STEP 1 -
The first step in DNA synthesis is to attach a protected deoxynucleoside to a silica $\left(\mathrm{SiO}_{2}\right)$ support by an ester linkage to the $3^{\prime}-\mathrm{OH}$ group of the deoxynucleoside. Both the $5^{\prime}-\mathrm{OH}$ group on the sugar and free $-\mathrm{NH}_{2}$ groups on the heterocyclic bases must be protected. Adenine and cytosine bases are protected by benzoyl groups, guanine is protected by an isobutyryl group, and thymine requires no protection. The deoxyribose $5^{\prime}-\mathrm{OH}$ is protected as its p-dimethoxytrityl (DMT) ether.

adenine

N-protected
guanine

N-protected cytosine

Thymine

STEP 2
The second step is removal of the DMT protecting group by treatment with dichloroacetic acid in $\mathrm{CH}_{2} \mathrm{Cl}_{2}$. The reaction occurs by an $\mathrm{S}_{\mathrm{N}} 1$ mechanism and proceeds rapidly because of the stability of the tertiary, benzylic dimethoxytrityl cation.

STEP 3
The third step is the coupling of the polymer-bonded deoxynucleoside with a protected deoxynucleoside containing a phosphoramidite group $\left[\mathrm{R}_{2} \mathrm{NP}(\mathrm{OR})_{2}\right]$ at its $3^{\prime}$ position. The coupling reaction takes place in the polar aprotic solvent acetonitrile, requires catalysis by the heterocyclic amine tetrazole, and yields a phosphite, $\mathrm{P}(\mathrm{OR})_{3}$, as product. Note that one of the phosphorus oxygen atoms is protected by a $\beta$-cyanoethyl group, $-\mathrm{OCH}_{2} \mathrm{CH}_{2} \mathrm{C} \equiv \mathrm{N}$. The coupling step takes place with better than $99 \%$ yield.


STEP 4 -
With the coupling accomplished, the phosphite product is oxidized to a phosphate by treatment with iodine in aqueous tetrahydrofuran in the presence of 2,6-dimethylpyridine. The cycle of (1) deprotection, (2) coupling, and (3) oxidation is then repeated until an oligonucleotide chain of the desired sequence has been constructed.

STEP 5
The final step is removal of all protecting groups and cleavage of the ester bond holding the DNA to the silica. All these reactions are done at the same time by treatment with aqueous $\mathrm{NH}_{3}$. Purification by electrophoresis then yields the synthetic DNA.

PROBLEM p-Dimethoxytrityl (DMT) ethers are easily cleaved by mild acid treatment. Show the mechanism of 28-11 the cleavage reaction.

PROBLEM Propose a mechanism to account for cleavage of the $\beta$-cyanoethyl protecting group from the 28-12 phosphate groups on treatment with aqueous ammonia. (Acrylonitrile, $\mathrm{H}_{2} \mathrm{C}=\mathrm{CHCN}$, is a byproduct.) What kind of reaction is occurring?


STEP 4 - 28.8 The Polymerase Chain Reaction
It often happens that only a tiny amount of DNA can be obtained directly, as might occur at a crime scene, so methods for obtaining larger amounts are sometimes needed to carry out sequencing and characterization. The invention of the polymerase chain reaction (PCR) by Kary Mullis in 1986 has been described as being to genes what Gutenberg's invention of the printing press was to the written word. Just as the printing press produces multiple copies of a book, PCR produces multiple copies of a given DNA sequence. Starting from less than 1 picogram ( pg ) of DNA with a chain length of 10,000 nucleotides ( $1 \mathrm{pg}=10^{-12} \mathrm{~g}$; about $10^{5}$ molecules), PCR makes it possible to obtain several micrograms ( $1 \mu \mathrm{~g}=10^{-6} \mathrm{~g}$; about $10^{11}$ molecules) in just a few hours.

The key to the polymerase chain reaction is Taq DNA polymerase, a heat-stable enzyme isolated from the thermophilic bacterium Thermus aquaticus found in a hot spring in Yellowstone National Park. Taq polymerase is able to take a single strand of DNA having a short, primer segment of complementary chain at one end and then finish constructing the entire complementary strand. The overall process takes three steps, as shown in FIGURE 28.10. More recently, improved heat-stable DNA polymerases have become available, including Vent polymerase and Pfu polymerase, both isolated from bacteria growing near geothermal vents in the ocean floor. The error rate of both enzymes is substantially less than that of Taq.

FIGURE 28.10 The polymerase chain reaction. Details are explained in the text.
STEP 1
The double-stranded DNA to be amplified is heated in the presence of Taq polymerase, $\mathrm{Mg}^{2+}$ ion, the four deoxynucleotide triphosphate monomers (dNTPs), and a large excess of two short oligonucleotide primers of about 20 bases each. Each primer is complementary to the sequence at the end of one of the target DNA segments. At a temperature of $95^{\circ} \mathrm{C}$, double-stranded DNA denatures, spontaneously breaking apart into two single strands.


STEP 4 - 28.8 The Polymerase Chain Reaction
STEP 2
The temperature is lowered to between 37 and $50^{\circ} \mathrm{C}$, allowing the primers, because of their relatively high concentration, to anneal by hydrogen-bonding to their complementary sequence at the end of each target strand.

STEP 3
The temperature is then raised to $72^{\circ} \mathrm{C}$, and Taq polymerase catalyzes the addition of further nucleotides to the two primed DNA strands. When replication of each strand is complete, two copies of the original DNA now exist. Repeating the denature-anneal-synthesize cycle a second time yields four DNA copies, repeating a third time yields eight copies, and so on, in an exponential series.

PCR has been automated, and 30 or so cycles can be carried out in an hour, resulting in a theoretical amplification factor of $2^{30}\left(\sim 10^{9}\right)$. In practice, however, the efficiency of each cycle is less than $100 \%$, and an experimental amplification of about $10^{6}$ to $10^{8}$ is routinely achieved for 30 cycles.


DNA Fingerprinting -
The invention of DNA sequencing has affected society in many ways, few more dramatic than those stemming from the development of DNA fingerprinting. DNA fingerprinting arose from the discovery in 1984 that human genes contain short, repeating sequences of noncoding DNA, called short tandem repeat (STR) loci. Furthermore, the STR loci are slightly different for everyone except identical twins. By sequencing these loci, a pattern unique to each person can be obtained.

Perhaps the most common and well-publicized use of DNA fingerprinting is that carried out by crime laboratories to link suspects to biological evidence-blood, hair follicles, skin, or semen-found at a crime scene. Many thousands of court cases have been decided based on DNA evidence.

For use in criminal cases, forensic laboratories in the United States have agreed on 13 core STR loci that are most accurate for the identification of an individual. Based on these 13 loci, a Combined DNA Index System (CODIS) has been established to serve as a registry of convicted offenders. When a DNA sample is obtained from a crime scene, the sample is subjected to cleavage with restriction endonucleases to cut out fragments containing the STR loci, the fragments are amplified using the polymerase chain reaction, and the sequences of the fragments are determined.

If the profile of sequences from a known individual and the profile from DNA obtained at a crime scene match, the probability is approximately 82 billion to 1 that the DNA is from the same individual. In paternity cases, where the DNA of male parent and offspring are related but not fully identical, the identity of the male parent can be established with a probability of around 100,000 to 1 . Even after several generations, paternity can still be inferred from DNA analysis of the $Y$ chromosome of direct male-line descendants. The most well-known such case is that of Thomas Jefferson, one of the nation's founders, who likely impregnated enslaved person Sally Hemings. Although Jefferson himself has no male-line descendants, DNA analysis of the male-line descendants of Jefferson's paternal uncle contained the same Y chromosome as a male-line descendant of Eston Hemings, the youngest son of Sally Hemings. Thus, a mixing of the two genomes is clear, although the male individual responsible for that mixing can't be identified with $100 \%$ certainty.


DNA Fingerprinting -
Among its many other applications, DNA fingerprinting is widely used for the diagnosis of genetic disorders, both prenatally and in newborns. Cystic fibrosis, hemophilia, Huntington's disease, Tay-Sachs disease, sickle cell anemia, and thalassemia are among the many diseases that can be detected, enabling early treatment of an affected child. Furthermore, by studying the DNA fingerprints of relatives with a history of a particular disorder, it's possible to identify DNA patterns associated with the disease and perhaps obtain clues for an eventual cure. In addition, the U.S. Department of Defense requires blood and saliva samples from all military personnel. The samples are stored, and DNA is extracted if the need for identification of a casualty arises.


Key Terms -
- anticodon
- antisense strand
- codon
- deoxyribonucleic acid (DNA)
- double helix
- 3' end
- 5' end
- messenger RNA (mRNA)
- nucleoside
- nucleotides
- polymerase chain reaction (PCR)
- replication
- ribonucleic acid (RNA)
- ribosomal RNA (rRNA)
- Sanger dideoxy method
- sense strand
- small RNAs
- transcription
- transfer RNA (tRNA)
- translation


Summary -
DNA (deoxyribonucleic acid) and RNA (ribonucleic acid) are biological polymers that act as chemical carriers of an organism's genetic information. Enzyme-catalyzed hydrolysis of nucleic acids yields nucleotides, the monomer units from which RNA and DNA are constructed. Further enzyme-catalyzed hydrolysis of the nucleotides yields nucleosides plus phosphate. Nucleosides, in turn, consist of a purine or pyrimidine base linked to the C1 of an aldopentose sugar-ribose in RNA and 2-deoxyribose in DNA. The nucleotides are joined by phosphate links between the $5^{\prime}$ phosphate of one nucleotide and the $3^{\prime}$ hydroxyl on the sugar of another nucleotide.

Molecules of DNA consist of two complementary polynucleotide strands held together by hydrogen bonds between heterocyclic bases on the different strands and coiled into a double helix. Adenine and thymine form hydrogen bonds to each other, as do cytosine and guanine.

Three processes take place in deciphering the genetic information of DNA:

- Replication of DNA is the process by which identical DNA copies are made. The DNA double helix unwinds, complementary deoxyribonucleotides line up in order, and two new DNA molecules are produced.
- Transcription is the process by which RNA is produced to carry genetic information from the nucleus to the ribosomes. A short segment of the DNA double helix unwinds, and complementary ribonucleotides line up to produce messenger RNA (mRNA).
- Translation is the process by which mRNA directs protein synthesis. Each mRNA is divided into codons, ribonucleotide triplets that are recognized by small amino acid-carrying molecules of transfer RNA (tRNA), which deliver the appropriate amino acids needed for protein synthesis.

Sequencing of DNA is carried out by the Sanger dideoxy method, and small DNA segments can be synthesized in the laboratory by automated instruments. Small amounts of DNA can be amplified by factors of $10^{6}$ using the polymerase chain reaction (PCR).


Additional Problems -
Visualizing Chemistry
PROBLEM Identify the following bases, and tell whether each is found in DNA, RNA, or both:
28-13 (a)

(b)

(c)

PROBLEM Identify the following nucleotide, and tell how it is used:


28-14 -
PROBLEM Amine bases in nucleic acids can react with alkylating agents in typical $\mathrm{S}_{\mathrm{N}} 2$ reactions. Look at the
28-15 following electrostatic potential maps, and tell which is the better nucleophile, guanine or adenine. The reactive positions in each are indicated.

9-Methylguanine

9-Methyladenine


Mechanism Problems -
PROBLEM The final step in DNA synthesis is deprotection by treatment with aqueous ammonia. Show the
28-16 mechanisms by which deprotection occurs at the points indicated in the following structure:

PROBLEM The final step in the metabolic degradation of uracil is the oxidation of malonic semialdehyde to
28-17 give malonyl CoA. Propose a mechanism.

PROBLEM One of the steps in the biosynthesis of a nucleotide called inosine monophosphate is the formation 28-18 of aminoimidazole ribonucleotide from formylglycinamidine ribonucleotide. Propose a mechanism.

PROBLEM One of the steps in the metabolic degradation of guanine is hydrolysis to give xanthine. Propose a 28-19 mechanism.

PROBLEM One of the steps in the biosynthesis of uridine monophosphate is the reaction of aspartate with
28-20 carbamoyl phosphate to give carbamoyl aspartate followed by cyclization to form dihydroorotate. Propose mechanisms for both steps.


General Problems -
PROBLEM Human brain natriuretic peptide (BNP) is a small peptide of 32 amino acids used in the treatment
28-21 of congestive heart failure. How many nitrogen bases are present in the DNA that codes for BNP?
PROBLEM Human and horse insulin both have two polypeptide chains, with one chain containing 21 amino
28-22 acids and the other containing 30 amino acids. They differ in primary structure at two places. At position 9 in one chain, human insulin has Ser and horse insulin has Gly; at position 30 in the other chain, human insulin has Thr and horse insulin has Ala. How must the DNA for the two insulins differ?

