C value, Cot Curve & Rot Curve L1-3.pdf

The lecture describes the basic concepts of C-value, Cot curve and Rot curve analysis, MCQ questions regarding the same. Queries are always welcome.... Dr. Nitin Wahi ([email protected]).

1. Dr. Nitin Wahi
L1-3 Lecture
Molecular Biology:
C-value
Cot curve
V/s
Rot curve

2. Genome
✓GenomeisdefinedasthetotalityofalltheDNAofanindividualoraspecies.
✓Anorganismreceiveshalfofitsgenome fromthepaternalside and
anotherhalffromthe maternalside.
✓TheamountofDNApresentinthehaploidgenomeofanspecieisitsC-value.
Itisspecificforeachspecies.Itsvalueinpcgm.(picogm)canbe<10
6
,inP.K
to10
11
,inAmphibians.
Amphibianshasthe
largestgenomeasthese
representsatransition
stateinb/wPiscesand
Reptiles.

3. C-value during cell cycle
✓C-valueoftheDNAvariesfromonestageofthecelldivisiontotheother.It
followsthefollowingsequentialchanges:
Changes in the C-value during Meiosis Cell division
G
1-PhaseS-PhaseG
2-Phase M-Phase (Meiosis)
2n/2C2n(2C→4C)2n/4C
Prophase IMetaphase IAnaphase ITelophase I
2n/4C 2n/4C 2n/4C n/2C and n/2C
Prophase IIMetaphase IIAnaphase IITelophase II
n/2C n/2C 2n/2C n/c and n/C
Changes in the C-value during Mitosis Cell division
G
1-PhaseS-PhaseG
2-Phase M-Phase (Mitosis)
2n/2C2n(2C→4C)2n/4C
Prophase Metaphase Anaphase Telophase
2n/4C 2n/4C 4n/4C 2n/2C and 2n/2C

4. C-value during cell cycle

5. Exercise
Q1.Anorganismhassixpairofchromosomes.Intheabsenceofcrossing
over,howmanydifferentchromosomalcombinationsarepossibleinthe
gametes?
Ans:Thenumberofcombinationsis2
n
=2
6
=64,wheren:no.ofdifferentchromosomesintheset.
Q2.Anorganismhas56chromosomesinitsdiploidstage.Indicatehowmany
chromosomesarepresentinthefollowing:
P.SomaticcellinGostage:56Chromosomes
Q.Metaphase(mitosis):56Chromosomes
R.Anaphase(mitosis):112Chromosomes
S.MetaphaseII(meiosis):28Chromosomes
T.Gametes:28Chromosomes

6. C-value paradox
❖InlowerE.KstheamountofDNAincreaseswithincreaseinthe
complexityofanorganism.HoweverinhigherE.K.thereisno
correlationb/wincreasedgenomicsizeandcomplexity.
❖Thislackofcorrelationinbetweentheincreasedgenomicsizeand
complexityiscalledasCvalueparadox.
Organism
type
Organism name Genomic
Size
Encoding
genes
Remark
Bacterium M. genitalium 580 kb 482 Smallest
Genome
Plant
Paris japonica
(Japanese flower)
149 billion--
Largest
Genome
Animal
Neoceratodusforsteri
(Australian lung fish)
43 billion--
Largest
genome
(Animals)

7. Genomic sequences in Eukaryotes
Genomicsequences
Nonrepetitive
sequences
Repetitive
sequences
Unique DNA seq.
> 2% of Genomic DNA
Up to 98%
of Genomic DNA
Moderately repetitive
Sequences (45%) Highlyrepetitive
sequences(Transposable elements)
MicrosatellitesMinisatellites
Ex: Telomeric DNAEx: CAG/ AT repeats
Transposones Retrotransposones
Ex: IS elements,
P elements
Ex: Alu sequence,
Ty elements

8. Genomic Complexity
❖Genomiccomplexityreferstothetotallengthofdifferentsequencesof
DNA.Itcanbemeasuredthroughtherenaturation(reassociation)
kineticsofdenaturedDNA.
Renaturationkineticsstudy
❖Genomiccomplexityreferstothetotallengthofdifferentsequencesof
DNA.Itcanbemeasuredthroughtherenaturation(reassociation)
kineticsofdenaturedDNA.
R.E
temp.
Allowed to
reanneal

9. cOtcurve
❖Renaturationkinetics(rateofrenaturation)canbeusedtodifferentiate
inbetweenunique,moderatelyrepeatedandhighlyrepeatedDNA
sequences.
❖HighlyrepetitiveDNArenaturesfaster whilenonrepetitiveDNA
renaturesslowly .
Rateofrenaturationα[Conc.ofcomplementaryseq.ofDNA]
2
❖Renaturationdependsuponrandom
collisionofcomplementarystrandsand
followsSecondOrderratekinetics.
❖DNArenaturationkineticsisdescribed
byCot
1/2.

