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Deming and Pardue studied the kinetics for the hydrolysis of \(p\) -nitrophenyl phosphate by the enzyme alkaline phosphatase. \({ }^{23}\) The reaction's progress was monitored by measuring the absorbance of \(p\) -nitrophenol, which is one of the reaction's products. A plot of the reaction's rate (with units of \(\mu \mathrm{mol} \mathrm{mL}^{-1} \mathrm{sec}^{-1}\) ) versus the volume, \(V\), in milliliters of a serum calibration standard that contained the enzyme, yielded a straight line with the following equation. $$ \text { rate }=2.7 \times 10^{-7} \mu \mathrm{mol} \mathrm{mL}^{-1} \mathrm{~s}^{-1}+\left(3.485 \times 10^{-5} \mu \mathrm{mol} \mathrm{mL}^{-2} \mathrm{~s}^{-1}\right) V $$ A 10.00 -mL sample of serum is analyzed, yielding a rate of \(6.84 \times 10^{-5}\) \(\mu \mathrm{mol} \mathrm{mL}^{-1} \mathrm{sec}^{-1}\). How much more dilute is the enzyme in the serum sample than in the serum calibration standard?

Short Answer

Expert verified
The enzyme in the serum is approximately 5.12 times more dilute than in the calibration standard.

Step by step solution

01

Understanding the Given Equation

The given equation is \( \text{rate} = 2.7 \times 10^{-7} + (3.485 \times 10^{-5}) V \), which is a linear equation with \( \text{rate} \) as the dependent variable and \( V \) as the independent variable. The slope, \( 3.485 \times 10^{-5} \), represents the change in rate per unit volume of the serum standard.
02

Substitute the Given Rate into the Equation

We know that the rate measured from a 10.00 mL sample of serum is \( 6.84 \times 10^{-5} \) \( \mu \text{mol} \text{mL}^{-1} \text{s}^{-1} \). First, substitute this value into the equation to find the corresponding volume \( V \):\[ 6.84 \times 10^{-5} = 2.7 \times 10^{-7} + (3.485 \times 10^{-5}) \cdot V \]
03

Solve for Volume \( V \)

To find \( V \), solve the equation for \( V \):\[ 6.84 \times 10^{-5} - 2.7 \times 10^{-7} = 3.485 \times 10^{-5} \cdot V \]\[ 6.813 \times 10^{-5} = 3.485 \times 10^{-5} \cdot V \]\[ V = \frac{6.813 \times 10^{-5}}{3.485 \times 10^{-5}} \approx 1.954 \text{ mL} \]
04

Determine Dilution Factor

The original sample volume was 10.00 mL, but the calculated volume \( V \) required to achieve the same rate using the standard is approximately 1.954 mL. To find how much more dilute the enzyme is in the serum sample compared to the calibration standard, calculate the dilution factor:\[ \text{Dilution factor} = \frac{10.00}{1.954} \approx 5.12 \]
05

Interpret the Result

The serum sample's enzyme concentration is \( 5.12 \) times more diluted than in the calibration standard because the 10.00 mL sample corresponds to a standard volume of approximately 1.954 mL, meaning the sample is diluted by this factor.

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Key Concepts

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

Enzyme Kinetics
Enzyme kinetics is a core concept in biochemistry which studies the rates at which enzymatic reactions occur. This helps chemists understand how different concentrations and conditions can affect the reaction rate. In the case of alkaline phosphatase acting on \( p \)-nitrophenyl phosphate, we're exploring how effectively the enzyme speeds up this particular reaction. This process involves measuring the speed of conversion of substrates to products. Understanding this rate is crucial for applications in research and industry, where enzymes often serve as catalysts in chemical reactions. By plotting reaction rates against the volume of an enzyme-containing solution, as was done in this exercise, investigators can derive key insights into enzyme behavior and activity.
Hydrolysis Reaction
Hydrolysis reactions are chemical processes where water molecules are used to break bonds in larger molecules, splitting them into smaller parts. In this context, \( p \)-nitrophenyl phosphate undergoes hydrolysis through the action of alkaline phosphatase, producing \( p \)-nitrophenol and inorganic phosphate. This reaction is key for detecting the presence and activity of alkaline phosphatase, because the product, \( p \)-nitrophenol, can be quantitatively measured. Hydrolysis reactions are essential for biological processes, such as digestion and metabolism, where complex molecules are broken down into simpler, more manageable forms.
Alkaline Phosphatase
Alkaline phosphatase is an enzyme that removes phosphate groups from various molecules, including proteins and nucleotides, in a process termed dephosphorylation. This enzyme operates optimally in alkaline conditions, which is why it is named 'alkaline.' It is commonly used in labs for biochemical reactions, where it helps with processes like DNA cloning and protein analysis. In the exercise discussed, alkaline phosphatase facilitates the breakdown of \( p \)-nitrophenyl phosphate during the reaction. Monitoring this enzyme's activity through various experimental conditions can provide insights into its functionality and structure, which may inform medical and industrial applications.
Absorbance Measurement
Absorbance measurement is a technique used in chemistry to determine the concentration of substances in a solution by measuring how much light is absorbed. In this exercise, absorbance is used to track the production of \( p \)-nitrophenol. The absorbance of light at a specific wavelength corresponds to the concentration of \( p \)-nitrophenol, allowing for the quantitative analysis of the enzymatic reaction's progress. This method is crucial in many biochemical assays as it provides a simple and direct way to monitor reactions as they occur. Absorbance measurements are typically carried out with spectrophotometers, instruments designed to measure the intensity of light passing through a sample and calculate its absorbance.
Dilution Factor
The dilution factor refers to how much a solution is diluted, either through adding solvent or increasing its volume. In enzyme reactions, knowing the dilution factor is important for determining the concentration of enzymes in a sample. In the original exercise, the dilution factor was calculated to understand how diluted the enzyme in a serum sample was relative to a calibration standard. It was determined by comparing the actual volume of the sample to the effective volume calculated from the reaction rate equation. This process ensures accuracy in experiments requiring precise enzyme concentrations, which is critical in both research and clinical diagnostics.