PROBLEM The DNA of sea urchins contains about $32 \%$ A. What percentages of the other three bases would you
28-23 expect in sea urchin DNA? Explain.
PROBLEM The codon UAA stops protein synthesis. Why does the sequence UAA in the following stretch of 28-24 mRNA not cause any problems?

PROBLEM Which of the following base sequences would most likely be recognized by a restriction 28-25 endonuclease? Explain.
(a) GAATTC
(b) GATTACA
(c) CTCGAG

PROBLEM For what amino acids do the following ribonucleotide triplets code?
28-26
(a) AAU
(b) GAG
(c) UCC
(d) CAU

PROBLEM From what DNA sequences were each of the mRNA codons in Problem 28-26 transcribed? 28-27

PROBLEM What anticodon sequences of tRNAs are coded for by the codons in Problem 28-26? 28-28

PROBLEM Draw the complete structure of the ribonucleotide codon UAC. For what amino acid does this 28-29 sequence code?

PROBLEM Draw the complete structure of the deoxyribonucleotide sequence from which the mRNA codon in 28-30 Problem 28-29 was transcribed.

PROBLEM Give an mRNA sequence that will code for the synthesis of metenkephalin. 28-31
Tyr-Gly-Gly-Phe-Met

PROBLEM Give an mRNA sequence that will code for the synthesis of angiotensin II. 28-32


General Problems -
PROBLEM Give an mRNA sequence that will code for the synthesis of angiotensin II. 28-32

> Asp-Arg-Val-Tyr-Ile-His-Pro-Phe

PROBLEM What amino acid sequence is coded for by the following DNA coding strand (sense strand)?
28-33
(5') CTT-CGA-CCA-GAC-AGC-TTT (3')

PROBLEM What amino acid sequence is coded for by the following mRNA base sequence?
28-34
(5') CUA-GAC-CGU-UCC-AAG-UGA (3')
PROBLEM If the DNA coding sequence -CAA-CCG-GAT- were miscopied during replication and became -CGA-28-35 CCG-GAT-, what effect would there be on the sequence of the protein produced?

PROBLEM Show the steps involved in a laboratory synthesis of the DNA fragment with the sequence CTAG. 28-36

PROBLEM Draw the structure of cyclic adenosine monophosphate (cAMP), a messenger involved in the 28-37 regulation of glucose production in the body. Cyclic AMP has a phosphate ring connecting the $3^{\prime}$ and 5 '-hydroxyl groups on adenosine.

PROBLEM Valganciclovir, marketed as Valcyte, is an antiviral agent used for the treatment of cytomegalovirus.
28-38 Called a prodrug, valganciclovir is inactive by itself but is rapidly converted in the intestine by hydrolysis of its ester bond to produce an active drug, called ganciclovir, along with an amino acid.

Valganciclovir
(a) What amino acid is produced by hydrolysis of the ester bond in valganciclovir?
(b) What is the structure of ganciclovir?
(c) What atoms present in the nucleotide deoxyguanine are missing from ganciclovir?
(d) What role do the atoms missing from deoxyguanine play in DNA replication?
(e) How might valganciclovir interfere with DNA synthesis?


CHAPTER 29 -
The Organic Chemistry of Metabolic Pathways

FIGURE 29.1 Acyl CoA dehydrogenase is an enzyme that catalyzes the introduction of a $\mathbf{C =} \mathbf{C}$ double bond into fatty acids during their metabolism. (credit: modification of image from the RCSB PDF (rcsb.org) of PDB ID 2WBI. Muniz, J.R.C., Guo, K., Savitsky, P., Roos, A., Yue, W., Pilka, E., Vondelft, F., Edwards, A.M., Bountra, C., Arrowsmith, C.H., Weigelt, J., Oppermann, U. Crystal structure of human Acyl-CoA dehydrogenase 11)


CHAPTER CONTENTS -
29.1 An Overview of Metabolism and Biochemical Energy
29.2 Catabolism of Triacylglycerols: The Fate of Glycerol
29.3 Catabolism of Triacylglycerols: $\beta$-Oxidation
29.4 Biosynthesis of Fatty Acids
29.5 Catabolism of Carbohydrates: Glycolysis
29.6 Conversion of Pyruvate to Acetyl CoA
29.7 The Citric Acid Cycle
29.8 Carbohydrate Biosynthesis: Gluconeogenesis
29.9 Catabolism of Proteins: Deamination
29.10 Some Conclusions about Biological Chemistry

WHY THIS CHAPTER? In this chapter, we'll look at some of the pathways by which organisms carry out their chemistry, focusing primarily on how they metabolize fats and carbohydrates. The treatment will be far from complete, but it should give you an idea of the kinds of processes that occur.

Anyone who wants to understand or contribute to the revolution now taking place in the biological sciences must first understand life processes at the molecular level. This understanding, in turn, must be based on a detailed knowledge of the chemical reactions and pathways used by living organisms. Just knowing what occurs is not enough; it's also necessary to understand how and why organisms use the chemistry they do.

Biochemical reactions are not mysterious. Even though the biological reactions that take place in living organisms often appear complicated, they follow the same rules of reactivity as laboratory reactions and they operate by the same mechanisms.

Some of the molecules we'll be encountering are substantially larger and more complex than those we've been dealing with thus far. But don't be intimidated; keep your focus on the parts of the molecules where changes occur, and ignore the parts where nothing changes. The reactions themselves are exactly the same additions, eliminations, substitutions, carbonyl condensations, and so forth, that we've been dealing with all along. By the end of this chapter, it should be clear that the chemistry of living organisms is organic chemistry.


CHAPTER CONTENTS - 29.1 An Overview of Metabolism and Biochemical Energy
The many reactions that occur in the cells of living organisms are collectively called metabolism. The pathways that break down larger molecules into smaller ones are called catabolism, and the pathways that synthesize larger biomolecules from smaller ones are known as anabolism. Catabolic reaction pathways are usually exergonic and release energy, while anabolic pathways are often endergonic and absorb energy. Catabolism can be divided into the four stages shown in FIGURE 29.2.

FIGURE 29.2 An overview of catabolic pathways for the degradation of food and the production of biochemical energy. The ultimate products of food catabolism are $\mathrm{CO}_{2}$ and $\mathrm{H}_{2} \mathrm{O}$, with the energy released in the citric acid cycle used to drive the endergonic synthesis of adenosine triphosphate (ATP) from adenosine diphosphate (ADP) plus hydrogen phosphate ion, $\mathrm{HOPO}_{3}{ }^{2-}$.

In the first catabolic stage, commonly called digestion, food is broken down in the mouth, stomach, and small intestine by hydrolysis of ester, acetal (glycoside), and amide (peptide) bonds to yield fatty acids, simple sugars,
and amino acids. These smaller molecules are then absorbed and further degraded in the second stage of catabolism to yield acetyl groups attached by a thioester bond to the large carrier molecule, coenzyme A. The resultant compound, acetyl coenzyme A (acetyl CoA), is a key substance in the metabolism of food molecules and in many other biological pathways. As noted in Section 21.8, the acetyl group in acetyl CoA is linked to the sulfur atom of phosphopantetheine, which is itself linked to adenosine 3',5'-bisphosphate.


Acetyl CoA-a thioester -
Acetyl groups are oxidized inside cellular mitochondria in the third stage of catabolism, the citric acid cycle, to yield $\mathrm{CO}_{2}$. (We'll see the details of the process in Section 29.7.) Like most oxidations, this stage releases a large amount of energy, which is used in the fourth stage, the electron-transport chain, to accomplish the endergonic phosphorylation of adenosine diphosphate (ADP) with hydrogen phosphate ion ( $\mathrm{HOPO}_{3}{ }^{2-}$, abbreviated $\mathrm{P}_{\mathrm{i}}$ ) to give adenosine triphosphate (ATP).

As the final result of food catabolism, ATP has been called the "energy currency" of the cell. Catabolic reactions "buy" ATP with the energy they release to synthesize it from ADP and hydrogen phosphate ion. Anabolic reactions then spend the ATP by transferring a phosphate group to another molecule, thereby regenerating ADP. Energy production and use in living organisms thus revolves around the ATP $\rightleftarrows$ ADP interconversion.

breaking a $\mathrm{C}-\mathrm{O}$ bond and forming a carboxylic ester, ROCOR' (Section 21.5), phosphoric acid anhydrides react with alcohols by breaking a $\mathrm{P}-\mathrm{O}$ bond and forming a phosphate ester, $\mathrm{ROPO}_{3}{ }^{2-}$. The reaction is, in effect, a nucleophilic acyl substitution at phosphorus. Note that phosphorylation reactions with ATP generally require the presence of a divalent metal cation in the enzyme, usually $\mathrm{Mg}^{2+}$, to form a Lewis acid-base complex with the phosphate oxygen atoms and to neutralize negative charge.

How does the body use ATP? Recall from Section 6.7 that the free-energy change $\Delta G$ must be negative and energy must be released for a reaction to be energetically favorable and occur spontaneously. If $\Delta G$ is positive, the reaction is energetically unfavorable and the process can't occur spontaneously.


Acetyl CoA-a thioester -
For an energetically unfavorable reaction to occur, it must be "coupled" to an energetically favorable reaction so that the overall free-energy change for the two reactions together is favorable. To understand what it means for reactions to be coupled, imagine that reaction 1 does not occur to any reasonable extent because it has a small equilibrium constant and is energetically unfavorable; that is, the reaction has $\Delta G>0$.

$$
(1) \mathbf{A}+m \longleftrightarrow \mathbf{B}+n \quad \Delta G>0
$$

where $\mathbf{A}$ and $\mathbf{B}$ are the biochemically "important" substances while $m$ and $n$ are enzyme cofactors, $\mathrm{H}_{2} \mathrm{O}$, or other small molecules.

Imagine also that product $n$ can react with substance $o$ to yield $p$ and $q$ in a second, highly favorable reaction that has a large equilibrium constant and $\Delta G \ll 0$.
(2) $n+o \rightleftarrows p+q \quad \Delta G \ll 0$

Taking the two reactions together, they share, or are coupled through, the common intermediate $n$, which is a product in the first reaction and a reactant in the second. When even a tiny amount of $n$ is formed in reaction 1 , it undergoes essentially complete conversion in reaction 2 , thereby removing it from the first equilibrium and forcing reaction 1 to continually replenish $n$ until reactant $\mathbf{A}$ is gone. That is, the two reactions added together have a favorable $\Delta G<0$, and we say that the favorable reaction 2 "drives" the unfavorable reaction 1 . Because the two reactions are coupled through $n$, the transformation of $\mathbf{A}$ to $\mathbf{B}$ becomes favorable.


Acetyl CoA-a thioester -
| (1) $\mathbf{A}+m \longleftrightarrow \mathbf{B}+\not h$ | $\Delta G>0$ |
| :--- | :--- |
| (2) $\not \propto+o \rightleftarrows p+q$ | $\Delta G \ll 0$ |
| Net: $\mathbf{A}+m+o \rightleftarrows \mathbf{B}+p+q$ | $\Delta G<0$ |

For an example of two reactions that are coupled, look at the phosphorylation reaction of glucose to yield glucose 6-phosphate plus water, an important step in the breakdown of dietary carbohydrates.

Glucose
Glucose 6-phosphate
The reaction of glucose with $\mathrm{HOPO}_{3}{ }^{2-}$ does not occur spontaneously because it is energetically unfavorable, with $\Delta G^{\circ{ }^{\prime}}=13.8 \mathrm{~kJ} / \mathrm{mol}$. (The standard free-energy change for a biological reaction is denoted $\Delta G^{\circ}$ and refers to a process in which reactants and products have a concentration of 1.0 M in a solution with $\mathrm{pH}=7$.) At the same time, however, the reaction of water with ATP to yield ADP plus $\mathrm{HOPO}_{3}{ }^{2-}$ is strongly favorable, with $\Delta G^{\circ}$
$=-30.5 \mathrm{~kJ} / \mathrm{mol}$. When the two reactions are coupled, glucose reacts with ATP to yield glucose 6-phosphate plus ADP in a reaction that is favorable by about $16.7 \mathrm{~kJ} / \mathrm{mol}(4.0 \mathrm{kcal} / \mathrm{mol})$. That is, ATP drives the phosphorylation reaction of glucose.

It's this ability to drive otherwise unfavorable phosphorylation reactions that makes ATP so useful. The resultant phosphates are much more reactive as leaving groups in nucleophilic substitutions and eliminations than the alcohols they're derived from and are therefore more chemically useful.


