How Fast Could DNA Be Replicated?
In the preceding sections we have seen that the bacterial chromosome is replicated by two synthesis forks moving away from a replication origin at a chain elongation speed of about 500 nucleotides per second for cells growing at 37°. How does this rate compare to the maximum rate at which nucleotides could diffuse to the DNA polymerase? This question is one specific example of a general concern about intracellular conditions. Often it is important to know an approximate time required for a particular molecule to diffuse to a site.
Consider a polymerase molecule to be sitting in a sea of infinite dimensions containing the substrate. Even though the polymerase moves along the DNA as it synthesizes, we will consider it to be at rest since the processes of diffusion of the nucleotides which we are consid-ering here are much faster. We will consider that the elongation rate of the enzyme is limited by the diffusion of nucleotides to its active site. Under these conditions, the concentration of substrate is zero on the surface of a sphere of radius r0 constituting the active site of the enzyme (Fig. 3.21). Any substrate molecules crossing the surface into this region disappear. At great distances from the enzyme, the concentration of substrate remains unaltered. These represent the boundary conditions of the situation, which requires a mathematical formulation to deter-mine the concentrations at intermediate positions.
The basic diffusion equation relates time and position changes in the concentration C of a diffusible quantity. As diffusion to an enzyme can be considered to be spherically symmetric, the diffusion equation can be written and solved in spherical coordinates involving only the radius r, the concentration C, the diffusion coefficient D, and time t:
Figure 3.21 A DNA polymerase possessing an active site of radiusr0into whichnucleotides disappear as fast as they can reach the enzyme and the expected nucleotide concentration as a function of distance from the enzyme.
The solution to this equation, which satisfies the conditions of being constant at large r and zero at r = r0, is
The flow rate J of substrate to the enzyme can now be calculated from the equation, giving the flux J:
The total flow through a sphere centered on the enzyme yields
Flow = 4πr2J
The final result shows that the flow is independent of the size of the sphere chosen for the calculation. This is as it should be. The only place where material is being destroyed is at the active site of the enzyme. Everywhere else, matter must be conserved. In steady state there is no change of the concentration of substrate at any position, and hence the net amount flowing through the surface of all spheres must be equal.
To calculate the flow rate of nucleoside triphosphates to the DNA polymerase, we must insert numerical values in the final result. The concentration of deoxynucleoside triphosphates in cells is between 1 mM and 0.1 mM. We will use 0.1 mM, which is 10- 4 moles per liter or 10-7 moles per cm3. Taking the diffusion constant to be 10-7 cm2/sec and r0to be 10Å, we find that the flow is 10-20moles/sec or about 6,000molecules/sec. The rate of DNA synthesis per enzyme molecule of about 500 nucleoside triphosphates per second is less than 10% of the upper limit on the elongation rate set by the laws of diffusion. Considering that the actual active site for certain capture of the triphosphate might be much smaller than 10 Å, the elongation rate of DNA seems remarkably high.
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