Essential Biochemistry, 3rd Edition Pratt, Cornely Solution Manual.pdf (22)

For exam and test prep, contact excellentessaywriters@gmail.com 59 · · · CUACCAGGAGCAAAGCUAAUGGCUUUA · · · 39 39 · · · UCCUC · · · 59 34. S1 helps to maintain mRNA in the single-stranded state and prevents it from forming a double-stranded structure that would block initiation of translation because the initiator tRNA would be unable to bind. 35. Colicin

Ch 22 Solutions

103

44. The number of phosphoanhydride bonds (about 30 kJ ? mol 1 each) that are cleaved in order to synthesize a 20-residue polypeptide can be calculated as follows (the relevant ATP- or GTP-hydrolyzing proteins are indicated in parentheses):

Aminoacylation (AARS) E3 is lethal to the cells because it prevents accurate and effi cient Translation initiation (IF-2) translation. Cleavage of the 16S rRNA at A1493 destroys the part of the 30S ribosomal subunit that verifi es codon–anticodon pairing. As a result, Positioning of each aminoacyl–tRNA (EF-Tu) the ribosome is less able to incorporate the correct aminoacyl group into a Translocation after each transpeptidation (EF-G) growing polypeptide. In addition, EF-Tu hydrolysis of GTP is slow because EF-Tu does not receive a signal from the ribosome that an mRNA–tRNA Termination (RF-3) match has occurred, so the speed of translation decreases. 36. In 16S rRNA, Total: 80 ATP equivalents

2 3 20 ATP 1 GTP 19 GTP 19 GTP 1 GTP

A1492 and A1493 act as a sensor to distinguish correctly and incorrectly paired codons and anticodons. tRNA binding triggers Thus, approximately 80 30 kJ ? mol 1, or 2400 kJ, is required. In a cell, a conformational change in the rRNA that allows A1492 and A1493 to form proofreading during aminoacylation and during translation requires the hydrogen bonds with an mRNA that has correctly base paired with a tRNA hydrolysis of additional phosphoanhydride bonds, making the cost of accuanticodon in the A site. Changing one of these two rRNA residues would inrately synthesizing the 20-residue polypeptide greater than 2400 kJ ? mol 1. activate the translational proofreading mechanism by eliminating the specifi c 45. In prokaryotes, both mRNA and protein synthesis take place in the hydrogen bonding between the rRNA and the mRNA. As a result, incorrectly paired tRNAs could not be distinguished from correctly paired cytosol, so a ribosome can assemble on the 5 end of an mRNA even while tRNAs, RNA polymerase is synthesizing the 3 end of the transcript. In eukaryotes, and the error rate of translation would increase. 37. The correctly charged RNA is produced in the nucleus, but ribosomes are located in the cytosol.

Because transcription and translation occur in separate compartments, they tRNAs (Ala–tRNAAla and Gln–tRNAGln) bind to EF-Tu with approximately the same affi nity, so they are delivered to the cannot occur simultaneously. A eukaryotic mRNA must be transported from the nucleus to the cytosol before it can be translated. ribosomal A site with the same effi ciency. The mischarged Ala–tRNAGln binds to EF-Tu more loosely, indicating that it may dissociate from EF-Tu 46. If a peptidyl–tRNA dissociates from the ribosome during translation, before it reaches the ribosome. The mischarged Gln–tRNAAla binds to EF- the hydrolase releases the peptide from the tRNA. Because peptide synthesis is prematurely terminated, the polypeptide is likely to be nonfunctional, and Tu much more tightly, indicating that EF-Tu may not be able to dissociate from its amino acids must be recycled. Similarly, the tRNA, once released from the it at the ribosome. These results suggest that either a higher or a lower bind- peptidyl group, can be reused. The essential nature of the peptidyl–tRNA ing affi nity could affect the ability of EF-Tu to carry out its function, which hydrolase suggests that ribosomes that have initiated translation sometimes would decrease the rate at which mischarged aminoacyl–tRNAs bind to the stop translating before reaching a stop codon. ribosomal A site during translation.

47. (a) The ribosome positions the peptidyl group for reaction with the incoming aminoacyl group, so a peptidyl group with a constrained 38. The ribosome minimizes the chances of misreading the A-site codon geometry, like Pro, is unable to react optimally. by binding the A-site tRNA with lower affi nity. If the tRNA bound with (b) Because Arg and Lys (both with positively charged side chains) react higher affi nity, it would be less likely to dissociate as part of the much faster than Asp (negatively charged side chain), the active site must be more accommodating of cationic groups than anionic groups. proofreading (c) Transpeptidation of Ala is faster than for Phe or Val, so for nonpolar mechanism. amino acids, small size is more favorable. [From Wohlgemuth, I., Brenner, 39. In a living cell, EF-Tu and EF-G enhance the rate of protein syntheS., Beringer, M., and Rodnina, M. V., J. Biol. Chem. 283, 32229–32235 sis by rendering various steps of translation irreversible. They also promote (2008).] the accuracy of protein synthesis through proofreading. In the absence of the elongation factors, translation would be too slow and too inaccurate to 48. During elongation, attack by the aminoacyl group attached to the support life. These constraints do not apply to an in vitro translation system, A-site tRNA breaks the ester bond linking the peptidyl group to the tRNA which can proceed in the absence of EF-Tu and EF-G. However, the result, the fMet–tRNA could ing protein is likely to contain more misincorporated amino acids than a in the P site. Bond cleavage by adding an amino group is aminolysis. During be delivered to the ribosomal protein synthesized in a cell. A site when a Met codon was positioned there. translation termination, a water molecule adds to the ester bond to remove However, transpeptidation could not occur because the amino group of fMet the peptidyl group from the tRNA. Bond cleavage by adding water is hy40.

If EF-Tu a complex fMet–tRNAMetf is blocked byformed the formyl group.with Polypeptide synthesis would be halted until drolysis.

Met et–tRNA in the A site. f the fMet–tRNAMet was replaced byM 49. (a) Transpeptidation involves the nucleophilic attack of the amino 41. The mRNA has the sequence group of the aminoacyl–tRNA on the carbonyl carbon of the peptidyl– tRNA (see Fig. 22-15). The higher the pH, the more nucleophilic the CGAUAAUGUCCGACCAAGCGAUCUCGUAGCA amino group (the less likely it is to be protonated).

The start codon and stop codon are highlighted. The encoded protein has 50. (a) the sequence Met–Ser– Asp –Gln–Ala – Ile – Ser.

NH

Peptidyl–tRNA

42. (a) Translation begins at the fi rst AUG codon (ATG in the DNA). The polypeptide sequence is Met–Val–His–Leu–Thr. (b) The mutated sequence has a T residue inserted in the second codon.

This is a frameshift mutation, so all codons following that point will be altered. The polypeptide sequence is Met–Val–Ala–Ser–Asp. 43. To encode 1480 amino acids, 1480 codons or 4440 nucleotides (1480 3) are needed, plus a stop codon. The 1686 additional mRNA nucleotides (6129 4443) include segments at the 3 and 5 ends where translation factors and the ribosome bind.

CHR O H

C N

O H

CHR Aminoacyl–tRNA

O

C O tRNA

tRNA

104 (b)

Ch 22 Solutions

30S subunits allows binding at one site to infl uence events at another site. [From Thompson, J., O’Connor, M., Mills, J. A., and Dahlberg, A. E., J. Mol. Biol. 322, 273–279 (2002).]

N HN 53. Anfi nsen’s ribonuclease experiment demonstrated that a protein’s primary structure dictates its three-dimensional structure. The purifi ed ribonuN clease was able to refold without the assistance of chaperones because other N cellular components were absent. Chaperones are required in vivo because they prevent the interaction and aggregation of the many proteins and other The protonated A2451 residue, which has a positive charge, could stabi- components that exist in the cell. Molecular chaperones assist in the proteinlize the negatively charged oxyanion of the tetrahedral reaction interme- folding process and do not contribute any additional information regarding diate. As the pH increases, A2451 would be less likely to be protonated, the tertiary structure of the protein; that information is contained within the primary structure of the protein, as demonstrated by Anfi nsen. so this catalytic mechanism would be less effective at higher pH. 54. Protein folding is driven by the hydrophobic effect (see Section 4-3), 51. (a) The mutation would allow the aminoacylated tRNA rather than a which directs nonpolar side chains to the interior of the folded protein berelease factor to enter the ribosome and pair with a stop codon. The cause these side chains do not interact favorably with water. The derivatized result would be incorporation of an amino acid into a polypeptide rather protein was able to fold as easily as the native protein because the lysine than translation termination, so the ribosome would continue to read residues are located on the surface of the protein and do not play a large role mRNA codons and produce elongated polypeptides. The inability of in the protein-folding process. the mutated tRNA to recognize its amino acid–specifying codons would 55. Mitochondria contain ribosomes that synthesize proteins encoded by have a minor impact on protein synthesis, since the cell likely contains mitochondrial DNA. Like cytosolic proteins, mitochondrial proteins require other isoacceptor tRNAs that can recognize the same codons. the assistance of chaperones for proper folding. Other proteins are synthe(b) Not all proteins would be affected. Only proteins whose genes include sized in the cytosol and are transported partially unfolded through pores in the stop codon that is read by the mutated tRNA would be affected. the mitochondrial membranes; these proteins also require the assistance of Proteins whose genes include one of the other two stop codons would be chaperones to fold properly once they reach their destination. synthesized normally. (c) Aminoacyl–tRNA synthetases usually recognize both the anticodon 56. Cells expressing growth factor receptors that have lost the ligand-binding and acceptor ends of their tRNA substrates. A mutation in the tRNA domain but retain the tyrosine kinase domain are transformed cells; that is, anticodon, such as a nonsense suppressor mutation, might interfere with they are cancerous cells (see Box 10-B) because they grow and proliferate in the tRNA recognition, so the mutated tRNA molecule might not undergo absence of extracellular signaling ligands. Hsp90 binds to the tyrosine kinase aminoacylation. This would minimize the ability of the mutated tRNA domain to stabilize it or help it fold. When geldanamycin is added to these to insert the amino acid at a position corresponding to a stop codon. transformed cells, Hsp90 can no longer function, the tyrosine kinase activity drops, and the cells no longer exhibit the characteristics of transformed cells.

52. (a) Introduction of a stop codon would terminate protein synthesis pre57. The different domains in a multidomain protein associate with one maturely, so no functional -galactosidase would be synthesized. If the stop codon were located near the C-terminus of the protein, the poly- another via van der Waals forces, since the domain interfaces eventually end peptide would be shorter than normal but might still retain activity if its up in the interior of the protein. A cagelike chaperonin structure allows these proteins to fold in a protected environment where the hydrophobic regions active site region were intact. (b) These results indicate that the oxazolidinone increases the ability of of the protein are not exposed to other intracellular proteins with which they ribosomes to overlook the stop codon and synthesize -galactosidase could potentially aggregate.

58. (a) The chains, when in excess, combine with all available chains to polypeptides of normal size. (c) -Galactosidase activity would be extremely low because a nucleoform functional 2 2 hemoglobin, thereby minimizing the formation of tide insertion or deletion changes the reading frame for translation. The nonfunctional 4 hemoglobin. resulting polypeptide would have a different amino acid sequence and (b) The protein helps prevent the precipitation of the chains that have not yet paired with chains. would therefore be nonfunctional. (d) These results indicate that the oxazolidinone promotes frameshift- (c) When a defi ciency of chains is coupled with an excess of chains, ing in the ribosome so that despite the insertion or deletion, the correct the chains precipitate and destroy the red blood cells, worsening the reading frame is occasionally translated and some functional -galacto- anemia that results from the lack of chains. (d) The imbalance between the amounts of and chains is minimized sidase is synthesized. (e) There are two possible Glu codons: GAA and GAG. Changing a when the synthesis of both globins is depressed due to mutations in both single base in these codons generates codons specifying four different an globin gene and a globin gene. amino acids. Substitutions at the second position yield codons for Ala (e) The process of initiating translation requires that eIF2 hydrolyze its (GCA and GCG), Gly (GGA and GGG), and Val (GUA and GUG). bound GTP to GDP and Pi. In order for the protein to participate in Substitution at the fi rst position yields codons for Gln (CAA and CAG). subsequent translation initiation events, its GDP must be replaced with (f) The oxazolidinone does not promote codon misreading, that is, in- GTP. If this exchange does not occur, reinitiation is not possible, and corporation of an amino acid other than the one specifi ed by a codon. protein synthesis comes to a halt.

If codon misreading were occurring, the encoded Ala, Gln, Gly, and (f) Heme prevents the phosphorylation of eIF2, so translation initiation Val codons would occasionally be read as Glu codons, and a functional can proceed. This mechanism regulates the level of globin synthesis acenzyme would result. cording to the availability of heme. Consequently, the cells can produce (g) The ability of the oxazolidinone to cause the ribosome to read functional hemoglobin, which contains globin polypeptides as well as through stop codons would result in the synthesis of longer-thanheme prosthetic groups. normal polypeptides. These proteins might not fold properly or might not be catalytically active, which would interfere with normal cellular 59. The basic residue is highlighted in gray; the hydrophobic core is unmetabolism. The ability of the oxazolidinone to promote frameshifting derlined. would result in polypeptides with garbled amino acid sequences, which would be nonfunctional and potentially toxic to the cell.