Acetyl CoA-a thioester -
PROBLEM One of the steps in fat metabolism is the reaction of glycerol (1,2,3-propanetriol) with ATP to yield 29-1 glycerol 1-phosphate. Write the reaction, and draw the structure of glycerol 1-phosphate.


Acetyl CoA-a thioester - 29.2 Catabolism of Triacylglycerols: The Fate of Glycerol
The metabolic breakdown of triacylglycerols begins with their hydrolysis in the stomach and small intestine to yield glycerol plus fatty acids. The reaction is catalyzed by a lipase, whose mechanism is shown in FIGURE 29.3. The active site of the enzyme contains a catalytic triad of aspartic acid, histidine, and serine residues, which act cooperatively to provide the necessary acid and base catalysis for the individual steps. Hydrolysis is accomplished by two sequential nucleophilic acyl substitution reactions, one that covalently binds an acyl group to the side chain -OH of a serine residue on the enzyme and a second that frees the fatty acid from the enzyme.


FIGURE 29.3 MECHANISM -
Mechanism for the action of lipase. The active site of the enzyme contains a catalytic triad of aspartic acid, histidine, and serine, which react cooperatively to carry out two nucleophilic acyl substitution reactions. Individual steps are explained in the text.
(1) The enzyme active site contains an aspartic acid, a histidine, and a serine. First, histidine acts as a base to deprotonate the -OH group of serine, with the negatively charged carboxylate of aspartic acid stabilizing the nearby histidine cation that results. Serine then adds to the carbonyl group of the triacylglycerol, yielding a tetrahedral intermediate.

This intermediate expels a diacylglycerol as leaving group in a nucleophilic acyl substitution reaction, giving an acyl enzyme. The diacylglycerol is protonated by the histidine cation.

3 Histidine deprotonates a water molecule, which adds to the acyl group. A tetrahedral intermediate is again formed, and the histidine cation is again stabilized by the nearby carboxylate.
4) The tetrahedral intermediate expels the serine as leaving group in a second nucleophilic acyl substitution reaction, yielding a free fatty acid. The serine accepts a proton from histidine, and the enzyme has now returned to its starting structure.

Tetrahedral intermediate
(2)

(3) $\downarrow \mathrm{H}_{2} \mathrm{O}$

Tetrahedral intermediate


(4) -
Enzyme
Fatty acid

Steps 1-2 of FIGURE 29.3: Acyl Enzyme Formation
The first nucleophilic acyl substitution step-reaction of the triacylglycerol with the active-site serine to give an acyl enzyme-begins with deprotonation of the serine alcohol by histidine to form the more strongly nucleophilic alkoxide ion. This proton transfer is facilitated by a nearby side-chain carboxylate anion of aspartic acid, which makes the histidine more basic and stabilizes the resultant histidine cation by electrostatic interactions. The deprotonated serine adds to a carbonyl group of a triacylglycerol to give a tetrahedral intermediate, which expels a diacylglycerol as a leaving group and produces an acyl enzyme. This step is catalyzed by a proton transfer from histidine to make the leaving group a neutral alcohol.

Steps 3-4 of FIGURE 29.3: Hydrolysis
The second nucleophilic acyl substitution step hydrolyzes the acyl enzyme and gives the free fatty acid by a mechanism analogous to that of the first two steps. Water is deprotonated by histidine to give hydroxide ion, which adds to the enzyme-bound acyl group. The tetrahedral intermediate then expels the neutral serine residue as the leaving group, freeing the fatty acid and returning the enzyme to its active form.

The fatty acids released on triacylglycerol hydrolysis are transported to mitochondria and degraded to acetyl CoA, while the glycerol is carried to the liver for further metabolism. In the liver, glycerol is first phosphorylated by reaction with ATP and then oxidized by $\mathrm{NAD}^{+}$. The resulting dihydroxyacetone phosphate (DHAP) enters the carbohydrate glycolysis pathway, which we'll discuss in Section 29.5.

You might note that C2 of glycerol is a prochiral center (Section 5.11) with two identical "arms." As is typical for enzyme-catalyzed reactions, the phosphorylation of glycerol is selective. Only the pro- $R$ arm undergoes reaction, although this can't be predicted in advance.


(4) -
Note also that the phosphorylation product is named $s n$-glycerol 3-phosphate, where the $s n$ - prefix means "stereospecific numbering." In this convention, the molecule is drawn as a Fischer projection with the -OH group at C2 pointing to the left and the glycerol carbon atoms numbered from the top.


(4) - 29.3 Catabolism of Triacylglycerols: $\boldsymbol{\beta}$-Oxidation
The fatty acids that result from triacylglycerol hydrolysis are converted into thioesters with coenzyme A and then catabolized by a repetitive four-step sequence of reactions called the $\boldsymbol{\beta}$-oxidation pathway, shown in FIGURE 29.4. Each passage along the pathway results in the cleavage of an acetyl group from the end of the fatty-acid chain, until the entire molecule is ultimately degraded. As each acetyl group is produced, it enters the citric acid cycle and is further degraded to $\mathrm{CO}_{2}$, as we'll see in Section 29.7.


FIGURE 29.4 MECHANISM -
The four steps of the $\boldsymbol{\beta}$-oxidation pathway, resulting in the cleavage of an acetyl group from the end of the fatty-acid chain. The key chain-shortening step is a retro-Claisen reaction of a $\beta$-keto thioester. Individual steps are explained in the text.

Fatty acyl CoA
(1) A conjugated double bond is introduced by removal of hydrogens from C2 and C3 by the coenzyme flavin adenine
dinucleotide (FAD).

$\alpha, \beta$-Unsaturated acyl CoA
(2) $\downarrow \mathrm{H}_{2} \mathrm{O}$ $\beta$-hydroxyacyl CoA.

$\beta$-Hydroxyacyl CoA
(3) The alcohol is oxidized by $\mathrm{NAD}^{+}$to give a $\beta$-keto thioester.

(4) Nucleophilic addition of coenzyme A to the keto group occurs, followed by a retro-Claisen condensation reaction. The products are acetyl CoA and a chain-shortened fatty acyl CoA.

Acetyl CoA

Step 1 of FIGURE 29.4: Introduction of a Double Bond
The $\beta$-oxidation pathway begins when two hydrogen atoms are removed from C 2 and C 3 of the fatty acyl CoA by one of a family of acyl-CoA dehydrogenases to yield an $\alpha, \beta$-unsaturated acyl CoA. This kind of oxidation-the introduction of a conjugated double bond into a carbonyl compound-occurs frequently in biochemical pathways and usually involves the coenzyme flavin adenine dinucleotide (FAD). Reduced $\mathrm{FADH}_{2}$ is the byproduct.


FIGURE 29.4 MECHANISM -
The mechanisms of FAD-catalyzed reactions are often difficult to establish because flavin coenzymes can follow both two-electron (polar) and one-electron (radical) pathways. As a result, extensive studies of the family of acyl-CoA dehydrogenases have not yet provided a clear picture of how these enzymes function. What is known is that: (1) The first step is abstraction of the pro- $R$ hydrogen from the acidic $\alpha$ position of the acyl CoA to give a thioester enolate ion. Hydrogen-bonding between the acyl carbonyl group and the ribitol hydroxyls of FAD increases the acidity of the acyl group. (2) The pro- $R$ hydrogen at the $\beta$ position is transferred to FAD. (3) The $\alpha, \beta$-unsaturated acyl CoA that results has a trans double bond.

One suggested mechanism has the reaction taking place by a conjugate nucleophilic addition of hydride, analogous to what occurs during alcohol oxidations with $\mathrm{NAD}^{+}$. Electrons on the enolate ion might expel a $\beta$ hydride ion, which could add to the doubly bonded N5 nitrogen on FAD. Protonation of the intermediate at N1 would give the product.

Step 2 of FIGURE 29.4: Conjugate Addition of Water
The $\alpha, \beta$-unsaturated acyl CoA produced in step 1 reacts with water by a conjugate addition pathway (Section 19.13) to yield a $\beta$-hydroxyacyl CoA in a process catalyzed by enoyl CoA hydratase. Water as nucleophile adds to the $\beta$ carbon of the double bond, yielding an intermediate thioester enolate ion that is protonated on the $\alpha$ position.


FIGURE 29.4 MECHANISM -
(3S)-Hydroxyacyl CoA
Step 3 of FIGURE 29.4: Alcohol Oxidation
The $\beta$-hydroxyacyl CoA from step 2 is oxidized to a $\beta$-ketoacyl CoA in a reaction catalyzed by one of a family of L-3-hydroxyacyl-CoA dehydrogenases, which differ in substrate specificity according to the chain length of the acyl group. As in the oxidation of $s n$-glycerol 3-phosphate to dihydroxyacetone phosphate mentioned at the end of Section 29.2, this alcohol oxidation requires $\mathrm{NAD}^{+}$as a coenzyme and yields reduced $\mathrm{NADH} / \mathrm{H}^{+}$as byproduct. Deprotonation of the hydroxyl group is carried out by a histidine residue at the active site.


Step 4 of FIGURE 29.4: Chain Cleavage -
Acetyl CoA is split off from the chain in the final step of $\beta$-oxidation, leaving an acyl CoA that is two carbon atoms shorter than the original. The reaction is catalyzed by $\beta$-ketoacyl-CoA thiolase and is mechanistically the reverse of a Claisen condensation reaction (Section 23.7). In the forward direction, a Claisen condensation joins two esters together to form a $\beta$-keto ester product. In the reverse direction, a retro-Claisen reaction splits apart a $\beta$-keto ester (or $\beta$-keto thioester in this case) to form two esters (or two thioesters).

The retro-Claisen reaction occurs by nucleophilic addition of a cysteine - SH group on the enzyme to the keto group of the $\beta$-ketoacyl CoA to yield an alkoxide ion intermediate. Cleavage of the $\mathrm{C} 2-\mathrm{C} 3$ bond then follows, with the expulsion of an acetyl CoA enolate ion that is immediately protonated. The enzyme-bound acyl group then undergoes nucleophilic acyl substitution by reaction with a molecule of coenzyme A, and the chain-shortened acyl CoA that results re-enters the $\beta$-oxidation pathway for further degradation.

Look at the catabolism of myristic acid shown in FIGURE 29.5 to see the overall results of the $\beta$-oxidation pathway. The first passage converts the 14 -carbon myristoyl CoA into the 12 -carbon lauroyl CoA plus acetyl CoA, the second passage converts lauroyl CoA into the 10-carbon caproyl CoA plus acetyl CoA, the third passage converts caproyl CoA into the 8 -carbon capryloyl CoA, and so on. Note that the final passage produces two molecules of acetyl CoA because the precursor has four carbons.

Myristoyl CoA

$$
\begin{aligned}
& \beta \text {-Oxidation } \\
& \text { (passage 1) }
\end{aligned}
$$


Step 4 of FIGURE 29.4: Chain Cleavage -
Myristoyl CoA

$$
\begin{aligned}
& \beta \text {-Oxidation } \\
& \text { (passage 1) }
\end{aligned}
$$

Lauroyl CoA
$\downarrow \begin{aligned} & \beta \text {-Oxidation } \\ & \text { (passage 2) }\end{aligned}$

Caproyl CoA

$$
\begin{aligned}
& \beta \text {-Oxidation } \\
& \text { (passage 3) }
\end{aligned}
$$


Capryloyl CoA -
FIGURE 29.5 Catabolism of the 14 -carbon myristic acid by the $\boldsymbol{\beta}$-oxidation pathway yields seven molecules of acetyl CoA after six passages.

Most fatty acids have an even number of carbon atoms, so none are left over after $\beta$-oxidation. Those fatty acids with an odd number of carbon atoms yield the three-carbon propionyl CoA in the final $\beta$-oxidation. Propionyl CoA is then converted to succinate by a multistep radical pathway, and succinate enters the citric acid cycle (Section 29.7). Note that the three-carbon propionyl group should technically be called propanoyl, but biochemists generally use the nonsystematic name.

PROBLEM Write the equations for the remaining passages of the $\beta$-oxidation pathway following those shown
29-2 in Figure 29.5.
PROBLEM How many molecules of acetyl CoA are produced by catabolism of the following fatty acids, and how 29-3 many passages of the $\beta$-oxidation pathway are needed?
(a) Palmitic acid, $\mathrm{CH}_{3}\left(\mathrm{CH}_{2}\right)_{14} \mathrm{CO}_{2} \mathrm{H}$
(b) Arachidic acid, $\mathrm{CH}_{3}\left(\mathrm{CH}_{2}\right)_{18} \mathrm{CO}_{2} \mathrm{H}$


Capryloyl CoA - 29.4 Biosynthesis of Fatty Acids
One of the most striking features of the common fatty acids is that they have an even number of carbon atoms (TABLE 27.1). This occurs because all fatty acids are derived biosynthetically from acetyl CoA by sequential addition of two-carbon units to a growing chain. The acetyl CoA, in turn, arises primarily from the metabolic breakdown of carbohydrates in the glycolysis pathway, which we'll see in Section 29.5. Thus, dietary carbohydrates consumed in excess of immediate energy needs are turned into fats for storage.

