Enzymes with a kcat/Km ratio of about 10^8 M-Is-1 are considered to show optimal
ID: 522545 • Letter: E
Question
Enzymes with a kcat/Km ratio of about 10^8 M-Is-1 are considered to show optimal catalytic efficiency. Fumarase, which catalyzes the reversible-dehydration reaction fumarate + H20 doubleheadarrow malate has a ratio of turnover number to the Michaelis-Menten constant, (kcat/Km) of 1.6 times 10^8 for the substrate fumarate and 3.6 times 10^7 for the substrate malate. Because the turnover number for both substrates is nearly identical, what factors might be involved that explain the different ratio for the two substrates?Explanation / Answer
Turn-over number=TON=kcat=Vmax/[E] ,where [E]=enzyme concentration,Vmax=maximum reaction rate when all the catalytic sites are occupied by substrate
but km=michaelis -menten constant=amount of substrate required by the enzyme to reach Vmax/2
enzyme-substrate binding
S+E<---->products ,forward rate constant=k1,reverse rate constant=k-1,kd=k-1/k1=dissociation constant
km=k-1+kcat/k1=Kd+kcat/k1
small kd value means strong substrate-enzyme binding,
smaller kd,leads to smaller km ,so kcat/km ratio is large(fumarate)
Thus kd is important in determining the ratio kcat/km or the catalytic efficiency
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