Solutions Manual Biochemistry 1st Edition by Roger L. Miesfeld, Megan M. McEvoy Test Bank

Solutions Manual Biochemistry 1st Edition by Roger L. Miesfeld, Megan M. McEvoy Test Bank

Solutions Manual Biochemistry 1st Edition by Roger L. Miesfeld, Megan M. McEvoy Test Bank

Solutions Manual Biochemistry 1st Edition by Roger L. Miesfeld, Megan M. McEvoy Test Bank

Solutions Manual Biochemistry 1st Edition by Roger L. Miesfeld, Megan M. McEvoy Test Bank

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Micro RNA regulates messenger RNA translation and thus gene expression.

Ribosomal RNA is contained in ribosomes. Transfer RNA molecules carry amino acids to the ribosome and are required for protein synthesis.

11. Mutations in a gene will result in changes to the mRNA transcript. When the altered mRNA is translated into a protein, the protein can have changes in its amino acid sequence. Addition of the wrong amino acid can lead to protein denaturation (unfolding) or loss of function, resulting in breakdown of biochemical pathways, which can lead to disease or death.

12. Gene duplication can lead to new genes or higher survivability of individuals containing duplicated genes. If doubling the amount of gene product (protein) increases the fitness of an organism, the second copy of the gene is maintained by evolution. However, if no fitness advantages accrue from doubling the amount of gene product, the duplicated gene is free to mutate into another gene that could confer new fitness advantages to an organism.

In other cases, this gene could be removed by natural selection if it confers, or is mutated to confer, deleterious qualities to an organism.

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Long Answers to Challenge Problems Chapter 1 1. (a) Pyruvate → acetaldehyde + CO2 (pyruvate decarboxylase). (b) Acetaldehyde + NADH + H+ → ethanol + NAD+ (alcohol dehydrogenase). (c) Acetaldehyde is a product in (a) and a reactant in (b).

2. Buchner (1) used a different strain of yeast than Pasteur used, (2) prepared the yeast extract using quartz mixed with diatomaceous earth (kieselguhr) rather than glass, and (3) his extract buffer contained glucose, which fortuitously served as the carbon source for fermentation.

3. The total number of dodecanucleotides that can be synthesized is 412 = 16,777,216.

The total number of tetrapeptides that can be encoded is 204 = 160,000.

4. Amylose is a polymer of glucose containing α(1→4) glycosidic bonds that can be cleaved by the human enzyme amylase to produce glucose. However, cellulose is a glucose polymer containing β(1→4) glycosidic bonds that are cleaved by the enzyme cellulase, an enzyme that humans do not have. The reason why horses can digest cellulose is that they have intestinal bacteria expressing cellulase that is secreted into the gut and aids the horse in digestion of plant material.

5. Liver cells and skeletal muscle cells both contain insulin receptors, but only liver cells contain glucagon receptors; skeletal muscle cells do not.

6.

mRNA: 5′ – AAAAAAUUUAAAUUU – 3′ Protein: NH3 +-Lys-Lys-Phe-Lys-Phe-COO−

7. Germ-line mutations that occur outside of protein-coding sequences can persist in the germ line if they do not cause a defect in genome stability or regulation; these are called neutral mutations. However, germ-line mutations that occur 3

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in protein-coding sequences will be selected against if they alter reproductive rates, and they tend to disappear from the population over long periods of time.

8. Paralogous genes are the result of gene duplication and can diverge within the same species in both structure and function over time, whereas orthologous genes encode proteins with the same structure and function in different species.

9. Comparing two amino acid sequences of <20% does not reveal much about the structure or function of proteins because different amino acid sequences could specify a similar protein structure, or a similar protein function, or neither. In contrast, two proteins that are >80% identical at the amino acid level are very likely to have the same structure and function.

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Long Answers to Review Questions Chapter 2 1. Chemical work is in the form of biosynthesis of new organic molecules; osmotic work is done by differential solute concentrations across biological membranes; and mechanical work is in the form of muscle contraction or flagellar rotation.

2. Redox reactions are a form of chemical work, which can be used to synthesize important biomolecules with energy made available through electron transfer. Chemical work is also vital for an organism to maintain homeostasis.

3. In photosynthesis, a series of coupled redox reactions result in an electrochemical proton gradient across the chloroplast membrane. The energy from this gradient is used to drive the phosphorylation of ADP to form ATP. The ATP formed in the culmination of these reactions can then be used to convert carbon dioxide into glucose, which can be used by plants or the animals that eat those plants.

4. The equation that describes the change in Gibbs free energy is ΔG = ΔH – TΔS, where ΔG defines the free energy change between reactants and products of a specific reaction, ΔH is the change in enthalpy, T is the temperature, and ΔS is the change in entropy. A favorable reaction is one in which the sign of ΔG is negative. This equation shows that both ΔH and ΔS contribute to the overall change in free energy and that the temperature (T ) amplifies or decreases the entropic change of the reaction.

5. A reaction that is unfavorable under standard conditions might still proceed in a living system, if the actual conditions make it possible. The spontaneity of a reaction is determined by the actual free energy change ΔG. The standard free energy change and the initial concentrations of reactants and products both contribute to the actual free energy change (see Equation 2.14). In cells, if product concentrations are low and reactant concentrations are high ([products]/[reactants] < 1), the second 1

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term of Equation 2.14 will be negative. If it is sufficiently negative to overcome a positive ΔG °′, then the reaction will be spontaneous with ΔG < 0.

6. Water is less dense as a solid than as a liquid. This means that ice will float.

If ice were denser, it would sink to the bottom of the ocean, resulting in an upwelling of cold water, which would also freeze and sink. This process would continue until all of the water was frozen solid. Water is in its liquid state for a wide range of temperatures, particularly those found on Earth.

This is critical to the oxygen concentration found on Earth because oxygen is largely dependent on the photosynthetic algae that flourish in the oceans.

Water is a fantastic solvent because of its polar nature and hydrogen-bonding abilities, which are important for biological systems.

7. (1) Hydrogen bonds form between a donor group, consisting of a hydrogen atom attached to an electronegative atom, and a hydrogen-bond acceptor, which is an electronegative atom. The most common electronegative atoms involved in hydrogen bonding are oxygen, nitrogen, and sulfur.

(2) Electrostatic ionic interactions occur between two atoms with opposite charges. The strength of the interaction depends upon the environment of the ions participating in the bond and the distance between the ions themselves. (3) van der Waals interactions occur between nonpolar molecules and arise from the creation of temporary dipoles due to fluctuation in electron clouds. If dipoles align with opposite signs at the appropriate distance, an interaction can occur. van der Waals interactions are quite weak but can occur simultaneously with a variety of atoms, thus having a strong cumulative effect. (4) The hydrophobic effect is driven by the tendency for hydrophobic molecules to pack close together in solution. Water molecules maintain an ordered structure around hydrophobic molecules, thus decreasing entropy. By packing close together, hydrophobic molecules decrease their overall surface area, which decreases the number of water molecules that are involved in an ordered structure. This increases the entropy of the system, making the formation of such structures favorable.

8. Colligative properties are properties of solutions related to numbers of solute particles. These properties are freezing point depression, boiling point elevation, vapor pressure lowering, and osmotic pressure.

9. The pH scale is a logarithmic scale. A value of 0–6.5 is considered acidic, whereas a value of 7.5–14 is considered basic. Anything in between these two ranges (that is, 6.5–7.5) is considered to be neutral.

10. Amphipathic molecules contain hydrophobic and hydrophilic regions.

Amphipathic lipids are crucial for life because they form biological membranes with a hydrophobic region that is relatively impermeable to polar molecules. These membranes are necessary for the separation of the inside of cells from the environment and for compartmentalization within cells where specific reactions can occur.

11. (1) Phospholipid monolayers have the polar head groups of their phospholipids pointed toward water with their tails pointed toward air or a hydrophobic

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environment. (2) Phospholipid bilayers create a hydrophobic barrier between two aqueous compartments, which is characteristic of biological membranes.

(3) Micelles have the hydrophobic tails of the lipids packed in the center of a globular sphere, with the polar head groups facing outward toward water.

(4) Liposomes are spherical structures formed by a lipid bilayer that surrounds an aqueous center.

The organization of these structures such that the hydrophobic regions interact with each other increases entropy, thus favoring their formation, which is consistent with the hydrophobic effect.

12. (1) The plasma membrane is the lipid bilayer that surrounds every cell.

(2) The endomembrane consists of structurally related intracellular membrane networks and vesicles. (3) Chloroplast and mitochondrial membranes enclose the enzymes that convert energy into a usable form for organisms. These membranes separate the inside of these organelles from the cytoplasm, which is a very different environment.

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Long Answers to Challenge Problems Chapter 2 1. Plants (at night) and animals (all of the time) metabolize nutrients (carbohydrates) by the process of aerobic respiration.

2. The second law of thermodynamics states that all natural processes in the universe tend toward disorder (entropy) in the absence of energy input. Many types of biochemical reactions in living organisms are maintained at a steady state that is far from equilibrium with the environment. This is done by photosynthetic organisms that harness energy released from the Sun (thermonuclear fusion reactions) and convert it to chemical energy in the form of ATP, NADPH, and glucose. Chemical energy is used by all organisms to restrain entropy and avoid reaching equilibrium with the environment (death).

3. a. The notation ΔG °′ refers to the biochemical standard free energy change of a reaction, which is experimentally determined under physiologic conditions (298 K, 1 atm, 1 M of each reactant and product, pH 7) by allowing the reaction to go to equilibrium and then measuring the steady-state concentrations of reactants and products. The difference in free energy under these two conditions (1 M concentration and at equilibrium) is expressed in units of kilojoules (kJ). The value of ΔG is the actual free energy change in a reaction that is calculated from the known ΔG °′ value and the initial concentrations of reactants and products. b. When a reaction is at equilibrium, ΔG = 0, and ΔG °′ is described by the equation: ΔG = ΔG °′ + RT ln

[B]actual

[A]actual [B]eq ΔG °′ = −RT ln [A]eq

=0

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c. Enzymes are catalysts and only affect the rates of chemical reactions. The change in free energy (ΔG ), change in standard free energy (ΔG °′), and the equilibrium constant (Keq) are all unchanged in the presence of an enzyme.

4. a. Keq = [B]/[A] = 1 × 105 The total number of moles of A plus B must stay constant throughout the reaction.

Therefore, [A] + [B] = 1.000 M + 0.001 M = 1.001 M. Keq = [B]/[A] = 1 × 105 = 100,000 Rearrange to get [B] = 100,000[A] and substitute for [B] in [A] +[B] = 1.001 M: [A] + 100,000[A] = 1.001 M 100,001[A] = 1.001 M [A] = 1.000989 × 10−5 M [B] = 1.000989 M b. At equilibrium, ΔG = 0 and therefore [B]eq ∆G = ∆G °′ + RT ln = 0, and [A] eq

ΔG °′ = –RT ln Keq ΔG °′ = –(8.3 × 10–3 kJ K–1 mol–1)(298 K) ln1 × 105 ΔG °′ = –28.5 kJ mol–1 c. ΔG = ΔG °′ + RT ln

[B]actual =0 [A]actual

–1

–3

ΔG = –28.5 kJ mol

–1

–1

+ (8.3 × 10 kJ K mol )(298 K) ln

15 × 10−2 M 5 × 10−5 M

5. a. Pathway 1: A m B m E m F Pathway 2: A m B m C + D m F b. Pathway 1 has a ΔG °′ = –10 kJ/mol, and pathway 2 has a ΔG °′ = +4 kJ/mol; therefore, pathway 1 is more likely to proceed based on a favorable ΔG °′ value.

6. a. Energy charge (EC) = =

[ATP] + 0.5 [ADP] [ATP] + [ADP] + [AMP] 1.25 mM + 0.510.35 mM2 1.25 mM + 0.35 mM + 0.125 mM

= 0.83

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b. Keq = Keq =

[ATP] [AMP] [ADP] [ADP] 11.25 mM 2 10.12 mM 2 10.35 mM 2 2

= 1.22

7. The energy charge of the cell refers to the relative concentrations of ATP, ADP, and AMP. When the energy charge is high, it means that ATP concentrations are high relative to AMP and ADP. Anabolic pathways (biosynthesis) require ATP; therefore, a high energy charge in the cell leads to increased flux through anabolic pathways. In contrast, when the energy charge is low, flux is increased through catabolic pathways to replenish ATP levels and achieve homeostasis.

8. Glucose + Pi m Glucose-6-phosphate

ΔG °′ = +13.8 kJ/mol

ATP m ADP + Pi

ΔG °′ = –30.5 kJ/mol

Glucose + ATP m Glucose-6-phosphate + ADP

ΔG °′ = –16.7 kJ/mol

9. The water molecules hydrogen bond with each other to form a cage-like structure around the hydrophobic limonene molecule. This process increases the order of the water molecules, which is an entropically unfavorable process, and therefore ΔS is negative.

10. Water molecules in ice are oriented such that maximum hydrogen bonding between water molecules occurs. Protons move through ice (and water) via proton hopping.

11. The four noncovalent interactions are (1) hydrogen bonds, (2) ionic interactions, (3) van der Waals interactions, and (4) hydrophobic effects.

Hydrogen bonds are directly formed between H2O and biomolecules and between H2O molecules themselves, whereas hydrophobic effects are indirectly caused by H2O through the promotion of “water-avoiding” interactions between nonpolar molecules.

12. a. Using the Henderson–Hasselbalch equation: [A−] pH = pKa + log [HA] [HCO 3− ] 7.4 = 6.1 + log [CO21 aq 2 ] [HCO 3 − ] 7.4 − 6.1 = 1.3 = log [CO2 1 aq 2 ] 1.3

10

= 10

[HCO3 –] [CO2(aq)]

[HCO 3 − ]

= 20 [CO2 1 aq 2 ] 1 b. 0.025 M = 2.5 × 10–2 M = [HCO3 − ] + [CO2(aq)] 2.5 × 10−2 M = [CO2(aq)] = 1.19 × 10–3 M 20 + 1 [HCO3 − ] = 2.5 × 10–2 M – 1.19 × 10–3 M = 2.38 × 10–2 M

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13. The spot disappears because the proteins laterally diffuse in the plane of the membrane. Therefore, eventually the bleached molecules diffuse out of the laser-treated area and are replaced by fluorescent molecules. The rate of lateral protein diffusion in the plasma membrane can be determined using this experimental approach.

14. For the fatty acid to form a micelle, it must have a charged polar head group. This will only occur when the carboxyl group has ionized, that is, COOH → COO− + H+, and this will only occur at pH values higher than the pKa.

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Long Answers to Review Questions Chapter 3 1. A nucleoside contains a ribose sugar and base, whereas a nucleotide contains a ribose sugar, base, and one or more phosphate groups.

2. Chargaff ’s rule states that in any segment of DNA, A = T and G = C. This discovery led Watson and Crick to model their DNA structure around the concept of base pairing. Combined with the structural data obtained through Rosalind Franklin’s X-ray crystallography work, the base pair rule reduced the number of possible structures and allowed Watson and Crick to manipulate models of the bases to identify a double helix that fit within these parameters.

3. A-, B-, and Z-form DNA have all been identified through crystallization of short regions of DNA. All three forms are antiparallel double helices. The covalent linkages between the nucleotides are the same in all three forms, though their conformations differ. The most obvious difference is in the width of the helices. A-form DNA is the widest helix, and Z-form DNA is the narrowest. As a result, the number of base pairs per helical turn is greater in A-DNA than in B-DNA. Although Z-DNA is more narrow, it is stretched in comparison with the other two, so it contains 12 bp per helical turn. A-DNA and B-DNA are both righthanded helices, whereas Z-DNA is a left-handed helix.

4. The three topological properties of DNA are linking number, twist, and writhe. Linking number = twist + writhe. If the DNA backbone is not cleaved, the linking number must remain constant, so twist and writhe are adjusted accordingly.

5. Primary structure is the nucleotide sequence of each DNA strand.

Secondary structure is the formation of a double helix.

6. Type I topoisomerases cleave only one strand of the DNA, whereas type II topoisomerases cleave both strands.

7. Chromatin formation and DNA replication both involve 1

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the winding and/or unwinding of the double helix, leading to positive supercoiling and topological strain in other regions of the molecule.

Topoisomerases are necessary to relieve the positive supercoiling so that the topological strain on the molecule does not inhibit further unwinding.

8. The 2′-hydroxyl group on the ribose of RNA can lead to an autocleavage reaction that breaks the phosphodiester backbone. DNA does not have this 2′ -hydroxyl group and is therefore not prone to autocleavage events.

9. The hyperchromic effect describes the increase in light absorbance as complementary strands of DNA separate. By monitoring the absorbance of DNA at 260 nm, the separation of DNA strands can be monitored under different reaction and/or temperature conditions.

10. N 2,N 2-dimethylguanosine, 1-methyladenosine, pseudouridine, and inosine are all examples of modified RNA bases (see Figure 3.25 for more examples).

These bases can help form the correct tertiary structures or influence function. Inosine is often found in the anticodon of tRNA and allows one tRNA to pair with three different nucleotide bases (uracil, adenine, cytosine) in the mRNA during the process of protein synthesis (see Chapter 22).

11. In the first stage of eukaryotic DNA condensation, DNA is wrapped around eight histone proteins and sealed by a linker histone, generating a nucleosome particle. Packing of nucleosomes leads to the formation of the 30-nm chromatin fiber.

12. Euchromatin is more loosely packed than heterochromatin. Euchromatin forms in regions of DNA that contain coding sequences; the structural freedom allows easier access by DNA binding proteins. Heterochromatin is most often formed in noncoding regions and is therefore more tightly packed.

13. Centromeres are an area of heterochromatin found in the center of the chromosome. During cell division, kinetochore proteins assemble on the centromere and attach to the mitotic spindle. These proteins assist in the separation of sister chromatids so that each daughter cell receives a single copy of the chromosome. Telomeres are short, repetitive sequences on the ends of chromosomes. DNA polymerase is not able to copy to the very end of the chromosome, so the presence of the telomeric regions ensures that the polymerase is able to finish replicating the entire chromosomal DNA. Without telomeres, a bit of the chromosomal end would remain unreplicated and would be lost with each round of cell division.

14. Prokaryote genes can be organized as monocistronic or polycistronic transcription units. Monocistronic genes contain a promoter followed by a single coding region; polycistronic genes contain a promoter followed by multiple coding regions. Polycistronic genes are usually related in their function or role in the cell.

15. Exons contain the coding sequences, and introns contain noncoding sequences.

16. Plasmids are circular, self-replicating DNA molecules. They often confer a survival advantage for bacteria by expressing one or more antibioticresistance genes.

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17. PCR is initiated by denaturation in the first stage, which separates the DNA strands. In the second stage, annealing allows the primers to match up to their complementary regions in each single strand. Finally in the third stage, primer extension is carried out by DNA polymerase, which binds to the short, double-stranded region generated by the annealed primer and synthesizes a complementary strand.

18. Blunt ends are generated by a restriction endonuclease that cleaves each strand at a complementary position, usually in the middle of the recognition sequence. Two types of overhangs can be generated by enzymes that cleave at non-complementary positions of the two strands: a 5′ or a 3′ overhang.

19. Dideoxynucleoside triphosphates can be incorporated into a growing DNA strand by DNA polymerase, but strand synthesis stops immediately after their incorporation because they lack a 3′-hydroxyl group. In a sequencing reaction, strand synthesis is sequentially stopped at each nucleotide position, leading to a number of products, each one base different in length. These fragments can be resolved by capillary electrophoresis and the sequence determined by the order of the dideoxynucleoside triphosphates on the end of each sequence.

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Long Answers to Challenge Problems Chapter 3 1. Hydrogen bonding occurs between the bases when they are within a helix. When the strands are denatured, the hydrogen bonds between the bases are replaced by hydrogen bonds between the bases and water. These hydrogen bonds are similar in strength to those formed between the bases. Thus, hydrogen bonding between the bases contributes little to overall helix stability.

2. Topoisomerase II inhibitors such as etoposide prevent the repair of double-strand breaks. Cancer cells are rapidly dividing and are therefore frequently replicating. Failure to repair breaks in these cells leads to cell death and a decrease in the number of cancer cells, which is why these inhibitors are very successful in eradicating certain types of cancer. However, the inhibitors also affect normal cells, and cell death is not always the result.

When these breaks occur, they can result in the loss of a portion of a chromosome or in the joining together of two different chromosomes. If this break occurs within the coding region of certain genes, the deletions and translocations can result in certain types of leukemia.

3. The function of the DNA or RNA binding protein determines if the interaction is dependent on a specific nucleotide sequence in DNA or RNA. Histone proteins must be able to bind all DNA in order to compact it into chromosomes. Because histones have a more general function in the cell, it is not necessary that they recognize and bind to specific nucleotide sequences. Instead, histone binding is mediated by a specific property of DNA; that is, the negatively charged backbone. By contrast, proteins such as transcription factors must bind to a specific nucleotide sequence within a promoter region to initiate transcription of a gene or set of genes. In this case, it is very important that the protein be selective in its interaction with DNA. 4. Every time a cell divides, the telomere length of each chromosome is reduced. Somatic cells have little to no telomerase activity, 4

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and therefore the telomere is not restored after cell division. Eventually, the telomeres are lost, and cell death occurs shortly after. The age of a cell is related to the length of its telomeres, with older cells having shorter telomeres. In this way, telomere length and maintenance is part of the aging process. Scientists feared that an embryo created from a somatic cell would have the telomere length and therefore “age” of the somatic cell.

5. The promoter is a region of DNA that contains recognition sequences for one or more transcription factors. Binding of these transcription factors recruits RNA polymerase, which initiates transcription on separated DNA strands. Regions of DNA that are rich in A-T require less energy to separate the two strands, so are often found in transcriptional start sites.

6. In eukaryotes, mRNA must be spliced to remove introns before translation can occur. As a result, transcription must be completed before translation can begin. Because the majority of prokaryotic genes do not contain introns, the mRNA does not need to be processed and can be translated as it is transcribed. A quicker path to protein production in prokaryotes leads to faster cell division and growth.

7. The human genome is full of random sequence mutations that have little to no effect on protein expression or function. Some of these, such as SNPs and STRs, are inherited; however, spontaneous events do occur. Without any prior knowledge of the gene where the mutation may be present, comparison of the DNA of two individuals would result in the identification of hundreds to thousands of differences, most of which are simple variations. If the affected gene is known, then the sequence of that gene from an affected individual can be compared to the sequence of that gene in unaffected individuals. This process would be most efficient if the unaffected individuals were parents or siblings so that any inherited variations could be eliminated as potential disease-causing mutations.

8. Like paternity testing, forensic analysis of DNA involves the comparison of common variable regions in the genome. Short tandem repeats (STRs) are inherited from maternal and paternal DNA. Siblings, parents, and children of the suspect would be expected to have a significant number of STRs in common with the suspect, and so any of these would be an ideal choice for comparison if the suspect’s DNA were not available. More distant relatives would be expected to have more STRs in common with the suspect than with the general population, but this would not make a convincing case before a judge and jury.

9. When a plasmid is transformed into a culture of bacterial cells, the process is not 100% efficient. Without selection, it would be impossible to determine which bacterial cells carried the plasmid without further investigation.

The use of antibiotic selection prevents bacteria without the plasmid from growing on agar plates or in liquid culture. Therefore, scientists can assume that a bacterial colony that is antibiotic resistant has resulted from a plasmid transformation in which the plasmid has been taken up successfully by the bacteria of that colony.

10. During the annealing process, the temperature is lowered to just below the Tm so that the primer can anneal to its complementary region. The

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G-C content of a primer contributes more to the Tm than does the A-T content because G-C base pairs require more heat to disrupt the duplex.

Therefore, the higher the G-C content, the higher the Tm and the annealing temperature.

11. The primer sequence has 6 A, 6 T, 10 C, and 4 G. The Tm calculated using this method is 2 °C (6 + 6) + 4 °C (10 + 4)= 80 °C. There are a variety of online calculators, but all use similar methods of determining Tm. The Tm using one of these calculators will be in the range 61–65 °C. Factors that influence this value are the number of base pairs compared with the %GC, the monovalent cation concentration, and the base stacking (nearest neighbor) interactions based on sequence.

12. Eukaryotic genes contain many introns that interrupt the coding regions.

Cloned genes from genomic DNA would include these intron sequences.

Because prokaryotic genes do not contain introns and lack splicing proteins, the introns are not removed in bacteria, and a random termination signal is likely contained in the intronic sequence that leads to abortive protein synthesis. Instead, the student needs to synthesize cDNA from mRNA, which will then only contain the gene-coding sequence because the introns were removed during the processing of precursor mRNA to mature mRNA. The cDNA can then be cloned into a plasmid for protein expression in bacteria.

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Long Answers to Review Questions Chapter 4 1. The isoelectric point of a protein is the pH at which the molecule carries no net charge. Amino acids that carry positive and negative charges, but no net charge, are called zwitterions.

2. Chemical properties of nearby functional groups in the protein alter the pKa of specific amino acids in order to maintain energetically favorable conditions. For example, the generation of positive or negative charges in the hydrophobic interior of a protein would be energetically unfavorable, and therefore the pKa value of a side chain may be altered to favor the neutral state over the charged state over a wider range of pH values.

3. The charged amino acids are aspartate, glutamate, lysine, arginine, and histidine. The hydrophobic amino acids are glycine, alanine, proline, valine, leucine, isoleucine, and methionine. The hydrophilic amino acids are serine, threonine, cysteine, asparagine, and glutamine. The aromatic amino acids are phenylalanine, tyrosine, and tryptophan.

4. A peptide bond is a covalent bond between the carboxylic acid group of one amino acid and the amine group of another amino acid. Peptide bonds are formed from condensation reactions that are catalyzed by ribosomes. The reaction is unfavorable and thus requires the hydrolysis of ATP. 5. Phi (ϕ) is the torsional angle between the α carbon and the amide nitrogen of an amino acid, whereas the psi (ψ) angle is the angle between the α carbon and the carbonyl carbon. The values of ϕ and ψ will affect the conformation of the polypeptide backbone.

1

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6. Primary structure is the specific amino acid sequence of a protein. The secondary structure of a protein refers to the regular repetitive arrangements of local regions of the polypeptide backbone. There are three major secondary structures: α helices, β strands, and β turns. The tertiary structure includes the spatial location of all atoms in a polypeptide chain. Quaternary structures are found in multi-subunit protein complexes. Subunits can involve multiple copies of the same polypeptide chain or one or more copies of different polypeptide chains.

7. Alpha helices can be amphipathic when residues that are either hydrophobic or hydrophilic are placed three to four amino acids away from each other.

This happens because there are 3.6 amino acids per turn of the α helix. Thus, amino acids that are three to four amino acids apart will be on the same side of the helix.

8. Secondary structures are common in proteins for two reasons. First, they are formed from combinations of ϕ and ψ angles that minimize steric hindrance between adjacent amino acids. Second, secondary structure conformations allow maximum hydrogen bonding in the polypeptide backbone.

9. Glutamate, alanine, and leucine are the three most common amino acids found in α helices. The three most common amino acids found in β strands are valine, isoleucine, and tyrosine. Glycine, asparagine, and proline are most commonly found in β turns.

10. Multiple subunits can provide structural properties not present in individual subunits. Also, having multiple subunits provides a mechanism for regulation of protein function through conformational changes that alter the protein subunit interfaces. Lastly, increased efficiency of biochemical processes can be achieved by bringing linked functional components into close proximity.

11. The folding process for proteins is thought to be dominated by preferred folding pathways, which must exist in order to avoid time-consuming sampling of unstable protein structures. Folding cannot be random because it would take billions of years for a protein of only 100 amino acids to sample three possible ϕ and ψ angles for every amino acid in the peptide.

12. Favorable changes in enthalpy upon protein folding result because there are more (or more favorable) noncovalent interactions being made in the folded state as compared to the unfolded state. These changes must be the result of noncovalent interactions or disulfide bond formation (in some proteins) because covalent bonds do not change during the folding process (with the exception of disulfide bond formation). Protein folding restricts the number of conformations that a polypeptide can take, which is entropically unfavorable. However, favorable entropic changes result from the increase in the disorder of the water molecules surrounding a protein.

13. One proposed mechanism is the hydrophobic collapse model. In this model, hydrophobic residues first form the interior of the protein due to the hydrophobic effect, which leads to a loosely defined tertiary structure called a molten globule. Then, the close proximity of residues in the molten globule leads to the formation of well-ordered secondary and tertiary structures through van der Waals interactions and hydrogen bonding. According to the framework model, local secondary structures form independently in the first

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phase of folding. Then, in the second phase of folding, the local secondary structures lead to the formation of tertiary structures. The third model is the nucleation model. According to the nucleation model, random interactions lead to a localized region of correct three-dimensional structure, which then serves to facilitate the formation of the surrounding tertiary and secondary structures.

14. One phenotypical consequence of protein-folding diseases is the degradation of a misfolded protein. This is known as a loss-of-function mutation phenotype because the activity of the particular protein is missing from the cell. The second is protein aggregation, known as gain-of-function mutations because the proteins add a process to the cell. Gain-of-function protein-folding mutations phenotypes can result from changes in the amino acid sequence due to mutation or from accumulation of misfolded wild-type proteins.

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Long Answers to Challenge Problems Chapter 4 1.

Functional group

Charge at pH 7

Charge at pH 11

N-terminal amino group of chain

+1

0

C-terminal carboxyl group of chain

−1

−1

Side chain of Lys

+1

0

Side chain of Tyr

0

−1

Side chain of Glu

−1

−1

Side chain of Asn

0

0

Side chain of His

0

0

Net charge on peptide

0

−3

2. a. It is not energetically favorable for the α amino group of the

protein to be charged in the hydrophobic environment, and therefore the pKa would be lower in the interior of the protein compared to a pKa of ∼8 on the surface. A lower pKa means that there will be a more significant population of the uncharged species (NH2) at neutral pH, which is more favourable in the hydrophobic environment. b. The pKa for this α amino group would go up relative to the situation where it is not forming an ionic bond in the hydrophobic interior of the protein. The pKa would more likely be close to the pKa of ∼8 seen in an aqueous environment. This occurs +charge and because this ionic bond neutralizes both the NH 3 the COO− charge in the protein interior, which reduces the effect of the hydrophobic environment on a charged chemical group. In fact, ionic bonds are stronger within a hydrophobic environment because it is unfavorable to not have compensated charges.

4

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3. In the closely packed interior of the protein structure, changing an Ala to

a Val would introduce a bulkier side chain, taking up more volume in the interior, which would necessitate some structural adjustments in the tertiary structure to make room for the extra atoms. In this case, the structural adjustments were serious enough to cause the enzyme to lose activity. The readjustment of the packing in the interior from replacement of an Ile residue with a Gly resulted in recovery of a tertiary structure close enough to the original structure for the enzyme to regain partial activity.

4. When the amide and carbonyl groups are involved in hydrogen bonding,

such as in α-helix and β-sheet structures, this serves to minimize the effects of these polar groups.

5. The three residues in the partial protein sequence are Lys-Ser-Phe. Three

potential hydrogen-bond donors and acceptors are boxed.

Lys 1

Ser 2 OH

Phe 3

CH2

Donor CH2

C

C

H

O

H

H

N

C CH2

Acceptor

CH2 +

NH3

6. The four types of noncovalent interactions that stabilize tertiary and quater-

nary protein structures are ionic interactions, hydrogen bonds, van der Waals interactions, and hydrophobic interactions.

7. The answers to (a)–(c) are shown on the tetrapeptide structure below.

For (d), based on the pKa values for the three ionizable residues (Arg, ∼12.5; Tyr, ∼10.9; His, ∼6.0) and the N-terminal (∼8.0) and C-terminal (∼3.1) ionizable groups, a pH of 11–12 would result in the charges shown.

Arg 1 NH2 +

H2N

His 3

Val 4

N

C N

H

Tyr 2

NH

Coplanar atoms

H

(CH2)3

H

H

O

N

C

C

N

C

C

H

H

O

CH2

Peptide bond

CH2 N

C

C

H

H

O

H

H

O

N

C

C

CH H3C

Peptide bond

O–

CH3

O–

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CHAPTER 4 LONG ANSW ERS TO CHALLENGE PROBLEMS

8. In an unfolded protein, the amide and carbonyl groups of the backbone can

form hydrogen bonds with water. When polar backbone groups are buried inside the protein, they “give up” their hydrogen bonds to H2O. The enthalpy change (ΔH ) for this process would be unfavorable (breaking bonds) if new hydrogen bonds were not made within the protein. In both α helices and β sheets, hydrogen bonds are satisfied for these groups, and therefore these secondary structures are enthalpically favored.

9. The polypeptide strands are antiparallel and together form an antiparallel

β sheet stabilized by two hydrogen bonds, as shown.

N

C H

R

O

N

C

C

H

H

H

N

C

C

H

R

O

R

N

C

O C

H

Hydrogen bond

Hydrogen bond R

C

C

N

O

H

H

O

H

H

C

C

N

R

R C

C

N

O H

H N

C

10. The zwitterion of Glu is represented by c. The letter(s) corresponding to

the ionic forms of Glu that occur at each pH value are written above the arrows. a a+c

d

d+b

b

10 8

pH

6

4

2

1

2

3

Equivalents OH–

11. Besides the combination of cysteine pairs Cys1–Cys3 and Cys2–Cys4, the two

other combinations of cysteine pairs that would result in two disulfide bonds are Cys1–Cys2 and Cys3–Cys4, and Cys1–Cys4 and Cys2–Cys3. However, the only combination of two cysteine pairs that can explain the observation that two oligopeptides are generated by trypsin cleavage under nonreducing conditions is Cys1–Cys2 and Cys3–Cys4, as the other two combinations would result in a polypeptide that was held together by disulfide bonds.

12. Urea is a denaturing agent that unfolds proteins by disrupting noncovalent

interactions, whereas β-mercaptoethanol is a reducing agent that breaks

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disulfide bonds. In the presence of sufficient amounts of both urea and β-mercaptoethanol, the ribonuclease protein was completely unfolded. The most important conclusion of Anfinsen’s experiment was that the primary structure of a protein has all the information necessary to specify the tertiary structure.

13. At pH 1, polylysine should be unfolded because of the charge repulsion from

the numerous positively charged α amino groups (pKa of the α amino group of Lys is ∼10.8). At pH 7, the structure of polylysine should still be mostly unfolded, because even though the pKa of the α amino groups would be lower than the normal ∼10.8 as the polypeptide begins to form secondary structures, it would not be as low as pH 7, and therefore still mostly protonated. Last, at a pH of 11, which is above the pKa of ∼10.8 for the α amino group, slightly more than half of the α amino groups would be deprotonated and uncharged, which would facilitate α-helical formation.

