Lactose Intolerance and the Reaction Kinetics of Lactase

Learning Objectives

ACS Examinations Institute General Chemistry ACCM concepts taught in this exemplar:

VII. E. 1. a. Catalysts can be either homogeneous or heterogeneous. 

VII. E. 2. a. Catalysts provide an alternative reaction pathway that lowers this activation energy. 

               b. Because a catalyst lowers the activation energy, it affects both forward and reverse reactions. 

X. D. 1. a. Students should be able to distinguish exothermic and endothermic reaction depictions from an energy profile representation. 

     b. Students should be able to identify the concept of activation energy within an energy profile representation.

Lactose Intolerance

Lactose is the natural sugar present in milk and all dairy products. Because lactose is a disaccharide, a large, complex sugar that the small intestine cannot properly digest, it must be broken down by the enzyme lactase into two monosaccharides, glucose and galactose (refer to figure 1).¹ Glucose and galactose, two essential nutrients, are then properly digested by being absorbed into the bloodstream. However, if there is not enough lactase, the lactose will proceed to the colon (large intestine), where bacteria in the colon will then digest the lactose, producing fatty acids and gasses like carbon dioxide, methane, and hydrogen.² In addition, the additional lactose in the large intestine will draw additional water and gas into the colon, resulting in symptoms like bloating, cramping, flatulence, and diarrhea.³

Those who cannot properly produce the lactase to digest lactose are deemed lactose intolerant. Most cases of lactose intolerance do not result in lactase not being produced at all, but a significantly reduced production of lactase after infancy (congenital lactose intolerance is extremely rare). Thus, those who are lactose intolerant take measures like limiting or completely avoiding consuming dairy products or consume lactase enzyme tablets.

The cause of this digestive disorder is a result of decreased expression of the LCR gene. Thus, lactose intolerance often runs in families as an inherited genetic fault. Lactose intolerance is a very common disease, as approximately 65% of the world population develops some level of lactose intolerance after infancy.⁴ It is most prevalent among those of East Asian descent, where up to 100% of certain communities report lactose intolerance.

Hydrolysis of Lactose

Hydrolysis refers to any organic chemical reaction involving a water molecule to break bonds in substrate to form one or more chemical bonds in products. Lactase is the enzyme that breaks down lactose (C₁₂H₂₂O₁₁), a disaccharide in milk and all dairy products, into glucose (C₆H₁₂O₆) and galactose.⁸

In the image below, we can observe the standard hydrolysis of lactose. A molecule of lactose, a disaccharide, is hydrolyzed in the presence of the enzyme lactase to yield two monosaccharides, one molecule of galactose and one molecule of glucose.

Fig 1: Lactose hydrolyzed into galactose and glucose⁵

Chemical Reactions and Reaction Favorability

The digestion of lactose is a chemical reaction. For chemical reactions to occur, molecules must collide with sufficient energy and correct orientation. In biological hydrolysis reactions, like that of lactose, a water molecule splits the glycosidic bond of the polysaccharide into multiple monosaccharides. In this case, lactose is hydrolyzed to produce two monosaccharides, glucose and galactose. We can thus write a simple chemical reaction for the hydrolysis of glucose (note that glucose and galactose are individually labeled as they are epimers).

C₁₂H₂₂O₁₁ + H₂O → C₆H₁₂O_{6 glucose} + C₆H₁₂O_{6 galactose}⁵

For the sake of calculations, we will assume this reaction occurs at standard state conditions, representing 1 atmosphere of pressure and 1 molar for all solutions. This is not actually true in the catalysis of lactose, but this is ultimately an approximation. All thermodynamic measurements taken at standard state conditions are indicated with the “ ° ” symbol.

