2
CHEMICAL FOUNDATIONS
REVIEW THE CONCEPTS
1. Less energy is required to form noncovalent bonds than covalent bonds, and the
bonds that stick the gecko’s feet to the smooth surface need to be formed and broken
many times as the animal moves. Since van der Waals interactions are so weak, there
must be many points of contact (a large surface area) yielding multiple van der Waals
interactions between the septae and the smooth surface.
2. a. These are likely to be hydrophilic amino acids, and in particular, negatively
charged amino acids (aspartate and glutamate), which would have an affinity
for K+ via ionic bonds.
b. Like the phospholipid bilayer itself, this portion of the protein is likely to be
amphipathic, with hydrophobic amino acids in contact with the fatty acyl chains
and hydrophilic amino acids in contact with the hydrophilic heads.
c, d. Since both the cytosol and extracellular space are aqueous environments,
hydrophilic amino acids would contact these fluids.
3. At pH = 7.0, the net charge is –1 because of the negative charge on the carboxyl resi-
due of glutamate (E). After phosphorylation by a tyrosine kinase, two additional nega-
tive charges (because of attachment of phosphate residues to tyrosines (Y)) would be
added. Thus, the net charge would be –3. The most likely source of phosphate is ATP
since the attachment of inorganic phosphate (Pi) to tyrosine is energetically highly
unfavorable, but when coupled to the hydrolysis of the high-energy phosphoan-
hydride bond of ATP, the overall reaction is energetically favorable.
1
,2 CHAPTER 2: Chemical Foundations
4. Disulfide bonds are formed between two cysteine residue side chains. The forma-
tion of disulfide bonds increases the order and therefore decreases the entropy
(S becomes more negative).
5. Stereoisomers are compounds that have the same molecular formula but are
mirror images of each other. Many organic molecules can exist as stereoisomers
because of two different possible orientations around an asymmetric carbon
atom (e.g., amino acids). Because stereoisomers differ in their three-dimensional
orientation and because biological molecules interact with one another based on
precise molecular complementarity, stereoisomers often react with different
molecules, or react differently with the same molecules. Therefore, they may
have very distinct physiological effects in the cell.
6. The compound is guanosine triphosphate (GTP). Although the guanine base is
found in both DNA and RNA, the sugar is a ribose sugar because of the 2′
hydroxyl group. Therefore, GTP is a component of RNA only. GTP is an impor-
tant intracellular signaling molecule.
7. At least three properties contribute to this structural diversity. First, monosaccha-
rides can be joined to one another at any of several hydroxyl groups. Second, the
C-1 linkage can have either an α or a β configuration. Third, extensive branching
of carbohydrate chains is possible.
8. What is the pH of 1 L of water? In all aqueous solutions, water spontaneously
dissociates into hydrogen and hydroxide ions according to the equilibrium reac-
tion H2O G H+ + OH–. The ionization constant for aqueous solutions at 25°C
is Kw = [H+][OH–] = 1 × 10–14 M2. In a solution of pure water, the production of
one H+ ion will always be accompanied by the production of one OH– ion. In
other words, [H+] = [OH–].
Kw = [H+][OH–] = [H+]2 = 1 × 10–14 M2
[H+] = 1 × 10–7 M
pH = –log10[H+] = –log10(1 × 10–7) = 7
What is the pH after 0.008 moles NaOH are added? NaOH (sodium hydroxide)
is a strong base. This means all the added NaOH ionizes to increase the [OH–]
concentration to 0.008 M.
[H+] = Kw/[OH–] = (1 × 10–14 M2)/(0.008 M) = 1.25 × 10–12 M
pH = –log10[H+] = –log10(1.25 × 10–12) = 11.903
What is the pH of the solution of 50 mM MOPS? MOPS is a weak acid. As such,
upon dissolving in water it will undergo partial dissociation yielding equal
concentrations of hydrogen ions and MOPS conjugate base according to the
equilibrium reaction:
MOPS(weak acid form) = H+ + MOPS(conjugate base form)
, CHAPTER 2: Chemical Foundations 3
The extent to which this reaction goes forward determines the relative strength
of the MOPS weak acid and is given by its acid dissociation equilibrium
constant:
Ka = ([H+][MOPS(conjugate base form)])/[MOPS(weak acid form)]
pKa = –log10Ka = 7.20
If the relative concentrations are known for the weak acid and conjugate base
forms of a dissolved weak acid at equilibrium in water, then the solution pH can
be determined according to the equation:
pH = pKa + log10([conjugate base]/[weak acid])
Therefore,
pH = 7.20 + log10(0.39/0.61) = 7.01
What is the pH after 0.008 moles NaOH are added to the MOPS buffer solution?
