Had some issues with a trim pot and would like some input.

[guitchess]

I recently built a guitar effects pedal.  On this occasion, I wanted to try my hand at making a pcb instead of using perf board.  I also decided to forego bread boarding the circuit as it has been built by many people since it is a very popular design.  After completing the build, it would not work.  After much troubleshooting and internet research, I found that the issue was that the trim pot was letting full voltage get to the jfet.  It read 8.4 volts on all three pins.  Of course, I removed the pot and it tested fine.  I ended up getting the effect to work by installing the pot as a voltage divider.  Pin 1 to ground. Pin 2 to the Drain of the FET.  Pin 3 to voltage.

Am I misunderstanding how the pot is connected in the schematic? Is there something else I doing wrong and using a voltage divider is just creating a work around?

I have attached images of the original, my schematic, and my layout.   Any and all input/advice/tips are appreciated. 

Thanks.

[digsys]

Fig2: I assume you mean R7 / Q5. Normally off, D will be at Vcc (8.4V), which is why you see Vcc on all 3 pins.
I can't work out how they expect to drive Q5 in the linear region though, without any other biasing and being AC coupled?
I'm sure you'll get better response when drunken NY members sober up :-)

[ttp]

Double check FET pinouts, I suspect you may have them wrong way around or FET is faulty. I would expect gate to be at 0V for a start, can't understand how are you getting 8.4V gate voltage in original circuit. Looking at DC circuit I would expect gate at 0V, some drain/source current causing voltage across R11 (approx. 0.4V to 0.8V) so the gate is negative relative to source.