MK WVTFISLLLLFSSAYSRGV (h) No; the oxazolidinone affects translational accuracy, which is primarily a function of the 30S ribosomal subunit, the location of the binding 60. During translation, the SRP must be positioned on the ribosome where sites for mRNA and the tRNA anticodon loop. A binding site near the the signal peptide emerges from the exit tunnel. The SRP then undergoes a peptidyl transferase site would be more consistent with an effect on pep- conformational change that temporarily pauses the translation process. In tide bond formation. Presumably, communication between the 50S and order to fulfi ll these roles, the SRP must be able to bind to the ribosome, most likely through interactions between the SRP RNA and ribosomal RNA.

NH2

Ch 22 Solutions

105

61. The hydrophobic cleft of this particular protein might allow it to (c) The acetyltransferase acetylates Lys residues in histones, neutralizing recognize the hydrophobic core of the signal sequence (see Problem 59). the positive charge of the Lys side chain and weakening its interaction Other proteins in the SRP might be involved in pausing translation and with the DNA, so transcriptional activity is increased. If the acetylpromoting translocation. transferase is inactive, the DNA will be less transcriptionally active and certain genes will not be expressed. The loss of transcriptional activ- ity 62. If the SRP waited to bind the polypeptide until translation was comcould contribute to the progression of the polyglutamine disease. [From plete, it is highly likely that the newly synthesized protein would have Pennuto, M., Palazzolo, I., and Poletti, A., Hum. Mol. Gen. 18, R40–R47 already misfolded into a conformation that included its signal peptide or (2009).] would have aggregated with other cellular proteins. By binding to the nascent polypeptide as soon as its signal sequence emerges from the ribosome, the SRP ensures the delivery of the ribosome to the ER membrane so that 66.

CHOH 2 the polypeptide can be translocated co-translationally. Chaperones within O the ER lumen then ensure that the protein folds properly.

HO H H H O CH3 NH 63. In the cell-free system, SRP can bind to the exit tunnel of the ribosome, OH NH H but translation is not arrested when no membrane is present. This indicates CH CH that the SRP must interact with both the nascent polypeptide and the ER C C O H membrane in order to pause translation. When microsomal membranes are subsequently added, the protein is not translocated, indicating that transloO cation must occur co-translationally, not post-translationally. Proteins that CH3 are not translocated retain their signal sequences because they do not have access to the signal peptidase, which is located in the microsomal lumen.

64. The SRP recognizes the N-terminal signal sequence as it emerges from 67.

O the ribosome, pauses translation, and escorts the entire complex to the ER H H3C (H2C)12 C N CH C membrane, where it docks with its receptor and translation resumes. GTP is required for this process. GTP hydrolysis could be part of a proofreading O H step by occurring in response to SRP conformational changes and by triggering additional changes in the SRP. This mechanism would ensure that only proteins with signal sequences were translocated into the ER. Cyto- 68. G proteins (see Section 10-2) have palmitoylated Cys residues and as such solic proteins lacking signal sequences would not trigger GTP hydrolysis, are classifi ed as lipid-linked proteins. The nonpolar palmitate acyl chain helps to anchor this protein to the membrane. and the SRP would not be able to dock with its ER receptor. (b) 65. (a) O O H N

C

CH CH2 O

O

P O

H N

C

CH

H N

CH2 Polyglutamine protein

CH2 O

OH

CH2 S

CH2 CH2 HN

O C

CH

O C O

Ubiquitin

C

(CH2)14

CH3

Case 2 Histidine-Proline-rich Glycoprotein as a Plasma pH Sensor Last modified 6 September 2010

Focus concept A histidine-proline-rich glycoprotein may serve as a plasma sensor and regulate local pH in extracellular fluid during ischemia or metabolic acidosis.

Prerequisites C C

Acidic/basic properties of amino acids Amino acid structure and protein structure

Background This study focuses on the characteristics of the abundant plasma protein referred to as histidineproline-rich glycoprotein (HPRG). HPRG is so named because it has a very high content of histidine (13 mol %). The human HPRG contains a pentapeptide GHHPH sequence repeated in tandem twelve times.

The authors of this study hypothesized that its high histidine content might allow HPRG to play a role in regulating local pH in the blood. The local pH in blood may drop a half a pH unit during lactic acidosis or even a full pH unit in hypoxia or ischemia. In the case presented here, the binding of HPRG to glycosaminoglycans was investigated. Glycosaminoglycans are anionic polysaccharides that are the major component of the ground substance that forms the matrix of the extracellular spaces of the connective tissue in blood vessel walls. In this study, the binding of HPRG to the glycosaminoglycan heparin was measured. Based on their results, the investigators propose a model which describes how binding of HPRG to glycosaminoglycans may allow HPRG to regulate local blood pH.

Questions 1. Binding studies were carried out in which heparin was immobilized on the surface of cuvettes. The pH dependence of HPRG binding to heparin was investigated and the results are shown in Figure 2.1. a. How is the binding of HPRG to heparin dependent on pH? Give structural reasons for the binding dependence. The structure of heparin is shown in Figure 2.2. NOTE: Figure 2.1 also serves as a titration curve for the His residues in Figure 2.1: The pH-dependence of HPRG binding to HPRG–what is the pK for these R unmodified heparin and heparin modified with DEPC. residues? Explain. b.The same binding studies were carried out in which HPRG was reacted with diethylpyrocarbonate (DEPC), a compound that specifically reacts with histidine residues. The reaction is shown in Figure 2.3. Explain the results.

1

CASE 2 CHistidine-Proline-rich Glycoprotein as a pH sensor

Figure 2.2: Repeating disaccharide unit of heparin

Figure 2.3: Reaction of histidine side chains with diethylpyrocarbonate

2

CASE 2 C Histidine-Proline-rich Glycoprotein as a pH sensor 2. The effect of transition metals on the binding of HPRG was investigated next. The ability of increasing concentrations of Cu2+ and Zn2+ to promote HPRG binding to heparin at pH = 7.3 was measured. The results are shown in Figure 2.4. In addition, the binding of HPRG to heparin in the presence of these ions was compared at various pH’s. Figure 2.5 shows the comparison of binding of various concentrations of zinc ions at pH values ranging from 6.0 to 7.4. What is your interpretation of these results?

Figure 2.4: Binding of HPRG to heparin in the presence of copper and zinc ions

Figure 2.5: Binding of HPRG to heparin in the presence of increasing concentrations of zinc ions, at pH values 6.0-7.4.

3

CASE 2C Histidine-Proline-rich Glycoprotein as a pH sensor 3. Local cellular pH can decrease from one-half to one pH unit depending on a variety of circumstances including ischemia, hypoxia, and inflammation due to lactic acidosis. In addition, metabolic acidosis is often one of the symptoms in complications following surgery. The investigators have proposed that HPRG acts to relieve the acidosis in these circumstances, possibly in synergy with Zn2+ ions. Propose a simple model that explains the mechanism of pH regulation by HPRG. 4.Other plasma proteins have been studied for their ability to bind to glycosaminoglycans. One such protein is kininogen, which is a lysine-rich protein. Like HPRG, kininogen is able to bind to glycosaminoglycans, but this binding is far less sensitive to small fluctuations in physiological pH. a.Why does kininogen bind to glycosaminoglycans easily? b.Why is the binding of kininogen less sensitive to physiological pH changes?

References Borza, D-B., and Morgan, W. T. (1998) J. Biol. Chem., 273, pp. 5493-5499.

Lundblad, R. (1995) Techniques in Protein Modification, CRC Press, Boca Raton, FL, p. 111.

4

Case 4 The Structure of Insulin Last modified 3 January 2013

Focus concepts

The three dimensional structure of insulin is examined and sequences of various animal insulins are compared.

Prerequisites   

Amino acid structure Protein architecture Basic immunology

Background

Diabetics lack the protein insulin, which is produced by the pancreatic β-cells of the islets of Langerhans. Insulin stimulates uptake of glucose from the blood into the tissues. Diabetes is treated by replacing the missing insulin. Human insulin is produced industrially by recombinant bacteria, but before this method was available, animal insulin was used instead.

Insulin consists of two polypeptide chains, an A chain and a B chain, joined together by disulfide bonds. The smaller of the two chains is referred to as the A chain and is 21 amino acids long in humans.

The second chain is referred to as the B chain and is 30 amino acids long in humans. Insulin from various animals is similar to, but not identical to human insulin, as illustrated in Table 4.1. A schematic diagram of the structure of insulin is shown in Figure 4.1.

Table 4.1: Variation in positions A8, A9, A10, B1, B2, B27 and B30 of insulin. (All other amino acids are the same.)

Species

A8

A9

A10

B1

B2

B27

B30

human

Thr

Ser

Ile

Phe

Val

Thr

Thr

cow

Ala

Ser

Val

Phe

Val

Thr

Ala

pig

Thr

Ser

Ile

Phe

Val

Thr

Ala

horse

Thr

Gly

Ile

Phe

Val

Thr

Ala

rabbit

Thr

Ser

Ile

Phe

Val

Thr

Ser

dog

Thr

Ser

Ile

Phe

Val

Thr

Ala

chicken

His

Asn

Thr

Ala

Ala

Ser

Ala

duck

Glu

Asn

Pro

Ala

Ala

Ser

Thr

1

Questions 1. What animals would serve as the best sources of insulin to be used for treating diabetics? Explain your answer.

2. Would the pI values of the animal insulins be the same as, greater than, or less than human insulin?

3. Some people developed allergies to the animal insulin because their immune systems recognized the proteins as foreign. Explain why the immune system would be able to distinguish animal insulin from human insulin.

4. An SDS-PAGE gel is run of proinsulin and insulin. Samples were treated with β-mercaptoethanol prior to electrophoresis. Draw a picture of the predicted results.

5. A denaturation/renaturation (similar to the one carried out by Anfinsen with ribonuclease) experiment was carried out using insulin. However, in contrast to Anfinsen's results, only less than 10% of the activity of insulin was recovered when urea and β-mercaptoethanol were removed by dialysis. (This is the level of activity you would expect if the disulfide bridges paired randomly.) In contrast, if the experiment is repeated with proinsulin, full activity is restored upon renaturation.

Explain these observations.

Figure 4.1: The structure of proinsulin. The arrows show the cleavage sites of the peptide bonds that are hydrolyzed when proinsulin is converted to insulin (from Voet, Voet, and Pratt, 2013).

6. Diabetics are treated with insulin, not proinsulin. Do you think that this is a good idea?

2

References Devlin, T. M., ed. (1997)Textbook of Biochemistry with Clinical Correlations , Wiley-Liss, NY, p. 41.

Steiner, D. F., and Rubenstein, A. H. (1997) Science 277, pp. 531-532.

Voet, D., Voet, J., and Pratt, C. W. (2013) Fundamentals of Biochemistry , John Wiley and Sons, New York.

3

Case 8 Hemoglobin, the Oxygen Carrier Focus concept A mutation in the gene for hemoglobin results in an altered protein responsible for the disease sickle cell anemia. An understanding of the biochemistry of the disease may suggest possible treatments.

Prerequisite Hemoglobin structure and function concepts.

Background Normal adult hemoglobin is called Hemoglobin A (Hb A). Ninety-eight percent of adult hemoglobin is Hb A and 2% is Hb A2. There are other forms of hemoglobin. For example, the developing fetus has a different kind of hemoglobin than most normal adults. Fetal hemoglobin (or Hemoglobin F) consists of two á chains and two ã chains, whereas adult hemoglobin (Hemoglobin A) consists of two á chains and two â chains. Fetal hemoglobin is synthesized beginning at the third month of gestation and continues up through birth. After the neonate is born, hemoglobin F synthesis declines (because synthesis of the ã chain declines) and hemoglobin A is synthesized (because synthesis of â chains begins). By the time the baby is six months old, 98% of its hemoglobin is Hemoglobin A. There is also a mutant form of hemoglobin called Hemoglobin S which is found in persons with the disease sickle cell anemia. The disease sickle cell anemia is one of the major health problems facing the African-American community. The World Health Organization estimates that 250,000 babies world wide are born with sickle cell anemia. Currently there is no cure. A person afflicted with sickle cell anemia has inherited a defective gene from each parent. (Parents who are carriers of the sickle cell gene are heterozygous AS, whereas the person afflicted with sickle cell anemia is SS; non-carriers are designated AA.) The defective gene is the one coding for the â-chain. The amino acid at position 6 on each â chain has been mutated from a glutamate to a valine. Normal á chains have a decreased affinity for the mutated â chains; thus assembly of the HbS tetramer is more difficult. Red blood cells containing HbS form a sickle shape because the Hb S molecules polymerize. Hb S molecules are more likely to polymerize when in the deoxygenated T form than in the oxygenated R form. The polymerized Hb deforms the normal discoid shape of the red blood cells, producing a sickle-shaped cell. The sickle shaped red blood cells become trapped in capillaries and organs, depriving the victim of adequate oxygen supply and causing chronic pain and organ damage.

In this case we will consider our patient, a 10-year-old black male child named Michael B., who was admitted to the hospital because he was experiencing severe chest pain. He had been hospitalized on several previous occasions for vaso-occlusive episodes that caused him to experience severe pain that could not be managed with non-prescription drugs such as ibuprofen. He was slightly jaundiced, short of breath and easily tired, and feverish. A chest x-ray was taken and was abnormal. An arterial blood sample showed a pO2 value of 6 kPa (normal is 10-13 kPa).

1

CASE 8 CHemoglobin, the Oxygen Carrier

Questions 1.You suspect that Michael has sickle cell anemia and you have ordered an isoelectric focusing analysis of the child’s lysed red blood cells. (Lysing the red blood cells releases the hemoglobin.) Draw a diagram of the predicted results. Why will this test allow you to diagnose this child’s disease?