Rather than simply increase the total [OH–] by 0.008 moles, addition of the strong
base shifts the equilibrium of the dissolved MOPS such that [MOPS(weak acid)]
decreases by 0.008 M and [MOPS(conjugate base)] increases by 0.008 M.
Before addition of 0.008 moles NaOH:
[MOPS(weak acid)] = 0.61(0.050 M) = 0.0305 M
[MOPS(conjugate base)] = 0.39(0.050 M) = 0.0195 M
After addition of 0.008 moles NaOH:
[MOPS(weak acid)] = 0.0305 M – 0.008 M = 0.0225 M
[MOPS(conjugate base)] = 0.0195 M + 0.008 M = 0.0275 M
The final pH after addition of 0.008 moles NaOH to 50 mM MOPS at pH 7.01 is,
therefore:
pH = 7.20 + log10(0.0275/0.0225) = 7.29
9. In the acidic pH of a lysosome, ammonia is converted to ammonium ion. Ammo-
nium ion is unable to traverse the membrane because of its positive charge and is
trapped within the lysosome. The accumulation of ammonium ion decreases the
concentration of protons within lysosomes and therefore elevates lysosomal pH.
At neutral pH, ammonia has little, if any, tendency to protonate to ammonium
ion and thus has no effect on cytosolic pH.
10. Keq = [LR]/[L][R]
Since 90% of L binds R, the concentration of LR at equilibrium is 0.9(1 × 10–3M) =
9 × 10–4 M. The concentration of free L at equilibrium is the 10% of L that remains
, 4 CHAPTER 2: Chemical Foundations
unbound, 1 × 10–4 M. The concentration of R at equilibrium is (5 × 10−2 M) –
(9 × 10–4 M) = 4.91 × 10–2 M. Therefore, [LR]/[L][R] = 9 × 10–4 M/ ((1 × 10–4 M)
(4.91 × 10–2 M)) = 183.3 M–1.
The equilibrium constant is unaffected by the presence of an enzyme.
Kd = 1/Keq = 5. 4 × 10−3 M.
11. ΔG = ΔGº’ + RTln [products]/[reactants]
For this reaction, ΔG = −1000 cal/mol + [1.987 cal/(degree · mol) × (298 degrees)
× ln (0.01 M/(0.01 M × 0.01 M))].
ΔG = −1000 cal/mol + 2727 cal/mol = 1727 cal/mol
To make this reaction energetically favorable, one could increase the concen-
tration of reactants relative to products such that the term RTln [products]/
[reactants] becomes smaller than 1000 cal/mol. One might also couple this
reaction to an energetically favorable reaction.
12. The presence of one or more carbon-carbon double bonds is indicative of an
unsaturated or polyunsaturated fatty acid. The term saturated refers to the fact
that all carbons, except the carbonyl carbon, have four single bonds. In a cis
unsaturated fatty acid, the carbon atoms flanking the double bond are on the
same side, thus introducing a kink in the otherwise flexible straight chain.
There is no such kink in a trans unsaturated fatty acyl chain.
13. Glutamate is the amino acid that undergoes γ-carboxylation, resulting in the
formation of a host of blood clotting factors. Warfarin inhibits γ-carboxylation of
glutamate. Thus, blood clotting is severely compromised. Patients prone to
forming clots (thrombi) in blood vessels might be prescribed warfarin in order
to prevent an embolism, which would result if the clot dislodged and blocked
another vessel elsewhere in the body. Patients at risk for heart disease due to
blockages in the coronary arteries are also often prescribed this drug.
ANALYZE THE DATA
1. Biomolecules are relatively easy to synthesize from inorganic starting materials,
which suggests that living and nonliving matter are not fundamentally different.
Living matter is subject to the same laws of physics and chemistry that govern
nonliving matter. The fact that biomolecules can be produced through nonliving,
chemical processes suggests that life itself could have evolved by similar means.
Biochemistry attempts to describe the mechanisms that give rise to living sys-
tems from the perspective of the molecules that make up living things. We can
often gain considerable insight into the properties of a living thing by studying
the structure and chemistry of its molecules.
