Metabolism: Energy, Heat, Work, and Power of the Body



Fig. 6.1
Energy flow into and from the body



Metabolic processes can be divided into catabolic and anabolic reactions. In catabolic reactions complex molecules are broken into simple ones, for purposes such as energy usage. In anabolic reactions simple molecules are combined to form complex ones, for purposes such as energy storage.

The body uses food to (1) operate organs, (2) maintain a constant temperature by using some of the heat that is generated by operating the organs (while the rest is rejected), (3) do external work, (4) build a stored energy supply (fat) for later needs, (5) grow to adulthood, and (6) help the fetus grow during pregnancy and then nurse infants. About 5–10% of the food energy intake is excreted in the feces and urine.

We will first consider the basics of the conservation of energy (thermodynamics). Then we will examine the energy content of food and the way it is stored in the body. We will see what the body’s metabolic rate needs to be to perform tasks. (This is technically the catabolic rate.) We will then analyze how the body loses energy as heat. These steps are interrelated in a complex feedback and control mode that is discussed in Chap. 13.

For excellent general discussions about metabolism see [10, 11, 19, 20, 35, 57, 89, 106].


6.1 Conservation of Energy and Heat Flow


Let us briefly review some of the basics of the thermodynamics and heat flow physics that we will use in this discussion.

The First Law of Thermodynamics is essentially the conservation of energy in any process. In reference to the body, it can be stated as


$$\begin{aligned} \Delta U=Q-W, \end{aligned}$$

(6.1)
where $$\Delta U$$ is the change in stored energy, Q is the heat flow to the body, and W is the mechanical work done by the body. The stored energy decreases, $$\Delta U<0$$, when there is heat flow from the body, $$Q<0$$, and work done by the body, $$W>0$$” src=”http://basicmedicalkey.com/wp-content/uploads/2017/06/A114622_2_En_6_Chapter_IEq4.gif”></SPAN>. This type of work is purely mechanical in nature, such as in moving and lifting items. Heat flow includes heat production from the metabolism <SPAN id=IEq5 class=InlineEquation><IMG alt= and heat loss $$(Q_{\mathrm {loss}})$$ from radiation, convection, conduction, and evaporation. We can express $$Q=Q_{ \mathrm {met}}+Q_{\mathrm {loss}}$$, where metabolic heat production is positive and a negative $$Q_{\mathrm {loss}}$$ indicates heat flow away from the body, so


$$\begin{aligned} \Delta U=Q_{\mathrm {met}}+Q_{\mathrm {loss}}-W. \end{aligned}$$

(6.2)
$$Q_{\mathrm {met}}$$ is called the metabolic rate (MR).

Relationships in thermodynamics involve amounts of energies changing in a process at equilibrium and not those changing per unit time, i.e., the kinetics of that process, which involves the rates of energy changes or flows. The study of the metabolism usually involves rates and therefore


$$\begin{aligned} \frac{\mathrm{d}U}{\mathrm{d}t}=\frac{\mathrm{d}Q_{\mathrm {met}}}{\mathrm{d}t}+\frac{\mathrm{d}Q_{\mathrm {loss}}}{\mathrm{d}t}-\frac{\mathrm{d}W }{\mathrm{d}t} \end{aligned}$$

(6.3)
is more appropriate. (We need to be careful about signs. The body increases its energy with terms such as $$\mathrm{d}Q_{\mathrm {met}}/\mathrm{d}t$$ that are positive and loses it with terms such as $$\mathrm{d}Q_{\mathrm {loss}}/\mathrm{d}t$$ that are negative. The amount of heat flowing from the body is $$-\mathrm{d}Q_{\mathrm {loss}}/\mathrm{d}t$$, which is a positive quantity.)

All types of energy have the same units, including heat (often expressed in terms of calories) and work (often expressed in terms of joules). One important conversion between units is


$$\begin{aligned} \mathrm {1\,calorie \;(cal) = 4.184\; joule \;(J)}. \end{aligned}$$

(6.4)
1 kilocalorie (1 kcal = 1,000 cal) is sometimes called 1 Cal, which is also known as a food calorie. The energy content of food is always expressed in terms of these Cal (kcal) units. The relations between the various units of power (energy/time) are given in Table 6.1.

One important concept in the thermal physics of the body is the heat capacity C, which is the energy (or more specifically, the heat) required to raise the temperature T of an object by $$1\,^{\circ }$$C. The heat capacity per unit volume or mass is the specific heat c. The heat capacity is an extensive property of a given object, while the specific heat is an intensive property of a material. (This classification is analogous to that for mechanical properties described in Chap. 4.) If the specific heat is expressed per unit mass, it can be converted to that per unit volume by multiplying it by the mass density $$\rho $$. The heat capacity C is the specific heat (expressed per unit mass) $$\times $$ the total object mass m, so $$C=mc$$.

The temperature rise $$\Delta T$$ of an object with a heat flow Q to the body is


$$\begin{aligned} \Delta T=\frac{Q}{mc}. \end{aligned}$$

(6.5)
For water, $$c_{\mathrm {water}}$$ = 1.0 cal/g-$$^{\circ }$$C = 1.0 kcal/kg-$$ ^{\circ }$$C. Even though the human body contains much water, the average specific heat of the body is a bit less, $$c_{\mathrm {b}}$$ = 0.83 cal/g-$$ ^{\circ }$$C = 0.83 kcal/kg-$$^{\circ }$$C. This means that it takes 83 kcal to raise the temperature of a 100 kg person by $$1^{\circ }$$C. This 83 kcal (83 food calories) is approximately the food energy content of a slice of bread. An obvious question arises: If most of our metabolized energy becomes heat, why does not our body temperature increase by 1.0 $$^{\circ }\text {C} = 1.8\,^{\circ }\text {F}$$ each time we eat and metabolize a slice of bread? We are very fortunate it does not. (The reason is heat loss by the body.)


Table 6.1
Units of power















1 watt (W) = 1 J/s

100 W = 1.43 kcal/min

1 horsepower (hp) = 746 W = 642 kcal/h

1 kcal/min = 69.7 W = 0.094 hp

1 kcal/h = 1.162 W

The heat capacity of an object describes how its temperature changes with time due to heat flow to and from the object. The thermal conductivity K describes how the temperature varies $$(\Delta T)$$ spatially due to the heat flow between different regions that are separated by a distance $$ \Delta x$$. (Conversely, it also describes how much heat flows due to this spatial variation in temperature.) This relation is


$$\begin{aligned} \frac{1}{A}\frac{\mathrm{d}Q}{\mathrm{d}t}=-K\frac{\mathrm{d}T}{\mathrm{d}x}\sim -K\frac{\Delta T}{\Delta x}. \end{aligned}$$

(6.6)
The left-hand side is the amount of heat that flows per unit area A per unit time, and is also called the heat flux. The minus sign indicates that heat flows from hotter regions to colder regions. When there is a well-defined distance $$d=\Delta x$$ between two regions of different but uniform temperature, say due to the thickness of clothing or an air boundary layer, we can define a heat transfer coefficient per unit area $$h=K/d$$ and then


$$\begin{aligned} \frac{1}{A}\frac{\mathrm{d}Q}{\mathrm{d}t}=-h\,\Delta T. \end{aligned}$$

(6.7)
Heat flow due to other mechanisms, such as due to radiation, can often be expressed in terms of (6.6) or (6.7).

One consequence of thermodynamics is that engines that convert chemical energy to heat and use that heat for mechanical work, so-called heat engines, have a limited efficiency to do such useful mechanical work. An ideal heat engine has a maximum efficiency of $$\epsilon =1-(T_{c}/T_{h})$$ when it operates at a temperature $$T_{h}$$ and rejects heat to a lower temperature $$T_{c}$$ (both expressed in K). Humans operate internally at about $$T_{h}$$= 310 K and reject heat to a $$T_{c}\simeq $$ 293 K ambient, so $$\epsilon $$ would be $$5.5\%$$ if we were heat engines. This is much less than the $$\sim $$ $$25\%$$ efficiency of humans converting chemical energy into mechanical work. This is not a contradiction because we use the chemical energy directly to do mechanical work, as seen in Chap. 5, and do not produce heat in an intermediate step.


6.2 Energy Content of Body Fuel


There is some similarity between metabolic oxidation and combustion, even though the body does not “burn” its fuels in oxygen. It is useful to learn about the combustion of these fuels because combustion tells you the maximum amount of energy that is available from breaking and rearranging bonds. Metabolic oxidation is a bit less efficient, as we will see. The combustion energies (enthalpies) are obtained from “bomb” calorimetry in which the materials are burned in heavy-walled vessels. The resulting energies are useful and approximately correct even though the conditions of this bomb—its operation at constant volume and the actual “bomb” temperature—are different than those in the body. For example, many processes in the body occur at constant pressure and not constant volume.

Glucose. The metabolism of glucose is representative of that of carbohydrates. If we start with 1 mol of glucose (180 g, $$6.02 \times 10^{23}$$ molecules) and oxidize it with 6 mol of oxygen molecules (192 g $$=$$ 134.4 L at standard conditions, given there are 22.4 L/mol at 1 atm. and $$0\,^{\circ }$$C), there are 6 mol of carbon dioxide, 6 mol of water (108 g, 108 mL), and 686 kcal of energy produced. This can be represented by


$$\begin{aligned} \mathrm {C}_{6}\mathrm {H}_{12}\mathrm {O}_{6}\mathrm { \;+\; 6 O}_{2} \; \rightarrow \; \mathrm { 6 CO}_{2}\mathrm { \;+\; 6 H}_{2}\mathrm {O \;+\; 686\,kcal}. \end{aligned}$$

(6.8)
The energy produced per mass of fuel is 686 kcal/180 g glucose $$=$$ 3.80 kcal/g glucose. We also define a calorific equivalent , the energy produced per liter of oxygen consumed, which is 686 kcal/134.4 L O$$ _{2} = 5.5$$ kcal/L O$$_{2}$$ here.

We will see that in metabolic oxidation the body combines this 1 mol of glucose with 30–32 mol of ADP and 30–32 mol of the phosphate group P$$_{\mathrm {i}}$$ to form 30–32 mol of the energy storage molecule ATP.

