MCB 502A-2014. Lecture #5DNA degradation (continued).DNA synthesis.— The rate of DNA degradation— Polarity of ss-specific exo— Polarity of ds-specific exo— Processivity of exonucleases— Measuring the processivity— Endonucleases. The cleavage specificity (position)— Substrate-specificity of endonucleases— The biological roles of exo- and endonucleasesDNA synthesis— The oops of Ochoa and Kornberg’s DNA polymerase— DNA depolymerization versus DNA degradation— The need for a primer— The ghost of the non-templated synthesis— Nick-translationMajor Nucleases of E. coli-1— For example, E. coli cells are home to the following major classesof nucleases, most of them containing more than one entry:DNA substrate ClassSSL SSC DSL DSC assignment>95 <5 <5 <5 ss-exo~15 <5 >95 <5 ds-exo>95 <5 >95 <5 ss/ds-exo>95 >95 <5 <5 ss-endo<5 <5 >95 >95 ds-endo>95 >95 >95 >95 ss/ds-endo%TCA-soluble substrate in reactions run to completionFrom this perspective, the cell definitely looks like a bag of DNA-degradation enzymes, rather than a bag of DNA-replicating enzymes.The rateThe rate of DNA degradation-1— For example, one wants to startcharacterizing the first class ofenzymes.— They all degrade only linearssDNA, therefore, they must be single-strand-specific exonucleases. Thereare at least five of those in E. coli.— One of the first things to learn abouta DNase is how fast the enzyme candegrade its substrate DNA.— To see how many nucleotides agiven ssDNA-specific exonuclease canremove in a unit of time, one needsonly a uniformly-labeled DNA strand.Rate of DNA degradation ismeasured in nt/sec releasedThe rate of DNA degradation-2— One important condition is that theenzyme has to be in excess over thesubstrate.— If the opposite is true, that is, thesubstrate is in excess over the enzyme, thesubstrate will be non-occupied part of thetime during the assay, which may decreasethe apparent reaction rate.The rate of DNA degradation-7— You may be curious why, when we measurethe rate of DNA degradation, we saturate thesubstrate with an excess of enzyme.— In a typical enzymatic reaction to determinethe enzyme rate, the reverse is true: theunlimited substrate minimizes the non-occupancy of the enzyme.— The difference is due to the relative sizes inthese enzyme-substrate pairs (and the textbookrepresentation is sometimes misleading).— A typical enzyme is 100-1,000 times biggerthan its substrate molecule: consider aconitase(MW 97,677 Da) versus citrate (MW 189 Da) orbeta-galactosidase (MW 116,276 Da) versuslactose (342 Da).The rate of DNA degradation-8— For the DNA-processing enzymes the reverse istrue: their substrates are sometimes 10,000-timesbigger than the enzymes themselves.— Since in the reaction the smaller componentshould be in excess, to facilitate saturation of thebigger component, and the rate is then calculatedfrom the amount of the limiting component, theexcess of enzyme is used in the DNA-containingreactions to determine the rate of DNAdegradation.— The amount of active enzyme is then expressedthrough the known amount of the limiting DNAsubstrate.The rate of DNA degradation-3— In principle, one could use excess of substrate-over-enzyme if the amount of active enzyme isknown precisely.— However, while it is straightforward to measurethe total amount of a pure protein, it is difficult todetermine the fraction of the active enzyme in thepreparation of this protein. Saturating the substratewith an enzyme is easier.— In fact, one does not have to know the exactamount of enzyme, as long as it is saturating in thereaction (when adding more enzyme does notaccelerate the reaction further).— When using saturating amounts of enzyme, wecan assume that the amount of active enzymeequals the (known) amount of the availablesubstrate.Enzyme (ng)Reaction ratetotalactivePrep #1Prep #2The rate of DNA degradation-4— For example, 100 pM of the substrateis reacted for 100 seconds with 1 nM ofthe protein (10-fold excess andsaturating for active enzyme for thisenzyme prep.).— In chemistry an amount of substanceis expressed in Moles (M) or fractions ofMole (1 mM = 10-3 M).— 1 M of any substance comprises 6.02x 1023 molecules (Avogadro's number).— The same is true in biochemistry,although since biopolymers are hugemolecules, µM (10-6 M), nM (10-9 M)and even pM (10-12 M) are used.moleµmolnmolpmolThe rate of DNA degradation-5— Under these conditions only one enzymemolecule out of several will have theopportunity to attack the substrate at any giventime.— Once one enzyme molecule falls off, otherswill be eager to take turns working on thesubstrate.— This is what the enzyme excess is for: to keepthe substrate occupied by enzyme at all times.— One stops the reaction, separates the substratefrom the products (released nucleotides), forexample by TCA precipitation, and quantifiesthe amount of the released radioactivity.The rate of DNA degradation-6— For example, one finds that, in 100 seconds,exactly 100 nM of nucleotides are released from thesubstrate by 100 pM of the active fraction ofenzyme (which equals the amount of the substrate).— Therefore, the rate of DNA degradation by thisnuclease is 10 nt/sec.— How come 100 nM of nucleotides are generatedout of 100 pM of the substrate?— Because 1 M of DNA substrate may contain 104-106 M worth of mononucleotides (DNA strands arelong!).— For example, if DNA strands were uniformly10,000 nt long, 100 nM of substrate comprises atotal of 1 mM of nucleotides!100 nM/100 sec = 1nm/sec1 nm/sec / 100 pMenzyme = 10 nt/secPolarity (Directionaliy)Polarity of ss-specific exo-1— Now that we determined the rate ofDNA degradation by our ssDNA-specific exonucleases, we can askwhether they degrade linear DNAstrands in any particular direction.— DNA strands have chemicalpolarities, determined by their sugars:on the one end, the sugar has its 3’-hydroxyl (exposed or phosphorylated);on the other end, the exposed orphosphorylated hydroxyl is 5’.— Therefore, the two directions withwhich an enzyme moves along a DNAstrand are either 5’—>3’ or 3’—>5’.5'—>3'3'—>5'Polarity of ss-specific exo-2— To test the suspected polarity(directionality) of DNA processing,radioactive labeling is again employed, butthis time it is an end-specific labeling.— For example, one can prepare a DNAstrand labeled throughout with 3H-Thy,… and then use the enzyme polynucleotidekinase (PNK) to label it, in addition, with
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