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Intracellular Concentration of Enzymes To approximate the concentration of enzymes in a bacterial cell, assume that the cell contains equal concentrations of 1,000 different enzymes in solution in the cytosol and that each protein has a molecular weight of 100,000 . Assume also that the bacterial cell is a cylinder (diameter \(1.0 \mu \mathrm{m}\), height \(2.0 \mu \mathrm{m}\) ), that the cytosol (specific gravity \(1.20\) ) is \(20 \%\) soluble protein by weight, and that the soluble protein consists entirely of enzymes. Calculate the average molar concentration of each enzyme in this hypothetical cell.

Short Answer

Expert verified
The average molar concentration of each enzyme is \(2.40 \times 10^{-6} \text{ M}\).

Step by step solution

01

Calculate the Volume of the Bacterial Cell

The bacterial cell is modeled as a cylinder with a diameter of \(1.0 \mu \mathrm{m}\) and a height of \(2.0 \mu \mathrm{m}\). The formula for the volume of a cylinder is given by \(V = \pi r^2 h\), where \(r\) is the radius and \(h\) is the height. \[V = \pi \left(\frac{1.0}{2}\right)^2 \times 2.0 = \pi \times 0.5^2 \times 2 \approx 1.57 \mu m^3\]
02

Convert Volume to Liters

Since 1 cubic meter (m^3) equals \(10^{15}\) cubic micrometers (\(\mu m^3\)), we convert the volume to liters (1 m^3 = 1000 L): \[1.57 \mu m^3 = 1.57 \times 10^{-15} m^3 = 1.57 \times 10^{-12} L\]
03

Calculate Total Mass of Soluble Proteins

Given that the cytosol has a specific gravity of 1.20, it has a density of 1.20 g/mL. Convert volume from L to mL: \[1.57 \times 10^{-12} L = 1.57 \times 10^{-9} mL\] The mass of the cytosol is calculated by multiplying the volume by the density:\[\text{Mass of cytosol} = 1.57 \times 10^{-9} \times 1.20 = 1.884 \times 10^{-9} \text{ g}\] Considering that the cytosol is 20% protein by weight:\[\text{Total mass of protein} = 1.884 \times 10^{-9} \times 0.20 = 3.77 \times 10^{-10} \text{ g}\]
04

Calculate Moles of Proteins

If each protein has a molecular weight of 100,000 g/mol, calculate moles:\[\text{Moles of protein} = \frac{3.77 \times 10^{-10}}{100,000} = 3.77 \times 10^{-15} \text{ mol} \]
05

Calculate Molar Concentration of Each Enzyme Type

Since there are 1,000 different enzymes, divide the total moles by 1,000 to find moles of each enzyme:\[\text{Moles of each enzyme} = \frac{3.77 \times 10^{-15}}{1,000} = 3.77 \times 10^{-18} \text{mol/enzyme type} \] Find the concentration by dividing the moles of each enzyme by the volume of the cell in liters:\[\text{Concentration of each enzyme} = \frac{3.77 \times 10^{-18}}{1.57 \times 10^{-12}} = 2.40 \times 10^{-6} \text{ M} \]

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Key Concepts

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

Molar Concentration
When we talk about molar concentration, we refer to the amount of a particular substance (or solute) present in a certain volume of solution. It's essentially a way of expressing concentration in terms of moles per liter (M). In our exercise, we were tasked to find the molar concentration of each enzyme within a bacterial cell. Here's how we approached it:
  • We started by calculating the moles of the soluble protein, using its molecular weight and the mass already determined.
  • Next, since the problem assumes equal concentrations of 1,000 different enzymes, each with a molecular weight of 100,000 g/mol, we divided the total moles by 1,000 to get the moles of each enzyme type.
  • Finally, using the bacterial cell's volume in liters, we divided the moles of one enzyme type by this volume to determine the molar concentration, which turned out to be approximately \( 2.40 \times 10^{-6} \text{ M} \).
Understanding molar concentration helps in visualizing how dense a solution is with respect to its solute, crucial for biological and chemical applications.
Bacterial Cell Volume
The volume of a bacterial cell is crucial for calculating concentrations within the cell. In our problem, the cell was shaped like a cylinder, which made the computation straightforward. Using the formula for the volume of a cylinder, \( V = \pi r^2 h \), we calculated the cell's volume based on the given diameter and height.
  • The diameter of the cell is \( 1.0 \mu m \), which we converted to a radius of \( 0.5 \mu m \) for our calculation.
  • The height given is \( 2.0 \mu m \), thus using the formula, the volume achieved was roughly \( 1.57 \mu m^3 \).
  • To facilitate the following calculations, we converted this volume into liters resulting in \( 1.57 \times 10^{-12} \text{ L} \).
Understanding the volume allows us to accurately determine the space within which the proteins are dissolved, critical for calculating concentrations.
Enzyme Molecular Weight
Enzyme molecular weight is a key factor in determining how much of the enzyme is present in terms of moles. In many biological studies, enzymatic activities are associated with their molecular weights, often expressed in g/mol.
  • In this particular exercise, a uniform molecular weight of 100,000 g/mol was assumed for each enzyme. This assumption simplified our calculations, allowing us to focus on deriving concentration rather than dealing with varying protein sizes.
  • Given the total mass of soluble protein, this molecular weight was essential to calculate the total moles of protein within the cell.
  • Understanding the molecular weight of enzymes aids in various applications, including drug formulation, enzyme assays, and metabolic engineering.
Enzyme molecular weights are generally experimentally determined, and this exercise highlights their importance in intracellular concentration calculations.
Soluble Protein Calculation
Soluble protein calculation involves finding out just how much protein is present within the cell. This is significant as enzymes, being largely proteins, dictate most biochemical reactions. Here's how we approached the calculation:
  • First, we used the cell's volume and the specific gravity of the cytosol to determine the mass of the cytosol.
  • Given that proteins make up 20% of this mass, we proceeded to find the exact mass of the soluble proteins.
  • This mass allowed us to calculate the total moles of protein by integrating molecular weight, providing a basis for eventually finding the concentration of individual enzymes.
Calculating soluble protein is fundamental in cell biology, as it impacts the understanding of metabolic capacities and the cell's enzymatic potential.