Palmitic Acid.The oxidation of palmitic acid is representative of that of fatty acids. Oxidation proceeds by


$$\begin{aligned} \mathrm {CH}_{3}(\mathrm {CH}_{2})_{14}\mathrm {COOH} + 23 \mathrm {O}_{2} \; \rightarrow \; \mathrm { 16 CO}_{2} + \mathrm {16 H}_{2}\mathrm {O} + 2,397\,\mathrm {kcal}, \end{aligned}$$

(6.9)
so 1 mol (256.4 g) of palmitic acid is burned by 23 mol (515.2 L) of oxygen to form 16 mol (358.4 L) each of carbon dioxide and water, and 2,397 kcal. The energy released is 2,397 kcal/256.4 g $$=$$ 9.3 kcal/g and the calorific equivalent is 2,397 kcal/515.2 L O$$_{2} = 4.7$$ kcal/L O$$_{2}$$. In metabolic oxidation the body uses this 1 mol of palmitic acid to combine 106 mol of both ADP and the phosphate group P$$_{\mathrm {i}}$$ to form 106 mol of ATP.

In comparing glucose C$$_{6}$$H$$_{12}$$O$$_{6}$$ with palmitic acid C$$_{16}$$H$$ _{32}$$O$$_{2}$$, both have the same 1:2 ratio of C and H, so after oxidation, equal numbers of moles of CO$$_{2}$$ and H$$_{2}$$O are formed in both cases. However, palmitic acid and other fats have much less oxygen per C and H than glucose and other carbohydrates, so a smaller mass of them is consumed per mole of CO$$ _{2} $$ and H$$_{2}$$O formed. (The molar mass of palmitic acid is 256 g, and this is much less than the 480 g molar mass of the “equivalent carbohydrate” C$$ _{16}$$H$$_{32}$$O$$_{16}$$.) Alternatively (and equivalently), we can say that fats are less oxidized, are more reduced, and have greater reducing power than carbohydrates.

The heats of combustion for carbohydrates, fats and protein differ slightly for different types of food. For carbohydrates they range from 3.7 to 4.4 kcal/g and they can be averaged and grouped in different food groups with 3.90, 3.95, 4.00, and 4.20 kcal/g respectively for animal products, meats, dairy products; sugars; fruits; cereals, legumes, starches, and vegetables. The heats of combustion per g for fats and oils varies from $$\sim $$9.0–9.5 kcal/g, and these group into 9.25, 9.30, and 9.50 kcal/g respectively for dairy products; cereals, vegetables, and fruits; and meat, fish and eggs. The comparison is more complex for nitrogeneous food components, which are generally called proteins in this field. This includes all compounds containing nitrogen and is usually determined from the analyzed mass of nitrogen in the food. For true proteins, sometimes called proteids in this field, there are 6.25 g of proteids per g of N (16% N), which have heats of combustion of $$\sim $$5.8 kcal/g protein. For other nitrogen-food components, sometimes called non-proteids, the N fraction and heats of combustion are different. For example, for gelatin, creatin (common in animals foods), and asparagin (vegetable foods) the heats combustion are respectively 5.27, 4.27, and 3.45 kcal/g, which is lower than that for proteids. Weighted averaging (using methods addressed in Problem 6.26) leads to average heats of combustion of 5.00, 5.20, 5.65, 5.70, 5.75 and 5.80 kcal/g respectively for the (general) proteins in vegetables (which are 60% proteids); fruits (70% proteids); dairy products and meat; legumes (96% proteids); eggs; and cereals (96% proteids) [68]. The heat of combustion for alcohol is 7.07 kcal/g. Ethanol has a high heat of combustion and, as we will see, provides much caloric input, which explains why it is easy to gain weight by drinking alcohol, such as beer. If your caloric input is mainly from alcohol, you can “starve” to death in the sense that your caloric needs are being met, but not your micronutrient needs, such as vitamins and minerals. (Note that some of the averaged values given here from [68] differ slightly from those listed in Table 6.2 from [35].)

These energies released per unit mass, or heats of combustion, are the maximum energies available, and are really called bomb calorimetry energies because they are often measured in laboratory-controlled explosions. The amount available to the body is somewhat less and is defined as the caloric value. Table 6.2 shows the average caloric values and calorific equivalents of the types of basic body fuels.


Table 6.2
Average caloric content of food














































Food
Net caloric value (kcal/g)
Bomb calorimetry energy (kcal/g)
Calorific equivalent (kcal/L O$$_{\mathbf {2}}$$)
CO$$_{\mathbf {2}}$$ production (kcal/L CO$$_{\mathbf {2}}$$)
RER (L CO$$_{\mathbf {2}}$$/L O$$_{\mathbf {2}}$$)

Carbohydrate
4.02
4.10
5.05
5.05
1.0

Protein
4.20
5.65
4.46
5.57
0.80

Ethanol
7.00
7.10
4.86
7.25
0.67

Fat
8.98
9.45
4.74
6.67
0.71


RER is the respiratory exchange ratio.

Using data from [35]

The caloric values are lower than the bomb calorimetry values because of losses during digestion, which average 2% for carbohydrates, 5% for fats, and 8% for proteins plus an additional loss of 17% of protein energy in the urine [68]. The small losses for ethanol, $$\sim $$2%, are in the urine and exhaled air. With these and related corrections, the average caloric value of carbohydrates and proteins is $$\simeq $$ $$4$$ kcal/g and that of fats is $$\simeq $$ $$9$$ kcal/g, which are known as the Atwater system values [68]. (Each kcal is a food calorie, or Cal, which is called a calorie on food package labels.) Using these values is often a very good approximation, but it is not perfect. Values for the digestible energy from these sources for specific foods are given in [100], along with the food nutrient content. A more general discussion of the energy and nutrient content is in [23], and a more detailed analysis about energy content is in [6, 68]. These caloric values can be compared to common heating and automotive fuels: gasoline, 11.4 kcal/g; coal, 8.0 kcal/g; and pine wood, 18.5 kcal/g. The energy storage densities of different chemical fuels are compared in Table 6.3.


Table 6.3
Energy storage density for chemical fuels
























































































Energy storage fuel
Storage density (J/m$$^{3}$$)
Storage density (J/kg)

ATP

1.4 $$\times $$ 10$$^{8}$$

1.0 $$\times $$ 10$$^{5}$$

H$$_{2}$$ gas, 10$$^{3}$$ atm.
4.9 $$\times $$ 10$$^{9}$$

1.2 $$\times $$ 10$$^{8}$$

Nitroglycerine
1.0 $$\times $$ 10$$^{10}$$

6.3 $$\times $$ 10$$^{6}$$

Glycine (amino acid)
1.0 $$\times $$ 10$$^{10}$$

6.5 $$\times $$ 10$$^{6}$$

Wood
1.1 $$\times $$ 10$$^{10}$$

1.9 $$\times $$ 10$$^{7}$$

Urea
1.4 $$\times $$ 10$$^{10}$$

1.1 $$\times $$ 10$$^{7}$$

Methanol
1.8 $$\times $$ 10$$^{10}$$

2.2 $$\times $$ 10$$^{7}$$

Vegetable protein
2.3 $$\times $$ 10$$^{10}$$

1.7 $$\times $$ 10$$^{7}$$

Acetone
2.4 $$\times $$ 10$$^{10}$$

3.1 $$\times $$ 10$$^{7}$$

Glucose
2.4 $$\times $$ 10$$^{10}$$

1.6 $$\times $$ 10$$^{7}$$

Glycogen (starch)
2.5 $$\times $$ 10$$^{10}$$

1.8 $$\times $$ 10$$^{7}$$

Animal protein
2.5 $$\times $$ 10$$^{10}$$

1.8 $$\times $$ 10$$^{7}$$

Carbohydrate
2.6 $$\times $$ 10$$^{10}$$

1.7 $$\times $$ 10$$^{7}$$

Gasoline
2.8 $$\times $$ 10$$^{10}$$

4.4 $$\times $$ 10$$^{7}$$

Butane
3.0 $$\times $$ 10$$^{10}$$

4.9 $$\times $$ 10$$^{7}$$

Fat
3.3 $$\times $$ 10$$^{10}$$

3.9 $$\times $$ 10$$^{7}$$

Cholesterol (lipid)
4.2 $$\times $$ 10$$^{10}$$

3.9 $$\times $$ 10$$^{7}$$

H$$_{2}$$ solid (10$$^{5}$$ atm.)
7.2 $$\times $$ 10$$^{10}$$

1.2 $$\times $$ 10$$ ^{8}$$

Diamond
1.2 $$\times $$ 10$$^{11}$$

3.3 $$\times $$ 10$$^{7}$$


Note that 10$$^{10}$$ J/m$$^{3}$$ = 2.39 kcal/cm$$^{3}$$ and 10$$^{7}$$ J/kg = 2.39 kcal/g.

Using data from [38]



Table 6.4
Caloric value of 1 rich frosted Entenmann’s$$^\mathrm{TM}$$ donut (in 2005)























18 g fat
$$\times $$9 kcal/g
$$= 162$$ kcal

29 g carbohydrate
$$\times $$4 kcal/g
$$= 116$$ kcal

2 g protein
$$\times $$4 kcal/g
$$= 8$$ kcal

49 g total
  
$$= 286$$ kcal

We can put the caloric value in perspective by developing a standard unit: The Donut. We will use 1 Rich Frosted (i.e., chocolate frosted) Entenmann’s$$^\mathrm{TM}$$ Donut as the standard donut. The package labeling says that a donut contains 18 g of fat, 29 g of carbohydrate, and 2 g of protein (in 2005), so we can estimate the caloric value of each donut, as in Table 6.4. We see that 49 g of the total 57 g mass of the donut has caloric content. Our calculation suggests there are 286 kcal (286 food calories) per donut. The package says there are 280 kcal per donut, which we will take as the caloric content of the standard donut. These values are very consistent with each other because we have used average caloric values for the fat, carbohydrate, and protein. Also, we have rounded off the caloric values and they have provided values of the mass of each body fuel in grams rounded off to the nearest integer (as required by the US FDA (Food and Drug Administration)). (For more on rounding off see Problems 6.11 and 6.12.) Approximately 57% of the calories (162 kcal/286 kcal) come from fat. We will soon determine how much physical exertion is needed to remove the “fattening” consequences of eating a standard donut. Also, we note that during the writing of this second edition (2015), the Rich Frosted Entenmann’s$$^\mathrm{TM}$$ Donut had 300 kcal. We will explore this aspect of our constantly changing world in Problem 6.15. In any case, we will still assume the standard donut has 280 kcal. (Problems 6.13 and 6.14 address other potential standard donuts.)

