Liver alcohol dehydrogenase (ADH) is relatively nonspecific and will oxidize ethanol or other alcohols, including methanol. Methanol oxidation yields formaldehyde, which is quite toxic, causing, among other things, blindness. Mistaking it for the cheap wine he usually prefers, my dog Clancy ingested about \(50 \mathrm{mL}\) of windshield washer fluid (a solution \(50 \%\) in methanol). Knowing that methanol would be excreted eventually by Clancy's kidneys if its oxidation could be blocked, and realizing that, in terms of methanol oxidation by ADH, ethanol would act as a competitive inhibitor, I decided to offer Clancy some wine. How much of Clancy's favorite vintage \((12 \% \text { ethanol })\) must he consume in order to lower the activity of his ADH on methanol to \(5 \%\) of its normal value if the \(K_{m}\) values of canine ADH for ethanol and methanol are 1 millimolar and 10 millimolar, respectively? (The \(K_{1}\) for ethanol in its role as competitive inhibitor of methanol oxidation by ADH is the same as its \(K_{m \cdot}\) ) Both the methanol and ethanol will quickly distribute throughout Clancy's body fluids, which amount to about 15 L. Assume the densities of \(50 \%\) methanol and the wine are both \(0.9 \mathrm{g} / \mathrm{mL}\).

Short Answer

Expert verified
The dog Clancy must consume 235 mL of wine to lower the activity of his ADH on methanol to 5% of its normal value.

Step by step solution

01

Determine the initial concentration of methanol in Clancy's body

This step requires the basic application of the principle of concentration (mass/volume). Given that Clancy has ingested 50 mL of a solution that is \(50 \%\) methanol and that the density of the solution is \(0.9 \mathrm{g}/ \mathrm{mL}\), this provides a starting point for determining the amount of methanol consumed. First, calculate the mass of 50mL of the solution: \(50 \mathrm{mL} \times 0.9 \mathrm{g}/ \mathrm{mL} = 45 \mathrm{g}\). Then, calculate the mass of methanol in this solution: \(45 \mathrm{g} \times 0.5 = 22.5 \mathrm{g}\). Convert the mass of methanol (which has a molar mass of 32 g/mol) to moles: \(22.5 \mathrm{g} \div 32 \mathrm{g}/ \mathrm{mole} = 0.703 \mathrm{mole}\). To get the concentration, divide the amount of methanol by the total volume of body fluids: \(0.703 \mathrm{mole} \div 15 \mathrm{L} = 0.0468 \mathrm{mol/L}=46.8 \mathrm{mM}\).
02

Determining the amount of ethanol to be consumed using the principles of competitive inhibition

Aiming for 5% of methanol to be metabolized, the effective \(K_{m_{methanol}}\) considering competitive inhibition is obtained by \(K_{m_{methanol}}'= K_{m_{methanol}} (1 + [I]/K_{i}) = 10 (1+ [ethanol]/1)\), where \(K_{m_{methanol}}'\) is the effective Michaelis constant considering the competitive inhibitor ethanol, [I] is the competitive inhibitor concentration ([ethanol]), and \(K_{i}\) is the inhibition constant (which is equal to the \(K_{m_{ethanol}}\) in this case). It can be rearranged as \([ethanol]= ( K_{m_{methanol}}' /K_{m_{methanol}} -1) \cdot K_{i}\). From Step 1, since we want the system to function at 5% activity, \(K_{m_{methanol}}'\) can be approximated by [methanol], the substrate concentration, because the Michaelis-Menten kinetics are in the limit of saturating substrate concentration. So, the ethanol concentration required is \([ethanol]= (46.8 /10 -1) \cdot 1 = 3.68 \mathrm{mM}\).
03

Determining the volume of wine to be consumed

The problem states that the wine Clancy is to consume has a \(12 \%\) ethanol content. Given the density of the wine is \(0.9 \mathrm{g}/ \mathrm{mL}\) and the molar mass of ethanol is 46 g/mol, the wine concentration in molarity can be calculated as: wine concentration \(= \frac{{12 \mathrm{g}}}{{100 \mathrm{g}}} \times \frac{{0.9 \mathrm{g}/ \mathrm{mL}}}{{46 \mathrm{g}/ \mathrm{mol}}} = 0.234 \mathrm{mol}/ \mathrm{L} = 234 \mathrm{mM}\). So the volume of wine \(V_{wine}[mL]\) to be consumed to achieve the required concentration can be found using \([ethanol] = \frac{{n_{ethanol}}}{V_{total}}\) where the total volume ( \(V_{total}\)) is 15 L: \(V_{wine} = \frac{{[ethanol] \cdot V_{total}}}{[wine] }= \frac{{3.68 * 15 * 10^{-3}}}{234 }= 0.235 \mathrm{L} = 235 \mathrm{mL}\). This is the volume of wine required to lower the activity of Clancy’s liver ADH on methanol to 5% of its normal value.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Alcohol Dehydrogenase
Alcohol dehydrogenase (ADH) is a crucial enzyme found primarily in the liver that plays a significant role in the metabolism of alcohols within the body. It catalyzes the oxidation of alcohols, such as ethanol, into aldehydes. This conversion is the first step in their metabolism and is essential because while alcohols can be relatively safe in moderate amounts, their aldehyde products can be toxic.

