Boy, this begs the question "how far will this thing talk" and nobody can answer that with any real number.
To get to basics you have to look at free space propagation (loss from transmitter to receiver). I penciled some numbers out, but be aware this is for a theoretical isotropic antenna on both ends. For the frequency listed there are two numbers, one for a distance of 1000 meters and the second of 100,000 meters. The loss is in dB.
154.600 MHz -76.2dB, and -116.2dB
462.600 MHz -85.7dB, and -125.7dB
You will see right off the bat, the lower frequency will "punch" through better than the high frequency.
If you check the difference between the two frequencies at both distances you will find 9.5dB and 9.5dB respectively. This means the higher frequency will always be at a constant lower level than the lower frequency as it moves farther out.
The propagation maps based on the Longley-Rice equation account for major topographic obstructions and frequency angles beyond line of site, but what if you are down at the bottom of a steep canyon trying to call to your buddy that your quad has broken down and you can't get out?
In the pacific northwest where the trees and ferns and moss will more or less suck up a higher frequency, I'd rely on VHF MURS to get out. If, however, this was a rocky canyon in the southwest, UHF GMRS may be a better bet to bounce your way out.
You can play with antenna gains and transmit power, but the deciding factor on communication distance is your frequency and environment.