The energy released per L O$$_{2}$$ (calorific equivalent) is fairly constant for all body fuels, ranging from $$\sim $$4.5 to 5.5 kcal/L O$$_{2}$$. For a “mixed diet” the calorific equivalent can be estimated to be 4.83 kcal/L O$$ _{2}$$ and the CO$$_{2}$$ production to be 5.89 kcal/L CO$$_{2}$$.

The respiratory exchange ratio (RER) (or respiratory quotient (RQ)) is another way to characterize metabolic processes. It is the number of moles of CO$$_{2}$$ produced/number of moles O$$_{2}$$ used, and is a measure of how much carbon dioxide needs to be released in respiration relative to how much oxygen the body needs to bring in by respiration. We see that RER $$=$$ 6 L CO$$ _{2}$$/6 L O$$_{2}$$ $$=$$ 1.0 for glucose oxidation and 16 L/23 L $$=$$ 0.7 for palmitic acid oxidation. It is typically 0.8 for protein oxidation.


6.2.1 Metabolizable Energy


We should be a bit more careful about defining how much of the energy from food is being metabolized and actually used, even though these relatively fine distinctions in body energetics may not be overly significant to us here. The rate of apparently digested energy is the difference between the rate of intake of dietary energy and the rate of loss of energy in the feces. The rate of obtaining metabolizable energy (ME), $$\mathrm{d}(\mathrm{ME})/\mathrm{d}t$$, is the difference between the rate of intake of dietary energy and the sum of the rates of loss of energy in the feces, urine, and combustible gas. This is the actual rate at which energy is being made available to the body [11].


Table 6.5
Components and energy (kcal) of edible parts of common foods






















































































































































































































































Food, serving size
Mass (g)
Energy (kcal)
Carb.$$^{\mathrm {a}}$$ (g)
Protein (g)
Fat (g)
Water (%)

Grains and cakes

Bread, white, 1 slice
25
67
12
2
1
37

Oatmeal, regular, 1 cup prepared
234
145
25
6
2
85

Yellow cake, chocolate frosting, 1 piece
64
243
35
2
11
22

Cheesecake, 1/6 of 17 oz cake
80
257
20
4
18
46

Dairy

Milk, whole, 1 cup
244
150
11
8
8
88

Butter, salted, 1/4 lb stick
113
813
Tr$$^{\mathrm {b}}$$

1
32
16

Cheddar cheese, 1 oz
28
114
Tr
7
9
37

Cottage cheese, 4%, 1 cup
225
233
6
28
10
79

Ice cream, chocolate, 1/2 cup
66
143
19
3
7
56

Eggs, raw, 1 large
50
75
1
6
5
75

Meat and fish

Chicken, meat only, roasted, 1/2 breast
86
142
0
27
3
65

Beef, ground, 79% lean, broiled, 3 oz
85
231
0
21
16
56

Salmon, broiled, 3 oz
85
184
0
23
9
62

Fruits, vegetables, nuts, and oils

Apple, raw, unpealed, 1 whole
138
81
21
Tr
Tr
84

Apricots, raw, without pits, 1 whole
35
17
4
Tr
Tr
86

Apricots, dried, sulfured, 10 halves
35
83
22
1
Tr
31

Orange, peeled, 1 whole
131
62
15
1
Tr
87

Carrots, raw, 7$$\frac{1}{2}$$ in long
72
31
7
1
Tr
88

Potato, baked, with skin
202
220
51
5
Tr
71

French fries, medium portion
134
458
53
6
25
35

Peanuts, dry roasted, 1 cup
146
854
31
35
73
2

Walnuts, 1 cup chopped
120
785
16
18
78
4

Canola oil, 1 cup
218
1,927
0
0
218
0

Beverages

Cola, 12 fl oz
370
152
38
0
0
89

Beer (regular), 12 fl oz
355
146
13
1
0
92

Gin, vodka, whiskey, 86 proof, 1.5 fl oz
42
105
Tr
0
0
64

Wine, red, 3.5 fl oz
103
74
2
Tr
0
89


$$^{\mathrm {a}}$$Carbohydrate.

$$^{\mathrm {b}}$$Trace.

Using data from [41]



Table 6.6
Food content of milk























































Species
Enthalpy of combustion (kcal/g)
Distribution of enthalpy (%)
Dry matter (%)

Fat
Protein
Carbohydrate

Human

0.69
54
7
39
12.4

Cow
0.71
48
26
26
12.4

Goat
0.69
50
22
27
12.0

Horse
0.51
23
22
54
10.5

Seal (northern fur)
5.09
88
11
0.1
61.0


Using data from [11]

Table 6.5 shows the caloric content and components of several types of food. The caloric content per unit mass depends on the relative amounts of carbohydrates, proteins, and fats, and ethanol for alcoholic drinks, and the amounts of water and “ash.” For comparison, our standard donut has 280 kcal, or 280 kcal/57 g $$=$$ 4.91 kcal/g. Table 6.6 shows the enthalpy of combustion for different types of milk; this constitutes consumed and not necessarily metabolizable energy. The enthalpy available from fat, protein, and carbohydrates is quite different for each source. With a 42% fat/18% protein diet, about 89.3% of the consumed food is typically metabolizable—meaning that this is the fraction of the energy content of food that is metabolizable energy. About 5.8% of the consumed energy is lost in the feces, 4.5% in the urine, and 0.4% lost as methane. The difference between the energy content using the Atwater factors and the consumable energy can be much higher than this in some foods. For example, it is $$\sim $$30% for almonds [76].

We spend relatively little time and energy in metabolizing food because of its high caloric value and digestibility (and so we have much time to engage in activities other than eating and digesting food). We often affect both by food processing, so the true caloric value or ME of a food depends not only on its heat of combustion and the metabolic losses described earlier, but on how we process it before consuming it [100]. Processing of foods by cooking them increases our caloric and nutrient intake after we eat them compared to foods that are eaten raw [14, 21, 56]. In some cases, caloric intake can be increased compared to raw foods by mechanical processing, such as by pounding, grinding, cutting the food into small pieces. The energy cost of metabolizing food decreases by $$\sim $$10% when it is processed into smaller components pieces (such as processing peanuts into peanut butter) [14, 56]. Cooking tubers (plant structures that expand to store energy, such as potatoes (stem tubers) and sweet potatoes and yams (root tubers)) converts starches from semicrystalline structures to amorphous materials that are easily hydrolyzed to sugars. Cooking meat increases caloric intake because it denatures proteins, making them more digestible, and eases the separation of muscle fibers, increasing the area exposed for digestive activities. Similarly, the digestibility of cooked eggs ($$\sim $$93%) exceeds that of raw eggs ($$\sim $$58%) [31, 32]. Cooking also reduces the energy needed to fight diseases that could be caused by eating some raw foods. The way food processing affects its mechanical properties is described in Chap. 4.

The average retention time of food in a human’s digestive track is 46 h; the first appearance of the residues of food appears in about one third this mean time and the last in about four times this time.

Very roughly 10 kcal of energy are needed to produce, ship and process each 1 kcal of food energy (and this amount is very different for different types of food) [56, 80].


6.2.2 Energy Storage


Energy can be retained or secreted (as milk ) by the body as the enthalpy of tissues R (which includes fat, protein, and CHO). (The enthalpy (or heat content) is the maximum thermal energy that is obtainable at constant pressure.) The rate at which enthalpy is retained $$\mathrm{d}R/\mathrm{d}t$$ is the difference between the rate of metabolizable energy input andheat production through the metabolism


$$\begin{aligned} \frac{\mathrm{d}R}{\mathrm{d}t}=\frac{\mathrm{d}(\mathrm{ME})}{\mathrm{d}t}-\frac{\mathrm{d}Q_{\mathrm {met}}}{\mathrm{d}t}. \end{aligned}$$

(6.10)
( This ignores lactation.) We combine this with (6.3) to give


$$\begin{aligned} \frac{\mathrm{d}R}{\mathrm{d}t}=\frac{\mathrm{d}(\mathrm{ME})}{\mathrm{d}t}+\frac{\mathrm{d}Q_{\mathrm {loss}}}{\mathrm{d}t}-\frac{\mathrm{d}W}{\mathrm{d}t}-\frac{ \mathrm{d}U}{\mathrm{d}t}. \end{aligned}$$

(6.11)
When food is not eaten $$(\mathrm{d}(\mathrm{ME})/\mathrm{d}t=0)$$, $$\mathrm{d}R/\mathrm{d}t$$ is negative and its magnitude is the heat of catabolism of body tissues. As more food is eaten, $$\mathrm{d}R/\mathrm{d}t$$ increases, and less stored (or retained) energy is used and there is increased heat production; this increased heat production is the specific dynamic or thermogenic effect of food. When energy retention is zero over several days (so no net stored energy is used or formed and $$\mathrm{d}R/\mathrm{d}t$$ averages to zero), the input dietary energy ME is that required for maintenance . For greater values of ME, energy is stored in the body.

Table 6.7 shows the results of calorimetric experiments for people who are fasting (for whom stored energy in proteins and fats is turned into heat), eating who do no exercise (who have net storage of body fat), and eating who do exercise (for whom a small amount of stored energy is turned into heat, even though there is more food intake).


