NH₃ and NH ⁺₄ permeability in aquaporin-expressing Xenopus oocytes

Abstract

We have shown recently, in a yeast expression system, that some aquaporins are permeable to ammonia. In the present study, we expressed the mammalian aquaporins AQP8, AQP9, AQP3, AQP1 and a plant aquaporin TIP2;1 in Xenopus oocytes to study the transport of ammonia (NH₃) and ammonium (NH ⁺₄ ) under open-circuit and voltage-clamped conditions. TIP2;1 was tested as the wild-type and in a mutated version (tip2;1) in which the water permeability is intact. When AQP8-, AQP9-, AQP3- and TIP2;1-expressing oocytes were placed in a well-stirred bathing medium of low buffer capacity, NH₃ permeability was evident from the acidification of the bathing medium; the effects observed with AQP1 and tip2;1 did not exceed that of native oocytes. AQP8, AQP9, AQP3, and TIP2;1 were permeable to larger amides, while AQP1 was not. Under voltage-clamp conditions, given sufficient NH₃, AQP8, AQP9, AQP3, and TIP2;1 supported inwards currents carried by NH ⁺₄ . This conductivity increased as a sigmoid function of external [NH₃]: for AQP8 at a bath pH (pH_e) of 6.5, the conductance was abolished, at pH_e 7.4 it was half maximal and at pH_e 7.8 it saturated. NH ⁺₄ influx was associated with oocyte swelling. In comparison, native oocytes as well as AQP1 and tip2;1-expressing oocytes showed small currents that were associated with small and even negative volume changes. We conclude that AQP8, AQP9, AQP3, and TIP2;1, apart from being water channels, also support significant fluxes of NH₃. These aquaporins could support NH ⁺₄ transport and have physiological implications for liver and kidney function.

Appendix

In aqueous solutions of NH₄Cl, NH ⁺₄ is present together with its conjugate base NH₃. Under equilibrium conditions, ${\rm NH}_{4}^{+} = {\rm NH}_{3} \times 10\,^{pK-pH}$ with pK=9.25. Thus, for the external solution:

$$\frac{-{\text{dpH}}} {2.3{\text{d}}t} = \frac{{\text{dH}}^{\text{+}}} {{\text{H}}^{\text{+}}{\text{d}}t} = \frac{- {\text{dNH}}_3 } {{\text{NH}}_3 {\text{d}}t} + \frac{{\text{dNH}}_4^{\text{+}}} {{\text{NH}}_4^{\text{+}} {\text{d}}t}.$$

(4)

Consequently, if the relative rate of influx of NH₃ is larger than that of NH ⁺₄ , the external solution acidifies. With ${\rm dNH}_{3}/{\text{d}}t = - P_{NH3} \times {\rm NH}_{3}$ and ${\rm dNH}_{4}^{+}/{\text{d}}t = - P_{NH4}^{+} \times {\rm NH}_{4}^{+},$ Eq. 4 becomes

$$\frac{-{\text{dpH}}} {2.3{\text{d}}t} = \frac{{\text{dH}}^ +} {{\text{H}}^ + {\text{d}}t} = {\text{P}}_{{\text{NH3}}} {\text{ - P}}_{{\text{NH4}}}^+.$$

(5)

In other words, if P_NH3 is larger than P_NH4+ the external solution will acidify. Note that P_NH4+ incorporates electrical terms.