A reaction must be thermodynamically and kinetically favored in order to produce any significant amount of product. In terms of thermodynamics, the Gibbs free energy of the reaction system, \(\Delta G_{rxn}^o\), dictates the favorability of the reaction. The Gibbs free energy can be found with the equation \(\Delta G^o_{rxn}=\Delta H^o_{rxn}-T \Delta S^o_{rxn}\), where ΔH represents the enthalpy change in \(\frac{\mathrm{kJ}}{\mathrm{mol}}\), T represents temperature in Kelvin (K), and ΔS represents the entropy change in \(\frac{\mathrm{J}}{\mathrm{K}}\). A \(\Delta G^o_{rxn}< 0\) means that the products are lower in energy than the reactants, making the reaction spontaneous and thermodynamically favored. Conversely, a \(\Delta G^o_{rxn}> 0\) means that the products have greater energy than the reactants and the reaction is not thermodynamically favored and not spontaneous.

This can be further visualized through the equation \(\Delta G^o=-RTln K_{eq}\), where R is a universal gas constant, \(8.314\frac{\mathrm{J}}{\mathrm{mol·K}}\), T represents temperature in Kelvin, and Keq represents the equilibrium constant. Notice that at 298 K, the standard room temperature used in reactions, a \(\Delta G^o_{rxn}< 0\) results in a \(K_{eq}> 1\), representing the reaction favoring the products at equilibrium.

Thermodynamic Favorability

To find the standard Gibbs free energy of the reaction, by the first law of thermodynamics, we can subtract the Gibbs free energy of the reactants from the Gibbs free energy of the products in the equation \(\Delta G_{rxn}^o=\sum \Delta G_{f_{\text {products }}^o}^o-\sum \Delta G_{f_{\text {reactants }}^o}^o\). (Note that any catalyst, like lactase, does not affect ΔG°rxn or any other thermodynamic values as they do not change the energy at the start and end point of the reaction. This is covered in the next section) The values of ΔG°product and ΔG°reactants can be found through online thermodynamic tables representing ΔG°formation of each molecule. In this case, we found the ΔG°formation of lactose, water, glucose and galactose, respectively, to be \(-1334.42\frac{\mathrm{kJ}}{\mathrm{mol}}\)¹¹, \(-228.80\frac{\mathrm{kJ}}{\mathrm{mol}}\)¹², \(-793.74\frac{\mathrm{kJ}}{\mathrm{mol}}\)¹³, and \(-793.74\frac{\mathrm{kJ}}{\mathrm{mol}}\)¹⁴, respectively. Referring back to the catalysis of lactose, C₁₂H₂₂O₁₁ + H₂O → C₆H₁₂O_{6 glucose} + C₆H₁₂O_{6 galactose}, to calculate ΔG°rxn:

\(\Delta G^o_{rxn}=\sum \Delta G_{f_{\text {products }}^o}^o-\sum \Delta G_{f_{\text {reactants }}^o}^o\)

\(\Delta G^o_{rxn}=\left[\left(-793.74 \frac{\mathrm{kJ}}{\mathrm{mol_{glucose}}}\right)\left(\frac{1 \mathrm{~mol_{glucose}}}{1 \mathrm{~mol_{rxn}}}\right)+\left(-793.74 \frac{\mathrm{kJ}}{\mathrm{mol_{galactose}}}\right)\left(\frac{1 \mathrm{~mol_{galactose}}}{1 \mathrm{~mol_{rxn}}}\right)\right]-\left[\left(-1334.42 \frac{\mathrm{kJ}}{\mathrm{mol_{lactose}}}\right)\left(\frac{1 \mathrm{~mol_{lactose}}}{1 \mathrm{~mol_{rxn}}}\right)+\left(-228.80 \frac{\mathrm{kJ}}{\mathrm{mol_{water}}}\right)\left(\frac{1 \mathrm{~mol_{water}}}{1 \mathrm{~mol_{rxn}}}\right)\right]\)

\(\Delta G^o_{rxn}=-24.26\frac{\mathrm{kJ}}{\mathrm{mol_{rxn}}}\)

\(\Delta G^o_{rxn}< 0\). Thus, the catalysis of lactose is thermodynamically favored and spontaneous.