14. a. Sequence 3 could include a β turn because it is the only one that contains

a Gly-Pro sequence. b. Sequence 2 could form a β strand, with one surface facing the interior of the protein and the other surface exposed to water, because it is the only one that contains alternating hydrophobic and hydrophilic side chains; that is, Ile, Met, Leu, Val, Leu, and Phe alternate with Asp, Glu, Lys, Asn, Asp, and Arg. c. Sequence 1 could form an α helix that would participate in a coiled coil structure within a protein because every third or fourth residue is Leu (heptad repeat motif ). d. Sequence 4 could form an amphipathic α helix with Phe, Val, Leu, Ile, Phe, Met, and Ala on the hydrophobic face and Asn, Ser, Gln, Asp, Glu, Gln, Ser, and Cys on the hydrophilic face.

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Long Answers to Review Questions Chapter 5 1. There are ∼20,000 protein-coding genes in the human genome. However, differences in RNA processing and protein modifications lead to many more proteins than ∼20,000.

2. Two properties that affect selection of a protein purification strategy are (1) the physical properties of the protein, such as the total number of amino acids and chemical properties of the amino acid side chains, and (2) the relative abundance of the protein in a specific cell type.

3. One method is called sonication. Sonication disrupts the cell membrane through the vibrational forces of ultrasonic waves.

A second method is called shearing, which uses a tight-fitting Teflon plunger in a glass vessel, a syringe, or a mechanical device called a French press to force cells through a small opening.

A third method is to incubate the cell sample with mild detergents that disrupt cell membranes.

4. Specific activity is the total amount of activity of a target protein divided by the total amount of protein in the fraction.

5. The three types of column chromatography are gel filtration chromatography, ion-exchange chromatography, and affinity chromatography. Gel filtration chromatography involves the use of porous carbohydrate beads to separate proteins on the basis of their size. The carbohydrate beads contain pores that allow small proteins to enter inside the beads. Larger proteins flow around the beads. The net result is that larger proteins flow through the column faster than smaller proteins.

Ion-exchange chromatography is a purification method that exploits the charge differences between proteins. Different proteins, with different charges, bind to the column matrix based on the charge of the matrix and the pH of the buffer. Negatively charged proteins bind to anion-exchange resins, whereas a cationexchange resin binds positively charged proteins. Once the protein 1

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CHAPTER 5 LONG ANSW ERS TO REVIEW QUESTIONS

sample has been loaded onto the column, oppositely charged proteins and resin interact in such a manner as to retain the target protein in the column.

The matrix is then extensively washed with loading buffer to remove unbound protein. To remove the bound protein, a high concentration of an appropriate competing ion is added to the column to displace the bound protein. The eluted fractions are assayed, pooled, and dialyzed.

Affinity chromatography exploits specific binding properties of the target protein to separate it from other cellular proteins. A high-affinity ligand for the target protein is covalently linked to a matrix bead. Then the protein sample is passed through the column under optimal binding conditions for the protein and ligand. The target protein specifically associates with the column matrix, whereas proteins with no affinity for the ligand do not.

Copious amounts of buffer containing competing ligand are added to the column in order to release the target protein from the column.

6. Polyacrylamide gel electrophoresis (PAGE) separates proteins on the basis of their mass and charge. When sodium dodecyl sulfate (SDS) is present, it interacts with proteins through noncovalent interactions, leading to a uniform negative charge across the protein. The magnitude of the negative charge is based on the length of the polypeptide chain, and thus the mass, of the protein. Samples of the protein–SDS complex are added to a gel in a vertical box, which contains buffer in the top and bottom chambers. This gel matrix is sandwiched between two pieces of glass. Current then flows between the two chambers, passing through the gel matrix. This results in the migration of the negatively charged proteins toward the anode side of the chamber. Smaller proteins migrate faster through the buffer-saturated polyacrylamide gel matrix than do larger proteins.

7. After each round of cleavage, a completely new sample of protein had to be added because the acid hydrolysis destroyed the sample.

8. First, the N-terminal amino acid is tagged using phenylisothiocyanate (PITC). The peptide is then treated with trifluoroacetic acid, which cleaves the peptide bond between the first and second amino acid, releasing a thiazolinone derivative and the original polypeptide minus one amino acid. The thiazolinone derivative is then isolated and converted to a phenylthiohydantoin (PTH) derivative. This PTH derivative is identified by chromatography against known PTH-amino acid standards.

9. Mass spectrometry measures the mass-to-charge ratio of molecules. The basic operating principle is that the mass of a charged molecule in an applied magnetic field, which exerts a force on the molecule, directly affects its acceleration. The relationship is described by Newton’s second law, F = ma.

Therefore, we can calculate the mass of the molecule by measuring its acceleration along a curved path and knowing the applied magnetic force.

10. First, the C-terminal amino acid is covalently attached to a resin molecule. This serves as the first amino acid in the polypeptide chain. The N-terminal blocking group of this first amino acid, abbreviated as Fmoc, is removed through treatment with a base. This allows it to react with an Fmoc-blocked amino acid that has been activated at the carboxyl group by dicyclohexylcarbodiimide (DCC). The side chains of the amino acids are

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chemically protected to prevent reactions with them. Therefore, the amino terminus of the resin-bound amino acid is the only group available to react with the incoming DCC-activated amino acid, and a peptide bond is formed.

11. X-ray crystallography is not as limited by constraints to protein size as is NMR spectroscopy, which is limited to proteins of less than ∼40 kDa. 12. An X-ray beam is aimed at a homogeneous protein crystal, and when it interacts with the electron-dense regions of the protein, the X-rays are diffracted and scatter in different directions. The intensities and directions of these diffracted X-rays are recorded by X-ray detectors. From these data, computational tools are used to calculate a map of the electron-dense regions, from which a model of the protein is then built.

13. Often, the two most difficult steps involved in performing X-ray crystallography are growing diffraction-quality crystals and determining the phase of the diffracted X-rays.

14. NMR spectroscopy takes advantage of the intrinsic magnetic properties of several types of atoms, most often 1H, 15N, and 13C, to derive the relative locations of atoms from highly concentrated solutions of purified proteins.

NMR instruments contain large magnetic fields that align the nuclear spins from NMR-active nuclei. Short radio-frequency pulses are used to perturb these nuclear spins, and information about their perturbations and return to the ground state is collected in order to interpret the environment of each nucleus.

15. Antibodies are large, multi-subunit proteins that bind antigen molecules, such as other proteins, with high specificity and affinity.

16. The two types of antibodies used in biochemical research are polyclonal and monoclonal antibodies. Polyclonal antibodies are a heterogeneous mixture of antibodies that recognize one or more epitopes on an antigenic protein.

Monoclonal antibodies are homogeneous and recognize only a single epitope on the antigenic protein.

17. Epitope tags are highly antigenic amino acid sequences created through the use of recombinant DNA technology. They are useful for analysis of geneencoded proteins.

18. A primary monoclonal antibody, which recognizes a single epitope on the antigenic protein, is covalently attached to the bottom of a 96-well microtiter plate. This antibody is referred to as the capture antibody. Dilutions of an aqueous sample are then added to each well and incubated for several hours to promote antigen–antibody binding. After washing away unbound proteins from the sample, a second primary monoclonal antibody, called the detection antibody, is added to the well. It is important that the detection antibody recognizes a distinct epitope on the antigenic protein in order for the molecular sandwich to form. Finally, an antibody that recognizes the detection antibody as an antigen, and which is linked to an enzyme such as horseradish peroxidase, is added to the wells, along with a chromogenic or fluorogenic substrate for the horseradish peroxidase enzyme assay. Samples that contain the antigenic protein are positive for the horseradish peroxidase reaction product, which can be detected using a spectrophotometric 96-well plate reader.

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Long Answers to Challenge Problems Chapter 5 1. a. The specific activity after the first step is 2 × 104 units/mg protein, and after the fourth step it is 2 × 108 units/mg protein. b. The total protein yield after the second purification step is 1 × 105 mg, and after the third purification step it is 1 × 103 mg. c. Percentage yield = 4 × 109 units/1 × 1010 units = 40% yield. d. The molecular mass of the purified protein based on SDSPAGE appears to be ∼25,000 kDa, indicating that the native protein is a homotetramer of four subunits totaling 100,000 kDa. 2. a. Protein C (smallest-mass native protein). b. Protein E (largest-mass denatured protein). c. Protein D (most negatively charged at pH 6.5, so highest affinity to DEAE). d. Protein D is most likely a homotrimer consisting of three subunits of 25,000 Da each.

3. Although both methods lead to a precipitated protein, the enzyme precipitated by heat denaturation will likely have very little, if any, activity upon resolubilization, whereas the protein precipitated by ammonium sulfate will likely retain most of its activity upon resolubilization. Because the heat-denatured protein will have lost most of its enzyme activity, it will be impossible to purify further because the activity assay will not work.

4. Size-exclusion column: The 80-kDa protein migrates faster than the 20-kDa protein because the 80-kDa protein has less interaction with the resin. SDS-PAGE: The 20-kDa protein migrates faster because it has a smaller frictional coefficient than that of the 80-kDa protein.

5. On the basis of the pI values given, at pH 7 the cytochrome c protein is positively charged, and the cytochrome c peroxidase protein is negatively charged. DEAE is an anion-exchange resin 4

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(it is positively charged). Because the resin and cytochrome c both have a positive charge, the protein cannot be directly bound to the resin; therefore, it must be bound to cytochrome c peroxidase, which is in turn bound to the resin. Increasing the salt concentration of the buffer weakens the interaction of cytochrome c with cytochrome c peroxidase, causing cytochrome c to elute first.

6. Run the protein mixture through a gel filtration column using a buffer at pH 6, which is a pH greater than the pI of protein A and of protein B, but below the pI of protein C. Also be sure the buffer has a low ionic strength (10 mM NaCl). Under these conditions, protein B will elute more rapidly from the column than protein A or C, which in all likelihood will co-elute together.

Because you have chosen a pH for the buffer midway between the pI of A and the the pI of C, protein A will have a net negative charge, whereas C will have a net positive charge. These two proteins can be loaded directly onto a DEAE column (positively charged), and you would expect to see protein A bind tightly to the column, while C should elute off in the flow-through or wash fractions. Protein A could then be eluted off either by increasing ionic strength or by increasing the pH of the buffer.

7. Trypsin cleaves on the carboxyl-terminal side of lysine or arginine residues.

Because the 1st or 14th amino acid must be Q (it cannot be the 7th or 8th amino acid based on the substrate specificity of trypsin), the 6th or 7th amino acid and the 13th or 14th amino acid must be K or R. However, because chymotrypsin cleaves on the carboxyl side of tyrosine, tryptophan, and phenylalanine residues and generated the tripeptide G-I-F, and cyanogen bromide cleaves on the carboxyl side of methionine and generated the tetrapeptide G-I-F-M, the N-terminal tetrapeptide sequence must be G1-F2-I3-M4, and the C-terminal amino acid must be Q14. Using the data collected from the V-8 protease cleavage analysis (V-8 cleaves on the carboxyl side of glutamate and aspartate), the C-terminal tripeptide must be P12-R13-Q14, which means the sixth amino acid must be K6, as only G is represented twice in the tetradecapeptide. In addition, the tetrapeptide produced by V-8 cleavage must be adjacent to this C-terminal tripeptide and correspond to the amino acid residues G8-Y9-N10-D11. Finally, the heptapeptide generated by V-8 protease must correspond to the first seven amino acids and have the sequence G1-F2-I3-M4-X5-K6-E7, in which X5 must correspond to cysteine (C5) as this is the only amino acid not yet accounted for based on the initial amino acid composition analysis. The illustration below summarizes these findings.

Chymotrypsin

Trypsin

Cyanogen bromide

Chymotrypsin

Trypsin V-8

V-8

G1 - I2 - F3 - M4 - C5 - K6 - E7 - G8 - Y9 - N10 - D11 - P12 - R13 - Q14 G -

I - F

G -

I - F - M

G - Y - N - D

P - R - Q

8. a. Structural heterogeneity of the protein can disrupt the formation of crystals, which are formed by regular repeating units.

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b. The main advantage of using orthologous proteins is that amino acid differences between closely related proteins can result in subtle changes in the chemical and physical properties of a protein, which may now form diffractable crystals and lead to a high-resolution molecular structure. A disadvantage is that the two proteins may have significant structure–function differences that limit the data interpretation, such as using similar structures to encode distinct functions or using distinct structures to encode similar functions (see Figure 1.30). c. Minimal changes in amino acid sequence, which do not alter protein function, can sometimes provide the necessary chemical and physical differences that permit protein crystallization. For example, a small number of amino acid changes that alter hydrophobic regions or ionic interactions near the surface of the protein can sometimes be enough to facilitate crystallization. These “mutations” are performed using recombinant DNA techniques (see Chapter 3). d. An alternative method to analyze protein structure is nuclear magnetic resonance (NMR) spectroscopy, which identifies the relative locations of atoms in a concentrated solution of a purified protein and thereby provides spatial information that can be used to decipher a set of three-dimensional structures. The main advantage of NMR over X-ray crystallography is that data collection uses soluble protein in its native form; however, a disadvantage is that there is an upper size limit to the proteins that can be studied.

9. A polyclonal antibody is a collection of immunoglobulin proteins that recognize different epitopes on the same antigen with various specificities and affinities. In contrast, a monoclonal antibody consists of a single immunoglobulin protein species that recognizes only one epitope on the antigen. Monoclonal antibodies, but not polyclonal antibodies, are suitable for human diagnostic and clinical applications for two main reasons: (1) the antibody–antigen interaction is highly specific and subject to quality control, and (2) monoclonal antibodies are secreted by immortalized cell lines and can be produced indefinitely.

10. SDS-PAGE analysis physically separates denatured proteins on the basis of molecular mass and is more likely to expose the antigen to the antibody during the Western blotting procedure. In contrast, immunofluorescent staining traps proteins in the cell under conditions that could mask the antigen site as a result of intramolecular or intermolecular interactions.

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Long Answers to Review Questions Chapter 6 1. The five major functional classes of proteins are metabolic enzymes, structural proteins, transport proteins, cell signaling proteins, and genomic caretaker proteins.

2. Enzymes increase the rate of a reaction by lowering the activation energy of that reaction. Enzymes do not change the equilibrium constant of the reaction.

3. The two most abundant cytoskeletal proteins are actin and tubulin. Actin polymers are called thin filaments, and tubulin polymers are called microtubules.

4. Passive transporter proteins are energy-independent transporters that function as membrane pores in response to chemical gradients. Active transporter proteins require energy from ATP hydrolysis or an ionic gradient to induce protein conformational changes that open and close a gated channel.

5. Genomic caretaker proteins ensure that the integrity of genomic DNA is maintained throughout the life of the cell and that the expression of specific genes is tightly controlled and reflects the biochemical needs of the organism.

6. The two oxygen-binding globin proteins in animals are myoglobin and hemoglobin. Hemoglobin transports heme-bound O2 from the lungs to tissues through the circulatory system. Myoglobin is concentrated in muscle tissue and functions as a storage depot for O2.

7. The affinity of carbon monoxide for the Fe2+ in hemoglobin is 200 times higher than that of oxygen.

8. The oxygen-binding curve for myoglobin is hyperbolic, while the oxygen-binding curve for hemoglobin is sigmoidal. The sigmoidal binding curve for hemoglobin is indicative of cooperative binding. This means that binding of the first molecular oxygen to the protein complex facilitates the binding of additional molecular oxygen.

1

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9. The three features of ligand–protein interactions are: (1) ligand binding is a reversible process involving weak noncovalent interactions; (2) ligand binding induces or stabilizes structural conformations in target proteins; (3) the equilibrium between ligand-bound protein and ligand-free protein can be altered by the binding of effector molecules, which induce conformational changes in the protein that increase or decrease ligand affinity.

10. Fetal hemoglobin consists of a tetramer complex in which the normal adult β subunit is replaced by a special fetal hemoglobin subunit called γ. In this subunit, His143 is replaced by a serine, which eliminates two of the six positive charges in the 2,3-BPG binding site, leading to a lower affinity for 2,3BPG. This decreased affinity facilitates transfer of oxygen from the mother’s hemoglobin, which is mostly in the T state, to the fetal hemoglobin, which is mostly in the R state.

11. Glutamic acid is changed to valine at position 6 of the β-globin polypeptide (β-E6V), which causes the recessive genetic disease sickle cell anemia. The hemoglobin tetramer containing the sickle cell β polypeptide is given the name HbS. The presence of HbS in red blood cells is pathologic under low O2 concentration, primarily in the tissues such as the kidney and the spleen where erythrocytes circulate more slowly and are often destroyed. It has been hypothesized that Plasmodium infection of human red blood cells leads to a reduction in pH, which induces aggregation of HbS tetramers and cell sickling. Because these mis-shapen cells are removed by the spleen, the Plasmodium-infected cells will be preferentially depleted from circulation, thereby protecting heterozygous HbS individuals from severe malaria.

12. Membrane receptor proteins, the first type of membrane proteins, are involved in transducing extracellular signals across the plasma membrane through ligand-induced conformational changes. The second type, membrane-bound metabolic enzymes, includes examples such as the inner mitochondrial and chloroplast thylakoid membrane enzymes involved in oxidation–reduction reactions and ATP synthesis. The third type, membrane transport proteins, facilitate the transfer of polar molecules across the hydrophobic membrane. Transporter proteins can be active transporters, requiring energy to transport molecules against their concentration gradient, or passive transporters, which do not require energy and transport molecules according to their concentration gradient.

13. The two types of active transporter proteins are primary active transporters and secondary active transporters. Primary active transporters depend directly on ATP hydrolysis to drive conformational changes that allow the protein to pump molecules across the membrane. Secondary active transporters, which are either symporters or antiporters, use energy available from a downhill concentration gradient of one molecule to transport a different molecule against its concentration gradient. This co-transport mechanism is usually coupled to an ATP-dependent primary active transport mechanism that establishes a concentration gradient.

14. Zoloft targets the serotonin transporter protein, which is a secondary active symporter of serotonin that couples the sodium ion transport with the reuptake of serotonin from the synaptic cleft into presynaptic neurons.

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Inhibition of the serotonin transporter results in elevated levels of serotonin in neuronal synapses and therefore elevated levels of serotonin receptor activation. This is associated with feelings of well-being and happiness.

15. In the presence of Ca2+, the troponin complex undergoes a conformational change that moves the TnI and TnC subunits and a large portion of the tropomyosin complex away from the myosin binding site on either side of the actin polymer duplex. This allows the myosin head domain to bind tightly to the actin filament in preparation for contraction.

16. The five steps of muscle contraction are as follows: (1) Ca2+ induces conformational changes in actin thin filaments that facilitates the binding of myosin heads containing ADP + Pi to actin subunits; (2) release of Pi from the myosin head causes a large conformational change in myosin (power stroke) that pulls the actin fiber along the myosin fiber; (3) ADP is released from the myosin head to free up the nucleotide binding site; (4) ATP binds to the myosin head causing myosin to disengage from the actin filament; (5) ATP hydrolysis induces the recovery conformation in the myosin head, and the ADP + Pi form of the protein is ready for another round of the reaction cycle.

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Long Answers to Challenge Problems Chapter 6 1. The three heterotropic effectors of hemoglobin are 2,3-bisphosphoglycerate (2,3-BPG), H+, and CO ,2 all of which are negative heterotropic effectors of O2 binding affinity for hemoglobin. O2 is considered a positive homotropic effector for hemoglobin because the binding of one O2 molecule increases the affinity for O2 of other sites on the same hemoglobin molecule.

2. The 2,3-BPG-depleted red blood cells contain Hb that binds O2 too strongly and would not deliver O2 to the tissues.

3. a. Inositol hexaphosphate, a negatively charged compound, is the most likely 2,3-BPG analog in birds.

CH2OH O

HO

OH

OH HO

OH Glucose

3

CH3 Choline

Indole

2–

+ 2– 2–

O–

2–

Inositol hexaphosphate

Histidine

b. It should bind in the central cavity between the four globin subunits and form ionic interactions with positively charged residues such as Lys and His. c. The 2,3-BPG analog should decrease O2 affinity by shifting the equilibrium to the T-state conformation.

4

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4. The four graphs below show the O2 binding curves of the various Hb proteins. b.

1.0 0.8

2,3-BPG

0.6

Hb reference pH 7.4

0.4 0.2 0.0 0

4

8

Fractional saturation (θ)

Fractional saturation (θ)

a.

1.0 0.8

Hb reference pH 7.2

0.6 0.4

Hb reference pH 7.4

0.2 0.0

12

0

pO2 (kPa)

8

12

pO2 (kPa)

1.0 0.8

Monomeric mutant

0.6 0.4

Hb reference pH 7.4

0.2 0.0 0

4

8

12

Fractional saturation (θ)

d.

c.

Fractional saturation (θ)

4

1.0 0.8

Fetal Hb

0.6

Hb reference pH 7.4

0.4 0.2 0.0 0

pO2 (kPa)

4

8

12

pO2 (kPa)

a. Hemoglobin with all of the 2,3-BPG removed from the O2 binding assay.

In absence of 2,3-BPG, the R-state–T-state equilibrium shifts toward the R state at lower concentrations of O2, and the curve is slightly more hyperbolic (less sigmoidal) because of reduced cooperativity of O2 binding. The P50 will be lower as the curve is shifted to the left. b. Hemoglobin at pH 7.2 rather than pH 7.4.

Higher H+ concentration (lower pH) shifts the R-state–T-state equilibrium toward the T state, so there is a higher P50 as the curve is shifted to the right. c. Hemoglobin mutant in which the predominant form of the protein is monomeric, with a very low binding affinity for O2.

Monomeric Hb would show no cooperativity, and the plot should be purely hyperbolic with a higher P50 as the curve is shifted to the right. d. Fetal hemoglobin in which His143 in the γ subunit is replaced by Ser143.

Fetal hemoglobin has one less positive charge on each γ subunit in the central cavity, so 2,3-BPG still binds, but less tightly, and more of the hemoglobin is in the R state. The curve is mostly sigmoidal and shifted to the left with a lower P50.

5. The α and β subunits of hemoglobin should have more hydrophobic amino acids than myoglobin because hydrophobic amino acids are needed for subunit interactions.

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CHAPTER 6 LONG ANSW ERS TO CHALLENGE PROBLEMS

6. This mutant hemoglobin would have a lower O2 binding affinity because the R-state–T-state equilibrium will be mostly shifted toward the T state.

7. HbF is a tetramer that contains two α chains and two γ chains (α2γ2), whereas HbA has two α chains and two β chains (α2β2). The γ chains in HbF, which substitute for the HbA β chains, contain a Ser143 residue in place of the His143 residue found in HbA β. With two fewer positively charged groups on HbF (the γ chains) to bind 2,3-BPG, the 2,3-BPG binds more weakly to HbF than to HbA, with the result that HbF has a higher O2 affinity than that of maternal HbA. At comparable 2,3-BPG concentrations in maternal and fetal blood, the HbF will bind O2 more tightly than HbA, and O2 can be released from maternal Hb and bound by fetal Hb in the placenta. Thus, maternal HbA can “deliver” O2 to the HbF through the placenta.

8. The most likely protein sequence is (d). Porin structure is a large antiparallel β barrel with an aqueous channel through its middle. The β strands must have alternating hydrophobic and hydrophilic side chains in order for the membrane side of the β barrel to be hydrophobic and the side of the β barrel facing the aqueous channel to be hydrophilic. Only answer (d) has alternating hydrophobic and hydrophilic residues.

9. The answer is 33.4 ms, as shown below: 2 × 105 monomers = 5 × 104 tetramers 4 monomers/tetramer (5 × 104 tetramers) (5 × 108 molecules/tetramer ⋅ s) = 2.5 × 1013 molecules/s Half volume = 2.5 × 10–11 mL = 2.5 × 10–14 L If the cell is all water, then (55.5 mol/L) (2.5 × 10–14 L) (6.02 × 1023 molecules/s) = 8.35 × 1011 molecules 8.35 × 101311 molecules = 3.34 × 10−2 s = 33.4 ms 2.5 × 10 molecules/s 10. The answer is ΔG = –10.1 kJ/mol as shown below: C2 ΔG = RT lna b + ZFΔV C1 Because glucose is an uncharged polar compound, C2 ΔG = RT lna b + 0 C1 C2 [glucose]inside = RT ln a b = RT lna b [glucose] C1 ouside 5 × 10−4 M –3 = (8.314 × 10 kJ/mol K) (310 K) lna b 1 × 10−3 M = –10.1 kJ/mol

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7

11. a. The answer is ΔG = +3.66 kJ as shown below: The Na+–K+ ATPase protein transports K+ ions into the cell, therefore C2 ΔG = RT lna b + ZFΔV C1 14 × 10−2 Mb = (8.314 × 10–3 kJ/mol K) (310 K) lna −3 5 × 10 M + [(+1) × 96.5 kJ/V mol × –0.070 V] (Note that the membrane potential was defined as negative inside relative to the outside. The membrane potential here is –0.070 V, thus making the electrical potential term, ZFΔV, negative. This negative value reflects the fact that it is favorable electrically to transport a positive ion into a negative environment.) ΔG = 8.59 kJ/mol – 6.76 kJ/mol = 1.83 kJ/mol To transport 2 mol of K+, ΔG = 1.83 kJ/mol × 2 mol = +3.66 kJ. b. The answer is ΔG = +41.22 kJ as shown below: The Na+–K+ ATPase protein transports Na+ ions out of the cell, therefore C2 ΔG = RT lna b + ZFΔV C1 15 × 10−2 Mb = (8.314 × 10–3 kJ/mol K) (310 K) lna −2 1 × 10 M + [(+1) × 96.5 kJ/V mol × 0.070 V] (Note that the membrane potential was defined as negative inside relative to the outside. The membrane potential here is 0.070 V, thus making the electrical potential term, ZFΔV, positive. This positive value reflects the fact that it is unfavorable electrically to transport a positive ion into a positive environment.) ΔG = 6.98 kJ/mol + 6.76 kJ/mol = 13.74 kJ/mol For 3 mol of Na+, ΔG = 13.74 kJ/mol × 3 mol = +41.22 kJ. c. The Na+–K+ ATPase transporter protein exchanges 2 mol of K+ imported into the cell for every 3 mol of Na+ exported out of the cell, and therefore hydrolysis of a single ATP is sufficient; that is, 50 kJ is greater than either 3.66 kJ for K+ transport or 41.22 kJ for Na+ transport.

12. The answer is a ratio of 47.9:1 as shown below: C2 ΔG = RT lna b + ZFΔV C1 Because lactose is an uncharged polar compound, ΔG = RT lna

C2

b+0

C1 = −10 kJ/mol = RT lna

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C2

[lactose]inside b = RT lna b C1 [lactose]outside

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CHAPTER 6 LONG ANSW ERS TO CHALLENGE PROBLEMS

= –10 kJ/mol = (8.314 × 10–3 kJ/mol K) (310 K) lna = –10 kJ/mol = 2.58 kJ/mol × lna

[lactose]inside

[lactose]inside

b [lactose]outside

b

[lactose]outside 10 kJ/mol = 3.87 2.58 kJ/mol a

[lactose]inside [lactose]outside

b = e3.87 = 47.9:1 maximum concentration ratio

13. Actin and ATP must bind to different conformational states of the myosin molecule, so actin binding stabilizes a conformational state with a very low affinity for ATP, whereas ATP binding stabilizes a conformational state with a very low affinity for actin.

14. Ca2+ binds to troponin C (TnC), which changes the conformation of the troponin complex that is bound to tropomyosin in such a way that the TnC and TnI subunits rotate away from the myosin binding site on actin. This Ca2+-induced structural change in the troponin complex permits myosin to bind actin and initiate the contraction cycle.

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Long Answers to Review Questions Chapter 7 1. The three structure–function relationships that make enzymes efficient biological catalysts are (1) substrates usually bind with high affinity and specificity to the enzyme active sites; (2) substrate binding to the enzyme active site often involves structural changes, which maximize the number of weak interactions between the substrate and the enzyme; and (3) enzyme activity is highly regulated in order to maximize energy balance between catabolic and anabolic pathways and to alter cell behavior in response to environmental stimuli.

2. Enzymes are regulated by increases or decreases in enzyme levels (bioavailability) and by changes in catalytic efficiency. Enzyme bioavailability is regulated at the level of gene expression and protein synthesis, whereas catalytic efficiency is regulated by posttranslational modifications (for example, phosphorylation) and by binding of allosteric regulators such as small metabolites and protein subunits.

3. Enzymes increase reaction rates by lowering the activation energy of a reaction. They do this by providing favorable reaction conditions that have the effect of lowering the transition energy barrier, thereby increasing the rate of the reaction without altering overall ΔG. 4. Enzyme cofactors are small molecules that aid in catalytic reaction mechanisms by providing additional chemical groups to supplement the chemistry of amino acid functional groups.

Enzyme cofactors include a variety of inorganic ions such as Fe2+, Mg2+, Mn2+, Cu2+, and Zn2+, as well as coenzymes, which are enzyme cofactors containing organic components.

Coenzyme molecules can be loosely associated with the enzymes or very tightly bound, even covalently attached to the enzyme.

Coenzymes provide additional chemical flexibility to the catalytic reaction, which the protein component of the enzyme may be unequipped to do. Coenzymes that are permanently associated 1

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with enzymes, such as the heme group of catalase, are called prosthetic groups. Most vitamins that people obtain from nutritional sources are coenzymes.

5. Enzymes lower the activation energy (ΔG ‡) of a reaction in three ways: (1) by providing an alternative path to product formation, which could involve forming stable reaction intermediates that are covalently attached to the enzyme; (2) by lowering the activation barrier through stabilization of the transition state; and (3) by orienting the substrates appropriately for the reaction to occur.

6. The catalytic triad is a set of three amino acids that form a hydrogen-bonded network required for catalysis. In chymotrypsin, these three amino acids are Ser195, His57, and Asp102.

7. In the first phase of the chymotrypsin reaction, a covalent acyl-enzyme intermediate is formed between Ser195 and the polypeptide substrate, which promotes cleavage of the scissile peptide bond to release the carboxyl-terminal polypeptide fragment. In the second phase, the enzyme is regenerated after a series of steps resulting in deacylation and release of the amino-terminal polypeptide fragment.

8. There are eight amino acids with ionizable side chains that can gain or lose a proton, depending on the local environment and effective pKa (see Figure 7.17). Changes in pH alter the chemistry of the active site by protonation and deprotonation of ionizable side chains, which can disrupt critical tertiary structures within the protein, altering enzyme activity.

9. The three classes of reversible inhibitors are competitive inhibitors, uncompetitive inhibitors, and mixed inhibitors. A competitive inhibitor is defined as a molecule that competes with substrate for binding to the active site (see Figure 7.45a). Because the inhibitor competes for binding to the active site, the apparent Km value for the enzyme is shifted to the right as a function of increasing inhibitor concentration. The vmax of the reaction, however, is not affected by a competitive inhibitor. Uncompetitive inhibitors bind to enzyme–substrate complexes and alter the active site conformation, thus rendering the enzyme less catalytically active (see Figure 7.45b). The net effect of uncompetitive inhibition is a decrease in both the Km and vmax, without altering the Km/vmax ratio. Mixed inhibitors are similar to uncompetitive inhibitors in that they bind to sites distinct from the active site. The main difference is that mixed inhibitors can bind to both the enzyme and the enzyme–substrate complex (see Figure 7.45c). A mixed inhibitor decreases vmax and may increase or decrease Km, depending on the relative KI and KI′ values.

10. Feedback inhibition occurs when the end product of a pathway functions as an inhibitor of the first enzyme in the pathway.

11. By far, the most common of these covalent modifications in enzymes is the phosphorylation of Ser, Thr, and Tyr residues by kinase enzymes, which adds negative charge to the enzyme through the addition of a phosphoryl group (PO32−).

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Long Answers to Challenge Problems Chapter 7 1. The substrate binding site and the proper orientation of the catalytic residues are the result of the tertiary structure of the enzyme. The bulk of the protein is necessary to produce the correct tertiary structure.

2. You would modify the enzyme so that the transition state bound more tightly, which would lower the activation energy of the reaction. If the substrate were to bind more tightly, the activation energy would be higher.

3. An enzyme cannot use an induced-fit mechanism to achieve catalytic perfection because some of the energy that could be used to bind the transition state and lower the activation energy is used to drive the conformational change in the protein upon substrate binding.

4. a. The rate enhancement is 5 × 108-fold enhancement: kF catalyzed Rate enhancement = kF uncatalyzed =

5 × 106 s−1 1 × 10−2 s−1

= 5 × 108-fold enhancement b. Catalysts do not change the Keq of a reaction, so the Keq in the absence or presence of enzyme is the same; in this case, 1 × 103.

5. The observation can be explained by the inhibitor losing its activity as a result of deprotonation at pH 8.5 relative to pH 6.5, on a group that is found only in the inhibitor, not in the substrate.

6. a. no effect b. decrease c. increase d. no effect e. decrease 3

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CHAPTER 7 LONG ANSW ERS TO CHALLENGE PROBLEMS

7. Because the transition state energy is unchanged, significant stabilization of ES decreases the rate of product formation because the activation energy increases.

8. a. What is Keq for the reaction? kF1uncat2 10−5 s−1 = −2 −1 = 10−3 1 unitless 2 Keq = kR1uncat2 10 s b. What is kR(cat) with enzyme catalyst? kF1cat2 10 −1 107 s−1 = kR1cat2 = −3 = 10 s Keq 10 c. What is the kcat value for this enzyme if the vmax value is 5 × 103 M s–1 at an enzyme concentration of 2 × 10–6 M? v max 5 × 103 M s−1 9 −1 = 2.5 × 10 s kcat = = −6 [Et ] 2 × 10 M 9. a. The imidazole of the active site His residue. b. The pKa is the pH at which [base] = [acid], so protein would be 50% active at pH 6.8.

100

Percent activity (%)

80 60 50 40 20

0 5.0

5.5

6.0

6.5

7.0

7.5

8.0

8.5

9.0

c. In the first catalytic step of the chymotrypsin mechanism, the active site His residue has to act as a general base by accepting a proton from the active site Ser. Therefore, the His residue has to start out in its unprotonated form to be a conjugate base.

10. a.

c.

d.

O– R

C

O NH

Enz

R

C

O

Enz

OH Enz

b.

First intermediate

O R

C

NH

e.