As a general rule in biological chemistry, the anabolic pathway by which a substance is made is not the reverse of the catabolic pathway by which the same substance is degraded. The two paths must differ in some respects for both to be energetically favorable. Thus, the $\beta$-oxidation pathway for converting fatty acids into acetyl CoA and the biosynthesis of fatty acids from acetyl CoA are related but are not exact opposites. Differences include the identity of the acyl-group carrier, the stereochemistry of the $\beta$-hydroxyacyl reaction intermediate, and the identity of the redox coenzyme. FAD is used to introduce a double bond in $\beta$-oxidation, while NADPH is used to reduce the double bond in fatty-acid biosynthesis.

In bacteria, each step in fatty-acid synthesis is catalyzed by a separate enzyme. In vertebrates, however, fattyacid synthesis is catalyzed by an immense, multienzyme complex called a synthase that contains two identical subunits of 2505 amino acids each and catalyzes all steps in the pathway. In fact, for an 18-carbon fatty acid, the synthase catalyzes 42 separate steps! An overview of fatty-acid biosynthesis is shown in FIGURE 29.6.


Steps 1-2 of FIGURE 29.6: Acyl Transfers -
The starting material for fatty-acid biosynthesis is the thioester acetyl CoA, the final product of carbohydrate breakdown, as we'll see in Section 29.6. The pathway begins with several priming reactions, which transport acetyl CoA and convert it into more reactive species. The first priming reaction is a nucleophilic acyl substitution reaction that converts acetyl CoA into acetyl ACP (acyl carrier protein).

Notice that the mechanism of the nucleophilic acyl substitution in step 1 can be given in an abbreviated form that saves space by not explicitly showing the tetrahedral reaction intermediate. Instead, electron movement is shown as a heart-shaped path around the carbonyl oxygen to imply the two steps of the full mechanism. Biochemists commonly use this kind of abbreviated format, and we'll also use it on occasion through the rest of this chapter.


FIGURE 29.6 MECHANISM -
The pathway for fatty-acid biosynthesis from the two-carbon precursor, acetyl CoA. Individual steps are explained in the text.

An acetyl group is transferred from CoA to ACP (acyl carrier protein).
(2) The acetyl group is
transferred again, from
ACP to a synthase enzyme.

Acetyl ACP


(7) -
(5) Claisen-like condensation of malonyl ACP with acetyl synthase occurs, followed by decarboxylation to yield acetoacetyl ACP, a $\beta$-keto thioester.
(6) Reduction of the ketone by NADPH yields the corresponding $\beta$-hydroxy thioester.

Dehydration of $\beta$-hydroxybutyryl ACP gives crotonyl ACP, an $a, b$-unsaturated thioester.

8 Reduction of the double bond yields the saturated, chain-elongated butyryl ACP.


Acetyl synthase -
©

Acetyl CoA

(3) Acetyl CoA is carboxylated to give malonyl CoA.
(4) The malonyl group is transferred from CoA to ACP.

Acetoacetyl ACP

$\boldsymbol{\beta}$-Hydroxybutyryl ACP

Crotonyl ACP

Butyryl ACP

Tetrahedral intermediate

In bacteria, ACP is a small protein of 77 residues that transports an acyl group from one enzyme to another. In vertebrates, however, ACP appears to be a long arm on a multienzyme synthase complex, whose apparent function is to shepherd an acyl group from site to site within the complex. As in acetyl CoA, the acyl group in acetyl ACP is linked by a thioester bond to the sulfur atom of phosphopantetheine. The phosphopantetheine is in turn linked to ACP through the side-chain -OH group of a serine residue in the enzyme.


Acetyl ACP -
Step 2, another priming reaction, involves a further exchange of thioester linkages by another nucleophilic acyl substitution and results in covalent bonding of the acetyl group to a cysteine residue in the synthase complex that catalyzes the upcoming condensation step.

Steps 3-4 of FIGURE 29.6: Carboxylation and Acyl Transfer
Step 3 is a loading reaction in which acetyl CoA is carboxylated by reaction with $\mathrm{HCO}_{3}{ }^{-}$and ATP to yield malonyl CoA plus ADP. This step requires the coenzyme biotin, which is bonded to the lysine residue of acetyl CoA carboxylase and acts as a carrier of $\mathrm{CO}_{2}$. Biotin first reacts with bicarbonate ion to give $N$-carboxybiotin, which then reacts with the enolate ion of acetyl CoA and transfers the $\mathrm{CO}_{2}$ group. Thus, biotin acts as a carrier of $\mathrm{CO}_{2}$, binding it in one step and releasing it in another.

The mechanism of the $\mathrm{CO}_{2}$ transfer reaction with acetyl CoA to give malonyl CoA is thought to involve $\mathrm{CO}_{2}$ as the reactive species. One proposal is that the loss of $\mathrm{CO}_{2}$ is favored by hydrogen-bond formation between the N -carboxybiotin carbonyl group and a nearby acidic site in the enzyme. Simultaneous deprotonation of acetyl CoA by a basic site in the enzyme gives a thioester enolate ion that can react with $\mathrm{CO}_{2}$ as it forms (FIGURE 29.7).


Mechanism of step 3 in FIGURE 29.6, the biotin-dependent carboxylation of acetyl CoA to yield malonyl -
CoA.
(1) A basic site in the enzyme deprotonates acetyl CoA.

(3) The enolate ion adds in an
bond of carbon dioxide, yielding malonyl CoA.

Malonyl CoA

Following the formation of malonyl CoA, another nucleophilic acyl substitution reaction occurs in step 4 to form the more reactive malonyl ACP, thereby binding the malonyl group to an ACP arm of the multienzyme synthase. At this point, both acetyl and malonyl groups are bound to the enzyme, and the stage is set for their condensation.


Step 5 of FIGURE 29.6: Condensation -
The key carbon-carbon bond-forming reaction that builds the fatty-acid chain occurs in step 5 . This step is simply a Claisen condensation between acetyl synthase as the electrophilic acceptor and malonyl ACP as the nucleophilic donor. The mechanism of the condensation is thought to involve decarboxylation of malonyl ACP to give an enolate ion, followed by immediate nucleophilic addition of the enolate ion to the carbonyl group of acetyl synthase. Breakdown of the tetrahedral intermediate then gives the four-carbon condensation product acetoacetyl ACP and frees the synthase binding site for attachment of the chain-elongated acyl group at the end of the sequence.

Steps 6-8 of FIGURE 29.6: Reduction and Dehydration
The ketone carbonyl group in acetoacetyl ACP is next reduced to the alcohol $\beta$-hydroxybutyryl ACP by $\beta$-keto thioester reductase and NADPH, a reducing coenzyme closely related to NADH. $R$ stereochemistry results at the newly formed chirality center in the $\beta$-hydroxy thioester product. (Note that the systematic name of a butyryl group is butanoyl.)

Subsequent dehydration of $\beta$-hydroxybutyryl ACP by an E1cB reaction in step 7 yields trans-crotonyl ACP, and the carbon-carbon double bond of crotonyl ACP is reduced by NADPH in step 8 to yield butyryl ACP. The doublebond reduction occurs by conjugate nucleophilic addition of a hydride ion from NADPH to the $\beta$ carbon of trans-crotonyl ACP. In vertebrates, this reduction occurs by an overall syn addition, but other organisms carry out similar chemistry with different stereochemistry.


Step 5 of FIGURE 29.6: Condensation -
The net effect of the eight steps in the fatty-acid biosynthesis pathway is to take two 2 -carbon acetyl groups and combine them into a 4-carbon butyryl group. Further condensation of the butyryl group with another malonyl ACP yields a 6-carbon unit, and still further repetitions add two carbon atoms at a time until the 16-carbon palmitoyl ACP is produced.

Palmitoyl ACP

Further chain elongation of palmitic acid occurs by reactions similar to those just described, but CoA rather than ACP acts as the carrier group, and separate enzymes are needed for each step rather than a multienzyme synthase complex.

PROBLEM Write a mechanism for the dehydration reaction of $\beta$-hydroxybutyryl ACP to yield crotonyl ACP in 29-4 step 7 of fatty-acid synthesis.

PROBLEM Evidence for the role of acetate in fatty-acid biosynthesis comes from isotope-labeling experiments.
29-5 If acetate labeled with ${ }^{13} \mathrm{C}$ in the methyl group $\left({ }^{13} \mathrm{CH}_{3} \mathrm{CO}_{2} \mathrm{H}\right)$ were incorporated into fatty acids, at what positions in the fatty-acid chain would you expect the ${ }^{13} \mathrm{C}$ label to appear?

PROBLEM Does the reduction of acetoacetyl ACP in step 6 occur on the Re face or the Si face of the molecule? 29-6


Step 5 of FIGURE 29.6: Condensation - 29.5 Catabolism of Carbohydrates: Glycolysis
Glucose is the body's primary short-term energy source. Its catabolism begins with glycolysis, a series of ten enzyme-catalyzed reactions that break down glucose into 2 equivalents of pyruvate, $\mathrm{CH}_{3} \mathrm{COCO}_{2}{ }^{-}$. The steps of glycolysis, also called the Embden-Meyerhoff pathway after its discoverers, are summarized in FIGURE 29.8.

The ten-step glycolysis pathway for catabolizing glucose to two molecules of pyruvate. Individual steps are described in the text.

(1) Glucose is phosphorylated by reaction with ATP to yield glucose 6-phosphate.

$\alpha$-Glucose 6-phosphate

Glucose 6-phosphate is isomerized to fructose 6-phosphate by ring opening followed by a keto-enol tautomerization.
(2)


$\alpha$-Fructose 6-phosphate -
Fructose 6-phosphate is phosphorylated by reaction with ATP to yield fructose

1,6-bisphosphate.

$\beta$-Fructose 1,6-bisphosphate

(4) Fructose 1,6-bisphosphate undergoes ring opening and is cleaved by a retro-aldol reaction into glyceraldehyde 3-phosphate and dihydroxyacetone phosphate (DHAP).
(5) DHAP then isomerizes to glyceraldehyde 3-phosphate.


Glyceraldehyde 3-phosphate -
6) Glyceraldehyde 3-phosphate is oxidized to a carboxylic acid and then phosphorylated to yield 1,3-bisphosphoglycerate.
$6 \curvearrowleft \mathrm{NAD}^{+}, \mathrm{P}_{\mathrm{i}}$

1,3-Bisphosphoglycerate ${ }^{2-} \mathrm{O}_{3} \mathrm{POCH}_{2} \mathrm{CHCO}_{2} \mathrm{PO}_{3}{ }^{2-}$
(7) A phosphate is transferred from the carboxyl group to ADP, resulting in synthesis of an ATP and yielding 3-phosphoglycerate.

3-Phosphoglycerate
8 Isomerization of 3-phosphoglycerate gives 2-phosphoglycerate.

2-Phosphoglycerate
9 Dehydration occurs to yield phosphoenolpyruvate (PEP).

Phosphoenolpyruvate
(10) A phosphate is transferred from PEP to ADP, yielding pyruvate and ATP.

Pyruvate

Steps 1-2 of FIGURE 29.8: Phosphorylation and Isomerization
Glucose, produced by the digestion of dietary carbohydrates, is phosphorylated at the C6 hydroxyl group by reaction with ATP in a process catalyzed by hexokinase. As noted in Section 29.1, the reaction requires $\mathrm{Mg}^{2+}$ as a cofactor to complex with the negatively charged phosphate oxygens. The glucose 6-phosphate that results is then isomerized by glucose 6-phosphate isomerase to give fructose 6-phosphate. This isomerization takes place by initial opening of the glucose hemiacetal ring to its open-chain form, followed by keto-enol tautomerization to a cis enediol, $\mathrm{HO}-\mathrm{C}=\mathrm{C}-\mathrm{OH}$. But because glucose and fructose share a common enediol, further tautomerization to a different keto form produces open-chain fructose, and cyclization completes the process (FIGURE 29.10).

