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What is the difference between $S_{0.5}$ values and $K_m$ values in enzyme kinetics?

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What do you mean by S0.5? $K_M$ is the concentration of substrate at the point when the rate of the reaction is 0.5 of the maximal rate ($V_{max}$). –  Bitwise Oct 24 '12 at 1:13
I edited the Mathjax into the title as well, as it seems to yield an appropriate URL slug: http://biology.stackexchange.com/questions/5004/s-0-5-vs-k-m-values-in-enzyme-k‌​inetics –  jonsca Oct 24 '12 at 7:29
Km applies to a data set exhibiting MM kinetics, i.e. it has a "mathematical" underpinning. Presumably S0.5 is simply the observed substrate concentration for 0.5 x Vmax regardless of whether the data fit MM? –  Alan Boyd Oct 24 '12 at 8:20
probably worth looking at biology.stackexchange.com/questions/3147/… –  bobthejoe Oct 24 '12 at 9:42
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1 Answer 1

This expands my comment on the question to an answer.

If an enzyme exhibits Michaelis-Menten kinetics, then it is valid to define a KM and this equates to the substrate concentration when reaction velocity is 0.5 * Vmax.

However, many enzymes do not exhibit Michaelis-Menten kinetics. One example is when the enzyme shows a co-operative response to substrate concentration. In these circumstances the substrate concentration when reaction velocity is 0.5 * Vmax is still a useful parameter, but there is no KM.

See this paper for a discussion of the kinetic properties of the glucokinase of pancreatic β cells, which acts as the glucose sensor. This enzyme shows positive co-operativity with respect to its substrate glucose, and the authors develop a kinetic analysis in terms of various parameters. These include an S0.5 value for glucose, but a KM value for ATP (presumably because the kinetics with respect to ATP are simple Michaelis-Menten).

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