> how does the size of the pull-up
> resistor come into play? Why does a
> 10k not cause this, but a 1M would?

10 kOhm and about 100 pF gives you a time constant of 1 uS, so at 12 MHz clock using 2-cycle read port instruction (2 uS), you may even not notice any delays. With 1 MOhm, time constant is 100 uS, and that is many, many more.

> Now where does speed come into play
> here? The more capacitance, the longer
> a change from low to high or
> vise-versa?

Right. The same with resistance, since time constant is equal to resistance times capacitance.

--
Tomas