CHEMICAL FOUNDATIONS
REVIEW THE CONCEPTS
1. Less energy is required to form noncovalent bonds than covalent bonds, and the
bonds that stick the gecko’s feet to the smooth surface need to be formed and broken
many times as the animal moves. Since van der Waals interactions are so weak, there
must be many points of contact (a large surface area) yielding multiple van der Waals
interactions between the septae and the smooth surface.
2. a. These are likely to be hydrophilic amino acids, and in particular, negatively
charged amino acids (aspartate and glutamate), which would have an affinity
for K+ via ionic bonds.
b. Like the phospholipid bilayer itself, this portion of the protein is likely to be
amphipathic, with hydrophobic amino acids in contact with the fatty acyl chains
and hydrophilic amino acids in contact with the hydrophilic heads.
c, d. Since both the cytosol and extracellular space are aqueous environments,
hydrophilic amino acids would contact these fluids.
3. At pH = 7.0, the net charge is –1 because of the negative charge on the carboxyl resi-
due of glutamate (E). After phosphorylation by a tyrosine kinase, two additional nega-
tive charges (because of attachment of phosphate residues to tyrosines (Y)) would be
added. Thus, the net charge would be –3. The most likely source of phosphate is ATP
since the attachment of inorganic phosphate (Pi) to tyrosine is energetically highly
unfavorable, but when coupled to the hydrolysis of the high-energy phosphoan-
hydride bond of ATP, the overall reaction is energetically favorable.
1
,2 CHAPTER 2: Chemical Foundations
4. Disulfide bonds are formed between two cysteine residue side chains. The forma-
tion of disulfide bonds increases the order and therefore decreases the entropy
(S becomes more negative).
5. Stereoisomers are compounds that have the same molecular formula but are
mirror images of each other. Many organic molecules can exist as stereoisomers
because of two different possible orientations around an asymmetric carbon
atom (e.g., amino acids). Because stereoisomers differ in their three-dimensional
orientation and because biological molecules interact with one another based on
precise molecular complementarity, stereoisomers often react with different
molecules, or react differently with the same molecules. Therefore, they may
have very distinct physiological effects in the cell.
6. The compound is guanosine triphosphate (GTP). Although the guanine base is
found in both DNA and RNA, the sugar is a ribose sugar because of the 2′
hydroxyl group. Therefore, GTP is a component of RNA only. GTP is an impor-
tant intracellular signaling molecule.
7. At least three properties contribute to this structural diversity. First, monosaccha-
rides can be joined to one another at any of several hydroxyl groups. Second, the
C-1 linkage can have either an α or a β configuration. Third, extensive branching
of carbohydrate chains is possible.
8. What is the pH of 1 L of water? In all aqueous solutions, water spontaneously
dissociates into hydrogen and hydroxide ions according to the equilibrium reac-
tion H2O G H+ + OH–. The ionization constant for aqueous solutions at 25°C
is Kw = [H+][OH–] = 1 × 10–14 M2. In a solution of pure water, the production of
one H+ ion will always be accompanied by the production of one OH– ion. In
other words, [H+] = [OH–].
Kw = [H+][OH–] = [H+]2 = 1 × 10–14 M2
[H+] = 1 × 10–7 M
pH = –log10[H+] = –log10(1 × 10–7) = 7
What is the pH after 0.008 moles NaOH are added? NaOH (sodium hydroxide)
is a strong base. This means all the added NaOH ionizes to increase the [OH–]
concentration to 0.008 M.
[H+] = Kw/[OH–] = (1 × 10–14 M2)/(0.008 M) = 1.25 × 10–12 M
pH = –log10[H+] = –log10(1.25 × 10–12) = 11.903
What is the pH of the solution of 50 mM MOPS? MOPS is a weak acid. As such,
upon dissolving in water it will undergo partial dissociation yielding equal
concentrations of hydrogen ions and MOPS conjugate base according to the
equilibrium reaction:
MOPS(weak acid form) = H+ + MOPS(conjugate base form)
, CHAPTER 2: Chemical Foundations 3
The extent to which this reaction goes forward determines the relative strength
of the MOPS weak acid and is given by its acid dissociation equilibrium
constant:
Ka = ([H+][MOPS(conjugate base form)])/[MOPS(weak acid form)]
pKa = –log10Ka = 7.20
If the relative concentrations are known for the weak acid and conjugate base
forms of a dissolved weak acid at equilibrium in water, then the solution pH can
be determined according to the equation:
pH = pKa + log10([conjugate base]/[weak acid])
Therefore,
pH = 7.20 + log10(0.39/0.61) = 7.01
What is the pH after 0.008 moles NaOH are added to the MOPS buffer solution?
