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Quiz (2/20/08, Chpt 3, ECB) 1. A chemical process where there is a net gain of electrons is called _______________. A chemical process where there is a.

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Presentation on theme: "Quiz (2/20/08, Chpt 3, ECB) 1. A chemical process where there is a net gain of electrons is called _______________. A chemical process where there is a."— Presentation transcript:

1 Quiz (2/20/08, Chpt 3, ECB) 1. A chemical process where there is a net gain of electrons is called _______________. A chemical process where there is a net loss of electrons is called _________________. 2. Enzymes are catalysts, they often _____________ the free energy of a reaction by favoring a transition state. 3. This nucleotide cofactor prominently featured as an electron carriers are _____________ and ______________. 4. This is the constant at which an enzyme is operating at half of its maximum speed. _____________________ 5. The maximum number of catalytic cycles an enzyme can perform per unit time is called the _____________. reduction oxidation lower NAD + NADP + Km, Michalis-Menten constant Turnover rate

2 Fluorescence Imaging with One Nanometer Accuracy (1.5 nm, 1-500 msec)

3 Techniques needed Specificity to look at heads Nanometer spatial localization Second temporal resolution Single Molecule sensitivity Single Molecule Photostability FluorescenceFluorescence How to get nanometer localization with visible photons? in out 8 nm

4 Imaging Single Molecules with very good S/N Total Internal Reflection Microscopy For water (n=1.33) to air (n=1.0): what is TIR angle? For glass (n=1.5), water (n=1.33): what is TIR angle? what is penetration depth? >57° d p =( /4  )[n 1 2 sin 2  i ) - n 2 2 ] -1/2 d p = 58 nm With d p = 58 nm, can excite sample and not much background. TIR- (  >  c ) Exponential decay

5 Experimental Set-up for TIR (2 set-ups) Wide-field Objective-TIR Laser Objective Filter Dichroic Sample CCD Detector Lens Wide-field, Prism-type, TIR Microscope Sample Laser Objective Filter CCD Detector Lens In one case, sample is “upside-down.” Does this make a difference? No!

6 Diffraction limited spot Width of /2 ≈ 250 nm center width Enough photons (signal to noise)…Center determined to ~ 1.3 nm Dye last 5-10x longer -- typically ~30 sec- 1 min. (up to 4 min) Accuracy of Center = width/ S-N = 250 nm / √10 4 = 2.5 nm= ± 1.25nm

7 How well can you localize? What does it depend on? (3 things) 1. # of Photons Detected (N) 3. Noise (Background) of Detector (b) (includes background fluorescence and detector noise) 2. Pixel size of Detector (a) = derived by Thompson et al. (Biophys. J.). center width

8 Experimental Setup: Imaging Single Molecules Cy3-DNA Immobilized on coverslip Cy-3 Dye Biotinylated DNA Streptavidin Biotinylated BSA coverslip DNA Sample 3’ 18mer Move stage In 8 nm, 16 nm, 37 nm increments

9 Data: Model System: 30 nm Artificial Steps 020406080100 0 200 400 600 800 Time (sec) 26283032 0 5 10 no. of steps poissonian step size (nm) 26283032 0 5 10 no. of steps regular step size (nm)  = 28.52nm  = 1.24nm  Ex. = 28.55nm  = 29.08nm  = 1.32nm  Ex. = 28.55nm Position (nm)

10 12 nm steps 01020304050 0 30 60 90 120 150 101112131415 0 1 2 3 4 5 no. of steps regular step size (nm) Time (sec) 1011121314 0 1 2 3 no. of steps poissonian step size (nm)  = 12.47nm  = 1.37nm  Ex. = 12.35nm  = 12.26nm  = 1.39nm  Ex. = 12.35nm Position (nm)

11  8 nm steps 01020304050 0 20 40 60 80 100 6789 0 1 2 3 4 5 Time (sec) Position (nm) 456789 0 1 2 3 4 5 no. of steps regular step size (nm)  = 7.63nm  = 0.93nm  Ex. = 7.45nm  = 6.42nm  = 1.09nm  Ex. = 6.52nm no. of steps poissonian step size (nm)

12 Center of mass 37/2 nm x 74 nm 37-2x 37 nm Myosin V Labeling on Light Chain: Expected Step Sizes Expected step size Hand-over-hand: Head = 2 x 37 nm= 74, 0, 74 nm CaM-Dye: 37-2x, 37+2x, … Inchworm: always S cm = 37 nm 0 nm 37+2x

13 A Single Myosin V moving 37 nm or 74 nm? 86 nm pixel [ATP] = 300 nM (Low)

14 higher [ATP] (>400 nM) [ATP] = 300 nM Myosin V steps: 74 nm +/- 5nm +/- 1.3 nm +/- 1.5 nm +/- 1.3 nm

15 52-23 nm steps

16 [ATP] = 300 nM 40-30 nm and 50-20 nm dwell time histogram k = 0.302 R 2 = 0.979 N mol =12 N step =191 74-0 nm dwell time histogram k 1 =0.3281 N mol =37 k 2 = 0.3278 N step =361 R 2 = 0.983 k 1 =k 2  k = 0.3279 ± 0.007 A observable! 74 nm (t sec) observable! k1k1 k2k2 B A x nm 74-x nm (t-u) secu sec A 74-0 nm Steps: Detecting 0 nm Intermediate by Kinetics

17 Class evaluation 1.What was the most interesting thing you learned in class today? 2. What are you confused about? 3. Related to today’s subject, what would you like to know more about? 4. Any helpful comments. Answer, and turn in at the end of class.


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