Table 6.7
Calorimetric experiments each averaged over several runs (in kcal/day)

















































Energy
Eating, without exercise
Eating, without exercise (bicycle)
Fasting

Intake energy
2,659
4,340
0

Feces energy
107
176
0

Urine energy
134
138
105

Change in body protein
$$-$$16
$$-$$57
$$-$$463

Change in body fat
176
$$-$$484
$$-$$1,892

Heat produced
2,270
4,554
2,187

Discrepancy
12
$$-$$13
$$-$$63


Using data from [11], from [6]



A114622_2_En_6_Fig2_HTML.gif


Fig. 6.2
Energy retention and heat production in the body versus food intake (based on [11])

Figure 6.2 shows the retention of energy and the heat production (thermogenic effect) versus food intake. A fraction of the metabolizable energy from food intake is retained in the body, K, and a fraction is lost as heat, $$1-K$$. The slope of the retention curve k is


$$\begin{aligned} k=\frac{\mathrm{d}R/\mathrm{d}t}{\mathrm{d}(\mathrm{ME})/\mathrm{d}t} \end{aligned}$$

(6.12)
and the slope of the heat production curve is $$1\,-\,k$$. K is an averaged value of k. Because the curves in Fig. 6.2 are not linear, k and K are usually not equal. k , also called the efficiency of the utilization of metabolizable energy, varies with the amount of food consumed. Below maintenance, k is $$\simeq $$ $$0.90$$—where it represents the efficiency of using stored energy—and above it is $$\simeq $$ $$0.75$$—where it really represents the efficiency of forming stored energy. This means that below maintenance only about 10% of food goes to heat and above it about $$\simeq $$25% goes into heat production—and so $$\simeq $$75% goes into making us fatter (unless we are building up our muscles). These numbers are for average diets. For carbohydrates, $$k = 0.94$$ below maintenance and 0.78 above it. For fats, $$k = 0.98$$ below maintenance and 0.85 above it. For proteins, $$k = 0.77$$ below maintenance and 0.64 above it. The body is more efficient in using stored energy than in depositing fat and proteins. Proteins are metabolized less efficiently than are carbohydrates and fat [11].

We have seen how metabolized carbohydrates, fats, and proteins contribute to our energy balance. However, the details of how each is used by the body is different [56], and so, aside from providing the energy you need, what you eat makes a difference. Proteins are used building and maintaining tissues but usually not directly as a fuel. Some fat is stored in the liver, some is burned, some is stored in muscles as glycogen, as we will soon see, and the rest is stored in fat cells. These cells are throughout the body under your skin (subcutaneous fat), in muscles and other organs, such as belly fat (visceral fat), which is thought to be particularly bad for one’s health. Glucose and fructose are important basic products in the metabolism of carbohydrates. Glucose can come from the decomposition of starches, and sucrose (table sugar) and lactose (milk sugar), which are both one half glucose. It is used directly for fuel or for glycogen, and insulin converts excess levels of blood glucose to fat (see below). Another common basic sugar is fructose, which is paired with glucose in sucrose (table sugar). If there is too much fructose to be metabolized by the liver, the liver converts the rest into fat, much into triglycerides (see Chap. 8) that are stored in the liver or released into the blood.

The glucose levels in your blood can become too high when you eat large quantities of foods containing much glucose or components that easily decompose into glucose (such as starches); these foods often called high glycemic index foods. Moreover, carbohydrates are metabolized even faster and glucose is released faster into the blood stream more after you eat many processed foods (that have been divided into smaller pieces, have had fiber removed from them, and so on). The pancreas responds to lower glucose levels that are too high by producing insulin, which converts excess glucose into fat for fat cells, much of it forming as visceral fat. Sometimes when glucose levels are very high the insulin level overshoots, and too much glucose is removed and you feel hungry. With long-term high levels of insulin production, the various stages of type 2 diabetes can begin and progress, first with the body becoming less sensitive to insulin and then with insulin production decreasing [56]. When blood glucose and glycogen levels are too low, fat is released for energy use. This is controlled by a range of hormones that are controlled by a variety of factors, including stress and exercise. The existence and subsequent release of visceral fat is thought to increase triglyceride levels. (See Chap. 8 for more on this and on foods that may help contribute to the potential for heart disease (such as red meat) or lessen it (fish, nuts).)

This leads us to a discussion about how energy is locally stored and used.


6.3 Energy Storage Molecules



6.3.1 How ATP Is Produced and Used as an Energy Source


Catabolism. ATP, adenosine triphosphate (a-duh’-nuh-seen), is the basic unit of energy storage in the body and it enables the rapid release of energy. Why does the body convert food fuel to ATP and not directly oxidize carbohydrates, fatty acids, and proteins? The use of ATP is more controllable. Also, the unit of energy provided by ATP is small enough to be useful.

A114622_2_En_6_Fig3_HTML.gif


Fig. 6.3
The structures of $$(\mathbf {a})$$ ATP and $$(\mathbf {b})$$ ADP, showing that ATP has one more inorganic phosphate group P$$_{\mathrm {i}}$$ than does ADP. From left to right are the $$(\mathbf {a})$$ three or $$(\mathbf {b})$$ two phosphate groups, the five-carbon sugar, ribose, and adenine, which is the double-ringed structure. The two unstable, high energy bonds linking the phosphate groups in ATP and the one similar unstable bond in ADP are denoted by arrows


A114622_2_En_6_Fig4_HTML.gif


Fig. 6.4
Schematic of the processes involved in the resynthesis of ATP from ADP and inorganic phosphate by anaerobic glycolysis in the cytosol (the intracellular fluid) and aerobic metabolism (or respiration) in the mitochondrion (based on [20, 66, 75])

The structure of ATP is shown in Fig. 6.3. It consists of a five-carbon sugar, ribose, that is linked to the aromatic base, adenine—forming adenosine, and three phosphate groups. The two bonds linking the phosphate groups are unstable, high-energy bonds; the leftmost interphosphate bond in Fig. 6.3 splits in the hydrolysis of ATP


$$\begin{aligned} \mathrm {ATP+H_{2}O} \; \rightarrow \; \mathrm {ADP} \; + \; \mathrm {inorganic \; phosphate + energy}, \end{aligned}$$

(6.13)
where ADP is adenosine diphosphate. This is the basic catabolic process for energy release, and was discussed as the driving force in the myosin power stroke in generating force in muscles in Chap. 5. The energy released, or more precisely the free energy, ranges from 7 to 14 kcal/mol of ATP, depending on conditions (see Problem 6.32). Under typical cellular conditions it can be $$\simeq $$12–14 kcal/mol [55, 63, 75].

Anabolism. After hydrolysis, ADP needs to be combined with a phosphate group to reform ATP for later use. On the average, each ATP molecule is recycled this way roughly every two minutes (see Problem 6.38). How does the body use food sources like glucose to do this? This occurs by a series of chemical steps that can proceed to a limited extent without oxygen (anaerobic glycolysis) and to a greater extent with oxygen (aerobic metabolism or respiration) [20, 35, 57, 75, 106]. This utilizes glucose in the blood stream and glycogen, (C$$_{6}$$H$$_{12}$$O$$_{6}$$)$$_{n}$$—a branched-chain polymerized sugar consisting of glucose molecules as monomers linked together with glycosidic (oxygen) bonds. Glycogen is stored in muscle cells, where it is used directly, and in the liver, where is it broken down into glucose by glycogenolysis, which is then delivered by the blood to the other cells. In aerobic metabolism, the products of the anaerobic steps are further metabolized (in the presence of oxygen) to complete the metabolism of stored glucose by a complex series of steps collectively called the Krebs cycle and the electron transfer system (ETS); they are shown in Fig. 6.4 and described in a bit more detail later. The overall result is that in aerobic metabolism 1 mol of glucose can produce approximately 30–32 mol of ATP (depending on the details of the membrane shuttle mechanism). (In fact, 1 mol of carbohydrates from muscle glycogen can produce approximately 31–33 mol of ATP, but energy is used in forming glycogen from glucose; we will consider only the metabolism of glucose.) Of these, 2 mol are produced by anaerobic processes and 28–30 additional moles are produced when there is sufficient oxygen. The bottom line is that the hydrolysis of these, say for now, 30 mol of ATP provided from glucose


$$\begin{aligned} \mathrm {30\,mol \;ATP/mol \;glucose }\times \mathrm { 14.0\,kcal/mol \;ATP = 420\,kcal/mol \;glucose}.\nonumber \\ \end{aligned}$$

(6.14)
This should be compared to the energy from the combustion of glucose


$$\begin{aligned} \mathrm {180\,g/mol \;glucose }\times \mathrm { 3.8\,kcal/g \;glucose = 686\,kcal/mol \;glucose}. \end{aligned}$$

(6.15)
This means that the efficiency of the body’s usage of glucose to form available energy in the form of ATP is 420 kcal/686 kcal $$=$$ 61%; for 32 ATP formed by glucose, this gives 448 kcal/686 kcal $$=$$ 65%. This efficiency range of $$\simeq $$61–65$$\%$$ is for conditions in muscle cells. For other specific conditions the free energy of ATP can be a bit lower. (This is explored in Problems 6.32 and 6.33.) In any case, this efficiency is pretty good, although it is not perfect. (It is much better than the efficiency of typical heat engines, $$\sim $$10–20$$\%$$.) The overall efficiency of using glucose to do mechanical work is much lower, because of other efficiency factors that we will discuss later.


6.3.2 How ATP Is Actually Used by the Body


ATP is the ultimate source of energy for muscular motion, but it is not always the locally stored source. The body has a mechanism for using it directly, and then several levels of mechanisms for transferring energy from other molecules to the formation of ATP (from ADP), which is then used directly. There are four levels of steps, each of which can be used for successively longer times, although at successively lower levels of activity (Table 6.8) [35, 66, 75].


Table 6.8
Estimated power and energy available from the body, for a 70 kg man with 30 kg of muscle, assuming 10 kcal/mol of ATP
























System
Maximum power (moles of ATP/min)
Maximum capacity (total moles of ATP)

Phosphagen (ATP-PC) system
3.6
0.7

Anaerobic glycolysis
1.6–2.5
1.2

Aerobic metabolism from glycogen
1.0
90.0


Using data from [35]

Step 1. Normally there is enough ATP in living skeletal muscles to supply energy for about 8 twitches. This may be enough for about 3 s or for about half of a 50 m dash. The energy comes from (Fig. 6.3)


$$\begin{aligned} \mathrm {ATP + H}_{2}\mathrm {O} \; \rightarrow \; \mathrm {ADP + P_{\mathrm {i}} + H + 14\,kcal/mol}. \end{aligned}$$

(6.16)
Step 2. Now more ATP is needed. It is resynthesized from the local phosphocreatine (or creatine phosphate) (PCr) (Fig. 6.5) reservoir by


$$\begin{aligned} \mathrm {ADP + PCr} \; \rightarrow \; \mathrm {ATP + Cr}, \end{aligned}$$

(6.17)
where Cr is creatine. This reaction is driven strongly to the right because it has an equilibrium constant $$>$$” src=”http://basicmedicalkey.com/wp-content/uploads/2017/06/A114622_2_En_6_Chapter_IEq233.gif”></SPAN> <SPAN id=IEq234 class=InlineEquation><IMG alt=$$20$$ src=. Muscle has enough PCr to provide ATP for about $$\sim $$ $$100$$ twitches, or for the first 50 m ($$\sim $$8–10 s) of a sprint.

