quiz (2/20/08, chpt 3, ecb)

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. 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 oxidati on lower NAD + NADP + Km, Michalis-Menten constant Turnover rate

Upload: lieu

Post on 14-Jan-2016

29 views

Category:

Documents


0 download

DESCRIPTION

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. - PowerPoint PPT Presentation

TRANSCRIPT

Page 1: Quiz  (2/20/08, Chpt 3, ECB)

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

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

Fluorescence Imaging with One Nanometer Accuracy

(1.5 nm, 1-500 msec)

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

Techniques neededSpecificity to look at heads

Nanometer spatial localization

Second temporal resolution

Single Molecule sensitivity

Single Molecule Photostability

Fluorescence

How to get nanometer localization with visible photons?

in

out

8 nm

Page 4: Quiz  (2/20/08, Chpt 3, ECB)

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°

dp=(/4)[n12sin2i) - n2

2]-1/2

dp = 58 nm

With dp = 58 nm , can excite sample and not much background.

TIR- (> c) Exponential decay

Page 5: Quiz  (2/20/08, Chpt 3, ECB)

Experimental Set-up for TIR(2 set-ups)

Wide-fieldObjective-TIR

Laser

Objective

Filter

Dichroic

Sample

CCDDetector

Lens

Wide-field, Prism-type, TIR Microscope

Sample

Laser

Objective

Filter

CCDDetector

Lens

In one case, sample is “upside-down.” Does this make a difference? No!

Page 6: Quiz  (2/20/08, Chpt 3, ECB)

Diffraction limited spotWidth 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 / √104 = 2.5 nm= ± 1.25nm

Page 7: Quiz  (2/20/08, Chpt 3, ECB)

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)

i

22

2422 812

Na

bs

N

a

N

s ii =

derived by Thompson et al. (Biophys. J.).

center

width

Page 8: Quiz  (2/20/08, Chpt 3, ECB)

Experimental Setup: Imaging Single MoleculesCy3-DNA Immobilized on coverslip

Cy-3 DyeBiotinylated DNA

Streptavidin

Biotinylated BSA

coverslip

coverslip

DNA Sample

3’

18mer

Move stage

In 8 nm, 16 nm, 37 nm increments

Page 9: Quiz  (2/20/08, Chpt 3, ECB)

Data: Model System: 30 nm Artificial Steps

0 20 40 60 80 1000

200

400

600

800

Time (sec)

26 28 30 320

5

10

no

. of

step

s

poissonian step size (nm)

26 28 30 320

5

10

no

. of

step

s

regular step size (nm)

= 28.52nm = 1.24nmEx.= 28.55nm

= 29.08nm = 1.32nmEx.= 28.55nm

Po

sitio

n (

nm

)

Page 10: Quiz  (2/20/08, Chpt 3, ECB)

12 nm steps

0 10 20 30 40 50

0

30

60

90

120

150

10 11 12 13 14 150

1

2

3

4

5

no

. of

step

s

regular step size (nm)

Time (sec)

10 11 12 13 140

1

2

3

no

. of

step

s

poissonian step size (nm)

= 12.47nm = 1.37nm

Ex.= 12.35nm

= 12.26nm= 1.39nmEx.= 12.35nm

Po

sitio

n (

nm

)

Page 11: Quiz  (2/20/08, Chpt 3, ECB)

8 nm steps

0 10 20 30 40 50

0

20

40

60

80

100

6 7 8 90

1

2

3

4

5

Time (sec)

Po

sitio

n (

nm

)

4 5 6 7 8 90

1

2

3

4

5

no

. of

step

s

regular step size (nm)

= 7.63nm= 0.93nmEx. = 7.45nm

= 6.42nm = 1.09nmEx. = 6.52nm

no

. of

step

s

poissonian step size (nm)

Page 12: Quiz  (2/20/08, Chpt 3, ECB)

Cys66

Cys73

Center of mass

37/2 nmx

74 nm

37-2x

37 nm

Myosin V Labeling on Light Chain: Expected Step Sizes

Expected step sizeHand-over-hand:

Head = 2 x 37 nm= 74, 0, 74 nmCaM-Dye: 37-2x, 37+2x, …

Inchworm: always Scm = 37 nm

0 nm

37+2x

Page 13: Quiz  (2/20/08, Chpt 3, ECB)

A Single Myosin V moving

37 nm or 74 nm?

86 nm pixel

[ATP] = 300 nM (Low)

Page 14: Quiz  (2/20/08, Chpt 3, ECB)

0 10 20 30 40 50 60 70 80 90 1000

100

200

300

400

500

600

700

800

900

1000

1100

1200

1300

60 65 70 75 80 85 900

10

20

30

40 = 74.08 nm = 5.25 nm

R2 = 0.99346

2 = 1.6742 N

mol = 31

Nstep

= 227

histogram of 74-0 nm steps

step size (nm)

num

ber o

f ste

ps

75.4 nm

82.4 nm

65.1 nm

83.5 nm

69.8 nm

64.2 nm

78.8 nm

70.0 nm

70.9 nm

79.1 nm

67.4 nm

68.8 nm

73.7 nm

70.7 nm

70.1 nm

68.9 nm

71.3 nm

72.0 nm

time (sec)

Posi

tion

(nm

)

higher [ATP](>400 nM)

[ATP] = 300 nM

Myosin V steps: 74 nm +/- 5nm

+/- 1.3 nm

+/- 1.5 nm

+/- 1.3 nm

Page 15: Quiz  (2/20/08, Chpt 3, ECB)

52-23 nm steps

0 10 20 30 40 50 60

0

100

200

300

400

500

600

700

800

900 s1= 23.08 nm =3.44 nm N

mol=6

s2= 51.65 nm = 4.11 nm N

step=92

73.4 nm

73.5 nm

69.2 nm

44.8 nm24.9 nm

74.4 nm

72.2 nm

25.6 nm58.1 nm

18.9 nm

45.5 nm

19.2 nm

27.2 nm

74.3 nm

58.6 nm18.7 nm

66.6 nm

82.1 nm

72.5 nm

56.6 nm

68.4 nm

73.5 nm54.1 nm

48.8 nm

15.4 nm48.3 nm

23.7 nm

25.8 nm

68.2 nm

20 30 40 50 60 70 800

2

4n

o.

of

ste

ps

step size (nm)

time (sec)

Pos

ition

(nm

)

Page 16: Quiz  (2/20/08, Chpt 3, ECB)

[ATP] = 300 nM0 2 4 6 8 10 12 14 16 18 20 22 24 26 28

0

5

10

15

20

25

30

35

40

duration between adjacent steps (sec)

num

ber

of s

teps

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

74 nm (t sec) observable!observable!

k1 k2

B A

x nm 74-x nm

tketg 2)( tketf 1)(

(t-u) secu sec

A

tkettPkk 21

A A t

duutguftP0

)()()(21

12

kk

ee tktk

74-0 nm Steps: Detecting 0 nm Intermediate by Kinetics

Page 17: Quiz  (2/20/08, Chpt 3, ECB)

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.