Shot 8.48 PM MacBook Air 8. (Bonus, 15 pts) Figure 4 below shows the UV/visible

Question

Shot 8.48 PM MacBook Air 8. (Bonus, 15 pts) Figure 4 below shows the UV/visible absorption spectra of a 23-mer target ssDNA (t-ssDNA, Figure 4a), Ru(bpy)32 labeled t- ssDNA (Figure 4c), and Ru(bpy)2" phosphoramidite tag itself (Figure 4b) in pH 7.4 PBS solutions with a 257 (0.972) 278 (0.905) 1.00 cm quartz cuvette 457 (0.089) 287 (I) If t-ssDNA has a molecular weight of 7039 g/mol, and one unit absorbance at 260 nm corresponds to 33 g/mL of DNA, what is the 0 747 245 (0268) concentration of this oligonucleotide? 457 (0.137) 260 (0.836) (2) Under the present experimental conditions, what is the molar extinction coefficient or molar absorptivity of Ru(bpy), 2 phosphoramidite at 457 nm? 240 280 320 360 400 440 80 Wavelength (nm) (3) Refer to Figure 4c, estimate the mole ratio of the Figure 4. uvivis absorption spectra DNA to the Ru (Il) tag within the complex. t-ssDNA (see main text), (b) 13.3 jiM Ru(bpy)s? phosphoramidite. and (c) 1-ssDNA tagged with Rulbpy's?" species in 0.10 M PBS (pH 7.4) using a 1-cm cuvette

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.

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