A114622_2_En_6_Fig5_HTML.gif


Fig. 6.5
The structures of $$(\mathbf {a})$$ phosphocreatine (PCr) and $$(\mathbf {b})$$ creatine (Cr), showing that PCr has an inorganic phosphate group P$$_{\mathrm {i}}$$. The unstable, high-energy bond linking the phosphate group in PCr is denoted by an arrow. The enzyme that catalyzes the breakdown of PCr to form ATP, which is the transfer of the phosphate group from PCr to ADP to form ATP (6.17), is called creatine kinase

Together, steps 1 and 2 (reactions (6.16) and (6.17)) constitute the phosphagen system or the ATP-PC system. It is the source of the highest peak power, with about 4 mol ATP used by a person every minute. It can be used for short times, 8–10 s, and at most for 100 m dashes or for limited jumping. PCr can be reformed from Cr and P$$_{\mathrm {i}}$$ by using ATP itself. For intense levels of exercise, this can occur only after the activity, and usually occurs by aerobic metabolism.

When more energy is needed than can be supplied by the phosphagen system alone, it is produced from glycogen that is stored in the muscle and from the metabolism of glucose and fatty acids in the blood stream. Two high-energy molecules are produced as intermediates during this formation of ATP (1) NADH (from NAD$$^{+}$$, nicotinamide adenine dinucleotide, a derivative of the vitamin niacin) and (2) FADH$$_{2}$$ (from FAD, flavin adenine dinucleotide, derived from riboflavin, a B vitamin). The modes of using these energy sources to form ATP are different for light and heavy activity.

Step 3. During “heavy” activity people need ATP fast and there is not enough oxygen for aerobic metabolism, so only anaerobic glycolysis (the Embden–Meyerhof pathway) occurs. One mole of glucose-1-phosphate ( obtained from 1 mol of carbohydrates from glycogen) can convert only 3 mol of ADP to ATP. In this process 1 mol of ATP is used to make fructose-1, 6-diphosphate from fructose-6-phosphate (which is produced from glucose-1-phosphate), and 2 mol of ATP are formed in each of the two subsequent steps of glycolysis, leading to a net of 3 mol of ATP. Metabolism of blood glucose yields a net of only 2 mol of ATP because it takes 1 mol of ATP to convert it into glucose-6-phosphate (see Problem 6.35). This all occurs in the intracellular fluid in the muscle cell (the cytosol).

This degradation of the glycogen or glucose is only partial, and without sufficient oxygen the potential of producing the other 28–30 mol of ATP by the other steps in aerobic metabolism is totally lost. At the end of anaerobic glycolysis, energy is still stored in the 2 pyruvic acid molecules (C$$_{3}$$H$$_{4}$$O$$_{3}$$) (pyruvate ions in solution) formed from each glucose molecule, which are converted to 2 lactic acid molecules (C$$_{3}$$H$$_{6}$$O$$_{3}$$) (lactate ions in solution) if they cannot be used in the Krebs cycle in aerobic metabolism. In addition, 2 NADH molecules are formed per glucose molecule in anaerobic glycolysis, and the energy from these high-energy molecules is also wasted because they are not metabolized by the ETS in aerobic metabolism (see Fig. 6.4). (However, the NADH can remain until the oxygen debt is recovered.)

The advantage of this anaerobic glycolysis or glycogen-lactic acid system is that it provides a medium level of power, $$ \sim $$2.5 mol ATP/min, which is $$\sim $$ $$60\%$$ of that of the phosphagen system. It provides power for about 600 twitches, for an intermediate time $$ \sim $$1.3–1.6 min—enough for a 400 m run. (The maximum anaerobic peak power is 2.1 hp for men and 1.7 hp for women. These values decrease dramatically after 25 years of age. ) Anaerobic glycolysis has disadvantages. In addition to being inefficient, it produces lactic acid, which causes discomfort from acidosis, and causes fatigue. Because the pyruvate-lactate reaction is reversible, the lactic acid from anaerobic glycolysis can be converted to form pyruvate when oxygen is available during the recovery from heavy exercise (but primarily in the liver and not the muscle).

Step 4. During “light” exercise there is enough oxygen for aerobic metabolism to complete the oxidation of the carbohydrates. Aerobic metabolism consists of two complex processes in addition to anaerobic glycolysis: the Krebs cycle (which is also known as the tricarboxylic acid (TCA) cycle or the citric acid cycle) and the electron transfer system (ETS) (which is also known as the cytochrome pathway or oxidative phosphorylation) (Fig. 6.4). They both occur in the cell mitochondrion. Only the ETS directly requires oxygen, but without oxygen even the benefits of the Krebs cycle are lost due to back reactions. (By the way, Hans Adolf Krebs shared the Nobel Prize in Physiology or Medicine in 1953 for his discovery of the citric acid cycle.)

The 2 mol of NADH from anaerobic glycolysis are shuttled in through the mitochondrion membrane. Two moles of pyruvate from anaerobic glycolysis serve as the starting material for the Krebs cycle. Before entering this cycle, each pyruvate produces an acetyl CoA molecule, which enters the Krebs cycle, and 1 NADH, so 2 acetyl CoA and 2 NADH are produced per glucose molecule. For each pyruvate, the Krebs cycle produces 1 ATP, 3 NADH, and 1 FADH$$_{2}$$, along with 2 CO$$_{2}$$ molecules, which are waste products, and so for each glucose molecule the Krebs cycle produces 2 ATP, along with 6 NADH and 2 FADH$$_{2}$$ molecules.

This means that for each mole of glucose, a total of 10 mol of NADH and 2 mol of FADH$$_{2}$$ enter the ETS, where they are converted to ATP and H$$_{2}$$O. (Alternatively, if the 2 mol of NADH from anaerobic glycolysis shuttled through the mitochondrion membrane are cycled through the membrane as 2 mol of FADH$$_{2}$$, a total of 8 mol of NADH and 4 mol of FADH$$_{2}$$ enter the ETS.) There are several series of alternative pathways in which these high-energy molecules are converted to ATP, which lead to effectively 2.5 ATP per NADH and effectively 1.5 ATP per FADH$$_{2}$$, or $$2.5 \times 10+1.5 \times 2=28$$ mol of ATP. There are also the 2 mol of ATP directly from anaerobic glycolysis and the 2 mol of ATP directly form the Krebs cycle, for a total of 32 mol of ATP. Overall, a total of 30–32 mol of ATP are produced from 1 mol of glucose (depending on the membrane shuttle mechanism). (This estimate of 30–32 mol of ATP per mole of glucose is cited in more current work [57, 75]; older estimates concluded that of 36–38 mol ATP are produced per mole of glucose [55].)

ATP is also formed by the oxidation of fatty acids and some amino acids and proteins. (Fats and protein can be metabolized only in the presence of oxygen.) The role of protein metabolism is very minor during rest and does not contribute more than 5–10$$\%$$ of the total energy supply during normal exercise. Each fatty acid molecule is activated using 2 ATP molecules, and is then metabolized in successive passes through a metabolic cycle within the mitochondrion in which two carbon atoms are lost in each pass to produce a molecule of acetyl-CoA, NADH, and FADH$$_{2}$$ by $$\beta $$ oxidation, until acetyl-CoA is left. For the 16-carbon chain palmitic acid, there are seven complete passes and so one mole of palmitic acid produces 8 mol of acetyl-CoA (including the one remaining after $$\beta $$ oxidation), 7 mol of NADH, and 7 mol of FADH$$_{2}$$. Each mole of acetyl-CoA is then oxidized in the Krebs cycle to give 3 mol of NADH, 1 mol of FADH$$_{2}$$, and 1 mol of ATP. The NADH and FADH$$_{2}$$ enter the ETS system. For palmitic acid, this means that 31 mol of NADH and 15 mol of FADH$$_{2}$$ enter the ETS system. Overall, $$2.5 \times 31+1.5 \times 15=100$$ mol of ATP are formed in the ETS. Including the other 8 mol of ATP, 108 moles of ATP are formed per mole of palmitic acid. Including the 2 ATP moles used for activation, a net of 106 moles of ATP are formed [75].

A healthy man with proper training can provide $$\sim $$50 mL oxygen/kg-min. This aerobic system delivers the lowest peak power, 1 mol ATP/min $$\sim $$ $$25\%$$ of the phosphagen system, but for a very long time. There is enough glycogen for $$\sim $$ 10,000 twitches. It is useful for long-distance, endurance activities, including marathon running. The glycogen in the muscle is depleted after several hours ($$\sim $$1.5–4.0 h) of this level of activity. ( Glucose in the blood can also be converted into pyruvic acid and then be used in the cell mitochondria in the presence of oxygen to form more ATP. This is an efficient, long-term process.)


Table 6.9
Estimated energy available from the body (per kg and also, in parentheses, total), for a 70 kg man with 30 kg of muscle, assuming 10 kcal/mol ATP










































System
Muscular amounts (in mmol/kg muscle$$^{\mathrm {a}}$$) (total)
Useful energy (in kcal/kg muscle) (total)

Phosphagen (ATP-PC) system

   ATP
4–6 (120–180)
0.04–0.06 (1.2–1.8)

   PC
15–17 (450–510)
0.15–0.17 (4.5–5.1)

   Total: ATP + PC
19–23 (570–690)
0.19–0.23 (5.7–6.9)

Anaerobic glycolysis

   ATP Formation
33–38 (1,000–1,200)
0.33–0.38 (10.0–12.0)

Aerobic metabolism

   From stored glycogen
13–15 g (400–450)
  

   ATP formation
2,800–3,200 (87,000–98,000)
28–32 (870–980)


$$^{\mathrm {a}}$$Unless otherwise noted.

