The given question is incomplete, the complete question is:
Consider a simplified metabolic model to describe nitrogen metabolism in the body. Nitrogen is ingested in the food (F), a fraction of which is actually absorbed (A), the remainder being lost in the feces (W). The absorbed nitrogen is taken up by the muscle to make protein. At small fraction of the stored muscle protein undergoes degradation and the liberated nitrogen (L) is released from the muscle. Of the liberated nitrogen, a fraction is recycled back to the muscle (R), while the rest is lost to excretion (E) in the urine. If the nitrogen ingested in food is 100 g/day, but only 80% of the food is actually absorbed, while the rest goes to waste, what is the amount of nitrogen lost by excretion? Assume steady-state. Of the nitrogen liberated from muscle, 90% is excreted, while the rest is recycled, what fraction of the nitrogen entering the muscle is from the absorbed food? (In other words, what is the ratio A/(A+R)). Again, assume steady-state
Answer:
The amount of nitrogen lost in the process is 20 grams in a day and the fraction of nitrogen entering the muscle is 0.909.
Explanation:
Based on the given question, the amount of nitrogen absorbed is 80 percent of 100 grams, that is,
100*(80/100) = 80 grams in a day.
The concentration of nitrogen lost in the process is,
Nitrogen ingested-nitrogen absorbed = 100-80 = 20 grams in a day.
Let us assume that all the ingested nitrogen is liberated from the muscle. Therefore, 80 grams of nitrogen is liberated. If 90 percent of the nitrogen liberated from the muscle get excreted, then the value of excreted nitrogen will be,
90% of 80 grams = 90/100*80 = 72 grams
Now the nitrogen recycled will be,
Liberated nitrogen - excreted nitrogen = 80-72 = 8 grams
From the absorbed food, the fraction of the nitrogen entering the muscle will be,
80/(80+8) = 80/88 = 0.909.