Enz Second intermediate

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CHAPTER 7 LONG ANSW ERS TO CHALLENGE PROBLEMS

5

11. a. Calculate Km for this substrate: −1 = −8.0 mM−1, therefore K m = 0.125 mM Km b. Calculate vmax for this substrate: 1 = 4 (mM min–1)–1, therefore v max = 0.25 mM min–1 vmax c. Calculate kcat for papain with this substrate: [E t] = concentration of total enzyme active sites = 1.2 × 10–7 M = 1.2 × 10–4 mM vmax 1 0.25 mM min− kcat = = [Et ] 1.2 × 10−4 mM 1 min 3 –1 = 34.7 s–1 kcat = 2.08 × 10 min × 60 s d. The fraction of total enzyme that has substrate bound (θ) is v0 [ES] [S] 2.5 × 10−4 M θ= = v = = [E ] max 11.25 × 10−4 M + 2.5 × 10−4 M 2 1 K + [S] 2 t

θ=

[ES] [Et ]

m

=

2.5 = 0.67 3.75

12. a. Yes. Because Y behaves as a competitive inhibitor of X, the apparent Km for X will be increased. b. No. The vmax for X will be unaffected. c. Yes. The pH dependence of vmax will reflect the ionization of residues involved in catalysis, whereas the pH dependence of vmax/Km will reflect the ionization state of residues involved in substrate binding.

13. a. In the absence of inhibitor: vmax = 0.1 μmol/min, using 1/vmax = 10 (μmol/min)–1 b. In the presence of inhibitor: vmax = 0.04 μmol/min, using 1/vmax = 25 (μmol/min)–1 c. In the absence of inhibitor: Km = 0.25 mM, using –1/Km = –4 (mM)–1 (± inhibitor) d. In the presence of inhibitor: Km = 0.25 mM, using –1/Km = –4 (mM)–1 (± inhibitor) e. This is an example of: mixed (noncompetitive) inhibition because vmax decreased in the presence of inhibitor, but Km stayed the same in the presence and absence of the inhibitor.

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CHAPTER 7 LONG ANSW ERS TO CHALLENGE PROBLEMS

14. See the accompanying graph.

1

E-mut

v0 vmax

0.8 0.6 0.5 0.4 0.2 0

[Substrate] (mM) Km for mutant

15. It would require 638 mL of whiskey: • methanol concentration in the body: (100 mL × 0.79 g/mL)/(32 g/mol × 40 L) = 0.062 M • ethanol will behave as a competitive inhibitor: v (uninhibited) = vmax[methanol]/Km+ [methanol]; v (inhibited) = vmax[methanol]/α Km + [methanol], where α = (1 + [ethanol]/KI) • need v (inhibited)/v (uninhibited) = 0.05 = (Km + [methanol])/ (α Km + [methanol]) • 0.05 = (0.01 + 0.062)/(0.01α + 0.062); α = 138 = [ethanol]/KI; [ethanol] = 0.137 M • 0.137 M in 40 L = 5.48 mol; 46 g/mol × 5.48 mol = 252 g; (252 g)/(0.79 g/mL) = 319 mL • whiskey is 50% ethanol, therefore the person needs to consume 638 mL of whiskey

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Long Answers to Review Questions Chapter 8 1. The six components of a typical signaling pathway are (1) first messengers, which can be hormones like insulin or epinephrine; (2) receptor proteins such as the insulin or adrenergic receptors; (3) upstream signaling proteins such as heterotrimeric G proteins and adenylate cyclase; (4) second messengers such as cyclic AMP, diacylglycerol, and calcium; (5) downstream signaling proteins, which can be kinases such as cAMP-dependent protein kinase A and protein kinase C or calcium binding proteins such as calmodulin; and (6) target proteins, which include metabolic enzymes such as phosphorylase kinase, transcription factors, cytoskeletal proteins, and transport proteins.

2. Endocrine hormones are produced by secretory glands that export hormones into the circulatory system, where they come in contact with target cells expressing appropriate receptors, many of which are on the cell surface. Paracrine signaling involves localized production of first messengers, such as nitric oxide or acetylcholine, which activate receptors on nearby cells without being transported through the circulatory system. Autocrine signaling refers to activation of receptors on the same cells that produce the first messenger; this is a common signaling mechanism in immune system cells.

3. Consider that activation of a single extracellular receptor by a first messenger molecule, such as epinephrine, results in the production of 100 second messengers by an upstream signaling enzyme such as adenylate cyclase or phospholipase C. If each of these second messengers activates a downstream signaling protein that has catalytic activity, such as a protein kinase capable of phosphorylating 100 protein targets, then the signal amplification up to this point would be 102 × 102 = 104. However, as most protein targets are themselves catalytic enzymes or transcription factors, the signal is amplified at least 100-fold more, which leads 1

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CHAPTER 8 LONG ANSW ERS TO REVIEW QUESTIONS

to a million-fold amplification of the initiating extracellular signal (104 × 102 = 106).

4. G protein–coupled receptors (GPCRs) are transmembrane receptors that undergo a conformational change in response to ligand binding. This results in dissociation of the heterotrimeric G protein complex (Gαβγ) and activation of enzymes that generate second messenger molecules. Receptor tyrosine kinases (RTKs) are enzymes that contain an extracellular ligand binding domain and an intracellular catalytic domain that phosphorylates tyrosine residues in both the receptor C-terminal tail and in receptorassociated target proteins. Tumor necrosis factor (TNF) receptors transmit extracellular signals by forming receptor trimers, which direct the association of cytosolic adaptor protein complexes. Nuclear receptors are ligandregulated transcription factors that modulate gene expression through protein–DNA and protein–protein interacting functions. Ligand-gated ion channels control the flow of K+, Na+, and Ca2+ ions across cell membranes in response to acetylcholine binding and control membrane depolarization in neuronal cells.

5. The Gα subunit is a GTPase enzyme containing a covalently attached lipid anchor tethering it to the cytosolic face of the plasma membrane.

Dissociation of the heterotrimeric G complex, following GPCR activation by extracellular ligand binding, leads to GDP–GTP exchange in the Gα subunit, which converts it from the inactive GDP-bound form to the active GTP-bound form. Stimulation of the intrinsic GTPase activity in the Gα subunit reverts it to the inactive GDP-bound form. The Gβ subunit is not membrane associated; however, it is tightly bound to the Gγ subunit, which is anchored to the plasma membrane through a covalently attached lipid moiety. The Gβγ heterodimeric complex functions to sequester Gα in the GDP-bound inactive form in the absence of GPCR stimulation. However, upon receptor stimulation, the Gβγ heterodimer functions as a membraneassociated signaling complex and activates downstream signals independently of the Gα subunit.

6. The human genome contains 17 Gα, 5 Gβ, and 12 Gγ genes, which are able to form distinct heterotrimeric G protein complexes depending on which subunits are present. These 34 genes can form ∼1,000 different Gαβγ complexes (17 × 5 × 12 = 970).

7. The two key differences in each of these neuronal signaling pathways are ligand-selective GPCRs expressed in visual, olfactory, and gustatory cells, and the association of these GPCRs with heterotrimeric G protein complexes containing one of three different Gα proteins. Light absorption by the retinal group in rhodopsin activates Gtα, which stimulates a phosphodiesterase and cGMP hydrolysis, causing ion channel closure and neuronal signaling. Odorant binding to olfactory receptors activates Gsα, which stimulates adenylate cyclase and cAMP generation, causing ion channel opening and neuronal signaling. Binding of sweeteners to taste receptors on tongue cells activates Gqα, which stimulates phospholipase C and the generation of DAG and IP3, causing ion channel closure and neuronal signaling.

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3

8. The efficiency of GPCR-mediated signaling is controlled by two types of Gα regulatory proteins (GEF and GAP/RGS) and by proteins that control the turnover of GPCRs on the surface of the cell (GRK and β-arrestin).

Guanine nucleotide exchange factors (GEFs) are proteins that stimulate activation of Gα proteins by promoting GDP–GTP exchange. This GEF activity can be mediated by the ligand-activated GPCR itself or by other proteins that associate with Gα. In contrast, GTPase activating proteins (GAPs) stimulate the intrinsic GTPase activity and thereby inhibit GPCRmediated signaling. GPCR-specific GAPs called regulators of G protein signaling (RGS) have been identified, and indeed, the target protein itself can function as a GAP. The third type of regulatory proteins are the G protein–coupled receptor kinases (GRKs), which phosphorylate the GPCR cytoplasmic domains on serine and threonine residues, marking them for recycling. GRK-mediated phosphorylation of GPCRs provides a docking site on the receptor for β-arrestin, which binds to the receptor and prevents it from reassociating with the Gαβγ complex.

9. Growth factor receptor–bound 2 (GRB2) is an EGFR adaptor protein containing two different types of protein–protein interaction domains. One is an SH2 domain that binds to phosphotyrosine residues on the cytosolic tail of ligand-activated EGFRs. The other type is an SH3 domain (there are two SH3 domains in GRB2), which binds to a downstream GEF protein called son of sevenless (SOS). The function of GRB2-activated SOS is to bind to and activate GDP–GTP exchange in a G protein called Ras, which then activates a downstream signaling kinase called Raf. Once this happens, a whole series of subsequent phosphorylation reactions mediated by the mitogen-activated protein kinase (MAP kinase) family of signaling proteins are stimulated.

10. The EGFR and insulin receptor pathways both stimulate the MAP kinase pathway through upstream signaling requiring GRB2–SOS–Ras. However, one difference is that the adaptor protein Shc is needed to link GRB2 to the ligand-activated insulin receptor through a phosphotyrosine binding (PTB) domain. Another difference is that the insulin receptor also stimulates a completely separate pathway through PTB-mediated binding of insulin receptor substrate (IRS-1), which is an adaptor protein that activates the downstream signaling protein phosphoinositide-3 kinase (PI-3K).

11. Trimerization of the tumor necrosis factor (TNF) receptor initiates the assembly of an intracellular adaptor protein complex. The complex contains upstream signaling proteins with a death domain (DD), which associate with two types of downstream adaptor proteins containing DD binding surfaces. In the cell death apoptotic pathway, the TNF-receptor DD sequences recruit an adaptor protein called TNF receptor–associated death domain (TRADD), which in turn recruits Fas-associated death domain (FADD). FADD contains a death effector domain (DED) that interacts with a DED binding surface on procaspase 8. Assembly of this DED-mediated adaptor complex results in conformational changes in procaspase 8 that stimulate autocleavage and generation of a proteolytically active CASP8 enzyme. Once activated, CASP8 cleaves and activates the executioner caspase, CASP3, which leads to cell death.

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CHAPTER 8 LONG ANSW ERS TO REVIEW QUESTIONS

In the cell survival pathway, TRADD recruits an adaptor protein called TNF receptor–associated factor 2 (TRAF2), as well as a downstream signaling kinase called receptor interacting protein (RIP). TRAF2 recruits another downstream signaling kinase called NFκB-inducing kinase (NIK), which along with RIP phosphorylates and activates a signaling kinase called IκBα kinase (IKK). IKK phosphorylates the inhibitor protein IκBα, leading to activation of the NFκB transcription complex, which translocates to the nucleus and induces the expression of anti-apoptotic genes. The molecular mechanism determining the cell death versus cell survival outcome in a TNF receptor–stimulated cell is the relative expression level of downstream signaling proteins in the pro-apoptotic and anti-apoptotic pathways.

12. Nuclear receptors are ligand-activated transcription factors that control a wide range of physiologic responses. The two major types of nuclear receptor proteins can be distinguished by their mode of DNA binding to cognate response element DNA sequences located near the promoters of nuclear receptor target genes. Steroid receptors represent one type of nuclear receptor, which bind specifically to inverted repeat DNA sequences as headto-head homodimers. Steroid receptors include those nuclear receptors that bind to cholesterol-derived hormones such as cortisol, estradiol, progesterone, and testosterone. Metabolite receptors are the other type of nuclear receptor, which bind as head-to-tail heterodimers to direct repeat DNA sequences. For example, peroxisome proliferator–activated receptors are metabolite receptors that selectively bind dietary unsaturated fatty acids and form heterodimers with retinoid X receptors. The three molecular determinants of nuclear signaling are (1) ligand bioavailability, (2) cellspecific expression of nuclear receptors and coregulatory proteins, and (3) genomic accessibility of target gene DNA sequences in chromatin.

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Long Answers to Challenge Problems Chapter 8 1. Second messengers are generated by enzymes that function catalytically to produce large numbers of second messenger molecules rapidly. Two examples of these enzymes are adenylate cyclase, which converts ATP to cAMP, and phospholipase C, which converts phosphatidylinositol-4,5-bisphosphate (PIP2) into diacylglycerol (DAG) and inositol-1,4,5-trisphosphate (IP3).

2. For a signaling pathway with three transmission processes, each of which amplifies the signal 250-fold, the net level of amplification would be 2503 = 15,625,000. If process A were defective and only amplified the signal by 100-fold, then the net level of amplification would be 100 × 2502 = 6,250,000, which is a reduction of 60%.

3. Sildenafil (Viagra) is a cGMP analog that inhibits the enzyme cyclic GMP phosphodiesterase (cGMP PDE), resulting in elevated levels of cGMP and stimulation of cGMP-dependent protein kinase G, the signaling molecule required for vasodilation.

Sildenafil only works if nitric oxide signaling is activated by neuronal input to the brain, which is necessary to activate guanylyl cyclase and produce cGMP. 4. Glucagon signals through a G protein–coupled receptor mechanism in which Gsα stimulates adenylate cyclase activity, resulting in elevated levels of cAMP and activation of protein kinase A (PKA) signaling. Epinephrine binding to the β2-adrenergic receptor in liver cells activates the same pathway through Gsα, an example of shared signaling pathways. In contrast, epinephrine binding to α1-adrenergic receptors signals through a G protein–coupled receptor mechanism in which Gqα stimulates phospholipase C activity, resulting in elevated levels of diacylglycerol (DAG), inositol-1,4,5-trisphosphate (IP3), and Ca2+. Epinephrine signaling through both the β2-adrenergic and α1-adrenergic receptors in the same cell types is an example of parallel signaling pathways.

5

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CHAPTER 8 LONG ANSW ERS TO CHALLENGE PROBLEMS

5. When you are hungry and want to eat a piece of cake, glucagon will stimulate cAMP production through activation of Gsα signaling, but it is unlikely that epinephrine will be released by the adrenal medulla. In contrast, when you almost step on a snake, acute stress will stimulate epinephrine signaling through both β2-adrenergic and α1-adrenergic receptors, which will lead to increased intracellular levels of the second messengers cAMP, DAG, IP 3, and Ca2+.

6. EGF signaling normally stimulates exchange of GDP for GTP to produce the active form of Ras (Ras–GTP), and cleavage of GTP by the intrinsic Ras GTPase converts it back to the inactive Ras–GDP form. However, mutations that decrease the activity of Ras GTPase—either mutations in the catalytic site itself or mutations that alter GAP protein binding to Ras–GTP—result in formation of a constitutive “activated” Ras–GTP even in the absence of upstream EGF signaling.

7. The G protein cycle for Ras illustrates that SOS functions as a GEF protein and RasGAP simulates the intrinsic GTPase activity of Ras. EGFR • EGF GRB2 • SOS • Ras–GDP

Ras–GDP Inactive

GTP

GDP SOS

RasGAP

Ras–GTP Active Pi

RasGAP • Ras–GTP

RasGAP

8. First, GTPase activating proteins (GAPs), which in the case of GPCR signaling pathways are called regulators of G protein signaling (RGS), stimulate the intrinsic GTPase activity of Gα to terminate signaling by converting the active form (Gα–GTP) into the inactive form (Gα–GDP).

Second, regulatory proteins called G protein–coupled receptor kinases (GRKs) phosphorylate serine and threonine residues in the GPCR cytoplasmic domain, which provides a docking site for an inhibitory protein called β-arrestin. These β-arrestin–GPCR complexes prevent reassociation of Gαβγ heterotrimers with the receptor, and moreover, β-arrestin–GPCR complexes are removed from the plasma membrane and translocated to endocytic vesicles. Once this happens, the GPCRs are either degraded by lysosomes or returned to the plasma membrane.

9. Several steps, each involving protein conformational changes, must be involved. First, EGF binding to the extracellular domain of an EGFR induces a conformational change in receptor monomers that facilitates dimerization with other EGF-bound receptor monomers. Second, receptor dimerization provides an opportunity for protein–protein interactions, which promotes activation of the receptor intrinsic tyrosine kinase activity and phosphorylation of the partnered receptor monomer. Assuming that a

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receptor monomer cannot phosphorylate its own cytoplasmic tail and that simultaneous intermolecular phosphorylation between the two receptors does not take place, initial intermolecular phosphorylation of one receptor by the other (EGFR1 phosphorylates EGFR2; see Figure 8.36) would have to occur. Once this happens, a third protein conformational change would be induced that permits reciprocal intermolecular phosphorylation of the initiating receptor monomer.

10. Answers: 7 PDK1 and Akt bind to PIP3 in the plasma membrane via PH domains.

2 Insulin receptor autophosphorylates tyrosine residues in the cytoplasmic tail.

10 Increased rates of glucose uptake and glycogen synthesis lower blood glucose.

4 Insulin receptor phosphorylates IRS proteins on tyrosine residues.

6 PI-3K phosphorylates PIP2 to generate PIP3.

1 Insulin binds to the insulin receptor and activates its intrinsic kinase activity.

3 IRS proteins bind to phosphotyrosines in the insulin receptor via PTB domains.

8 Akt is phosphorylated and activated by the serine/threonine kinase activity of PDK1.

5 PI-3K binds to phosphotyrosines on IRS proteins via SH2 domains.

9 Akt dissociates from PIP3 and phosphorylates downstream target proteins.

11. The five DD-containing proteins involved in TNF receptor signaling are as follows: (1) the TNF receptor’s cytoplasmic region contains a DD that initiates adaptor complex assembly; (2) SODD, an inhibitory protein that binds to the DDs of the TNF receptor in the absence of ligand to prevent adaptor complex formation; (3) TRADD, the primary adaptor protein, has a DD and a TRAF interaction domain; (4) FADD, the DD-containing protein that links TRADD and the TNF receptor to the cell death pathway; and (5) RIP, a DD-containing kinase that activates a phosphorylation cascade leading to cell survival.

12. There are several good answers to the question; here are a few. (1) Proteases destroy proteins, the workhorses of the cell—without proteins, the cell cannot survive. (2) Proteases are enzymes that function catalytically and therefore can do a lot of damage in a very short amount of time. (3) Proteases must be activated by proteolytic cleavage to convert the inactive zymogen to the active form—this provides a means to control protease activity by regulating the initiating cleavage reaction. (4) Proteases, especially caspases, have preferred substrate recognition sites for cleavage and therefore can preferentially degrade other proteins without undergoing high rates of autocleavage (caspases lack accessible substrate recognition sequences and are protected from degradation).

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13. Steroid receptors bind as homodimers to inverted repeat DNA sequences separated by N3 nucleotides, whereas metabolite receptors bind as heterodimers to direct repeat DNA sequences separated by N1–5 nucleotides.

Because metabolite receptors are heterodimers and bind two different ligands, one of which is usually 9-cis-retinoic acid (binds to RXR), there is potential for tighter control over ligand-dependent activation compared to that of steroid receptors, which, as homodimers, bind only one type of ligand.

In addition, the metabolite receptors bind to direct repeat DNA sequences separated by 1–5 nucleotides (depending on the type of heterodimer), which provides higher specificity in the target gene binding site.

14. Cell-specific glucocorticoid signaling must be due to one or all of these parameters: (1) cell-specific expression of coregulatory proteins, (2) cellspecific bioavailability of glucocorticoids, and (3) cell-specific chromatin accessibility of target gene DNA sequences to glucocorticoid receptor binding.

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Long Answers to Review Questions Chapter 9 1. Catabolic pathways capture energy in the form of ATP and reducing potential (NADH and FADH2) by degrading macromolecules obtained in food into small metabolites. In contrast, anabolic pathways build macromolecules by polymerizing small metabolites using chemical energy in the form of ATP and reducing ability (NADPH).

2. Metabolic flux is a term used to describe the overall reaction rates in catabolic and anabolic pathways. Metabolic flux is primarily controlled by the amount of enzyme activity available, which depends on both the enzyme abundance and catalytic activity.

In addition, metabolic flux is controlled by the availability of substrates and utilization of products.

3. Before breakfast, glucagon signaling in the liver elevates glucose efflux by increasing flux through liver glycogen degradation and glucose synthesis (gluconeogenesis). However, after breakfast, insulin signaling stimulates glucose influx into the liver by increasing flux through the glycogen synthesis and glucose degradation (glycolysis) pathways.

4. The key macromolecules are proteins, nucleic acids, carbohydrates, and lipids. The primary metabolites are amino acids, nucleotides, fatty acids, glucose, pyruvate, and acetyl-CoA. The key small molecules are NH +, CO , NADH, FADH , O , ATP, and H O. 4

2

2

2

2

5. Whether a reaction is favorable or unfavorable under cellular conditions depends on the sign of ΔG. ΔG is the sum of the standard free energy change (ΔG °′) and RT × ln Q, where Q is the mass action ratio (the ratio of actual product concentrations over actual substrate concentrations). An unfavorable (positive) ΔG °′ may be overcome by the RT × ln Q term if it is sufficiently less than zero, resulting in an overall negative ΔG. 6. Three reasons why glycolysis is considered a core metabolic pathway are as follows: (1) glycolytic enzymes are evolutionarily 1

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conserved and found in essentially all living organisms; (2) glycolysis is the only anaerobic pathway for ATP generation; and (3) glycolytic intermediates are shared metabolites in numerous pathways.

7.

H H

2

H

C

OH

C

H

C

OH

5 6

H

6 CH OH 2

4

H

Fructose (ketose sugar) H

O

1C

3

HO

Glucose (aldose sugar)

C

CH2

OH OH

H 4

HO

5

H OH

O H

3

2

H

OH

C

OH 6

H 1

1

HO H

OH H

3

1

HOCH2 C

4

C

5

C

6

O

H OH

5

H OH

H

HO

4

3

OH

H

CH2OH 2

OH

CH2OH

8. The net reaction of the glycolytic pathway is Glucose + 2 NAD+ + 2 ADP + 2 P → i 2 Pyruvate + 2 NADH + 2 H+ + 2 ATP + 2 H 2O The ATP investment phase (stage 1) of the glycolytic pathway consists of five reactions and requires the input of 2 ATP to generate fructose1,6-bisphosphate. Cleavage of fructose-1,6-biphosphate, followed by an isomerization reaction, yields two molecules of glyceraldehyde-3-phosphate for each molecule of glucose metabolized by the pathway. The ATP earnings phase (stage 2) consists of five reactions and generates a total of four molecules of ATP for each molecule of glucose. Because 2 ATP were invested in stage 1, the net ATP yield of glycolysis is 2 ATP per glucose.

9. Substrate-level phosphorylation refers to metabolic reactions in which ATP is generated by transferring a phosphoryl group from a phosphorylated metabolite directly to ADP, independent of the ATP synthase complex (see Chapter 11). The phosphorylated metabolite must have a more negative ΔG °′ for hydrolysis than that of ATP hydrolysis (ΔG °′ = −30.5 kJ/mol) for this reaction to be energetically favorable. The substratelevel phosphorylation reactions in glycolysis are reaction 7, catalyzed by phosphoglycerate kinase, and reaction 10, catalyzed by pyruvate kinase.

10. Hexokinase and glucokinase both phosphorylate glucose on the C-6 position in the first reaction of the glycolytic pathway, but they differ significantly in the following ways. Hexokinase has a very high affinity for glucose (100 times higher than that of glucokinase); is able to phosphorylate fructose; is allosterically inhibited by glucose-6-phosphate; and is expressed in all tissues. Glucokinase is highly specific for glucose but has a low affinity for this substrate; is not allosterically inhibited by glucose-6-phosphate; and is expressed primarily in liver and pancreas tissues. Moreover, glucokinase functions as a glucose sensor in pancreatic β cells.

11. Phosphofructokinase-1 (PFK-1) is a homotetrameric protein that is present in the cell in two conformational structures, called the T state (inactive) and the R state (active). Allosteric regulation of PFK-1 activity shifts the equilibrium between these two conformational states. PFK-1 activity is stimulated by a low energy charge in the cell (low ATP relative to AMP

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and ADP), whereas high energy charge inhibits PFK-1 activity. PFK-1 is allosterically activated by AMP, ADP, and fructose-2,6-bisphosphate binding and is allosterically inhibited by ATP and citrate binding.

12. Maltose is a disaccharide of glucose, which is cleaved by the enzyme maltase to yield two molecules of glucose that are directly metabolized by the glycolytic pathway. Lactose is cleaved by the enzyme lactase to yield glucose and galactose, of which galactose must first be phosphorylated by galactokinase and converted to glucose-1-phosphate before it is isomerized and enters the glycolytic pathway as glucose-6-phosphate. Sucrose is cleaved by sucrase to generate glucose and fructose. The fructose in muscle cells is directly phosphorylated by hexokinase and enters the pathway as fructose-6phosphate, whereas in liver cells, fructose is phosphorylated by fructokinase to yield fructose-1-phosphate, which is then cleaved to yield eventually two molecules of glyceraldehyde-3-phosphate.

13. Pyruvate is decarboxylated and converted to acetyl-CoA under aerobic conditions in all organisms that form citrate from oxaloacetate and acetylCoA in the citrate cycle. This yields the maximum number of ATP from glucose oxidation. Under anaerobic conditions, most organisms convert pyruvate to lactate using the enzyme lactate dehydrogenase in order to regenerate NAD+ for the glyceraldehyde-3-phosphate dehydrogenase reaction in glycolysis. Fermenting organisms, such as yeast, are also able to convert pyruvate to CO2 and ethanol under anaerobic conditions.

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Long Answers to Challenge Problems Chapter 9 1. Before breakfast, glucose levels are low in the blood due to a night of fasting, which triggers the release of glucagon from the pancreas. Glucagon receptors are activated in liver cells, leading to an increase in glycogen degradation to produce free glucose, as well as glucose synthesis from pyruvate (gluconeogenesis).

Together, these two metabolic pathways increase glucose export to maintain appropriate blood glucose levels for brain function (∼5 mM). After breakfast, blood glucose levels rise, leading to insulin release from the pancreas and stimulation of insulin receptor signaling in liver cells. Under these conditions, the metabolic flux of glucose is shifted toward glucose influx in the liver, where it is stored as glycogen and used for energy conversion by the glycolytic pathway.

2. Fructose is converted to glucose by the base (OH−) in Benedict’s reagent, and glucose is an aldose that is readily oxidized by Cu2+.

Fructose, a ketose, gives a false-positive result in Benedict’s test.

3. No. The two substrate-level phosphorylation reactions in glycolysis (reactions 7 and 10) can occur under anaerobic conditions when pyruvate is converted to lactate.

4. Reaction: 2-phosphoglycerate ↔ phosphoenolpyruvate ΔG °′ = +1.7 kJ/mol ΔG = (1.7 kJ/mol) + (8.314 × 10−3 kJ/mol K) (298 K) aln

1

b 10

ΔG = (1.7 kJ/mol) + (8.314 × 10−3 kJ/mol K) (298 K) (−2.3) ΔG = (1.7 kJ/mol) + (−5.69 kJ/mol) = −4 kJ/mol The ΔG value remains the same; enzymes only change the reaction rate, not the ΔG. 5. ΔG °′ = −RT ln Keq ΔG °′= −(8.314 × 10−3 kJ/mol K) (298 K) (ln 6.8 × 10−5) ΔG °′= +23.8 kJ/mol 4

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There is no effect on the Keq value when twice as much aldolase is added to the reaction; enzymes are catalysts and don’t change the equilibrium constant of a reaction.

6. a.

Glucose HO H 4

HO

Pyruvate

Pyruvate

CH2

6

5

H OH

O

H

H

1

3

2 OH

H

OH

–O

O

–O

1C

+ 2 NAD+ + 2 ADP + 2 Pi

2C

O C (4)

O

+

C

(5)

CH3 3

O

+ 2 NADH + 2 H+ + 2 ATP + 2 H2O

CH3 (6)

b. 2 Glyceraldehyde-3-P + 2 Pi + 2 NAD+ m 2 1,3-Bisphosphoglycerate + 2 NADH + 2 H+ 2 2-Phosphoglycerate m 2 Phosphoenolpyruvate + 2 H2O 7. Net reaction: 4 ATP (equivalents) + 4 H2O → 4 ADP (equivalents) + 4 Pi Enzyme regulation is critical in controlling these two opposing pathways; otherwise, ATP would be continually hydrolyzed, leading to futile cycling and heat production. Coordinate control of glycolytic and gluconeogenic enzymes shifts the metabolic flux in the appropriate direction in response to glucose availability to prevent futile cycling.

8. Key differences include (a) hexokinase is in all cells, whereas glucokinase is liver specific; (b) glucokinase has a much lower affinity (higher Km) for glucose than that of hexokinase; and (c) hexokinase is feedback inhibited by glucose-6-P, whereas glucokinase is not inhibited.

9. Pyruvate kinase converts phosphoenolpyruvate and ADP to pyruvate and ATP in the last step of the glycolytic pathway. When blood glucose levels are high, cells metabolize glucose to form glycolytic intermediates such as fructose-1,6-bisphosphate, which in turn activates pyruvate kinase to stimulate metabolic flux in the direction of pyruvate formation.

10. Complete loss of PFK-1 activity is probably lethal at early stages of embryonic development because this enzyme is critical to maintaining flux through the glycolytic pathway. This would result in a miscarriage early in the pregnancy.

11. a. The lack of phosphoglycerate mutase blocks production of pyruvate from glucose. During intense exercise, lactate is normally produced from pyruvate, which replenishes the necessary NAD+ required for the energy production phase of anaerobic glycolysis. b. No. Because the individual is not producing enough pyruvate, there will not be very much, if any, lactate.

12. Inherited fructose intolerance is due to defects in aldolase B (fructose1-P aldolase), which prevents efficient conversion of fructose-1-P to dihydroxyacetone-P and glyceraldehyde. Because this traps available phosphate in fructose-1-P, which cannot be further metabolized, ADP + Pi → ATP reactions are inhibited. The resulting low energy charge in liver cells shuts down membrane pumps that prevent the buildup of toxins.

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CHAPTER 9 LONG ANSW ERS TO CHALLENGE PROBLEMS

High-fructose foods and beverages need to be avoided or consumed in limiting quantities. These include processed foods and soda or fruit drinks containing “fructose sweetener, ” which comes from corn syrup; also foods and beverages containing sucrose or honey, both of which contain fructose.

13. The enzyme lactase converts lactose to glucose and galactose. Lactose metabolism results in the production of four pyruvate molecules, two each from glucose and galactose. Anaerobic metabolism would generate four net ATP from lactose, and aerobic metabolism would generate up to 64 ATP. 14. a. During the first phase of beer production, the yeast need to be grown under aerobic conditions to facilitate rapid growth from the yeast starter culture; growth under anaerobic conditions is much slower than that under aerobic conditions because of differences in energy conversion efficiency (ATP generation). Once the yeast have grown sufficiently, the beer mixture is shifted to anaerobic conditions to initiate the fermentation process, using the sugars in the added plant materials. b. To bottle the beer, the fermentation vessel must be opened, and the CO2 is lost. However, because the original yeast culture is no longer undergoing rapid growth and consuming sugars to make CO2, a fresh yeast culture needs to be added, along with additional glucose, to start the fermentation process again in the capped bottles.

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Long Answers to Review Questions Chapter 10 1. Commercial-grade citric acid can be purified from both lemon extracts and fermentation cultures of Aspergillus niger.

2. The citrate cycle accomplishes four goals: (1) it decarboxylates citrate to remove two carbons contributed by acetyl-CoA; (2) it transfers 8 e− to 3 NAD+and 1 FAD; (3) it generates 1 GTP (converted to ATP by diphosphate kinase) by substrate-level phosphorylation; and (4) it regenerates oxaloacetate in the last reaction of the cycle.

3. The name citrate cycle reflects that all three carboxylate groups on citrate are deprotonated at physiologic pH. Citrate, which predominates at physiologic pH, is the conjugate base of citric acid. The name Krebs cycle pays tribute to Hans Krebs, but it is not very descriptive of the actual pathway, especially as Krebs also discovered the urea cycle.

4. The energy available from redox reactions is due to differences in the electron affinity of two reactants, with each redox reaction consisting of two half-reactions containing a conjugate redox pair. Compounds that accept electrons are called oxidants and are reduced in the reaction, whereas compounds that donate electrons are called reductants and are oxidized in the reaction.

The standard reduction potential E°′ of a conjugate redox pair has units of volts and is measured experimentally against hydrogen, which has a E°′ value of 0 V. Oxidants with a higher affinity for electrons than that of H+ have positive E°′ values, and oxidants with a lower affinity for electrons than that of H+ have negative E°′ values. The amount of energy available from a coupled redox reaction, ΔE°′, is determined by subtracting the E°′ of the reductant (e– donor) from the E°′ of the oxidant (e– acceptor).

Using the ΔE°′ value for a coupled redox reaction, we can calculate the change in free energy using the equation ΔG°′ = –nFΔE°′.

1

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5. The net pyruvate dehydrogenase reaction is Pyruvate + CoA + NAD+ → Acetyl-CoA + CO2 + NADH The pyruvate dehydrogenase enzyme complex requires the chemical contributions of five coenzymes: (1) NAD+ (nicotinamide adenine dinucleotide), derived from the vitamin niacin (B3), which is contained in poultry, fish, and vegetables; (2) FAD (flavin adenine dinucleotide), derived from riboflavin (B2), which is contained in dairy products, almonds, and asparagus; (3) CoA (coenzyme A), derived from the water-soluble vitamin pantothenic acid (B5), which is contained in brown rice, lentils, chicken, yogurt, and avocados; (4) TPP (thiamine pyrophosphate), derived from thiamine (B1), which can be obtained from cooked brown rice and thiamine-fortified processed foods; and (5) α-lipoic acid (lipoamide), which is synthesized in plants and animals and therefore is not an essential nutrient like vitamins.

Foods rich in α-lipoic acid include broccoli, liver, spinach, and tomatoes.

6. The five steps are as follows. Step 1: The E1 subunit (pyruvate dehydrogenase) binds pyruvate and catalyzes a decarboxylation reaction resulting in the formation of hydroxyethyl-TPP and the subsequent release of CO2. Step 2: E1 reacts with the disulfide of the lipoamide group on the N-terminal domain of the E2 subunit (dihydrolipoyl acetyltransferase) to generate acetyl-dihydrolipoamide through a thioester bond. Step 3: The E2 lipoamide domain carries the acetyl group from the E1 catalytic site to the E2 catalytic site, where it reacts with CoA to yield acetyl-CoA. Step 4: The E2 lipoamide domain swings over to the E3 subunit (dihydrolipoyl dehydrogenase) where it is reoxidized and transfers 2 e– and 2 H+ to the E3-linked FAD moiety. Step 5: The E3-FADH2 coenzyme intermediate is reoxidized and transfers 2 e – to NAD+ to form NADH + H+.