Using data from [35]



Table 6.10
Body stores of fuel and energy, for a 65 kg (143 lb) person with 12$$\%$$ body fat











































 
Amount (g)
Energy (kcal)

Carbohydrates

   Liver glycogen
110
451

   Muscle glycogen
500
2,050

   Glucose in body fluids
15
62

   Carbohydrates total
625
2,563

Fat

   Subcutaneous and visceral
7,800
73,320

   Intramuscular
161
1,513

   Fat total
7,961
74,833


Using data from [106]



Table 6.11
Energy systems used in sports





























































Mostly phosphagen system

   100-m dash

   Jumping

   Weightlifting

   Diving

   Football dashes

Phosphagen and glycogen–lactic acid systems

   200-m dash

   Basketball

   Baseball home run (running around the bases)

   Ice hockey dashes

Mainly glycogen–lactic acid system

   400-m dash

   100-m swim

   Tennis

   Soccer

Glycogen–lactic acid and aerobic systems

   800-m dash

   200-m and 400-m swim

   1,500-m skating

   Boxing

   2,000-m rowing

   1,500-m/1-mile run

Aerobic systems

   10,000-m skating

   Cross-country skiing

   Marathon run (26.2 miles, 42.2 km)

   Jogging


Using data from [45]

Negligible oxygen is stored in the body. It must be brought in by the lungs and transferred to the blood in the arteries continuously. The oxygen is transferred to the cells, leaving oxygen-depleted blood in the veins. We can track the conservation of oxygen molecules in the body. The rate of body consumption of O$$_{2}$$, $$\mathrm{d}V_{\mathrm {O}_{2}}/\mathrm{d}t$$ equals the product of the cardiac output (blood flow rate, $$Q_{\mathrm {t}}$$, see Chap. 8) and the difference in the oxygen content (oxygen partial pressure) in the arteries and veins, $$p_{\mathrm {a}}-p_{\mathrm {v}}$$:


$$\begin{aligned} \frac{\mathrm{d}V_{\mathrm {O}_{2}}}{\mathrm{d}t}=Q_{\mathrm {t}}(p_{\mathrm {a}}-p_{\mathrm {v}}). \end{aligned}$$

(6.18)
During aerobic exercise, $$\mathrm{d}V_{\mathrm {O}_{2}}/\mathrm{d}t$$ increases linearly with $$Q_{ \mathrm {t}}$$. For a person with average fitness, the maximum oxygen use and blood flow rates are $$(\mathrm{d}V_{\mathrm {O}_{2}}/\mathrm{d}t)_{\mathrm {max}}\approx $$ 2.8 L/min (L of oxygen gas at an atmosphere) and $$(Q_{\mathrm {t}})_{\mathrm {max}} \approx $$ 19 L/min, and for a highly fit person they are $$\approx $$4 L/min and 25 L/min, respectively.

Table 6.9 gives the concentrations and total amounts of energy storage molecules and the energy available from them for a 70 kg man with 30 kg of muscle. Women have about the same concentrations of ATP and PC per kg muscle as do men, but have less overall muscle. The typical specific chemical energy resources stored in the body are listed in Table 6.10.

Table 6.11 lists the energy system that is primarily used in various sports activities. This is explored in more detail in Table 6.12, which gives the fraction of each system used in different sport activities. These fractions are given for shorter to longer distances in ice speed skating, swimming, andrunning in Tables 6.13, 6.14 and 6.15. In each case, short distance sprints use the ATP-PC and anaerobic systems, while the longer distance events primarily use the aerobic system.


Table 6.12
Percent emphasis of energy systems


































































































































Sport or activity
ATP-PC and anaerobic glycolysis
Anaerobic glycolysis and aerobic
Aerobic

Aerobic dance
5
15–20
75–80

Baseball
80
15
5

Basketball
60
20
20

Diving
98
2
Negligible

Fencing
90
10
Negligible

Field hockey
50
20
30

Football
90
10
Negligible

Golf
95
5
Negligible

Gymnastics
80
15
5

Ice hockey

   Forward, defense
60
20
20

   Goalie
90
5
5

Lacrosse

   Goalie, defense, attacker
50
20
30

   Midfielders, man-down
60
20
20

Rowing
20
30
50

Soccer

   Goalie, wings, strikers
60
30
10

   Halfbacks or sweeper
60
20
20

Stepping machine
5
25
70

Tennis
70
20
10

Field events, in track and field
95–98
2–5
Negligible

Volleyball
80
5
15

Walking
Negligible
5
95

Wrestling
90
5
5


Using data from [35, 36]



Table 6.13
Percent emphasis of energy systems for a range of distances in ice speed skating













































Activity
ATP-PC and anaerobic glycolysis
Anaerobic glycolysis and aerobic
Aerobic

Ice speed skating

500 m
80
10
10

1,000 m
35
55
10

1,500 m
20–30
30
40–50

5,000 m
10
25
65

10,000 m
5
15
80

In-line skating

$$>$$” src=”http://basicmedicalkey.com/wp-content/uploads/2017/06/A114622_2_En_6_Chapter_IEq289.gif”></SPAN>10 km</DIV></TD><br /> <TD align=left><br /> <DIV class=SimplePara>5</DIV></TD><br /> <TD align=left><br /> <DIV class=SimplePara>25</DIV></TD><br /> <TD align=left><br /> <DIV class=SimplePara>70</DIV></TD></TR></TBODY></TABLE><br /> <DIV class=TableFooter><br /> <DIV class=SimplePara>Using data from [<CITE><A href=35, 36]



Table 6.14
Percent emphasis of energy systems for a range of distances in swimming







































Swimming
ATP-PC and anaerobic glycolysis
Anaerobic glycolysis and aerobic
Aerobic

50 m
90
5
5

100 m
80
15
5

200 m
30
65
5

400 m
20
40
40

1,500 m
10
20
70


Using data from [35, 36]



Table 6.15
Percent emphasis of energy systems for a range of distances in running






















































Running
ATP-PC and anaerobic glycolysis
Anaerobic glycolysis and aerobic
Aerobic

100, 200 m
95–98
2–5
Negligible

400 m
80
15
5

800 m
30
65
5

1,500 m (or mile)
20–30
20–30
40–60

3,000 m (or 2 miles)
10
20
70

5,000 m (or 3 miles)
10
20
70

10,000 m (or 6 miles)
5
15
80

Marathon
Negligible
5
95


Using data from [35, 36]



Table 6.16
Recovery times after exhaustive exercise







































 
Minimum
Maximum

Restoration of phosphagen (ATP + PC)
2 min
5 min

Muscle glycogen replenishment
5–10 h
24–46 h

Liver glycogen replenishment
Unknown
12–24 h

Restoration of O$$_{2}$$ in plasma and myoglobin
10–15 s
1 min

Duration of fast component of O$$_{2}$$ recovery
3 min
6 min

Duration of slow component of O$$_{2}$$ recovery
30 min
1 h

Reduction of lactic acid in blood and muscle             
30–60 min$$^{\mathrm {a}}$$

1–2 h$$^{\mathrm {a}}$$


$$^{\mathrm {a}}$$Faster recovery with exercise and slower recovery with rest.

Using data from [35]

The metabolic steps involved in these systems are all regulated by a feedback and control mechanism (Chap. 13), and are followed by a recovery step. Table 6.16 gives the recovery time for returning resting energy reserves and O$$_{2}$$ and for reducing lactic acid after exercise. Anaerobic exercise is limited by the maximal lactic acid tolerance, which is about 2.0–2.3 g/kg muscle, and so this is 60–70 g for 30 kg of muscle. You can learn a bit about the recovery step by running as fast as you can for as long as you can. You will then be huffing and puffing, breathing in air as fast as possible. This is part of recovery. (Consult with your physician before attempting this demonstration.)


6.4 Metabolic Rates


There are variations among people’s metabolic rates (MR) due to their different weights, genes, etc. Metabolic needs over the years have changed, in part due to changing body heights and weights, and levels of activity [34]. First, we will discuss the minimum metabolic rates for people and then address how activity increases the metabolic rate. The metabolic rate can also depend on food intake. For example, the MR is known to decrease under fasting conditions and this change counters some of the expected effects of dieting. Although widely called metabolic rates, these are more precisely catabolic rates. Sometimes the MR is called the total energy expenditure, TEE, which might make it clearer that caloric intake that is not expended as energy is stored by the body.


6.4.1 Basal Metabolic Rate


The basal metabolic rate (BMR) is that of an inactive, awake body. The BMR for a 70 kg person is about 1,680 kcal/day $$\simeq $$70 kcal/h $$\simeq $$81 W. This means that even at rest each of us gives off almost the same amount of heat as a 100 W incandescent light bulb. As we will see later, the heavier the person the higher the BMR. Approximately 85$$\%$$ of people of the same gender and weight have a BMR within 6–10$$\%$$ of the mean for their classification (Table 6.14).

The exact numbers one finds for the BMR sometimes vary with the source. This may be due in part to the exacting conditions used to define the BMR and the lack of specification of the person’s gender, age, weight, etc. To have your BMR measured you must:

(1)

Have eaten no food for at least 12 h

 

(2)

Have had a night of restful sleep and no strenuous activity thereafter

 

(3)

Be resting completely in a reclining position for at least 30 min

 

(4)

Be experiencing no excitement from psychic or physical factors

 

(5)

Be in a room with a temperature from 20 to 27$$\,^{\circ }$$C (68 to 80$$\,^{\circ }$$F).

 
The BMR is greater than the metabolic rate during sleeping (when there is minimal digestion of food). Because of the stringent nature of BMR testing, “background” or “minimal” metabolic rates are sometimes tested under somewhat less exacting conditions and these measured rates are not cited as BMRs, but are termed differently although not always in a uniform way. The basal energy expenditure (BEE) is often defined to be equivalent to the BMR. The resting metabolic rate (RMR) is similar to the BMR, but is measured under somewhat less restrictive conditions—early in the morning after an overnight fast and 8 h of sleep—so it is easier to achieve and thus it is often used. The resting energy expenditure (REE) is usually measured after 4 h of sitting and is a bit larger than the BMR. The BMR and metabolic rates during activities are often determined by measuring oxygen intake, because they are proportional to each other for aerobic activities.

