366 JOURNAL OF THE SOCIETY OF COSMETIC CHEMISTS To determine the equilibrium constant, K, and the amount of cystine per gram of hair, ICy], from the equilibrium concentration of mono-DNP-cystine, [C], eq. 5 may be rearranged to [C] 2 0.5 m - K'[Cy] + [•yy] [C] (eq. 9) Eq. 9 is a simple equation of m expressed in quadratic form of [C] with 1/K'[Cy] and 1/ICy] as constant parameters. Therefore, ICy] and K' (or K) can be determined using a least squares regression method with several values of m and [C] determined exper- imentally. Figure 1 shows the relationship between the equilibrium concentration of mono-DNP-cystine, [C], obtained and the amount of hair per liter of hydrolysate (m). The regression curve of eq. 9 is shown as the solid line in Figure 1 with [Cy] = 6.54 X 10 -4 mole/g dry hair and K' = 3.33 X 10 -3. The half-cystine content in Caucasian virgin hair obtained here (1308 }x mole/g hair) is well within the range of the literature 1.5 .5 0 .5 i 1.5 mono-DNP CYSTINE CONCENTRATION (m mole/L) Figure 1. The relationship between yield of mono-DNP-cystine at equilibrium and concentration of hair in hydrolysate used in disulfide interchange reaction. The closed circles represent experimental data and the curve was calculated from eq. 5 with K' = 3.33 X 10 -3 and ICy] = 6.54 X 10 -4 mole/g hair.

HAIR BIS-DINITROPHENYL CYSTINE INTERCHANGE 367 values (1269- 1650 Ix mole/g hair) (5). The equilibrium constant K for the interchange reaction in strong acidic medium (•9.6N) was determined to be 2.56. DETERMINATION OF THE REACTION RATE CONSTANTS The rate equation for the interchange reaction is d[C] - k•[A][B] - k2[C] 2 dt or (eq. 10) 1 d[C] k2 dt - K'(m[Cy] - 0.5[C]) - [C] • (eq. 11) The solution for eq. 11, i.e, the formation of mono-DNP-cystine (C) as a function of time (t) is E o, z ! o o E .5 o o 5 10 15 20 25 TIME (days) Figure 2. Formation of mono-DNP-cystine as a function of time for various concentrations of hair hy- drolysate from intact hair. The curves were calculated from eq. 12 with K' = 3.3 X 10 -3, k2 = 5.0 hr -•, and ICy] = 6.54 x 10 -4 mole/g hair. Key: (2) = 0.5 g ß -- 0.2 g ß -- 0.05 g of hair in 250 ml of 9.6 N HCI.