ADH is not entirely specific to ethanol and can oxidize other alcohols too. For instance, it can metabolize methanol into formaldehyde, which can have harmful effects on health, including the potential to cause blindness. Understanding the nonspecific nature of ADH is important because it opens the door to competitive inhibition, a concept employed when treating poisoning from substances like methanol.

Competitive inhibition occurs when two different substances (both of which can be acted upon by the same enzyme), compete for the active site of that enzyme. If a non-toxic substance, such as ethanol, is present in high enough concentrations, it can hinder the enzyme's interaction with the toxic substance, in this case, methanol, thereby slowing down its metabolism. In a practical sense, this can buy time for the toxic substance to be excreted by the kidneys before it accumulates to dangerous levels in the body.
Methanol Toxicity
Methanol toxicity is a dangerous and potentially fatal condition that occurs when methanol, a toxic alcohol, is ingested. Methanol is found in many industrial solvents, windshield washer fluid, and some homemade alcoholic beverages. It is metabolized by the liver enzyme ADH, initially into formaldehyde, which is then further converted into formic acid. Both of these metabolites are toxic and are responsible for the adverse effects associated with methanol poisoning, such as damage to the optic nerve leading to blindness, central nervous system depression, and even death.

To treat methanol toxicity, competitive inhibitors of ADH like ethanol are used, which have a higher affinity for ADH and can outcompete methanol for the enzyme's active site. By doing so, the conversion of methanol into its toxic metabolites is slowed, and the time is gained to allow for methanol's elimination from the body through excretion. Treatment with ethanol is a delicate process and requires careful calculation and administration to ensure that the toxic effects of methanol are mitigated without causing further harm from ethanol itself.

Understanding methanol’s distribution in the body, its rate of metabolism by ADH, and the concentration at which it becomes toxic requires precise knowledge of the body’s physiology and biochemistry. Calculating the right amount of ethanol to administer takes into account factors such as ADH's affinity for both substances and the total body fluid volume in which these alcohols dissolve.
Michaelis-Menten Kinetics
Michaelis-Menten kinetics is a theoretical framework used to understand the kinetics of enzyme-catalyzed reactions. It provides insight into how substrate concentration affects the rate of reaction, and it defines two important parameters: the Michaelis constant (Km) and the maximum reaction rate (Vmax).

The Michaelis constant (Km) represents the substrate concentration at which the reaction rate is half of Vmax. A low Km indicates high affinity between enzyme and substrate, meaning that the enzyme can achieve a high rate even in low substrate concentration. Conversely, a high Km indicates lower affinity and hence requires a higher substrate concentration to achieve the same reaction rate. The concept of Km is particularly useful when considering competitive inhibition, as the presence of an inhibitor increases the apparent Km for the substrate.

In the context of methanol toxicity and ADH, where ethanol is used as a competitive inhibitor, the Michaelis-Menten equation can be adapted. It helps predict how the enzyme's activity will be affected by different concentrations of methanol and ethanol. This understanding is vital in the medical treatment of poisoning, allowing clinicians to calculate how much of an inhibitor must be given to decrease the enzyme’s activity on the toxic substrate to a level where the patient can safely process and excrete the toxin.

In simpler terms, by using the principles of Michaelis-Menten kinetics, one can effectively assess the interplay between enzymes, substrates, and inhibitors to optimize the body's biochemical responses during medical treatments.

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Most popular questions from this chapter

Acetylcholinesterase catalyzes the hydrolysis of the neurotransmitter acetylcholine: Acetylcholine \(+\mathrm{H}_{2} \mathrm{O} \longrightarrow\) acetate \(+\) choline The \(K_{m}\) of acetylcholinesterase for its substrate acetylcholine is \(9 \times 10^{-5} M .\) In a reaction mixture containing 5 nanomoles/mL of acetylcholinesterase and \(150 \mu M\) acetylcholine, a velocity \(v_{\mathrm{o}}=\) \(40 \mu \mathrm{mol} / \mathrm{mL} \cdot\) sec was observed for the acetylcholinesterase reaction. a. Calculate \(V_{\max }\) for this amount of enzyme. b. Calculate \(k_{\text {cat }}\) for acetylcholinesterase. c. Calculate the catalytic efficiency \(\left(k_{\mathrm{cat}} / K_{m}\right)\) for acetylcholinesterase. d. Does acetylcholinesterase approach "catalytic perfection"?

