucleases(RNase).HighconcentrationoftheseenzymesincreasethemRNAdecayrate,kmRNA

decay .Thedecayrateistheinverseofthehalf-lifeof

mRNA,kmRNAdecay = ln2/τ1/2mRNA .FormRNAinlivingcellsτ1/2

mRNA =3min to10Hr NOTE:DNAaredegradedbyproteinscalleddeoxyribonuclease(DNase)

Steady-State(SS)SolutionAfterasufficientlylongperiod,thesystemreachesasteadystate(SS)ofconstantmRNAnumber,NmRNA =NmRNA

SS :

dNmRNA

dt=0→0= rmRNA −kmRNAdecayNmRNA

SS →NmRNAss = rmRNA /kmRNAdecay

ExactSolutionformRNA:Thesolutionof

dNmRNA

dt= rmRNA −kmRNA

decayNmRNA is

NmRNA =NmRNASS 1−exp −kmRNA

decayt( )( ) .Afteralongperiod, exp −kmRNAdecayt( )∼0→

NmRNA =NmRNAss = rmRNA /kmRNAdecay .

ProteinsinlivingcellsareproducedbyRibosomesthat“reads”themRNAcode.Assuch,theproteinproductionrateisproportionaltothenumberofmRNA,NmRNA .

Therateofincrease(ordecrease)ofproteinsis

dNprotein

dt= rproteinNmRNA −kprotein

decay Nprotein ,

whereNprotein isthenumberofproteins,

rprotein istherateofproteinproducedby

translationpermRNA,andhasunitofs−1 ,kproteindecay istherateofdecay.

Howareproteinsdegraded?Inlivingcells,proteinsaredegradedbyenzymes(proteins),suchasproteases.Highconcentrationoftheseenzymesincreasestheproteindecayrate,

kproteindecay .Thedecayrateistheinverseoftheofproteins,

kproteindecay = ln2/τ1/2protein .FormRNAinlivingcellsτ1/2

protein =30min todays .Steady-State(SS)SolutionAfterasufficientlylongperiod,thesystemreachesasteadystate(SS)ofconstantproteinnumber,

Nprotein =NproteinSS :

dNprotein

dt=0→0= rproteinNmRNA −kprotein

decay NproteinSS →Nprotein

ss = rproteinNmRNA /kproteindecay

ExactSolutionformRNA:Thesolutionof

dNprotein

dt= rproteinNmRNA −kprotein

decay Nprotein is

Nprotein =NproteinSS 1−exp −kprotein

decay t( )( ) .Afteralongperiod, exp −kproteindecay t( )∼0→

Nprotein =Nproteinss = rproteinNmRNA /kproteindecay .

WHATISTHEPROTEINTOmRNARATIO?http://book.bionumbers.org/how-many-proteins-are-made-per-mrna-molecule/FromthelastsectionweshowedthatatsteadystateNprotein

ss = rproteinNmRNA /kproteindecay ,ans

theratioofProteinstomRNAinacellisNprotein /NmRNA = rprotein /kdecayprotein .ForE.Coli

rprotein isthetranslationratepermRNAwhichcanbearguedtobe rprotein ∼1s

−1 ,and

kdecayprotein =1/τ proteindecay ,wheretheaverageproteindecaytimeis τ protein

decay ∼1000s ,whichgives Nprotein /NmRNA ∼1000 ,andindeed,experimentsusuallyobservethatthenumberofproteinsisabout1000timesofthenumberofmRNA.HOWFASTDOPROTEASOMESDEGRADEPROTEINS?http://book.bionumbers.org/how-fast-do-proteasomes-degrade-proteins/Proteasomes(orproteases)comprisedabout1%oftotalbulkproteins.Oneproteasomedegrades ∼0.05−5proteins/min ,whichisequivalenttoabout 1aa i s−1 .However,thisdependsonproteinconcentrationandtemperatureIn-VitroProteinsInvitro(testtube)proteinsolutionsusuallycontainproteasecontaminantsthatwilldegradeproteins.Butunlikeinlivingcells,therearenoribosomestoproducenewproteinstoreplacethedegradedproteins.Therateequationbecomes:

dNprotein

dt= −kprotein

decay Nprotein ,withsolutionNprotein =N0 exp −kproteindecay t( ) ,whereN0 isthe

numberofproteinsatt=0.Wecancalculatethehalf-life,τ1/2protein ,ofproteins,which

isthetimeittakesforthepopulationtodecreaseby1/2,

N =N0 /2= exp −kprotein

decay τ1/2protein( )→τ1/2

protein = ln2kproteindecay = 0.693

kproteindecay

Example1:Foratypicalinvivosystem,theproteinconcentrationis4×106 protein

µm3 ,

andthehalf-lifeisaboutτ1/2protein = 40min ,or

kproteindecay = 0.693

τ1/2protein ,kprotein

decay =0.0173min−1 .The

timeittakesfortheconcentrationtodecreaseto10%isfoundby

N =0.1N0 =N0 exp −kproteindecay t( ) ,whichgives

t = − ln 0.1( )/kproeindecay = − ln 0.1( )/0.0173min−1 =133min .Theabovecalculationisdoneforatypicalinvivoproteinconcentrationof4mM,atroomtemperatureT=300K,andproteaseconcentrationofabout1%ofproteins,or