At standard room temperature conditions, 298K, we can find the equilibrium constant \(K_{eq}> 1\) to find where the equilibrium of the reaction lies :

\(\Delta G^o=-RTln K_{eq}\)

\(-24.26\frac{\mathrm{kJ}}{\mathrm{mol_{rxn}}}=-(8.314\frac{\mathrm{J}}{\mathrm{mol·K}})(\frac{\mathrm{1kJ}}{\mathrm{1000J}})(298K)lnK_{eq}\)

\(K_{eq}=1.789·10^3\)

\(K_{eq}> 1\). Thus, the reaction favors the products.

Again, the calculated \(\Delta G^o_{rxn}\) and \(K_{eq}\) values are only estimates under the assumption that the reaction occurs at standard conditions (1 molar solutions, 1 atm of pressure). This is not actually true in the human gut, where the lactose is hydrolyzed in the digestive system. However, \(\Delta G_{rxn}\) still <0 and \(K_{eq}\) still >>1 in the gut, and the reaction is still thermodynamically favored.

Catalysts and Kinetic Favorability

If the conversion of lactose to glucose and galactose is thermodynamically favored, why is lactase needed?

Chemical reactions must be both thermodynamically and kinetically favored to produce product significantly. Every chemical reaction has an activation energy (Fig 2), the minimum amount of energy the reactants must possess, in order to react. Many reactions have activation energies that are so high that reactions proceed extremely slowly. Take, for instance, the reaction that turns diamond into graphite. It is thermodynamically favored, but we know it takes enormous amounts of energy to create diamond, hence the great activation energy that prevents diamond rings and jewelry from decomposing into graphite (it will take billions of years for ~1 gram to decompose at room temperature).

For many reactions, certain substances lower the activation energy of the reaction, decreasing the energy barrier and allowing the reaction to proceed at a much higher rate. These substances are called catalysts. Humans create thousands of enzymes, biological catalysts in order to support chemical reactions for human function. Lactase is one of them.

Catalysis refers when the rate of reaction is increased by adding catalysts. Catalysts are substances that influence the rate of a chemical reaction without going through any chemical changes itself by providing an alternative pathway of reaction which significantly lowers activation energy.⁶ This would make it more kinetically favorable and not appear in the net reaction equation. Additionally, catalysts affect the forward and reverse rates equally as it lowers the height of the energy barrier. This means that catalysts have no effect on the equilibrium constant and thus has no effect on the composition of the equilibrium state. A catalyst only affects Ea.

Thus, theoretically, an individual who could not produce any lactose at all would properly break down lactose in their body. However, the rate at which this is done is so low that they cannot consume any significant amount of dairy. The presence of lactase during digestion lowers the activation energy to the point where the digestion of lactose is kinetically favorable. Scientific studies found that the actual activation energy of the catalyzed hydrolysis of lactose to be a relatively low 39.1kJ/mol.¹⁵ Without lactase, this figure is estimated to be several thousand-fold higher.

We can use the Arrhenius equation, \(k=Ae^{Ea/RT}\) to analyze the effects of this several-thousand fold increase on k, the reaction rate constant. A several thousand-fold increase in reaction rate will decrease reaction constant to such a great extent that the reaction rate will be completely negligible when lactase is not present.

Fig 2: potential energy diagram for a reaction in presence and absence of catalyst⁶

There are two types of catalysis reaction: homogenous and heterogenous. In homogeneous catalysis, the reactants and catalyst are in the same state of matter. In heterogeneous catalysis, the reactants and catalyst are not in the same state of matter.⁷ Lactase is a homogeneous catalyst as during the reaction, lactose (substrate) and lactase (enzyme) are both in aqueous solution dissolved in the small intestine during digestion, where the catalysis occurs. Many biochemical reactions include enzyme catalysis. The structure of the enzyme and the ligand have a significant impact on the specific mechanisms for enzyme activity. One of the models utilized for explaining substrate binding during enzyme catalysis is the Michaelis-Menten mechanism. 

