Is there a simple way to explain quantised conductance? I am biologist and I need to pass a nanotechnology class. I am stuck with this term of quantum resistance. Basically this appears to me as normal Ohm's law, except that takes into account Heisenberg uncertainty principle. Is this an OK way of thinking, or have I misunderstood?
 A: Not quite. Ohm's law concerns electric conduction in the scattering regime, when resistivity develops due to the scattering of electrons on the material's ions/atoms , impurities, defects, etc. In this case the mean free path of an electron in the (semi)conductor is much shorter than the dimensions/length of the conductor itself. 
Quantum conductance characterizes the ballistic regime, when the size of the nanoscale conductor (molecule, nanotube, quantum dot, etc) is much smaller than the electron mean free path. In this case electrons pass through the nanoconductor essentially without any interaction, and resistivity arises from scattering on end (metal) contacts that act largely like walls to a box. The electrons become quantum particles in a box under applied voltage, while conductance becomes independent of the size of the conductor and quantized, with a conductance quantum of $2e^2/h$.
Since you mention the uncertainty principle, you may want to check the simple derivation in this Wikipedia paragraph. Also useful: 
• Wikipedia page on Ballistic Conduction 
• Beginner friendly Tutorial on (nano)Electronic Transport 
Note added after comment
Here is another take at the argument for the conductance quantum: 
Current in a nanoconductor may be viewed as carried through a limited number of discrete energy levels or channels. Consider a single one of these channels. In the absence of coupling to contacts, it corresponds to a single state at energy E. But with contacts and a voltage $V$ applied across the contacts, it undergoes a broadening to a continuous band of width $\Delta E \sim eV$. If an electron takes an average time $\tau$ to traverse the conductor through this broadened level, we can say that it produces a current $I = \text{charge through conductor per unit of time}\sim \frac{e}{\tau}$.  The conductance enabling this current is then $G_0 = \frac{I}{V} \sim \frac{e}{\tau} \frac{e}{\Delta E} = \frac{e^2}{\Delta E\; \tau}$. But the average transit time $\tau$ is on the order of the level's lifetime, that is, the average time it takes an electron to transition to another level. On the other hand, the time-energy uncertainty principle relates energy broadening and lifetime as $\Delta E \;\tau \approx \frac{\hbar}{2}$. Substituting this in the expression for the conductance leaves $G_0 = \frac{2 e^2}{\hbar}$. 
