Explain this circuit to me
With the negative supply disabled, i.e. with the collector resistor's lower leg grounded, you are driving the BJT in saturation. That's because in this circuit, due to R2 feedback, \$I_C\$ is essentially constant between the two situations. In fact the base divider is very stiff and provides about 1.4V at the base. Since \$V_{BE} \approx 0.7V\$ is essentially constant, the emitter voltage is ~2.1V. Hence the voltage across R2 is always 5V-2.1=~2.9V. Hence the current through R2 is always \$I_E \approx I_C \approx ~9mA\$, regardless of the negative supply.

That current must flow also through R4, giving rise to a ~2.2V drop across it. Sum this 2.2V with the 2.9V across R2 and you end up with less than 1V of \$V_{CE}\$: that means a saturated BJT.

With the negative supply present you have 5V more headroom to be spent across collector and emitter, hence the BJT remains in the active region, where the relation between \$I_C\$ and \$I_B\$ is linear (\$I_C = h_{FE} I_B\$) and \$I_B &lt;&lt;I_C\$. 

In the saturation region \$I_C\$ doesn't depend on \$I_B\$ any longer and this latter is no longer a small fraction of the former.

EDIT (prompted by a comment)

Look at the datasheet of the 2N3906 (what you call \$\beta\$ is called \$h_{FE}\$ on datasheets):



with the negative supply connected the BJT operates in its active region and the relation between \$I_C\$ and \$I_B\$ is approximately linear, as I already said above. If you invert that relation you get:

$$
I_B = \dfrac{I_C}{h_{FE}} \approx \dfrac{9mA}{300} \approx 30 \mu A &lt;&lt; I_C = 9 mA
$$

If you don't understand or know what the different operating regions of a BJT you should review your basics about BJTs.