7. Allosteric control of pyruvate dehydrogenase activity is in response to energy charge, which controls the rate of pyruvate conversion to acetylCoA for the purpose of oxidation in the citrate cycle or fatty acid synthesis in the cytosol. Pyruvate dehydrogenase activity is allosterically inhibited by high ratios of NADH to NAD+ and acetyl-CoA to CoA, as well as by serine phosphorylation, which inhibits pyruvate dehydrogenase activity.

The regulatory enzyme pyruvate dehydrogenase kinase phosphorylates and inactivates pyruvate dehydrogenase in response to elevated NADH, acetylCoA, and ATP, whereas the enzyme pyruvate dehydrogenase phosphatase-1 dephosphorylates pyruvate dehydrogenase and activates the enzyme in response to elevated Ca2+ levels. Lastly, pyruvate dehydrogenase kinase activity is inhibited by NAD+, CoA, ADP, and Ca2+, which maintains pyruvate dehydrogenase in the active dephosphorylated conformation.

8. The net reaction of the citrate cycle is Acetyl-CoA + 3 NAD+ + FAD + GDP + Pi + 2 H2O → CoA + 2 CO2 + 3 NADH + 3 H+ + FADH2 + GTP The citrate cycle consists of eight enzymatic reactions: (1) citrate synthase catalyzes a condensation reaction forming citrate from oxaloacetate and acetyl-CoA; (2) aconitase isomerizes citrate to form isocitrate; (3) isocitrate dehydrogenase catalyzes an oxidative decarboxylation reaction

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converting isocitrate to α-ketoglutarate, generating CO2 and NADH in the process; (4) α-ketoglutarate dehydrogenase catalyzes a second oxidative decarboxylation reaction, which requires CoA and converts α-ketoglutarate to succinyl-CoA, yielding CO2 and NADH; (5) succinyl-CoA synthetase catalyzes a substrate-level phosphorylation reaction generating GTP and succinate with release of CoA; (6) succinate dehydrogenase oxidizes succinate to form fumarate and reduces FAD to form FADH2; (7) fumarase hydrates fumarate to generate malate; and (8) malate dehydrogenase oxidizes malate to form oxaloacetate and generates a third NADH. 9. The three main control points within the citrate cycle are regulation of citrate synthase, isocitrate dehydrogenase, and α-ketoglutarate dehydrogenase activities. All three of these enzymes are inhibited by high NADH-to-NAD+ ratios, which signal that the electron transport system is not working at full efficiency because of elevated ATP-to-(ADP + AMP) ratios. Citrate synthase, isocitrate dehydrogenase, and α-ketoglutarate dehydrogenase are inhibited by NADH and ATP, whereas citrate synthase and α-ketoglutarate dehydrogenase are both inhibited by succinyl-CoA. Citrate synthase is also potently inhibited by citrate, which is exported out of the mitochondria when its levels are too high.

10. The citrate cycle is an amphibolic pathway because it provides biosynthetic precursors to both catabolic and anabolic pathways. The pyruvate carboxylase reaction is anaplerotic because it provides oxaloacetate to the citrate cycle; that is, it fills up the citrate cycle with intermediates. Other anaplerotic reactions supplying oxaloacetate to the citrate cycle are those catalyzed by phosphoenolpyruvate carboxylase and malic enzyme.

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Long Answers to Challenge Problems Chapter 10 1. a. Malate dehydrogenase reaction: Malate + NAD+ → Oxaloacetate + NADH + H+ ΔE°′ = (ΔE°′e– acceptor) – (ΔE°′e– donor) = (–0.32 V) – (–0.17 V) = –0.15 V ΔG°′ = –nFΔE°′ = –(2)(96.48 kJ/mol V)(–0.15 V) = +28.9 kJ/mol Yes, this value is within ∼1 kJ/mol of the value given in Table 10.2. ΔG < 0 kJ/mol for this reaction because oxaloacetate is converted to citrate in the citrate synthase reaction; that is, the mass action ratio (Q) is less than 1. b. Succinate dehydrogenase reaction: Succinate + FAD → Fumarate + FADH2 ΔE°′ = (+0.03 V) – (0.0 V) = +0.03 V ΔG°′ = –nFΔE°′ = –(2)(96.48 kJ/mol V)(+0.03 V) = –5.8 kJ/mol No, this value is not within ∼1 kJ/mol of the value given in Table 10.2. One explanation for the difference in ΔG°′ values of –5.8 kJ/mol calculated here and +0.4 kJ/mol given in Table 10.2 is that the E°′ value for FAD bound to succinate dehydrogenase is more positive than the value of ∼0 V listed in Table 10.1. In order for the ΔG°′ value for the isocitrate dehydrogenase reaction to be +0.4 kJ/mol, the E°′ value for succinate dehydrogenase-bound FAD would have to be close to 0.03 V. ΔG is ∼0 kJ/mol for this reaction.

4

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5

2. Pyruvate is decarboxylated by the pyruvate dehydrogenase reaction to yield 14CO and acetyl-CoA. However, the reverse reaction is highly unfavorable 2 and does not occur in the cell.

3. • Thiamine pyrophosphate (TPP): decarboxylation of pyruvate using the thiazolium ring to create a Schiff base.

• Lipoic acid/lipoamide: oxidation and reduction. It is reduced as it accepts the acetyl group from TPP and oxidized as it passes electrons to FAD, one electron at a time. It holds the acetyl group in place until it is replaced by CoA. • FAD/FADH2: electron carrier. It accepts electrons one at a time from reduced lipoamide and passes electrons to NAD+.

4. a. Individuals with beriberi cannot readily metabolize pyruvate to acetylCoA because pyruvate dehydrogenase is inactive without thiamine pyrophosphate. This leads to a buildup of pyruvate in the blood under conditions when glycolysis is stimulated. b. Thiaminase degrades thiamine pyrophosphate, which is a required coenzyme in the pyruvate dehydrogenase reaction. Cooking denatures the thiaminase and inactivates it. c. A defect in pyruvate dehydrogenase phosphatase would result in a dehydrogenase that is never dephosphorylated. Therefore, the enzyme would always be inhibited as a result of phosphorylation by pyruvate dehydrogenase kinase. Thiamine supplementation would have no effect in this case because thiamine deficiency is not related to the symptoms.

5. a. The first CO2 arises from C-6 of the citrate molecule. b. Isocitrate dehydrogenase is the enzyme. c. The carbons present in oxaloacetate after the first turn are C-1, C-2, C-3, and C-4 of the original citrate molecule.

6. a. The radioactive carbon became incorporated into all of the cycle intermediates at atom positions that could only be explained by products also serving as substrates. The metabolism of oxaloacetate in one round of the cycle produces a different radioactively labeled product depending on the number of cycles. If the citrate cycle were a linear pathway, only one type of radioactively labeled intermediate would be observed. b. Inorganic phosphate is required by one of the citrate cycle reaction steps (succinyl-CoA synthetase) that couples GTP synthesis to thioester bond cleavage. Without Pi, this enzyme reaction is inhibited, and radioactive carbon would only be found in cycle intermediates that precede this reaction step.

7. The rate of oxygen consumption is limited by the rate of acetyl-CoA metabolism in the citrate cycle. However, by increasing the concentration of citrate cycle intermediates through the addition of fumarate, it increased the capacity of the cycle to oxidize acetyl-CoA, resulting in a large increase in oxygen consumption.

8. The addition of citrate increased the capacity of the citrate cycle to metabolize acetyl-CoA by increasing the concentration of all cycle intermediates—most

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CHAPTER 10 LONG ANSWERS TO CHALLENGE PROBLEMS

important, the concentration of oxaloacetate. The addition of acetyl-CoA to this system had little effect because oxaloacetate levels were limiting and glucose metabolism was already at a maximum rate. The cycle was working at full capacity, and having more substrate available for the citrate synthase reaction had little effect on the rate of glucose metabolism.

9. Pyruvate dehydrogenase reaction: 1 NADH Citrate cycle: 3 NADH + 1 FADH2 + 1 GTP (ATP) 4 NADH total = ∼10 ATP 1 FADH2 = ∼1.5 ATP 1 GTP = 1 ATP Total = ∼12.5 ATP (∼13 ATP) 10. If pyruvate carboxylase is unable to be activated by acetyl-CoA, then the oxaloacetate required for citrate formation becomes rate limiting, and the excess acetyl-CoA is not fully used for energy conversion reactions in the citrate cycle.

11. a. When blood glucose levels are low, the oxaloacetate is converted to phosphoenolpyruvate and used in the gluconeogenic pathway to produce glucose for release into the blood. b. When mitochondrial acetyl-CoA levels are high, the oxaloacetate is converted to citrate in the citrate cycle to meet the energy needs of the cell.

12. It would disrupt the coordinate regulation of the citrate cycle and glycolysis, because normally citrate is exported from the mitochondrion when citrate levels are high due to high citrate cycle activity. In the cytosol, citrate inhibits phosphofructokinase, thereby shutting down glycolysis when citrate cycle activity is high because there is less need for glycolytic energy production.

13. The defective enzyme is fumarase. The high lactate and low citrate are a consequence of low citrate cycle activity. The lack of fumarase causes a buildup of fumarate and succinate—the intermediates before the block— and a lack of malate, which is the intermediate after the block.

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Long Answers to Review Questions Chapter 11 1. The chemiosmotic theory was proposed by Peter Mitchell.

It states that energy conversion for redox reactions or light absorption is used to form an electrochemical proton gradient across a proton-impermeable organelle membrane. This electrochemical proton gradient is used to generate protonmotive force, which consists of both a chemical gradient (ΔpH) and a membrane potential (Δψ). The flow of protons down this electrochemical proton gradient is harnessed by the ATP synthase complex enzyme to drive protein conformational changes that catalyze the reaction ADP + Pi → ATP. 2. Data supporting the chemiosmotic theory were published in 1973 by Racker and Stoeckenius. They used reconstituted membrane vesicles containing a light-activated bacteriorhodopsin proton pump from bacteria and an ATP synthase complex from animal mitochondria. In the presence of light, bacteriorhodopsin translocated protons into the vesicle interior, which generated an electrochemical proton gradient across the vesicle membrane.

When the protons crossed the membrane through the ATP synthase complex in response to the energetically favorable electrochemical proton gradient, ATP was generated on the outside of the vesicle, exactly as Mitchell predicted should happen.

3. Two mechanisms have been proposed to account for proton translocation across the inner mitochondrial membrane. One mechanism, called the redox loop mechanism, involves a redox reaction in which the resulting H+ and e− are separated within the transmembrane electron transport system protein and deposited on opposite sides of the inner mitochondrial membrane.

The e− is used to reduce an oxidant in the mitochondrial matrix, and the H+ is released into the intermembrane space. The Q cycle in complex III is thought to use the redox loop mechanism.

A second mechanism is based on redox-driven conformational changes in the transmembrane electron transport system protein.

1

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In this mechanism, matrix-derived protons are translocated across the inner mitochondrial membrane by altering pKa values of functional groups located on the inner and outer faces of the membrane. Protein complexes I and IV in the electron transport system have properties that are consistent with such a mechanism.

4. Three of the six electron carriers in the mitochondrial electron transport system are membrane-spanning protein complexes that translocate protons across the inner mitochondrial membrane (complexes I, III, IV); one is a membrane protein that transfers electrons directly from the citrate cycle to the electron transport system (complex II); and two are mobile electron carriers (cytochrome c and coenzyme Q ). Complex I (NADH– ubiquinone oxidoreductase) oxidizes NADH and translocates 4 H+ from the mitochondrial matrix to the intermembrane space. Complex III (ubiquinone–cytochrome c oxidoreductase) transfers electrons from ubiquinol (Q H 2) to cytochrome c and translocates 4 H+ across the inner mitochondrial membrane. Complex IV (cytochrome c oxidase) transfers electrons from cytochrome c to oxygen to form water and translocates 2 H+ across the inner mitochondrial membrane. Complex II (succinate dehydrogenase) is a citrate cycle enzyme that transfers electrons from FADH2 to Q to form Q H2 without translocating any protons across the inner mitochondrial membrane. Cytochrome c is localized to the intermembrane space and transfers 1 e− from complex III to complex IV using an iron-containing heme prosthetic group. Coenzyme Q (ubiquinone) is a hydrophobic, membrane-localized electron carrier that transfers electrons from complexes I and II to complex III. 5. The ATP synthase complex contains (using the naming of the yeast system): (1) a protein rotor consisting of the c ring and the γ, δ, and ε subunits, which rotates as protons enter and exit the c ring; (2) a protein catalytic headpiece made from the α3β3 hexamer, with each β subunit containing a catalytic site for ATP synthesis; and (3) a protein stator made from the a, b, d, h, and OSCP subunits that functions to stabilize the entire complex and provides the proton half-channels needed to drive rotation of the rotor.

6. The binding change mechanism of ATP synthesis was first proposed by Boyer. It explains how conformational changes in β subunits of the ATP synthase complex could control ATP production. The model proposes that the rotation of the γ subunit changes the conformation of the β subunits and sequentially alters nucleotide binding affinities with the catalytic active sites. The protein contacts between the γ subunit and each β subunit are unique and mediate three distinct protein conformations, referred to as loose (L), tight (T), and open (O). The occupancy of the enzyme active site in each of these protein conformations is ADP + Pi bound (L), ATP bound (T state), and unbound (O). Proton flow through the c ring of the ATP synthase complex induces rotation of the γ subunit, such that with each 120° rotation, ∼3 H+ cross the inner mitochondrial membrane, and the β subunits sequentially undergo a conformational change from L → T → O → L. A 360° rotation of the γ subunit produces ∼3 ATP and requires the translocation of ∼9 H+ from the intermembrane space to the mitochondrial matrix.

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7. One protein is the ATP/ADP translocase, also called the adenine nucleotide translocase, which functions to export 1 ATP for every ADP that is imported. The second translocase protein is the phosphate translocase, which translocates 1 P i and 1 H+ into the matrix by an electrically neutral import mechanism. The phosphate translocase can have either symporter or antiporter functions. For example, when H PO − accompanies an H+ across 2

4

the inner mitochondrial membrane in response to the proton gradient, it is acting as a symporter because both molecules are translocated in the same direction. This is an electrically neutral translocation because the two charges (H PO − and H+) cancel each other out.

2

4

8. Cytosolic NADH cannot cross the inner mitochondrial membrane and must donate electrons via one of two shuttle systems to regenerate NAD+ in the cytosol to maintain flux through the glycolytic pathway. The malate–aspartate shuttle functions in liver cells and reduces oxaloacetate in the cytosol with NADH to generate NAD+ and malate, which is shuttled into the matrix and oxidized to produce NADH and oxaloacetate.

The glycerol-3-phosphate shuttle operates within the mitochondrial membrane of muscle and brain cells and catalyzes a redox reaction that oxidizes cytosolic NADH and reduces FAD to yield cytosolic NAD+ and mitochondrial FADH2. Oxidation of FADH2 within the inner mitochondrial membrane transfers an electron pair directly to coenzyme Q .

The malate–aspartate shuttle in liver cells provides the maximum amount of redox energy for ATP synthesis, as 1 NADHcytosol is converted to 1 NADHmatrix. In contrast, the glycerol-3-phosphate shuttle in muscle and brain cells provides less ATP because oxidation of FADH2 in the electron transport system results in the translocation of fewer H+ across the inner mitochondrial membrane.

9. The ATP currency exchange ratio is based on the number of H+ translocated across the inner mitochondrial membrane and the number of H+ that must flow through the ATP synthase complex to generate each ATP. In the case of complex I–mediated oxidation of mitochondrial NADH and subsequent electron transfer through complexes III and IV, a total of 10 H+ are translocated from the matrix into the intermembrane space (4 H+ + 4 H+ + 2 H+ = 10 H+). However, because FADH 2 oxidation donates a pair of electrons directly to coenzyme Q and bypasses complex I, a total of only 6 H+ are translocated by the combined action of complexes III and IV (4 H+ + 2 H+ = 6 H+). The currency exchange ratio states that 1 ATP is synthesized for every 4 H+ that are translocated back into the mitochondrial matrix from the intermembrane space (3 H+ for each 120° rotation and 1 H+ for the P itranslocase = 4 H+ per ATP). Therefore, ∼2.5 ATP per NADH are generated (10 H+/4 H+ = 2.5), and ∼1.5 ATP + + per FADH2 are generated (6 H /4 H = 1.5).

10. Three classes of inhibitors decrease rates of mitochondrial ATP synthesis.

The first is those that inhibit electron flow through the electron transport system. These include rotenone (inhibits e− flow through complex I), antimycin (inhibits e− flow through complex III), and cyanide (inhibits e− flow through complex IV). The second class of inhibitors is uncouplers of the electrochemical proton gradient, which allow protons to cross the

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mitochondrial membrane without going through the ATP synthase complex; for example, 2,4-dinitrophenol. The third class of inhibitors is those that interfere with ATP synthesis, either by interfering with the activity of the ATP synthase complex (oligomycin) or by translocation of ATP and ADP by the ATP/ADP translocase (bongkrekic acid).

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Long Answers to Challenge Problems Chapter 11 1. Cytochrome c is required for electron transport and is localized to the inner mitochondrial membrane; it is not supposed to be in the cytosol. The presence of high levels of cytochrome c in the cytosol signals that the mitochondria are not functioning properly and induces apoptosis.

2. The phosphate translocase requires that 1 H+ accompanies each Pi brought into the matrix.

3. The calculated membrane ΔV is 140 mV, as follows: ΔG = RT ln(C2/C1) + ZFΔV ΔG – RT ln(C2/C1) = ZFΔV ΔG − RT ln(C2/C1)

= ΔV ZF 121.8 kJ/mol 2 − [(8.314 × 10−3 kJ/mol K 2 1298 K 2 ln 1252 ] 1 +12 196.48 kJ/mol V 2

= ΔV

ΔV = 0.14 V = 140 mV 4. When the H+ binds to a single c subunit, it causes a rotation of the c ring of 30° (1/12 of a circle). The three-sided γ subunit rotates along with the c ring and interacts directly with the three β subunits. For every 120° rotation of the c ring, the γ subunit induces the L, T, or O conformation in each of the β subunits.

Because an ATP is synthesized every time the γ subunit completes a turn of 120°, 4 H+ must cross through the ATP synthase complex to synthesize 1 ATP. 5. a. Because there is no membrane or proton-motive force in this system, and the ATP synthase complex functions as an ATPase when unconstrained, the direction of the headpiece rotation would be in the direction opposite of (clockwise) the intact cells when viewed from the same orientation.

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b. ATP levels would decrease as a result of ATP hydrolysis mediated by the ATP synthase complex.

6. a. Cyanide blocks electron transfer in the electron transport system, resulting in the loss of the proton-motive force; therefore, ATP synthesis rates decrease. Adding 2,4-dinitrophenol to this cyanide-inhibited mitochondrial suspension has no effect on ATP synthesis rates because the electron transport system is blocked and there is no proton-motive force to drive ATP synthesis. b. Oligomycin blocks proton flow through the ATP synthase complex, resulting in a buildup of the electrochemical proton gradient, which eventually inhibits additional proton translocation, and O2 consumption decreases. Adding 2,4-dinitrophenol to this oligomycin-inhibited system leads to increased rates of O2 consumption because the electron transport system is able to resume as protons travel down the gradient via 2,4-dinitrophenol.

7.

Inhibitor(s) added

Electron transport system activity?

ATP synthesis?

Myxothiazol

None

Decrease

Electron transport system is blocked, and H+ gradient is depleted

FCCP

Increase

Decrease

FCCP is an uncoupler, and H+ flow through ATP synthase complex decreases

Explain

Venturicidin

Decrease

None

ATP synthase complex is inhibited, and electron transport system activity is low due to excessive H+ gradient

Venturicidin + FCCP

Increase

None

FCCP is an uncoupler independent of ATP synthase complex activity

8. Thermogenin uncouples the electron transport system from oxidative phosphorylation. Heat created by an uncoupled electron transport system in these special cells keeps blood flowing to the brain and keeps the animal alive even in the middle of winter.

9. Cytosolic NADH in muscle cells donates 2 e− to FADH2 by the glycerol phosphate shuttle, and mitochondrial FADH2 oxidation results in the production of 1.5 ATP per NADH, or 3 ATP per 2 NADH. In contrast, liver cells use the malate–aspartate shuttle, consisting of complex redox reactions to transfer 2 e− from cytosolic NADH to mitochondrial NAD+, producing mitochondrial NADH, which is used to generate 2.5 ATP per NADH, or 5 ATP per 2 NADH. 10. The radioactive 3H is transported inside the mitochondria by shuttle systems (either the glycerol-3-phosphate shuttle or the malate shuttle), whereas NAD cannot cross the inner mitochondrial membrane, and therefore the 14C label does not get into the matrix.

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11. The mitochondrial matrix produces 25 ATP from oxidation of two molecules of pyruvate, which does not include any energy conversion reactions from glycolysis.

8 NADH → 20 ATP 2

FADH2 →

3

ATP

2

GTP (ATP)

25

Total ATP

12. (a-4) The canister shell represents the proton-impermeable inner mitochondrial membrane. (b-3) The spray nozzle represents the ATP synthase complex because useful work is accomplished from the pressure in the canister. (c-2) The pump handle represents reductants that provide redox energy to establish the gradient, which is analogous to internal pressure in the canister. (d-1) The release valve represents an uncoupler that permits protons to cross the membrane in response to the electrochemical gradient without passing through the ATP synthase complex.

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Long Answers to Review Questions Chapter 12 1. Photosynthesis is the energy-converting process that generates ATP and NADPH using the energy obtained from light absorption. Photosynthetic organisms are autotrophs, meaning that they are completely self-sufficient and are able to survive using stored chemical energy when light energy is not available; for example, at night. Photosynthetic autotrophs do this by converting the chemical energy available in ATP and NADPH into carbohydrates (sucrose and starch) using a process called carbon fixation, which in plants takes place inside chloroplasts. In the absence of light, photosynthetic autotrophs metabolize the stored carbohydrates using aerobic respiration, a process that takes place in plant mitochondria. Heterotrophs cannot convert light energy into chemical energy and depend on consuming carbohydrates stored in autotrophs for nutrients that the heterotrophs metabolize by aerobic respiration to generate ATP. Heterotrophs also must consume autotrophs to obtain essential amino acids, fatty acids, and coenzymes (vitamins). Taken together, autotrophs are completely self-sufficient and use both photosynthesis and aerobic respiration, whereas heterotrophs are completely dependent on consuming autotrophs and using aerobic respiration as their only means of energy conversion.

2. Chloroplasts are the photosynthetic organelles in eukaryotic cells that are thought to have evolved from an endosymbiotic relationship between an ancestral eukaryotic cell and a photosynthetic bacterium. The two key metabolic pathways contained in chloroplasts are the photosynthetic electron transport system, including the chloroplast ATP synthase, and the Calvin cycle, which converts CO2 into glyceraldehyde-3-phosphate using ATP and NADPH as a source of chemical energy. Chloroplasts contain three membranes: (1) the outer chloroplast membrane, which is permeable to most metabolites; (2) the inner chloroplast membrane, which contains metabolite transporter proteins; and 1

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(3) the thylakoid membrane, which is a proton-impermeable membrane and the site of photosynthetic electron transport and ATP synthesis. The thylakoid membrane is analogous to the inner mitochondrial membrane with regard to redox energy conversion and ATP synthesis. The stacks of thylakoid membranes are called grana. The chloroplast compartment outside the grana is the stroma, and it contains all of the enzymes required for the Calvin cycle. Similar to mitochondria, chloroplasts also form a barrier between two compartments within the chloroplast. These compartments are the thylakoid lumen, which is a continuous aqueous chamber inside the thylakoid membrane, and the stroma. Similar to the inner mitochondrial membrane, chloroplasts contain their own DNA and carry out protein synthesis within the organelle.

3. Light absorption by plant pigments, such as chlorophyll, results in electron excitation, which lifts the electron to a higher orbital. If the photon is absorbed by pigments in the PSII or PSI reaction centers, then an electron is transferred to a nearby electron acceptor. The process is called photooxidation because the chlorophyll molecule is oxidized after excitation by light and the acceptor molecule is reduced. Oxidation of H2O by the O2evolving complex replaces the lost electron in chlorophyll, which returns this light-absorbing pigment to the ground state. In contrast to photooxidation, absorption of light by chlorophylls in the light-harvesting complexes results in energy transfer to a nearby chlorophyll molecule by a process called resonance energy transfer. There are ∼300 energy transfer reactions by lightharvesting complexes for every photooxidation reaction at PSII and PSI reaction centers.

4. The net equation for photosynthetic electron transport is 2 H O + 8 photons + 2 NADP+ + ∼3 ADP + ∼3 P → 2

i

O2 + 2 NADPH + ∼3 ATP

The photosynthetic electron transport system consists of three transmembrane proteins (PSII, cytochrome b6f, and PSI) and three soluble proteins (plastocyanin, ferredoxin-NADP+ reductase, and ferredoxin). PSII contains 20 protein subunits, ∼40 pigment molecules, two plastoquinones (PQ A and PQ B), and the O2-evolving complex containing a single Mn4 cluster.

PSII undergoes a photooxidation reaction in which it absorbs 4 photons at 680-nm wavelength and transfers electrons to pheophytin. The cytochrome b6 f complex functions as the proton pump in the system and transfers 8 H+ from the stromal side of the thylakoid membrane to the thylakoid lumen.

Plastocyanin carries the electrons from cytochrome b6f to PSI. PSI contains 18 protein subunits, ∼190 pigment molecules, two phylloquinones, and three 4 Fe–4 S clusters. PSI undergoes a photooxidation reaction in which it absorbs 4 photons at 700 nm and transfers electrons to nearby chlorophyll molecules.

5. In cyclic photophosphorylation, the electron normally transferred from PSI to ferredoxin in standard photophosphorylation is transferred instead to plastoquinone (PQ). Cyclic photophosphorylation provides for the maintenance of an electrochemical proton gradient by cytochrome b6f and ATP synthesis; however, less NADPH is generated and the ATP-to-NADPH ratio

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increases. Altering the balance of electron transfer through cyclic photophosphorylation and standard photophosphorylation facilitates metabolic regulation of biosynthetic pathways, depending on their need for ATP and NADPH. 6. The PSI and ATP synthase complexes are concentrated in unstacked lamellar regions of the thylakoid membranes, whereas PSII complexes are mostly localized to the stacked membranes (grana). When PSI is not being oxidized by light absorption, then plastocyanin and PQ BH2 levels build up because PSI remains in the reduced state. In order to more readily oxidize PSI to stimulate electron flow, light-harvesting complex II (LHC II) is phosphorylated, which results in the redistribution of LHC II molecules to the lamellae, where they can harvest photons for PSI. Increased levels of photooxidation by PSI complexes not only stimulates electron flow through PSII by oxidizing plastocyanin but also leads to higher levels of cyclic photophosphorylation and ATP synthesis when NADP+ levels become limiting.

7. It is incorrect to refer to the Calvin cycle as the dark reactions because flux through the Calvin cycle pathway is severely restricted during the night in the absence of light. The reason is that photosynthetic electron transport is required to supply the Calvin cycle with sufficient ATP and NADPH to maintain flux through this energy-dependent carbon fixation pathway. Moreover, light is required to activate many of the enzymes in the Calvin cycle. Therefore, in the absence of light, flux through the Calvin cycle is severely restricted because of limiting ATP and NADPH, as well as low enzyme activity.

8. Stage 1 consists of a CO2 fixation reaction by the enzyme ribulose-1,5-bisphosphate carboxylase/oxygenase (rubisco), in which ribulose-1,5-bisphosphate is combined with CO2. In stage 2, the 3-phosphoglycerate is reduced to glyceraldehyde-3-phosphate. In stage 3, the ribulose-1,5bisphosphate starting material is regenerated through a series of carbon shuffle reactions that recombine C3 molecules to produce C5 molecules with no net loss of carbon atoms.

9. Calvin cycle enzyme activity is controlled by (1) inhibitor proteins that are present at higher levels in the dark than in the light; (2) elevated pH and increased Mg2+ levels in the stroma, which stimulate enzyme activity when photosynthetic electron transport is fully active; and (3) thioredoxin-mediated activation of Calvin cycle enzymes through reduction of disulfide bridges.

10. The rubisco oxygenase reaction is wasteful because it produces 2-phosphoglycolate, which must be converted to 3-phosphoglycerate through the glycolate pathway. The glycolate pathway requires numerous enzyme reactions in chloroplasts, peroxisomes, and mitochondria, which convert 2-phosphoglycolate into 3-phosphoglycerate with the investment of 1 ATP.

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11. The term C4 plant refers to plants that store CO2 in the form of malate, a C4 metabolite, under conditions when rubisco oxygenase rates are elevated because of high temperatures and increased relative levels of soluble O2 compared to soluble CO2 (a high O2-to-CO2 ratio). C4 plants have adapted to growing in high-temperature climates by trapping CO2 in the form of malate, and thereby limiting the exposure of rubisco to elevated O2. When conditions are favorable for carbon fixation by rubisco, malate is decarboxylated, and the released CO2 is used by rubisco to generate 3-phosphoglycerate. Two variations of C4 metabolism in plants have been described: (1) the Hatch–Slack pathway in tropical plants that use two separate cell types, one for CO2 uptake and the other for rubisco-mediated carboxylation; and (2) the CAM pathway found in desert succulents, which captures CO2 in the form of malate at night when transpiration rates are low, and then releases the CO2 during the day when the Calvin cycle is favored.

The key to CAM metabolism is that plant stomata are open at night to capture the CO2 but are closed during the day to prevent O2 from entering the plant cell and stimulating the oxygenase reaction.

12. The glyoxylate cycle is a metabolic pathway in plants that converts acetyl-CoA into succinate, which serves as a carbon source for glucose biosynthesis. Animals do not have glyoxylate cycle enzymes and cannot convert acetyl-CoA into glucose. The glyoxylate cycle provides a mechanism for fats stored in plant seeds to be converted to sucrose for export to developing plant tissues prior to the onset of photosynthesis. Two unique glyoxylate enzymes, isocitrate lyase and malate synthase, are localized to cell organelles called glyoxysomes in plant cells and are also present in bacteria. Succinate serves as the central metabolic intermediate in this pathway by transporting the four carbons obtained from two acetyl-CoA molecules to the mitochondria where succinate is converted to malate, which is then exported to the cytosol and oxidized to oxaloacetate, a substrate for glucose synthesis by gluconeogenesis.

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Long Answers to Challenge Problems Chapter 12 1. a. Both PSI and PSII and the mitochondrial electron transport system capture redox energy and convert it to proton-motive force. b. ATP and NADPH. c. CO2. d. Plant cells do contain mitochondria because they need to generate high levels of ATP at night to sustain metabolic activity.

Mitochondria produce more ATP (30 ATP) than glycolysis (2 ATP) for every glucose metabolized.

2. The diagram below shows seven boxes for mitochondrion labels and seven ovals for chloroplast labels.

Cytochrome c Plastocyanin

D Stroma

F

Lumen Matrix

Intermembrane space

3. The oxidation of 2 H2 O to form O2 + 4 H+ + e−.

4. Plastocyanin must be oxidized by the redox reactions of PSI in response to light. Therefore, if electron flow through cytochrome 5

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b6f is inhibited because no oxidized plastocyanin is available, then proton translocation is also inhibited.

5. The oxidation of 2 H2O generates 4 e−; however, 4 photons are required at each of the PSII and PSI reaction centers to move the electrons through the photosynthetic electron transport system.

6. The electron carrier molecule that accepts the electron from chlorophyll must have a higher reduction potential (the standard reduction potential is more positive than chlorophyll).

7. The oxidation of H2O generates the electron that reduces chlorophyll.

8. Paraquat functions as a strong oxidant (electron acceptor) that prevents electrons from reaching the final electron acceptor in the PSI electron transport system (ferredoxin-NADP+ reductase).

9. The C5 substrate is ribulose-1,5-bisphosphate, and the C3 products are two molecules of 3-phosphoglycerate. Rubisco catalyzes the majority of carbon fixation reactions, and its abundance reflects its importance and the fact that it only slowly catalyzes this reaction. It is found in plants as well as algae and cyanobacteria.

10. One mechanism is increased stromal pH and Mg2+ concentration, which activates some Calvin cycle enzymes as a result of increased flux through the photosynthetic electron transport system. A second mechanism is thioredoxin-mediated reduction of disulfide bonds.

11. Photorespiration is a “wasteful” reaction because 2-phosphoglycolate must be salvaged by the glycolate pathway at the expense of ATP hydrolysis.

Therefore, the C4 plant has an advantage in the heat of summer because it can minimize loss of carbon to photorespiration when CO2-to-O2 ratios are low (elevated soluble O2). However, the C4 pathway requires an investment of ATP to regenerate metabolites in a reaction that is not required by C3 plants. Therefore, at low temperatures, when the CO2-to-O2 ratio is still relatively high, this extra ATP investment by C4 plants comes at an energy cost, thereby giving C3 plants a growth advantage.

12. The two reasons are that ATP and NADPH reserves will be quickly depleted, and the Calvin cycle enzymes, most importantly rubisco, are inactivated in the dark.

13. The disulfide bridges are spontaneously oxidized in the absence of reduced thioredoxin.

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Long Answers to Review Questions Chapter 13 1. Glycans can be divided into three major groups: (1) simple sugars consisting of monosaccharides, disaccharides, and oligosaccharides, which primarily function as readily available chemical energy when obtained as dietary nutrients or as the sugar moiety in nucleotides, including ATP; (2) polysaccharides consisting of glucose homopolymers or disaccharide heteropolymers containing a hexosamine, which primarily function as forms of stored energy (starch and glycogen) or as structural elements in multicellular organisms (cellulose and chitin); and (3) glycoconjugates of proteins or lipids with covalently attached glycans, which are part of the proteoglycan cell walls of bacteria and extracellular matrix materials in animals (aggrecan, syndecan).

