Biological Physics: Energy, Information, Life

10. Problems[[Student version, January 17, 2003]] 407

byadding a second reaction (compare Figure 10.13),

ES+I

cIk 3




k- 3

ESI.

Here “E” is the enzyme, “S” the substrate, “P” the product, and “ES” the enzyme-substrate
complex. The inhibitor “I” is a second substance which, like “E,” is not used up. “I” can bind
to the enzyme-substrate complex “ES” to create a dead-end complex “ESI,” which cannot process
substrate due to an allosteric interaction. Eventually, however, ESI spontaneously dissociates back
to ES+I and the enzyme goes back to work. This dead-end branch slows down the reaction. As
usual, assume a large reservoir of “S” and “I,” no product initially, and a small amount of enzyme.
a. Find the steady-state reaction ratevin terms ofcS,cI,the total enzyme concentrationcE,tot,
and the rate constants.
b. Consider the dependence ofvoncS,holdingcE,totandcIfixed. Can you express your answer as
aMichaelis–Menten function, wherevmaxandKMare functions ofcI?
c. Regardless of your answer to (b), find the saturating valuevmaxascSincreases at fixedcE,tot
andcI. Comment on why your answer is physically reasonable; if you did Problem 10.4, contrast
to the case studied there.

10.6 T 2 Generality of MM kinetics
In this problem you’ll see that the MM formula is really more generally applicable than the discus-
sion in the text may have made it seem.
The enzyme chymotrypsin catalyzes the hydrolysis of peptides (short protein fragments). We
will denote the enzyme’s original state as EOH to emphasize one key hydroxyl group on one of its
residues. We will also represent a peptide generically by the symbol RCONHR′,where the central
atoms CONH indicate one particular peptide bond (see Figure 2.15 on page 43) and R, R′denote
everything to the left and right respectively of the bond in question.
The enzyme operates as follows: A noncovalent complex (EOH·RCONHR′)forms rapidly be-
tween the enzyme EOH and the peptide substrate RCONHR′,which we will call S. Next EOH gives
up a hydrogen and bonds covalently to one half of the peptide, breaking the bond to the other half,
which is released. Finally, the remaining enzyme-peptide complex splits a water molecule to restore
EOH to its original form and release the other half of the peptide:

EOH+S+H 2 O

cSKeq,S


 EOH·S+H 2 O

k 2

⇀EOCOR + NH 2 R′+H 2 O

k 3

⇀EOH + RCO 2 H+NH 2 R′.

Assume as shown that the last step is irreversible.
Assume that the first reaction is so fast that it’s practically in equilibrium, with equilibrium
constantcSKeq,S.Apply the steady-state assumption tocEOCORto show that the overall reaction
velocity of the scheme described above is of Michaelis–Menten form. Find the effective Michaelis
constant and maximum velocity in terms ofKeq,S,the rate constantsk 2 ,k 3 ,and the total concen-
trationcE,totof enzyme.

10.7 T 2 Invertase
Earlier problems discussed two distinct forms of enzyme inhibition: competitive and unncompet-
itive. More generally, “noncompetitive inhibition” refers to any mechanism not obeying the rule
youfound in Problem 10.4. The enzyme invertase hydrolyzes sucrose (table sugar). The reaction is