GATE 2026 BT Q.62: solution and answer

GATE 2026 · BTQ.62 · 2 marks · NAT

It is desired to cultivate 150 mg (dry weight) microbial cells with an empirical formula of $C_{5} H_{7} O_{2} N$ . If the phosphorus requirement for the cells is 20% of the nitrogen requirement (on a weight/weight basis), the minimum amount of phosphorous required to be added to the cultivation medium is ________ mg. (rounded off to one decimal place)

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Answer and solution

ANSWER

The minimum phosphorus to add is about 3.7 mg.

Official answer: 3.6 to 3.8Source: Official answer key, GATE 2026 BT (IIT Guwahati)

Molar mass of the biomass formula $C_{5} H_{7} O_{2} N$ (atomic masses C = 12, H = 1, O = 16, N = 14): $M = 5 (12) + 7 (1) + 2 (16) + 14 = 113 g mol^{- 1}$ .
Nitrogen mass fraction of the cells, $w_{N}$ (mass of N per unit dry mass of cells): $w_{N} = \frac{14}{113} = 0.1239$ .
The minimum nitrogen that must be supplied is the nitrogen that ends up inside 150 mg of cells (elemental balance, assuming all nitrogen supplied is taken up into biomass): $m_{N} = 150 \times \frac{14}{113} = 18.58 mg$ .
The phosphorus requirement is 20% of the nitrogen requirement on a weight/weight basis: $m_{P} = 0.20 \times 18.58 = 3.72 mg$ .
Rounded to one decimal place: $m_{P} \approx 3.7 mg$ . (With precise atomic masses, $M = 113.12 g mol^{- 1}$ and $m_{P} = 3.715 mg$ , which also rounds to $3.7$ .)

KEY IDEA

This is an elemental mass balance. The formula $C_{5} H_{7} O_{2} N$ gives the mass fraction of nitrogen in the cells, $14/113$ . That fraction times the cell mass is the least nitrogen the medium must supply. Phosphorus is then 20% of that nitrogen, by weight.

WATCH OUT FOR

Taking 20% of the cell mass (30 mg) instead of 20% of the nitrogen mass. Another slip is to work on a mole basis, where 0.2 mol P per mol N would give $0.2 \times 150/113 \times 31 \approx 8.2 mg$ . The question says weight/weight, so the 20% applies to the mass of nitrogen.

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GATE 2026 BT Q.62: solution and answer

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