Transistor base-emitter voltage pulled down

Hi,

I have a question related to the circuit in the attachment down below. This is a circuit where LED's glows in music rhythm received from smartphone speaker (or any other source) that fall on electric mic. When there is no sound signal (or any other vibration near mic), Q1 transistor is on, and other transistors are off because there is not enough voltage to the bases of other transistors. With my pocket oscilloscope, I noticed that when there is a sound wave to the mic, voltage between base and emitter of Q1 is pulled down, thus transistor Q1 is opened (no current from collector to emitter) and other transistors are conducting, so LED's glows. My question is: Why is this so? Why is base pulled low when there is sound wave?

When there is sound, the left side of the C1 goes up and down with the sound, but the average voltage of that point doesn't change. When the left side goes up, so does the right side. However, the right side can't go up much because the base-emitter junction of Q1 conducts so C1 is charged and the voltage across C1 is more positive to the left. When the left side of C1 goes down, so does the right side, which causes Q1 to turn off.

The other way of looking at it is that the ac amplitude will be about the same on each side of C1, but the base-emitter junction of Q1 prevents the peak voltage being above 0.7 V.

Thank you for clear explanation!
I have another question, maybe a silly one. What is a purpose of resistor R1?

The microphone has a built in amplifier. (one transistor amplifier) It needs power.