Question

VO vs [S] 0.0000014 y -0.0698x2 +0.0006x- 2E-08 0.0000013 0.0000012 0.0000011 0.000001 0.0000009 vo 0.0000008 Series2 0.0000007 0.0000006 ...Log.(VO) ...Poly.(V0) 0.0000005 y-0.0074x2 + 0,0001x- 6E-09 Poly.(Series2) 0.0000004 0.0000003 0.0000002 0.0000001 0.01 [S]

What is the Vmax and Km on this graph for both lines?

Answer 1

This is not the best plot for determining V_max and K_M. A Lineweaver-Burke plot would be better. But still, these values can be approximated from them

V_max is the highest rate the reaction can reach. For the blue line, the value seems to be around 0.0000013 (you have not used units in your graph, but whichever are your units for rate). K_M is defined as the substrate concentration that results in a rate the is half the V_max. In this case, half of V_max would be 0.00000065, and the concentration corresponding to that value is, apporximately, 0.0015 (concentration units are not given). Please bear in mind that this is a very wide approximation, because the x axis is not divided in many values.

For the orange curve, the V_max seems to tend to 0.00000045 and K_M would be the concentration of substrate corresponding to a rate of 0.000000225, which is approximately 0.003.