4×10−2mM = 40µM .Toestimatetherateatdifferenttemperaturesand

concentrations,weuseArrheniusEquation:Arrheniusequationforreactionratereactionratekprotein

decay = Aexp −ΔE /kBT( ) ,whereA= A cprotein ,D( ) dependsonthecollisionfrequency,whichinturnsdependontheproteinconcentration,

cprotein andthediffusioncoefficient,D.Inclass,itisshown

thatwecanexpressed A=

A0Tℓ2

,whereA0 isaconstant,Tisthetemperature, ℓ isthemean,oraverage,separationdistancebetween2adjacentproteins.Intheequation,ΔE isthereactionfree-energybarrier,whichwewillmakeaneducatedguessequalstoabouttheenergyreleasedbyasingleATPhydrolysis,

ΔE = 4.8×10−20 J DiversiononATPhydrolysis

ATP +H2O→ ADP + phosphate+energy ,energyreleasedisΔHATP =29

kJmol

29 kJmol

× 1000 J /kJ6.023×1023ATP imol−1

→4.8×10−20 JATP

Example2:Supposethetemperatureisloweredfromroomtemperature,Ti=300K(27C),tojustabovefreezingTF=277K(4C).Wecancalculatethelowtemperature(Final,F)ratekprotein ,F

decay fromtheinitial(i)roomtemperaturerate,

kprotein ,idecay =0.0173min−1 ,bywriting:

kprotein ,idecay =

A0Tiℓ2exp − ΔE

kBTi

⎛

⎝⎜⎞

⎠⎟and

kprotein ,Fdecay =

A0TFℓ2

exp − ΔEkBTF

⎛

⎝⎜⎞

⎠⎟,andtakingtheratio:

kprotein ,Fdecay

kprotein ,idecay =

TFTi

exp − ΔEkBTF

⎛

⎝⎜⎞

⎠⎟

exp − ΔEkBTi

⎛

⎝⎜⎞

⎠⎟

=TFTiexp − ΔE

kB

1TF

− 1Ti

⎛

⎝⎜⎞

⎠⎟⎛

⎝⎜

⎞

⎠⎟ ,wehaveassumedthat ℓ

remainsthesame.Thisgiveskprotein ,Fdecay = kprotein ,i

decay TFTiexp − ΔE

kB

1TF

− 1Ti

⎛

⎝⎜⎞

⎠⎟⎛

⎝⎜

⎞

⎠⎟

kprotein ,Fdecay =0.0173min−1 277K

300K exp − 4.8×10−20 J1.381×10−23 J iK −1

1277K − 1

300K⎛⎝⎜

⎞⎠⎟

⎛

⎝⎜⎞

⎠⎟

kprotein ,Fdecay =6.1×10−3min−1

Let’scalculatethetimeittakesaninvitroproteinofthesameconcentrationasexample1todecayto10%itsoriginalvalue,N =0.1N0 :

N =0.1N0 =N0 exp −kprotein ,Fdecay t( ) ,whichgives

t = − ln 0.1( )/kproeinFdecay = − ln 0.1( )/6.1×10−3min−1 =377min ,whichislonger,butnotsubstantiallylongerforabiochemisttryingtokeepaproteinsolutionstable.Example3:Supposethetemperatureiskeptatroomtemperature,buttheproteinwerereducedby100times.Assumethattheoriginalproteinconcentrationis

NV

⎛⎝⎜

⎞⎠⎟ i= 4×106 proteins

µm3 ,asshowninclasstheaverageproteinseparationis

ℓi =

NV

⎛⎝⎜

⎞⎠⎟ i

⎡

⎣⎢⎢

⎤

⎦⎥⎥

−1/3

= 4×106 proteinsµm3

⎡

⎣⎢

⎤

⎦⎥

−1/3

=6.3×10−3µm .Thefinaldilutedconcentration

is100timesmoredilute,NV

⎛⎝⎜

⎞⎠⎟ F

= 1100

NV

⎛⎝⎜

⎞⎠⎟ i= 4×104 proteins

µm3 .Thisgivesafinal

averageproteinseparationof ℓF =

NV

⎛⎝⎜

⎞⎠⎟ i

⎡

⎣⎢⎢

⎤

⎦⎥⎥

−1/3

= 4×104 proteinsµm3

⎡

⎣⎢

⎤

⎦⎥

−1/3

=2.9×10−2µm .

kprotein ,idecay =

A0Tℓi2 exp − ΔE

kBT⎛

⎝⎜⎞

⎠⎟and

kprotein ,Fdecay =

A0TℓF2 exp − ΔE

kBT⎛

⎝⎜⎞

⎠⎟,andtakingtheratio:

kprotein ,Fdecay

kprotein ,idecay =

ℓi2

ℓF2 =

6.3×10−3µm( )22.9×10−2µm( )2

→

kprotein ,Fdecay =0.0173min−1 6.3×10

−3µm( )22.9×10−2µm( )2

=8.16×10−4min−1 .

Let’scalculatethetimeittakesaninvitroproteinofthesameconcentrationasexample1todecayto10%itsoriginalvalue,N =0.1N0 :

N =0.1N0 =N0 exp −kprotein ,Fdecay t( ) ,whichgives

t = − ln 0.1( )/kproeinFdecay = − ln 0.1( )/8.16×10−4min−1 =2820min ,orabout47hr.HowmanyribosomesinacellReadthefollowinghttp://book.bionumbers.org/how-many-ribosomes-are-in-a-cell/