Question

Our body deals with excess nitrogen by excreting it in the form of urea, NH2CONH2. The reaction producing it is the combination of arginine (C6H14N4O2) with water to give urea and ornithine (C5H12N2O2).

C6H14N4O2 + H2O → NH2CONH2 + C5H12N2O2, [molar masses: 174.2, 18.02, 60.06, 132.2 respectively]. If you excrete 95 mg of urea, what quantity of arginine must have been used?
a) 75.0 mg
b) 80.0 mg
c) 85.0 mg
d) 90.0 mg

(ii) 0.28 g arginine what quantity of ornithine, in moles, must have been produced?
a) 0.0024 mol
b) 0.0032 mol
c) 0.0028 mol
d) 0.0040 mol

Answer 1

To calculate the amount of arginine required to produce 95 mg of urea, and to determine the moles of ornithine produced from 0.28 g of arginine, stoichiometric relationships of the substances involved in the urea cycle are applied.

Step-by-step explanation:

The question involves a stoichiometric calculation based on the biochemistry of nitrogen waste excretion through urea production from the urea cycle. In particular, it relates to the conversion of arginine to urea and ornithine. To find the quantity of arginine used:

1. First, establish the molar mass ratio between urea and arginine.
2. Then, convert the mass of urea excreted to moles.
3. Using stoichiometry, calculate the moles and subsequently the mass of arginine required.

For the second query, given the mass of arginine, use the stoichiometric relationship between arginine and ornithine to find the moles of ornithine produced:

1. Convert the given mass of arginine to moles using its molar mass.
2. As the reaction shows a 1:1 molar ratio between arginine and ornithine, the moles of ornithine produced will be the same as the moles of arginine used.