The variation of the rate of an enzyme catalyzed reaction with substrate concentration is correctly represented by graph :

Solution

Chemistry ma’am · 3 steps

An enzyme has only so many active sites, so the rate climbs with substrate and then starts to level off.

1. Low substrate. When there is little substrate, most active sites on the enzyme are empty. Adding more substrate fills more sites, so the rate rises steeply, almost in proportion to [S].
2. High substrate. As [S] grows, the active sites fill up and the enzyme approaches saturation. Extra substrate then adds less and less to the rate, so the curve bends over towards a maximum rate Vₘₐₓ. This is the Michaelis–Menten shape.

   v = Vₘₐₓ[S]/(Kₘ + [S])

   Graph (option)	Shape	Matches enzyme kinetics?
   Graph (a), option (a)	straight line rising from O	no, the rate never levels off
   Graph (b), option (b)	straight line falling	no, more substrate does not slow it
   Graph (c), option (d)	rises from O, steep first, then flattens	yes
   Graph (d), option (c)	curve falling	no, the rate does not drop

3. So. The rate starts at zero, rises steeply and then bends towards a limiting value. Only graph (c) has this shape, which is option (d).

The idea to keep. Enzyme rate against substrate always rises and then saturates: nearly first order at low [S], nearly zero order at high [S]. Look for the curve that bends over, never a straight line.

Answer from the NTA final answer key. Solution by PrepWiser.