MCB 502A-2014. Lecture #5DNA degradation (continued). DNA synthesis. The rate of DNA degradationFor example, one wants to start characterizing the first class of enzymes. They all degrade only linear ssDNA, therefore, they must be single-strand-specific exonucleases. There are at least five of those in E. coli. One of the first things to learn about a DNase is how fast the enzyme can degrade its substrate DNA. To see how many nucleotides a given ssDNA-specific exonuclease can remove in a unit of time, one needs only a uniformly-labeled DNA strand. One important condition is that the enzyme has to be in excess over the substrate. If the opposite is true, that is, the substrate is in excess over the enzyme, the substrate will be non-occupied part of the time during the assay, which will decrease the apparent reaction rate. You may be curious why, when we measure the rate of DNA degradation, we saturate the substrate with an excess of enzyme. In a typical enzymatic reaction to determine the enzyme rate, the reverse is true: the unlimited substrate minimizes the non-occupancy of the enzyme. Thedifference is due to the relative sizes in these enzyme-substrate pairs. A typical enzyme is 100-1,000 times bigger than its substrate molecule: consider aconitase (MW 97,677 Da) versus citrate(MW 189 Da) or beta-galactosidase (MW 116,276 Da) versus lactose (342 Da). For the DNA-processing enzymes the reverse is true: their substrates are sometimes 10,000-times bigger than the enzymes themselves. Since in the reaction the smaller component should be in excess, to facilitate saturation of the bigger component, and the rate is then calculated from the amount of the limiting component, the excess of enzyme is used in the DNA-containing reactions to determine the rate of DNA degradation. The amount of active enzyme is then expressed through the known amount of the limiting DNA substrate.In principle, one could use excess of substrate-over-enzyme if the amount of active enzyme is known precisely. However, while it is straightforward to measure the total amount of a pure protein, it is difficult to determine the fraction of the active enzyme in the preparation of this protein. Therefore, saturating the substrate with an enzyme is easier. In fact, one does not have to know the exact amount of enzyme, as long as it is saturating in the reaction (when addingmore enzyme does not accelerate the reaction further). When using saturating amounts of enzyme, we can assume that the amount of active enzyme equals the (known) amount of the available substrate. For example, 100 pM of the substrate is reacted for 100 seconds with 1 nM of the protein (10-fold excess and saturating for active enzyme for this enzyme prep.). [In chemistry an amount of substance is expressed in Moles (M) or fractions of Mole (1 mM = 10-3 M). 1 M of any substance comprises 6.02 x 1023 molecules. The same is true in biochemistry, although since biopolymers are huge molecules, µM (10-6 M), nM (10-9 M) and even pM (10-12 M) are used.] Under these conditions only one enzyme molecule out of several will have the opportunity to 1attack the substrate at any given time. Once one enzyme molecule falls off, others will be eager to take turns working on the substrate. This is what the enzyme excess is for: to keep the substrate occupied by enzyme at all times. One stops the reaction, separates the substrate from the products (released nucleotides), for example by TCA precipitation, and quantifies the amount of the released radioactivity. For example, one finds that, in 100 seconds, exactly 100 nM of nucleotides are released from the substrate by 100 pM of the active fraction of enzyme (which equals the amount of the substrate). Therefore, the rate of DNA degradation by this nuclease is 10 nt/sec. How come 100 nM of nucleotides are generated out of 100 pM of the substrate? Because 1 M of DNA substrate may contain 104-106 M worth of mononucleotides. For example, if DNA strands were uniformly 10,000 nt long, 100 nM of substrate comprises a total of 1 mM of nucleotides. Polarity of ss-specific exoNow that we determined the rate of DNA degradation by our ssDNA-specific exonucleases, we can ask whether they degrade linear DNA strands in any particular direction. DNA strands have chemical polarities, determined by their sugars: on the one end, the sugar has its 3’-hydroxyl (exposed or phosphorylated); on the other end, the exposed or phosphorylated hydroxyl is 5’. Therefore, the two directions with which an enzyme can move along a DNA strand are called 5’—>3’ or 3’—>5’. To test the suspected polarity (directionality) of DNA processing, radioactive labeling is again employed, but this time it is an end-specific labeling. For example, one can prepare a DNA strand labeled throughout with 3H-Thy, and then use the enzyme polynucleotide kinase (PNK) to label it, in addition, with 32P-phosphate at the 5’-end, or use a different enzyme, terminal nucleotidyl-transferase (TNT), to label it with 32P-nucleotide at the 3’-end. The released 3H or 32P labels from the same DNA molecule can be counted separately in thesame reaction, by a scintillation counter.For example, you have three different ss-exo enzymes to test with the two end-labeled substrates. With the 5’-32P-labeled substrate, enzyme #1 removes the 32P label after degrading most of the 3H label, while enzymes #2 and #3 remove the 32P label before degrading most of the 3H label. Therefore you conclude that enzyme #1 degrades DNA strands in the 3’—>5’ direction,while enzymes #2 and #3 degrade DNA strands in the 5’—>3’ direction. You can verify these conclusions with the 3’-32P-labeled substrate. You find that enzymes #1 and #3 now remove the 32P label from the 3’-labeled substrate before degrading most of the 3H label, while enzyme #2 removes the 32P label after degrading most of the 3H label. Thus, you conclusions about enzymes #1 and #2 are verified: #1 degrades 3’—>5’, while #2 degrades 5’—>3’. Enzyme #3, apparently, is able to degrade DNA strands from both ends. Polarity of ds-specific exoIs a similar experimental strategy applicable to the characterization of polarity of dsDNA-specific exonucleases? You must be surprised: what kind of polarity could be there? After all, it was Rosalind Franklin who first showed that, in the DNA duplex, the two strands run antiparallel,2and if an enzyme moves along DNA duplex, it will move along one of the DNA strands
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