# User:Allison K. Alix/Notebook/Thesis Research/2013/10/29

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## Current revision

October 29th 2013 Main project page
Previous entry      Next entry

## To Do

• Calculate percentage of MB open using FCS curves
• Run 10000X green beads to obtain average diffusion time.

## Calculating %MB open

Example:

N = 7.9 particles

7.9 particles x 1 mol/6.02x1023 particles x 1/6.93 x 10-16L = 1.89x10-8M

NOTE: 6.93x10-16 is the focal volume of the spot, previous calculated

1.89x10-8M x 1nM/10-9M = 18.9nM

Actual concentration of this sample was 22.4nM (According to dilution calculations)

%MB open = 18.9nM/22.4nM x 100% = 84.4nM

• This is a very large percentage that is open. All other calculations show similar results with the lowest %open being 69%. Because the beacon is specifically designed NOT to fluoresce when closed, there may be something else contributing to these high percentages. First, the width of the spot needs to be recalculated as 10-16 is a bit small (usually around 10-15). To recalculate this, 9 samples of 10000X diluted green beads will be run on the FCS for 3 minutes at a time to obtain an average diffusion time. This can then be substituted into the equation relating ω0 and τD.
• The second thing to check is that the known concentration of MB is actually what was calculated. Although these samples were made straight from a stock solution with a known concentration, there could have been error during preparation, that is affecting the actual concentration that is not being accounted for. To check the concentrations, new MB samples will be prepared, and their UV-Vis spectra will be obtained

## Data

The above image is an FCS curve showing the average diffusion time of 9 10000X diluted green bead samples. τD was calculated to be 58.7ms

## Calculations

radius of the spot

τD = (ω023ηπd)/4kT

where τD is the diffusion time of the sample

ω0 is the radius of the spot

η is the viscosity of the solvent (water), 0.001028kg/m s

d is the diameter of the particle

k is Boltzmann's constant = 1.3806488 × 10-23 m2 kg s-2-2 K-1

T is the temperature of the room (18.7 °C, 291.7K)

ω0 = sqrt([(0.05874s(4)(1.3806488x10-23m2kg/s2K)(291.7K)]/[3(0.001028kg/m s)(π)(2x10-7m)])

ω0 = 0.698μm

volume of the spot

V =4/3(π)r3

= 4/3(π)(6.98x10-7)3

= 1.42x10-18 x 1000L/1m3

= 1.42x10-15L

[MB](nM)  %MB open
22.4 39.3
17.9 26.2
14.9 30.6
12.8 34.1
11.2 50.7
9.9 40.1
8.9 48.2
8.2 62.3