2. Milk oligosaccharides contain a core lactose disaccharide and between 3 and 22 sugar residues. Milk oligosaccharides are degraded by bacterial flora in the infant’s intestinal tract and are thought to function as either probiotics by supporting the growth of beneficial bacteria or as soluble glycan decoys, which are structurally related to glycoconjugates on the surface of intestinal epithelial cells and prevent pathogenic bacteria from binding to these cells. A second abundant oligosaccharide in nature is the raffinose series of plant oligosaccharides—raffinose, stachyose, and verbascose—all of which are derived from sucrose. Most mammals cannot digest raffinose oligosaccharides because they lack the necessary α-galactosidase enzyme needed to hydrolyze the α-1,6 glycosidic bond. Therefore, eating foods high in raffinoseseries oligosaccharides can lead to gastrointestinal discomfort because the undigested carbohydrates are metabolized by intestinal bacteria, which do contain α-galactosidase and ferment the glycans to produce methane, carbon dioxide, and hydrogen gases.

3. Plant cell walls consist of cellulose, which is a glucose polysaccharide linked together by β-1,4 glycosidic bonds.

1

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Individual cellulose strands are extensively hydrogen bonded to build a polysaccharide fibril that is hydrogen bonded to the polysaccharides hemicellulose and pectin. Hemicellulose is a branched polysaccharide consisting of up to six different sugar residues, whereas pectin is a homopolymer of galacturonic acid.

4. Starch and glycogen contain linear branched glucose polymers consisting of α-1,4 and α-1,6 glycosidic bonds. Plants synthesize two forms of starch: amylose, a linear polysaccharide containing ∼100 glucose units linked by α-1,4 glycosidic bonds, and amylopectin, an α-1,6–branched glucose polymer that has the same molecular structure as glycogen. Amylose is a linear homopolymer of glucose that forms a stable left-handed helix stabilized by intrastrand hydrogen bonds. The helical amylose structure has six glucose residues per turn. The presence of α-1,6 bonds in amylopectin and glycogen creates branch points that greatly increase the number of free ends in the homopolymeric molecule. Amylopectin contains an α-1,6 glycosidic bond about once every 30 glucose units, whereas glycogen contains the same α-1,6 glycosidic bond about once every 10 residues. Moreover, whereas amylopectin contains one free glucose at the reducing end, glycogen contains a covalently linked protein anchor called glycogenin.

5. Lectins are proteins that recognize and bind to glycan groups on glycoconjugates. Lectins mediate two types of glycoconjugate binding interactions in human immune cells: (1) intrinsic glycoconjugate binding between glycans and lectins on human cells; and (2) extrinsic glycoconjugate binding between glycans and lectins on human cells and pathogen cells.

Intrinsic glycoconjugate interactions mediate cell recognition functions on immune cells and also facilitate cell attachment and cell migration.

Extrinsic glycoconjugate interactions between immune cells and invading pathogens, such as bacteria, viruses, or fungi, protect host cells from infection by inducing immune responses that lead to the release of antibodies and hydrolytic enzymes that neutralize or kill the invading pathogen.

6. Carbohydrate linkages to glycoproteins occur through either the amide nitrogen atom of asparagine, leading to the generation of N-linked oligosaccharides, or through the oxygen atom of serine or threonine residues, resulting in O-linked oligosaccharides. The most common N-glycosidic bond in glycoproteins is between asparagine and GlcNAc, although not all asparagine residues in proteins can be N-glycosylated. O-glycosidic bonds do not require a preferred amino acid recognition sequence, and the most common monosaccharide used to create the O-glycosidic bond is GalNAc. Both N-linked and O-linked glycoproteins contain one of several core glycan structures, which serve as scaffolds for the addition of a variety of monosaccharides using different glycosidic bonds.

7. The human ABO blood groups are determined by genetic variants of the glycosyltransferase enzymes, which attach either a GalNAc or Gal sugar residue to a glycan subgroup called the O antigen, present on glycoproteins and glycolipids on red blood cells. The α-1,3-Nacetylgalactosaminyltransferase (GTA) variant of the enzyme attaches GalNAc to the O-antigen glycan group, an N-acetyllactosamine

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disaccharide, giving rise to A-type blood, whereas the α-1,3galactosyltransferase (GTB) variant attaches Gal in the same position to specify B-type blood. If the glycosyltransferase enzyme is absent or nonfunctional in an individual, then the O-antigen glycan group is not modified, and the person has O-type blood.

If an individual inherits one copy of the GTA variant from one parent and one copy of the GTB variant from the other parent, then that person expresses the AB blood group on their red blood cells.

8. Penicillin blocks bacterial cell wall biosynthesis by inhibiting the enzyme transpeptidase, which is required to form the oligopeptide linkages between hexosamine polysaccharide chains in the peptidoglycan layer. The mechanism of action of penicillin involves the formation of a suicide inhibitor complex between a serine residue in transpeptidase and a carbonyl carbon in the β-lactam ring of penicillin. Without sufficient amounts of enzymatically active transpeptidase to support cell wall biosynthesis during cell division, the penicillin-treated bacteria die. Some bacteria are resistant to penicillin because they produce an enzyme called β-lactamase, which hydrolyzes the β-lactam ring so that it is no longer a suicide inhibitor substrate for transpeptidase. Methicillin is a penicillin derivative that is not hydrolyzed by β-lactamase and is therefore a potent β-lactam antibiotic even in bacteria that have developed resistance to penicillin. Methicillin resistance is due to expression of a variant transpeptidase enzyme that does not bind methicillin.

9. The two primary objectives of glycobiology research are (1) identification of glycan group structures on purified glycoproteins using liquid chromatography and mass spectrometry, and (2) applications of high-throughput, array-based screening assays to identify biologically relevant glycan binding interactions.

10. Structural characterization of glycans on glycoconjugates is technically challenging because of three primary factors: (1) glycans have multiple bonding arrangements between as many as 11 different sugars; (2) the presence of sugar stereoisomers with identical masses; and (3) glycan structures can differ in subtle ways between identical classes of glycoconjugates. The two approaches commonly used to address these challenges are liquid chromatography in combination with specific glycosidase enzymes and mass spectrometry, which is able to detect mass differences to very high accuracy.

11. Lectin arrays contain hundreds of different lectin proteins covalently attached to a solid support and arranged in a grid.

Three types of experimental samples are routinely screened with lectin arrays: (1) fluorescently labeled glycoproteins isolated from cell extracts; (2) fluorescently labeled glycoconjugates on the surface of eukaryotic viruses; and (3) bacteria expressing a

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CHAPTER 13 LONG ANSWERS TO REVIEW QUESTIONS

fluorescent protein (GFP). Laser scanning of the array after incubation with the experimental sample generates a pattern of positive and negative signals that are interpreted on the basis of the known location of specific lectins. Antibody arrays can be used in glycobiology to identify specific glycan groups or the protein moiety of glycoproteins. When glycan binding antibodies are used to detect fluorescently labeled glycoproteins in an experimental sample, the antibody array functions in much the same way as a lectin array. Antibodies that recognize the protein component of a glycoprotein, rather than the glycan groups, can be used to investigate glycan modifications on glycoproteins when combined with fluorescently labeled lectins of known specificity. Glycan arrays are used to analyze the complete set of cellular glycan groups in a sample and are constructed using HPLC fractions of cell extracts that are covalently linked to solid surfaces through protein-specific linkages. The associated glycans are recognized by fluorescently labeled lectins or glycan-specific antibodies. Glycan arrays can also be constructed using chemically synthesized glycan groups of known structure. A synthetic glycan array is also useful for investigating the glycan specificity of purified lectins and monoclonal antibodies on the basis of their affinities for different glycan groups on the array.

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Long Answers to Challenge Problems Chapter 13 1. Water disrupts the intrafiber hydrogen bonds that hold the cellulose fibers together, whereas oil has no effect on hydrogen bonds.

2. The trisaccharide has many more possible structures. For example, 3 amino acid residues in a tripeptide can have a maximum of 6 possible sequences (3 × 2 × 1), all in a linear chain connected by peptide bonds. In contrast, a trisaccharide can have a minimum of 384 different structures. Consider that the trisaccharide has 6 possible linear sequences, but with multiple arrangements of glycosidic bonds possible at the two linkage sites.

Specifically, each of the two glycosidic bonds can be an α or β linkage at C-1, which is connected to the hydroxyl group at C-2, C-3, C-4, or C-6. In total, 6 linear sugar arrangements multiplied by 2 bond arrangements (α or β) of the first glycosidic bond, which can be made with connections to any one of 4 carbons of the second sugar, and multiplied by 2 bond arrangements (α or β) of the second glycosidic bond, which can be made with connections to any one of 4 carbons of the third sugar, gives us 6 × 2 × 4 × 2 × 4 = 384.

3. a. Pigs and other mammals lack the enzyme α-galactosidase and therefore cannot break the glycosidic bonds in raffinose oligosaccharides to metabolize this carbohydrate fully. By predigesting the raffinose oligosaccharides with α-galactosidase, the feed has more nutritional value (broken down to simple sugars), and the pigs gain more weight in less time. b. The flatulence comes from intestinal bacteria that digest the raffinose oligosaccharides contained in broccoli, cabbage, and soybeans using their own α-galactosidase enzyme. Some of the products of anaerobic bacterial metabolism are hydrogen and methane gas. In contrast, potatoes, squash, and corn contain 5

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low amounts of raffinose but high amounts of starch, which is easily metabolized by animals using the enzyme α-amylase.

4. a.

CH2OH

CH2OH

H

HO

CH2

CH2OH

O H

b.

0.5 g 162 g/mol

HO

O H

HO

H

HO

= 3.0 × 10−3 mol of glucose in sample

0.025 g = 1.2 × 10−4 mol of 2,3-dimethylglucose in sample 208 g/mol 1.2 × 10−4 mol × 100% = 4.0% of the sample is 2,3-dimethylglucose 3.0 × 10−3 mol 100% = 25:1 α-1,4 glycosidic bonds compared to α-1,6 glycosidic bonds 4% c. 3,000 glucose residues with a branch point every 25 residues = 120 nonreducing ends.

5. Catalytic activity is not the only consideration for determining the in vivo function of an enzyme. First, the glycan groups may be required for targeting the enzyme to the proper secretion or folding pathway inside the cell, which would not be required in an in vitro protein synthesis system.

Second, the glycan groups may be required for functional associations with the extracellular matrix, which were not tested in the in vitro enzyme assay.

Third, the glycan groups may determine serum survival time for the enzyme, and again, this was not tested.

6. a. The A blood group contains a terminal GalNAc residue on the core glycan group called the O antigen. The B blood group contains a terminal Gal residue in place of GalNAc, whereas half of the AB blood group glycans contain a terminal GalNAc residue and the other half contain a terminal Gal residue. Lastly, the O blood group glycan lacks a terminal residue on the O-antigen glycan group. b. Humans contain three functional variants of the glycosyltransferase enzyme that differ in key amino acid residues in the enzyme active site.

The α-1,3-N-acetylgalactosaminyltransferase (GTA) variant of the enzyme binds UDP-GalNAc in the enzyme active site and catalyzes a reaction linking GalNAc to the O-antigen glycan group; the α-1,3galactosyltransferase (GTB) variant of the enzyme binds UDP-Gal in the enzyme active site and catalyzes a reaction linking Gal to the O-antigen

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glycan group; and the third variant of the enzyme is defective, which gives rise to the O blood group consisting of the O-antigen glycan group.

Individuals containing paternal and maternal alleles corresponding to the GTA and GTB variants have the AB blood group. c.

Blood group

Allele 1 (paternal or maternal)

Allele 2 (paternal or maternal)

A blood group

GTA

Defective enzyme or GTA

B blood group

GTB

Defective enzyme or GTB

AB blood group

GTA

GTB

O blood group

Defective enzyme

Defective enzyme

d. Type O blood group is the most useful at blood banks because type O red blood cells lack both the GTA and GTB glycan groups, so they do not stimulate an immune reaction when used in otherwise compatible transfusions. In contrast, type AB blood group is the best to have when in need of a transfusion because serum in type AB individuals lacks anti-A and anti-B antibodies, so blood can be donated from any group and it won’t stimulate an immune reaction.

7. a. Heparin binds with high affinity to antithrombin, which activates the inhibitory activity of antithrombin. This leads to inhibition of the bloodclotting cascade through antithrombin inactivation of thrombin and Factor Xa. b. Protamine sulfate is positively charged and binds to the negatively charged heparin, which prevents heparin from binding to and activating antithrombin. This leads to inactivation of antithrombin and normal regulation of the blood-clotting cascade.

8. a. Fleming noticed that bacteria failed to grow in an area surrounding a large mold colony on the plate, and he hypothesized that the mold secreted an antibacterial compound. Fleming identified the compound and named it penicillin after the mold that produces the antibiotic: Penicillium notatum. b. Penicillin targets the enzyme transpeptidase, which is required for bacterial cell wall synthesis. If bacteria cannot synthesize a functional cell wall at each cell division, the cell cannot divide and proliferate. c. The most common mechanism of penicillin resistance in bacteria is expression and secretion of the enzyme β-lactamase, which cleaves the lactam ring of penicillin and destroys its transpeptidase-inhibiting activity. d. MRSA is resistant to a wide range of antibiotics, including methicillin, which until recently was one of the most potent antibiotics available.

Methicillin resistance in MRSA is due to expression of a functional transpeptidase enzyme that does not bind methicillin or penicillin. This methicillin-resistant transpeptidase was acquired by lateral gene transfer from another bacterial species.

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9. After releasing the N-linked glycan group from the antibody-bound glycoprotein with PNGaseF, the purified glycan would be labeled with the fluorescent dye 2-AB for use in HPLC analysis. In this method, the glycan is treated sequentially with different glycosidases having known specificities, and the reaction products after each enzyme digestion are then separated

by HPLC. The elution profiles of the glycan products are compared to the elution profiles of glucose polymers of known sizes. By combining the information gained from glycosidase specificities with the size of digestion products as determined by elution profiles, a probable glycan structure is proposed.

A second method to determine the glycan structure is to use mass spectrometry to determine the precise mass of the glycan before and after various treatments, which could be enzyme digestion or, more likely, fragmentation by electrospray ionization (ESI) in combination with collision-induced dissociation (CID). On the basis of predicted molecular weights of various glycan derivatives compared to observed molecular weights, a probable glycan structure is proposed and compared to that predicted from HPLC. 10. a. A synthetic glycan array could be constructed containing variations of common glycan groups most likely to be present on uroplakin. By comparing the binding capacity of fluorescently labeled pathogenic (expresses FimH) and nonpathogenic E. coli strains to the glycan array, it should be possible to identify one or more glycan groups on the array with selective binding to FimH-expressing pathogenic E. coli strains. If glycan structural analysis of uroplakin were available using HPLC or mass spectrometry, the predicted glycan structure should be consistent with results from the unbiased glycan array. b. A glycan array could be constructed containing the FimH target glycan, as well as other glycans that serve as positive and negative controls.

Positive controls would be glycans known to be binding sites for E. coli lectins not associated with pathogenicity. This would control for overall binding of bacteria to the array in case the drug candidate nonselectively blocked all binding. In contrast, negative controls would be structurally related glycans that are known to have only very low affinity for FimH, which should not be affected by the presence of drug candidates in the binding reaction.

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Long Answers to Review Questions Chapter 14 1. Although NADH and NADPH each function as 2 e− carriers in metabolic reactions, the phosphate group on C-2 of the ribose ring of NADPH distinguishes it from NADH and is recognized by enzymes in redox reactions involved in biosynthetic pathways and detoxification. NADH is more often used as a strong reductant in energy-converting reactions and other catabolic pathways.

2. The three conditions are (1) low NADPH, (2) low nucleotides, and (3) low ATP. 3. Glucose-6-P + NADP+ m 6-Phosphogluconolactone + NADPH + H+. Glucose-6-phosphate dehydrogenase catalyzes this reaction, which has ΔG °′ = –17.6 kJ/mol.

4. There are 12 molecules of NADPH generated from 6 molecules of glucose-6-P in the oxidative phase of the pentose phosphate pathway, which gives rise to 5 molecules of glucose-6-P in the nonoxidative phase. The six carbons are lost as CO2 during the oxidative phase.

5. The enzyme glucose-6-phosphate dehydrogenase is activated by NADP+, signaling a need for increased NADPH production. The same enzyme is also inhibited by NADPH, which decreases flux through this rate-limiting reaction.

6. Reduced glutathione protects red blood cells from oxidizing agents by functioning as a strong reductant in redox reactions such as those catalyzed by glutathione reductase. Reduction of glutathione by the enzyme glutathione reductase requires NADPH, which is primarily generated in red blood cells by the glucose-6-phosphate dehydrogenase reaction in the pentose phosphate pathway.

Individuals with a glucose-6-phosphate dehydrogenase deficiency are sensitive to primaquine because it elevates reactive oxygen species (ROS) in red blood cells beyond a safe level, as their 1

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glutathione detoxification mechanisms are compromised. However, elevated basal levels of ROS in individuals with a glucose-6-phosphate dehydrogenase deficiency inhibit the growth of malarial parasites in red blood cells.

7. Each mole of glucose generated by the gluconeogenic pathway requires 2 moles of pyruvate. Because dephosphorylation of GTP is counted as ATP dephosphorylation, the total number of ATP required to convert 2 pyruvate to 2 phosphoenolpyruvate is 4 ATP. 8. The pyruvate carboxylase reaction is responsible for maintaining high oxaloacetate levels in the citrate cycle under conditions of high acetyl-CoA metabolism when energy charge is low.

9. Phosphofructokinase-1 is a cytosolic enzyme in the glycolytic pathway that is allosterically inhibited by high levels of cytosolic citrate, which makes sense because it means that energy charge in the cell is high and citrate is being shuttled out of the mitochondrial matrix into the cytosol. At the same time, this high cytosolic citrate activates flux through the gluconeogenic pathway by activating the enzyme fructose-1,6-bisphosphatase-1. Reciprocal regulation of these two enzymes by citrate prevents futile cycling and net consumption of ATP. 10. Fructose-2,6-BP is a potent allosteric regulator of flux through both the gluconeogenic and glycolytic pathways and is not a metabolite in either pathway. Similar to citrate, fructose-2,6-BP is a reciprocal regulator of the gluconeogenic and glycolytic pathways; however, in this case, fructose2,6-BP activates the glycolytic enzyme phosphofructokinase-1 and inhibits the gluconeogenic enzyme fructose-1,6-bisphosphatase-1.

11. The dual-function enzyme phosphofructokinase-2/fructose-2,6bisphosphatase is a single polypeptide encoding two catalytic activities. The kinase activity encoded by the phosphofructokinase-2 domain of the enzyme is inhibited by phosphorylation on a serine residue (Ser32), whereas the phosphatase activity encoded by the fructose-2,6-bisphosphatase domain is activated by Ser32 phosphorylation.

12. Glycolysis in muscle cells generates 2 ATP from every glucose molecule that is metabolized to 2 lactate under anaerobic conditions. Conversion of 2 lactate into 1 glucose in liver cells requires the investment of 6 ATP, however, and therefore the net ATP cost of running the Cori cycle between muscle and liver cells is 4 ATP. 13. The frequency of α-1,6 glycosidic bonds is about once every 10 glucose residues, therefore, the best answer is ∼2,500 ends.

14. The high physiologic concentration of inorganic phosphate (HPO42−) in cells leads to a mass action ratio (Q) that is much less than 1 and an ln Q value less than zero. Therefore, the positive ΔG °′ value of +3.1 kJ/mol gives rise to a negative ΔG value of −6 kJ/mol using the Gibbs free energy equation expressed as ΔG = ΔG °′ + RT ln Q. 15. Glucose-6-phosphatase dephosphorylates glucose-6-P generated by the gluconeogenic pathway so that glucose can be exported to other tissues.

Muscle cells lack the enzyme glucose-6-phosphatase and therefore cannot export glucose.

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16. Generation of glucose-1-P by the glycogen phosphorylase reaction does not require an ATP investment because the phosphate group is derived from inorganic phosphate (Pi).

Moreover, because the phosphoglucomutase reaction that converts glucose-1-P to glucose-6-P is energy independent, the metabolism of glucose-6-P derived from glycogen degradation yields 3 net ATP in the glycolytic pathway compared to only 2 ATP generated by the metabolism of dietary glucose. The reason is that dietary glucose needs to be phosphorylated by hexokinase to yield glucose-6-P, which requires ATP investment.

17. Glucagon signals low blood glucose levels and activates glycogen degradation by stimulating protein kinase A signaling and subsequent phosphorylation of phosphorylase kinase, which phosphorylates and activates glycogen phosphorylase. Because protein kinase A also phosphorylates and inactivates glycogen synthase, futile cycling is prevented, and there is a net efflux of glucose out of liver cells to increase blood glucose levels.

Insulin signals high blood glucose levels and activates glycogen biosynthesis by stimulating protein phosphatase 1 activity, which dephosphorylates and activates glycogen synthase and at the same time dephosphorylates and inactivates glycogen phosphorylase.

The net result is glucose influx into liver and muscle cells to lower blood glucose levels.

18. The inability to add branches to glycogen chains drastically reduces the number of nonreducing ends, making it extremely difficult to store and retrieve glucose in liver cells to maintain safe blood glucose levels. In contrast, lack of debranching enzyme has no effect on the ability to store glucose and only a minor effect on the ability to retrieve glucose from glycogen particles, as 90% of the glucose in glycogen is linked by α-1,4 glycosidic bonds, the substrate for glycogen phosphorylase.

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Long Answers to Challenge Problems Chapter 14 1. a. Glucose-6-phosphate dehydrogenase is required in the pentose phosphate pathway to generate reducing equivalents in the form of NADPH, which is required to sustain a high level of reduced glutathione in red blood cells. Fava beans contain a toxic compound (vicine) that must be reduced by glutathione in red blood cells. Because NADPH is required to keep glutathione in the reduced state, the red blood cells in these individuals are highly susceptible to favism, a diet-induced form of anemia. b. First, geographic overlap in Africa of the regions of glucose-6phosphate dehydrogenase deficiencies and malarial resistance suggest a correlation. Second, growth of the malarial parasite Plasmodium is inhibited by an oxidizing environment in host cells. Because glucose-6-phosphate dehydrogenase deficiencies result in elevated reactive oxygen because of decreased levels of reduced glutathione, individuals with a glucose-6-phosphate dehydrogenase deficiency likely have red blood cells that are unsuitable for growth of Plasmodium.

2. Glucose-1-P is converted to glucose-6-P, which then enters the glycolytic pathway and bypasses the first investment step (hexokinase). Therefore, only 1 ATP is invested, which means that the net ATP is 3 considering that 4 total ATP are generated in glycolysis.

3. Glucose m glucose-6-phosphate: Enzymes are hexokinase and glucose-6-phosphatase; controlled by substrate availability and feedback inhibition of hexokinase. Also note most tissues lack the phosphatase.

Fructose-6-phosphate m fructose-1,6-bisphosphate: Enzymes are phosphofructokinase-1 and fructose-1,6-bisphosphatase.

Phosphofructokinase-1 is inhibited by ATP and citrate and is activated by AMP and fructose-2,6-BP; fructose-1,64

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bisphosphatase is inhibited by fructose-2,6-BP and AMP. PEP to pyruvate in glycolysis and PEP to OAA then pyruvate in gluconeogenesis. Acetyl-CoA, ATP, alanine, and cAMP-dependent phosphorylation inhibit pyruvate kinase during gluconeogenesis. AcetylCoA stimulates pyruvate carboxylase during gluconeogenesis.

4. The finding that glycerol addition to the biopsied liver sample leads to glucose production indicates that the gluconeogenic enzymes fructose1,6-bisphosphatase and glucose-6-phosphatase are normal. Therefore, the defect could be in either the gluconeogenic enzymes phosphoenolpyruvate carboxykinase or pyruvate carboxylase. The finding that malate, which is converted to oxaloacetate by malate dehydrogenase, does not lead to glucose production suggests that the defective gluconeogenic enzyme is phosphoenolpyruvate carboxykinase. If the defect were instead in pyruvate carboxylase, then malate addition would lead to glucose production.

5. This high level of NADH produced by the alcohol dehydrogenase reaction drives the lactate dehydrogenase reaction in the direction of lactate production, leading to depletion of pyruvate. Moreover, because NAD+ becomes limiting for the glyceraldehyde-3-phosphate dehydrogenase reaction, the glycolytic pathway is inhibited, further depleting pyruvate pools, and thereby inhibiting gluconeogenesis. In addition, excess NADH stimulates the malate dehydrogenase reaction in the direction of malate formation, depleting the gluconeogenic substrate oxaloacetate.

Ethanol O

Acetaldehyde

H

H

C

H

H

C

H

H Alcohol dehydrogenase

O

C

H

C

H

H

NAD+ NADH + H+

H

COO–

NADH +H+

CH3

Lactate dehydrogenase

Pyruvate

O

COO– NADH +H+

NAD+

CH2 COO– OAA

CH3 Lactate

COO– C

COO–

NAD+

HC

OH

CH2 Malate dehydrogenase

COO– Malate

6. a. The net yield is 3 ATP because one additional net ATP is produced, as isomerization of glucose-1-P to glucose-6-P generates a phosphorylated glycolytic metabolite without the investment of 1 ATP by the hexokinase reaction.

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b. This reaction requires 2 ATP equivalents because the glucose must first be phosphorylated by hexokinase to generate glucose-6-P, which is then isomerized to glucose-1-P and activated by UTP to form UDPglucose. To regenerate UTP, a second ATP is required in the nucleotide diphosphate kinase reaction. c. The overall ATP investment is 5 ATP per glucose using glycogen in muscle cells: +3 ATP yield per glucose derived from glycogen to lactate in muscle cells −6 ATP to convert 2 lactate to 1 glucose in liver cells −2 ATP to add back each glucose into glycogen in muscle cells −5 ATP net cost 7. Liver cells use glycogen as a source of glucose for export, and after isomerization of glucose-1-P to glucose-6-P, the phosphate must be removed before it can be exported. Muscle cells, however, use the glucose-6-P derived from glycogen degradation and isomerization as a source of metabolic energy for muscle contraction and need it to remain phosphorylated so that it is not exported.

8. In individuals with von Gierke disease, glucose-6-P accumulates as a result of stimulated glycogen breakdown (glucagon signals low blood sugar), but because it can’t leave the liver, glucose-6-P stimulates glycogen synthase.

Moreover, glucose-6-P derived from gluconeogenesis (also stimulated by low blood sugar) is converted to glycogen by the same mechanism (stimulation of glycogen synthase). Glucagon stimulates glycogen breakdown, and the product of glycogen debranching enzyme is free glucose, which is released into the blood (∼10% of available glucose in glycogen is contained in α-1,6 branch points).

9. Muscle glycogen phosphorylase is required to release glucose from stored muscle glycogen, which is used for energy conversion processes (ATP generation) needed to sustain muscle contraction.

10. Muscle phosphorylase is activated by AMP, which signals a low energy state in the cell. Glucose inhibits liver phosphorylase by feedback inhibition, signaling that glucose is not being exported at a high enough rate to require more glycogen degradation. Muscle phosphorylase is not inhibited by glucose because muscle cells use glucose for energy production. Degradation of glycogen in the liver is independent of the energy needs of the liver cell.

11. a. Protein kinase A activation results in the phosphorylation and inactivation of glycogen synthase, and, at the same time, the phosphorylation and activation of glycogen phosphorylase through the phosphorylation and activation of phosphorylase kinase. b. Protein kinase A activation results in decreased amounts of stored glycogen because glycogen synthase is inactivated and glycogen phosphorylase is activated. c. Glucagon signals a need for glucose in between meals, and epinephrine signals a need for glucose when there is imminent danger or need for

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quick energy. In contrast, insulin signals the fed state and a need to replenish glycogen stores by converting glucose to glycogen.

12. a. Phosphorylase kinase and glycogen phosphorylase are both activated by phosphorylation, a covalent modification. Other enzymes in the cascade are not modified covalently. b. The theoretical number of glucose-1-phosphate molecules can be calculated from: 1 glucagon receptor activates 10 Gsα−GTP 10 Gsα−GTP subunits activate 10 adenylate cyclase enzymes 10 adenylate cyclase enzymes generate 10,000 cAMP per second, but 1,000 cAMP per second are degraded by cAMP phosphodiesterase 9,000 cAMP activate 4,500 protein kinase A catalytic subunits per second 4,500 protein kinase A catalytic subunits activate 4,500,000 phosphorylase kinase subunits per second 4,500,000 phosphorylase kinase subunits per second activate 4,500,000,000 glycogen phosphorylase subunits per second 4,500,000,000 glycogen phosphorylase subunits per second generate 4,500,000,000,000 glucose-1-phosphate per second = 4.5 × 1012 glucose-1-phosphate formed per second

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Long Answers to Review Questions Chapter 15 1. The three biological roles of lipids are energy storage (fatty acids and triacylglycerols), the hydrophobic barrier of cell membranes (glycerophospholipids and sphingolipids), and endocrine signaling through receptor-mediated signaling pathways (steroids and eicosanoids).

2. The answers are contained in the following table.

Carbon bonds

Molecular formula

Common name at pH 3.0

Common name at pH 7.2

cis 18:1(Δ9)

CH3(CH2)7CH=CH(CH2)7COOH

Oleic acid

Oleate

18:0

CH3(CH2)16COOH

Stearic acid

Stearate

cis 16:1(Δ9)

CH3(CH2)5CH=CH(CH2)7COOH

Palmitoleic acid

Palmitoleate

16:0

CH3(CH2)14COOH

Palmitic acid

Palmitate

cis 18:2(Δ9,12)

CH3(CH2)4CH=CHCH2CH=CH(CH2)7COOH

Linoleic acid

Linoleate

trans 18:1(Δ9)

CH3(CH2)7CH=CH(CH2)7COOH

Elaidic acid

Elaidate

3. Triglycerides have a higher stored energy potential per gram because fats are essentially unhydrated compared to glycogen, which is very hydrated.

4. The three major lipids in beeswax are palmitic acid ester of myricyl alcohol, hexaeicosanoic acid ester of myricyl alcohol, and palmitic acid ester of cetyl alcohol. The fully saturated long hydrocarbon chains of wax esters in beeswax pack tightly together, leading to an elevated melting point. In contrast, jojoba oil consists of large amounts of erucic acid ester of erucyl alcohol, which contains monounsaturated hydrocarbon chains that pack less tightly, resulting in a much lower melting point.

5. Examples of glycerophospholipids in eukaryotic plasma membranes are phosphatidylcholine, phosphatidylserine, phosphatidylethanolamine, and phosphatidylinositol. Examples 1

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of sphingolipids in eukaryotic plasma membranes are sphingomyelin, cerebrosides, and gangliosides. The cholesterol content of most plasma membranes is 25% to 40% of the membrane lipid. Cholesterol is a small membrane lipid that disrupts the packing of membrane lipids with long hydrophobic tails. Low concentrations of cholesterol in lipid bilayers increase membrane fluidity, whereas at very high cholesterol concentrations, membrane fluidity decreases because of the packing of cholesterol with itself.

6. Glycerophospholipids are important sources of fatty acid–derived signaling molecules, which are released from membrane lipids by hydrolytic cleavage reactions catalyzed by phospholipase enzymes. For example, phospholipase C cleaves the ester bond at C-3 in phosphatidylinositol-4,5bisphosphate (PIP2) to generate the signaling molecules inositol-1,4,5trisphosphate (IP3) and diacylglycerol (DAG). In addition, phospholipase A2 cleaves glycerophospholipids containing arachidonate, which is an important mediator of inflammatory signaling.

7. Ceramide is the precursor biomolecule of all sphingolipids and consists of sphingosine containing an attached fatty acid. The linkage of phosphocholine to ceramide generates the abundant membrane lipid sphingomyelin, which is technically a phospholipid chemically similar to phosphatidylcholine.

8. The glycan structures of ABO gangliosides and glycoproteins are attached to membrane lipids and proteins, respectively, by the same two variant GTA and GTB glycosyltransferase enzymes. Therefore, the blood group antigen profile on erythrocytes in any one individual consists of the same four combinations (A, B, AB, O) regardless of the membrane-anchoring biomolecule.

9. Progesterone prepares the uterus for implantation of an ovum; androgens regulate development of male secondary sex characteristics; estrogens regulate development of female secondary sex characteristics; glucocorticoids regulate gluconeogenesis and fat mobilization; and mineralocorticoids regulate salt balance and kidney functions.

10. Drugs that function as steroid agonists bind to steroid hormone receptors and activate the same biological response as the natural steroid hormone.

Drugs that function as steroid antagonists inhibit the biological response by binding and inactivating the steroid receptor or competing for binding with the natural steroid. Nandrolone is an example of an androgen agonist that mimics dihydrotestosterone and stimulates muscle growth. Bicalutamide is an example of an androgen antagonist that blocks dihydrotestosterone signaling in prostate cancer growth.

11. Ultraviolet light activates a pathway converting 7-dehydrocholesterol present in skin cells to vitamin D3, also called cholecalciferol. Humans also obtain vitamin D3 directly in their diet from vitamin D–fortified foods, as well as from fish and dairy products, which are rich in this nutrient. Regardless of the source of vitamin D3, it is first converted to 25-hydroxyvitamin D3 by the

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enzyme P450C25 hydroxylase in liver cells, and then the active form of vitamin D, called 1,25-hydroxyvitamin D3, is synthesized in kidney cells by the enzyme P450C1 hydroxylase.

12. The four major classes of arachidonate-derived eicosanoids are prostaglandins, prostacyclins, thromboxanes, and leukotrienes.

Prostaglandins play a primary role in regulation of blood flow by modulating smooth muscle contraction and relaxation, and they also stimulate inflammatory responses and modulate the secretion of proteoglycans that protect the stomach lining from the effects of low pH. Prostacyclins control platelet aggregation and blood clot formation and stimulate vasodilation. Thromboxanes regulate blood vessel constriction. Leukotrienes are pro-inflammatory molecules that also regulate smooth muscle contraction.