2.Why do you think that Hb S molecules would be likely to clump together whereas Hb A molecules do not?

3. In the emergency room, oxygen (100%) was administered to the patient. (Inspired air normally is about 20% oxygen.) Why was this an effective treatment?

4.You recall reading in the medical literature about a dramatic new drug treatment for sickle cell anemia, and you’d like to try it on this patient. The drug is hydroxyurea, and is thought to function by stimulating the afflicted person’s synthesis of fetal hemoglobin. Exactly how hydroxyurea stimulated fetal hemoglobin synthesis is unclear, but it is believed that hydroxyurea is metabolized to NO, which binds to a soluble guanylate cyclase enzyme which then catalyzes the synthesis of a second messenger, cyclic GMP (cGMP). The cGMP interacts with transcription factors in a manner that is not completely understood to induce the transcription (and then translation) of the fetal hemoglobin gene. a.In a clinical study, patients who took hydroxyurea showed a 50% reduction in frequency of hospital admissions for severe pain, and there was also a decrease in the frequency of fever and abnormal chest x-rays. Why would increasing the synthesis of fetal hemoglobin result in alleviating the symptoms of sickle cell anemia? b.Medical practioners who used hydroxyurea as a treatment for sickle cell anemia noted that their patients seemed to benefit from the administration of the drug long before the synthesis of fetal hemoglobin had time to take effect. It has recently been determined that hydroxyurea can react directly with the iron ion of oxy- and deoxyHb to form iron nitrosyl hemoglobin (HbNO). Why would this be of benefit to the sickle-cell anemic patient? c. And finally, hydroxyurea has been shown to produce NO directly, in less than an hour after hydroxyurea administration. It’s also possible that HbNO could produce NO, either directly or indirectly. NO is an important second messenger which, even in nanomolar amounts, stimulates vasodilation. How could NO production help the sickle-cell anemic patient?

5. A year ago, at a conference, one of your colleagues told you that she had “cured” a patient of sickle cell anemia by performing a bone marrow transplant. Why would this procedure “cure” sickle cell anemia?

6. The patient’s parents tell you that they are planning on having another child and that they are confident that subsequent children will not have sickle cell anemia, since they already have a child with the disease. What will you tell them?

2

CASE 8 CHemoglobin, the Oxygen Carrier

References Glew, R. H., and Ninomiya, Y. (1997) Clinical Studies in Medical Biochemistry, Oxford University Press, pp. 78-90.

Glover, R. E., Ivy, E. D., Orringer, E. P., Maeda, H., and Mason, R. (1999) Molecular Pharmacology 55, pp. 1006-1010.

Huang, J., Hadimani, S. B., Rupon, J. W., Ballas, S. K., Kim-Shapiro, D. B., and S. Bruce King (2002) Biochemistry 41, pp. 2466-2474.

S. Bruce King (2003) J.Clin. Invest., 111, pp. 171-172.

Acknowledgment Thanks to Richard Pedro ‘05 for his assistance with this case study.

3

Case 12 Production of Methanol in Ripening Fruit Focus concept The link between the production of methanol in ripening fruit and the activity of pectin methylesterase, the enzyme responsible for methanol production, is examined in wild type and transgenic tomato fruit.

Prerequisite Enzyme kinetics and inhibition.

Background Methanol is produced by fruit during the ripening process. The major source of this methanol is likely to be from the enzyme pectin methylesterase (PME) which acts upon pectin methyl esters to de-esterify them and produce methanol and other pectic substances. One might assume that the production of methanol is correlated with PME activity, but this has not been firmly established. In this case, the investigators at the Plant Science Department at Rutgers University developed a transgenic tomato which was deficient in PME activity. By comparing the behavior of the transgenic tomato with the wild type tomato, the production of methanol in the ripening tomato fruit may be better understood.

Questions 1. The investigators measured PME enzyme activity in immature green (IMG), mature green (MG), breaker (Br), turning (Tu) and red ripe (RR) tomatoes. They also determined the amount of PME protein present by using a Western blot assay, which uses a specific antibody to detect the PME protein. The results are shown in Figure 12.1. What is your interpretation of these results?

A

B

Figure 12.1: PME protein levels (A) and activity (B) in wild-type and transgenic tomatoes.

(Modified from Frenkel, et al., 1998.)

1

CASE 12 CProduction of Methanol in Ripening Fruit 2. The investigators next wanted to study the correlation between PME activity and methanol levels.

They measured methanol content of tomato tissue at each developmental stage of ripening as described in Question 1. The results are shown in Figure 12.2. a. Compare the results in Figure 12.2 with those presented in Figure 12.1. Is methanol production correlated with PME activity? Be specific. b.What is responsible, at least in part, for the production of methanol in the ripening fruit?

Figure 12.2: Methanol content in ripening wild-type and transgenic tomato fruit (based on Frenkel, et al., 1998).

3.Next, the investigators measured the ethanol content in the ripening fruit for the wild type and transgenic tomatoes. The results are shown in Figure 12.3. Compare the results in Figure 12.2 with the results presented in Figure 12.3. What is your interpretation of these results?

Figure 12.3: Ethanol content in ripening wild-type and transgenic tomato fruit. (Based on Frenkel, et al., 1998.)

2

CASE 12 CProduction of Methanol in Ripening Fruit 4.Methanol is produced from the action of PME on pectin methyl esters, as described above. The only route for ethanol production in the plant is the alcohol dehydrogenase-catalyzed reduction of acetaldehyde to ethanol. Write a balanced equation for the production of ethanol, including structures and cofactors.

5. In order to understand the mechanism for ethanol production, the investigators measured the activity of the alcohol dehydrogenase (ADH) enzyme in both the wild type and transgenic tomatoes at each stage of the fruit-ripening process. The results are shown in Figure 12.4. What factor or factors might influence ADH activity in the ripening tomato?

Figure 12.4: Alcohol dehydrogenase activity in ripening wild-type and transgenic tomato fruit (based on Frenkel, et al., 1998).

Reference Frenkel, C., Peters, J. S., Tieman, D. M., Tiznado, M. E., and Handa, A. K. (1998) J. Biol. Chem., 273, pp. 4293-4295.

3

Case 13 Inhibition of Alcohol Dehydrogenase Last modified 12 July 2012

Focus concept The inhibition of the alcohol dehydrogenase by a formamide compound is examined.

Prerequisites C C

Principles of enzyme kinetics Identification of inhibition via Lineweaver-Burk plots

Background Alcohol dehydrogenase (ADH) is the enzyme that is responsible for converting ethanol to acetaldehyde (the reaction is shown in Figure 13.1). It is the enzyme responsible for the metabolism of ethanol in the alcoholic beverages we consume. Five different isozymes of ADH have been identified, and it has been shown that the enzyme has a rather broad substrate specificity and can oxidize aldehydes as well as primary and secondary alcohols. For example, ADH can also oxidize methanol (wood alcohol) and ethylene alcohol (antifreeze). The poisonous nature of these compounds results from the ADH-catalyzed conversion of these compounds to toxic products. For example, ADH converts methanol to formaldehyde, which is toxic to the optic nerve and can produce blindness. In high doses, formaldehyde may be fatal.

In this study, the authors investigated the ability of formamide compounds to inhibit alcohol dehydrogenase. Only a portion of their data is presented here. The authors were able to propose a mechanism for the inhibition from the extensive data they collected using a wide variety of formamide compounds. The mechanism is shown in Figure 13.2.

Figure 13.1: ADH-catalyzed oxidation of ethanol.

Figure 13.2: Mechanism of ADH1. The inhibitor binds as an

aldehyde analog.

Questions 1. Certain individuals are more sensitive to alcohol than others. For example, women are more sensitive to alcohol than men–even when body weight and % body fat are taken into account, women become more intoxicated than men consuming an identical amount of alcohol. Using what we have learned in the enzyme chapters, give biochemical reasons that would explain why women become more intoxicated than men when consuming an equal amount of alcohol.

2.A treatment for methanol poisoning is to have the victim drink large amounts of ethanol. Why might this be an effective treatment?

3.The authors of this study studied the ability of N-1,5-dimethylhexylformamide to inhibit mouse ADH1. The activity of the enzyme was measured in the absence of inhibitor, and in the presence of 1.0 ìM inhibitor. The data are presented in Table 13.1.

Table 13.1: Inhibition of mouse ADH1 by N,1-5-dimethylhexylformamide

Ethanol Concentration, mM

ADH1 velocity, Ä NADH absorbance/min (without inh ibit or)

ADH1 velocity, Ä NADH absorbance/min (with inh ibit or)

0.20

0.036

0.022

0.25

0.042

0.024

0.36

0.048

0.027

0.60

0.065

0.029

2.00

0.075

0.033

a. Construct a Lineweaver-Burk plot using the above data. What are the KM and Vmax values for ADH in the absence of inhibitor? in the presence of the inhibitor? Show your calculations. b Calculate the values of á and/or á’, if they are significantly different from 1. . What kind of inhibitor is N-1,5-dimethylhexylformamide? Explain, using the data you c. collected in the first part of this problem. What are the characteristics of this type of inhibit o r ? d. Calculate the Ki and/or Ki’ (whichever is appropriate) for N-1,5-dimethylhexylformamide (Hint: You can obtain these values from á and á’, respectively).

4. The authors describe the mechanism of ADH as an “ordered bi-bi” mechanism, or an ordered bisubstrate mechanism. The mechanism is diagramed in Figure 13.2; use the diagram to write a description of the mechanism. How does N-1,5-dimethylhexylformamide inhibit the activity of the ADH enzyme? How does N-1,5-dimethylhexylformamide differ from the “classic” inhibitors of this type that are described in our textbook?

5. The authors found that a class of compounds called pyrazoles were also inhibitors of ADH. These inhibitors bind to the E-NAD+ complex. What kind of inhibitor are pyrazoles? Are these inhibitors the same or different as the formamides? Explain.

6. a.Would N-1,5-dimethylhexylformamide be an effective alternative for the treatment of methanol and ethylene glycol poisoning, assuming that it is non-toxic itself (and as an alternative to getting the patient drunk, as described in Question 2)? (Hint: There are two considerations here (1) substrate concentration and (2) affinities of the substrate and inhibitor for the enzyme). b.The compound 4-methyl pyrazole is currently being used as a treatment for methanol poisoning. How would the effectiveness of 4-methyl pyrazole compare with the effectiveness of a formamide treatment?

Reference Venkataramaiah, T. H., and Plapp, B. V. (2003) J. Biol. Chem. 278, pp. 36699-36706.

Case 14 Shavings from the Carpenter’s Bench: The Biological Role of the Insulin C-peptide Focus concept Recent experiments indicate that the insulin C-peptide, which is removed upon conversion of proinsulin to insulin, may have biological activity in its own right.

Prerequisites C C C

Amino acid structure.

Principles of protein folding.

Membrane transport proteins.

Background The protein hormone insulin is synthesized as a prohormone and is processed by cleavage of two peptide bonds to yield the mature hormone, which consists of two separate polypeptide chains, A, and B, linked by disulfide bonds. The spliced out portion is referred to as the C peptide and, although the excised C peptide is secreted along with the mature insulin, scientists believed that the C peptide was biologically inactive. However, recent evidence indicates that the C peptide may have biological activity in its own right. This discovery has important implications for the treatment of diabetes.

Pharmacological doses of C-peptide administered to diabetic rats increased the level of activity of the Na+K+ATPase pump, improved nerve conduction, vascular permeability and blood flow, but did not decrease hyperglycemia, nor did it prevent the glycosylation of many proteins seen in diabetics.

Interestingly, a synthetic C-peptide made up of D-amino acids was able to bring about these same effects.

Rat and human C-peptides showed the biological effects just described while C-peptides from pig and cow did not.

The authors proposed that the C-peptides were able to alter biological activity of cells by becoming integrated into the cell membrane, forming an ion channel and altering cell function. However, other researchers think that this hypothesis is unlikely.

1

CASE 14 CShavings from the Carpenter’s Bench: The Biological Role of the Insulin C-peptide

Table 14.1:

Sequences of human and animal insulins. The amino acid residues that make up the C peptide are underlined.

Rat 1 16 31 46 61 76

1 2 3 4 6 5 Leu P h e Asn V aGlnl HisCys Gly T y r I l Val e Leu S e r A r Glu g Arg Gly Val Pro Glu G l y A Gly l a Arg Gln Ala V a l A l a Ala Arg C y s Leu S eTyrr Gln Glu

7 CysGly 8 9 10 11 12 GluArg SerHis LeuVal AspPro GlyPhePhe Tyr GlnVal GlyGln ThrLeu Gly Ser GluGln CysThr Ile Val GlnGln Cys Asn Asn Tyr

13 14 15 Glu Ala Leu Thr Pro Met Val Glu Leu Ala Leu Glu Ser Gly Ile

Human 1 16 31 46 61 76

1 2 3 Asn4Gln His 5 Phe Val Tyr Leu Val Cys Gly ArgArg Glu Ala Pro Glu Gly Gly Gly Ala Ser Leu Gln Lys Arg Cys Ser Leu Tyr Gln

6 Leu Glu Asp Gly Gly Leu

7 8 9 10 11 12 13 14 15 CysGly SerHis LeuVal Glu Ala Leu Arg Gly PhePhe Tyr ThrPro Lys Thr LeuGln Val Gly GlnVal Glu Leu Gly LeuAla Leu Glu Gly Ser Leu GlnPro CysCys Thr Ser Ile Ile Val GluGln Glu Asn Tyr Cys Asn

Cow 1 16 31 46 61 76

1 2 3 Asn4Gln His 6 5 Leu Glu Phe Val Gly Gly Tyr Leu Val Cys Gly Cys Cys ArgArg Glu Pro ValGlyAsn Glu Ala Gly Gly Val Glu Gln Gly Ile Leu Glu Asn Tyr

7 8 CysGly Arg Pro Gly Gly Gln CysAla Leu

9 10 11 12 13 14 15 SerHis LeuVal Glu Ala Leu PhePhe Tyr ThrPro Lys Ala Val Gly AlaLeu Glu Leu Ala GluGly ProPro Gln Lys Arg SerVal CysSer Leu Tyr Gln

1 2 3 Asn4Gln His 5 Phe Val Tyr Leu Val Cys Gly ArgArg Glu Ala Glu Gly Gly Leu Gly Gly Gln Lys Arg Gly Ile Leu Tyr Gln Leu Glu

7 8 9 10 11 12 13 14 15 CysGly SerHis LeuVal Glu Ala Leu Arg Gly PhePhe TyrThr Pro Lys Ala Pro Gln AlaGly AlaVal Glu Leu Gly Gln Ala LeuAla LeuGlu Gly Pro Pro Glu Gln CysCysThr Ser Ile Cys Ser Tyr Cys Asn

Pig 1 16 31 46 61 76

6 Leu Glu Asn Leu Val Asn

2

CASE 14 CShavings from the Carpenter’s Bench: The Biological Role of the Insulin C-peptide

Questions 1.

