GATE 2022 BT Q.57: solution and answer

GATE 2022 · BTQ.57 · 2 marks · NAT

Figure below depicts simplified metabolic and transport reactions taking place in the production of B from A in a cell. The subscript ' $i$ ' refers to intracellular metabolites. $r_{j}$ is the $j^{th}$ reaction flux in $\frac{g}{( g dry mass ) h}$ . Under pseudo-steady-state condition, the following reaction fluxes are available.

$r_{1} = 4$ , $r_{3} = 1$ and $r_{6} = 1$ .

The transport flux of B, $r_{4}$ , is ________ $\frac{g}{( g dry mass ) h}$ .

GATE 2022 BT question 57: A shaded oval labelled 'Cell' represents the cell. — Figure from the GATE 2022 BT paper. It is part of the question.

Answer and solution

ANSWER

The transport flux of B is $r_{4} = 2 g (g dry mass)^{- 1} h^{- 1}$ .

Official answer: 2Source: Official answer key, GATE 2022 BT (IIT Kharagpur)

Pseudo-steady state for an intracellular metabolite means it does not accumulate, so at every internal node $\sum (fluxes in) = \sum (fluxes out)$ . All fluxes $r_{j}$ are in the same unit, $g (g dry mass)^{- 1} h^{- 1}$ , and each arrow converts one metabolite into the next, so the fluxes at a node add directly.
Balance on $A_{i}$ (in: $r_{1}$ ; out: $r_{2}$ and $r_{3}$ ): $r_{1} = r_{2} + r_{3} \Rightarrow r_{2} = 4 - 1 = 3$
Balance on $B_{i}$ (in: $r_{2}$ ; out: the transport flux $r_{4}$ across the membrane and the conversion $r_{6}$ to $D_{i}$ ): $r_{2} = r_{4} + r_{6} \Rightarrow r_{4} = 3 - 1 = 2$
Check on $C_{i}$ : $r_{5} = r_{3} = 1$ . Of the $r_{1} = 4$ units of A taken up, $r_{4} = 2$ leave the cell as B, $r_{5} = 1$ leaves as C, and $r_{6} = 1$ is drawn off to $D_{i}$ . No reaction out of $D_{i}$ is drawn, so no balance is written on it: $r_{6}$ is simply a drain on $B_{i}$ . The totals agree, $4 = 2 + 1 + 1$ , so the network closes.
Therefore the transport flux of B is $r_{4} = 2 \frac{g}{( g dry mass ) h}$ .

KEY IDEA

In metabolic flux analysis at pseudo-steady state, no intracellular metabolite accumulates, so at each branch point the fluxes in equal the fluxes out. Work node by node, from the uptake flux towards the product: $A_{i}$ gives $r_{2} = r_{1} - r_{3}$ , then $B_{i}$ gives $r_{4} = r_{2} - r_{6}$ .

WATCH OUT FOR

Forgetting the second branch at $B_{i}$ and taking $r_{4} = r_{2} = 3$ . Part of $B_{i}$ is diverted to $D_{i}$ by $r_{6}$ , so it never leaves the cell as B. Another mistake is to treat $D_{i}$ as a balanced node. No reaction out of it is shown, so it only acts as a drain.

From the GATE 2022 BT paper. Every question of this paper · Download the full paper · All 24 Growth kinetics questions, with notes

GATE 2022 BT Q.57: solution and answer

Answer and solution

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