10. cOtcurve
❖IflargeDNAisshearedintouniformfragmentsandallowedtoreanneal,
thenrenaturationrate,isexpressedas:
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=??????&#3627408490;
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❖At,t=t
1/2,i.e.,whenhalfoftheDNAhasbeenrenatured,so,C/C
o=½
❖TheproductofC
oxtisc/aC
ot.Itisinverselyproportionaltotherateconstant‘k’.
Where,
k:2
nd
orderrateconstant
C:Conc.ofssDNAattime‘t’
Where,
C
o:Conc.ofssDNAattime‘t
o’
C:Conc.ofssDNAattime‘t’
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??????
=
&#3627409359;
(&#3627409359;+??????.&#3627408490;
&#3627409358;.??????)
C
o.t
1/2 =
&#3627409359;
&#3627408420;

11. cOtcurve
❖SinceCot
1/2istheproduceofconc.ofssDNA(Co)andt
1/2,soa
repetitiveDNAwillrenaturefasterandwillhavelowC
ot
1/2valuewhilea
nonrepetitiveDNAwillrenatureslowlywithagreaterC
ot
1/2value.
❖Thus,agraphofthefractionofass-DNAreannealed(1-C/C
o)asa
functionofC
otonasemilogarithmicplotisreferredtoasC
otcurve.
DNA2
abab ab ab ab
cdefghijklmnopqrstuv
Izyajczkblqfreighttrainrunninsofastelizabethcottonqw
Unique DNA
abab ab ab ab
abab ab ab ab
abab ab ab ab
abab ab ab ab
abab ab ab ab
ababababab
ababababab
etc.
cdefghijklmnopqrstuv
cdefghijklmnopqrstuv
cdefghijklmnopqrstuv
DNA1 DNA3
Repetitive DNA

12. cOtcurve
❖Thus,Cot
1/2describestherateofrenaturationofads-DNAthathas
beensheerdegradedintofragments,denaturedandthenallowed
toreanneal.
❖C
ot
1/2describestotallengthofdifferentsequenceofDNAandis
referredtoaskineticcomplexity.
❖C
ot
1/2comparisonisusedtoaccessDNAcomplexity.
R.E
temp.
Allowed to
reanneal

13. Exercise
Q1.Whatrelativeratesofrenaturationareexpectedforsequencesrepeated1000
timesinthegenomecomparedtogeneswithonlyasinglecopy?
DNA renaturation follows Second order rate kinetics.
Rate of renaturation α[Conc. of complementary seq. of DNA]
2
Rate of renaturation α[1000]
2
Rate of renaturation α10
6
Ans:Thusasequencerepeated1000timeswouldrenaturewitharateof10
6
times

14. Exercise
Q2.Thefigurebelowisagraphofthereannealing(reformingofds-DNAmolecules)oftotalgenomic
DNAfromtwodifferentorganisms.DNAwasextractedfromeachorganism,shearedtoauniformsize,and
thendenaturedbyheating.Anidenticalamountofeachpreparationofsheared,ss-DNAwasallowedto
reannealunderidenticalconditionsofpH,buffer,ionicstrength,andtemperature.Thereannealingcurves
shownwereobtainedbyplottingthepercentageofDNAremainingsingle-strandedversustheproductof
thetotalDNAconcentration(C
o),inmolesofnucleotideperlitre,andtherenaturationtime(t),inseconds.

15. ROtcurve
❖RotcurveisameasureofthekineticcomplexityofRNAbasedonits
hybridizationabilitywiththeDNAmolecules.
❖TheexpressionofRNAvaries[↑↓↔]indifferentcelltypes,cellstages,
lifespanandresponseagainststimulus.
❖TypesofRNA:t-RNA;Sc/SnoRNAetc.,representtheirowngeneswhile
mRNArepresentsanexpressedgene.
❖EstimationofeachkindcanbemadebyRNAre-associationkinetics.
❖DifferenttypesofRNA(mRNA,etc)canbereversetranscribedinto
cDNA,whichcanbehybridizedtoexcessamountofnon-repeatedDNA
(UniqueDNA).
❖TheDNA:RNAhybridcanberecoveredfromnitrocellulosemembrane
andthesamecanbequantifiedusingre-associationkinetics.