Michaelis-Menten Rate Equation and Steady State Approximations: Theory and Derivation

 

The steady state approximation is a technique for calculating the total reaction rate of a multi-step reaction. In a multi-step reaction, it is assumed that the rate of change of intermediate concentration is constant. In an intermediate-forming sequential reaction, this approach can only be used when the first step of the reaction is much slower than the subsequent phase. Michaelis-Menten Rate Equation is a mathematical model depicting a 2-step reaction mechanism of enzymes in biological systems including enzymatic hydrolysis of lactose.⁹ The first step known as Binding involves binding of the substrate to enzyme while the second step, catalysis, involves the conversion of substrate to product which is then released, The equation is based on several assumptions: 

• Assumption 1: The substrate concentration does not have any traces of products at the start of the reaction. Hence, we overlook the possibility of the reverse reaction of [E+P] to [ES]. 
• Assumption 2: The reaction can achieve equilibrium ratios of [E], [S], and [ES] soon since the binding step is a fast one. In contrast, the catalysis step is rate-limiting as it moves more slowly.
• Assumption 3: Concentration of total enzymes in the system is equal to concentration of free enzymes and enzyme-substrate complexes.

1. We can use the Michaelis-Menten model as a simple model for enzyme kinetics. As enzyme-substrate-product reaction can be written as:

E + S ⇌ ES → E + P

Where we consider the enzyme E and substrate S reacting together to form the enzyme-substrate complex, ES, in a rapid equilibrium reaction. The ES then reacts irreversibly to generate the product P and regenerate the free enzyme, E, in a slow, rate-determining reaction.

Thus, since the second step of the reaction is the slow step, we can derive the rate law of the reaction as:

\(\frac{d[P]}{d t}=rate=k_2[ES]\)

We can thus assume that [ES], the concentration of the enzyme-substrate complex, reaches a steady state.

\(\frac{d[E S]}{d t}=k_1[E]_t[S]-k_1[E S]-k_2[E S]=0\)

Since the enzyme is conserved in the reaction, we can conclude that the total enzyme concentration E is equal to the concentration of the enzyme-substrate complex plus the concentration of free enzyme at a given time, E_t.

\([E] = [ES] + [E]_t\)

Solving for [E]_t:

\([E]_t = [E] - [ES]\)

We can substitute this equation into the steady-state equation to get: 

\(\frac{d[E S]}{d t}=k_1([E]-[ES])[S]-k_1[ES]-k_2[ES]=0\)

\(k_1([E]-[ES])[S]=(k_1+k_2)[ES]=0\)

Solving for [ES]:

\([ES]=\frac{\left[E\right][S]}{\left(\frac{k_{-1}+k_2}{k_1}\right)+[S]}\)

\([ES]=\frac{\left[E\right][S]}{\left(\frac{k_{-1}+k_2}{k_1}\right)+[S]}\)

\(\left(\frac{k_{-1}+k_2}{k_1}\right)\) can be simplified into a a single observed rate constant k_m. This is the Michaelis-Menten rate constant. Thus the equation can be rewritten as:

\([ES] = \frac{\left[E_T\right][S]}{k_m+[S]}\)

Plugging this into the differential rate law:

\(\frac{d[P]}{d t}=rate=k_2(\frac{\left[E_T\right][S]}{k_m+[S]})\)

Since V_max is the reaction velocity at the maximum saturating substrate concentration, it is equal to k₂ [ES] when [ES] = [E]. Since k₂ is the rate constant directly involved in the rate law, it is often notated as k_cat, representing the rate constant of catalysis. Thus, the reaction can be rewritten as:

\(\frac{d[P]}{d t}=rate=k_2(\frac{\left[ES]\right][S]}{k_m+[S]})\)

