Question

Consider a chain reaction with the sequence S1 to S2 to S3 to S4, where S4 is converted to P1 and P1 is converted to P2. The enzymes catalyzing the three reactions S1 to S2, S2 to S3, and S3 to S4 are EA, EB, and EC respectively. Enzyme EA is under positive feedback and enzyme EB is under negative feedback. If enzyme EC is absent, which of the following is true?

• A. S1 accumulates
• B. S2 accumulates
• C. P1 accumulates
• D. P2 accumulates

Hint:

Loss of EC lets S3 rise, which switches off EB by negative feedback; EA keeps feeding S2, so S2 backs up.

Answer 1

The Correct Option is B

Step 1: Map out the chain.
The pathway order is $S1 \to S2 \to S3 \to S4 \to P1 \to P2$, run by enzymes $EA$ (first step), $EB$ (second step), and $EC$ (third step). $EA$ is turned up by positive feedback, $EB$ is turned down by negative feedback.

Step 2: Remove EC and watch S3.
$EC$ normally clears $S3$ forward into $S4$. Take $EC$ away and $S3$ has no exit, so it starts to build.

Step 3: Let the built-up S3 talk to EB.
Because $EB$ answers to negative feedback, and $S3$ is exactly the molecule $EB$ makes, the rising $S3$ signals back and switches $EB$ down. This closes the door between $S2$ and $S3$, so $S3$ stops rising further and $S2$ stops draining away.

Step 4: Check the supply side.
$EA$ has no such brake, in fact it is under positive feedback, so it keeps converting $S1$ into $S2$ without slowing.

Step 5: See where the traffic jam forms.
$S2$ keeps arriving from $EA$ but cannot leave through the now silenced $EB$. That mismatch, still coming in, blocked from going out, is what makes $S2$ the molecule that piles up, not $S1$, $P1$, or $P2$.

Step 6: Conclude.
$\boxed{\text{S2 accumulates}}$