$\alpha$-Fructose
6-phosphate


Glyceraldehyde 3-phosphate -
$\alpha$-Fructose
6-phosphate

FIGURE 29.10 Mechanism of step 2 in glycolysis, the isomerization of glucose 6-phosphate to fructose 6-phosphate.
Step 3 of FIGURE 29.8: Phosphorylation
Fructose 6-phosphate is converted in step 3 to fructose 1,6-bisphosphate (FBP) by a phosphofructokinasecatalyzed reaction with ATP (recall that the prefix bis- means two). The mechanism is similar to that in step 1 , with $\mathrm{Mg}^{2+}$ ion again required as cofactor. Interestingly, the product of step 2 is the $\alpha$ anomer of fructose 6-phosphate, but it is the $\beta$ anomer that is phosphorylated in step 3 , implying that the two anomers equilibrate rapidly through the open-chain form. The result is a molecule ready to be split into the two three-carbon intermediates that will ultimately become two molecules of pyruvate.

Step 4 of FIGURE 29.8: Cleavage
Fructose 1,6-bisphosphate is cleaved in step 4 into two 3-carbon pieces, dihydroxyacetone phosphate (DHAP) and glyceraldehyde 3-phosphate (GAP). The bond between C3 and C4 of fructose 1,6-bisphosphate breaks, and a $\mathrm{C}=\mathrm{O}$ group is formed at C 4 . Mechanistically, the cleavage is the reverse of an aldol reaction (Section 23.1) and is catalyzed by an aldolase. A forward aldol reaction joins two aldehydes or ketones to give a $\beta$-hydroxy carbonyl compound, while a retro-aldol reaction, as in this case, cleaves a $\beta$-hydroxy carbonyl compound into two aldehydes or ketones.


Glyceraldehyde 3-phosphate -
Two classes of aldolases are used by organisms for catalysis of the retro-aldol reaction. In fungi, algae, and some bacteria, the retro-aldol reaction is catalyzed by class II aldolases, which function by coordination of the fructose carbonyl group with $\mathrm{Zn}^{2+}$ as Lewis acid. In plants and animals, the reaction is catalyzed by class I aldolases and does not take place on the free ketone. Instead, fructose 1,6 -bisphosphate undergoes reaction with the side-chain $-\mathrm{NH}_{2}$ group of a lysine residue on the aldolase to yield a protonated enzyme-bound imine (Section 19.8), which is often called a Schiff base in biochemistry.

Because of its positive charge, the iminium ion is a better electron acceptor than a ketone carbonyl group. Retroaldol reaction ensues, giving glyceraldehyde 3-phosphate and an enamine, which is protonated to give another iminium ion that is hydrolyzed to yield dihydroxyacetone phosphate (FIGURE 29.11).

FIGURE 29.11 Mechanism of step 4 in FIGURE 29.8, the cleavage of fructose 1,6-bisphosphate to yield glyceraldehyde 3-phosphate and dihydroxyacetone phosphate. The reaction occurs through an iminium ion formed by reaction with a lysine residue in the enzyme.

Step 5 of FIGURE 29.8: Isomerization
Dihydroxyacetone phosphate is isomerized in step 5 by triose phosphate isomerase to form a second equivalent of glyceraldehyde 3-phosphate. As in the conversion of glucose 6-phosphate to fructose 6-phosphate in step 2, the isomerization takes place by keto-enol tautomerization through a common enediol intermediate. A base deprotonates C 1 and then reprotonates C2 using the same hydrogen. The net result of steps 4 and 5 together is the production of two glyceraldehyde 3-phosphate molecules, both of which pass down the rest of the pathway. Thus, each of the remaining five steps of glycolysis takes place twice for every glucose molecule entering at step 1.


Steps 6-7 of FIGURE 29.9: Oxidation, Phosphorylation, and Dephosphorylation -
Glyceraldehyde 3-phosphate is oxidized and phosphorylated in step 6 to give 1,3-bisphosphoglycerate (FIGURE 29.12). The reaction is catalyzed by glyceraldehyde 3-phosphate dehydrogenase and begins by nucleophilic addition of the -SH group of a cysteine residue in the enzyme to the aldehyde carbonyl group to yield a hemithioacetal (Section 19.10), the sulfur analog of a hemiacetal. Oxidation of the hemithioacetal -OH group by $\mathrm{NAD}^{+}$then yields a thioester, which reacts with phosphate ion in a nucleophilic acyl substitution step to yield 1,3-bisphosphoglycerate, a mixed anhydride derived from a carboxylic acid and a phosphoric acid.

FIGURE 29.12 Mechanism of step 6 in FIGURE 29.9, the oxidation and phosphorylation of glyceraldehyde 3-phosphate to give 1,3-bisphosphoglycerate. The process occurs through initial formation of a hemiacetal that is oxidized to a thioester and converted into an acyl phosphate.
Like all anhydrides (Section 21.5), the mixed carboxylic-phosphoric anhydride is a reactive substrate in nucleophilic acyl (or phosphoryl) substitution reactions. Reaction of 1,3-bisphosphoglycerate with ADP occurs in step 7 by substitution on phosphorus, resulting in transfer of a phosphate group to ADP and giving ATP plus 3-phosphoglycerate. The process is catalyzed by phosphoglycerate kinase and requires $\mathrm{Mg}^{2+}$ as cofactor. Together, steps 6 and 7 accomplish the oxidation of an aldehyde to a carboxylic acid.


Steps 6-7 of FIGURE 29.9: Oxidation, Phosphorylation, and Dephosphorylation -
Step 8 of FIGURE 29.9: Isomerization
3-Phosphoglycerate isomerizes to 2-phosphoglycerate in a step catalyzed by phosphoglycerate mutase. In
plants, 3-phosphoglycerate transfers its phosphoryl group from its C3 oxygen to a histidine residue on the enzyme in one step and then accepts the same phosphoryl group back onto the C2 oxygen in a second step. In animals and yeast, however, the enzyme contains a phosphorylated histidine, which transfers its phosphoryl group to the C2 oxygen of 3-phosphoglycerate and forms 2,3-bisphosphoglycerate as intermediate. The same histidine then accepts a phosphoryl group from the C3 oxygen to yield the isomerized product plus the regenerated enzyme. As explained in Section 29.4, we'll occasionally use an abbreviated mechanism for nucleophilic acyl substitution reactions to save space.

Steps 9-10 of FIGURE 29.9: Dehydration and Dephosphorylation
Like most $\beta$-hydroxy carbonyl compounds, 2-phosphoglycerate undergoes a ready dehydration in step 9 by an E 1 cB mechanism (Section 23.3). The process is catalyzed by enolase, and the product is phosphoenolpyruvate, abbreviated PEP. Two $\mathrm{Mg}^{2+}$ ions are associated with the 2-phosphoglycerate to neutralize the negative charges.

Transfer of the phosphoryl group to ADP in step 10 then generates ATP and gives enolpyruvate, which tautomerizes to pyruvate. The reaction is catalyzed by pyruvate kinase and requires that a molecule of fructose 1,6-bisphosphate also be present, as well as 2 equivalents of $\mathrm{Mg}^{2+}$. One $\mathrm{Mg}^{2+}$ ion coordinates to ADP, and the other increases the acidity of a water molecule necessary for protonation of the enolate ion.

The overall result of glycolysis can be summarized by the following equation:

Glucose

PROBLEM Identify the two steps of glycolysis in which ATP is produced. 29-7


Steps 6-7 of FIGURE 29.9: Oxidation, Phosphorylation, and Dephosphorylation -
The overall result of glycolysis can be summarized by the following equation:

Glucose

PROBLEM Identify the two steps of glycolysis in which ATP is produced. 29-7

PROBLEM Look at the entire glycolysis pathway, and make a list of the kinds of organic reactions that take 29-8 place-nucleophilic acyl substitutions, aldol reactions, E1cB reactions, and so forth.


Steps 6-7 of FIGURE 29.9: Oxidation, Phosphorylation, and Dephosphorylation - 29.6 Conversion of Pyruvate to Acetyl CoA
Pyruvate, produced by catabolism of glucose (and by degradation of several amino acids), can undergo several further transformations depending on the conditions and on the organism. In the absence of oxygen, pyruvate can either be reduced by NADH to yield lactate $\left[\mathrm{CH}_{3} \mathrm{CH}(\mathrm{OH}) \mathrm{CO}_{2}{ }^{-}\right.$] or, in yeast, fermented to give ethanol. Under typical aerobic conditions in mammals, however, pyruvate is converted by a process called oxidative decarboxylation to give acetyl CoA plus $\mathrm{CO}_{2}$. (Oxidative because the oxidation state of the carbonyl carbon rises from that of a ketone to that of a thioester.)

The conversion occurs through a multistep sequence of reactions catalyzed by a complex of enzymes and cofactors called the pyruvate dehydrogenase complex. The process occurs in three stages, each catalyzed by one of the enzymes in the complex, as outlined in FIGURE 29.13. Acetyl CoA, the ultimate product, then acts as fuel for the final stage of catabolism, the citric acid cycle.

Step 1 of FIGURE 29.13: Addition of Thiamin Diphosphate
The conversion of pyruvate to acetyl CoA begins by reaction of pyruvate with thiamin diphosphate, a derivative of vitamin $B_{1}$. Formerly called thiamin pyrophosphate, thiamin diphosphate is usually abbreviated as TPP. The spelling thiamine is also correct and frequently used.


Steps 6-7 of FIGURE 29.9: Oxidation, Phosphorylation, and Dephosphorylation - 29.6 Conversion of Pyruvate to Acetyl CoA
The key structural element in thiamin diphosphate is the thiazolium ring-a five-membered, unsaturated heterocycle containing a sulfur atom and a positively charged nitrogen atom. The thiazolium ring is weakly acidic, with a $\mathrm{p} K_{\mathrm{a}}$ of approximately 18 for the ring hydrogen between N and S . Bases can therefore deprotonate thiamin diphosphate, leading to formation of an ylide much like the phosphonium ylides used in Wittig reactions (Section 19.11). As in the Wittig reaction, the TPP ylide is a nucleophile and adds to the ketone carbonyl group of pyruvate to yield an alcohol addition product.

$$
\text { Pyruvate } \begin{gathered}
\text { Thiamin } \\
\text { diphosphate } \\
\text { ylide }
\end{gathered}
$$


FIGURE 29.13 MECHANISM -
Mechanism for the conversion of pyruvate to acetyl CoA through a multistep sequence of reactions that requires three different enzymes and four different coenzymes. The individual steps are explained in the text.
(1) Nucleophilic addition of thiamin diphosphate (TPP) ylide to pyruvate gives an alcohol addition product.

Decarboxylation occurs in a step analogous to the loss of $\mathrm{CO}_{2}$ from a $\beta$-keto acid, yielding the enamine hydroxyethylthiamin diphosphate (HETPP).
(3) The enamine double bond attacks a sulfur atom of lipoamide and carries out an $\mathrm{S}_{\mathrm{N}}{ }^{2-}$ like displacement of the second sulfur to yield a hemithioacetal.
(4) Elimination of thiamin diphosphate ylide from the hemithioacetal intermediate yields acetyl dihydrolipoamide . . .
5. . . which reacts with coenzyme A in a nucleophilic acyl substitution reaction to exchange one thioester for another and give acetyl CoA plus dihydrolipoamide.

Acetyl CoA
$+$

Dihydrolipoamide

Step 2 of FIGURE 29.13: Decarboxylation
The TPP addition product, which contains an iminium ion $\beta$ to a carboxylate anion, undergoes decarboxylation in much the same way that a $\beta$-keto acid decarboxylates in the acetoacetic ester synthesis (Section 22.7). The $\mathrm{C}=\mathrm{N}^{+}$bond of the pyruvate addition product acts like the $\mathrm{C}=\mathrm{O}$ bond of a $\beta$-keto acid to accept electrons as $\mathrm{CO}_{2}$ leaves, giving hydroxyethylthiamin diphosphate (HETPP).


FIGURE 29.13 MECHANISM -
Step 3 of FIGURE 29.13: Reaction with Lipoamide
Hydroxyethylthiamin diphosphate is an enamine $\left(\mathrm{R}_{2} \mathrm{~N}-\mathrm{C}=\mathrm{C}\right)$, which, like all enamines, is nucleophilic (Section 23.11). It therefore reacts with the enzyme-bound disulfide lipoamide by nucleophilic attack on a sulfur atom, displacing the second sulfur in an $\mathrm{S}_{\mathrm{N}} 2$-like process.

Step 4 of FIGURE 29.13: Elimination of Thiamin Diphosphate
The product of the HETPP reaction with lipoamide is a hemithioacetal, which eliminates thiamin diphosphate ylide. This elimination is the reverse of the ketone addition in step 1 and generates acetyl dihydrolipoamide.

Step 5 of FIGURE 29.13: Acyl Transfer
Acetyl dihydrolipoamide, a thioester, undergoes a nucleophilic acyl substitution reaction with coenzyme A to yield acetyl CoA plus dihydrolipoamide. The dihydrolipoamide is then oxidized back to lipoamide by FAD (Section 29.3), and the $\mathrm{FADH}_{2}$ that results is in turn oxidized back to FAD by $\mathrm{NAD}^{+}$, completing the catalytic cycle.

