How do the pKa and pH indicate to the proper environment of protonation? I have been reading an abstract of a paper for a couple of minutes (reference below) but I cannot understand a piece of it. The abstract:


  The ionization state and $\mathrm pK_\mathrm a$ of the inhibitor 6,8-dimethyl-N5-deazapterin bound to the recombinant human dihydrofolate reductase (rhH2folate reductase) complex with NADPH was determined by a spectrofluorimetric method. The excitation spectra for bound ligand as a function of pH from 6.1 to 9.7 indicated it was the same cationic form as for unbound ligand, which is protonated on N3. However, the lower limit for the $\mathrm pK_\mathrm a$ of the bound form was determined to be 9.1, a value about pH 2.5 higher than that for free ligand, indicating that ligand bound to the enzyme is protonated at neutral pH. The excitation spectra for bound ligand as a function of pH were generated by computer simulation by employing corrections for the pH dependence of the proportion of bound ligand (variable $K_\mathrm d$; ligand-dissociation constant) and taking account of the different $\mathrm pK_\mathrm a$ values for bound and unbound forms. A plot of $K_\mathrm d$ values against pH showed a bell-shaped curve indicating that 6,8-dimethyl-N5-deazapterin bound to rhH2folate reductase. NADPH to form a ternary complex of ionised enzyme with protonated ligand and/or protonated enzyme with unprotonated ligand; the spectrofluorimetric results are consistent with the first alternative.


I made the not-understandable part (for me) in bold. Could some one help me to understand how do the previous sentences help to indicate to this sentence? Maybe I am missing something in the previous sentences.



Jeong, S.-S.; Gready, J. E. Ionization state and pka of pterin-analogue ligands bound to dihydrofolate reductase. Eur. J. Biochem. 1994, 221 (3), 1055–1062. DOI: 10.1111/j.1432-1033.1994.tb18824.x.
The sentence immediately before the bolded one indicates that the $\mathrm pK_\mathrm a$ value of the ligand bound to the enzyme was determined to be $9.1$. It also indicates that the $\mathrm pK_\mathrm a$ value o the free ligand is 2.5 logarithmic units less, i.e. $\mathrm pK_\mathrm a(\ce{Lig}) \approx 6.6$.

The human cell is well-buffered to ensure that the pH stays within a very narrow range. There are enough buffering compounds around, that we never need to use any Henderson-Hasselbalch equation or other difficult calculation methods to determine the protonation state of a given compound. We can simply examine its $\mathrm pK_\mathrm a$ value; if that is lower than the cell’s pH value, the compound is deprotonated. In all other cases, it is not.

The $\mathrm pK_\mathrm a$ value of the free ligand indicates it should be deprotonated in the cell. However, upon binding to the protein with the $\mathrm pK_\mathrm a$ value rising by $2.5$  logarithmic units, the ligand suddenly becomes protonated. This is what the bolded sentence implies.