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Cell bio calculation, Study notes of Cell Biology

Broward CollegeCell Biology

Some calculations you may need

Typology: Study notes

2022/2023

Uploaded on 05/25/2023

unknown0785
unknown0785 🇺🇸

2 documents

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bg1
#3
AminoacidD
Ratio
[D-]/[D]
practice
test
PK
=
4.7
pH
=
b.7
Henderson
Hasselbranch
Equation
[conjugate
base]
(forweakacid)
pH
=
pKa+log
(weakacid]
=
+
log
109
=
2
Ontaking
Antilog
(Base
10)
5
=
10
2
=
100
Amino
acid
D
Ratio
(D-]/[D]
pK
=
4.7
pH
=
4.7
[
pH=
pKa+log
"attac
another
ex.
=
+
109
1095
=
0
-
100
=
1
antilog
(base
(0)
5
=
10
=
=
10
pf3
pf4
pf5

Partial preview of the text

Download Cell bio calculation and more Study notes Cell Biology in PDF only on Docsity!

#3 AminoacidD Ratio [D-]/[D]

practice test

PK

pH = b. Henderson (^) Hasselbranch (^) Equation [conjugate base]^ (forweakacid) pH = pKa+log (weakacid]

+log

109 =^2

Ontaking Antilog^

(Base 10)

= 102 =

Amino (^) acid (^) D Ratio (D-]/[D] pK = 4.7 (^) pH= 4.

[

pH=pKa+log

"attac another ex.

= 0 -^100

antilog (base (^) (0)

=

==

In class assignment

units- name abbrev^ concentration millimolar mm 10-

micromolar NM 10

  • b nanomolar nM 10
  • 9 G-alpha-130nm 130 x^ 10-9 => 0.

GTP->5mm 5x

  • 3

     100

=

38,4612 GTP/G-Alpha

1H-bond =-1 kcallmole

  • = logK 0.1= logk

100.7 = k

= k, CoV2 -^ makes^4 bonds^ (12) 2 = 109K 2.81 = 1092

182.81 = k

645 = k

close ans

1284

nM=nanomolar (10-9) A + B (^) - AB

[A] = (B)

=

[AB] = 1 nM

nM=1x10-

[P]

[ABBS^ Seduct^3

[R] Reactant Keg = - 9M

  • 9MXIX109M Keg

= 109 M-

Easy way^ to^ solvethis^ problem: Think of^ what nM

equalsto

nm= 10-9^ mol/L

AtB -^ since it's

positive ans would^ be 10"^ moll 1H-bond =^1 kcal/mole Given= each H bond contributes = 4.2K)Imole / key at^

lower conc." =>

Keq,ahigher

k conc." x = 183

x=^1000 48 =

about 4-5 bonds

not sure

plasma membrane^

= 5.4 nM 3.6 amino (^) acid = 0.54 nM ->>

=^ amino^ acid^ can^ take^10 turns ----- around

0.54x

nM

3.6 x 10 =^36 2um ->^2000 nm

2 => 1000