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Most popular questions from this chapter

Perturbed \(\mathbf{p} \boldsymbol{K}_{\mathrm{a}}\) Values in Enzyme Active Sites Alanine racemase is a bacterial enzyme that converts \(\mathrm{L}\)-alanine to \(\mathrm{D}\) alanine, which is needed in small amounts to synthesize the bacterial cell wall. The active site of alanine racemase includes a Tyr residue with a p \(K_{\mathrm{a}}\) value of \(7.2\). The \(\mathrm{p} K_{\mathrm{a}}\) of free tyrosine is 10 . The altered \(\mathrm{p} K_{\mathrm{a}}\) of this residue is due largely to the presence of a nearby charged amino acid residue. Which amino acid(s) could lower the \(\mathrm{p} K_{\mathrm{a}}\) of the neighboring Tyr residue? Explain your reasoning.

The Effects of Reversible Inhibitors The MichaelisMenten rate equation for reversible mixed inhibition is written as $$ V_{0}=\frac{V_{\max }[\mathrm{S}]}{\alpha K_{\mathrm{m}}+\alpha^{\prime}[\mathrm{S}]} $$ Apparent, or observed, \(K_{\mathrm{m}}\) is equivalent to the [S] at which $$ V_{0}=\frac{V_{\max }}{2 \alpha^{\prime}} $$ Derive an expression for the effect of a reversible inhibitor on apparent \(K_{\mathrm{m}}\) from the previous equation.

Dlinical Application of Differential Enzyme Inhibition Human blood serum contains a class of enzymes known as acid phosphatases, which hydrolyze biological phosphate esters under slightly acidic conditions \((\mathrm{pH} .0)\) : Acid phosphatases are produced by erythrocytes and by the liver, kidney, spleen, and prostate gland. The enzyme of the prostate gland is clinically important, because its increased activity in the blood can be an indication of prostate cancer. The phosphatase from the prostate gland is strongly inhibited by tartrate ion, but acid phosphatases from other tissues are not. How can this information be used to develop a specific procedure for measuring the activity of prostatic acid phosphatase in human blood serum?

Estimation of \(V_{\max }\) and \(\boldsymbol{K}_{\mathrm{m}}\) by Inspection Graphical methods are available for accurate determination of the \(V_{\max }\) and \(K_{\mathrm{m}}\) of an enzyme-catalyzed reaction. However, these quantities can sometimes be estimated by inspecting values of \(V_{0}\) at increasing [S]. Estimate the \(V_{\max }\) and \(K_{m}\) of the enzyme-catalyzed reaction for which the data in the table were obtained. \begin{tabular}{cc} {\([\mathbf{S}](\mathrm{M})\)} & \(V_{0}(\mu \mathrm{M} / \mathrm{min})\) \\ \hline \(2.5 \times 10^{-6}\) & 28 \\ \(4.0 \times 10^{-6}\) & 40 \end{tabular} \begin{tabular}{ll} \(1 \times 10^{-5}\) & 70 \\ \(2 \times 10^{-5}\) & 95 \\ \(4 \times 10^{-5}\) & 112 \\ \(1 \times 10^{-4}\) & 128 \\ \(2 \times 10^{-3}\) & 140 \\ \(1 \times 10^{-2}\) & 139 \\ \hline \end{tabular}

Applying the Michaelis-Menten Equation I An enzyme has a \(V_{\max }\) of \(1.2 \mu \mathrm{M} \mathrm{s}^{-1}\). The \(K_{\mathrm{m}}\) for its substrate is \(10 \mu \mathrm{M}\). Calculate the initial velocity of the reaction, \(V_{0}\), when the substrate concentration is a. \(2 \mu \mathrm{M}\) b. \(10 \mu_{M}\) c. \(30 \mu_{\mathrm{M}}\).

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