The functioning of several organs contributes to this BMR in a resting person, as is seen from Table 6.17. Three-fourths of the metabolic activity takes place in organs with a total mass of 5 kg, which is $$\sim $$8$$\%$$ of the total body mass. We will estimate these metabolic rates of the heart and lungs in Chaps. 8 and 9.

Human brains and gastro-intestinal systems (which are not listed in Table 6.17) have about same mass and metabolic rates, and so they require about the same energy per unit mass, $$\sim $$11–12 W/kg. Our brains are much larger and our gastro-intestinal systems are much smaller than those in primates and other mammals as expected from allometric mass scaling (though our livers and kidneys are of the expected sizes). Our smaller than expected, and consequently less energy demanding, gastro-intestinal tracts are sufficient because humans can easily acquire and eat high caloric-value foods that we can digest rapidly, in part because they do not have extremely high fiber content and can be processed (i.e., cooked). Our brains are larger than expected, and consequently more energy demanding. Perhaps, because of the smaller gastro-intestinal tracts, humans can energetically “afford”to have very large brains [3, 56] (and, in balance, still have overall basal metabolic needs consistent with those expected from the allometric scaling of the BMR with body mass, which we will soon discuss).


Table 6.17
Metabolism of a resting person

















































System
Percentage of BMR
Met. rate (kcal/min)
Organ mass (kg) for a 65 kg man

Liver and spleen
27
0.33
–

Brain
19
0.23
1.40

Skeletal muscle
18
0.22
28.0

Kidney
10
0.13
0.30

Heart
7
0.08
0.32

Remainder
19
0.23
–
   
Sum $$=$$ 1.22
  


Using data from [19]

The BMR is a function of mass, height, and gender, as we will see later in this chapter. It is also a function of body temperature. It changes by about 10$$\%$$ (some say 13$$\%$$) per 1$$\,^{\circ }$$C in body temperature. Consequently, the BMR increases by 30$$\%$$ (or 39$$\%$$) when the body temperature increases from the normal 37$$\,^{\circ }$$C (98.6$$\,^{\circ }$$F) to 40$$\,^{\circ }$$C (104.0$$\,^{\circ }$$F) and decreases by 30$$\%$$ (or 39$$\%$$) when the body temperature decreases from normal to 34$$\,^{\circ }$$C (93.2$$\,^{\circ }$$F).

Scaling of the BMR

Before addressing how the BMR and MR depend on gender, weight, age, etc. we will temporarily expand our discussion of the BMR from that for humans to that for all mammals. The larger the animal the higher the metabolic (catabolic) rate, which is as expected. Moreover, there is a way that the BMR scales with a physical attribute of size, specifically the body mass $$m_{\mathrm {b}}$$.

Table 6.18 shows that the BMR of mammals increases sublinearly with $$m_{\mathrm {b} } $$. Many think the variation for mammals is as the 3/4 power:


$$\begin{aligned} \mathrm {BMR}=cm_{\mathrm {b}}^{3/4}, \end{aligned}$$

(6.19)
where $$c\approx $$ 90 kcal/kg$$^{3/4}$$-day. This is known as Kleiber’s Law [51–53], and is an example of an allometric relationship, as described in Table 1.​13. It is valid from mice to elephants and some say to whales. (Steinbeck’s novel was valid only from mice to men. Regarding whales: there is great uncertainty about the BMR of whales. How do you measure it?) While this scaling relation was actually developed for comparing different species, it works fairly well intraspecies for humans, from children to adults. Others claim that the BMRs of mammals increase as mass to the 2/3 power. Those believing in Kleiber’s Law point out that this other scaling rule is off by a factor of two from mice to cows.


Table 6.18
BMR determined for several mammals




























Species
Mass
BMR (kcal/day)

Mouse
20 g
3

Reference woman (25 years)
55 kg
1,260

Reference man (25 years)
65 kg
1,500

Elephant
5,000 kg
70,000


See, for example [66]

Both dependences give essentially the same prediction over the mass range of humans, so it matters little which we use for analysis. Adherents of both rules claim that the available BMR data support them. In any case, it is interesting to examine the physical reasons put forth to support the $$m_{\mathrm {b}}^{2/3}$$ and $$m_{\mathrm {b}}^{3/4}$$ scaling rules [27, 42]. There is both some sound and some fairly questionable physical reasoning in both sets of arguments.

Supporting Kleiber’s $$m^{3/4}$$ Scaling Rule

The metabolic rate, and the BMR in particular, are supposed to scale as the maximum power output of muscles [65, 66]. Power usage is proportional to Fv, where F is the muscle force and v is the speed of muscle contraction. The muscle force $$F=\sigma $$(PCA), where $$\sigma $$ is the muscle force per unit area and PCA is the muscle cross-sectional area. Therefore,


$$\begin{aligned} \mathrm {BMR}\propto Fv=\sigma (\mathrm {PCA})v. \end{aligned}$$

(6.20)
Studies show that for muscles of all species and size, v and $$\sigma $$ do not vary substantially, so


$$\begin{aligned} \mathrm {BMR}\propto \mathrm {PCA}. \end{aligned}$$

(6.21)
If the limbs of mammals, which of course contain many of these muscles, have a characteristic width d and length L, this area and the mass of the mammal scale $$m_{\mathrm {b}}$$ as


$$\begin{aligned} \mathrm {PCA} \propto d^{2} \end{aligned}$$

(6.22)



$$\begin{aligned} m_{\mathrm {b}}\propto d^{2}L. \end{aligned}$$

(6.23)
To relate the BMR to the mass, we need to relate L to d. Four arguments have been put forth suggesting that L scales with $$d^{2/3}$$. First, experimental data on primates show that L varies as d this way. Second, this scaling law also relates the height of a tree L with the width of the tree trunk d, and so this scaling law seems to be universal. Third, $$ L\propto d^{2/3}$$ is also the scaling law for the condition for which a long vertical column of height L and width d buckles under its own weight when displaced from the vertical orientation.

A fourth argument is based on self-similarity, using the expression we derived earlier, (4.​46). The end of the beam of length L and lateral dimension d in Fig. 4.​40 deflects down by a distance $$ y(L)=-FL^{3}/3YI_{\mathrm {A}}$$ when a force F is applied at its end. Even though the area moment of inertia is different for different cross-sections—$$I_{\mathrm {A} }=\pi d^{4}/64$$ for cylinders (from $$I_{\mathrm {A}}=\pi a^{4}/4$$ for a circular cross-section of radius $$a=d/2$$ using (4.​42)) and $$ d^{4}/12$$ for rectangular solids with a square cross-section (from $$I_{\mathrm {A}}=wh^{3}/12$$ for a rectangular cross-section of width w and height h—here with $$d=w=h$$, using (4.​41))—it makes no difference which we use because both have the same $$d^{4}$$ dependence. (We will formally use the circular cross-section.)

This downward angle of deflection $$\theta $$ is $$\sim $$ $$|y|/L$$, giving


$$\begin{aligned} \theta =\frac{|y|}{L}=\frac{FL^{2}}{3YI_{\mathrm {A}}}. \end{aligned}$$

(6.24)
Self-similarity suggests that $$\theta $$ would be same for all species, and consequently equal to a constant. If F is due to gravity, then $$F=m_{\mathrm { b}}g=\rho Vg$$, where $$\rho $$ is the beam mass density and V is its volume $$ =(\pi d^{2}/4)L$$. If this force acted entirely at the end, then


$$\begin{aligned} \theta =\frac{(\rho ((\pi d^{2}/4)L)g)L^{2}}{3Y(\pi d^{4}/64)}=\frac{16}{3} \frac{\rho g}{Y}\frac{L^{3}}{d^{2}}. \end{aligned}$$

(6.25)
If $$\theta $$ were independent of L and d, we again get $$L\propto d^{2/3}$$.

In any case, given that $$m_{\mathrm {b}}\propto d^{2}L$$ and $$L\propto d^{2/3}$$, we see that $$m_{ \mathrm {b}}\propto d^{2}(d^{2/3})= d^{8/3}$$. Therefore $$d\propto m_{\mathrm {b} }^{3/8}$$. With BMR $$\propto $$ PCA and PCA $$\propto d^{2}$$, we find that BMR $$\propto d^{2}$$. Using $$d\propto m_{\mathrm {b}}^{3/8}$$, we see that


$$\begin{aligned} \mathrm {BMR}\propto (m_{\mathrm {b}}^{3/8})^{2}=m_{\mathrm {b}}^{3/4}, \end{aligned}$$

(6.26)
which is Kleiber’s Law.

There are many gaps in this reasoning. For example, the power needed to operate skeletal muscles is actually a small fraction of the BMR (Table 6.17), so the scaling laws based on this power may not represent those of the whole BMR.

Supporting the $$m^{2/3}$$ Scaling Rule

The previous argument assumed that the BMR scaled as the metabolic power needs. This line of reasoning says that the BMR is limited by the rate of heat loss from the mammal. A mammal has a cross-sectional area that scales as $$L^{2}$$, where L is its dimension, while its volume and therefore mass $$m_{ \mathrm {b}}$$ scale as $$L^{3}$$. Heat loss scales as the available surface area A, which clearly varies as $$m_{\mathrm {b}}^{2/3}$$, and so


$$\begin{aligned} \mathrm {BMR}\propto m_{\mathrm {b}}^{2/3}. \end{aligned}$$

(6.27)
BMR Scaling in Humans

The BMR in humans depends on mass as well as other factors, such as gender, height, and age. The BMRs for 20–39 year old men in moderate temperature climates ranges between 1,350 and 2,000 kcal/day for common ranges of heights and weights [34]. The Harris–Benedict equations are commonly used to characterize measured human BMRs:


$$\begin{aligned} \mathrm {For \;men{:}\quad BMR}= & {} 66.4730+13.7516\mathrm { }m_{\mathrm {b}}+5.0033\mathrm { }H \mathrm { }-\mathrm { }6.75505\mathrm { }Y, \end{aligned}$$

(6.28)



$$\begin{aligned} \mathrm {For \;women{:}\quad BMR}= & {} 655.0955+9.5634\mathrm { }m_{\mathrm {b}}+1.8496\mathrm { }H\mathrm { }-\mathrm { }4.6756\mathrm { }Y, \end{aligned}$$

(6.29)
where the BMR is in kcal/day, $$m_{\mathrm {b}}$$ is the body mass in kg, H is the height in cm, and Y is the age in yr. This can also be expressed as


$$\begin{aligned} \mathrm {For \;men{:}\,\, BMR}= & {} 71.2\mathrm { }m_{\mathrm {b} }^{3/4}[1+0.004(30-Y)+0.010(S-43.4)], \quad \end{aligned}$$

(6.30)



$$\begin{aligned} \mathrm {For \;women{:}\,\, BMR}= & {} 65.8\mathrm { }m_{\mathrm {b} }^{3/4}[1+0.004(30-Y)+0.010(S-43.4)], \end{aligned}$$

(6.31)
where S is the specific stature $$=$$ $$H$$(in cm)/$$m_{\mathrm {b}}^{1/3}$$($$m_{ \mathrm {b}}$$ in kg). These relations track averages, and not necessarily how the BMR changes when a given individual becomes heavier or older.