What are the physiological symptoms of untreated diabetes?

2.

What is the structural purpose of the C peptide? In other words, why is insulin synthesized as a prohormone rather than being synthesized as separate A and B chains?

3.

Prohormone convertase enzymes convert proinsulin to insulin. What do you note about the substrate specificity of the convertases?

4.

In the human disease familial hyperinsulinemia, the arginine at position 89 has been mutated to a histidine. What are the biochemical consequences of this mutation?

5.

Why do you think that a synthetic C-peptide consisting of D-amino acids (rather than the naturally occurring L amino acids) is as biologically active as the natural C-peptide which consists of Lamino acids? What can you conclude about the structural requirements for C-peptide biological activity?

6.

Why do you think that the rat and human C-peptides had biological activity whereas the porcine and bovine C-peptides did not?

7.

Evaluate the hypothesis that the C-peptide functions as a channel-former.

8.

How is diabetes currently treated in this country? Based on the information presented here, would you modify the protocol in any way?

References Ido, Y., Vindigni, A., Chang, K., Stramm, L., Chance, R., Heath, W. F., DiMarchi, R. D., DiCera, E., Williamson, J. R.(1997) Science 277, pp. 563-566.

Steiner, D. F., and Rubenstein, A. H. (1997) Science 277, pp. 531-532.

Wahren, J., Johansson, B.-L., and Wallberg-Henriksson, H. (1994) Diabetologica [Suppl 2] 37, pp. S99S107.

3

Case 15 Site-Directed Mutagenesis of Creatine Kinase Focus concept Site-directed mutagenesis is used to create mutant proteins so that the role of a single reactive cysteine in binding and catalysis can be assessed for the enzyme creatine kinase.

Prerequisites C C C C

Amino acid structure Protein architecture Enzyme kinetics and inhibition Basic enzyme mechanisms

Background The enzyme creatine kinase is important in energy metabolism and catalyzes the reaction shown in Figure 15.1. The reaction is readily reversible in vitro. Enzyme activity can be assayed by measuring hydrogen ion release (as determined by pH) in the forward reaction or hydrogen ion consumption in the reverse reaction.

In order to understand a particular enzyme’s mechanism, scientists rely on a variety of techniques.

One preferred technique is X-ray crystallography, which gives a three-dimensional picture of what the protein molecule looks like. X-ray crystallographic data are not always available however, since it is difficult to prepare the protein crystals that are used in carrying out the analysis. In the absence of crystallographic data, scientists use other experimental methods to determine the important features of an enzyme’s catalytic mechanism. One technique is chemical modification in which various reagents that might derivatize specific amino acid side chains are added to an enzyme solution. Then the experimenter measures the activity of the enzyme using a specific assay, and uses sequencing techniques to determine which amino acid has been modified. If a chemical modification results in inactivation of the protein, it can be inferred that that amino acid is essential for the enzyme’s activity in some way.

Another technique which is especially powerful is site-directed mutagenesis in which genetic engineering techniques are used to create mutant proteins with a single amino acid substitution. In this manner, the role of specific amino acids in the catalytic mechanism can be ascertained.

1

CASE 15 CSite-Directed Mutagenesis of Creatine Kinase

Figure 15.1: Reaction catalyzed by creatine kinase

Figure 15.2: Reaction of Cys in proteins with NEM

Figure 15.3: Reaction of Cys with iodoacetate and iodoacetamide

2

CASE 15 CSite-Directed Mutagenesis of Creatine Kinase

Questions 1. Sulfhydryl groups react with the alkylating reagent N-ethylmaleimide (NEM) in chemical modification experiments. When NEM is added to a purified solution of creatine kinase, Cys 278 is alkylated, but no other Cys residues in the protein are modified. What can you infer about the Cys 278 residue based on this observation?

2. Based on the result described in Question 1, the investigators used the technique of site-directed mutagenesis to synthesize five mutant creatine kinase proteins in which the Cys 278 was replaced with either a Gly, Ser, Ala, Asn or Asp residue. These mutants were termed C278G, C278S, C278A, C278N and C278D, respectively, to indicate the exact position of the amino acid change. The activities of the mutant enzymes were measured in the presence and absence of specific cofactors. a. All of the mutants had decreased creatine kinase activity as compared to the wild-type enzyme.

What information does this give you about the wild-type enzyme mechanism? b.The activity of the mutant enzymes C278D and C278N were compared and it was found that the activity of the C278D mutant was 12-fold greater than the activity of the C278N mutant (although both mutants had lower enzyme activities as compared to the wild-type). Suggest an explanation for this observation. c. The activities of the mutant enzymes (although decreased from the wild-type) were enhanced when either chloride or bromide ions were added to the assay mixture. (An exception was the C278D mutant). Why do you think that the ions were able to enhance enzyme activity? d.The C278D mutant was an exception to the observation described in Question 2c above. This mutant did not show an enhancement of enzyme activity in the presence of chloride and bromide ions; in fact its minimal enzyme activity decreased somewhat in the presence of these ions. Explain why.

3.The results of the study presented here provided greater understanding for previous chemical modification experiments that had been carried out with creatine kinase. These reactions are shown in Figure 15.3. Some investigators modified the Cys 278 with iodoacetamide and found that the enzyme activity was abolished as a result. They concluded that the Cys 278 residue was absolutely essential for enzymatic activity. But other investigators modified the Cys 278 with iodoacetate and found that activity was decreased but not abolished, leading them to conclude that the Cys 278 was not essential.

Taking these results together with the results of the current study, can you suggest an explanation that will clear up the confusion?

3

CASE 15 CSite-Directed Mutagenesis of Creatine Kinase 4. The investigators next carried out kinetic studies in which they measured the ability of a second substrate to bind once the first substrate had bound. The data are shown in Table 15.1. Kd refers to the constant for the binding of substrate to free enzyme and KM refers to the constant for binding of that same substrate to the enzyme when the other substrate is already bound to the enzyme. The results for the wild type and two of the mutants are shown in Table 15.1. a.Compare the Kd and KM values for creatine and ATP for the wild-type enzyme. What does a comparison of the Kd and KM values tell you about the ability of each substrate to bind to the enzyme alone? to the enzyme when the other substrate is present? b.Similarly, compare the Kd and KM values for the two mutant enzymes. Again, what does a comparison of the Kd and KM values tell you about the ability of each substrate to bind to the enzyme alone? to the enzyme when the other substrate is present? c. Consider your answers to 4a and 4b and consider the Vmax values for both the wild type and mutant enzymes. Assess the role of Cys 278 in the binding of creatine and ATP substrates to the creatine kinase enzyme.

Table 15.1: Binding constants for phosphocreatine synthesis from creatine and ATP (modified from Furter, et al., 1993).

Creatine Enzyme

ATP

Vmax, ìmol/min

Kd, mM

KM, mM

Kd, mM

KM, mM

Wild-type

19.6

8.9

0.70

0.32

60.7

C278G

64

273

0.27

1.13

6.0

C278S

92

209

0.31

0.70

2.0

Reference Furter, R., Furter-Graves, E. M., and Wallimann, T. (1993) Biochemistry 32, pp. 7022-7029.

4

Case 16 Allosteric Regulation of ATCase Focus concept An enzyme involved in nucleotide synthesis is subject to regulation by a variety of combinations of nucleotides.

Prerequisites C C

Properties of allosteric enzymes.

Basic mechanisms involving regulation of metabolic pathways.

Background Aspartate transcarbamoylase (ATCase) catalyzes an early step in the synthesis of the pyrimidine nucleotides UTP and CTP. The enzyme catalyzes the condensation of carbamoyl phosphate and aspartate to form carbamoyl aspartate. The reaction pathway is shown in Figure 16.1. The enzyme has been fairly well characterized. It is known to consist of six regulatory subunits and six catalytic subunits.

In this case, we examine the properties of ATCase isolated from E. coli to illustrate some of the important regulatory properties of multi-subunit enzymes. As an early enzyme in a multi-step pathway, the ATCase reaction is a logical one to regulate the synthesis of pyrimidine nucleotides. Both purine nucleotides and pyrimidine nucleotides are needed in roughly equal amounts as substrates for DNA synthesis in rapidly dividing cells. The regulation of the ATCase enzyme ensures a proper balance of purine and pyrimidine pools in E. coli. The goal in this case was to identify the cellular metabolites that serve as activators and inhibitors of ATCase.

Figure 16.1: Pyrimidine synthetic pathway.

1

CASE 16 C Allosteric Regulation of ATCase

Questions 1.Gerhart and Pardee measured ATCase activity in the presence of a variety of purine and pyrimidine derivatives. Their results are presented in Table 16.1. What compound(s) were the most effective inhibitors? activators? Explain the significance of the metabolites that served as inhibitors or activators in the context of the biosynthetic pathway presented in Figure 16.1.

Table 16.1: Effect of nitrogen bases, nucleosides and nucleotides on ATCase activity. *Indicates stimulation (based on Gerhart and Pardee, 1962).

Compound

Inhibition, % (Conc = 2 mM)

Cytosin

0

e

24

Cytidin

38

e CMP

48

dCMP

86

CTP

88

dCTP

8

UTP

35

GTP

31

dGTP

-180*

ATP

-162*

dATP

2

CASE 16 C Allosteric Regulation of ATCase

Figure 16.2: Kinetics of ATCase in the presence of ATP and CTP (based on Gerhart and Pardee, 1962).

2. The kinetics of the ATCase reaction were examined using increasing concentrations of aspartate, in the presence and absence of CTP and ATP as shown in Figure 16.2. a.What information can you obtain by looking at the shapes of the curves in this figure? b.What kinetic parameter(s) change in the presence of CTP? What parameter(s) do not change?

What is the significance of these observations? c. Answer question 2b for ATP. 3. The investigators examined the behavior of the ATCase enzyme in the presence of CTP, and in the presence of both CTP and ATP. The concentration of CTP is 0.1 mM and the concentration of ATP is 2 mM. The results are shown in Figure 16.3. What is the significance of these observations?

Figure 16.3: Percent ATCase activity in the presence of CTP, and in the presence of both CTP and ATP (based on Gerhart and Pardee, 1962).

3

CASE 16 C Allosteric Regulation of ATCase 4. Back in 1962, Gerhart and Pardee developed a model for the regulation of the activity of the ATCase enzyme by CTP and ATP, using the pathway given in Figure 16.1. Describe that model, using information presented here as well as what you have learned about allosteric enzymes. Be sure to include a sentence explaining the physiological significance of your model.

5.Many years later, in 1989, Wild, et al. revisited the idea of allosteric control of ATCase by CTP. They noted that CTP did indeed inhibit ATCase, but that the inhibition was always incomplete, even at high concentrations of CTP. They hypothesized that perhaps CTP did not act alone, but in combination with some other nucleotide. They tested the activity of ATCase in the presence of several nucleotide combinations. The results are shown in Table 16.2. a.What combination gives the most effective inhibition? b.What is the physiological significance of this combination? c. Revise Figure 16.1 to include this new information. d.Redraw Figure 16.2 to include this new information. How does the KM of the nucleotide combination compare with the values for the nucleotides alone?

Table 16.2: Relative specific activities for combinations of nucleotide effectors. A value greater than 1 indicates stimulation, a value less than 1 indicates inhibition.

Aspartate Concentration Nucleotide Effector

2.5 mM

5.0 mM

ATP

1.86

1.35

CTP

0.31

0.43

GTP

0.57

0.71

UTP

0.95

0.95

ATP/CT

0.70

0.85

P

1.98

1.58

ATP/GT

1.96

1.52

P

0.41

0.58

ATP/UT

0.05

0.06

P

0.66

0.84

CTP/GT P CTP/UT P GTP/UT P

4

CASE 16 C Allosteric Regulation of ATCase

References Gerhart, J. C., and Pardee, A. B. (1962) J. Biol. Chem. 237, pp. 891-896.

Voet, D., and Voet, J. (1995) Biochemistry, John Wiley & Sons, New York, p. 191-194.

Wild, J. R., Loughrey-Chen, S. J., and Corder, T. S. (1989) Proc. Natl. Acad. Sci. 86, pp. 46-50.

