1 / 17

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.

lieu
Download Presentation

Quiz (2/20/08, Chpt 3, ECB)

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  1. 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 _____________. Quiz (2/20/08, Chpt 3, ECB) reduction oxidation lower NAD+ NADP+ Km, Michalis-Menten constant Turnover rate

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

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

  4. Imaging Single Moleculeswith very good S/N Total Internal Reflection Microscopy TIR- (q > qc) Exponential decay dp=(l/4p)[n12sin2i) - n22]-1/2 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° dp = 58 nm With dp = 58 nm , can excite sample and not much background.

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

  6. center width Diffraction limited spot Width of l/2 ≈250 nm Accuracy of Center = width/ S-N = 250 nm / √104= 2.5 nm= ± 1.25nm 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)

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

  8. DNA Sample 18mer Biotinylated DNA Cy-3 Dye 3’ Streptavidin Biotinylated BSA coverslip coverslip Move stage In 8 nm, 16 nm, 37 nm increments Experimental Setup: Imaging Single MoleculesCy3-DNA Immobilized on coverslip

  9. m = 28.52nm 10 800 s = 1.24nm m = 28.55nm Ex. no. of steps 5 600 0 26 28 30 32 poissonian step size (nm) m = 29.08nm s = 1.32nm 400 m = 28.55nm Ex. 10 200 no. of steps 5 0 26 28 30 32 regular step size (nm) 0 0 20 40 60 80 100 Time (sec) Data: Model System: 30 nm Artificial Steps Position (nm)

  10. 12 nm steps m = 12.47nm s = 1.37nm m = 12.35nm Ex. 5 150 4 3 no. of steps 2 120 1 0 10 11 12 13 14 15 regular step size (nm) m = 12.26nm 90 s = 1.39nm m = 12.35nm Ex. 60 3 2 no. of steps 30 1 0 10 11 12 13 14 poissonian step size (nm) 0 0 10 20 30 40 50 Time (sec) Position (nm)

  11.  8 nm steps m = 7.63nm s = 0.93nm 5 m = 7.45nm Ex. 4 100 3 2 1 0 80 6 7 8 9 60 m = 6.42nm Position (nm) s = 1.09nm m = 6.52nm Ex. 40 5 4 3 no. of steps 20 2 1 0 4 5 6 7 8 9 regular step size (nm) 0 0 10 20 30 40 50 Time (sec) no. of steps poissonian step size (nm)

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

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

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

  15. 52-23 nm steps

  16. 74-0 nm Steps: Detecting 0 nm Intermediate by Kinetics B A k1 k2 A u sec (t-u) sec 74-x nm x nm 74 nm (t sec) observable! 40-30 nm and 50-20 nm dwell time histogram k = 0.302 R2 = 0.979 Nmol=12 Nstep=191 74-0 nm dwell time histogram k1 =0.3281 Nmol=37 k2 = 0.3278 Nstep=361 R2 = 0.983 k1=k2 k = 0.3279 ± 0.007 A [ATP] = 300 nM

  17. Class evaluation • 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.

More Related