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Long Answers to Challenge Problems Chapter 15 1. a. Amount of stored fat = 90 kg × 0.15 = 13.5 kg (13.5 kg) × (1,000 g/kg) × (37 kJ/g) = 5 × 105 kJ b. If this amount of energy were stored as hydrated glycogen, the man would weigh: (5 × 105 kJ) × (g/6 kJ) = 8 × 104 g = 80 kg of glycogen – 13.5 kg of fat = 66 kg of extra weight, which makes him weigh 90 kg + 66 kg = 156 kg = 344 lb 2. a. The molecular mass of KOH is 56 g/mol, therefore the number of moles of KOH required is 193.2 g 56 g/mol

= 3.5 mol

Because 3 mol of KOH are required to saponify each mole of triacylglycerol, there are three times more moles of KOH needed than there are moles of triacylglycerols. Therefore, the number of moles of triacylglycerols in the sample of beef fat must be 3.5 mol = 1.2 mol of triacylglycerols 3 The average molecular mass (MM) of the triacylglycerols in 1 kg of beef fat is MM =

Mass in grams 1,000 g = = 833 g/mol Number of moles 1.2 mol

b. Start with the molecular masses of the fatty acids and glycerol component: MM of palmitate = 255.4 g/mol MM of stearate = 283.5 g/mol

4

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5

MM of oleate = 281.5 g/mol MM of glycerol component of triacylglycerol (C3H5) = 41 g/mol Because a triacylglycerol contains three fatty acids, you need to calculate the total molecular mass of four triacylglycerol molecules containing the following distribution of fatty acids: 6 oleate : 3 palmitate : 3 stearate, which reduces to 2 oleate : 1 palmitate : 1 stearate. triacylglycerol 1 = Oleate + Stearate + Palmitate + C3H5 = 281.5 + 283.5 + 255.4 + 41 = 861.4 g/mol triacylglycerol 2 = Oleate + Palmitate + Oleate + C3H5 = 281.5 + 255.4 + 281.5 + 41 = 859.4 g/mol triacylglycerol 3 = Oleate + Palmitate + Stearate + C3H5 = 281.5 + 255.4 + 283.5 + 41 = 861.4 g/mol triacylglycerol 4 = Oleate + Stearate + Oleate + C3H5 = 281.5 + 283.5 + 281.5 + 41 = 887.5 g/mol Add these molecular masses and divide by 4 to get the average molecular mass: triacylglycerol 1 + triacylglycerol 2 + triacylglycerol 3 + triacylglycerol 4 3,469.7 = = 867.4 g/mol 4 c. Yes, these calculations are consistent with the measured values listed in Table 15.1 showing that oleate is the most abundant fatty acid in beef tallow at 43%, with 24% palmitate and 19% stearate. The average molecular mass of ∼870 g/mol based on saponification (part 2b) is similar to an average molecular mass based on an approximate fatty acid ratio of 2 oleate : 1 palmitate : 1 stearate.

3. Hepatocytes synthesize fatty acids from excess dietary carbohydrate and protein-derived metabolites and assemble them into triacylglycerols using three fatty acids and glycerol. These neutral lipids are packaged into very-low-density lipoproteins and exported into the circulatory system, where they are used as an energy source by the peripheral tissues. In contrast, adipocytes store triacylglycerols in lipid droplets and release free fatty acids into the circulatory system in response to hormonal signaling. The free fatty acids are transported to peripheral tissues by serum albumin protein.

4. The bacteria grown at higher temperature will have the higher ratio of saturated-to-unsaturated fatty acids. They need more saturated fatty acids to raise the transition temperature of their membrane lipids because of the higher ambient temperature.

5. Moles of glycerophospholipids = 3.7 × 10−4 mol Moles of sphingoglycolipids = 1.2 × 10−4 mol Moles of sphingomyelin = 7.3 × 10−5 mol Moles of cholesterol = 4.0 × 10−4 mol

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Sample Calculation The relative amounts of lipid from Figure 15.25 is equal to 17 + 3 + 8 + 9 = 37.

Therefore, the amount of glycerophospholipids = 17/37 = 0.46 × 650 mg = 299 mg.

The average molecular mass (MM) of glycerophospholipid is 800 g/mol (calculated average from MM of phosphatidylcholine, phosphatidylethanolamine, phosphatidylinositol, and phosphatidylserine).

Therefore, the number of moles of glycerophospholipid in the 650 mg membrane fraction sample is (0.299 g)/(800 g/mol) = 3.7 × 10−4 mol of glycerophospholipids. The same calculations can be used to determine the number of moles for other lipids.

Sphingoglycolipid: 8/37 = 0.22 × 650 mg = 140.5 mg of lipid (0.141 g)/(1,192 g/mol) using average MM of cerebroside and GM1 ganglioside = 1.2 × 10−4 mol of sphingoglycolipid Sphingomyelin: 3/37 = 0.22 × 650 mg = 52.7 mg of lipid (0.053 g)/(731 g/mol) using average MM of cerebroside and GM1 ganglioside = 7.3 × 10−5 mol of sphingomyelin Cholesterol: 9/37 = 0.22 × 650 mg = 158.1 mg of lipid (0.158 g)/(387 g/mol) using average MM of cerebroside and GM1 ganglioside = 4.0 × 10−4 mol of sphingomyelin 6. a. The hydrocarbon chain would span the bilayer with the free hydroxyl groups on either side. b. The compound would be stable at high temperature because of extensive van der Waals interactions and the fact that the hydrocarbon chain spanning the bilayer leads to greater stability; the ether linkages are chemically stable to hydrolysis.

7. a. The phospholipase C cleavage products are IP3 and DAG, which induce glucose export from liver cells by stimulating glycogen degradation and inhibiting glycogen synthesis (see Figure 8.29).

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O

O

acylglycerol (DAG) H2C

O

H2C

O

HC

O

HC

O

O

CH2

CH2

O

OH

P

+

O–

OPO 32–

O OPO32–

OH OH

+

H2O

OH

OPO 32– Phosphatidylinositol-4,5bisphosphate (PIP2)

OPO 32–

OH Phospholipase C (PLC)

OH

OH

OPO 32– Inositol-1,4,5-trisphosphate (IP3)

b. Cleavage of the phosphatidylinositol-4,5-bisphosphate (PIP2) membrane lipid shown in Figure 15.30 by phospholipase A2 releases arachidonate, which is the precursor molecule for the synthesis of inflammatory eicosanoids (see Figure 15.41).

8. a. Tay–Sachs disease is a neurologic disorder caused by a deficiency in the enzyme hexosaminidase A, which is required for sphingolipid metabolism.

The disease phenotype is caused by a buildup of the ganglioside GM2 substrate, which is not degraded. b. The probability of having a child with Tay–Sachs disease if both parents are heterozygous carriers is 25%. The probability of having a child that is a heterozygous carrier of the Tay–Sachs gene mutation if one parent is heterozygous and one parent is homozygous normal is 50%. c. Tay–Sachs disease is rarely seen today in the clinic because of prenatal diagnosis, which is able to detect fetuses with the disease prior to birth and provide an opportunity to terminate the pregnancy. Moreover, individuals who are identified by genetic screening as Tay–Sachs carriers often choose not to have children.

9. a. Taurocholate is an emulsifying agent in the digestive process. It works by associating with hydrophobic molecules in the diet, such as triacylglycerols and other oils. b. The structures of taurocholate and glycocholate are shown below. You can see that they are both amphipathic molecules with a nonpolar hydrophobic cholesterol ring structure and a polar head group.

This structure facilitates their function as the body’s soap by forming micelle-like structures around hydrophobic compounds in food.

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CHAPTER 15 LONG ANSWERS TO CHALLENGE PROBLEMS Glycocholate Polar O

Nonpolar OH

NH O

H OH H HO

H OH

H

Taurocholate Polar O

Nonpolar OH

NH

H O H HO

H

H

S

O

OH

OH

c. Bile soap would be more effective at removing Italian dressing stains than removing red wine stains because salad dressing contains a large amount of olive oil, which is a lipid that can be emulsified by bile acids and soap.

In contrast, stains caused by red wine are best removed with oxidizing agents such as hydrogen peroxide.

10. a. A block in the conversion of progesterone to cortisol and corticosterone increases the production of androstendione in the adrenal cortex, and in turn adrenal androgens, through both substrate availability and increased stimulation of adrenal steroidogenesis due to elevated adrenocorticotropic hormone (in response to decreased circulating cortisol). Therefore, females overproduce androgens and develop masculine features. A deficiency in 21-hydroxylase also decreases aldosterone (mineralocorticoid) production, which causes kidney malfunction and electrolyte imbalance. This condition is observed in both males and females. b. The enzyme 5α-reductase is required to convert testosterone to dihydrotestosterone, which is required during prenatal reproductive development in males and is a more potent androgenic hormone than testosterone. However, at puberty, sufficient amounts of testosterone are produced by the testes to result in an androgen surge that promotes masculinization. Females are asymptomatic because dihydrotestosterone is apparently not required for female reproductive development.

11. The growth of prostate cells is stimulated by androgens. Therefore, by blocking the activity of 5α-reductase with finasteride, production of the physiologic androgen dihydrotestosterone is inhibited and the prostate shrinks in size to provide relief from benign prostatic hypertrophy. In

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addition, androgens inhibit the growth of hair follicles, and therefore by blocking dihydrotestosterone production through inhibition of 5α-reductase activity with finasteride, androgen signaling is decreased and hair is stimulated to grow.

12. a. NSAIDs such as aspirin are generally small molecules that bind to the active sites of both COX-1 and COX-2 enzymes. In contrast, celecoxib is a large molecule that only binds to the COX-2 enzyme active site.

Because COX-1 inhibition is associated with stomach bleeding, this NSAID side effect is not seen with celecoxib. b. The rofecoxib molecule is big enough to partially bind to the prostacyclin synthase active site and inhibit enzyme activity. Aspirin does not bind to or inhibit prostacyclin synthase (the aspirin molecule is too small).

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Long Answers to Review Questions Chapter 16 1. Fatty acids are amphipathic molecules that must be transported across the inner mitochondrial membrane by the carnitine cycle.

Activation of fatty acids by the enzyme fatty acyl-CoA synthetase, in a coupled reaction with ATP, neutralizes the carboxyl group and provides a high-energy thioester linkage to facilitate covalent attachment to carnitine.

2. The four reactions needed in the mitochondrial β-oxidation pathway to remove a C2 acetyl group from a fatty acylCoA substrate are (1) an oxidation catalyzed by acyl-CoA dehydrogenase; (2) a hydration catalyzed by enoyl-CoA hydratase; (3) an oxidation catalyzed by 3-L-hydroxyacyl-CoA dehydrogenase; and (4) thiolysis catalyzed by β-ketoacyl-CoA thiolase.

3. Each molecule of stearyl-CoA yields 9 acetyl-CoA, 8 FADH2, and 8 NADH. The total yield of ATP generated from stearylCoA is 90 ATP from oxidation of acetyl-CoA in the citrate cycle and electron transport system (ETS), 12 ATP from oxidation of FADH2 in the ETS, and 20 ATP from oxidation of NADH in the ETS, which equals 122 ATP. Subtracting the 2 ATP equivalents needed to activate stearate in the cytosol with coenzyme A results in a net ATP yield of 120 ATP. 4. X-ALD is caused by a defect in a peroxisomal transport protein needed to translocate very-long-chain fatty acids (VLCFAs) across the peroxisomal membrane so they can be oxidized by the peroxisomal β-oxidation pathway. Accumulation of VLCFAs in the cytosol results in export to the circulatory system, where they cause damage to the myelin sheath surrounding neuronal cells. Lorenzo’s oil is a 4:1 mixture of glycerol derivatives of the monounsaturated fatty acids oleate and erucidate, which is prepared from olive and rapeseed oils. It is thought to decrease production of VLCFAs and thereby limit myelin sheath damage in the early stages of the disease.

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5. Ketoacidosis is caused by increased flux through the ketogenic pathway under conditions of low carbohydrate availability and a demand for energy conversion from fatty acid degradation. Abnormally high production of acetoacetate and D-β-hydroxybutyrate causes low blood pH, hypoglycemia, delirium, nausea, vomiting, and a fruity odor on the breath from high acetone levels in the blood.

6. Both pathways require a four-step reaction cycle that removes (fatty acid degradation) or adds (fatty acid synthesis) C2 units attached to coenzyme A. Fatty acid degradation differs from fatty acid synthesis by its cellular location (mitochondrial matrix), use of coenzymes (FAD and NAD+ are oxidants), large number of enzymes required, use of CoA as the acetyl group anchor, and its control by rate of fatty acyl-CoA transport across the inner mitochondrial membrane. Unique features of fatty acid synthesis are that it occurs in the cytosol, uses NADPH as a reductant, is catalyzed by a multifunctional enzyme, uses acyl carrier protein (ACP) as the hydrocarbon anchor, and its rate-limiting step is production of malonyl-CoA by the enzyme acetyl-CoA carboxylase.

7. Each cycle of the fatty acid synthesis pathway involves four reactions: (1) a condensation reaction catalyzed by β-ketoacyl-ACP synthase (KS); (2) a reduction catalyzed by β-ketoacyl-ACP reductase (KR); (3) a dehydration catalyzed by β-hydroxyacyl-ACP dehydratase (DH); and (4) a reduction catalyzed by enoyl-ACP reductase (ER).

8. The priming reaction in the first step of palmitate synthesis uses acetyl-CoA to generate the KS enzyme–bound acetyl group, which then condenses with malonyl-ACP in a KS-catalyzed decarboxylation reaction. The production of C16 palmitate requires six more malonyl-CoA to extend the C4 butyryl-ACP moiety, bringing the total malonyl-CoA required to seven.

9. Palmitate is the substrate for elongation and desaturation reactions localized to the endoplasmic reticulum. The elongation reactions use coenzyme A as the carrier molecule rather than ACP and begin with palmitoyl-CoA, which is extended with C2 groups donated by malonyl-CoA. The desaturating enzymes, called mixed-function oxidases, use molecular oxygen (O2) as the oxidant. For example, elongation of palmitate, 16:0, generates stearate, 18:0, which is then desaturated to form oleate, 18:1(Δ9).

10. Sphingomyelin is a sphingophospholipid, which makes it both a phospholipid and a sphingolipid. The phosphate group in sphingomyelin comes from attachment of phosphocholine to ceramide.

11. High levels of intracellular glucose lead to conversion of glucose to acetylCoA by the glycolytic pathway and the pyruvate dehydrogenase reaction.

Because the citrate cycle is inhibited by high energy charge, the excess acetylCoA is combined with oxaloacetate to generate citrate by the citrate synthase reaction, which is exported to the cytosol by the citrate shuttle. Citrate lyase in the cytosol reverses the citrate synthase reaction to yield acetyl-CoA and oxaloacetate. The acetyl-CoA is first converted to malonyl-CoA by acetylCoA carboxylase, and then the malonyl-CoA is used to synthesize palmitate by the fatty acid synthase complex. Triacylglycerol synthesized in the cytosol

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by glycerol metabolizing enzymes and acyltransferases is packaged into VLDL particles and exported. In order to maintain flux through these pathways, the oxaloacetate product of the citrate lyase reaction is converted to malate, which is transported back into the mitochondrial matrix, where it is used to replenish oxaloacetate. Cytosolic malate can also be converted to pyruvate by malic enzyme in an NADP+-dependent reaction, and this pyruvate is then transported into the mitochondrial matrix and converted to oxaloacetate by the enzyme pyruvate carboxykinase.

12. The four stages of de novo cholesterol biosynthesis are (1) synthesis of mevalonate; (2) conversion of mevalonate to isopentenyl diphosphate and dimethylallyl diphosphate; (3) formation of squalene from isopentenyl diphosphate and dimethylallyl diphosphate; and (4) squalene cyclization to form cholesterol.

13. The primary metabolic fates of liver cholesterol are (1) it is stored as cholesterol esters in lipid droplets; (2) it is packaged into lipoproteins and exported to the circulatory system; and (3) it is converted into bile acids, which are secreted into the small intestine through the bile duct.

14. Steady-state levels of circulating cholesterol in humans are determined by three major processes: (1) cholesterol input (diet and de novo biosynthesis); (2) cholesterol recycling (returning tissue cholesterol to the liver); and (3) cholesterol output (excretion of bile acids).

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Long Answers to Challenge Problems Chapter 16 1. The child might have a defect in the ability of chylomicrons to unload triacylglycerols in the circulatory system. The two most likely protein defects would be a defect in lipoprotein lipase, which is needed to unload triacylglycerols from chylomicrons, or a defect in the chylomicron surface protein apolipoprotein CII, which is needed to activate lipoprotein lipase.

2. Without carnitine acyltransferase, no fatty acid oxidation takes place and no acetyl-CoA is produced for ketogenesis.

Hypoglycemia results from a lack of ATP from fatty acid oxidation needed to drive gluconeogenesis during starvation.

3. Fat mobilized from adipose tissue is transported as free fatty acids bound to serum albumin, whereas dietary triacylglycerols are transported by chylomicrons.

4. Palmitate has 31 hydrogens, of which 21 are lost during β oxidation. The hydrogens that remain are the three on the methyl carbon and one each on the other even-numbered carbons, which become the methyl carbons of acetyl-CoA. As a result, the maximum expected specific radioactivity would be (10/31) × (3.0 × 108 cpm/μmol) = 0.97 × 108 cpm/μmol 5. The likely site of inhibition is carnitine acyltransferase I. Without this activity, palmitoyl-CoA cannot be converted into palmitoylcarnitine for transport into the mitochondrion by the carnitine transporter. In contrast, added palmitoylcarnitine can be directly transported.

6. The molecular mass of palmitic acid is 256 g/mol, and therefore 1 g × 1 mol/256 g = 3.9 × 10−3 mol palmitate 3.9 × 10−3 mol palmitate × 130 mol H 2O/mol palmitate = 0.5 mol H2O 0.5 mol × 18 g/mol = 9.0 g H2O. The density of water is 1 g/mL at 4 °C, therefore 9 mL of water will be generated per gram of palmitate.

4

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7. The product of odd-chain fatty acid degradation, propionyl-CoA, is converted to succinyl-CoA, which gives rise to the gluconeogenic precursor oxaloacetate via citrate cycle reactions.

8. Each mole of triglyceride contains 1 mol of glycerol, which can be converted to dihydroxyacetone phosphate, a metabolite in gluconeogenesis. Humans do not contain the necessary enzymes to convert fatty acids to gluconeogenic intermediates.

9. Palmitoyl-CoA is not transferred directly into the mitochondria; it is first converted to a carnitine ester, which then enters the mitochondria. There, the carnitine ester is replaced by mitochondrial acetyl-CoA. The radioactive CoA in the cytosol never enters the mitochondria.

10. Four key steps are required to completely oxidize pentanoyl-CoA: Step 1: Pentanoyl-CoA → Propionyl-CoA + Acetyl-CoA (4 ATP; one cycle of β oxidation → 1 FADH2 and 1 NADH) Step 2: Propionyl-CoA → Succinyl-CoA (cost is 1 ATP) Step 3: Succinyl-CoA → 4 CO2 (20 ATP; two turns of the TCA cycle) Step 4: Acetyl-CoA → 2 CO2 (10 ATP; one turn of the TCA cycle) The total is 33 net ATP (34 ATP yield − 1 ATP invested in step 2).

11. a. The liver contains only a limited amount of CoA, and therefore when β oxidation is producing acetyl-CoA faster than the citrate cycle is consuming it, the hepatocytes quickly run out of CoA, and rates of β oxidation decrease. The purpose of ketogenesis is to maintain flux through the β-oxidation pathway and ensure that cells obtain the energy from acetyl-CoA in the form of ketone bodies. b. Ketogenesis results from high rates of fatty acid oxidation in the absence of sufficient levels of carbohydrates needed to supply citrate cycle intermediates for the complete oxidation of acetyl-CoA in the mitochondria. Ketogenesis is due to an overflow of acetyl-CoA in the mitochondria of the liver. c. A high-fat and low-carbohydrate diet leads to accumulation of acetylCoA from fatty acid degradation because carbohydrates are limiting and the cell is unable to maintain oxaloacetate levels via carboxylation of pyruvate (pyruvate carboxylase reaction). The result is increased flux through the ketogenic pathway.

12. a. Hydrolysis of ATP to form ADP and Pi. b. Release of CO2 coupled with hydrolysis of a high-energy thioester bond. c. Reduction of β-ketoacyl-ACP by NADPH and reduction of enoyl-ACP

by NADPH. 13. a. When excess acetyl-CoA is available (beyond the requirements for energy production), mitochondrial citrate is exported to the cytoplasm.

Cytoplasmic citrate is converted to acetyl-CoA for use in fatty acid biosynthesis. NADPH is produced by the malic enzyme as part of the citrate transport mechanism. Increasing levels of citrate in the cytoplasm activate fatty acid synthesis by stimulating acetyl-CoA carboxylase activity through conversion of the monomer (inactive) to the polymerized (active)

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CHAPTER 16 LONG ANSWERS TO CHALLENGE PROBLEMS

form, resulting in malonyl-CoA production. Moreover, citrate inhibits the activity of phosphofructokinase-1 in the glycolytic pathway, which shifts carbohydrate metabolism toward the pentose phosphate pathway to generate the additional NADPH needed for fatty acid synthesis. b. Glucagon signaling activates AMP-activated protein kinase (AMPK), which phosphorylates acetyl-CoA carboxylase and stimulates depolymerization to the inactive monomeric form.

14. Under conditions where energy levels in the cell are high, excess citrate is transported into the cytosol, where it is cleaved by citrate lyase into oxaloacetate and acetyl-CoA. The acetyl-CoA is a substrate for fatty acid synthesis. Citrate is an allosteric activator of acetyl-CoA carboxylase, which stimulates the synthesis of malonyl-CoA from acetyl-CoA. Moreover, the citrate shuttle provides net NADPH for fatty acid synthesis by the cytosolic reactions converting oxaloacetate → malate → pyruvate.

15. (1) Malonyl-CoA links energy charge to this anabolic pathway, as the energy for activation is derived from ATP. (2) The carboxylation reaction to malonyl-CoA provides a pathway intermediate that is separate from β oxidation and conserves the energy in the reactive carboxyl group of malonyl-CoA. (3) Malonyl-CoA is an inhibitor of the carnitine transacetylase reaction (cytosolic side), such that β oxidation is slowed when flux through the fatty acid synthesis pathway is high.

16. (1) Inability of the LDL receptors to remove LDL from the serum. LDL is the major source of serum cholesterol. (2) Cholesterol biosynthesis (also VLDL production) in the liver is higher than normal because the rate-limiting enzyme HMG-CoA reductase is not feedback-inhibited by intracellular cholesterol levels.

17. a. There will be no apoB-containing lipoproteins, chylomicrons, VLDL, or LDL. b. Cholesterol biosynthesis is upregulated because there is no LDL to deliver cholesterol to the cells. They must make all the cholesterol they require. c. Carbohydrate is converted to fatty acids in the liver and then to triacylglycerol, but these patients cannot package the lipid into VLDL particles, so it accumulates.

18. a. The test depends on the fact that fatty acids are oxidized to CO2, but in order for that to occur, the fed triacylglycerol must be digested and the fatty acids absorbed. b. No. It can tell whether the fatty acid is malabsorbed, but it cannot distinguish between these two possible mechanisms. To make that distinction would require assessing the chemical nature of the fat in the stool.

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Long Answers to Review Questions Chapter 17 1. Nitrogen is incorporated into the biosphere by three processes: (1) biological nitrogen fixation catalyzed by the enzyme nitrogenase, contained in microorganisms that live in soil and aquatic environments; (2) industrial nitrogen fixation by the Haber process, in which N2 and H2 gases are reacted under conditions of extreme temperature and pressure to produce NH3; and (3) atmospheric nitrogen fixation, when energy from lightning combines N2 with O2 to form nitrogen oxides that are dissolved in rain and fall to Earth.

2. Each time the Fe protein in nitrogenase is reduced by ferredoxin, 2 ATP are required for conformational changes in this redox protein. Because a total of 6 e− are required to convert N → 2 NH , 2

3

and each round of nitrogen reduction by the MoFe protein transfers 1 e−, then 12 ATP should be needed. The nitrogenase reaction is inefficient, however, and 2 e− are lost to a side reaction that converts 2 H+ → H .2Therefore, an additional 2 e− and 4 ATP are required in the nitrogen fixation cycle, bringing the total to 16 ATP. 3. The three mechanisms that nitrogen-fixing microorganisms use to limit exposure to the inhibitory effects of O2 are as follows: (1) Klebsiella pneumoniae only synthesizes the protein components of the nitrogenase complex when it is living in an anaerobic environment; (2) Azotobacter vinelandii decreases local O2 concentrations by increasing flux through the electron transport system to reduce O2 to H2O rapidly; and (3) Sinorhizobium meliloti invades the roots of leguminous plants, which synthesize a heme-containing protein called leghemoglobin that sequesters O2 away from the nitrogenase complex.

4. The nitrogen cycle redistributes soil NH 4+, NO 2−, and NO −3 between plants and the atmosphere through decomposition, nitrification, and denitrification processes. Beginning with decomposition of dead plant and animal materials by soil bacteria and fungi, organic nitrogen is converted to NH 4+, 1

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CHAPTER 17 LONG ANSWERS TO REVIEW QUESTIONS

which is oxidized by Nitrosomonas bacteria to generate NO2− (nitrite).

Bacteria in the Nitrobacter genus convert NO2 − to NO 3− (nitrate), which together with NH 4+ are reincorporated into plants through the soil to complete the nitrogen cycle. Some of the nitrogen is returned to the atmosphere as N2 through denitrification, which is carried out by Pseudomonas (NO3 − → N2 ) bacteria.

5. The combined action of glutamine synthetase and glutamate synthase is a +. net reaction that generates glutamate from α-ketoglutarate and NH 4 + + ATP → Glutamine + ADP + P Glutamine synthetase: Glutamate + NH 4 i

Glutamate synthase: Glutamine + α-Ketoglutarate + NAD(P)H + H+ → Glutamate + Glutamate Net: α-Ketoglutarate + NAD(P)H + H+ + NH4+ + ATP → Glutamate Glutamate dehydrogenase catalyzes the same reaction, but only under conditions of very high NH 4+ levels, such as after application of agricultural fertilizers, as the reaction is unfavorable (ΔG°′ = +30 kJ/mol). Under physiologic conditions, this reaction deaminates glutamate.

6. E1 enzymes attach ubiquitin to E2 enzymes, E2 enzymes conjugate ubiquitin to target proteins, and E3 enzymes recognize target proteins and facilitate ubiquitination by interacting directly with E2–ubiquitin and the target protein. Because E3 proteins must be able to recognize a large number of target proteins, there are hundreds of different E3 genes in the human genome. In contrast, E1 only needs to recognize ubiquitin and E2, and therefore only two genes are needed.

7. The urea cycle and citrate cycle are metabolically linked through the shared intermediate fumarate in a set of reactions called the Krebs bicycle.

The metabolic function of this shunt pathway is to recycle the carbon backbone of aspartate, which donates one of the two nitrogens to urea.

This is done by incorporating fumarate produced by the urea cycle into the citrate cycle, which is used to generate oxaloacetate for the aspartate aminotransferase reaction. The net result is that the amino group of glutamate is transferred to urea via aspartate and fumarate.

8. Glucogenic amino acids are metabolized to form pyruvate or citrate cycle intermediates, which are precursors in the gluconeogenic pathway. Ketogenic amino acids are converted into acetyl-CoA or acetoacetyl-CoA, which give rise to ketone bodies. The amino acids isoleucine, phenylalanine, threonine, tryptophan, and tyrosine are considered to be both glucogenic and ketogenic amino acids.

9. “Inborn errors of metabolism” was a term first used by Sir Archibald Garrod to describe the disease alkaptonuria, which is a genetically inherited defect in the enzyme homogentisate-1,2-dioxygenase. Homogentisate is a metabolite in the tyrosine degradation pathway that accumulates in alkaptonuria and is responsible for the name black urine disease. Two other genetic diseases related to the phenylalanine degradation pathway are phenylketonuria, a defect in the enzyme phenylalanine hydroxylase, and albinism, a defect in the enzyme tyrosinase.

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10. Essential amino acids are the 10 amino acids we must obtain from the diet because humans do not have the necessary enzymes to synthesize them.

Nonessential amino acids are those 10 amino acids we can synthesize and do not necessarily need in the diet. This distinction can be explained in evolutionary terms by recognizing that plants and bacteria must be able to synthesize all 20 amino acids because they cannot reliably obtain them from their environment. In contrast, humans eat diverse foods, so many human amino acid requirements can be met by dietary proteins. In general, the amino acid biosynthetic pathways that have been lost in humans are those that require multiple steps and are the most complex and energy demanding; that is, pathways for the 10 essential amino acids.

11. a. Glyphosate is an inhibitor of the enzyme 5-enolpyruvylshikimate3-phosphate (EPSP) synthase, which is required to convert shikimate-3-phosphate to EPSP in the chorismate biosynthetic pathway. Animal cells do not contain the enzymes required for chorismate biosynthesis, and therefore enzyme inhibitors of this pathway are used as animal-safe herbicides. b. It takes 7 days to observe the effects of glyphosate on plants because it takes time for the plant to use up all available phenylalanine and become dependent on phenylalanine biosynthesis. c. Roundup Ready soybeans contain a bacterial gene coding for the enzyme CP4 EPSP synthase, which does not bind glyphosate and thereby provides glyphosate resistance to the plant.

12. a. A recessive genetic mutation in a protein-coding gene results in loss of protein function, either through loss of protein expression (increased degradation) or through a deleterious functional mutation that renders the protein inactive. In contrast, a dominant mutation in a protein-coding gene leads to a gain of protein function; for example, unregulated activity (not subject to negative control) or relaxed substrate specificity. Recessive loss-of-function phenotypes usually require homozygous mutations (both copies of the gene are defective), whereas a dominant gain-of-function phenotype only involves one mutated gene because the variant protein can function even in the presence of wild-type protein. b. A homozygous with phenylketonuria father and a homozygous normal mother will produce offspring that are all phenylketonuria carriers because they will each inherit a defective gene copy from the father; that is, a probability of 100%. c. A heterozygous mother with acute intermittent porphyria will have a 50% chance of having a child with the disease because she can donate either a normal copy of the gene or a mutant copy encoding a gain-of-function mutation.

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Long Answers to Challenge Problems Chapter 17 1. Nitrogen fixation converts atmospheric N2 into NH3 (ammonia or NH +). Nitrogen assimilation incorporates NH (ammonia or 4

3

NH 4+) into amino acids.

2. a. NH 4+ is preferred because it is already reduced, so no additional energy is required to convert NO − to NH +, using 3

4

nitrite reductase and nitrate reductase enzyme reactions, before it can be assimilated into amino acids. b. After nitrogen starvation, the plant needs nitrogen, so it adapts to obtaining it from NO −3 even at the expense of the additional redox energy needed to convert NO − to NH +.

3

4

c. Any type of NH 4+-based fertilizers would be the better choice because plants use it as a nitrogen source more efficiently than NO 3−-based fertilizers.

3. Root nodules are the sites of nitrogen fixation, so large amounts of ammonia accumulate that need to be assimilated into glutamine by this enzyme.

4. One mechanism of glutamine synthetase regulation is cumulative allosteric inhibition by compounds that are end products in pathways in which glutamine is a precursor; for example, AMP, tryptophan, carbamoyl phosphate, CTP, histidine, and glucosamine-6-phosphate. Alanine and glycine are also allosteric inhibitors of the enzyme, which is related to the central role of glutamine in amino acid metabolism. A second mechanism of glutamine synthetase regulation is covalent modification by adenylylation and deadenylylation. Adenylylation, which inhibits the enzyme, is indirectly stimulated by glutamine and Pi and inhibited by α-ketoglutarate and ATP. 5. a. Aspartate is one of the most common amino group donors in transamination reactions. It can participate as an amino donor in the following reaction: Aspartate + α-Keto acid → Oxaloacetate + α amino acid 4

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Because the α-keto acid in the equation can represent any amino acid carbon skeleton, the 15N in the aspartate amino group is rapidly transferred to other amino acids. b. The enzyme cofactor is pyridoxal phosphate. A Schiff base must be formed between the cofactor carbonyl carbon and either an enzyme Lys amino group or an amino acid amino group.

6. a. Fasting lowered blood glucose, which meant that the fed amino acids would be deaminated to form substrates for gluconeogenesis. b. Arginine is required for the urea cycle. Therefore, the control animals, but not the experimental animals, were able to remove ammonia via the urea cycle.

7. Because tadpoles live in water, they can excrete ammonia directly. But frogs that live on land need to conserve water and therefore excrete ammonia as urea. Arginase is essential for the production of urea from arginine in the urea cycle.

8. If the enzyme defect were in carbamoyl phosphate synthetase, the urea cycle would stop at the first step of citrulline production from ornithine and carbamoyl phosphate. In this case, the concentrations of alanine and glutamine would be higher than normal because they cannot be metabolized to produce urea. Moreover, the concentration of arginine would be lower than normal because the urea cycle is not functioning and arginine would not be produced by the argininosuccinase reaction. If elevated blood ammonia were due to a defective arginase enzyme, however, then arginine would accumulate and the concentrations of alanine and glutamine would be closer to normal, assuming that excess nitrogen could be excreted as argininosuccinate and arginine.

9. a. Ammonia is removed from muscle tissue by the glucose–alanine cycle, which converts pyruvate to alanine using glutamate as the nitrogen donor. b. Ammonia is removed from liver tissue by the production of urea in the urea cycle using glutamate and glutamine as nitrogen donors. c. Ammonia is removed from all other tissues by the export of glutamine produced in the glutamine synthetase reaction.

10. Because these individuals cannot synthesize urea to remove excess ammonia, they need an alternative way to excrete nitrogen-rich products. Glycine depletion by formation of hippuric acid shifts more ammonia into amino acid synthesis reactions to replace the lost glycine, thereby lowering the level of free ammonia.

11. With amino acids as the only source of carbon for energy production, there is increased production of urea to remove the ammonia resulting from amino acid deamination. This deamination is necessary to generate pyruvate and the citrate cycle metabolites oxaloacetate and α-ketoglutarate to replace the missing carbohydrate from the diet. Because a large volume of water is required to excrete urea, individuals become dehydrated and weigh less in the short term if they do not drink large amounts of water to replace what is lost in the urine.

12. Glutamine is the compound that carries nitrogen from most tissues to the liver for disposal after transamination reactions, and alanine is the compound

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that carries nitrogen from muscle to the liver for disposal. The alanine– glucose cycle is responsible for removing excess muscle nitrogen that is due to protein degradation.

13. a. Kwashiorkor is caused by a lack of protein in the diet. The decreased pigmentation is caused by a deficiency in tyrosine, which is the precursor to melanins that serve as pigments. Moreover, because phenylalanine is converted to tyrosine by the phenylalanine hydroxylase reaction, deficiencies in both tyrosine and phenylalanine contribute to the skin and hair condition. b. If an increase of protein in the diet has no effect on skin and hair pigmentation, it is possible that the individual has a defect in one of the enzymes required to convert tyrosine to melanin.

14. a. Because phenylalanine is an essential amino acid, the level of phenylalanine in the body can be controlled by diet. There is no safe way to continually add pigments to cells through diet or drugs. b. Individuals with phenylketonuria have tyrosinase and obtain tyrosine in their diet.