Rather than simply increase the total [OH–] by 0.008 moles, addition of the strong
base shifts the equilibrium of the dissolved MOPS such that [MOPS(weak acid)]
decreases by 0.008 M and [MOPS(conjugate base)] increases by 0.008 M.
Before addition of 0.008 moles NaOH:
[MOPS(weak acid)] = 0.61(0.050 M) = 0.0305 M
[MOPS(conjugate base)] = 0.39(0.050 M) = 0.0195 M
After addition of 0.008 moles NaOH:
[MOPS(weak acid)] = 0.0305 M – 0.008 M = 0.0225 M
[MOPS(conjugate base)] = 0.0195 M + 0.008 M = 0.0275 M
The final pH after addition of 0.008 moles NaOH to 50 mM MOPS at pH 7.01 is,
therefore:
pH = 7.20 + log10(0.0275/0.0225) = 7.29
9. In the acidic pH of a lysosome, ammonia is converted to ammonium ion. Ammo-
nium ion is unable to traverse the membrane because of its positive charge and is
trapped within the lysosome. The accumulation of ammonium ion decreases the
concentration of protons within lysosomes and therefore elevates lysosomal pH.
At neutral pH, ammonia has little, if any, tendency to protonate to ammonium
ion and thus has no effect on cytosolic pH.
10. Keq = [LR]/[L][R]
Since 90% of L binds R, the concentration of LR at equilibrium is 0.9(1 × 10–3M) =
9 × 10–4 M. The concentration of free L at equilibrium is the 10% of L that remains
, 4 CHAPTER 2: Chemical Foundations
unbound, 1 × 10–4 M. The concentration of R at equilibrium is (5 × 10−2 M) –
(9 × 10–4 M) = 4.91 × 10–2 M. Therefore, [LR]/[L][R] = 9 × 10–4 M/ ((1 × 10–4 M)
(4.91 × 10–2 M)) = 183.3 M–1.
The equilibrium constant is unaffected by the presence of an enzyme.
Kd = 1/Keq = 5. 4 × 10−3 M.
11. ΔG = ΔGº’ + RTln [products]/[reactants]
For this reaction, ΔG = −1000 cal/mol + [1.987 cal/(degree · mol) × (298 degrees)
× ln (0.01 M/(0.01 M × 0.01 M))].
ΔG = −1000 cal/mol + 2727 cal/mol = 1727 cal/mol
To make this reaction energetically favorable, one could increase the concen-
tration of reactants relative to products such that the term RTln [products]/
[reactants] becomes smaller than 1000 cal/mol. One might also couple this
reaction to an energetically favorable reaction.
12. The presence of one or more carbon-carbon double bonds is indicative of an
unsaturated or polyunsaturated fatty acid. The term saturated refers to the fact
that all carbons, except the carbonyl carbon, have four single bonds. In a cis
unsaturated fatty acid, the carbon atoms flanking the double bond are on the
same side, thus introducing a kink in the otherwise flexible straight chain.
There is no such kink in a trans unsaturated fatty acyl chain.
13. Glutamate is the amino acid that undergoes γ-carboxylation, resulting in the
formation of a host of blood clotting factors. Warfarin inhibits γ-carboxylation of
glutamate. Thus, blood clotting is severely compromised. Patients prone to
forming clots (thrombi) in blood vessels might be prescribed warfarin in order
to prevent an embolism, which would result if the clot dislodged and blocked
another vessel elsewhere in the body. Patients at risk for heart disease due to
blockages in the coronary arteries are also often prescribed this drug.
ANALYZE THE DATA
1. Biomolecules are relatively easy to synthesize from inorganic starting materials,
which suggests that living and nonliving matter are not fundamentally different.
Living matter is subject to the same laws of physics and chemistry that govern
nonliving matter. The fact that biomolecules can be produced through nonliving,
chemical processes suggests that life itself could have evolved by similar means.
Biochemistry attempts to describe the mechanisms that give rise to living sys-
tems from the perspective of the molecules that make up living things. We can
often gain considerable insight into the properties of a living thing by studying
the structure and chemistry of its molecules.
