pH of an amino acid solution
I think the professor is right. I would solve it like any other acid-base problem. Glycine (G, neutral as zwitterion) can either protonate to HG+; or deprototonate to G-. The mass balance and charge balance equations are:

C$_G$=1.0=[HG+]+[G]+[G-] (Mass balance for glycine)

[H+]+[HG+]=[OH-]+[G-]    (Charge balance)

Given these two equations, and the three for $K_{a1}$, $K_{a2}$ and $K_{w}$, I could solve exactly the system of five equations with five unknowns. But that would be overkill, since we can do some approximations:


Since G is both a (not so) weak acid and a weak base, we can assume that it doesn't dissociate that much to HG+ or G-. So, the predominant form would be G, which (given the $pK_a$'s) is predominant in the 3.5 to 8.5 pH range (aprox, it depends on the errors that we tolerate).
In this pH range, and given the high glycine concentration and $pK_a$ values, [HG+]>>[H+] and [G-]>>[OH-]; so that, from the charge balance:


[HG+]=[G-]

This equality solves the problem since, combining the equations for $K_{a1}$, $K_{a2}$, we get:

$K_{a1}K_{a2}=\frac{[G-][H+]^2}{[HG+]}=[H+]^2$ so $[H+]=\sqrt{K_{a1}K_{a2}}$

$pH=\frac{1}{2}log(pK_{a1}+pK_{a2})=6.0$

You can check that at this pH, [G]=1,[HG+]=$10^{-3.2}$,[G-]=$10^{-3.6}$, so that the approximations [HG+]>>[H+] and [G-]>>[OH-] are valid.

(For more details see 6.7.5 at https://chem.libretexts.org/Textbook_Maps/Analytical_Chemistry_Textbook_Maps/Map%3A_Analytical_Chemistry_2.0_(Harvey)/06_Equilibrium_Chemistry/6.07%3A_Solving_Equilibrium_Problems)