5

Case 17 A Possible Mechanism for Blindness Associated with + Diabetes: Na-Dependent Glucose Uptake by Retinal Cells Focus concept Glucose transport into cells can influence collagen synthesis, which causes the basement membrane thickening associated with diabetic retinopathy.

Prerequisites  Transport proteins. + +  NaKATPase and active transport.

Background In the retina of the eye, there are equal amounts of endothelial cells and cells called pericytes. The two types of cells are associated with one another and may communicate with each other. The investigators were particularly interested in pericytes because it had been observed that pericyte damage and basement membrane thickening occur during the early stages of diabetic retinopathy. Diabetic retinopathy eventually leads to blindness.

The investigators were interested in the factors responsible for causing pericyte damage. They noted that cells in the kidney that were similar in function to the pericytes possessed a sodium-coupled glucose transporter (SGLT), and they hypothesized that the pericytes might also have such a transporter. They hypothesized that such a transporter, if it existed, would facilitate and perhaps regulate the entry of glucose into the cell. Next, the investigators determined if entry of glucose into the cell was correlated with collagen synthesis. They also investigated several inhibitors of the glucose transport process.

Questions 1. The investigators cultured pericytes and endothelial cells separately and measured their ability to take up glucose. Glucose uptake was measured both in the presence of sodium ions (provided by sodium chloride) and in the absence of sodium (choline chloride was added in order to maintain constant osmolarity). The results are shown in Figure 17.1. What is your interpretation of these results?

1

+

Figure 17.1: Effect of Na on glucose uptake by two different types of retinal cells. The concentrations of both NaCl and choline chloride were 145 mM (based on Wakisaka, et al., 1997).

2. Next, the investigators measured the effect of increasing sodium concentration on glucose uptake in both pericytes and endothelial cells. The results are shown in Figure 17.2. What is your interpretation of these results? What information is conveyed by the shape of the curves?

Glucose uptake (nmol/mg/hr)

1.6 1.2 0.8 0.4 0 0

50

100

150

200

+

[Na], mM endothelial cells

p e ri cyte s

Figure 17.2: A dose-dependent response for the sodiumdependent uptake of glucose by pericytes and endothelial cells (based on M. Wakisaka, et al., 1997).

2

.1/v, mghr/nmol

3. Transport kinetics through protein transporters can be described using the language of MichaelisMenten. The transported substance binding to its protein transporter is analogous to the substrate binding to an enzyme. Thus, KM and vmax values can be determined for protein transporters. Use the information provided in Figure 17.3 to estimate the KM and vmax for glucose uptake by pericytes in the presence and absence of sodium. What information is conveyed by these values? Have the investigators demonstrated convincingly that a SGLT exists in pericytes? What about endothelial cells? Explain, using the data to support your answers.

0.5 0.4 0.3 0.2 0.1 0 -0.5

0

0.5

1

[Glucose]

-1 , mM-1

1.5

2

without Na+ ions

with Na+ ions

Figure 17.3: Lineweaver-Burk plots for the uptake of glucose by pericytes in the presence (145 mM NaCl) or absence (145 mM choline chloride) of sodium ions (based on Wakisaka, et al., 1997).

3

4. The activity of the SGLT in pericytes was investigated in the presence of inhibitors. Radioactively labeled glucose was added to cultured pericytes, in the absence (control) or presence of several potential inhibitors. The transport experiments were carried out both in the presence and absence of sodium ions. The results are shown in Figure 17.4. a. What is the effect of sugars such as galactose and 2-deoxyglucose on glucose transport? b. What is the effect of 0.2 mM phlorizin on glucose transport?

Figure 17.4: Effect of inhibitors on glucose uptake by pericytes in the presence and absence of sodium ions. The sugars were added at a concentration of 5 mM, while the phlorizon was added at a concentration of 0.2 mM (based on Wakisaka, et al., 1997).

4

5. Next, the investigators wanted to explore the relationship between glucose uptake and collagen synthesis. They added increasing amounts of glucose to cultured pericytes and then measured the relative density of the cells as a result. (They confirmed in a separate immunoblotting experiment that increased collagen synthesis was responsible for the increased cellular density.)

The results are shown in Figure 17.5. How do you interpret these data?

Relative density

2.5 2.0 1.5 1.0 0.5 5

10

15

20

25

30

Glucose concentration, mM Type IV

Type VI

Figure 17.5: Effect of increasing glucose concentration of the synthesis of Type IV and Type VI collagens (based on Wakisaka, et al., 1997)

5

6. The effect of the drug phlorizin on glucose consumption and relative cellular density (compared with 5 mM glucose) was examined. The results are shown in Figure 17.6. a. What effect does phlorizin have on glucose consumption at the two different concentrations of glucose? b. What effect does phlorizin have on relative density of the cells at the two glucose concentrations?

Figure 17.6: Effect of phlorizin on glucose consumption (left) and collagen synthesis (right). Based on Wakisaka, et al., 1997.

7.

Consider the diabetic vs. the nondiabetic patient. Why do you think that retinopathy is more likely to occur in the diabetic patient? Describe the sequence of biochemical events leading to pericyte cell damage. What steps might be taken to decrease the incidence of diabetic retinopathy?

Explain your answer, using the experimental results presented here.

Reference Wakisaka, M., Yoshinari, M., Yamamoto, M., Nakamura, S., Asano, T., Himeno, T., Ichikawa, K., Doi, Y., and Fujishima. M. (1997) Biochim. Biophys. Acta 1362, 87-96.

6

Case 19 Purification of Rat Kidney Sphingosine Kinase Focus concept The purification and kinetic analysis of an enzyme that produces a product important in cell survival is the focus of this study.

Prerequisites  

Basic enzyme kinetics Protein purification techniques and protein analytical methods

Background Sphingolipid metabolites have recently been shown to be involved in cell viability. For example, ceramide has been shown to stimulate apoptosis, or programmed cell death, whereas sphingosine-1phosphate (SPP) promotes cell survival. A number of different factors influence the levels of ceramide and SPP in the cell, as shown in Figure 19.1.

Clearly, the balance of ceramide and SPP will be one of the many factors that influences whether a cell lives or dies. An understanding of the biochemical processes involved in regulating these sphingolipids has important applications in the treatment of cancer. We will concentrate on sphingosine1-phosphate (SPP).

SPP is synthesized in the cell from sphingosine and ATP as shown in Figure 19.2. The enzyme that catalyzes the reaction is sphingosine kinase. In this case, we will study the purification of the sphingosine kinase enzyme.

Anticancer drugs TNF-α Radiation

Ceramide

Programmed Cell Death (Apoptosis)

Platelet-derived growth factor Other growth factors

Spingosine-1-phosphate (SPP)

Cell Survival

Figure 19.1: Factors that influence cellular levels of ceramide and spingosine-1-phosphate.

OO HO CH2

OH CH

CH

NH+ 3

CH

P

OOH

O

sphingosine kinase + ATP

CH2

CH

CH

NH+ 3

CH

CH

CH

(CH2)12

(CH2)12

CH3

CH3

+ ADP

Figure 19.2: Synthesis of sphingosine-1-phosphate from sphingosine and ATP.

Questions 1. The investigators purified the enzyme sphingosine kinase using rat kidneys as a source of enzyme since it had been shown previously that kidneys contain a high concentration of SPP. Kidneys were homogenized in buffer, filtered, and then fractionated by ammonium sulfate precipitation. The ammonium sulfate precipitate was dissolved in buffer and then applied to a DEAE-cellulose (anion exchange) column. The investigators eluted the protein using Tris CH2OH buffer at pH = 7.4 with increasing concentrations of sodium chloride (from 0 to 0.50 M). Fractions were collected and assayed for sphingosine kinase in a procedure that involved the addition of C CH2OH H2N sphingosine to radioactively labeled (γ-32P) ATP and the subsequent quantitation of 32P in the SPP product. The results are shown in Figure CH2OH 19.3.

Compared to the other proteins in the rat kidney, what statements can Figure 19.4: Structure of Tris you make about the sphingosine kinase enzyme, based on its elutionconjugate base. behavior from the DEAE column?

2. The investigators reported that the sphingosine kinase enzyme bound to the DEAE column when Tris was used as a buffer, but not when a phosphate buffer was used. The structure of the conjugate base of Tris buffer is shown in Figure 19.4. (Both buffers were adjusted to an identical pH of 7.4.)

Explain why.

2

0.5 M NaCl

15000

25 20 15

10000 10 5000

5

0 M NaCl

0 0

Protein, mg

Kinase activity, units

20000

50

100

150

0 200

Fraction number Kinase activity, units

Protein, mg

Figure 19.3: Elution profile of rat kidney homogenate on DEAE-cellulose (based on Olivera, et al., 1998).

3. The investigators carried out several additional chromatographic steps and finally obtained a fraction that contained pure sphingosine kinase, as determined by Mwt markers SDS-PAGE. The molecular weight was determined by 1 2 3 (Kd) comparing the migration of the enzyme with the migration of 102 standard molecular weight markers in Lane 1. The investigators also carried out an additional experiment in which one batch of the enzyme was treated with the reducing 78 agent β-mercaptoethanol (Lane 2). A second batch was not treated (Lane 3). The results are shown in Figure 19.5.

50 a. Estimate the molecular weight of sphingosine kinase. b. Interpret the results of the experiment described above.

What can you say about the structure of sphingosine 34.2 kinase?

28.3

Figure 19.5: SDS-PAGE gel of purified sphingosine kinase.

4. The investigators next carried out simple kinetic analyses using sphingosine and ATP as substrates in separate experiments. Michaelis-Menten plots and the corresponding Lineweaver-Burk plots are shown in Figures 19.6 and 19.7. a. Calculate the KM and Vmax values for both substrates. b. The activity of sphingosine kinase was measured in the presence of threo-dihydrosphingosine, a stereoisomer of sphingosine. The Lineweaver-Burk plot obtained from a kinetic analysis of sphingosine kinase activity carried out in the presence of this inhibitor is shown in Figure 19.6.

Calculate the KM and Vmax values in the presence of the inhibitor. What kind of an inhibitor is threo-dihydrosphingosine? Explain your answer completely.

Figure 19.6: Lineweaver-Burk analysis of sphingosine kinase activity using varying amounts of sphingosine substrate, in the presence and in the absence of the inhibitor threodihydrosphingosine (based on Olivera, et al., 1998).

4

Figure 19.7: Lineweaver-Burk analysis of sphingosine kinase activity using varying amounts of ATP substrate (based on Olivera, et al., 1998)

Reference J. Biol. Chem.273, 12576-12583.

Olivera, A., Kohama, T., Tu, Z., Milstein, S., and Spiegel, S. (1998)

Case 20 NAD+-dependent Glyceraldehyde-3-phosphate Dehydrogenase from Thermoproteus tenax Last modified 3 January 2013

Focus concept Glycolytic enzymes from Thermoproteus tenax are regulated in an unusual manner.

Prerequisites C C C

The glycolytic pathway.

Enzyme kinetics and inhibition.

The cooperative nature of regulated enzymes.

Background Carbohydrate metabolism in the thermophilic archaeal bacterium Thermoproteus tenax is rather peculiar compared to the types of organisms usually studied in introductory biochemistry. For example, the phosphofructokinase reaction in T. tenax is reversible, and is dependent upon pyrophosphate rather than ATP. In addition, T. tenax has two different glyceraldehyde-3-phosphate dehydrogenase (GAPDH) isoenzymes. One is well known and, although it requires NADP+ as a cofactor instead of NAD+, it resembles the GAPDH enzyme we studied in class and is referred to as the “phosphorylating GAPDH”.

In contrast, the second isoenzyme is irreversible and requires NAD+ as a cofactor and is referred to as the “nonphosphorylating GAPDH”. In this case, we will consider the properties of the latter enzyme. The balanced equation of the reaction catalyzed by the nonphosphorylating NAD+-dependent glyceraldehyde3-phosphate dehydrogenase is shown below.

Figure 20.1: Non-phosphorylating NAD+-dependent GAPDH in T. tenax.

T. tenax stores energy in the form of glycogen, which is degraded to glucose-1-phosphate. The glucose-1-phosphate is then converted to glucose-6-phosphate and then enters the glycolytic pathway.

The two GAPDH enzymes are probably differently regulated in T. tenax. The authors of this study propose that “phosphorylating, NADP+-dependent” GAPDH is involved in efficient ATP production whereas the “non-phosphorylating, NAD+-dependent” GAPDH is somewhat involved in ATP production 1

CASE 20 CNAD+-dependent Glyceraldehyde-3-phosphate Dehydrogenase from Thermoproteus tenax but is also involved in providing intermediates for cellular biosynthetic reactions.

The gene for the non-phosphorylating, NAD+-dependent GAPDH was cloned and sequenced and its kinetic characteristics were studied. Summary information is presented in Table 20.1.

Table 20.1: Kinetic properties of NAD+-dependent GAPDH isolated from T. tenax.

NAD+ saturation Without AMP Vmax, units/mg KM, mM With AMP Vmax, units/mg KM, mM Molecular Mass Subunit (kD) Native (kD)

36.5 3.3 37.0 1.4 55,000 220,000

Questions 1.Name the three enzymes that catalyze irreversible, regulated reactions in glycolysis as studied in class.

2.What is the significance of the GAPDH reaction in E. coli to glycolysis?

3.How does the reaction catalyzed by GAPDH from T. tenax presented here differ from the reaction carried out in E. coli?

4. The activity of the GAPDH enzyme was assayed in the presence of a constant amount of glyceraldehyde-3-phosphate and an increasing amount of NAD+. The activity of the control was compared to the activity in the presence of various metabolites. The results are shown in Figure 20.2.