Simplifying the equation as \(rate=k_2[ES]=k_{cat}[ES]\), to get the Michaelis-Menten Equation:

\(\frac{d[P]}{d t}=rate=\frac{v_{max}[S]}{k_m+[S]}\)

Note that with other substitutions, the equation k_m, the Michaelis-Menten rate constant represents the reaction rate as exactly half of the maximum rate, v_max.¹⁰

Fig 3: Michaelis-Menten Curve¹⁶

Below are some exercises to test your understanding of Michaelis-Menten lactase enzyme kinetics

Exercise 1

For lactase that follows Michaelis Menten kinetics equation, what is the maximum catalysis rate of lactase (V_max) in μmoles/min if V = 65 moles/ min when [S] = 0.5 km and k_m = 1M

Answer

We have to use the Michaelis–Menten Equation derived above for this problem. Through a simple substitution the given values for each of the variables into the equation, we get: 

\begin{aligned}
& V=\frac{V_{\max }[S]}{K_m+[S]} \\
& 65=\frac{V_{\max }\left[S\right]}{K_m+\left[S\right]} \\
& 65=\frac{V_{\max }[S]}{1+[S]} \Rightarrow 97.5 \\
& \frac{97.5}{0.5}=V_{\max } \Rightarrow V_{\max }= 195  μmoles/min
\end{aligned}

Exercise 2

At what substrate concentration does the lactase reaction reach 90% of its maximum reaction rate (V_max)? 

Answer

Knowing that V, the reaction rate, is 90% of the maximum reaction rate, V_max, we can conclude that V=0.9Vmax and make a substitution.

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

\(0.9 \times V_{\max }=\frac{V_{\max }[S]}{K_m+[S]} \Rightarrow 0.9=[\mathrm{S}]\)

\(0.9(K_m+[S])=[\mathrm{S}]\)

\(0.9 K_m+0.9[S]=[S]\)

\(0.9 K_m=[S]-0.9[S]\)

\(0.9 K_m=(1-0.9)[S] \Rightarrow \frac{0.9}{0.1} K_m=[S]\)

\({[S]=9 K_m}\)

Exercise 3

(3.1) Determine the initial rate of lactase when conditions are such that the substrate concentration of lactose used is enough to produce half of lactase’s maximum rate, 60 µM/s.

(3.2) Determine the k_cat, if the concentration of lactase is 20 mM.

Answer

(3.1)
$$
V_0=\frac{V_{\max }[S]}{k_m+[S]} \quad \text { where }[S]=\begin{gathered}
\text { substrate } \\
\text { concentration of lactose }
\end{gathered}
$$
We know that [S] at \(\frac{\mathrm{1}}{\mathrm{2}}v_{max}=k_m\)
$$
V_0=\frac{1}{2} V_{\text {max }}
$$
$$
\begin{aligned}
& V_0=\frac{V_{\max }[S]}{1[S]+1[S]} \Rightarrow \frac{V_{\max }[S]}{2[S]} \\
& \Rightarrow V_0=\frac{1}{2} V_{\text {max }}=\frac{1}{2}\left(60 \frac{\mu M}{s}\right)=30 \frac{\mu M}{s}
\end{aligned}
$$
(3.2)
V_max = k_cat[E]_t where [E]_t is the lactase concentration at time t
Dividing [E]_t on both sides
$$
\frac{V_{\max }}{[E]_t}=\frac{[E]_t k_{\text {cat }}}{[E]_t} \Rightarrow k_{\text {cat }}=\frac{V_{\text {max }}}{[E]_t}
$$
It is given that [E] is 20mM = 20·10³µM
$$
k_{\text {cat }}=\frac{60 \frac{µM}{\mathrm{~s}}}{2 \times 10^4µM }=30 \times 10^{-4} \mathrm{~s}^{-1}=3 \times 10^{-3} \mathrm{~s}^{-1}
$$

References

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