PROBLEM Which carbon atoms in glucose end up as $-\mathrm{CH}_{3}$ carbons in acetyl CoA? Which carbons end up as 29-9 $\mathrm{CO}_{2}$ ?


FIGURE 29.13 MECHANISM - 29.7 The Citric Acid Cycle
The initial stages of catabolism result in the conversion of both fats and carbohydrates into acetyl groups that are bonded through a thioester link to coenzyme A. Acetyl CoA then enters the next stage of catabolism-the citric acid cycle, also called the tricarboxylic acid (TCA) cycle, or Krebs cycle, after Hans Krebs, who unraveled its complexities in 1937. The overall result of the cycle is the conversion of an acetyl group into two molecules of $\mathrm{CO}_{2}$ plus reduced coenzymes by the eight-step reaction sequence shown in FIGURE 29.14.

As its name implies, the citric acid cycle is a closed loop of reactions in which the product of the final step (oxaloacetate) is a reactant in the first step. The intermediates are constantly regenerated, flowing continuously through the cycle, which operates as long as the oxidizing coenzymes $\mathrm{NAD}^{+}$and FAD are available. To meet this condition, the reduced coenzymes NADH and $\mathrm{FADH}_{2}$ must be reoxidized via the electron-transport chain, which in turn relies on oxygen as the ultimate electron acceptor. Thus, the cycle is dependent on the availability of oxygen and on the operation of the electron-transport chain.

FIGURE 29.14 MECHANISM

The citric acid cycle is an eight-step series of reactions that results in the conversion of an acetyl group into two molecules of $\mathbf{C O}_{2}$ plus reduced coenzymes. Individual steps are explained in the text.

Oxaloacetate
8 Oxidation of (S)-malate gives oxaloacetate, completing the cycle.

(S)-Malate
(L-malate)
(7) Fumarate undergoes conjugate addition of water to its double bond to give (S)-malate.

Fumarate
(6) Succinate is dehydrogenated by FAD to give fumarate.

Citrate
2

Citrate is isomerized by dehydration and rehydration to give isocitrate.

Isocitrate

$\alpha$-Ketoglutarate

Succinate


FIGURE 29.13 MECHANISM - 29.7 The Citric Acid Cycle
Citrate
2

Citrate is isomerized by dehydration and rehydration to give isocitrate.

Isocitrate

$\alpha$-Ketoglutarate

Succinate

Succinyl CoA
Succinyl CoA is converted to
succinate in a reaction coupled to the phosphorylation of GDP to give GTP.

Step 1 of FIGURE 29.14: Addition to Oxaloacetate
Acetyl CoA enters the citric acid cycle in step 1 by nucleophilic addition to the oxaloacetate carbonyl group, to give (S)-citryl CoA. This addition is an aldol reaction and is catalyzed by citrate synthase, as discussed in Section 26.11. (S)-Citryl CoA is then hydrolyzed to citrate by a typical nucleophilic acyl substitution reaction with water, catalyzed by the same citrate synthase enzyme.

Note that the hydroxyl-bearing carbon of citrate is a prochirality center and contains two identical arms. Because the initial aldol reaction of acetyl CoA to oxaloacetate occurs specifically from the Si face of the
ketone carbonyl group, the pro- $S$ arm of citrate is derived from acetyl CoA and the pro- $R$ arm is derived from oxaloacetate.


FIGURE 29.13 MECHANISM - 29.7 The Citric Acid Cycle
Citrate
Step 2 of FIGURE 29.14: Isomerization
Citrate, a prochiral tertiary alcohol, is next converted into its isomer, ( $2 R, 3 S$ )-isocitrate, a chiral secondary alcohol. The isomerization occurs in two steps, both of which are catalyzed by the same aconitase enzyme. The initial step is an E1cB dehydration of a $\beta$-hydroxy acid to give cis-aconitate, the same sort of reaction that occurs in step 9 of glycolysis (FIGURE 29.8). The second step is a conjugate nucleophilic addition of water to the $\mathrm{C}=\mathrm{C}$ bond (Section 19.13). The dehydration of citrate takes place specifically on the pro- $R$ arm-the one derived from oxaloacetate-rather than on the pro- $S$ arm derived from acetyl CoA.

Step 3 of FIGURE 29.14: Oxidation and Decarboxylation
( $2 R, 3 S$ )-Isocitrate, a secondary alcohol, is oxidized by $\mathrm{NAD}^{+}$in step 3 to give the ketone oxalosuccinate, which loses $\mathrm{CO}_{2}$ to give $\alpha$-ketoglutarate. Catalyzed by isocitrate dehydrogenase, the decarboxylation is a typical reaction of a $\beta$-keto acid, just like that in the acetoacetic ester synthesis (Section 22.7). The enzyme requires a divalent cation as a cofactor to polarize the ketone carbonyl group and make it a better electron acceptor.


FIGURE 29.13 MECHANISM - 29.7 The Citric Acid Cycle
Step 4 of FIGURE 29.14: Oxidative Decarboxylation
The transformation of $\alpha$-ketoglutarate to succinyl CoA in step 4 is a multistep process just like the transformation of pyruvate to acetyl CoA that we saw in FIGURE 29.13. In both cases, an $\alpha$-keto acid loses $\mathrm{CO}_{2}$ and is oxidized to a thioester in a series of steps catalyzed by a multienzyme dehydrogenase complex. As in the conversion of pyruvate to acetyl CoA, the reaction involves an initial nucleophilic addition reaction of thiamin diphosphate ylide to $\alpha$-ketoglutarate, followed by decarboxylation. Reaction with lipoamide, elimination of TPP ylide, and finally a transesterification of the dihydrolipoamide thioester with coenzyme A yields succinyl CoA.

Step 5 of FIGURE 29.14: Acyl CoA Cleavage
Succinyl CoA is converted to succinate in step 5. This reaction is catalyzed by succinyl CoA synthetase and is coupled with phosphorylation of guanosine diphosphate (GDP) to give guanosine triphosphate (GTP). The overall transformation is similar to that of steps 6 through 8 in glycolysis (FIGURE 29.8), in which a thioester is converted into an acyl phosphate and a phosphate group is then transferred to ADP. The overall result is a "hydrolysis" of the thioester group without involvement of water.

Step 6 of FIGURE 29.14: Dehydrogenation
Succinate is dehydrogenated in step 6 by the FAD-dependent succinate dehydrogenase to give fumarate. This process is analogous to what occurs during the $\beta$-oxidation pathway of fatty-acid catabolism (Section 29.3). The reaction is stereospecific, removing the pro- $S$ hydrogen from one carbon and the pro- $R$ hydrogen from the other.

Succinate

Fumarate


FIGURE 29.13 MECHANISM - 29.7 The Citric Acid Cycle
Succinate

Fumarate

Steps 7-8 of FIGURE 29.14: Hydration and Oxidation
The final two steps in the citric acid cycle are the conjugate nucleophilic addition of water to fumarate to yield $(S)$-malate and the oxidation of $(S)$-malate by $\mathrm{NAD}^{+}$to give oxaloacetate. The addition is catalyzed by fumarase and is mechanistically similar to the addition of water to cis-aconitate in step 2 . This reaction occurs through an enolate-ion intermediate, which is protonated on the side opposite the OH , leading to a net anti addition.

The final step is the oxidation of $(S)$-malate by $\mathrm{NAD}^{+}$to give oxaloacetate, a reaction catalyzed by malate dehydrogenase. The citric acid cycle has now returned to its starting point, ready to revolve again. The overall result of the cycle is

PROBLEM Which of the substances in the citric acid cycle are tricarboxylic acids, thus giving the cycle its 29-10 alternative name?

PROBLEM Write mechanisms for step 2 of the citric acid cycle, the dehydration of citrate and the addition of
29-11 water to aconitate.
PROBLEM Is the pro- $R$ or pro- $S$ hydrogen removed from citrate during the dehydration in step 2 of the citric 29-12 acid cycle? Does the elimination reaction occur with syn or anti geometry?


FIGURE 29.13 MECHANISM - 29.8 Carbohydrate Biosynthesis: Gluconeogenesis
Glucose is the body's primary fuel when food is plentiful, but in times of fasting or prolonged exercise, glucose stores can become depleted. Most tissues then begin metabolizing fats as their source of acetyl CoA, but the brain is different. The brain relies almost entirely on glucose for fuel and is dependent on receiving a continuous supply in the blood. When the supply of glucose fails, even for a brief time, irreversible damage can occur. Thus, a pathway for synthesizing glucose from simple precursors is crucial.

Higher organisms are not able to synthesize glucose from acetyl CoA but must instead use one of the threecarbon precursors lactate, glycerol, or alanine, all of which are readily converted into pyruvate.

Pyruvate then becomes the starting point for gluconeogenesis, the 11-step biosynthetic pathway by which organisms make glucose (FIGURE 29.15). The gluconeogenesis pathway is not the reverse of the glycolysis pathway by which glucose is degraded. As with the catabolic and anabolic pathways for fatty acids (Section 29.3 and Section 29.4), the catabolic and anabolic pathways for carbohydrates differ in some details so that both are energetically favorable.


FIGURE 29.15 MECHANISM -
The gluconeogenesis pathway for the biosynthesis of glucose from pyruvate. Individual steps are explained in the text.

Pyruvate

(1) Pyruvate undergoes a biotin-dependent carboxylation on the methyl group to give oxaloacetate...

Oxaloacetate

2. . . which is decarboxylated and then phosphorylated by GTP to give phosphoenolpyruvate.

Phosphoenolpyruvate
(3) Conjugate nucleophilic addition of water
to the double bond of phosphoenolpyruvate 3 (
gives 2 -phosphoglycerate $\ldots$.

2-Phosphoglycerate

4. . . which is isomerized by transfer of the phosphoryl group to give 3-phosphoglycerate.

3-Phosphoglycerate

5 Phosphorylation of the carboxyl group by reaction with ATP yields 1,3-bisphosphoglycerate.

1,3-Bisphosphoglycerate

(6) Reduction of the acyl phosphate gives glyceraldehyde 3-phosphate, which
(7) undergoes keto-enol tautomerization to yield dihydroxyacetone phosphate.

Dihydroxyacetone phosphate

$\stackrel{7}{4}$

Glyceraldehyde 3-phosphate

$\downarrow$

Step 1 of FIGURE 29.15: Carboxylation
Gluconeogenesis begins with the carboxylation of pyruvate to yield oxaloacetate. The reaction is catalyzed by pyruvate carboxylase and requires ATP, bicarbonate ion, and the coenzyme biotin, which acts as a carrier to transport $\mathrm{CO}_{2}$ to the enzyme active site. The mechanism is analogous to that of step 3 in fatty-acid biosynthesis (FIGURE 29.6), in which acetyl CoA is carboxylated to yield malonyl CoA.


FIGURE 29.15 MECHANISM -
Step 2 of FIGURE 29.15: Decarboxylation and Phosphorylation
Decarboxylation of oxaloacetate, a $\beta$-keto acid, occurs by the typical retro-aldol mechanism like that in step 3 in the citric acid cycle (FIGURE 29.14), and phosphorylation of the resultant pyruvate enolate ion by GTP occurs concurrently to give phosphoenolpyruvate. This reaction is catalyzed by phosphoenolpyruvate carboxykinase.

Steps 3-4 of FIGURE 29.15: Hydration and Isomerization
Conjugate nucleophilic addition of water to the double bond of phosphoenolpyruvate gives 2-phosphoglycerate by a process similar to that of step 7 in the citric acid cycle. Phosphorylation of C3 and dephosphorylation of C2 then yields 3-phosphoglycerate. Mechanistically, these steps are the reverse of steps 9 and 8 in glycolysis (FIGURE 29.8), which have equilibrium constants near 1 so that substantial amounts of reactant and product are both present.

Steps 5-7 of FIGURE 29.15: Phosphorylation, Reduction, and Tautomerization
Reaction of 3-phosphoglycerate with ATP generates the corresponding acyl phosphate, 1,3-bisphosphoglycerate, which binds to the glyceraldehyde 3-phosphate dehydrogenase by a thioester bond to a cysteine residue. Reduction of the thioester by $\mathrm{NADH} / \mathrm{H}^{+}$yields the corresponding aldehyde, and keto-enol tautomerization of the aldehyde gives dihydroxyacetone phosphate. All three steps comprise a mechanistic reversal of the corresponding steps 7,6 , and 5 of glycolysis and have equilibrium constants near 1.


