Question

4. A 64.3 mg sample of a protein (MW = 58,600) was treated with 2.00 mL of 0.0487 M sodium periodate (NaIO4) to react all of the serine and threonine residues. The solution was then treated with excess iodide ion to convert the unreacted periodate into iodine. IO4- + 3I- + H2O →IO3- + I3- + OH- Titration of the iodine required 823 µL of 0.0988 M thiosulfate. Calculate the number of serine plus threoine residues per molecule of protein. Avogadro’s number = 6.02 × 1023mol-1 2S2O32- + I3- →3I- + S4O62- (Note: that in the presence of excess iodide ion, iodine is rapidly interconverted to triiodide ion; I2 + I- → I3-)

Answer 1

To find the number of serine and threonine residues per protein molecule, the moles of thiosulfate used in the titration of iodine are calculated, which are then used to find the moles of periodate reacted. The number of protein molecules is determined, and from this, the number of residues per protein molecule is calculated to be approximately 37057.

Step-by-step explanation:

Calculating the Number of Serine and Threonine Residues in a Protein:

To calculate the number of serine plus threonine residues per molecule of protein, we need to understand the stoichiometry of the reactions involved. In this case, the titration of iodine with thiosulfate is essential. From the given information, we can calculate the number of moles of thiosulfate used to titrate the iodine.

First, determine the number of moles of thiosulfate (S2O32-):

• Volume of thiosulfate = 823 µL = 0.000823 L

• Concentration of thiosulfate = 0.0988 M
• Moles of thiosulfate = Volume x Concentration = 0.000823 L x 0.0988 mol/L = 8.13 x 10-5 mol

Since two moles of thiosulfate react with one mole of triiodide (I3-), the moles of I3- are half:

• Moles of I3- = 8.13 x 10-5 mol / 2 = 4.065 x 10-5 mol

Each mole of I3- came from one mole of periodate (IO4-), therefore moles of reacted IO4- are the same as moles of I3-.

• Moles of periodate = 4.065 x 10-5 mol

Knowing the moles of periodate reacted allows us to find the moles of serine and threonine residues, as each serine or threonine residue reacts with one mole of periodate:

• Molecules of protein in the sample = 64.3 mg / 58,600 g/mol = 1.097 x 10-9 mol
• Residues per molecule of protein = Moles of periodate / Molecules of protein = 4.065 x 10-5 mol / 1.097 x 10-9 mol = 37057 residues

Therefore, there are approximately 37057 serine and threonine residues per molecule of protein.