Figure 4 below shows the UV/visible absorption spectra of a 23-mer target ssDNA

Question

Figure 4 below shows the UV/visible absorption spectra of a 23-mer target ssDNA (t-ssDNA, Figure 4a), Ru(bpy)^2+_3 labeled t- ssDNA (Figure 4c), and Ru(bpy)^2+_3 phosphoramidite tag itself (Figure 4b) in pH 7.4 PBS solutions with a 1.00 cm quartz cuvette. If t-ssDNA has a molecular weight of 7039 g/mol, and one unit absorbance at 260 nm corresponds to 33 mu g/mL of DNA, what is the concentration of this oligonucleotide? Under the present experimental conditions, what is the molar extinction coefficient or molar absorptivity of Ru(bpy)^2+_3 phosphoramidite at 457 nm? Refer to Figure 4c, estimate the mole ratio of the DNA to the Ru (II) tag within the complex.

Explanation / Answer

Part 1:

Calculating molar extinction coefficient(epsilon) from the data in Part 1 we have;

for unit absorbance at 260 nm corresponds to 33 microgram/mL for 0.1 dm cuvette.

Therefore, from Beer - Lambert's law;

A = epsilon * C * L [where C is concentration and L is length of cuvette]

or 1 = epsilon * (33 microgram/mL) * (0.1dm)

or, 1 = epsilon * (33 * 10-3g/L) * (0.1dm)

or, 1 = epsilon * (33*10-3/7039 mol/L) * (0.1 dm)      [because molecular weight = 7039 g/mol]

Or, epsilon = (7039*104/33 dm2/mol)

Now, for absorption(A) = 0.836 (figure 4a), we need to find the concentration of the oligonucleotide(i.e. t-ssDNA).

So, the Beer Lambert equation gives us;

0.836 = (7039*104/33 dm2/mol) * C * (0.1dm) [where C is our required concentration]

Or, C = (0.836*33/7039*103 mol/L) = 0.0276 mol/L = 27.6 microgram/mL.

Part 2:

From the diagram index, given concentration of Ru(bpy)32+ phosphoramidite is 13.3 micromolar.

13.3 micromolar = 13.3*10-6 mol/L

Given from figure 4b, A = 0.137

So, required is the molar extinction coefficient given by the Beer Lambert law;

epsilon = A/C*L = 0.137/((13.3*10-6 mol/L) *(0.1dm)) = 1.03*105 dm2/mol

Part 3:

From the analysis of figures 4a and 4b, the peak at 257nm corresponds to absorption by t-ssDNA and peak that at 278nm corresponds to Ru(bpy)32+ . The shifts in peaks are because they are the two entities are coupled together and absorb at slightly different wavelengths.

From figure 4c, absorptions corresponding to t-ssDNA and Ru(bpy)32+ are 0.972 and 0.908 respectively.

So, let C1 be the concentration of t-ssDNA and C2 be the concentration of Ru(bpy)32+ in the solution.

From Beer Lambert law for evaluating C1 and C2 and using epsilon values from 1st and 2nd parts, we have;

0.972 = (7039*104/33 dm2/mol) * C1 * (0.1dm)

Or, C1 = 4.56*10-6 mol/L

Similarly, 0.908 = (1.03*105) * C2 * (0.1dm)

Or, C2 = 8.81*10-5 mol/L

Therefore the mole ratio is given by C1/C2 = (4.56*10-6 mol/L)/( 8.81*10-5 mol/L ) = 0.0517.

Related Questions

Navigate

• Browse (All)
• Browse F
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.