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The uncatalyzed rate of amide bond hydrolysis in hippurylphenylalanine was measured by monitoring the formation of the product, phenylalanine. A \(30 \mathrm{mM}\) solution of hippurylphenylalanine was monitored for amide bond hydrolysis for 50 days. At the end of this time, \(25 \mu \mathrm{M}\) phenylalanine was detected in the solution. What is the rate of formation of phenylalanine (in units of \(\mathrm{M} \cdot \mathrm{s}^{-1}\) ) in this reaction?

Short Answer

Expert verified
The rate of formation of phenylalanine is approximately \(5.787 \times 10^{-12}\) M鈰卻鈦宦.

Step by step solution

01

Identify Known Quantities

We are given the following values: the total concentration of hippurylphenylalanine is 30 mM, phenylalanine concentration formed is 25 渭M, and the reaction time is 50 days.
02

Convert Units

We need to convert all quantities into consistent units. 30 mM = 30 x 10^-3 M (molarity) and 25 渭M = 25 x 10^-6 M. Additionally, convert 50 days to seconds: 50 days = 50 x 24 x 60 x 60 seconds.
03

Calculate Total Reaction Time in Seconds

Calculating 50 days in seconds: \[50 \times 24 \times 60 \times 60 = 4,320,000 \text{ seconds}\]
04

Determine Rate of Reaction

The rate of reaction can be calculated using the formula: \[ \text{Rate} = \frac{\text{Change in Concentration of Product}}{\text{Time}}\]Insert the known values: \[ \text{Rate} = \frac{25 \times 10^{-6} \text{ M}}{4,320,000 \text{ s}}\]
05

Perform the Calculation

Calculate the rate:\[ \text{Rate} = \frac{25 \times 10^{-6}}{4,320,000} = 5.787 \, \times 10^{-12} \, \text{M} \, \text{s}^{-1}\]

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Key Concepts

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

Reaction Rate Calculation
Understanding how to calculate the reaction rate is crucial in biochemistry, especially when studying enzyme activity and chemical reactions. The reaction rate tells us how quickly a product is formed or a reactant is used up in a chemical process. In the context of amide bond hydrolysis, this is about calculating how fast the product phenylalanine is produced.For this calculation, we use the formula:\[\text{Rate} = \frac{\text{Change in Concentration of Product}}{\text{Time}}\]To find the change in concentration, we calculate the final concentration of phenylalanine, which is known as 25 渭M, and since it starts at 0, the change is simply this final concentration. The time taken is 50 days, which needs to be converted into seconds for unit consistency. Once everything is set, inserting these values into our formula gives us the exact rate of the reaction.Remember, while this specific exercise is about an uncatalyzed reaction, similar methods can apply to enzyme-catalyzed situations, where understanding reaction rates can help decipher enzyme efficiency or substrate viability.
Unit Conversion
Performing accurate unit conversion is key in scientific calculations to ensure consistency and correct results. In this exercise, we need to convert several units to keep everything in molarity (M) and time in seconds so that they align properly for our calculations.
  • Converting Concentrations: Initial hippurylphenylalanine concentration is given as 30 mM, which we convert to molarity as 30 x 10鈦宦 M. Similarly, the phenylalanine concentration must be converted from 25 渭M to 25 x 10鈦烩伓 M.
  • Converting Time: The reaction time spans 50 days. Since a day has 24 hours and each hour has 3600 seconds, we find the total seconds by multiplying: 50 x 24 x 60 x 60, totaling 4,320,000 seconds.
Understand that consistent units are vital for making computation straightforward. Having everything in the correct units simplifies complex problems and minimizes errors. Once the units align, plugging values into equations becomes much simpler, allowing for straightforward computation of reaction rates or any other dynamic meters.
Biochemistry Problem-Solving
Problem-solving in biochemistry combines understanding theoretical concepts with practical application of chemical principles. This involves breaking down complex problems, like this exercise on amide bond hydrolysis, into manageable steps. One key skill is the methodical approach. Start by identifying what you know: the given concentrations and the total time taken for the reaction. Then, move to conversion 鈥 getting everything into consistent units. Whether it鈥檚 dealing with millimoles, micromoles, seconds, or minutes, conversion ensures that all aspects integrate seamlessly. Finally, apply known formulas like the reaction rate equation to tie together the data you鈥檝e organized. Troubleshooting comes naturally when these basics are mastered. Errors can usually be traced to missteps in unit conversion or plugging values into the wrong parts of your equation. Successful biochemistry problem-solving is largely about precision and practice. The more problems you tackle, the more adept you become at recognizing essential conversion steps and the application of biochemical equations.

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Most popular questions from this chapter

What relationship exists between \(K_{\mathrm{M}}\) and [S] when an enzymecatalyzed reaction proceeds at a. \(75 \% V_{\max }\) and b. \(90 \% V_{\max }\) ?

Protein phosphatase 1 (PP1) helps regulate cell division and is a possible drug target to treat certain types of cancer. The enzyme catalyzes the hydrolysis of a phosphate group from myelin basic protein (MBP). The activity of PP1 was measured in the presence and absence of the inhibitor phosphatidic acid (PA). a. Use the data to construct a Lineweaver-Burk plot for the PP1 enzyme in the presence and absence of PA. What kind of inhibitor is PA? b. Report the \(K_{\mathrm{M}}\) and \(V_{\max }\) values for PP1 in the presence and absence of the inhibitor. [MBP] (mg 路 mL鈥1) v0 without PA (nmol 路 mL鈥1 路 min鈥1) v0 with PA (nmol 路 mL鈥1 路 min鈥1) $$ \begin{array}{lll} 0.010 & 0.0209 & 0.00381 \\ 0.015 & 0.0355 & 0.00620 \\ 0.025 & 0.0419 & 0.00931 \\ 0.050 & 0.0838 & 0.01400 \end{array} $$

Inhibitor A at a concentration of \(2 \mu \mathrm{M}\) doubles the apparent \(K_{\mathrm{M}}\) for an enzymatic reaction, whereas inhibitor \(B\) at a concentration of \(9 \mu \mathrm{M}\) quadruples the apparent \(K_{\mathrm{M}}\). What is the ratio of the \(K_{\mathrm{I}}\) for inhibitor B to the \(K_{\mathrm{I}}\) for inhibitor \(\mathrm{A}\) ?

A species of Shewanella bacteria contains an enzyme that catalyzes the dehalogenation of tetrachloroethene. The \(K_{\mathrm{M}}\) is \(120 \mu \mathrm{M}\) and the \(V_{\max }\) is \(1.0 \mathrm{nmol} \cdot \mathrm{min}^{-1} \cdot \mathrm{mg}^{-1}\). What is the substrate concentration when the velocity is \(0.75 \mathrm{nmol} \cdot \mathrm{min}^{-1} \cdot \mathrm{mL}^{-1}\) ?

The rate of hydrolysis of trehalose to its constituent glucose monomers in the absence of a catalyst is even slower than hydrolysis of sucrose (see Problem 3) and may indicate that the two sugars are hydrolyzed by different mechanisms. If the initial concentration of trehalose is \(0.050 \mathrm{M}\), it takes \(6.6 \times 10^{6}\) years for the concentration to decrease by half. What is the rate of disappearance of trehalose in the absence of a catalyst?

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