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Chapter 20: Problem 25

The weight of proteins or nucleic acids in solution is commonly determined by UV spectroscopy using the Beer-Lambert law. For example, the \(\varepsilon\) of double-stranded DNA at \(260 \mathrm{~nm}\) is \(6670 \mathrm{M}^{-1} \mathrm{~cm}^{-1}\). The formula weight of the repeating unit in DNA (650 Daltons) can be used as the molecular weight. What is the weight of DNA in \(2.0 \mathrm{~mL}\) of aqueous buffer if the absorbance, measured in a 1 -cm cuvette, is \(0.75\) ?

Short Answer

Expert verified
Answer: The weight of DNA in the 2.0 mL aqueous buffer solution is approximately \(1.46 \times 10^{-4}\mathrm{g}\).

Step by step solution

01

Write down the Beer-Lambert law formula

The Beer-Lambert law states that the absorbance (A) of a solution is directly proportional to its concentration (c) and the path length (l). It can be written as follows: A = \(\varepsilon \cdot l \cdot c\) where A is the absorbance, \(\varepsilon\) is the molar absorptivity in \(\mathrm{M}^{-1} \mathrm{~cm}^{-1}\), l is the path length in cm, and c is the molar concentration of the solution in \(\mathrm{M}\) (moles per liter).
02

Rearrange the formula for concentration

To find the concentration of DNA in the solution, we need to rearrange the Beer-Lambert law formula for c: c = \(\frac{A}{\varepsilon \cdot l}\)
03

Plug in the given values and calculate concentration

Now, we can plug the given values into the rearranged formula and calculate the concentration of DNA: c = \(\frac{0.75}{6670 \mathrm{M}^{-1}\mathrm{~cm}^{-1} \cdot 1 \mathrm{~cm}} = 1.12 \times 10^{-4} \mathrm{~M}\)
04

Convert volume from milliliters to liters

We are given the volume of the solution in milliliters, but we need it in liters to be consistent with the concentration units. To convert 2.0 mL to liters, we divide by 1000: 2.0 mL = 0.002 L
05

Calculate the number of moles of DNA

To calculate the number of moles of DNA in the solution, multiply the concentration by the volume: moles of DNA = \(c \cdot V = 1.12 \times 10^{-4} \mathrm{M} \cdot 0.002 \mathrm{L} = 2.24 \times 10^{-7} \mathrm{~mol}\)
06

Calculate the weight of DNA

Finally, to find the weight of DNA, multiply the number of moles by the molecular weight of DNA (650 Daltons in this case): Weight of DNA = number of moles \(\times\) molecular weight = \(2.24 \times 10^{-7} \mathrm{mol} \cdot 650 \mathrm{\frac{g}{mol}} = 1.46 \times 10^{-4}\mathrm{g}\) The weight of DNA in the 2.0 mL aqueous buffer solution is approximately \(1.46 \times 10^{-4}\mathrm{g}\).

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Most popular questions from this chapter

1,3-Butadiene is a gas at room temperature and requires a gas-handling apparatus to use in a Diels-Alder reaction. Butadiene sulfone is a convenient substitute for gasheated above its boiling point of \(110^{\circ} \mathrm{C}\), decomposes by a reverse Diels-Alder reaction to give s-cis-1,3-butadiene and sulfur dioxide. Draw Lewis structures for butadiene sulfone and \(\mathrm{SO}_{2}\); then, show by curved arrows the path of this reaction, which resembles a reverse Diels-Alder reaction.

One of the published syntheses of warburganal begins with the following Diels- Alder reaction. Propose a structure for compound A.

If an electron is added to 1,3 -butadiene, into which molecular orbital does it go? If an electron is removed from 1,3 -butadiene, from which molecular orbital is it taken?

Write the frontier molecular orbital analysis for a [3,3]-sigmatropic shift in the analogous fashion as presented in the chapter except using a geometry that would lead to a boatlike conformation for the transition state. As a hint, you should find that the reaction is allowed. However, why would this geometry be less favorable?

Estimate the stabilization gained as a result of conjugation when 1,4 -pentadiene is converted to trans-1,3-pentadiene. Note that the answer is not as simple as comparing the heats of hydrogenation of 1,4 -pentadiene and trans-1,3-pentadiene. Although the double bonds are moved from unconjugated to conjugated, the degree of substitution of one of the double bonds is also changed, in this case from a monosubstituted double bond to a trans disubstituted double bond. To answer this question, you must separate the effect that is the result of conjugation from that caused by a change in the degree of substitution.
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