This question is answered using the Michaelis-Menten equation:

\(\displaystyle V_o = \frac{V_{max}[S]}{ [S] + K_M}\)

Rearrange the equation to find the ratio of interest.

\(\displaystyle \frac{V_o}{V_{max}} = \frac{[S]}{ [S] + K_M}\)

Plug in \(\displaystyle 5K_M\) for \(\displaystyle [S]\) and simplify.

\(\displaystyle \frac{V_o}{V_{max}}=\frac{5}{6}\)

In this model, an intermediate is formed when the substrate binds to an enzyme.  The intermediate decomposes to an enzyme and a product, not just the product alone. The model requires enzyme-catalyzed reactions that include a rate limiting step of the enzyme-substrate complex breaking down into the enzyme and product.

To solve this, we need the solve for \(\displaystyle [S]\) in the Michaelis-Menten equation:

\(\displaystyle V_o= \frac{V_{max} [S]}{K_m+[S]}\)

We know the following information:

\(\displaystyle K_M=0.5mM\)

\(\displaystyle V_{max}= 200 \frac{mmol}{s}\)

\(\displaystyle V_o=\frac{1}{4}V_{max}= 50 \frac{mmol}{s}\)

Plug in these numbers and solve for substrate concentration.

\(\displaystyle 50= \frac{200[S]}{.5+[S]}\)

\(\displaystyle \frac{1}{4}= \frac{[S]}{.5+[S]}\)

\(\displaystyle .5+[S]=4([S])\)

\(\displaystyle .5=3[S]\)

\(\displaystyle \frac{1}{6}mM=[S]\approx 0.17mM\)

This question is asking us to determine the fractional saturation of a solution containing enzyme and substrate. Let's start by considering what we have, and what we are trying to solve for.

We know that the substrate concentration is \(\displaystyle 4.5*10^{-6}M\) and we also know that this enzyme has a Michaelis constant of \(\displaystyle 5.0*10^{-6}M\).

Also, let's consider what fractional saturation is. Fractional saturation refers to the proportion of enzyme molecules in a solution that are bound to substrate. This value can range from \(\displaystyle 0\%\) (all enzymes are not bound to substrate) to \(\displaystyle 100\%\) (all enzymes are bound to, or saturated with, substrate).

So, we are looking to calculate the number of enzyme-substrate complexes divided by the total amount of enzyme in solution, which we can express as \(\displaystyle \frac{\left [ ES\right ]}{\left [ E\right ]_{T}}\). We can also relate these terms by the following equations.

\(\displaystyle V_{0}=k_{cat}\left [ ES\right ]\)

\(\displaystyle V_{max}=k_{cat}\left [ E\right ]_{T}\)

Dividing these equation by each other, we obtain:

\(\displaystyle \frac{V_{0}}{V_{max}}=\frac{[ES]}{[E]_{T}}\)

Furthermore, we can relate these terms by considering the Michaelis-Menten equation:

\(\displaystyle V_{0}=\frac{V_{max}[S]}{K_{M}+[S]}\)

Or, written another way:

\(\displaystyle \frac{V_{0}}{V_{max}}=\frac{[S]}{K_{M}+[S]}\)

And combining everything we have done so far, we have:

\(\displaystyle \frac{V_{0}}{V_{max}}=\frac{[ES]}{[E]_{T}}=\frac{[S]}{K_{M}+[S]}\)

And plugging in values, we can solve:
\(\displaystyle Fractional\:Saturation=\frac{[ES]}{[E]_{T}}=\frac{4.5*10^{-6}\:M}{5.0*10^{-6}\:M+4.5*10^{-6}\:M}=47.37\: \%\)

The Michaelis-Menten equation can be expressed as: 

\(\displaystyle V=\frac{k{_{cat}}[E_{0}][S]}{K_{m}+[S]}\) 

The velocity is therefore proportional to the enzyme concentration \(\displaystyle E_{0}\), not inversely so. \(\displaystyle k_{cat}\) is also referred to as the turnover number. As the substrate concentration \(\displaystyle [S]\) keeps increasing, then we end up with a steady state in which all the enzyme is bound. At this point, the maximum velocity, \(\displaystyle V_{max}\), has been reached.

For this question, we're provided with the michaelis constant for an enzyme, as well as the reaction rate for that enzyme at a particular substrate concentration. We're asked to determine the maximum possible reaction rate that this enzyme can achieve when it is saturated with substrate.

To solve this problem, we'll need to use the michaelis-menten equation, which is expressed as follows.

\(\displaystyle V_{o}=\frac{V_{max}[S]}{K_{M}+[S]}\)

Then, we can rearrange the equation above in order to isolate the \(\displaystyle V_{max}\) term.

\(\displaystyle V_{max}=\frac{V_{o}(K_{M}+[S])}{[S]}\)

Now, we can plug in the values given to us in the question stem in order to solve for our answer.

\(\displaystyle V_{max}=\frac{(200\: \frac{mmol}{min})(5\: mmol+0.5\: mmol)}{0.5\: mmol}\)

\(\displaystyle V_{max}=2200\: \frac{mmol}{min}\)

Reaction rate of an enzymatic reaction can be calculated using the Michaelis-Menten equation.