15. Porphyrias are genetically inherited diseases caused by defects in enzymes required for heme biosynthesis. For example, acute intermittent porphyria is a genetic defect in the heme biosynthetic enzyme porphobilinogen deaminase. Jaundice is characterized by yellow skin and is caused by a buildup of bilirubin, a product in the heme degradation pathway. Because some of the intermediate metabolites in the heme biosynthetic pathway are chemically related to products in the heme degradation pathway, defects in heme biosynthesis can lead to a buildup of substrates that are converted to bilirubin and cause jaundice. Not all porphyrias lead to an accumulation of metabolites that are converted to bilirubin, therefore not all porphyrias are associated with jaundice.

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Long Answers to Review Questions Chapter 18 1. The four biochemical processes that nucleotides participate in are energy conversion reactions, signal transduction pathways, coenzyme-dependent reactions, and genetic information storage and transfer. Guanine has a role in energy conversion reactions in the form of GTP, which is generated by the succinyl-CoA synthetase reaction in the citrate cycle. Signal transduction pathways depend on cyclic GMP as a second messenger, and G protein signaling activities are regulated by GDP–GTP exchange. All of the common coenzymes described in Chapter 10 are based on adenine (NAD+, FAD, CoA); however, the biosynthesis of coenzyme A requires CTP as a phosphoryl donor in the phosphopantothenate–cysteine ligase reaction (see Figure 10.9).

Genetic information and storage requires all five of the common nucleotide bases: adenine, cytosine, guanine, uracil, and thymine.

2. The names for these nucleotide structures are (a) uridine, (b) deoxyadenosine-5′-triphosphate, (c) cytidine-5′-diphosphate, (d) thymine, and (e) guanylate (or guanosine-5′-monophosphate).

3. Most nucleotide bases recovered in salvage pathways come from degradation of cellular RNA, which is a transient nucleic acid product required for gene expression. Salvaged dTMP (thymidylate) from DNA is the only source of thymine and thymidine, as this pyrimidine base is not present in RNA. 4. Purine bases are synthesized directly on the ribose sugar, whereas pyrimidine bases are first synthesized as a closed ring before attaching to the ribose sugar.

5. The four nitrogen atoms in purine bases are derived from aspartate (N-1), glycine (N-7), and two glutamines (N-3 and N-9). The five carbons come from glycine (C-4 and C-5), HCO 3− (C-6), and two molecules of N 10-formyl-tetrahydrofolate (C-2 and C-8). The name of the metabolic precursor to adenylate and guanylate is inosine-5′-monophosphate.

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6. Purinosomes are multifunctional protein complexes in eukaryotic cells containing all of the catalytic activities required for a 10-step reaction leading to the synthesis of IMP. The existence of purinosomes was proposed on the basis of experiments in which green fluorescent protein (GFP) fusion proteins containing coding sequences of human enzymes in the de novo purine biosynthetic pathway were observed to form fluorescent cytosolic protein aggregates in purine-depleted cell cultures. To prove the existence of purinosomes under physiologic conditions, the complexes need to be purified and characterized. One approach would be to use chemical cross-linking, coupled with protein purification and mass spectrometry, to characterize “nearest neighbor” proteins, which should be other enzymes in the same pathway.

7. Flux through the purine biosynthetic pathway is primarily controlled by inhibition of the PRPP synthetase and glutamine-PRPP amidotransferase reactions. The balance of AMP and GMP synthesis is controlled by both feedback inhibition of the individual branches in the pathway and by ATP and GTP cross-talk regulation.

8. Excess purines are degraded into uric acid by all of these organisms.

Primates (which includes humans), as well as insects (cockroaches), reptiles (rattlesnakes), and birds (flamingos), all excrete the uric acid directly without further degradation. In contrast, nonprimate mammals (Siberian tigers) metabolize uric acid to allantoin, which is excreted. Bony fish (tuna) convert allantoin to allantoic acid for excretion, whereas cartilaginous fish (sharks) and amphibians (salamanders) convert allantoic acid to urea that is excreted.

Lastly, marine invertebrates (lobsters) use the enzyme urease to break down urea into NH 4+ as the end product of purine degradation.

9. The biochemical basis for ADA–SCID is defects in the adenosine deaminase (ADA) gene, which lead to adenosine accumulation and elevated levels of dATP. Because dATP is an inhibitor of the enzyme ribonucleotide reductase, rapidly dividing B and T cells are starved of deoxyribonucleotides and fail to proliferate, resulting in severe immunodeficiency. Treatment for ADA–SCID patients involves frequent intravenous infusions of immunoglobulins or recombinant ADA enzyme. The only curative treatment for ADA–SCID is a bone marrow transplant, which usually requires a sibling match to avoid tissue rejection. Human gene therapy using a virus containing the human ADA gene is a promising experimental procedure.

10. Pyrimidine biosynthesis is regulated by both feedback inhibition and allosteric activation in E. coli and human cells. Aspartate transcarbamoylase is the key regulated enzyme in the pyrimidine biosynthetic pathway in E. coli cells, which is activated by ATP and inhibited by CTP. In contrast, flux through the pyrimidine biosynthetic pathway in human cells is controlled by allosteric regulation of the trifunctional CAD enzyme and by regulation of UMP synthase and CTP synthetase.

11. Dihydropyrimidine dehydrogenase enzyme deficiencies are fairly common (∼5% of the human population) and can cause 5-fluorouracil toxicity in cancer patients undergoing chemotherapy because the dose of 5-fluorouracil

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is too high, owing to its reduced degradation by dihydropyrimidine dehydrogenase in the liver. Normally, ∼80% of 5-fluorouracil is degraded by dihydropyrimidine dehydrogenase, which means that the effective concentration of 5-fluorouracil is only ∼20% of the supplied dose. But in dihydropyrimidine dehydrogenase–deficient individuals, this same dose of 5-fluorouracil is potentially toxic because the effective dose is much higher and the drug harms healthy cells. Cancer patients being considered for 5-fluorouracil treatment are now routinely screened for dihydropyrimidine dehydrogenase deficiencies so that patients who demonstrate a deficiency may have their drug dose adjusted to maximize benefit (kill cancer cells) and decrease risk (toxic side effects).

12. The RNA world hypothesis proposes that RNA was the first genetic molecule used in early life on Earth; moreover, that RNA-based reaction catalysts called ribozymes mediated reactions, rather than proteinbased reaction catalysts, which arose later in evolutionary history. After the emergence of ribonucleotide reductase, deoxyribonucleotides were polymerized into DNA, which is better suited to function as genetic material. The best evidence to date supporting the RNA world hypothesis is the finding that the catalytic activity in ribosomes responsible for peptide bond formation during protein synthesis is in fact due to RNA, not protein.

13. Ribonucleotide reductase reduces C-2′ on nucleoside diphosphates to convert NDPs to dNDPs in a reaction that requires a pair of electrons donated from NADPH and, in the process of NDP reduction, produces H2O. The reduction of these sulfhydryls in ribonucleotide reductase is not done by NADPH directly, but rather by a redox circuit requiring intermediary proteins (thioredoxin or glutaredoxin). Ribonucleotide reductase contains two subunits, R1 and R2, that function as a tetrameric complex (R12R22). Substrate specificity of ribonucleotide reductase is regulated by allosteric binding of dTTP, dGTP, dATP, or ATP to the R1 substrate specificity site. The overall activity of ribonucleotide reductase is regulated by allosteric binding of ATP and dTTP to the regulatory site, such that ATP is an activator and dTTP is an inhibitor.

14. Deoxyuridylate (dUMP) is derived from three sources in E. coli cells: (1) dephosphorylation of dUTP by the enzyme dUTP diphosphohydrolase; (2) deamination of dCTP by dCTP deaminase to generate dUTP, which is then converted to dUMP by dUTP diphosphohydrolase; or (3) phosphorylation of deoxyuridine by the enzyme thymidine kinase to yield dUMP. 15. Three mechanisms of methotrexate drug resistance have been characterized in human cancer cells: (1) point mutations in the dihydrofolate reductase coding sequence that lower methotrexate binding affinity; (2) dihydrofolate reductase gene amplification, which increases expression of the dihydrofolate reductase enzyme; and (3) overexpression of the multidrug resistance protein, which rapidly exports methotrexate from the cell.

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Long Answers to Challenge Problems Chapter 18 1. PRPP is required for (1) the nucleotide salvage pathway (base + PRPP forms nucleoside monophosphates); (2) de novo purine biosynthesis (IMP is synthesized beginning with PRPP); and (3) de novo pyrimidine biosynthesis (orotate + PRPP forms OMP).

2. In defect A, large amounts of 15N should be incorporated in uric acid because glycine contributes a nitrogen to the purine ring in the de novo synthesis pathway. In defect B, the uric acid should be relatively free of 15N because much of the excess uric acid would be due to purines in the diet and not a result of de novo purine biosynthesis.

3. A deficiency of HGPRT prevents salvage biosynthesis of purines; therefore, de novo synthesis is stimulated to maintain sufficient purines for cell function.

4. a. The defective enzyme is HGPRT, and the ∼50% level of radioactive DNA reflects that this individual has a heterozygotic HGPRT genotype; that is, he or she is an asymptomatic carrier of the HGPRT defect. b. The cell donor is female. Because Lesch–Nyhan syndrome is an X-linked recessive genetic disorder, females can be heterozygous asymptomatic, whereas males are either normal (for a male) or are symptomatic for Lesch–Nyhan syndrome. c. A single copy of HGPRT is sufficient to remove excess metabolic intermediates and protect the individual from the clinical symptoms of Lesch–Nyhan syndrome. This is true in males, which have only a single X chromosome. d. On the basis of the genetic inheritance pattern of an X-linked recessive disorder such as Lesch–Nyhan syndrome, a daughter will not develop Lesch–Nyhan syndrome but does have a 50% chance of being a carrier, whereas a son has a 50% chance of developing Lesch–Nyhan syndrome.

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5. The synthesis of AMP from IMP (see Figure 18.11) generates fumarate from the carbon skeleton of aspartate, which provides increased citrate cycle intermediates to stimulate flux through this energy conversion pathway in muscle cells. The production of fumarate from aspartate as a consequence of purine metabolism can be sustained by the enzyme AMP deaminase, which deaminates AMP to regenerate IMP (see Figure 18.14).

6. a. Deficiencies in adenosine deaminase (ADA) cause the human disease ADA–SCID (SCID: severe combined immunodeficiency disease), which is characterized by a defective immune system, resulting from extremely low levels of infection-fighting B and T cells. b. A defect in adenosine deaminase leads to accumulation of dATP, which inhibits ribonucleotide reductase, thereby inhibiting deoxyribonucleotide biosynthesis. This is only a problem in rapidly dividing human cell types, such as immune cells that must be able to proliferate quickly in response to antigen-mediated stimulation (B and T cells).

7. a. Orotate phosphoribosyl transferase is required in the de novo pyrimidine biosynthesis pathway, and the reaction it catalyzes is Orotate + PRPP → OMP + PPi b. OMP is the precursor to UTP and CTP, which are not produced when orotate phosphoribosyl transferase is defective; however, uridine and cytidine can be used to synthesize UTP and CTP by the salvage pathway and provide nucleotides for the synthesis of blood cell precursors to alleviate anemia. The decrease in orotic acid excretion may occur because the fed nucleotides inhibit the de novo pathway of pyrimidine biosynthesis by feedback inhibition, preventing further accumulation of free orotate.

8. The substrate for thymidylate synthase is dUMP, which is derived from dUDP. 9. Cells that lack thymidine kinase activity are not able to phosphorylate 5-bromodeoxyuridine, which prevents 5-bromodeoxyuridine incorporation and mutation of A-T base pairs into G-C base pairs.

10. a. Xanthine oxidase inactivates 6-mercaptopurine by converting it to 6-thiouric acid, which lowers its effective concentration as a competitive inhibitor of adenylosuccinate synthetase and IMP dehydrogenase in leukemic cells. Therefore, by including allopurinol in the treatment, inactivation of 6-mercaptopurine by xanthine oxidase is repressed, leading to a higher effective concentration. b. If a leukemia patient had a xanthine oxidase deficiency, then the treatment dose of 6-mercaptopurine would need to be reduced to avoid possible toxic side effects. Moreover, the addition of allopurinol to the treatment regimen would likely have no effect and should not be part of the treatment. Treating a cancer patient with 6-mercaptopurine who has a deficiency in xanthine oxidase is similar to treating a cancer patient with 5-fluorouracil who has a deficiency in dihydropyrimidine dehydrogenase, as illustrated in Figure 18.26.

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11. Addition of high amounts of thymidine to cultured cells leads to the accumulation of dTTP through the salvage pathway. High levels of dTTP bound to the specificity site of ribonucleotide reductase inhibits reduction of CDP to dCDP, resulting in the inhibition of DNA replication due to insufficient amounts of dCTP. 12. Addition of methotrexate to normal cells inhibits the enzyme dihydrofolate reductase, which blocks the production of new tetrahydrofolate, a critical metabolite for numerous pathways in the cell. Because thymidylate synthase is active in these cells, the remaining tetrahydrofolate pool is quickly depleted. Adding thymidine to the culture does not rescue the cells because it is the low levels of tetrahydrofolate that cause cell death, not lack of thymidine. In contrast, cells that are deficient in thymidylate synthase do not use up the tetrahydrofolate pool once the methotrexate is added, and the other tetrahydrofolate-dependent pathways are able to function for a time. Moreover, because added thymidine rescues the thymidylate synthase deficiency in these mutant cells, they are able to survive longer than normal cells.

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Long Answers to Review Questions Chapter 19 1. The liver maintains safe blood glucose levels of 4.5 mM by releasing glucose between meals to elevate blood glucose levels; the drop is due to glucose uptake by glucose-dependent brain cells and erythrocytes. Glucose efflux from the liver is the result of liver glycogen breakdown and gluconeogenesis. After a meal, the liver helps lower blood glucose levels that spike from carbohydrate-rich foods by promoting glucose influx through the GLUT2 glucose transporter and glucose phosphorylation by hexokinase and glucokinase, which trap glucose-6-P inside liver cells. If glucose levels fall to 3.5 mM, a person feels light-headed; if they fall to 2.5 mM, the person becomes hypoglycemic (fainting, perspiring); and if they fall to 2.2 mM, the person can go into a coma and die.

2. The key metabolite in liver cells is glucose-6-P, which has four metabolic fates: (1) glucose for release into the blood; (2) glucose-1-P for use in glycogen synthesis; (3) 6-phosphogluconolactone to generate NADPH by the pentose phosphate pathway; and (4) fructose-6-P, which is used in the glycolytic pathway to produce pyruvate. Metabolic flux of glucose-6-P among these pathways is controlled by the activity of rate-limiting enzymes and metabolite concentrations.

3. The largest amount of muscle in the human body is skeletal muscle, which uses free fatty acids, glucose, and ketone bodies for metabolic fuel. The other major muscle type is cardiac muscle (heart), which primarily uses fatty acids and ketone bodies for metabolic fuel.

4. The two primary types of adipose tissue are visceral fat and subcutaneous fat. The primary function of both types of fat is energy storage in the form of triacylglycerol; however, visceral fat also functions as an endocrine organ that releases peptide hormones called adipokines in response to metabolic cues.

Visceral fat lies deep within the abdominal cavity (apple body 1

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type), whereas subcutaneous fat is located just below the skin surface in the thighs, buttocks, arms, and face (pear body type).

5. The three parameters are (1) body mass index (BMI), which is calculated by dividing a person’s weight in kilograms by the square of his or her height in meters; (2) waist-to-hip ratio, which is based on the circumferences of an individual’s waist and hips; and (3) percent body fat, which can be determined using whole-body bioelectrical impedance. A retrospective clinical study uses data that has already been collected and applies statistical analysis to identify disease associations. Because BMI only takes into account height and weight, it cannot distinguish between muscular, apple shape, and pear shape body types. The waist-to-hip ratio can distinguish body types if the ratios are significantly higher or lower than 1.0, but cannot distinguish between a muscular individual with small hips and an obese individual with large hips and waist. The best approach is to combine percent body fat with BMI and waist-to-hip ratio to get an accurate profile of the few “outlier” individuals that might affect the statistical analysis, considering that the patients are not available for additional measurements, as they would be in a prospective study.

6. The triacylglycerol cycle exchanges triacylglycerols synthesized in the adipose tissue with those synthesized in liver cells using the same pool of fatty acids. The systemic component of the triacylglycerol cycle exchanges free fatty acids and triacylglycerols between adipose tissue and liver cells, whereas the intracellular component of the triacylglycerol cycle interconverts pools of free fatty acids and triacylglycerols in adipocytes.

Most of the fatty acids and triacylglycerols are recycled back to the adipose tissue, unless needed by the peripheral tissues for energy conversion reactions or membrane biosynthesis. When dietary glucose levels are low, the glyceroneogenic pathway in adipocytes and hepatocytes synthesizes glycerol-3-P from dihydroxyacetone phosphate, which is generated from amino acids and lactate. Glycerol-3-P is required to maintain triacylglycerol synthesis using free fatty acids.

7. Insulin and glucagon are hormones with opposing actions that are secreted from the pancreas and function to regulate blood glucose levels by activating cognate receptors on multiple target tissues. Insulin is called the “I just ate” hormone because insulin levels rise in response to high blood glucose levels after a carbohydrate-rich meal, whereas glucagon levels rise in response to low blood glucose levels, which usually occur between meals. Insulin signaling in liver, skeletal muscle, and adipose tissue stimulates glucose uptake and storage of glycogen and lipid, whereas insulin signaling in the brain stimulates the anorexigenic neurons that decrease appetite and increase energy expenditure. Glucagon signaling in liver and adipose tissue stimulates glycogen and triacylglycerol degradation. Skeletal muscle and brain cells do not express glucagon receptors.

8. PPARs regulate gene expression in response to the binding of low-affinity, fatty acid–derived nutrients such as polyunsaturated fatty acids and eicosanoids. This ligand binding function of PPARs makes them ideal metabolic sensors of lipid homeostasis and results in long-term control

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of pathway flux by directly altering the steady-state levels of key proteins.

PPAR-regulated metabolic genes control fatty acid oxidation, fatty acid synthesis, adipose differentiation, energy uncoupling, and insulin sensitivity.

9. The human body adapts to starvation conditions by altering the flux of metabolites between tissues in order to extend life. The two metabolic objectives are to supply (1) the brain with glucose to maintain ATP-dependent ion pumps and (2) the heart with fatty acids and ketone bodies to maintain rhythmic contraction and blood circulation. The four major changes in metabolic flux under starvation conditions are (1) increased triacylglycerol hydrolysis in adipose tissue, (2) increased gluconeogenesis in liver and kidney cells, (3) increased ketogenesis in liver cells, and (4) protein degradation in skeletal muscle tissue.

10. The thrifty gene hypothesis states that humans contain metabolic gene variants that provide protection against famine by maximizing fat storage during times of feast. These same gene variants contribute to the epidemic of obesity and diabetes in developed countries, where high-Calorie food is readily available and a sedentary lifestyle is common. The thrifty gene hypothesis is consistent with the observed differences in rates of obesity and type 2 diabetes in two populations of genetically related Pima Indians. The Pima Indians of southern Arizona have a sedentary lifestyle and a highCalorie diet, which leads to higher rates of obesity and diabetes than those found for the Pima Indians of northern Mexico, who cultivate their fields using manual labor.

11. Leptin is produced in adipose tissue at a level that is directly related to the amount of stored lipid; that is, a high amount of stored fat leads to a high amount of secreted leptin. Leptin functions by binding to leptin receptors on first-order POMC and NPY/AGRP neurons, which produce neuropeptides (α-MSH, NPY, and AGRP) that bind to their cognate receptors on anorexigenic and orexigenic second-order neurons. The net effect of leptin signaling is to eat less and metabolize more. Insulin is secreted from the pancreas in response to elevated blood glucose and has the same effect on neuronal signaling as leptin; that is, eat less, metabolize more. Ghrelin is synthesized at high levels in the stomach between meals and binds to ghrelin receptors on NPY/AGRP first-order neurons that promote signaling in second-order neurons to eat more, metabolize less. PYY3-36 is synthesized in the colon and counters the effect of ghrelin by sending signals that food is in the digestive tract and it is time to eat less, metabolize more.

12. Type 1 diabetes is due to insufficient insulin production by the pancreatic β cells and is treatable with insulin injections. In contrast, insulin-resistant type 2 diabetes is characterized by high levels of circulating insulin and desensitization of insulin receptor signaling in muscle, liver, and adipose tissue. The symptoms of diabetes at initial diagnosis include frequent urination, thirst, and blurry vision, all of which are directly related to the osmotic effects of high glucose concentrations. A glucose tolerance test is used to confirm a diabetes diagnosis by measuring glucose clearance from the blood after ingestion of a glucose solution: Unlike individuals with normal insulin signaling, type 1 and type 2 diabetics are unable to lower

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blood glucose levels within 2 hours. Type 1 and type 2 diabetes can be distinguished by an insulin sensitivity test, which shows a decrease in blood glucose levels for type 1 diabetics but not for type 2 diabetics.

13. The four major classes of drugs that have been developed to treat type 2 diabetes are (1) α-glucosidase inhibitors (miglitol); (2) sulfonylurea drugs that inhibit the pancreatic ATP-dependent K+ channel (glipizide); (3) drugs that stimulate the activity of AMPK (metformin); and (4) ligand agonists of the nuclear receptor PPARγ (pioglitazone).

14. Four classes of compounds have been developed as weight-loss drugs: (1) Ephedrine is a stimulant that increases basal metabolic rates. Side effects include nausea, nervousness, and insomnia. (2) Lorcaserin targets neuronal signaling in the brain to control appetite (serotonin receptors). Side effects are mild and include headache, dizziness, and nausea. (3) Orlistat inhibits the activity of pancreatic lipase in the small intestine. Side effects include stomach pain, fecal urgency, and nausea. (4) Olestra is a zero-Calorie food substitute containing fatty acid side chains covalently linked to sucrose and is not a substrate for digestive lipases. Side effects include intense diarrhea and anal leakage. A drug-free alternative to weight loss is to eat less and exercise more.

15. AMPK-mediated phosphorylation of serine or threonine residues on metabolic target proteins in muscle cells leads to a net increase in ATP concentration by three mechanisms: (1) stimulation of flux through glycolysis; (2) stimulation of flux through fatty acid oxidation; and (3) increased oxidative phosphorylation.

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Long Answers to Challenge Problems Chapter 19 1. a. Assuming all 56 g of starch is converted to 56 g of glucose: 56 g × 3.7 kcal/g = 207.2 kcal 207.2 kcal × 4.18 kJ/kcal = 866 kJ 866 kJ × mol ATP/30.5 kJ = 28.3 mol ATP × 0.65 = 18.5 mol of ATP (∼19 mol) b. (Recall that 1 Calorie equals 1 kcal.) Amount of time a female would need to garden: 207.2 Calories × 1 min gardening/4 Calories = 51.8 minutes of gardening Amount of time a male would need to vacuum: 207.2 Calories × 1 min vacuuming/4.3 Calories = 48.1 minutes of vacuuming 2. a. Lower blood glucose levels b. Lower blood glucose levels c. Higher blood glucose levels d. Higher blood glucose levels e. Lower blood glucose levels 3. Eating six donuts will cause a sharp rise in insulin levels, which in turn will stimulate fatty acid synthesis in adipose tissue and inhibit fatty acid release. Therefore, the short-lived glucose rise at the beginning of the race is detrimental later, when muscle and liver glycogen levels are depleted and fat utilization from adipose tissue is needed to sustain the energy needs of muscle cells.

4. The pattern is related to the need to maintain blood glucose levels during the first few days of the fast. This is initially accomplished by glycogen degradation, but once this source of glucose is 5

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depleted, gluconeogenesis in the liver and kidneys is required, using amino acids from protein hydrolysis as the source of carbon. But after the brain adapts to using ketones as a component of its energy needs, flux through gluconeogenesis decreases, and protein is spared until survival is the highest priority.

5. a. Yes, this would be a candidate thrifty gene because acetyl-CoA carboxylase would synthesize malonyl-CoA to higher levels than normal (no feedback inhibition) and thereby increase the total amount of stored triacylglycerol in adipose tissue. b. Yes, this would be a candidate thrifty gene because even at low insulin levels, fatty acid synthesis and lipogenesis would increase fat stores. c. No, this would not be a candidate thrifty gene because uncoupling protein overexpression would lead to more thermogenesis and high rates of fatty acid oxidation, which would raise basal metabolic rates. d. Yes, this would be a candidate thrifty gene because a hyperactive lipoprotein lipase would scavenge as much lipid as possible from circulating lipoprotein particles to increase fat storage in adipose tissue. e. No, this would not be a candidate thrifty gene because a hyperactive hormone-sensitive lipase would continually hydrolyze triacylglycerol in adipose tissue and make it difficult to accumulate stored fat.

6. The uncoupler protein dissipates the proton gradient in the mitochondria.

This permits higher rates of fatty acid oxidation than would be possible if the mitochondria were coupled. Such high rates of oxidation would use some of the excess fatty acids, reducing the amount stored in nonadipose tissue.

7. OB mutant mice lack leptin, and when it is provided to them through injection, they are able to better maintain energy balance and metabolic homeostasis. In contrast, most obese individuals have chronically high leptin levels rather than a mutation in the leptin gene like the OB mice. In fact, studies have shown that most obese individuals are leptin insensitive. This is analogous to type 2 diabetics at initial diagnosis, who often have high levels of insulin but are insulin insensitive.

8. Eating slowly gives ghrelin and PYY3-36 a chance to tell your brain that food is in the stomach and you feel full. If you eat too quickly, you eat more than you need.

9. a. Because the Km of fructokinase for fructose is 20 times lower than the Km of glucokinase for glucose, fructose is efficiently converted to fructose1-P in liver cells and converted to acetyl-CoA, which is the substrate for fatty acid synthesis in the liver. In contrast, glucokinase requires a higher concentration of glucose to achieve the same flux to acetyl-CoA, and moreover, glucose is converted to glucose-6-P, which can be used for glycogen synthesis (after conversion to glucose-1-P). An equimolar amount of sucrose contains 50% fructose and 50% glucose, which is 5% less fructose than that in high-fructose corn syrup.

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b. 110 g carbohydrate/soda + 56 g carbohydrate/bag of chips = 166 g carbohydrate × 4 Calories/g carbohydrate = 664 Calories 664 Calories/2,000 Calories = 33% of recommended Calories in this late-night snack.

10. The use of the sulfonylurea is predicated on the assumption that the insulin receptor is less sensitive to insulin in type 2 diabetes; therefore, increasing the insulin concentration should increase saturation of the receptor and upregulate the insulin pathway and blood glucose uptake. The use of metformin is predicated on the assumption that decreased flux through gluconeogenesis lowers blood glucose levels.

11. Diabetics have lost the ability to respond to high glucose levels in the blood because of the inability to produce or respond to insulin. Under these conditions, the body responds to a decreased level of cellular glucose by increasing fatty acid degradation. The excess acetyl-CoA is converted to ketone bodies by enzymes in the liver.

12. Diet no. 1 is the most complete because proteins are required (nine essential amino acids), and fats can be synthesized from carbohydrates, but carbohydrates cannot be synthesized from fats.

13. In the absence of sufficient dietary carbohydrate, oxaloacetate, α-ketoglutarate, and pyruvate are derived from amino acid degradation. The acetone breath indicates that these reactions are not sufficient to keep the citrate cycle functioning at full capacity and acetyl-CoA builds up, leading to the production of ketone bodies, which can produce acetone.

14. Dinitrophenol is a chemical uncoupler of mitochondrial ATP synthesis, leading to high rates of fatty acid oxidation to compensate for a low energy charge induced by dinitrophenol. Although dinitrophenol can lead to weight loss because it mimics the “starved” state and fat is catabolized at a high rate, it is a hydrophobic compound that accumulates in mitochondrial membranes. It has been reported that long-term dinitrophenol use causes fatal liver damage due to hepatocyte apoptosis.

15. High-glycemic-index foods (jelly beans) lead to a rapid rise in insulin levels, which stimulates fatty acid synthesis. In contrast, low-glycemic-index foods (kidney beans) release carbohydrates slowly, and therefore insulin levels are gradually elevated, resulting in a more balanced metabolic response.

16. The terms “good fat” and “bad fat” came from epidemiologic studies, which found an association between improved cardiovascular health and diets low in saturated fats and high in unsaturated fats. The opposite is true for people who favor bad fats over good fats in terms of cardiovascular disease and death.

17. Muscle cells use the oxygen to replenish ATP by oxidative phosphorylation so that phosphocreatine can be resynthesized. Liver cells use the oxygen to replenish ATP, which is used to convert lactate to glucose via gluconeogenesis.

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Long Answers to Review Questions Chapter 20 1. Semiconservative replication results in two duplex DNA molecules, each containing one original strand and one newly synthesized DNA strand. Conservative replication would produce DNA duplexes in which one is composed of the two original strands and the other contains two newly synthesized strands.

Dispersive replication would give two duplex DNA molecules, each containing portions of the original DNA and newly synthesized DNA mixed throughout the chromosome.

2. Okazaki fragments are segments of DNA produced by discontinuous replication of the lagging strand template. An Okazaki fragment is started each time a new β clamp is loaded onto the lagging strand template and is completed when DNA polymerase encounters the 5′ end of the preceding Okazaki fragment.

3. The 3′-hydroxyl of the previously added nucleotide is generally the nucleophile for DNA polymerase. Initiation requires a primer because without the 3′-hydroxyl, DNA polymerization cannot proceed. Generally, the primer is supplied by a short stretch of RNA produced by the enzyme primase.

4. In prokaryotic cells, DNA polymerase III is the primary replicative polymerase. In eukaryotic cells, DNA polymerase δ is responsible for lagging strand synthesis, and DNA polymerase ε is responsible for leading strand synthesis.

5. HIVRT uses the single-stranded RNA of the HIV genome as a template to first produce a hybrid DNA–RNA duplex. The 3′ end of the host cell tRNA for Lys is used as the primer for the polymerization process. HIVRT contains RNase activity that is used to degrade the RNA after one strand of DNA is produced.

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RNA fragments serve as the primer for synthesis of the complementary DNA strand, thus producing double-stranded DNA. 6. Ahead of the replication fork, topoisomerase introduces negative supercoils to relieve the positive supercoiling added by separation of the double helix.

Helicase and primase are coupled together at the front of the replication fork to separate the double helix into single-stranded DNA and to synthesize RNA primers, respectively. Single-stranded DNA binding proteins keep the DNA in a single-stranded state and prevent damage to it. DNA polymerase III is tethered to helicase and consists of a core enzyme and a β clamp, which is critical to increase the processivity of the polymerase.

DNA polymerase I and ligase follow the replication fork to remove the RNA primers and seal the nicks in the newly formed DNA. 7. Binding of DnaA begins the process of unwinding the DNA and guides helicase to the site of single-stranded DNA (with the help of DnaC). Gyrase activity allows further opening of the double helix by relieving torsion created by opening of the single-strand region, and single-stranded DNA binding proteins bind to keep the DNA in a single-stranded form. DNA Pol III binds, releasing DnaA and starting the replication process.

8. Telomerase contains a tandem repeat of RNA with a 3′ end that is complementary to the end of the chromosome. In human telomerase, the RNA molecule is 451 nucleotides in length. After association of the 3′ end of the chromosome with the 3′ end of the RNA, it extends the chromosome, using the RNA as a template. When the end of the template is reached, telomerase releases from the chromosome and can reposition itself to start the process again.

9. The Ames test uses a strain of Salmonella that has been mutated such that it cannot produce the amino acid histidine. It is used to test the mutagenicity of a variety of agents. The molecule to be tested is incubated with both the Salmonella and liver extract to simulate mammalian metabolism. If colonies are produced on minimal-histidine media, then a back mutation has occurred that has mutated the histidine synthesis genes back to functional genes. The test has been very successful at predicting the carcinogenicity of substances that are mutagenic to bacteria and are often carcinogenic to humans.

10. Cytosine often undergoes spontaneous deamination, producing uracil. Uracil is removed from DNA by uracil DNA glycosylase (UNG) to produce an abasic site. Repair of the abasic site is achieved by base excision repair, where a nick is created in the DNA by AP endonuclease (AP1). This is followed by replacement of the nucleotide by short-patch repair or replacement of a short section of nucleotides by long-patch repair.

11. Both cyclobutane pyrimidine dimers and cross-linked guanine residues distort the helical structure of DNA and present a roadblock for any process that must access the individual bases. During replication, the replication fork stalls at such lesions or skips over them and attempts to put an incorrect nucleotide in place. During transcription, RNA polymerase either stalls or skips over the lesion, again putting a potentially incorrect nucleotide at the site.

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12. A silent mutation does not change the amino acid coded by the DNA sequence due to the redundancy of the genetic code. A nonsense mutation results in a premature stop codon that generally makes for a nonfunctional protein. A missense mutation incorporates the incorrect amino acid into the protein. Both nonsense and missense mutations can be very detrimental to a cell, depending on what protein is altered and how drastic the alteration is.

13. The methylation product is shown in Figure 20.42. O6-methylguanine-DNA methyltransferase (MGMT) is the only way to remove O6-methylguanine.

The protein is not a true enzyme, as it becomes irreversibly modified during the repair process. A cysteine residue in the active site performs a nucleophilic attack upon the methyl group to remove it from the guanine residue. This converts the cysteine residue to homocysteine. The protein must then be degraded to metabolize the homocysteine.

14. Within the oriC region, 11 sites are methylated by Dam methylase to distinguish the template strand from the newly synthesized strand. This area of high methylation serves as a binding site for SeqA. The SeqA– oriC complex is sequestered at the cell membrane, preventing further initiation events from occurring until the current replication process is at least a third of the way completed. In DNA repair, the methylation sites allow recognition of the template strand for mismatch repair by the MutS–MutL–MutH system. These proteins locate mismatches and the methylation site to determine the template strand, thus enabling repair of the mismatched nucleotide.

15. In meiotic homologous recombination (MHR), the process exerts a great deal of control over where strand breaks occur, whereas in double-strand breaks induced by DNA damage, the double-stranded break might occur anywhere within the chromosome. During MHR, the break in DNA is catalyzed by Spo11, producing ends that are fairly easily repaired.

After DNA damaging events, the ends of the DNA must be extensively processed, often involving resection of the DNA to produce ends that can be repaired.

