`
`
`1. Determining molarity of protein in example IA
`
`
`Given that there is approximately 5 milligrams of protein per milliliter in the eluate
`(Ex. 1003 at 21), there is approximately 75 grams of protein in the 15 liters of
`eluate.
`- 5 mg mAb/mL × (15L ×(1000ml/1 L) ×(1 g/ 1000 mg)= 75 g
`
`
`Given that Monomeric IgG proteins have a molecular weight of 150 kiloDaltons
`(kDa) or 150,000 grams per mole, the total molarity of the protein in the eluate can
`be calculated at 0.03 mM.
`-
`(75 g mAb / (150,000 g/mole)) = 0.0005 mol
`-
`(0.0005 mol / 15 L) × (1000 mmol/ mol) = 0.03 mM mAb
`
`
`
`2. Determining molarity of neutralized eluate solution in example IA
`
`
`The eluate of 15 L is produced by using 15-20 L of 25 mM Citrate elution buffer
`with a pH of 3.5. The volume of 2.5 M HCl needed to adjust the pH of the eluate
`to 3.5 is therefore minimal.
`
`15L of 25 mM Citrate contains 375 mmol Citrate.
`- 15 L× 25 mM Citrate = 375 mmol
`
`
`Subsequent adjustment to pH 5.5 requires the addition of 350 mL of 1 M Tris.
`
`350 mL of 1 M Tris contains 350 mmol Tris.
`- 350 mL × (1 L/1000 mL) × 1 M Tris × (1000 mmol/1 mol) = 350 mmol
`
`
`
`Total volume of the pH adjusted eluate is 15.35 L.
`- 15L + (350 mL × (1L/1000 mL)) = 15.35 L
`
`
`Thus, the total molarity of the pH adjusted eluate is 47.2 mM.
`-
`(375 mmol Citrate + 350 mmol Tris)/15.35 L = 47.2 mM
`
`
`
`3. Determining volume of HCl added to adjust pH of eluate to 3.5
`
`
`Assuming eluate pH at 3.7.
`
`
`1
`
`PFIZER, INC., IPR2017-01357, Ex. 1007, p. 1 of 3
`
`
`
`Buffer capacity of 25 mM citrate is extrapolated as 17.5 µEq/mL. Or, in more
`detail, (34 – 16.5 Eq/mL)/(5.0 – 4.0 pH) = (17.5 Eq HCl/mL)/pH from figure
`2A of Anne R. Karow et al., “Buffer Capacity of Biologics-From Buffer Salts to
`Buffering by Antibodies,” (2013) (“Karow”) Ex. 1012, 5.
`
`Buffer capacity of monoclonal antibodies (mAb) is extrapolated as 0.18 µEq/mg.
`Or, in more detail, (2.0 – 0.2 Eq/mL)/(5.0 – 4.0 pH) = (1.8 Eq HCl/mL)/pH
`from figure 9A of Karow for a 10 mg mAb/mL mAb solution. Since buffer
`capacity scales with mAb concentration, based on the data for 10 mg/mL and 100
`mg/mL solutions in figure 9A, we can divide the buffer capacity by the mAb
`concentration to express buffer capacity per milligram mAb: ((1.8 Eq
`HCl/mL)/pH)/(10 mg mAb/mL) = (0.18 Eq HCl/mg mAb)/pH.
`
`The HCl added to bring 25 mM Citrate down from pH 3.7 to 3.5 is 3.5 µEq/mL.
`- 0.2 pH × (17.5 µEq HCl/mL)/pH = 3.5 µEq HCl/mL
`
`Given that there is approximately 5 milligrams of protein per milliliter in the
`neutralized eluate, the HCl added to bring a 5 mg/mL mAb solution down from pH
`3.7 to 3.5 is 0.18 µEq/mL.
`- 0.2 pH × 5 mg mAb/mL × (0.18 µEq HCl/mg mAb)/pH = 0.18 µEq
`HCl/mL
`
`
`The estimated total volume of 2.5 M HCl added to bring the 15 liters of eluate
`containing both 25 mM Citrate and 5 mg/mL protein to pH 3.5 is 22.08 mL.
`-
` (0.18 µEq HCl/mL + 3.5 µEq HCl/mL) × 15 L × (1000 mL/ L) × (1 mol
`HCl/ 1000000 µ Eq HCl)/(2.5 M HCl × ((0.001 mol/mL)/M)) = 22.08 mL
`
`
`Factoring in the HCl, total volume of the pH adjusted eluate (including the 350 mL
`of Tris) would be 15.372 L.
`- 15.35 L + (22.08 mL × (1L/1000 mL)) = 15.372 L
`
`Volume of HCl added as a percentage of the total volume of the pH adjusted eluate
`is less than 0.2%.
`-
`(22.08 mL/(15.372 L × (1000 mL/1 L)) × 100% = 0.144%
`
`
`22.08 mL of 2.5 M HCl adds 55.2 mmol of HCl to the solution.
`-
`((22.08 mL× (1 L/1000 mL)) × (2.5 M × (1000 mmol/1 mol))= 55.2 mmol
`
`
`
`2
`
`PFIZER, INC., IPR2017-01357, Ex. 1007, p. 2 of 3
`
`
`
`Thus, the total molarity of the pH adjusted eluate after factoring in HCl is 50.8
`mM.
`-
`
`(375 mmol Citrate + 350 mmol Tris + 55.2 mmol HCl)/(15 L + 0.350 L +
`0.02208 L) = 50.8 mM
`
`
`
`3
`
`PFIZER, INC., IPR2017-01357, Ex. 1007, p. 3 of 3
`
`
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