Question

Using the total structure weight (= molecular weight) given in the PDB entry, calculate the molarity of a 0.3 mg/mL solution of a lactate dehydrogenase solution. You must state your chosen PDB entry ID

Answer #1

Lactic Dehydrogenase (LDH) has a total molecular weight of
**140 kDa** and is composed of 4 subunits which are
designated M subunit (muscle) and H subunit (heart). These subunits
may be mixed in any of 5 combinations (M4, M3H1, M2H2, MH3, and
H4).

0.3 mg/mL = 0.3 g/L because: 1 g/L = 1000 mg/1000 mL = 1 mg/mL.

The molar amount of 0.3 g protein in 1 L (1 dm^{3}) of
solution is:

0.3/140000 = 0.00000214285 mol = 2.14 x 10^{-6}
mol/dm^{3}

Therefore, the concentration is 2.14 x 10^{-6}
mol/dm^{3} (= 2.14 x 10^{-6} M).

Molarity of NaOH and molecular weight of the unknown
acid
Part A
Run 1
Run 2
Mass of H2C2O4*2H20 used in
grams
0.2127
0.2124
Moles of H2C2O4*2H2O
(mol)
1.69 *10^-5
1.69*10^-5
Number of protons
available for reaction with OH-
2H
2H
Moles of OH- which
reacted (mol)
2
2
Volume of
NaOH solution used (mL)
18.71
19.15
Molarity of
NaOH soultion (M)
.12
.12
Average
molarity of NaOH
(M)
Part B with unkown
sample
Run 1
Run2
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