Additional data are given in Table 20.2. a. Use the data in Figure 20.2 to estimate a KM value for the enzyme in the presence of these metabolites. Classify the metabolites listed in Table 20.2 as inhibitors or activators. Fill in your answers in the table provided. Explain how you decided whether these metabolites are inhibitors or activators, based on the graph. b. How would you classify NADH, ADP and ATP? (These data are not presented in the graph). Are they inhibitors or activators? Add this information to Table 20.2. c. Explain the physiological significance of your answers to questions 4a and 4b.

2

CASE 20 CNAD+-dependent Glyceraldehyde-3-phosphate Dehydrogenase from Thermoproteus tenax Table 20.2: Effect of various metabolites on the activity of NAD+-dependent GAPDH isolated from T. tenax. (Based on Brunner, et al., 1998.)

Metabolite

Apparent KM, mM

Inhibitor or activator?

None NADP+ Glucose-1-ph osph ate AMP NADH ATP ADP

5. The Hill coefficients for NAD+ binding to the T. tenax GAPDH in the presence and absence of NADP+ were measured and are shown on the graph in Figure 20.2. What is the significance of the change in the value of the Hill coefficient? Is this consistent with (a) the shape of the curve and (b) the information given in the background on the enzyme’s structure?

6.What is the ATP yield for one mole of glucose oxidized by the pathway that uses the nonphosphorylating GAPDH enzyme?

Figure 20.2: Cosubstrate saturation of NAD+-dependent GAPDH of T. tenax in the presence of various effectors.

Assay conditions were the following: 90 mM HEPES, pH = 7, 160 mM KCl, 4 mM DL-GAP (based on Brunner, et al.,

1998).

Reference Brunner, N. A., Brinkmann, H., Siebers, B., and Hensel, R. (1998) J. Biol. Chem. 273, pp. 6149-6156.

3

Case 21 Characterization of Pyruvate Carboxylase from Methanobacterium thermoautotrophicum Focus concept Pyruvate carboxylase is discovered in a bacterium that was previously thought not to contain the enzyme.

Prerequisites C C

Tricarboxylic acid cycle reactions and associated anapleurotic reactions.

Glyoxylate cycle reactions.

Background As one of the reactants in the first reaction of the tricarboxylic acid cycle, oxaloacetate (OAA) is an important cellular metabolite. The concentrations of oxaloacetate are tightly regulated.

Different organisms employ different mechanisms to obtain oxaloacetate. In mammals and yeast, OAA is the product of the pyruvate carboxylase (PYC) reaction. In E. coli, the enzyme phosphoenolpyruvate carboxylase (PPC) provides oxaloacetate from phosphoenolpyruvate obtained from glucose oxidation in the glycolytic pathway. If glucose is absent and E. coli is using acetate as a carbon source, the glyoxylate pathway serves to generate the needed oxaloacetate. Usually an organism will employ PPC or PYC, but not both.

Detectable levels of PYC in the methanogenic bacterium Methanobacterium thermoautotrophicum had previously not been found, and since PPC had been detected, it was believed that M. thermoautotrophicum did not possess PYC. However, in the case described here, microbiologists found that if they added biotin to cultures of the methanogenic bacterium, pyruvate carboxylase activity could be detected.

This was a surprising finding, especially since it is known that the methanogen can synthesize its own biotin. However, having identified the presence of the PYC enzyme, the investigators set out to isolate, purify, and characterize the enzyme. Purification of the PYC was rather straightforward since the enzyme is soluble and hydrophilic. In addition, the investigators were able to make use of the protein avidin, which binds to biotin with high affinity and specificity.

Questions 1.Write the balanced reaction catalyzed by pyruvate carboxylase (PYC), including structures and cofactors.

2.Write the balanced reaction catalyzed by phosphoenolpyruvate carboxylase (PPC), including st ruct ures.

3.Draw a diagram that describes how E. coli obtains oxaloacetate via the glyoxylate cycle when acetate, not glucose, is provided as a food source.

1

CASE 21 CCharacterization of Pyruvate Carboxylase from Methanobacterium thermoautotrophicum

4.Why is it so important that oxaloacetate be generated in the cell?

5.Why did the investigators add biotin to the methanogen cultures in an attempt to detect pyruvate carboxylase activity?

6. Following purification of the PYC enzyme by avidinSepharose affinity chromatography, the investigators carried out several experiments to characterize the enzyme. First they ran samples of the enzyme on denaturing and non-denaturing gels. The results are shown in Figure 21.1. In addition, they ran the protein through a calibrated gel filtration column. The results from the gel filtration column indicated that the PYC enzyme had a molecular weight of 540 kilodaltons. Use this information to determine the structure of the PYC enzyme.

Figure 21.1: Gel electrophoresis of pyruvate carboxylase purified from M. thermoautotrophicum.

(Based on Mukhopadhyay, et al., 1998.)

7. The catalytic properties of the pyruvate carboxylase enzyme were assessed following purification. The activity of the enzyme was assayed in the presence of ATP, pyruvate, bicarbonate, and Mg2+ ions as a control. In addition, the dependence of the enzyme on these various metabolites was tested by replacing them with similar compounds. The activity of the enzyme in the presence of these various effectors is shown in Table 21.1. a. Explain why PYC had no activity in the presence of avidin. b.Is PYC dependent on ATP for activity? Can other nucleotides substitute for ATP? What is the effect if other nucleotides are added to the assay mixture in addition to ATP? c.What is the effect of other tricarboxylic acid metabolites on PYC activity in the methanogen? d.What ion or ions are required for PYC activity?

8. The sequence of PYC from the methanogen was compared to other pyruvate carboxylase enzymes and it was discovered that the lysine at position 534 is strictly conserved. Why is this the case?

2

CASE 21 CCharacterization of Pyruvate Carboxylase from Methanobacterium thermoautotrophicum Table 21.1: Activity of pyruvate carboxylase enzyme (PYC) isolated from M. thermoautotrophicum in the presence of various effectors (based on Mukhopadhyay, et al., 1998).

Effector

Activity (% of control)

Co nt ro l

100

Avidin

0

Alternate nucleotides replacing ATP AMP ADP CTP GTP ITP UTP

0 0 0 0 0 0

Nucleotides in addition to ATP AMP ADP CTP GTP ITP UTP

104 73 106 94 80 105

Tricarboxylic acid cycle-related compounds 84 Acetyl-CoA 91 Aspart at e 95 Glutamate 73 á-ket o glut arat e Divalent cations replacing Mg2+ Mn2+ Co2+ Zn2+

17 46 0

Reference Mukhopadhyay, B., Stoddard, S. F., and Wolfe, R. S. (1998) J. Biol. Chem. 273, pp. 5155-5166.

3

Case 22 Carrier-mediated Uptake of Lactate in Rat Hepatocytes Focus concept The structural characteristics of the lactate transport protein in hepatocytes are determined.

Prerequisites C Transport proteins.

C Major carbohydrate metabolic pathways including glycolysis and gluconeogenesis.

Background The goal of the investigation presented in this case was to determine the mechanism for lactate transport in the liver. Lactate is a major substrate for gluconeogenesis, but its transport from the blood to liver cells is incompletely understood. Lactate transport is physiologically important because the conversion of lactate to glucose in the liver as part of the Cori cycle requires that lactate released from the muscle during anaerobic fermentation be transported to the liver via the bloodstream. Previous studies have indicated that the transport of lactate is probably mediated by a carrier protein, but it was not understood exactly how the carrier worked. For example, it was not known whether lactate transport was linked to the transport of other ions such as H+ or OH-. In the experiments presented here, investigators measured the rate of uptake of [14C]-labeled lactate in rat hepatocytes under a variety of different physiological conditions. In this manner, the characteristics of the lactate transporter were determined.

Questions 1.

Transport of lactate from the blood to the liver is part of the Cori cycle.

Draw a diagram of the Cori cycle, and describe how it works. Under what physiological conditions would you expect the Cori cycle to be active?

2.

Consider the structure of lactate–do you think that lactate would easily pass through the plasma membrane by simple diffusion? Explain why or why not.

3.

The time course for the uptake of Dlactate and L-lactate by cultured rat Figure 22.1: D- and L-lactate transport in rat hepatocytes is shown in Figure 22.1. hepatocytes (based on Fafournoux, et al., 1985).

What conclusions about lactate transport can you draw from looking at these curves? Explain.

1

CASE 22 CCarrier-mediated Uptake of Lactate in Rat Hepatocytes 4.

The investigators next measured the transport of lactate in hepatocytes as a function of extracellular pH. Their results are shown in Figure 22.2. What is your interpretation of these results? Do these results give you more information about the characteristics of the lactate carrier protein?

Figure 22.2: Lactate transport at various extracellular pH values (based on Fafournoux, et al., 1985).

5.

The investigators considered the possibility that lactate did not have its own transporter, but instead used the general anionic transporter in liver cells. In order to test this hypothesis, they measured the uptake of 0.4 mM [14C] lactate in the presence of 10 mM Na2SO4 and found that lactate transport was inhibited 11% in the presence of the Na2SO4. (The sulfate ion is transported by the general anionic transporter.) Do these results indicate that lactate utilizes the general anionic transporter? Explain.

Reference

Fafournoux, P., Demigné, C., and Rémésy, C. (1985) J. Biol. Chem. 260, pp. 292-299.

2

Case 28 The Bacterium Helicobacter pylori and Peptic Ulcers Last modified 3 January 2013

Focus concept The entire genome of Helicobacter pylori has recently been sequenced. This will allow biochemists to examine the organism’s proteins in detail. In this case, mechanisms employed by Helicobacter pylori that allow it to survive in the acidic environment of the stomach will be examined.

Prerequisites C C

Protein structure and function.

Basic metabolic pathways up through amino acid metabolism.

Background For the past hundred years or so, the cause of peptic ulcers has been attributed to excess acid production by stomach parietal cells. The production of acid was thought to be influenced by environmental factors such as diet and stress. But in 1983, Australian physician Robin Warren described a spiral-shaped bacterium that he had observed on the gastric epithelium of his peptic ulcer patients, and the following year, he and biochemist Barry Marshall put forth the hypothesis that this as-yet unidentified spiral-shaped bacterium was the causative agent of peptic ulcers.

Although the scientific community was initially skeptical, further study of the bacterium identified subsequently as Helicobacter pylori (or Campylobacter pylori) supported this hypothesis. The complete genome sequence of H. pylori, reported in 1997, will allow further detailed studies of the proteins of this bacterium and will provide information that can be used to develop drugs to treat peptic ulcers.

H. pylori has the unusual ability to colonize host cells in the extremely acidic environment of the gastric mucosa, where the pH is typically around 2. H. pylori probably employs several different mechanisms to ensure its survival in this environment: • • •

The bacterium has the ability to establish a positive inside-membrane potential.

The bacterium may release factors that decrease the secretion of acid by gastric parietal cells.

The protein urease is essential to the H. pylori’s ability to survive in the gastric environment.

1

CASE 28 CThe Bacterium Helicobacter pylori and Peptic Ulcers

Questions 1. The proteins found in H. pylori have different characteristics than those of bacteria that colonize cells in the usual physiological environment where pH = 7.4. Describe these proteins–what kinds of amino acids would you expect to find in abundance? What would the pI of these proteins be?

2.H. pylori synthesizes the enzyme urease, which converts urea to carbon dioxide and ammonia in the aqueous medium of the cell. Write a balanced equation for this reaction. How would the products of this reaction contribute to the survival of H. pylori in an acidic environment?

3. A current hypothesis is that H. pylori has the ability to establish a positive inside-membrane potential. This means that the interior of the bacterium is more positively charged than its exterior.

What kinds of mechanisms might H. pylori employ in order to accomplish this? List as many as you can think of.

4. The following information concerning the metabolism of H. pylori was obtained from the complete genomic sequence: C Glucose is the sole source of carbohydrate, and enters the cell via a transporter that brings glucose and Na+ along in symport.

C Once glucose enters the bacterial cell, it is acted upon by the enzyme glucokinase (rather than the less specific hexokinase).

C Glycolysis, gluconeogenesis, and the pentose phosphate pathway are all active. Lactate dehydrogenase is present.

C Peptidoglycans (for cell walls) are synthesized from fructose-6-phosphate, phospholipids from glyceraldehyde-3-phosphate and aromatic amino acids from phosphoenolpyruvate.

C The conversion of pyruvate to acetyl CoA is accomplished by the enzyme pyruvate ferrodoxin oxidoreductase (POR) instead of pyruvate dehydrogenase.

The glyoxylate cycle is absent.

C The tricarboxylic acid cycle is incomplete, ending at isocitrate, which is converted to glutamate in C a two-step process.

Fatty acid synthetic processes are present.

C Glutamine synthetase and glutamate dehydrogenase are present.

C Glutamine is the nitrogen donor in pyrimidine biosynthesis.

C The urea cycle is present, and provides urea for the urease reaction described in Question 2.

C Using the above information, construct a diagram which integrates the major metabolic pathways found in H. pylori.

References Doolittle, R. F. (1997) Nature 388, pp. 515-516.

Tomb, J.-F., et al. (1997) Nature 388, 539-547.

2

Case 32 Glucose-6-phosphate dehydrogenase activity and cell growth Last updated 11 July 2012

Focus concept The activity of the pentose phosphate pathway enzyme glucose-6-phosphate dehydrogenase has been found to be important in the regulation of cell growth.

