Thanks for the link to the schematic.
To answer your question, I don't believe that Q2 in the linked schematic actually does much of anything useful. This device will conduct at all times. It will act as a relatively low value resistor. ( Small relative to the series 39k and 200k resistors.)
The author talks of Q2 functioning as a low leakage diode, but it is not connected correctly to do this.
This sort of error is not uncommon in published circuits.
Since the gate is strapped to the source, the gate cannot be driven negative relative to the source to get this depletion mode device turned off. So the channel resistance will remain small with both high and low control voltages. This will be true as long as the circuit current is much smaller than the transistor's Idss.
The circuit may work fairly well anyway. No diode is actually required. The low impedance of the U2-B op-amp output will absorb any current driven into the gate of Q3 when the control voltage is positive. When the gate of Q3 is driven negative, the gate is reverse biased and the leakage current is small.
The error introduced by the Q2 circuit might be reduced by eliminating Q2 and R14, while increasing the value of R17. A pull down resistor on 7 of U5-B might reduce leakage through Q1 when it is supposed to be off.
I think you have already mentioned good results after adding a large value resistor in series with the Q3 gate terminal. The main drawback to increasing the resistance there is that it makes the sample switch slower to turn on and turn off. For this application, pretty slow is still fast enough.
Ted