FIGURE 29.15 MECHANISM -
Step 8 of FIGURE 29.16: Aldol Reaction
Dihydroxyacetone phosphate and glyceraldehyde 3-phosphate, the two 3-carbon units produced in step 7, join by an aldol reaction to give fructose 1,6-bisphosphate, the reverse of step 4 in glycolysis (FIGURE 29.11). As in glycolysis, the reaction is catalyzed in plants and animals by a class I aldolase and takes place on an iminium ion formed by reaction of dihydroxyacetone phosphate with a side-chain lysine $-\mathrm{NH}_{2}$ group on the enzyme. Loss of a proton from the neighboring carbon then generates an enamine, an aldol-like reaction ensues, and the product is hydrolyzed.

Steps 9-10 of FIGURE 29.16: Hydrolysis and Isomerization
Hydrolysis of the phosphate group at C1 of fructose 1,6-bisphosphate gives fructose 6-phosphate. Although the result of the reaction is the exact opposite of step 3 in glycolysis, the mechanism is not. In glycolysis, phosphorylation is accomplished by reaction of fructose with ATP, with formation of ADP as by-product. The reverse of that process, however-the reaction of fructose 1,6-bisphosphate with ADP to give fructose 6-phosphate and ATP-is energetically unfavorable because ATP is too high in energy. Thus, an alternative pathway is used in which the C1 phosphate group is removed by a direct hydrolysis reaction, catalyzed by fructose 1,6-bisphosphatase.

Following hydrolysis, keto-enol tautomerization of the carbonyl group from C 2 to C 1 gives glucose 6-phosphate. The isomerization is the reverse of step 2 in glycolysis.


FIGURE 29.15 MECHANISM -
Following hydrolysis, keto-enol tautomerization of the carbonyl group from C 2 to C 1 gives glucose 6-phosphate. The isomerization is the reverse of step 2 in glycolysis.

Step 11 of FIGURE 29.16: Hydrolysis
The final step in gluconeogenesis is the conversion of glucose 6-phosphate to glucose by a second phosphatasecatalyzed hydrolysis reaction. As just discussed for the hydrolysis of fructose 1,6-bisphosphate in step 9, and for the same energetic reasons, the mechanism of the glucose 6-phosphate hydrolysis is not the exact opposite
of the corresponding step 1 in glycolysis.
Interestingly, however, the mechanisms of the two phosphate hydrolysis reactions in steps 9 and 11 are not the same. In step 9, water is the nucleophile, but in the glucose 6-phosphate reaction of step 11, a histidine residue on the enzyme attacks phosphorus, giving a phosphoryl enzyme intermediate that subsequently reacts with water.

The overall result of gluconeogenesis is summarized by the following equation:

PROBLEM Write a mechanism for step 6 of gluconeogenesis, the reduction of 3-phosphoglyceryl phosphate 29-13 with $\mathrm{NADH} / \mathrm{H}^{+}$to yield glyceraldehyde 3-phosphate.


FIGURE 29.15 MECHANISM - 29.9 Catabolism of Proteins: Deamination
The catabolism of proteins is much more complex than that of fats and carbohydrates because each of the 20 $\alpha$-amino acids is degraded through its own unique pathway. The general idea, however, is that (1) the $\alpha$ amino group is first removed as ammonia by a deamination process, (2) the ammonia is converted into urea, and (3) the remaining amino acid carbon skeleton (usually an $\alpha$-keto acid) is converted into a compound that enters the citric acid cycle.


Transamination -
Deamination is usually accomplished by a transamination reaction in which the $-\mathrm{NH}_{2}$ group of the amino acid is exchanged with the keto group of $\alpha$-ketoglutarate, forming a new $\alpha$-keto acid plus glutamate. The overall process occurs in two parts, is catalyzed by aminotransferases, and involves participation of the coenzyme pyridoxal phosphate, abbreviated PLP, a derivative of pyridoxine (vitamin $\mathrm{B}_{6}$ ). The aminotransferases differ in their specificity for amino acids, but the mechanism remains the same.

The mechanism of the first part of transamination is shown in FIGURE 29.17. The process begins with reaction between the $\alpha$-amino acid and pyridoxal phosphate, which is covalently bonded to the aminotransferase by an imine linkage between the side-chain $-\mathrm{NH}_{2}$ group of a lysine residue in the enzyme and the PLP aldehyde group. Deprotonation/reprotonation of the PLP-amino acid imine effects tautomerization of the imine $\mathrm{C}=\mathrm{N}$ bond, and hydrolysis of the tautomerized imine gives an $\alpha$-keto acid plus pyridoxamine phosphate (PMP).

FIGURE 29.17 MECHANISM

Mechanism for the enzyme-catalyzed, PLP-dependent transamination of an $\alpha$-amino acid to give an $\boldsymbol{\alpha}$-keto acid. Individual steps are explained in the text.
(1) An amino acid reacts with the enzyme-bound PLP imine by nucleophilic addition of its $-\mathrm{NH}_{2}$ group to the $\mathrm{C}=\mathrm{N}$ bond of the imine, giving a PLP-amino acid imine and releasing the enzyme

PLP-amino acid imine (Schiff base)

(2) Deprotonation of the acidic $\alpha$ carbon of the amino acid gives an intermediate $\alpha$-keto acid imine . . .


Transamination -
PLP-amino acid imine (Schiff base)

(2) Deprotonation of the acidic $\alpha$ carbon of the amino acid gives an intermediate $\alpha$-keto acid imine . . .

(3).. that is reprotonated on the PLP carbon. The net result of this deprotonation/reprotonation sequence is tautomerization of the imine $C=N$ bond.
$\alpha$-Keto acid imine tautomer
(4) Hydrolysis of the $\alpha$-keto acid imine by nucleophilic addition of water to the $\mathrm{C}=\mathrm{N}$ bond gives the transamination products pyridoxamine
ـ
phosphate (PMP) and $\alpha$-keto acid.

Pyridoxamine phosphate
(PMP)

Step 1 of FIGURE 29.17: Transimination
The first step in transamination is transimination-the reaction of the PLP-enzyme imine with an $\alpha$-amino acid
to give a PLP-amino acid imine plus expelled enzyme as the leaving group. The reaction occurs by nucleophilic addition of the amino acid $-\mathrm{NH}_{2}$ group to the $\mathrm{C}=\mathrm{N}$ bond of the PLP imine, much as an amine adds to the $\mathrm{C}=\mathrm{O}$ bond of a ketone or aldehyde in a nucleophilic addition reaction (Section 19.8). The protonated diamine intermediate undergoes a proton transfer and expels the lysine amino group in the enzyme to complete the step.


Transamination -
Steps 2-4 of FIGURE 29.17: Tautomerization and Hydrolysis
Following formation of the PLP-amino acid imine in step 1, a tautomerization of the $\mathrm{C}=\mathrm{N}$ bond occurs in step 2. The basic lysine residue in the enzyme that was expelled as a leaving group during transimination deprotonates the acidic $\alpha$ position of the amino acid, with the protonated pyridine ring of PLP acting as the electron acceptor. Reprotonation occurs on the carbon atom next to the ring, generating a tautomeric product that is the imine of an $\alpha$-keto acid with pyridoxamine phosphate, abbreviated PMP.

Hydrolysis of this PMP- $\alpha$-keto acid imine then completes the first part of the transamination reaction. This is the mechanistic reverse of imine formation and occurs by nucleophilic addition of water to the imine, followed by proton transfer and expulsion of PMP as leaving group.


Regeneration of PLP from PMP -
With PLP plus the $\alpha$-amino acid now converted into PMP plus an $\alpha$-keto acid, PMP must be transformed back into PLP to complete the catalytic cycle. The conversion occurs by another transamination reaction, this one between PMP and an $\alpha$-keto acid, usually $\alpha$-ketoglutarate. The products are PLP plus glutamate, and the mechanism is the exact reverse of that shown in FIGURE 29.17. That is, PMP and $\alpha$-ketoglutarate give an imine; the PMP- $\alpha$-ketoglutarate imine undergoes tautomerization of the $\mathrm{C}=\mathrm{N}$ bond to give a PLP-glutamate imine; and the PLP-glutamate imine reacts with a lysine residue on the enzyme in a transimination process to yield PLP-enzyme imine plus glutamate.

PROBLEM Write all the steps in the transamination reaction of PMP with $\alpha$-ketoglutarate plus a lysine residue
29-14 in the enzyme to give the PLP-enzyme imine plus glutamate.
PROBLEM What $\alpha$-keto acid is formed on transamination of leucine?
29-15
PROBLEM From what amino acid is the following $\alpha$-keto acid derived?
29-16


Regeneration of PLP from PMP - 29.10 Some Conclusions about Biological Chemistry
As promised in the chapter introduction, the past few sections have been a fast-paced tour of a large number of reactions. Following it all undoubtedly required a lot of work and a lot of page-turning to look at earlier sections.

After examining the various metabolic pathways, perhaps the main conclusion about biological chemistry is the remarkable similarity between the mechanisms of biological reactions and the mechanisms of laboratory reactions. If you were to look at the steps of vitamin $B_{12}$ biosynthesis, you would see the same kinds of reactions we've been seeing throughout the text-nucleophilic substitutions, eliminations, aldol reactions, nucleophilic acyl substitutions, and so forth. There are, of course, some complexities, but the fundamental mechanisms of organic chemistry remain the same, whether in the laboratory with smaller molecules or in organisms with larger molecules.

Succinyl CoA
$+\quad \longrightarrow$

Glycine

Vitamin $B_{12}$-cyanocobalamin

So what is there to be learned from studying metabolism? One good answer is given in the following Chemistry Matters, which relates the story of how knowledge of a biosynthetic pathway led to the design of new drugs that have saved millions of lives.


Statin Drugs -
Coronary heart disease-the buildup of cholesterol-containing plaques on the walls of heart arteries-is the leading cause of death for people older than 20 in industrialized countries. It's estimated that up to one-third of women and one-half of men will develop the disease at some point in their lives.

The onset of coronary heart disease is directly correlated with blood cholesterol levels (see the Chapter 27 Chemistry Matters), and the first step in disease prevention is to lower those levels. It turns out that only about $25 \%$ of your blood cholesterol comes from what you eat; the remaining $75 \%$-about 1 gram each day-is biosynthesized in your liver from dietary fats and carbohydrates. Thus, any effective plan for lowering your cholesterol level means limiting the amount that your body makes, which is where a detailed chemical knowledge of cholesterol biosynthesis comes in.

FIGURE 29.18 The buildup of cholesterol deposits inside arteries can cause coronary heart disease, a leading cause of death for adults in industrialized nations.

We saw in Section 27.5 and Section 27.7 that all steroids, including cholesterol, are biosynthesized from the triterpenoid lanosterol, which in turn comes from acetyl CoA through isopentenyl diphosphate. If you knew all the mechanisms for all the chemical steps in cholesterol biosynthesis, you might be able to devise a drug that would block one of those steps, thereby short-circuiting the biosynthetic process and controlling the amount of cholesterol produced.

But we do know those mechanisms! Look back at the pathway for the biosynthesis of isopentenyl diphosphate from acetyl CoA, shown in FIGURE 27.8. It turns out that the rate-limiting step in the pathway is the reduction of 3-hydroxy-3-methylglutaryl CoA (abbreviated HMG-CoA) to mevalonate, brought about by the enzyme HMGCoA reductase. If that enzyme could be stopped from functioning, cholesterol biosynthesis would also be stopped.


Statin Drugs -
To find a drug that blocks HMG-CoA reductase, chemists did two simultaneous experiments on a large number of potential drug candidates isolated from soil microbes. In one experiment, the drug candidate and mevalonate were added to liver extract; in the second experiment, only the drug candidate was added without mevalonate. If cholesterol was produced only in the presence of added mevalonate but not in the absence of mevalonate, the drug candidate must have blocked the enzyme for mevalonate synthesis.

The drugs that block HMG-CoA reductase, and thus control cholesterol synthesis in the body, are called statins. In just the 10-year period following their introduction in 1994, the death rate from coronary heart disease decreased by $33 \%$ in the United States.

Like many drugs, statins don't come without risks and have some serious side effects that people considering their use should be aware of, but approaching 30 years later, they remain among the most widely prescribed drugs in the world, with estimated annual sales of $\$ 14$ billion. Atorvastatin (Lipitor), simvastatin (Zocor), rosuvastatin (Crestor), pravastatin (Pravachol), and lovastatin (Mevacor) are some examples. An X-ray crystal structure of the active site in the HMG-CoA reductase enzyme is shown in the accompanying graphic, along with a molecule of atorvastatin (purple) that is tightly bound in the active site and stops the enzyme from functioning. A good understanding of organic chemistry certainly paid off in this instance.