Another study re-examined the BMR in term of age groups and gender, as presented in Table 6.19. Table 6.20 gives the BMRs for adults of given heights as predicted by Table 6.19. The person’s mass is chosen to make the BMI (or the Quételet’s index Q) equal to $$m_{\mathrm {b}}/H^{2}$$ ($$m_{\mathrm {b}}$$ is the body mass (in kg) and H is the height (in m)) equal to 22 for men and 21 for women.


Table 6.19
BMR (kcal/day) for different age groups




































Age group (years)
BMR (males)
BMR (females)

Under 3
59.5$$m_{\mathrm {b}}$$ $$-$$ 30
58.3$$m_{\mathrm {b}}$$ $$-$$ 31

3–10
22.7$$m_{\mathrm {b}}$$ $$+$$ 504
20.3$$m_{\mathrm {b}}$$ $$+$$ 486

10–18       
17.7$$m_{\mathrm {b}}$$ $$+$$ 658
13.4$$m_{\mathrm {b}}$$ $$+$$ 693

18–30
15.1$$m_{\mathrm {b}}$$ $$+$$ 692
14.8$$m_{\mathrm {b}}$$ $$+$$ 487

30–60
11.5$$m_{\mathrm {b}}$$ $$+$$ 873
8.1$$m_{\mathrm {b}}$$ $$+$$ 846

Over 60
11.7$$m_{\mathrm {b}}$$ $$+$$ 588
9.1$$m_{\mathrm {b}}$$ $$+$$ 658


$$m_{\mathrm {b}}$$ is the body mass in kg.

Using data from [11], using [90]



Table 6.20
BMR (kcal/day) for adult men and women of different ages, assuming the BMI (or Quételet’s index Q) is 22 for men and 21 for women and the relations in Table 6.19
























































































Height (m) (also ft, in)
Mass (kg) (also lb)
Age (yr)

18–30
30–60
Over 60

Men

1.5 (4$$^{\prime }$$11$$^{\prime \prime }$$)
49.5 (109)
1,440
1,450
1,150

1.6 (5$$^{\prime }$$3$$^{\prime \prime }$$)
56.5 (124)
1,540
1,530
1,250

1.7 (5$$^{\prime }$$7$$^{\prime \prime }$$)
63.5 (140)
1,650
1,620
1,350

1.8 (5$$^{\prime }$$11$$^{\prime \prime }$$)
71.5 (157)
1,770
1,710
1,450

1.9 (6$$^{\prime }$$3$$^{\prime \prime }$$)
79.5 (175)
1,900
1,800
1,560

2.0 (6$$^{\prime }$$7$$^{\prime \prime }$$)
88.0 (194)
2,030
1,900
1,670

Women

1.4 (4$$^{\prime }$$7$$^{\prime \prime }$$)
41.0 (90)
1,100
1,190
1,030

1.5 (4$$^{\prime }$$11$$^{\prime \prime }$$)
47.0 (104)
1,190
1,240
1,090

1.6 (5$$^{\prime }$$3$$^{\prime \prime }$$)
54.0 (119)
1,290
1,300
1,160

1.7 (5$$^{\prime }$$7$$^{\prime \prime }$$)
61.0 (134)
1,390
1,360
1,230

1.8 (5$$^{\prime }$$11$$^{\prime \prime }$$)
68.0 (150)
1,500
1,420
1,310


Using data from [11], from [33]

These relations show that younger people have higher BMRs than older people, which is true because they have more lean body mass. Similarly, tall thin people have higher BMRs and overweight people have lower BMRs. Pregnant women have higher BMRs than predicted above, as do people with fevers, those under stress, and those in hot and cold conditions. The BMR is lower than is predicted here for people who are fasting, starving, and malnourished.

During pregnancy, basic dietary energy needs increase by an average of roughly 200–250 kcal/day, with most of the increase occurring in the second half of gestation [18, 58, 73, 85]. Of the total extra estimated caloric needs a woman has during pregnancy, about half goes to increased metabolic needs, as expected from a larger body mass, and about half to building protein and fat in added tissue (for the fetus, uterus, blood, placenta and breasts) [18]. Consequently, a very large fraction of this increase can be considered an increase in BMR and a smaller fraction can be attributed to body activity, which causes the MR to exceed the BMR. In developed countries the extra caloric needs are $$\sim $$90,000 kcal for a total weight gain of $$\sim $$13.8 kg (30.4 lb). (An earlier estimate of the average extra energy cost was $$\sim $$80,000 kcal, with a weight gain during pregnancy of $$\sim $$12.5 kg and median infant birth weight of $$\sim $$3.3  kg [33].) In developed countries, the BMR increases by $$\sim $$5, 11, and 24% during the first, second and third trimesters of pregnancy [18]. The USDA recommends no increase in the caloric intake rate in the first trimester of pregnancy and 340 and 450 kcal/day increase in the second and third trimester. In postpartum, lactation increases energy needs by $$\sim $$500 kcal/day [73], and again this increase can be considered an increase in overall BMR. This increase in caloric intake can be smaller in those with more stored fat.

Tissue growth is important during pregnancy, normal growth from infancy through adolescence, and “catch-up” growth in malnourished children. During pregnancy the added tissue is estimated to be $$\sim $$5–8% protein and $$\sim $$25–31% fat, with the balance being water [18, 85]. The energy cost to produce new tissue can be determined from the heats of combustion for protein, fat, and carbohydrate are 5.7, 9.3, and 4.3 kcal/g, which are the Atwater factors excluding losses as to urine and feces. For tissue growth in children it is estimated to be roughly 5 kcal per g of (dry) tissue growth; in malnourished children it is 4.4 kcal/g [94]. This extra energy cost is only $$<$$5% of the total metabolic needs from age 1 through adolescence including the growth spurt [17], but can be very significant during fetus growth (as noted above) and from infancy until 12 months.

See the problems at the end of the chapter that address the BMR and tissue growth for pregnant women.

The variations in the BMR and MR among humans are discussed below.


6.4.2 Metabolic Rates During Common Activities


Fortunately, we spend most of our time under conditions less restrictive than those needed to measure the BMR. Metabolic rates during activity are commonly expressed in several ways. (1) The total metabolic rate (MR) (including that due to the BMR) is expressed either as (a) a cumulative value (BMR + the additional MR due to activities), (b) a factor times the BMR—the factor being called the activity factor f or the physical activity ratio (PAR) (which is commonly used in the health community), or (c) a factor times the RMR, with that factor called the MET, the metabolic equivalent (which is commonly used in the exercise community). The RMR is more frequently used as the reference because it is much easier to measure than the BMR. (For our purposes we can assume that the MET and the slightly larger f are equal.) (2) Only the net increase in metabolic rate over the BMR (or RMR) is given, such as MR$$-$$BMR.


Table 6.21
Approximate total metabolic rates (MR) and oxygen consumption for different levels of activity for an average 70 kg person

















































































Activity
Equivalent heat production (kcal/h)
(W)
O$$_{2}$$ consumption (L/min)

Very low level activity

   Sleeping
71
83
0.24

   Sitting at rest
103
120
0.34

   Standing relaxed
108
125
0.36

Light activity

   Walking slowly, 5 km/h
228
265
0.76

Moderate activity

   Cycling, 15 km/h
344
400
1.13

   Moderate swim
400
465
1.32

Heavy activity

   Soccer
500
580
1.65

Quite heavy activity

   Climbing stairs, 116 steps/min
589
685
1.96

   Cycling, 21 km/h
602
700
2.00

   Basketball
688
800
2.28

Extreme activity

   Racing cyclist
1,400
1,600
4.62


From [9] and [19]



A114622_2_En_6_Fig6_HTML.gif


Fig. 6.6
In a calorimetric chamber for direct calorimetry, heat generated by the person is transferred to the air and chamber walls, and the heat produced—and therefore the metabolic rate—are determined by the measured temperature change in the air and the water flowing through the chamber (From [106]. Used with permission)

Of course, each method should provide the same total MR. It is not surprising that the published values of MR expressed in these different ways are often somewhat inconsistent; this is true for several reasons. The exact metabolic rates may not be known very well, the details of the specific activity may not be well defined, the differences in rates among people of different genders, ages, and weights may not have been accounted for properly, and the exact differences between the reference metabolic rates may not be clear. For example, the differences in the BMR and RMR may not have been considered well. Also, sometimes the MET is defined relative to the RMR as defined earlier, right after sleep, and sometimes it is referenced to the MR for sitting. One more subtle reason is that the BMR varies as $$m_{\mathrm {b} }^{3/4}$$, while the increase in metabolic rate due to activity is thought to depend linearly on $$m_{\mathrm {b}}$$ (and this may not always be exactly true). Consequently, scaling factors, such as the activity factor, cannot be rigorously correct. Nevertheless, examining the total MR is intriguing and we can learn much from it, even amidst these potential inconsistencies.

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Related posts:

  1. Statics of the Body
  2. Muscles
  3. Mechanical Properties of the Body
  4. Cardiovascular System

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Jun 11, 2017 | Posted by in GENERAL & FAMILY MEDICINE | Comments Off on Metabolism: Energy, Heat, Work, and Power of the Body

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