Prerequisites ∙ ∙ ∙ ∙

Pentose phosphate pathway (PPP) reactions SDS-PAGE analysis DNA replication Signal transduction pathways involving tyrosine phosphorylation

Background Numerous studies have shown that the intracellular redox potential of the cell is important to cell growth. Cellular redox potential can be determined by the amounts of the reduced coenzyme NADPH, a principal product of the oxidative branch of the pentose phosphate pathway (PPP).

The investigators in the study presented here sought to demonstrate links between the activity of the enzyme glucose-6-phosphate dehydrogenase (G6PD) activity, cellular NADPH concentrations, and rates of cell growth. Previous studies have shown that the glucose-6phosphate dehydrogenase enzyme can be activated on the order of minutes or even seconds, possibly through the action of growth factors that release a bound, inactive G6PD to the cytosol, where, via a mechanism that might involve tyrosine phosphorylation of a membrane-bound receptor, the unbound G6PD translocates to the cytosol and becomes active.

NADPH is important to the cell in a variety of ways. The reduced coenzyme can react with potential damaging oxidizing agents, ridding the cell of these agents before they can damage important cellular components. For example, hydrogen peroxide, H2O2, may be reduced to water with concomitant oxidation of NADPH, as shown in Figure 32.1.

Notes on experimental technique: 1. Enzyme activity was measured by noting the increase in absorbance at 341 nm, as NADPH + absorbs light at this wavelength and NADP does not.

2. The Western blot is used to detect the presence (and relative amount) of a particular protein.

Samples are first subjected to separation by SDS-PAGE, and then the proteins in the gels are transferred to nitrocellulose paper. The paper is incubated with a specific antibody to the protein of interest; the antibody is tagged in some way to allow detection of the specific protein, which is visualized as a band. The darkness of the band is proportional to the amount of protein detected.

3 3. Incorporation of [H]-thymidine into DNA is used as a measure of cell growth, since rapidly growing cells will replicate their DNA before undergoing cell division.

Figure 32.1: Detoxification of oxidizing species. Hydrogen peroxide is reduced to water by glutathione peroxidase with concomitant oxidation of reduced glutathione (GSH). The oxidized glutathione (GSSG) is converted back to the reduced form by glutathione reductase; NADPH is a cofactor.

Hydrogen peroxide can also be reduced to water via the catalase reaction (not shown); this reaction also requires NADPH. (Adapted from Nelson, D. L., and Cox, M. M. (2000) Lehninger Principles of Biochemistry, Worth Publishers, NY, p. 560).

Questions 1. a. Draw the reaction of the PPP catalyzed by glucose-6-phosphate dehydrogenase (G6PD).

Why is this reaction important to the PPP? b. Draw a diagram of the branch of the pentose phosphate pathway which involves G6PD. What are the two main products of this branch, and what is the importance of these products to the cell?

2. List general cellular mechanisms that would affect the activity of G6PD. Which of these mechanisms would result in long-term (on the order of minutes or hours) vs. short term (minutes or seconds) regulation of the enzyme’s activity?

2

3. In the first experiment reported in this study, the authors stimulated cell growth through a variety of means; then measured corresponding G6PD activity of cultured fibroblasts. (Cells grown in culture are typically grown in a medium containing 10% fetal calf serum, a medium rich in growth factors.) The investigators grew fibroblasts in the absence of serum (serumstarved) to serve as a control. Cells were then treated with 10% fetal calf serum followed by the serum-starvation treatment; other cells were not starved. The results are shown in Figure 32.2. After the enzyme activity was measured, the cells were lysed and analyzed by Western blotting using a G6PD antibody. The results are of the Western blot are shown in Figure 32.3. What is your interpretation of these results? Be quantitative in your assessment of the data for both the activity assays and the Western blot. Why was the enzyme activity of PGD (6-phosphogluconate dehydrogenase) also investigated under the same conditions?

Serum starved Plus Serum

yes no

no yes

yes yes

Figure 32.2: Fibroblasts were (1) serum-starved for 48 hours (2) serum-starved for 48 hours and then stimulated with 10% serum, and (3) actively grown in medium containing 10% fetal calf serum.

Enzyme activity was measured in lysates containing the same number of cells (from Tian, et al, 1998).

Figure 32.3: Western blot analysis of the same cell lysates shown in Figure 32.2. Lane 1 shows serumstarved cells; Lane 2 is cells grown in the presence of serum (from Tian, et al., 1998).

1

2 G6PD

3

4. It has been shown that dehydroepiandrosterone (DHEA) is an inhibitor of G6PD. The investigators incubated cells with DHEA, then lysed the cells and measured G6PD and PGD enzymatic activity in terms of the change in absorbance at 341 nm. In a separate experiment, the incorporation of3[H]-thymidine was measured in the presence of PGDF (platelet derived growth factor, which stimulates cell growth) and increasing concentrations of DHEA. The results are shown in Figures 32.4 and 32.5, respectively. A Western blot analysis was also carried out (data not shown), in which the investigators found that there was no change in enzyme protein in cells exposed to either PGDF, or DHEA, or both. What is your interpretation of the results shown in these two figures? Again, be quantitative in your assessment of the data.

Figure 32.4: Dose-response curves of the effect of DHEA on G6PDH and PGD activity. Cell lysates were analyzed for enzyme activity in the presence of varying amounts of DHEA (from Tian, et al., 1998).

Figure 32.5: Serum-starved cells were treated with DHEA in the presence of PGDF for 12 hours, then [3H]-thymidine was added and

the cells were incubated for an additi4on al 12 hours (from Tian, et al., 1998).

+ 5. Predict the effect of the NADPH/NADP ratio (as compared to controls) in cells under the following circumstances: a. Withdrawal of serum. b. Addition of DHEA. c. Addition of H2O2. d. Withdrawal of serum and addition of H2O2.

6. In order to more conclusively show a link between G6PD and cell growth, the investigators transfected a cell line, COS-7, with the G6PD gene. The transfected COS-7 cells showed a 2-3-fold increase in G6PD activity as compared to cells that hadn’t been transfected with the gene. The authors of the study then investigated3 the effect of the inhibitor DHEA on these transfected cells by measuring the amount of [H]-thymidine incorporation into COS-7 and control cells. The results are shown in Figure 32.6. What is your interpretation of these results? Again, be quantitative.

Figure 32.6: The effect of DHEA, as measured by [3H]-thymidine incorporation, was determined in COS-7 cells transfected with the G6PD gene, and compared to control cells (from Tian, et al., 1998).

7. The authors of this study have shown that platelet derived growth factor (PGDF) stimulates the release of G6PD to the cytosol, and that this release depends on tyrosine phosphorylation of PGDF. They believe that it is likely that phosphatidylinositol-3-kinase (PI-3-kinase) and phospholipase C-γ are involved. Draw a diagram that shows the sequence of events involved in the release and subsequent activation of G6PD. Reference Tian, W.-N., Braunstein, L. D., Pang, J., Stuhlmeier, K. M., Xi, Q.-C., Tian, X., and Stanton, R. (1998) J. Biol. Chem., 273, pp. 10609-10617.

5

Case 4 The Structure of Insulin 1. Pig insulin is a good choice to replace human insulin. There is only one amino acid variation in B30.

Cow insulin has two differences in A8 and B30, and one difference in A10 (although these are conservative changes). Pig and cow are the best sources because these animal products are readily available from slaughterhouses, whereas horse, dog and rabbit are not. Chicken and duck have seven and six amino acid differences as compared to human insulin and wouldn’t be good choices. The primary sequence of amino acids influences the tertiary structure of proteins. Thus it is possible that a change in the protein sequence could alter the tertiary structure, thus giving the molecule a different shape, which means it might have difficulty interacting with receptors.

2. The pI values for all of the animal insulins would be similar to human insulin (with the exception of chicken and duck) because the amino acid changes all involved neutral amino acids. Chicken insulin would be more basic (because of the histidine substitution) whereas duck would be more acidic (because of the glutamate substitution.)

3. As discussed above, a change in the primary structure will lead to a change in tertiary structure. The immune system produces antibodies to proteins that are non-self and these non-self proteins are recognized because their shape differs from self-proteins. Some people developed allergies whereas others didn’t because some people’s immune systems are more sensitive.

4. The predicted results are shown in the drawing. Insulin has two bands because the mercaptoethanol reduced the disulfide bridges, Molecular causing the A and B chains to separate.

Proinsulin

5. With the loss of the C-peptide, critical information is lost that is essential to the proper folding of insulin.

When the C peptide is removed, two separate chains result, and it is much more difficult for two chains to resume their native conformation than one chain.

Proinsulin has no difficulty resuming its native B chain conformation in a denaturation/renaturation experiment because the proinsulin consists of only A chain one peptide chain. When proinsulin is converted to insulin, two peptide bonds are cleaved and two chains result.

Insulin

mass, kD 10 8 6 4

2

6. Scientists used to think so, since insulin is the active form of the protein. The C peptide was not believed to have any biological function. But this might change as a result of a recent study in which diabetic rats were injected with high doses of human C peptide and leaky arteries and nerve damage associated with diabetes were prevented. This might not mean that the C peptide ultimately prevents the complications associated with diabetes, but clearly more research needs to be done in this area.

Case 8 Hemoglobin, the Oxygen Carrier

1. The two β chains have a substitution of the

-

nonpolar (uncharged) valine for glutamate (negatively charged). Thus Hb A is more negatively charged than Hb S, and it therefore has a lower pI than Hb S, because a lower pH is required to neutralize the negatively charged carboxyl groups. Hb A consequently migrates more quickly toward the positive pole.

Hb A

Hb S pH gradient

2. The amino acid valine is nonpolar, whereas glutamate is polar. If glutamate on the surface of the protein is changed to a nonpolar amino acid, the hemoglobin molecules will show more of a tendency to polymerize so that the nonpolar valine can minimize its interaction with water. This occurs more readily in the deoxygenated form than the oxygenated form, since in the T form, the Hb has a nonpolar pocket on its surface in which the valine on a neighboring Hb molecule fits very well. The nonpolar pocket is not present on the oxygenated form of Hb. 3. Since Hb S sickles more readily when it’s in the T (deoxygenated) form, if oxygen is administered, the partial pressure of oxygen will increase and then more of the Hb S can be converted to the R form and it’s less likely to sickle.

4. a. The hydroxyurea increases the patient’s blood concentration of Hemoglobin F. Hemoglobin F has two γ chains and two α chains. The difference between Hb F and Hb A is the presence of the γ chain instead of the β chain. Since the β chain has been mutated in the patient suffering from sickle cell anemia, the increased synthesis of Hb F allows the patient to use Hb F to transport oxygen instead of the defective Hb S. In effect, the defective β chains are being replaced with healthy γ chains. This has a solublizing effect on the hemoglobin present in the red blood cell, because in effect, the increased synthesis of HbF dilutes the concentration of deoxygenated HbS. When the concentration of deoxy HbS decreases, sickling is less likely to occur. b. Conversion of deoxyHb to HbNO decreases the concentration of deoxyHb. Since it’s the deoxyHb that sickles, a decrease in its concentration may slow down polymerization time and decrease sickling. It’s also possible that HbNO is more soluble than HbS, and if this is the case, polymerization will occur to a lesser extent, which will decrease sickling. c. Vasodilation results in improved blood flow, which would decrease the trapping of red blood cells in the capillaries, relieving pain and improving oxygen delivery to the patient’s cells.

5. Red blood cells are made in the bone marrow. A bone marrow transplant from a healthy person without sickle cell anemia would provide reticulocytes with a healthy gene for the β chain, so the correct β chain would be synthesized, producing Hb A instead of Hb S.

1

6. The expected frequencies of the offspring of this couple are shown in the Punnett square. Because they have already had a child with sickle cell anemia (SS), the parents must both be heterozygous carriers (AS). Their chances of having a second child with sickle cell anemia are 25%; which is the same odds as with the first child, since each child is an independent event.

There is a 25% chance that the second child will be normal (AA) and a 50% chance that the child will not have the sickle cell disease, but will be a carrier for the disease (AS).

A S

2

A

AAAS

S

ASSS

Case 12 Production of Methanol in Ripening Fruit 1. There was an increase in PME activity as the fruit ripened for the wild-type tomato. For the PMEdeficient tomato, there was very little PME activity. The activity was correlated with total protein, as shown by the Western blot–low activity meant that not much protein was being synthesized whereas higher activities were the result of increased protein synthesis. (The exception to this correlation occurred at the RR stage, where protein content remained the same as in the TU stage but PME activity decreased.) Therefore, PME is regulated at the level of translation rather than being inhibited or stimulated by allosteric effectors.

2. a. There is an overall general correlation of PME activity and methanol production. In the wildtype tomato, as PME increases, methanol concentration increases as well. However, there is a short lag time between PME activity and methanol production between the IMG and MG stages where PME activity increased, but methanol production remained the same. There is also an increase in methanol between the TU and RR ripening stages, but the PME activity actually decreases between these two stages. For the transgenic tomato, the methanol content and PME activity both remain low. b. These results indicate that the PME enzyme is responsible for methanol production in ripening fruit.

3. Ethanol production in the wild-type tomato is low in all stages of the ripening process, and decreases to even lower levels as the ripening progresses. But in the transgenic tomato, ethanol concentration increases dramatically between the TU and RR stages. The production of ethanol and methanol appear to be inversely correlated. In the wild-type tomato, production of ethanol decreases as methanol increases. In the transgenic tomato, ethanol production increases in the absence of methanol production.

4.

