This question is sort of like asking how far a rifle can shoot.
Wood is generally a crappy building material, form an engineering standpoint, because of the variability in mechanical properties. However, there are some guidelines and rules-of-thumb.
No. 1 Douglas Fir (Southern) has an extreme fiber stress (Fb) of about 1300 psi. Fb is the quotient of Moment divided by Section Modulus (Fb=M/S). The section modulus of a 2x12 dimensional lumber section is 31.64 in^3, so the maximum allowable moment is 1300lb/in^2 x 31.64in^3 = 41,132in-lb.
For a uniformly loaded, simply supported span, M=(1/8)w*l^2, where w=load per unit length and l=length of span. We can solve this equation for w, since we know l=48". In this case w=142.8 lb/in. If the beams are spaced at 12" on center, then the maximum allowable floor load would be 142.8lb/in*(144in^2/ft^2)/(12"spacing) = 1714 pounds per square foot!
However, the maximum compression perpendicular to the grain, Fc, is 625 psi. Assuming 3.5" of end bearing, the bearing area would be 1.5"X3.5"=5.25in^2 * 625psi = 3281 lb * 2 ends = 6563 lb total load. Divide that by 4 feet, again assuming 12" spacing of the beams, and the maximum allowable load is 1630 psf.
But wait, there's more! A previous poster mentioned deflection, and rightly so. For a uniformly loaded, simple span, the maximum deflection, at the center of the span, is the quantity:
d=(5 * w * l^4)/(384 * E * I), where w and l are as above, E is the Modulus of Elasticity of the material (we will use 1,500,000 psi), and I is the Moment of Inertia of the section, which for a 2x12 is 178.0 in^4.
Deflection is generally limited to l/360, but for temporary loading, l/240 is OK. So 48"/240 = 0.20". We can now back-solve for w in the previous formula. In this example, w=773 lb/in. This value is much greater than previous results, so deflection does not govern the design.
Now the practical consideration of transferring such massive loads is really the limiting factor. A 2x12 is really quite a beefy member, and as such, will have much greater stiffness than any deck or secondary members that frame into it. So while the 2x12 itself can support a great amount of load, it's going to be tough to really get that amount of load to the 2x12. Also, such high loading would probably cause local crushing of fibers at the point of loading, well in excess of the compression limits previously stated. This is a good example of where the formulas sort of break down, primarily because of the very short span relative to the depth of the member.
I would probably limit the load to 300 pounds per square foot. The actual allowable load is dependant upon many factors, including type of lumber, lateral bracing, connections, bearing details, manner and duration of loading, condition of lumber and other things.