16. ROtcurve
❖IfexcessRNAorDNAisinvolvedinthehybridizationreactiontheseare
expressedintermsofRotorCotvalues.
❖Rotcurvecanbeobtainedbyplottingthefractions(%ofRNA:cDNA
hybrids)oftheRNA:DNA hybridrecoveredasthefunctionofRot
values(Ro:Conc.ofssRNAw.r.t‘t’time).
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=&#3627408420;??????&#3627408424;??????
??????
????????????
&#3627409359;/&#3627409360;=
&#3627409360;
??????
❖At,t=t
1/2,i.e.,halfoftheDNAhasbeenrenatured,so,D
o/D=½.R
oxtisc/aR
ot.
Where,
D:Conc.ofssDNA(nt./lit)
R:Conc.ofexcessmRNA
k:rateconstantforthereaction
R.E
temp.
Allowed
to
reanneal

17. ROtcurve Analysis
❖ThegraphicalexpressionoftheR
otcurvecouldbeusedtopredict
fastest renaturation;frommoderatelyrenaturedRNAandslowly
renaturedone .
❖Fromthegraphthefirstcomponentisthefastest
expressinglargestamountofRNA,whilethefinal
componentisslowestexpressinglowestRNA.
❖Rotcurvere-associationanalysiscanbeusedto
determinemRNAsizefromanRNAmixture.
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=
&#3627408453;
????????????
1/2(??????&#3627408453;????????????ɑ)
&#3627408453;
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1/2(??????&#3627408453;??????????????????)

18. ROtcurve application
❖RotcurvecanbeusedtodifferentiateinbetweenHousekeepingand
Luxurygenes.
❖TheexpressionofRNAvaries[↑↓↔]underdifferentcellularstimulus
couldbeeffectivelydetermined.
➢Housekeeping GenesV/sLuxury Genes:
✓Housekeepinggenesaretheonewhichareactivelytranscribedand
translatedintoproteins.Theproductofsuchgenesarenecessary
requiredtomaintainthecellularhomeostasis.Ex:Glycolyticgenes.
✓Luxurygenesaretheonewhichgetstranscribedandtranslatedinto
proteinsonlyunderspecializedenvironmentalconditions.Ex:Lac
operon.

19. Exercise
Q1.OvalbuminmRNAof2000nt.longwashavingaR
ot
1/2valueof4x10
-3
M
-1
sec
-1
.
ForprolactinaR
ot
1/2wasfoundtobe1.8x10
-3
M
-1
sec
-1
.Calculatethelengthof
mRNAforprolactin?
Ans: &#3627408454;????????????&#3627408466;??????&#3627408467;??????&#3627408453;????????????ɑ
&#3627408454;????????????&#3627408466;??????&#3627408467;??????&#3627408453;??????????????????
=
&#3627408453;
????????????
1/2(??????&#3627408453;????????????ɑ)
&#3627408453;
????????????
1/2(??????&#3627408453;??????????????????)
2000
&#3627408454;????????????&#3627408466;??????&#3627408467;??????&#3627408453;??????????????????
=
4x10
−3
1.8x10
−3
Size of mRNAx= (1.8 x 10
-3
) x (2000)/ (4 x 10
-3
) = 900 nt.
The length of mRNA for prolactin is 900 nt.

20. Exercise
Q2.Whichofthefollowingtechniquecouldbeusedtopredictaboutthe
complexityofthegenome?
a.C
otanalysisb.R
otanalysisc.BothC
otandR
otanalysisd.Noneoftheabove
Q2.Whichofthefollowingtechniquecouldbeusedtodifferentiatethelevel
ofaspecifictranscriptinbetweendifferentcells?
a.C
otanalysisb.R
otanalysisc.BothC
otandR
otanalysisd.Noneoftheabove
Q3.Whichofthefollowingtechniquecouldbeusedtodifferentiateinbetweena
strongandaweakpromoterforthesamegenetranscript?
a.C
otanalysisb.R
otanalysisc.BothC
otandR
otanalysisd.Noneoftheabove
Q4.Whichofthefollowingtechniquecouldbeusedtodifferentiateinbetweena
housekeepingandaluxurygenes?
a.C
otanalysisb.R
otanalysisc.BothC
otandR
otanalysisd.Noneoftheabove

21. R-looptechnique
❖EvidencesfortheexistenceofintronswasprovidebyR-loop
technique.
✓TechniqueinvolvedincubationofanmRNAwithitsencodingds-DNAin
thepresenceofFormamide.
✓Thebase-pairedcomplexisthenvisualizedunderanSEM.
✓RNAdisplacesds-DNAwhereverthetwosequencesmatchestoforma
RNA-ds-DNA,suchregionswerereferredtoasExons.
✓TheregionofDNAwithnomatchinthe
correspondingmRNAareclearlyvisibleintheformof
loopsofds-DNA,suchregionswerereferredtoas
Introns.

22. Exercise
Q1.IncaseoftheR-looptechniquetheloopsareformedby?
a.DNA b.RNA c.Both d.Noneoftheabove
Q2.ThechemicalusedinR-looptechniqueis?
a.Formate b.Formicacid c.Formaldehyde d.Formamide