Equation 13.9 presents the simple Michaelis-Menten situation where the reaction is considered to be irreversible ([P] is negligible). Many enzymatic reactions are reversible, and \(P\) does accumulate. a. Derive an equation for \(v\), the rate of the enzyme-catalyzed reaction \(\mathrm{S} \rightarrow \mathrm{P}\) in terms of a modified Michaelis-Menten model that incorporates the reverse reaction that will occur in the presence of product, \(\mathbf{P}\) b. Solve this modified Michaelis-Menten equation for the special situation when \(v=0\) (that is, \(S \rightleftharpoons P\) is at equilibrium, or in other words, \(\left.K_{\mathrm{eq}}=[\mathrm{P}] /[\mathrm{S}]\right)\) (J. B. S. Haldane first described this reversible Michaelis-Menten modification, and his expression for \(K_{\mathrm{eq}}\) in terms of the modified M-M equation is known as the Haldane relationship.)

The following kinetic data were obtained for an enzyme in the absence of any inhibitor \((1),\) and in the presence of two different inhibitors (2) and (3) at \(5 \mathrm{m} M\) concentration. Assume \(\left[\mathrm{E}_{T}\right]\) is the same in each experiment. $$\begin{array}{cccc} & (1) & (2) & (3) \\ {[\mathrm{S}]} & v(\mu \mathrm{mol} / & v(\mu \mathrm{mol} /) & v(\mu \mathrm{mol} /) \\ (\mathrm{m} M) & \mathrm{mL} \cdot \mathrm{sec}) & \mathrm{mL} \cdot \mathrm{sec}) & \mathrm{mL} \cdot \mathrm{sec} \\ 1 & 12 & 4.3 & 5.5 \\ 2 & 20 & 8 & 9 \\ 4 & 29 & 14 & 13 \\ 8 & 35 & 21 & 16 \\ 12 & 40 & 26 & 18 \end{array}$$ Graph these data as Lineweaver-Burk plots and use your graph to find answers to a. and b. a. Determine \(V_{\max }\) and \(K_{m}\) for the enzyme. b. Determine the type of inhibition and the \(K_{1}\) for each inhibitor.

Measurement of the rate constants for a simple enzymatic reaction obeying Michaelis-Menten kinetics gave the following results: \(k_{1}=2 \times 10^{8} M^{-1} \sec ^{-1}\) \(k_{-1}=1 \times 10^{3} \sec ^{-1}\) \(k_{2}=5 \times 10^{3} \mathrm{sec}^{-1}\) a. What is \(K_{\mathrm{S}},\) the dissociation constant for the enzyme- substrate complex? b. What is \(K_{m},\) the Michaelis constant for this enzyme? c. What is \(k_{\text {cat }}\) (the turnover number) for this enzyme? d. What is the catalytic efficiency \(\left(k_{\mathrm{cat}} / K_{m}\right)\) for this enzyme? e. Does this enzyme approach "kinetic perfection"? (That is, does \(k_{\mathrm{cat}} / K_{m}\) approach the diffusion-controlled rate of enzyme association with substrate? f. If a kinetic measurement was made using 2 nanomoles of enzyme per \(\mathrm{mL}\) and saturating amounts of substrate, what would \(V_{\max }\) equal? g. Again, using 2 nanomoles of enzyme per mL of reaction mixture, what concentration of substrate would give \(v=0.75 V_{\max } ?\) h. If a kinetic measurement was made using 4 nanomoles of enzyme per \(\mathrm{mL}\) and saturating amounts of substrate, what would \(V_{\max }\) equal? What would \(K_{m}\) equal under these conditions?

Enzyme A follows simple Michaelis-Menten kinetics. a. The \(K_{m}\) of enzyme A for its substrate \(\mathrm{S}\) is \(K_{m}^{\mathrm{S}}=1 \mathrm{m} M .\) Enzyme \(\mathrm{A}\) also acts on substrate \(\mathrm{T}\) and its \(K_{m}^{\mathrm{T}}=10 \mathrm{m} M .\) Is \(\mathrm{S}\) or \(\mathrm{T}\) the preferred substrate for enzyme A? b. The rate constant \(k_{2}\) with substrate \(\mathrm{S}\) is \(2 \times 10^{4} \mathrm{sec}^{-1}\); with substrate \(\mathrm{T}, k_{2}=4 \times 10^{5} \mathrm{sec}^{-1} .\) Does enzyme A use substrate S or substrate T with greater catalytic efficiency?

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