Enzymolysis with the following mechanism : \[ E + S \xrightleftharpoons[k_{-1}]{k_1} ES \] \[ ES \xrightarrow{k_2} E + P \] \[ E + I \rightleftharpoons EI \] \[ \therefore\ (I \equiv \text{inhibitor}) \quad I_1 < I_2 < I_3 \] The following behaviors are noted. Orders of the reaction, at point `A' and point `B' marked on the graph would be : 
One and zero
The stem carries a drawn graph; the four options are text and are reproduced in full above.
The Michaelis-Menten rate law is \(v = \frac{V_{max}[S]}{K_m + [S]}\), and its order in substrate is not fixed — it changes along the curve. That is exactly what the two marked points test.
At low substrate concentration (\([S] \ll K_m\)), the denominator is dominated by \(K_m\), so
\(v \approx \frac{V_{max}}{K_m}[S]\).
The rate is directly proportional to [S], so the reaction is first order. Physically, most of the enzyme is sitting free and unoccupied, so adding more substrate immediately increases the rate. This is the steeply rising, near-linear part of the curve.
At high substrate concentration (\([S] \gg K_m\)), the \(K_m\) in the denominator becomes negligible and [S] cancels:
\(v \approx V_{max}\).
The rate no longer depends on [S] at all, so the reaction is zero order. Physically the enzyme is saturated — every active site is occupied, and the turnover step \(k_2\) is now rate limiting. This is the flat plateau of the curve.
So a point on the rising portion is first order and a point on the plateau is zero order, giving one and zero. At \([S] = K_m\) exactly, the rate is half \(V_{max}\) and the order is intermediate, between one and zero — which is why no option offers a fractional value for both points.
Per the official final answer key the answer is one and zero.
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