The order of an enzyme-catalyzed reaction with respect to substrate concentration ($[S]$) under Michaelis-Menten kinetics is not constant. It depends on the specific substrate concentration ($[S]$) relative to the Michaelis constant ($K_m$).
When the substrate concentration is much lower than $K_m$ ($[S] \ll K_m$), the Michaelis-Menten equation simplifies. The rate ($v$) becomes directly proportional to $[S]$:
$v \approx \frac{V_{max}}{K_m}[S]$
In this scenario, the reaction is considered first order with respect to the substrate ($n=1$).
When the substrate concentration is much higher than $K_m$ ($[S] \gg K_m$), the equation simplifies differently. The rate ($v$) approaches the maximum velocity ($V_{max}$) and becomes independent of $[S]$:
$v \approx V_{max}$
In this scenario, the reaction is considered zero order with respect to the substrate ($n=0$).
As the substrate concentration ($[S]$) increases from very low levels to very high levels, the reaction order ($n$) transitions from 1 down to 0.
Therefore, for any given substrate concentration in a reaction following Michaelis-Menten kinetics, the order ($n$) must lie within the inclusive range of 0 to 1.
The graph below shows the activity of enzyme pepsin in the presence of inhibitors aliphatic alcohols (P) or N-acetyl-1-phenylalanine (Q). Which ONE of the following represents the nature of inhibition by P and Q, respectively?

The following plot represents the Lineweaver-Burk equation of an enzymatic reaction both in the presence and the absence of inhibitor. Here, V is the velocity of reaction and S is the substrate concentration.

The nature of inhibition shown in the plot is