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Most popular questions from this chapter

To improve the sensitivity of a FIA analysis you might do any of the following: inject a larger volume of sample, increase the flow rate, decrease the length and the diameter of the manifold's tubing, or merge separate channels before injecting the sample. For each action, explain why it leads to an improvement in sensitivity.

Ramsing and co-workers developed an FIA method for acid-base titrations using a carrier stream that is \(2.0 \times 10^{-3} \mathrm{M} \mathrm{NaOH}\) and that contains the acid-base indicator bromothymol blue. \({ }^{25}\) Standard solutions of \(\mathrm{HCl}\) were injected, and the following values of \(\Delta t\) were measured from the resulting fiagrams. $$ \begin{array}{cccc} {[\mathrm{HCl}](\mathrm{M})} & \Delta t(s) & {[\mathrm{HCl}](\mathrm{M})} & \Delta t(s) \\ \hline 0.008 & 3.13 & 0.080 & 7.71 \\ 0.010 & 3.59 & 0.100 & 8.13 \\ 0.020 & 5.11 & 0.200 & 9.27 \\ 0.040 & 6.39 & 0.400 & 10.45 \\ 0.060 & 7.06 & 0.600 & 11.40 \end{array} $$ A sample with an unknown concentration of \(\mathrm{HCl}\) is analyzed five times, giving values of \(7.43,7.28,7.41,7.37,\) and \(7.33 \mathrm{~s}\) for \(\Delta t .\) Determine the concentration of \(\mathrm{HCl}\) in the sample.

Fernández-Abedul and Costa-García developed an FIA method to determine cocaine in samples using an amperometric detector. \(^{27}\) The following signals (arbitrary units) were collected for 12 replicate injections of a \(6.2 \times 10^{-6} \mathrm{M}\) sample of cocaine, \(\mathrm{C}_{17} \mathrm{H}_{21} \mathrm{NO}_{4}\). \(\begin{array}{lll}24.5 & 24.1 & 24.1 \\ 23.8 & 23.9 & 25.1 \\ 23.9 & 24.8 & 23.7 \\ 23.3 & 23.2 & 23.2\end{array}\) (a) What is the relative standard deviation for this sample? (b) The following calibration data are available $$ \begin{array}{cc} \text { [cocaine] }(\mu \mathrm{M}) & \text { signal (arb. units) } \\ \hline 0.18 & 0.8 \\ 0.36 & 2.1 \\ 0.60 & 2.4 \\ 0.81 & 3.2 \\ 1.0 & 4.5 \\ 2.0 & 8.1 \\ 4.0 & 14.4 \\ 6.0 & 21.6 \\ 8.0 & 27.1 \\ 10.0 & 32.9 \end{array} $$ In a typical analysis a 10.0 -mg sample is dissolved in water and diluted to volume in a \(25-\mathrm{mL}\) volumetric flask. A \(125-\mu \mathrm{L}\) aliquot is transferred to a \(25-\mathrm{mL}\) volumetric flask and diluted to volume with a \(\mathrm{pH} 9\) buffer. When injected into the flow injection apparatus a signal of 21.4 (arb. units) is obtained. What is the \(\% \mathrm{w} / \mathrm{w}\) cocaine in the sample?

The concentration of \(\mathrm{Ni}\) in a new alloy is determined by a neutron activation analysis. A 0.500 -g sample of the alloy and a 1.000 -g sample of a standard alloy that is \(5.93 \% \mathrm{w} / \mathrm{w} \mathrm{Ni}\) are irradiated with neutrons in a nuclear reactor. When irradiation is complete, the sample and the standard are allowed to cool and their gamma ray activities measured. Given that the activity is \(1020 \mathrm{cpm}\) for the sample and \(3540 \mathrm{cpm}\) for the standard, determine the \(\% \mathrm{w} / \mathrm{w} \mathrm{Ni}\) in the alloy.

Analytes \(A\) and \(B\) react with a common reagent \(R\) with first-order kinetics. If \(99.9 \%\) of \(A\) must react before \(0.1 \%\) of \(B\) has reacted, what is the minimum acceptable ratio for their respective rate constants?

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