Chemical Foundations
bonds increases the order and therefore decreases the entropy (S becomes more negative).
Less energy is required to form noncovalent bonds than covalent bonds, and the bonds that stick the
Stereoisomers are compounds that have the same
gecko’s feet to the smooth surface need to be
molecular formula but are mirror images of each
formed and broken many times as the animal
other. Many organic molecules can exist as
moves. Since van der Waals interactions are so
stereoisomers because of two different possible
weak, there must be many points of contact (a
orientations around an asymmetric carbon atom
large surface area) yielding multiple van der Waals
(e.g. amino acids). Because stereoisomers differ in
interactions between the septae and the smooth
their three dimensional orientation and because
biological molecules interact with one another based on precise molecular complementarity, stereoisomers often react with different molecules,
2a. These are likely to be hydrophilic amino acids,
or react differently with the same molecules.
and in particular, negatively charged amino acids
(aspartate and glutamate), which would have an
The compound is guanosine triphosphate (GTP).
2b. Like the phospholipid bilayer itself, this portion of
Although the guanine base is found in both DNA
the protein is likely to be amphipathic, with hydrophobic amino acids in contact with the fatty
and RNA, the sugar is a ribose sugar because of
acyl chains and hydrophilic amino acids in contact
the 2′ hydroxyl group. Therefore, GTP is a component of RNA only. GTP is an important
2c,d. Since both the cytosol and extracellular space are
At least three properties contribute to this
aqueous environments, hydrophilic amino acids
structural diversity. First, monosaccharides can be
joined to one another at any of several hydroxyl groups. Second, the C-1 linkage can have either an
At pH = 7.0, the net charge is –1 because of the
α or a β configuration. Third, extensive branching
negative charge on the carboxyl residue of
glutamate (E). After phosphorylation by a tyrosine kinase, two additional negative charges (because
In the acidic pH of a lysosome, ammonia is
of attachment of phosphate residues to tyrosines
converted to ammonium ion. Ammonium ion is
(Y)) would be added. Thus, the net charge would be –3. The most likely source of phosphate is ATP
unable to traverse the membrane because of its
since the attachment of inorganic phosphate (P
positive charge and is trapped within the
lysosome. The accumulation of ammonium ion
to tyrosine is energetically highly unfavorable, but
decreases the concentration of protons within
when coupled to the hydrolysis of the high-energy
lysosomes and therefore elevates lysosomal pH. At
phosphoanhydride bond of ATP, the overall
neutral pH, ammonia has little, if any, tendency to
protonate to ammonium ion and thus has no effect on cytosolic pH.
Disulfide bonds are formed between two cysteine residue side chains. The formation of disulfide
straight chain. There is no such kink in a trans unsaturated fatty acyl chain.
Since 90% of L binds R, the concentration of LR at equilibrium is 0.9(1 × 10−3M) = 9 × 10−4 M. The
13. Glutamate is the amino acid that undergoes
concentration of free L at equilibrium is the 10%
g-carboxylation, resulting in the formation of a
of L that remains unbound, 1 × 10−4 M. The
host of blood clotting factors. Warfarin inhibits
concentration of R at equilibrium is (5 × 10−2 M)
g-carboxylation of glutamate. Thus, blood
– (9 × 10−4 M) = 4.91 × 10−2 M. Therefore,
clotting is severely compromised. Patients prone
[LR]/[L][R] = 9 × 10−4 M/ ((1 × 10−4 M) (4.91 ×
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 equilibrium constant is unaffected by the
the body). Patients at risk for heart disease due to
blockages in the coronary arteries are also often prescribed this drug.
10. The pH of cytosol is 7.2, which is the pKa for the
Therefore, phosphoric acid will exist as a mixture of H
phosphoric acid equals the optimal pH of cytosol, phosphoric acid serves as a biological buffer to help the cell maintain a constant pH.
11. ∆G = ∆Gº′+ RTln [products]/[reactants]
For this reaction, ∆G = −1000 cal/mol + [1.987 cal/(degree·mol) × (298 degrees) × ln (0.01 M/ (0.01 M × 0.01 M))].
∆G = −1000 cal/mol + 2727 cal/mol = 1727 cal/mol
To make this reaction energetically favorable, one could increase the concentration 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
Read Cussler Ch. 5-6 Problem 1 Problem 2 Problem 3 Problem 4 Problem 5 Problem 6 Problem 7 The bioavailability of drugs delivered orally is often low. Following absorption from the digestive tract into the blood, many drugs are quickly converted by the liver to an inactive form before they can reach the rest of the circulation. One way to circumvent
Preventive Effects of Rosiglitazone on Restenosis after Coronary Stenting in Patients with Type 2 Diabetes Donghoon Choi, MD, PhD Cardiology Division Yonsei University College of Medicine, Background 1. Cardiovascular disease is one of the important leading cause of deaths in Type 2 diabetic patients. 2. As a result of dramatic increase in implantation numbers, in