16. Infection begins with the phage binding to the cell and injecting its DNA genome into the cell. The DNA is circularized and integrates into the host chromosome with the help of integrase and integration host factor. Once the phage genome is integrated into the chromosome, bacterial replication passively replicates the phage DNA as well. If the cell is exposed to a high dose of UV light, a lytic cycle is initiated due to the damage to the E. coli DNA. The prophage excises from the host DNA using integrase, integration host factor, and Xis, which is an excision-specific phage protein. The phage DNA then replicates many times, and phage proteins are synthesized. Phage virions assemble, and the cell is lysed to release the virions.

17. V(D)J recombination allows for the production of the heavy chains of antibodies. By having 51 V segments, 27 D segments, and 6 J segments, more than 8,000 different heavy chains can be made by combining one of each segment. Combining this with a light chain produces even greater

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variety in possible antibodies. To combine the three segments, first the D and J segments are combined with all intervening segments removed.

Next, the V segment is combined with the DJ recombination product, once again removing the intervening segments. Removal of the remaining V and J segments is not done on a DNA level, but is instead performed by splicing of the mRNA.

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Long Answers to Challenge Problems Chapter 20 1. The typical size of an Okazaki fragment is 2,000 nucleotides, and the E. coli chromosome size is 4,638 kb. Therefore, about 2,319 Okazaki fragments form during the replication process.

2. The events up to the actual polymerization process would most likely occur as they do in a typical replication. Following the core polymerase, a pair of helicases would be required to separate the template and daughter strands, followed by a protein with the ability to rejoin the template strands and join the daughter strands together for double-strand formation. Before the separation of the strands of the template–daughter double helix, mismatch repair would need to occur. Ultimately, this shows the increased complexity of conservative replication over the semiconservative process.

3. Any decrease in the error rate of HIVRT would result in a virus with a much lower mutation rate. Thus, ironically, a mutation would ultimately bring about an overall decrease in the rate of viral mutations. Ultimately, this would probably not be an effective mutation for the virus, as the relatively high error rate often helps to thwart the use of antiviral drugs.

4. Two Mg2+ ions are complexed to aspartate residues in the active site of DNA polymerase. When a dNTP binds, the Mg2+ withdraws electron density from the phosphate groups, making the α phosphate more susceptible to nucleophilic attack by the 3′-hydroxyl. Electron density is also withdrawn from the β and γ-phosphoryl groups, making the pyrophosphate a better leaving group. Absence of Mg2+ would probably significantly decrease DNA polymerase activity.

5. Because helicase requires 1 ATP for each two bases, at 4.6 × 106 bases, helicase requires 2.3 × 106 molecules of ATP for the replication process.

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6. The average Okazaki fragment is 2,000 bases and requires 2 seconds for synthesis, at 1,000 bases/second for a typical Pol III. A typical primer is about 10–12 bases; thus, primase would need a rate of 5–6 nucleotides/ second to avoid being the rate-limiting step of the replication process.

7. Yeast cells contain an average of 80 copies of a sequence that is, on average, about 10 nucleotides in length, giving an average maximum telomere size of 800 nucleotides. If each Okazaki fragment is about 200 nucleotides in length, then 200 nucleotides are lost from the telomere with each replication.

Thus, after four replications, the telomeres would disappear.

8. The results indicate that the suspected mutagen is indeed mutagenic without liver metabolism but becomes even more mutagenic after metabolism by liver enzymes.

9. The guanine would be converted into 8-hydroxyguanine and its tautomer, 8-oxoguanine. The modified guanine base-pairs with adenine, resulting in a G-A mismatch that most likely will not be repaired by normal DNA repair mechanisms. After a second round of replication, the result would be a G→C substitution in one of the daughter strands.

First round of replication

5′-AACGC-3′

5′-AACGC-3′

(8-hydroxyguanine = G):

3′-TTGAG-5′

3′-TTGCG-5′

Second round of replication:

5′-AACCC-3′

5′-AACGC-3′

3′-TTGAG-5′

3′-TTGCG-5′

5′-AACGC-3′

5′-AACGC-3′

3′-TTGAG-5′

3′-TTGCG-5′

10. 5′-ACGTCAGTTACGTACTGACGT. The TC photoproduct is likely to be a (6-4) photoproduct, which can result in stalled replication forks if not repaired. The TT and CT photoproducts would most likely be pyrimidine dimers. If these are unrepaired prior to replication, most likely adenine would be placed on the daughter strand. Although the TT photoproduct would not result in a mutation, the CT photoproduct would result in a mismatch.

Depending on when repair occurred, the potential for a mutation is high. If the daughter strand were used to correct the photoproduct lesion, then the C would be converted to a T. 11. Because the Xis protein is required for excision of the λ phage DNA, a mutation resulting in a nonfunctional protein would prevent excision and therefore prevent initiation of a lytic cycle.

12. An RNase inhibitor would prevent the degradation of the RNA–DNA hybrid that is initially formed by reverse transcriptase. This would prevent the production of the double-stranded DNA that is required for insertion into the host genome and thus prevent an infection by HIV. One problem with an RNase inhibitor is that it might prevent the degradation of human RNA. The recycling of RNA nucleotides provides a cell with ample quantities of the building blocks for RNA. Without the ability to degrade RNA that is no longer needed, the energy demands placed upon the cell would be tremendous.

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Long Answers to Review Questions Chapter 21 1. The three forms found in all organisms are mRNA, tRNA, and rRNA. These three are common to all organisms because they are essential for converting the genetic code into proteins.

2. Ribosomes are the site of protein synthesis, and rRNA is an essential structural and catalytic component. Each ribosome is capable of translating only one mRNA molecule at a time; the more ribosomes present in the cell, the more proteins can be synthesized simultaneously. More complex organisms have a more complex genome, and a greater number of proteins are expressed.

Therefore, these more complex organisms (that is, yeast, humans) require more active ribosomes than simple organisms such as bacteria.

3. The antisense strand is used as the template for transcription; a complementary strand is synthesized. Therefore, the sequence of mRNA would be CGCGGAUCCUUGAAUUCUAAAUAAACC AUUUACCACCAUGACC 4. Co-transcriptional folding of rRNA is necessary because of the size of the transcripts. Because the 5′ end begins to fold as soon as it exits the polymerase complex, the number of structural variations are decreased and therefore it is less likely that the RNA would become trapped in an unfavorable conformation.

The secondary structure also generates binding sites for ribosomal proteins. Interactions with these proteins during transcription help “lock” the structures into place and protect the RNA from destruction.

5. The −35 and −10 boxes are underlined in this sequence.

Transcription would start at the ATG after the −10 box (in gray).

CAGGCGATGATCAAATCTCCGTTGTACTT TGTTTCGCGCGTTGGTATAATCGCT GGGGTCAAGATGAGT 1

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6. Bacteria contain multiple σ factors because this is the transcription factor that directs RNA polymerase to specific gene promoters. The holoenzyme can interact with any DNA strand, and the σ factor is required to increase the affinity of the holoenzyme for the −10 and −35 boxes. If the σ factor were general, then RNA polymerase would transcribe all genes equally.

Bacteria, however, need to respond to changing nutrient and environmental conditions. Having σ factors that preferentially bind to a specific set of genes allows transcription of only those needed genes.

7. The active site of RNA polymerase contains two Mg2+ ions that coordinate the incoming NTP by interacting with the negatively charged oxygens on the α and β phosphoryl groups and lowering the pKa of the 3′-OH of the last ribose in the existing transcript. These interactions position the NTP correctly and assist in deprotonating the 3′-OH such that nucleophilic attack can occur.

8. In Rho-dependent termination, Rho destabilizes the transcription bubble by binding to a C-rich sequence in mRNA, resulting in transcriptional pausing and termination. In Rho-independent termination, the interaction between RNA polymerase and the transcript is also destabilized. Instead of interactions with a protein leading to this destabilization, however, structural elements within the mRNA interfere with the transcription bubble.

Conserved sequences at the 3′ end of the transcript fold into a hairpin loop, which leads to release of the polymerase and transcription termination.

9. TATA-less promoters often contain the downstream promoter element that may interact with TBP. 10. Phosphorylation of the conserved heptapeptide repeat in the CTD is important to transcription because it mediates the interaction of capping and splicing factors, as well as the activity of the polymerase. During initiation and elongation, different serine residues are phosphorylated. Upon termination, the CTD is dephosphorylated, which facilitates release of the RNA polymerase II complex from the nascent transcript.

11. The 5′ m7G cap and poly(A) tail protect mRNA from exonuclease digestion.

During mRNA decay, removal of either (or both) allows exonucleases to digest the mRNA, resulting in its destruction. Although RNA decay mechanisms do exist where the transcript is first cleaved in the middle of the sequence, exonuclease digestion is more common.

12. Recall from previous chapters that most biological reactions require a catalyst to ensure that they occur in a time frame suitable for the survival of the organism. The hypothesis of an RNA world relies on the ability of RNA to catalyze reactions in the absence of protein. The hammerhead ribozyme and group I introns are examples of catalytic RNA that can function without the assistance of any protein components, suggesting that life is possible without proteins.

13. In both group I and group II introns, splicing is initiated by a nucleophilic attack on the 3′ splice site. In group I introns, however, this attack occurs from an essential guanosine cofactor, whereas in group II introns it is initiated from a conserved adenosine in the branch site between the 5′ and

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3′ splice sites. The released group II intron has a lariat structure, in contrast to the linear form of the released group I intron. The group II intron mechanism closely resembles a spliceosome-catalyzed reaction, and there are also sequence similarities between the splice and branch sites.

14. Base modifications in tRNA are used to help adopt the correct tertiary structure and to help in recognition of specific tRNAs by the aminoacyltRNA synthetases. The conserved primary sequence and secondary structure found in all tRNAs would make it difficult for the individual aminoacyltRNA synthetases to recognize their cognate tRNA if modifications did not exist to differentiate the molecules. By contrast, rRNA contains enough sequence and structural variation to facilitate interaction with the appropriate ribosomal proteins.

15. The U1 and U2 snRNAs in the U1 and U2 snRNPs recognize and bind the 5′ splice site and branch point through base pairing, allowing for assembly of the remainder of the spliceosome complex.

16. Point mutations can lead to alternative splicing through different mechanisms. Elimination of a splice site through mutation in one of the conserved U1 or U2 recognition sequences can result in loss of U1 or U2 binding and retention of the intron. A point mutation can also introduce a cryptic 5′ or 3′ splice site, leading to splicing within an exon.

17. RNAi and miRNA have several common features; however, they differ primarily in the structure of the initiating double-stranded RNA and in the mechanism of translational repression. The siRNAs in RNAi are perfectly matched double-stranded RNA fragments, whereas miRNAs often contain loops and bulges. This difference extends to the interaction with the target mRNA. Short interfering RNAs are generally a perfect match to their target, which leads to mRNA cleavage. Micro RNAs, however, are often an imperfect match with mRNA, resulting in translational repression but not immediate cleavage.

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Long Answers to Challenge Problems Chapter 21 1. Bacterial and eukaryotic promoters have different recognition sequences for RNA polymerase and associated transcription factors. A eukaryotic gene under the control of its own promoter would not be transcribed because the σ factor would most likely not bind the promoter sequences, and therefore RNA polymerase would not become associated with the DNA. Because mRNA would not be produced, the protein would not be expressed.

Cloning the gene of interest into a plasmid containing a bacterial promoter sequence could solve this problem.

2. If ribosomes were the source of genetic material, then the bacteriophage would synthesize its own ribosomes soon after infection as a way of propagating its DNA. This synthesis would occur in the normal medium, and therefore these ribosomes would not contain any of the heavy isotopes. But if ribosomes were only sites of protein synthesis, then the viral mRNA would be translated by the existing bacterial ribosomes, which were synthesized while the heavy isotopes were present in the medium.

Because the ribosomes did contain these isotopes, Jacob and Meselson were able to confirm that ribosomes were not carrying the viral genetic material.

3. Continued association with σ factor would make RNA polymerase unable to continue down the template strand.

Dissociation of σ factor is necessary to increase the affinity of the holoenzyme for general DNA. With σ factor still a part of the complex, the enzyme would have a greater affinity for the –35 and –10 boxes and would not be able to move down the template.

4. RNaseP recognizes the secondary structure of tRNA and rRNA rather than the primary structure. Because these structures form co-transcriptionally, the primary transcript can be cleaved on the basis of the presence of these structures, not any individual sequences within the transcripts.

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5. The structure of amylopectin is very similar to that of glycogen, as both contain an α(1→4) chain of glucose molecules with α(1→6) branches. The only difference between the two is the spacing between branches. Therefore, it is reasonable to conclude that starch-branching enzymes and glycogenbranching enzymes contain some degree of similarity, as both catalyze the formation of the same bond. If siRNA directed toward repression of starchbranching enzyme was transmitted to mammals, it could result in repression of glycogen-branching enzyme. The storage of only straight-chain glycogen would decrease the speed of glycogen breakdown and reduce the amount of glycogen that could be stored. Both of these effects would have a negative impact on ATP production and could lead to symptoms similar to those observed in individuals with glycogen storage disorders.

6. Bacterial RNA polymerase is highly conserved; therefore, a molecule able to inhibit its activity would likely have the same effect in all bacteria. This widespread effect means that the antibiotic can be used when the type of bacteria causing the infection is not known, which is an advantage because most bacterial infections require treatment to begin as soon as possible. We know from Section 21.2 that the structure of RNA polymerase is conserved from bacteria to eukaryotes. Because these antibacterial agents do not inhibit eukaryotic RNA polymerase, it is known that these enzymes do not share a high degree of sequence conservation. The interaction between the antibacterial molecule and RNA polymerase depends on specific base contacts. Even though all RNA polymerases share structural features, differences at the primary sequence level means that the eukaryotic RNA polymerases do not form the same contacts, and therefore the inhibitor has no effect.

7. Both snRNA and snoRNA are small noncoding RNAs with roles in RNA processing. Both are part of protein complexes and are a critical part of the activity of these complexes. In addition, both bind to their target RNA to facilitate the processing reaction. These RNAs, however, differ in their cellular location and RNA target. Small nuclear RNAs are part of the snRNP complexes of the spliceosome, located in the nucleus. Each spliceosome subunit (that is, U1, U2, and so forth) is named after the snRNA bound within the complex. The U1 and U2 snRNAs are responsible for binding the intron sequences. By contrast, snoRNAs are located in the nucleolus because of their role in rRNA processing. There are two main classes of snoRNA: (1) those that facilitate base modification of rRNAs, and (2) those that are bound to ribonucleoprotein complexes that associate with rRNA as it is transcribed.

8. The crystal structure of mouse guanylyltransferase (Mce1) bound to an 18-AA CTD phosphopeptide shows that the phosphoryl group bound to Ser5 interacts with Arg386, Arg330, and Lys331. The addition of these amino acid contacts increases the affinity of Mce1 for the RNA polymerase II CTD by strengthening the association between the two proteins. Elimination of these contacts would not result in a complete lack of interaction because Mce1 contains additional residues (Cys383, Val372, and Cys375) whose interaction does not depend on the phosphorylation state of Ser5. In the crystal structure, the phosphoryl group on Ser2 is pointed

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away from the binding pocket; therefore, it does not contribute to binding.

If it were pointed toward the binding site, it would not increase the affinity because interactions between the Ser2 phosphoryl group and Cys375 would replace existing van der Waals interactions between Cys375 and Cα of Ser2 and Cδ of Pro3, not add new ones.

9. U2 and U6 snRNA base pair, leaving a bulge in the center of the structure.

The residues in this bulge are then free to base pair with the mRNA intron.

U2 snRNA binds to the conserved RY sequence in the branch point, leaving the adenine necessary for catalysis unpaired and extruding from the structure. The 5′ splice site is stabilized through base pairing interactions with the conserved U6 snRNA sequence ACAGAGA. This positioning brings the adenine in the branch point in close proximity to the 5′ splice site for the first transesterification reaction.

10. Both DNA and RNA polymerase read 3′ to 5′, resulting in a 5′ to 3′ polymerization reaction. It is possible then to have both a replication and transcription complex proceeding down the same region of DNA. Because transcription is a much slower process than replication, the DNA polymerase can overtake RNA polymerase, resulting in a codirectional collision.

11. Poly(A) polymerase interacts with the 2′-OH in the three adenines forming the end of the existing poly(A) tail (positions −1, −2, and −3). Lack of these 2′-OH in a single-stranded DNA template would therefore decrease the affinity of the enzyme for the substrate.

12. RNA synthesis requires NTPs, and the first nucleoside added therefore contains three phosphoryl groups. The reaction to add each subsequent nucleoside is initiated by a nucleophilic attack on the α phosphoryl, resulting in cleavage and release of PPi. 13. Strong bacterial promoters are those that differ only slightly from the conserved sequences found at the −35 and −10 boxes, whereas weak promoters can contain several base pair changes. Because these are the sites of σ-factor binding, the closer the −35 and −10 boxes are to the conserved sequences, the stronger the association between RNA polymerase and the promoter region. Therefore, strong promoters are more likely to be bound by the σ factor and transcribed, whereas weak promoters are bound more infrequently.

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Long Answers to Review Questions Chapter 22 1. Francis Crick did not support models for protein synthesis in which DNA or mRNA directly bound amino acids because eukaryotic DNA is found in the nucleus, whereas proteins are very abundant in the cytosol. In addition, nucleotide structure seemed to lack the complexity to act as a direct template for 20 amino acids that can have quite varying properties.

2. There are three possible amino acids that can be produced from the RNA sequence: Pro (CCA), Gln (CAA), or Lys (AAA).

3. The wobble hypothesis states that noncanonical base pairs can exist between the third (3′) position in the codon and the first (5′) position in the anticodon. Wobble base pairing explains how 64 triplet codons can be used to code for 20 amino acids. Some mRNA codons that differ by the nucleotide at the third position can bind to the same tRNA. Wobble base pairing is facilitated by the presence of nucleotides with modified bases such as inosine at the wobble position. Inosine can form hydrogen bonds with A, C, or U. 4. tRNA forms a cloverleaf secondary structure with three arms, known as the D arm, TΨC arm, and the anticodon arm. The top stem of tRNA is the acceptor site for amino acid attachment.

The D loop and TΨC loop further interact to form the L-shaped tertiary structure.

5. Aminoacylation is a two-stage reaction. In the first stage, the amino acid is activated through the formation of an aminoacylAMP moiety. In the second stage, the amino acid is transferred to the acceptor stem of tRNA. Class I aminoacyl-tRNA synthetases transfer the amino acid to the 2′-hydroxyl on the last conserved adenosine residue in the acceptor stem. Class II enzymes transfer the amino acid to the 3′-hydroxyl of this nucleotide residue.

6. The three tRNA binding sites on the ribosome are the aminoacyl (A), peptidyl (P), and exit (E) sites. The A site is the 1

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entry point during elongation for aminoacylated tRNAs in the ribosome.

The P site is the site of peptide bond formation between the amino acid bound to the tRNA in the A site and the growing polypeptide chain attached to the P-site tRNA. The uncharged tRNA leaves the ribosome from the E site.

7. When a termination codon on the mRNA is in the A site, a release factor binds to the A site on the ribosome. Binding of this protein prevents further tRNA binding and leads to dissociation of the ribosomal subunits.

8. Puromycin is not a useful antibiotic for the treatment of bacterial infections in humans because it affects both bacterial and eukaryotic protein synthesis.

It would lead to cell death in humans as it would in bacteria.

9. Proteins that contain a specific N-terminal signal sequence are directed to the endoplasmic reticulum during translation. During translation, the N-terminal signal sequence is recognized and bound by the signal recognition particle. Translation halts until the ribosomal complex is associated with the endoplasmic reticulum. When the signal recognition particle binds to a receptor on the endoplasmic reticulum, this allows the nascent polypeptide to enter and translation to resume.

10. Methylation and acetylation have an opposite effect on histones.

Methylation leads to tighter compaction of histones and is often found in transcriptionally inactive regions. By contrast, acetylation is often found in histones that are not as tightly condensed and are therefore more accessible by the transcriptional machinery. Because these two modifications have the opposite effect, they are not found on the same region of histone proteins.

11. Coatomer proteins include COPI, COPII, and clathrin. COPI vesicles mediate transport within the Golgi apparatus, COPII vesicles mediate transport between the endoplasmic reticulum and the Golgi apparatus, and clathrin vesicles are involved in transport between the Golgi apparatus and plasma membrane.

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Long Answers to Challenge Problems Chapter 22 1. Gamow concluded that three nucleotides were the minimum number required for each codon by calculating the possible combinations from codons containing different numbers of nucleotides. A two-nucleotide codon would result in only 42, or 16, possible combinations. Because there are 20 standard amino acids, a two-nucleotide codon would have required more than one amino acid to be associated with the same codon. In order for each amino acid to be associated with a unique combination of nucleotides, a codon of at least three nucleotides were necessary, giving 43, or 64, possible combinations.

2. Khorana’s method of nucleotide synthesis was the first procedure that allowed directed synthesis. Past methods of synthesis were only able to produce random combinations of nucleotides. The ability to synthesize specific combinations of nucleotides was essential to cracking the genetic code, as it allowed Khorana and others to test specific triplet combinations in filter-binding and in vitro translation experiments.

3. Under normal conditions, the mRNA sequence AUGCACAGU is translated to produce the tripeptide Met-Thr-Gly. It would be possible to change the sixth or ninth nucleotides without changing the amino acid sequence, as there are four codons for Thr and Gly that only differ at the third position. It would not be possible to change the third nucleotide because there is only one Met codon.

4. Mitochondria likely have a unique genetic code because of their ancestral origins. It is widely believed that mitochondria evolved from bacteria taken up by early nucleated cells. Because mitochondrial DNA has different evolutionary origins from chromosomal DNA, it contains a separate protein translation system, including different tRNA genes than those found in the nucleus.

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5. All tRNA molecules must be able to bind to elongation factors and the A, P, and E sites on the ribosome. These interactions are not dictated by the particular amino acid charged to the tRNA, so all tRNAs must have a similar structure to enable these functions.

6. The initiator tRNA is charged with a methionine that begins the protein sequence. During the process of elongation, a peptide bond is formed between the amino acids in the P and A sites. Because Met is the first amino acid, it must bind directly to the P site so that a second tRNA can bind to the A site and the first peptide bond can be formed.

7. A bacterial mRNA may encode more than one polypeptide. Having multiple Shine–Dalgarno sequences allows multiple positions for the start of translation to occur.

8. Translation initiation is different in prokaryotes and eukaryotes because the structure and organization of mRNA is different. Transcription and translation can occur simultaneously in prokaryotes, and the mRNA is generally unprocessed and does not contain any significant secondary structure. By contrast, eukaryotic mRNA must travel to the cytosol from the nucleus, so it contains modifications to ensure its stability, such as the 5′ cap and the 3′ poly(A) tail. It also may contain secondary structures that must be removed for efficient translation. Prokaryotic and eukaryotic mRNA also vary in the size of the 5′ untranslated region. These differences require additional factors in eukaryotic systems to form the pre-initiation complex.

Once translation has begun, however, the process is very similar.

9. Proteins cannot be transported to the nucleus during translation because there are no complexes analogous to the endoplasmic reticulum SRP–SRP receptor complex that work co-translationally for nuclear transport.

10. The ribosome is considered a ribozyme because it contains an RNA component (23S rRNA in E. coli) that acts as an enzyme in catalyzing the formation of a peptide bond between amino acids.

11. Proteins that are associated with a membrane often have lipid modifications to allow strong association of the protein and membrane. As soluble proteins fold, the hydrophobic regions are mostly sequestered in the interior of the protein, while hydrophilic regions remain exposed to the polar, aqueous environment of the cell. Membranes are arranged in a similar fashion, with the hydrophobic fatty acid tails arranged in the center of the membrane. In order for a protein to be strongly associated with the membrane, there needs to be a hydrophobic interaction between the two. The addition of multiple lipids to the exterior of a protein adds hydrophobic regions that can insert within the membrane and therefore firmly associate with the membrane.

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Long Answers to Review Questions Chapter 23 1. a. Transcription initiation: lac operon, yeast galactose genes, Drosophila eve gene. b. Protein stability: LexA protein. c. Protein posttranslational modification: many examples of metabolic regulation through phosphorylation, methylation, or acetylation.

2. Trans-acting factors can operate on DNA sites located anywhere in the cell or nucleus. They are not restricted to the chromosome that encodes them. Cis-acting sites are binding sites for transacting factors. They are almost always located at or near the genes that are affected by the trans-acting factors. Binding of these factors is usually highly specific, thereby restricting the action of the trans-acting factors to these few locations.

3. Recognition of a specific DNA sequence by a DNA binding protein usually is through interaction of the protein with the major groove of the DNA. The edges of the base pairs present patterns of hydrogen-bond donors and acceptors into the major and minor grooves. In the major groove, all four base pairs have different chemical patterns; in the minor groove, AT and TA cannot be distinguished by the pattern of hydrogen-bond donors and acceptors, and neither can GC and CG. Most DNA binding proteins interact with DNA through the major groove because there is more information to distinguish the bases in the major groove. Additionally, α helices in proteins are of an appropriate diameter to fit into the major groove; the minor groove is too small to accommodate an α helix. Therefore, some of the side chains projecting from the α helix can make specific contacts with the edges of the base pairs in the major groove. Some protein–DNA interactions involve less direct contact; often, a well-positioned water molecule can form hydrogen-bonding networks between the protein and the DNA. Some protein–DNA 1

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interactions involve van der Waals interactions, but these are usually less specific.

4. a. Protein is blocked by an inhibitor: yeast GAL4. b. Protein requires a binding partner: mammalian Oct4 and Sox2. c. Active protein is unmodified: yeast PHO4. d. Active protein is a ligand-bound form: CRP, Trp repressor. e. Active protein is a form not bound to ligand: lac repressor.

5. In positive regulation, expression of a gene is stimulated by the action of regulatory proteins.

Prokaryotes: activation of the lac operon by cAMP–CRP; activation of bacteriophage λ cI by CI. Eukaryotes: activation of GAL genes by GAL4; activation of eve by Bicoid and Hunchback.

In negative regulation, expression of a gene is inhibited by the action of regulatory proteins.

Prokaryotes: repression of the lac operon by the lac repressor, the trp operon by the Trp repressor, the SOS regulon by LexA. Eukaryotes: inhibition of TR target genes in the absence of ligand by recruited SMRT complex; inhibition of long segments of DNA by HP1; inhibition of eve stripe 2 by Giant and Krüppel.

6.

Prokaryotes

Eukaryotes

No nucleus

Eukaryotes have a nucleus

Transcription and translation can be coupled

Transcription and translation occur in separate compartments

One RNA polymerase

Three types of RNA polymerase

RNA polymerase binds promoter directly

Promoter binding specificity is not in RNA polymerase but in general transcription factors, which recruit RNA polymerase II

Very little RNA processing

Extensive RNA processing, especially splicing

No chromatin

Chromatin

Activators and repressors work directly on RNA polymerase

Activators and repressors work indirectly by effects on chromatin and recruiting other factors

Enhancers rare

Enhancers very common, especially in higher eukaryotes

Default state is that genes are accessible

Default state is that genes are inaccessible due to chromatin

7. In prokaryotes, activators generally work by directly interacting with RNA polymerase, either by recruiting it to a promoter or by increasing the rate of open complex formation. In eukaryotes, activators work on other components of the transcription machinery, mostly if not entirely by recruiting proteins or protein complexes to particular genes. Either they open

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up the chromatin by various means or they recruit components of the general transcription machinery.

8. In prokaryotes, repressors affect RNA polymerase directly. They usually block access of RNA polymerase to the promoter, although they can affect later steps in transcription. In eukaryotes, repressors work indirectly to set up repressive states of chromatin. For instance, they recruit HDACs, which deacetylate histones, or they recruit histone methyltransferases, which in turn lead to recruitment of HP1.

9. The first role is to antagonize the repressive effects of chromatin. Examples of this are recruitment of HATs, which acetylate histones and open up the chromatin. Acetylated histones also recruit other components. Activators can also recruit chromatin remodeling complexes, which move nucleosomes around or remove them, exposing other cis-acting sites for additional factors. The second role is to recruit components of the general transcription machinery. Common examples of these are the large protein complexes termed mediator and TFIID, although other components can also be recruited. For these complexes to operate, the promoter region must be relatively free of nucleosomes.

10. An epigenetic state is a long-lasting pattern of gene expression, almost always defined as being passed from one generation to the next through cell division. It does not involve a change in the DNA sequence. It may involve modification of the DNA base cytosine to form 5-methylcytosine, which pairs as cytosine during DNA replication.

11. Cooperative DNA binding is the ability of one regulatory protein bound to its site to increase the occupancy of another molecule at a different cisacting site, usually lying close to the first one. This usually operates directly by protein–protein interaction between the two bound molecules; however, it can occur by dimerization in solution, followed by binding of the dimer, or more indirectly by one protein making the binding site for the second one available, for example by fluctuations in nucleosome structure (termed “breathing”). One property of cooperative DNA binding is that occupancy of the cis-acting sites goes from near zero to nearly complete over a much narrower range of concentrations of the trans-acting factors. This confers switch-like behavior on the regulatory circuit, and this can help stabilize epigenetic states.

12. Regulatory systems often save energy by making the gene products for biosynthesis or catabolism only when they are needed. If the end product of the biosynthetic pathway is present in the environment, there is no need to make more. Conversely, if a metabolite is not available, there is no need to make the enzymes to break it down to products that enter intermediary metabolism. Examples: Anabolic pathway: tryptophan or histidine biosynthesis.

Catabolic pathway: lactose metabolism in E. coli; galactose metabolism in yeast.

13. Pioneer factors are the first trans-acting factors to bind to an enhancer or promoter for a particular gene, despite the presence of nucleosomes in those regions. Binding can lead in various ways to making other cis-acting sites

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available. Hence, they set a long chain of events in motion that leads to assembly of a pre-initiation complex and subsequent transcription. They may be able to bind because their binding site is available, either because it is not covered by a nucleosome or because the nucleosomes have variable positions and the site is sometimes free. Alternatively, they may be able to bind to DNA while it is wrapped around the histone core. Finally, their site might become transiently available because of fluctuations in nucleosome structure.

Examples of pioneer factors are yeast PHO4, whose binding site is available, and Oct4 and Sox2, which can likewise convert a range of cell types to iPS cells.

14. iPS cells are induced pluripotent stem cells. They have the potential to develop into almost any cell type, given the right conditions and stimuli.

They can be produced from a wide range of cell types by adding the genes for four transcription factors, Sox2, Oct4, Klf4, and c-Myc, either as free DNA or as viruses. A small fraction of cells treated in this way are converted to iPS cells.

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Long Answers to Challenge Problems Chapter 23 1. The E. coli pathway would certainly not work in eukaryotes because transcription occurs in the nucleus and translation occurs in the cytoplasm. This mechanism depends on both processes occurring simultaneously.

2. Cytosine residues undergo deamination at a very low rate.

Deamination of cytosine forms uracil, which will lead to a U-G base pair. Because U is not a normal component of DNA, this would quickly be repaired to a C-G base pair by the action of uracil N-glycosylase and subsequent events. Deamination of 5-MeC would lead to formation of a T-G base pair, and the cell would not rapidly repair this, as there is no way to be sure which base is the correct base. If DNA replication occurs before the mismatch is repaired, it fixes a C-to-T transition mutation in one of the progeny duplexes.

3. a. Mutation in region 4 so it cannot pair with region 3: This would be constitutive, because the terminator cannot form. b. Mutation in region 3 so it can pair with region 4 but not with region 2: This would be noninducible. Because the antiterminator cannot form in the absence of Trp–tRNATrp, the terminator would always form. c. Change of the two Trp codons to glycine codons: Noninducible, because the ribosome would not pause at the location of the Gly codons in the presence of low Trp–tRNATrp. It might pause if glycine levels are low, but that is unlikely because the cell contains abundant sources of glycine from metabolic pathways. d. Change of the AUG codon of the leader peptide to AUA: Noninducible, because the ribosome would not be able to stabilize the antiterminator. e. Combination of the first mutation (a) with each of the other three (b, c, or d). Constitutive, because if the terminator cannot form, it does not matter what happens upstream. For the combination with mutation (d), the pausing at the 1–2 stem– loop is too brief to reduce the rate of transcription significantly.

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This last type of test is termed an “epistasis” test by geneticists, and it is often used to order the events in a regulatory pathway.

4. To validate your hypothesis, you would use recombinant DNA to place the proposed enhancer upstream of a reporter gene and introduce this artificial construct into Drosophila embryos. Several constructs should be prepared, with the enhancer in either orientation and at a variable distance from the promoter for the reporter. If the DNA region includes an enhancer, you should see expression of the reporter.

Note that this was the approach taken by Levine and colleagues to analyze the eve stripe 2 enhancer (see Figure 23.54).

5. a. A lysogen present in a λ plaque can be superinfected by other λ virions, but the lysogen already contains CI protein, which will bind to the lytic promoters of the incoming viral DNA, repressing them and preventing the incoming virus from growing lytically. This property is termed “immunity.” b. Clear plaques would carry mutations in the cI gene; these cannot make CI protein and cannot establish or maintain the lysogenic state. Indeed, analysis of clear plaques led to the discovery of this gene, as well as two others; hence the name of the gene cI as clear plaque, class I. 6. Another possible explanation for your colleague’s findings is that the purified protein is an alternate σ factor. Recall that σ determines the promoter binding specificity and that prokaryotes contain multiple σ factors. To distinguish this model from the activator model, the in vitro transcription system could be changed so that the RNA polymerase used in the assay did not include the usual σ70 factor. This form of RNA polymerase is called the “core.” The new-sigma model predicts that the target gene would be transcribed; the activator model predicts no transcription.

7. Binding of the green factor would greatly enhance the ability of the red factor to bind, as its binding site is now accessible rather than being bound up in the nucleosome. This is another form of cooperativity, which can also operate by protein–protein interactions between the two bound proteins or by interaction of them in solution.

8. No, you cannot be certain what the effect of protein binding will be.

Numerous examples indicate that the effect of a bound trans-acting factor depends on its context. Bound bacteriophage λ CI protein and CRP–cAMP both can act as either activators or repressors. The mammalian Oct4–Sox2 heterodimer activates several ES-specific genes and works to repress developmental genes.

9. Three models have been suggested for this correlation, which are not mutually exclusive. First, acetylation neutralizes the charge on the lysine side chain, which may weaken interaction of the histone tails with the DNA in a nucleosome. Second, charge neutralization may weaken interaction between nucleosomes, disrupting higher-order structure. Third, acetylated histones serve as docking sites for other proteins and complexes. Although these interactions do not necessarily favor activation, this is found to be true in many cases.

10. GAL3 and GAL4 mutants will not express the target genes in the presence of galactose. A GAL80 mutant and a GAL3–GAL80 double mutant will express the target genes in the absence of galactose.

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