\(\displaystyle V = \frac{Vmax [S]}{K_m + [S]}\)

where \(\displaystyle V\) is the reaction rate, \(\displaystyle V_m_a_x\)is the maximum reaction rate, \(\displaystyle [S]\) is the substrate concentration and \(\displaystyle K_m\) is the Michaelis constant.

Recall that adding a competitive inhibitor will increase the \(\displaystyle K_m\) but will not alter the \(\displaystyle V_m_a_x\). From the equation, we can see that increasing \(\displaystyle K_m\) will decrease the reaction rate; therefore, adding a competitive inhibitor will decrease reaction rate.

Substrate concentration is found in both the numerator and denominator of the equation; however, the substrate concentration is multiplied in the numerator whereas it is added in the denominator. This means that increasing substrate concentration will increase the numerator more than the denominator; therefore, increasing substrate concentration will increase reaction rate.

Affinity between enzyme and substrate is determined by \(\displaystyle K_m\). This constant is defined as the substrate concentration required to reach half the \(\displaystyle V_m_a_x\). Lowering \(\displaystyle K_m\) suggests that a lower substrate concentration is needed to reach the same half \(\displaystyle V_m_a_x\) (lower substrate concentration is needed because the affinity between enzyme and substrate increases and a more efficient reaction is carried out). This implies that lowering \(\displaystyle K_m\) increases the affinity between substrate and enzyme; therefore, \(\displaystyle K_m\) and affinity are inversely related. Decreasing the affinity will increase the \(\displaystyle K_m\) and, subsequently, decrease reaction rate.

The trick to this question is to notice that the \(\displaystyle K_m\) and the substrate concentration are the same. Recall that \(\displaystyle K_m\) is the substrate concentration required to reach half the maximum reaction rate. Since we are told that the substrate concentration and \(\displaystyle K_m\) are the same, we can conclude that that reaction rate of \(\displaystyle 15\frac{mol}{s}\) is half the maximum reaction rate; therefore, the maximum reaction rate is \(\displaystyle 30\frac{mol}{s}\).

To answer this question, it is essential to have an understanding of the Michaelis-Menten kinetics of enzymes.

When plotting a graph of initial reaction rate as a function of substrate concentration, the resulting plot shows a hyperbolic relationship. At first, the rate increases linearly with a fairly steep slope. But as substrate concentration rises, the graph begins to level off.

The question, however, is to determine the order of the reaction with respect to substrate (the effect substrate has on reaction rate). One helpful way to determine this is to make use of the Michaelis-Menten equation.

\(\displaystyle V_{0}=\frac{V_{max}[S]}{K_{M}+[S]}=\frac{k_{cat}[E]_{T}[S]}{K_{M}+[S]}\)

With this equation in mind, we can make predictions on how substrate concentration will affect the reaction order.

Low Substrate Concentrations:

When substrate concentration is very low, we can assume that \(\displaystyle K_{M}>>[S]\). As a result of this, we can essentially say that \(\displaystyle K_{M}+[S]\approx K_{M}\). Thus, at low concentrations of \(\displaystyle S\), the equation can change as follows.

\(\displaystyle V_{0}=\frac{V_{max}[S]}{K_{M}}=\frac{k_{cat}[E]_{T}[S]}{K_M}=(\frac{k_{cat}}{K_{M}})[E]_{T}[S]\)

As we can see above, under conditions of low substrate concentration, the reaction is first-order with respect to enzyme and first-order with respect to substrate. This means that any slight change in the concentration of substrate will proportionately affect the reaction rate (for instance, if the substrate concentration increases by \(\displaystyle 20\:\%\), then the reaction rate will also increase by \(\displaystyle 20\:\%\), assuming that enzyme concentration is held constant).

*** As a side note, it's also worth noting that the above expression also includes a new reaction rate constant, one that is a ratio of two other constants, \(\displaystyle \frac{k_{cat}}{K_{M}}\). This term is actually what is used to assess an enzyme's efficiency, since it states the reaction rate under conditions of very low substrate concentration.

High Substrate Concentrations:

Now, let's see what happens when substrate concentration is high. When this happens, we can say that \(\displaystyle [S]>>K_{M}\). Just as we made an estimation before, we can do so once again by stating that \(\displaystyle [S]+K_{M}\approx [S]\) under such conditions. Next, we can take a look at how this changes the original equation.

\(\displaystyle V_{0}=\frac{V_{max}[S]}{[S]}=\frac{k_{cat}[E]_{T}[S]}{[S]}=k_{cat}[E]_{T}=V_{max}\)

As we can see, under high substrate concentrations (saturating conditions), the reaction is at its maximum value. Furthermore, notice that the \(\displaystyle [S]\) term cancels out of the expression. This means that, under saturating conditions, the reaction is zero-order with respect to substrate and first-order with respect to enzyme. The consequence of this is that a change in substrate concentration is unable to change the reaction rate; only a change in enzyme concentration (or temperature) can affect the rate at saturating substrate concentrations.

With Michaelis-Menten Kinetics, The Vmax is the maximum rate of the enzyme mediated reaction. When a system has a concentration of substrate well above Km (which is the concentration of substrate at which the reaction is proceeding at one-half Vmax), then it is said that the system is saturated. In this state, the reaction is occurring basically at Vmax, since there is plenty of substrate to react with the enzyme, and adding more substrate would not increase the reaction rate. Rather, the concentration of enzyme is the limiting factor of how much reaction product will be produced per unit time.