Question

Metabolism: If a subject who is 6’ 2” tall and weighs 220 lbs. consumes 1250 ml
of oxygen in a 5 min period, calculate his BMR with the following data. In
addition, calculate his heat production per meter of skin.
xx grams of protein
xxx grams of fat
xxx grams of carbohydrates

Answer 1

Ans:

BMR (kcal/day)= 10 x weight(kg)+ 6.25x height(cm)– 5 xage (years)+s(kcal/day)

Where "s" is +5 for males and -161 for females.

BMR(kcal/day)= 2028 Calories/day

Since rate of energy production = rate of oxygen consumption (1 litre O2 = 20,000 joules)

1cal = 4.184 joules

Therefore

4.184 joules =1cal

20,000 joules (1 litre of O2 consumption) = 20,000/4.184 calories

1000 ml of O2 consumption = 4780.11 calories

Therefore 1280 ml of O2 consumption = 1280 x 4780.11 calories/1000

1280 ml of O2 consumption = 6118.54 calories

Now 6’ 2” tall subject = 1.88 meters

Heat production = rate of heat production on O2 consumption

Since 1.88 meter tall subject produces = 6118.54 calories

1 meter = 6118.54 calories/1.88

1 meter =3254.54 calories

Therefore heat production per meter of body = 3254.54 calories

Since energy produced on oxidation of 1gram of protein= 4000 calories (Calorific value of protein)

Thus 4000 calories = 1gram

6118.54 calories = 6118.54 calories x 1gram/4000 calories

6118.54 calories = 1.523 grams

1.523 grams of protein

Since energy produced on oxidation of 1gram of fat= 9000 calories (Calorific value of fat)

Thus 9000 calories = 1gram

6118.54 calories = 6118.54 calories x 1gram/9000 calories

6118.54 calories = 0.68 grams

0.68 grams of fat

Since energy produced on oxidation of 1gram of carbohydrate = 4000 calories (Calorific value of carbohydrate)

Thus 4000 calories = 1gram

6118.54 calories = 6118.54 calories x 1gram/4000 calories

6118.54 calories = 1.523 grams

1.523 grams of carbohydrate