Atorvastatin (Lipitor)


Key Terms -
- anabolism
- gluconeogenesis
- $\beta$-oxidation pathway
- glycolysis
- catabolism
- metabolism
- citric acid cycle
- Schiff base
- deamination
- transamination


Summary -
Metabolism is the sum of all chemical reactions in the body. Reactions that break down large molecules into
smaller fragments are called catabolism, and those that build up large molecules from small pieces are called anabolism. Although the details of specific biochemical pathways are sometimes complex, all the reactions that occur follow the normal rules of organic chemical reactivity.

The catabolism of fats begins with digestion, in which ester bonds are hydrolyzed to give glycerol and fatty acids. The fatty acids are degraded in the four-step $\boldsymbol{\beta}$-oxidation pathway by removal of two carbons at a time, yielding acetyl CoA. Catabolism of carbohydrates begins with the hydrolysis of glycoside bonds to give glucose, which is degraded in the ten-step glycolysis pathway. Pyruvate, the initial product of glycolysis, is then converted into acetyl CoA. Acetyl CoA next enters the eight-step citric acid cycle, where it is further degraded into $\mathrm{CO}_{2}$. The cycle is a closed loop of reactions in which the product of the final step (oxaloacetate) is a reactant in the first step.

Catabolism of proteins is more complex than that of fats or carbohydrates because each of the 20 different amino acids is degraded by its own unique pathway. In general, though, the amino nitrogen atoms are removed and the substances that remain are converted into compounds that enter the citric acid cycle. Most amino acids lose their nitrogen atom by transamination, a reaction in which the $-\mathrm{NH}_{2}$ group of the amino acid trades places with the keto group of an $\alpha$-keto acid such as $\alpha$-ketoglutarate. The products are a new $\alpha$-keto acid and glutamate.

The energy released in catabolic pathways is used in the electron-transport chain to make molecules of adenosine triphosphate, ATP. ATP, the final result of food catabolism, couples to and drives many otherwise unfavorable reactions.


Summary -
Biomolecules are synthesized as well as degraded, but the pathways for anabolism and catabolism are not the exact reverse of one another. Fatty acids are biosynthesized from acetate by an 8 -step pathway, and carbohydrates are made from pyruvate by the 11-step gluconeogenesis pathway.


Visualizing Chemistry -
PROBLEM Identify the amino acid that is a catabolic precursor of each of the following $\alpha$-keto acids:
29-17 (a)

(b)

PROBLEM Identify the following intermediate in the citric acid cycle, and tell whether it has $R$ or $S$
29-18 stereochemistry:

PROBLEM The following compound is an intermediate in the biosynthesis of one of the 20 common $\alpha$-amino
29-19 acids. Which one is it likely to be, and what kind of chemical change must take place to complete the biosynthesis?

PROBLEM The following compound is an intermediate in the pentose phosphate pathway, an alternative route 29-20 for glucose metabolism. Identify the sugar it is derived from.


Mechanism Problems -
PROBLEM In the pentose phosphate pathway for degrading sugars, ribulose 5-phosphate is converted to
29-21 ribose 5-phosphate. Propose a mechanism for the isomerization.

Ribulose 5-phosphate

Ribose 5-phosphate

PROBLEM Another step in the pentose phosphate pathway for degrading sugars (see Problem 29-21) is the
29-22 conversion of ribose 5-phosphate to glyceraldehyde 3-phosphate. What kind of organic process is occurring? Propose a mechanism for the conversion.

PROBLEM One of the steps in the pentose phosphate pathway for glucose catabolism is the reaction of 29-23 sedoheptulose 7-phosphate with glyceraldehyde 3-phosphate in the presence of a transaldolase to yield erythrose 4-phosphate and fructose 6-phosphate.

(a) The first part of the reaction is the formation of a protonated Schiff base of sedoheptulose 7-phosphate with a lysine residue in the enzyme followed by a retro-aldol cleavage to give an enamine plus erythrose 4-phosphate. Show the structure of the enamine and the mechanism by which it is formed.
(b) The second part of the reaction is a nucleophilic addition of the enamine to glyceraldehyde 3 -phosphate followed by hydrolysis of the Schiff base to give fructose 6-phosphate. Show the mechanism.

PROBLEM One of the steps in the pentose phosphate pathway for glucose catabolism is the reaction of xylulose
29-24 5-phosphate with ribose 5-phosphate in the presence of a transketolase to give glyceraldehyde 3 -phosphate and sedoheptulose 7-phosphate.


Mechanism Problems -
(a) The first part of the reaction is nucleophilic addition of thiamin diphosphate (TPP) ylide to xylulose 5-phosphate, followed by a retro-aldol cleavage to give glyceraldehyde 3-phosphate and a TPP-containing enamine. Show the structure of the enamine and the mechanism by which it is formed.
(b) The second part of the reaction is addition of the enamine to ribose 5-phosphate followed by loss of TPP ylide to give sedoheptulose 7-phosphate. Show the mechanism.

PROBLEM The amino acid tyrosine is biologically degraded by a series of steps that include the following
29-25 transformations:

The double-bond isomerization of maleoylacetoacetate to fumaroylacetoacetate is catalyzed by practically any nucleophile, : $\mathrm{Nu}^{-}$. Propose a mechanism.

PROBLEM Propose a mechanism for the conversion of fumaroylacetoacetate to fumarate plus acetoacetate 29-26 (see Problem 29-25).

PROBLEM Propose a mechanism for the conversion of acetoacetate to acetyl CoA (see Problem 29-25). 29-27

PROBLEM Design your own degradative pathway. You know the rules (organic mechanisms), and you've seen 29-28 the kinds of reactions that occur in the biological degradation of fats and carbohydrates into acetyl CoA. If you were Mother Nature, what series of steps would you use to degrade the amino acid serine into acetyl CoA?

PROBLEM The amino acid serine is biosynthesized by a route that involves reaction of 29-29 3-phosphohydroxypyruvate with glutamate to give 3-phosphoserine. Propose a mechanism.

3-Phosphohydroxypyruvate


Mechanism Problems -
3-Phosphohydroxypyruvate

3-Phosphoserine
PROBLEM The amino acid leucine is biosynthesized from $\alpha$-ketoisocaproate, which is itself prepared from
29-30 $\alpha$-ketoisovalerate by a multistep route that involves (1) reaction with acetyl CoA, (2) hydrolysis, (3) dehydration, (4) hydration, (5) oxidation, and (6) decarboxylation. Show the steps in the transformation, and propose a mechanism for each.

PROBLEM The amino acid cysteine, $\mathrm{C}_{3} \mathrm{H}_{7} \mathrm{NO}_{2} \mathrm{~S}$, is biosynthesized from a substance called cystathionine by a 29-31 multistep pathway.

(a) The first step is a transamination. What is the product?
(b) The second step is an E1cB reaction. Show the products and the mechanism of the reaction.
(c) The final step is a double-bond reduction. What organic cofactor is required for this reaction, and what is the product represented by the question mark in the equation?


Enzymes and Coenzymes -
PROBLEM What chemical events occur during the digestion of food?
29-32

PROBLEM What is the difference between digestion and metabolism? 29-33

PROBLEM What is the difference between anabolism and catabolism? 29-34

PROBLEM Draw the structure of adenosine $5^{\prime}$-monophosphate (AMP), an intermediate in some biochemical 29-35 pathways.

PROBLEM Cyclic adenosine monophosphate (cyclic AMP), a modulator of hormone action, is related to AMP 29-36 (Problem 29-35) but has its phosphate group linked to two hydroxyl groups at C3' and C5' of the sugar. Draw the structure of cyclic AMP.

PROBLEM What general kind of reaction does ATP carry out? 29-37

PROBLEM What general kind of reaction does $\mathrm{NAD}^{+}$carry out? 29-38

PROBLEM What general kind of reaction does FAD carry out? 29-39

PROBLEM What enzyme cofactor is associated with each of the following kinds of reactions?
29-40 (a) Transamination
(b) Carboxylation of a ketone
(c) Decarboxylation of an $\alpha$-keto acid

PROBLEM Lactate, a product of glucose catabolism in oxygen-starved muscles, can be converted into pyruvate 29-41 by oxidation. What coenzyme do you think is needed? Write the equation in the normal biochemical format using a curved arrow.

Lactate


Metabolism -
PROBLEM Write the equation for the final step in the $\beta$-oxidation pathway of any fatty acid with an even 29-42 number of carbon atoms.

PROBLEM Show the products of each of the following reactions:
29-43
(a)

(b)
Product of $(\mathbf{a})+\mathrm{H}_{2} \mathrm{O} \xrightarrow{\begin{array}{l}\text { Enoyl-CoA } \\ \text { hydratase }\end{array}}$ ?
(c)
Product of (b)

PROBLEM Why aren't the glycolysis and gluconeogenesis pathways the exact reverse of each other? 29-44

PROBLEM How many moles of acetyl CoA are produced by catabolism of the following substances?
29-45 (a) 1.0 mol of glucose
(b) 1.0 mol of palmitic acid
(c) 1.0 mol of maltose

PROBLEM How many grams of acetyl CoA (MW $=809.6 \mathrm{amu})$ are produced by catabolism of the following 29-46 substances? Which substance is the most efficient precursor of acetyl CoA on a weight basis?
(a) 100.0 g of glucose
(b) 100.0 g of palmitic acid
(c) 100.0 g of maltose

PROBLEM What is the structure of the $\alpha$-keto acid formed by transamination of each of the following amino 29-47 acids?
(a) Threonine
(b) Phenylalanine
(c) Asparagine

PROBLEM The glycolysis pathway shown in Figure 29.8 has a number of intermediates that contain phosphate 29-48 groups. Why can 3-phosphoglyceryl phosphate and phosphoenolpyruvate transfer a phosphate group to ADP while glucose 6-phosphate cannot?

PROBLEM Write a mechanism for the conversion of $\alpha$-ketoglutarate to succinyl CoA in step 4 of the citric acid 29-49 cycle (Figure 29.14).


Metabolism -
PROBLEM Write a mechanism for the conversion of $\alpha$-ketoglutarate to succinyl CoA in step 4 of the citric acid 29-49 cycle (Figure 29.14).

PROBLEM In step 2 of the citric acid cycle (Figure 29.14), cis-aconitate reacts with water to give ( $2 R, 3 S$ )-29-50 isocitrate. Does -OH add from the Re face of the double bond or from the Si face? What about -H? Does the addition of water occur with syn or anti geometry?


General Problems -
PROBLEM In glycerol metabolism, the oxidation of sn-glycerol 3-phosphate to give dihydroxyacetone
29-51 phosphate is catalyzed by sn-glycerol-3-phosphate dehydrogenase, with NAD ${ }^{+}$as cofactor. The reaction is stereospecific, occurring exclusively on the Re face of the nicotinamide ring.

$N A D^{+}$

NADH

Which hydrogen in the NADH product comes from sn-glycerol 3-phosphate? Does it have pro- $R$ or pro- $S$ stereochemistry?

PROBLEM The primary fate of acetyl CoA under normal metabolic conditions is degradation in the citric 29-52 acid cycle to yield $\mathrm{CO}_{2}$. When the body is stressed by prolonged starvation, however, acetyl CoA is converted into compounds called ketone bodies, which can be used by the brain as a temporary fuel. Fill in the missing information indicated by the four question marks in the following biochemical pathway for the synthesis of ketone bodies from acetyl CoA:

Acetone

3-Hydroxybutyrate

Ketone bodies
PROBLEM The initial reaction in Problem 29-52, conversion of two molecules of acetyl CoA to one molecule of 29-53 acetoacetyl CoA, is a Claisen reaction. Assuming that there is a base present, show the mechanism
of the reaction.
PROBLEM In step 6 of fatty-acid biosynthesis (Figure 29.6), acetoacetyl ACP is reduced stereospecifically by 29-54 NADPH to yield an alcohol. Does hydride ion add to the Si face or the Re face of acetoacetyl ACP?

Acetoacetyl ACP

$\beta$-Hydroxybutyryl ACP

PROBLEM In step 7 of fatty-acid biosynthesis (Figure 29.6), dehydration of a $\beta$-hydroxy thioester occurs to give 29-55 trans-crotonyl ACP. Is the dehydration a syn elimination or an anti elimination?


General Problems -
PROBLEM In step 8 of fatty-acid biosynthesis (Figure 29.6), reduction of trans-crotonyl ACP gives butyryl ACP.
29-56 A hydride from NADPH adds to C3 of the crotonyl group from the Re face, and protonation on C2 occurs on the Si face. Is the reduction a syn addition or an anti addition?


CHAPTER 30 -
Orbitals and Organic Chemistry: Pericyclic Reactions

FIGURE 30.1 It may look easy, but these sunbathers are working hard to carry out a pericyclic reaction. (credit: modification of work "Tanning" by Meraj Chhaya/Flickr, CC BY 2.0).