O C

OH H

+ NADH + H

+

ADH CH2

+ NAD

+

5. It is likely that methanolCH3 serves as an inhibitor for ADH, since ethanol levels are decreased in the CH3 Acetaldehyde

Ethanol

wild-type tomato when methanol levels are at their highest. In the transgenic tomato, there is no methanol production in the absence of PME, so there is nothing to inhibit the ADH enzyme and consequently ethanol levels increased.

1

Case 13 Inhibition of Alcohol Dehydrogenase 1. Women have lower levels of alcohol dehydrogenase than men. Either the enzyme is lower in concentration, or the enzyme has a lower activity than in men. Decreased enzyme activity means that less substrate (ethanol) is converted to acetaldehyde. If the enzyme concentration is decreased then the substrate concentration is greater than the enzyme concentration, so the rate of conversion of ethanol to acetaldehyde depends on the enzyme concentration. Ethanol is the intoxicating agent, so if less substrate is converted to product, more substrate remains in the bloodstream. If ethanol levels in the blood remain higher for a longer period of time, intoxication results. Women also have a higher percentage of body fat and lower percentage of body water, so the alcohol is more concentrated in women than men. However, women become more intoxicated than men even when these differences are taken into account, so the decreased enzyme activity is likely responsible.

2. The conversion of methanol to formaldehyde is toxic to the optic nerve and is potentially fatal. In order to avoid these disastrous effects, this conversion must be prevented. Adding ethanol to the system provides an alternative substrate and high levels of ethanol saturate the enzyme with ethanol so that methanol is less likely to bind to ADH. This provides time for methanol to be excreted from the system. In a way, ethanol is acting as a competitive inhibitor, although the ethanol isn’t strictly an inhibitor since it is acted upon by the enzyme to produce a product.

3. The Lineweaver-Burk plot is shown below. The x- and y-intercepts have changed in the presence of the inhibitor while the slopes are nearly the same. The appearance is that of an uncompetitive inhibitor.

1

a.

Absence of inhibitor Vmax = 1/y-intercept Vmax = 1 11.1 min/Δ NADH A Vmax = 0.0902 Δ NADH A/min

slope = KM/Vmax KM = slope × Vmax )(0.090 ΔA NADH /min) KM = (3.29min/Δ NADH A/mM-1 KM = 0.297 mM

Presence of inhibitor Vmax = 1/y-intercept Vmax = 1 28.6 min/Δ NADH A Vmax = 0.0350 Δ NADH A/min

slope = KM/Vmax KM = slope × Vmax KM= (3.33 min/Δ NADH A/mM-1 )(0.035 Δ NADH A/min) KM = 0.116 mM

b. The value of α = 1, since the inhibitor is uncompetitive. The inhibitor binds to a site on the enzyme distinct from the active site. In the presence and in the absence of an uncompetitive inhibitor, the slopes in the Lineweaver-Burk plot are the same. The value of α’ can be determined from either the x- or the y-intercepts. Both calculations are shown below. Within experimental error, the values are the same.

'

(w/oI) = y − intercept (w/I) Vmax   28.6 min  0.090 Ä NADH absorbance units    min   Ä NADH absorbance units    '= 2.58    ' − K M ( w / o= Ix) − intercept (w/I)

 '=

 ' =(- 0.30 mM ) (- 8.64 mM -1 )  ' =2.57 c. The inhibitor is uncompetitive. The slopes are nearly the same because KM and Vmax have both decreased by the factor α’. The KM value decreases in the presence of the inhibitor (from 0.297 mM to 0.116 mM); the Vmax also decreases (from 0.0902 to 0.0350 Δ NADH absorbance/min). The inhibitor binds to the enzyme-substrate complex but not to free enzyme. It does not compete with substrate for binding to the enzyme, but deforms the enzyme in some way so that the substrate cannot be converted to product (although as we will see below, the inhibitor does not quite act in this “classically” uncompetitive manner).

2

d. The value of Ki’ is obtained from α’ as shown below.

[I] ' I

 ' = 1 +K 2.57=1+

1.0 M KI'

KI ' = 0.60 M 4. According to the mechanism in Figure 13.2, the reactants bind in a specific sequence. The first substrate, NAD+, binds to the enzyme, next the second substrate, the alcohol, binds to the enzyme. The alcohol is oxidized to the aldehyde and the NAD+ is reduced to NADH. The aldehyde product is released first; then the NADH product is released. However, if an inhibitor is present, it acts as an aldehyde analog and binds to the E-NADH complex. This prevents the enzyme from being regenerated. Uncompetitive inhibitors are classically defined as those which bind to the ES complex and not free enzyme. The formamide does not bind to the free enzyme, but it does bind to a complex that is not really an ES complex, since E-NADH is really an enzyme-product complex. So the formamide displays the kinetics and behavior of an uncompetitive inhibitor, (i. e., it does not bind to free enzyme), but it binds to an EP complex rather than an ES complex.

5. As described in the above question, the formamide compounds bind to the E-NADH complex as aldehyde analogs. If pyrazoles bind to the E-NAD+ complex, then the mechanism of the pyrazoles differs from that of the formamides, which are noncompetitive inhibitors. Since NAD+ is a substrate, if the pyrazoles bind to the E-NAD+ complex, it’s possible that pyrazoles prevent ethanol from binding. If this is true, then pyrazoles are competitive inhibitors, since they compete with ethanol for binding to the E-NAD+ complex (although again, not classically, since E-NAD+ is an ES complex, and competitive inhibitors classically bind to free enzyme).

6. a. Yes, N-1,5-dimethylhexylformamide would be an effective alternative for the treatment of methanol and ethylene glycol poisoning, and according to the authors of the study, this compound is relatively nontoxic. The KI’ for the N-1,5-dimethylhexylformamide is 0.60 μM, while the KM for ethanol is 0.30 mM. We don’t know the KM for methanol, but the affinity of ADH for methanol is less than that for ethanol, so we can assume that the KM for methanol is greater than 0.30 mM. Therefore the enzyme binds the inhibitor at lower concentrations of the inhibitor than it binds substrate. Since formamide compounds are uncompetitive inhibitors, enzyme can bind both the inhibitor and the substrate simultaneously.

Because the binding site for the inhibitor and the substrate are distinct, the inhibitor will inhibit ADH activity even when concentrations of alcohol are saturating. Since the inhibitor and the enzyme do not compete for the same binding site, it’s not possible to overcome inhibition by adding a large amount of substrate, as is the case for a competitive inhibitor. Even at saturating concentrations of alcohol, the inhibitor can still work effectively at inhibiting the reaction because it binds to the E-NADH complex and prevents regeneration of the enzyme. Therefore, formamides are good choices for the treatment of methanol and ethylene glycol poisoning because these inhibitors work over a wide range of substrate concentrations. b. The 4-methyl pyrazole might be expected to be less effective, since it is a competitive inhibitor. The 4methyl pyrazole would be effective at low concentrations of alcohol, but at high concentrations, the alcohol substrate will “win” the competition and the enzyme will not be inhibited. So competitive 3

inhibitors are expected to be less effective than noncompetitive inhibitors which are effective at a wide range of alcohol concentrations. Of course, there are other factors, such as toxicity of the inhibitor, and its rate of elimination from the body, to consider, and we also do not have information about the relative affinities of the pyrazoles and alcohols for the enzyme.

4

Case 14 Shavings from the Carpenter’s Bench: The Biological Role of the Insulin C-peptide 1. Untreated diabetics have high blood glucose levels. Insulin stimulates uptake of glucose by cells, so in its absence glucose will be found circulating in the blood. High blood glucose levels lead to high concentrations of glucose in the urine. Untreated diabetics are frequently hungry, because their cells are literally starved of glucose, and thirsty, because the increased concentration of glucose in the urine requires additional water for dilution.

2. The prohormone insulin consists of a single polypeptide chain that allows insulin to fold into its correct biologically active conformation. It would be extremely difficult for the cell to assemble two separate A and B chains correctly. Once the proinsulin is processed to remove the C peptide, the A and B chains, linked by disulfide bonds, are arranged in the correct conformation.

3. There are pairs of basic residues at the CA and BC junctions for each of the C-peptides (with the exception of rat, which has only one basic residue at the CA junction). The convertases therefore appear to require a sequence of two basic residues (either Lys or Arg, which are positively charged at physiological pH).

Species Amino acids at BC and CA junctions, respectively R32-R33 and A65-R66 Rat R31-R32 and K64-R65 Human R31-R32 and K59-R60 Cow R31-R32 and K62-R63 Pig 4. Because the sequence of two arginines is lost, the convertase enzymes will not recognize the cleavage site and it’s possible that the conversion of proinsulin to insulin will be impaired.

5. The C-peptide is glycine-rich. Glycine is the only amino acid that is not chiral, thus it does not exist in D and L forms. Thus chirality does not appear to be important to the biological activity of the Cpeptide.

1

6.

Rat R R E V E D P Q V G Q V E L G A G P G A G S E Q T L A L E V A R Q A R Human R R E A E D L Q V G Q V E L G G G P G A G S L Q P L A L E G S L Q K R Cow R R E V E G P Q V G A L E L A G G P G A G G L E G P P Q K R Pig R R E V E G P Q V G A L E L A G G P G A G G L E G P P Q K R

The sequences of the C-peptide for all four species are compared. Amino acid differences among the four C-peptides are underlined. Rat and human closely resemble one another in primary sequence, and cow and pig resemble each other in primary sequence. For example, there are 9 amino acid differences out of 35 for rat and human. The cow and pig C-peptides are shorter than the rat and human C-peptides (only 30 amino acids instead of 35) and their sequences are identical. There are many more differences between the two groups. For example, between cow and human there are 14 amino acid differences. Although there is not enough information contained in the primary sequence to explain why rat and human C-peptides have biological activity whereas cow and pig do not, it is evident that rat and human C-peptides will resemble one another, and cow and pig C-peptides will resemble one another.

7. The C-peptide is unlikely to serve as a channel former because it contains a high percentage of polar (charged) amino acids. In order to integrate into a membrane, the peptide would have to have significant nonpolar character to span the lipid bilayer.

8. Diabetes is treated by insulin replacement therapy in which the mature hormone is injected subcutaneously. Perhaps in the future the preparation should include a mixture of the mature hormone as well as the C-peptide.

2

Case 15 Site-Directed Mutagenesis of Creatine Kinase 1. The Cys 278 is highly exposed and unusually reactive as compared to other cysteines in the creatine kinase enzyme. The Cys 278, because of its high reactivity, is probably one of the catalytic residues in the enzyme. The other cysteine residues are probably not as reactive because they are not directly involved in catalysis and/or because they are shielded in some way which prevents them from reacting with the NEM reagent.

2. a. Since Gly, Ser, Ala, Asn and Asp all lack the sulfhydryl group, the sulfhydryl group (or some chemical property that the sulfhydryl group possesses) must be essential to the activity of the enzyme. b. The difference between C278D and C278N is that the mutants contain Asp and Asn residues, respectively. The main difference between these two amino acid side chains is the negative charge on the Asp, while the Asn is neutral. Since the C287D mutant has a greater activity, the negative charge must be essential to the enzyme’s catalytic activity. c. Chloride and bromide ions are negatively charged. It’s possible that a negative charge at the 278 site is important to the catalytic mechanism of the creatine kinase enzyme. The cysteine could provide this negative charge through deprotonation of the -SH group to yield -S-. Without the Cys residue, a negative charge cannot form, so it’s possible that the chloride and bromide ions provide this needed negative charge, which would account for the enhancement seen in the enzymatic activity. d. The C278D mutant contains a negatively charged Asp residue. The negatively charged bromide and chloride ions would repel the side chain on the Asp residue and cause the enzyme activity to decrease.

3. The iodoacetate-modified enzyme preserves the negative charge whereas in the iodoacetamidemodified enzyme, the negative charge is abolished. Since the current study indicates that the negative charge is essential, the iodoacetate-modified enzyme retains some enzymatic activity because of the presence of the negative charge on the iodoacetate-modified cysteine side chain.

4. a. The wild-type enzyme has the highest maximal velocity of the three enzymes presented here and it also has the highest affinity for both creatine and ATP substrates. The wild-type enzyme has a higher affinity for ATP as a substrate than for creatine. For both substrates, the KM values are lower than the Kd values. This indicates that, for both creatine and ATP, the binding of the second substrate is facilitated by the binding of the first substrate. Thus, the binding of substrates is synergistic–the binding of one substrate allows the second substrate to bind more easily. b. For both the C278G mutant and the C278S mutant, the KM and Kd values are higher for both creatine and ATP than the wild-type, indicating that the affinity for both substrates is less for the mutant than for the wild-type enzyme. The mutant binds ATP with greater affinity than creatine.

However, when the Kd and KM values are compared, the opposite pattern is observed as 1

compared to the wild-type: For both substrates, the KM values are greater than the Kd values.

This indicates that the binding of the second substrate is not facilitated; in fact it’s the opposite.

Thus when Cys 278 is mutated, the synergistic binding of the two substrates is lost, indicating that it’s likely that the Cys278 is responsible for the synergistic effect. Perhaps the binding of the first substrate causes a conformational change in the mutant protein which makes the binding of the second substrate unfavorable. c. The Vmax values are decreased in the mutant enzymes, and the synergistic effect of substrate binding is lost when the Cys 278 is mutated. Thus, the sulfhydryl group on the side chain of the cysteine may play a role in the synergistic effect of substrate binding, and this binding is essential to the optimum functioning of the enzyme.

2

Case 16 Allosteric Regulation of ATCase 1. The results in Table 16.2 show that CTP functions as an inhibitor. Cytosine alone cannot inhibit, and cytidine and deoxycytidine, which contain the sugar, show a slight amount of inhibition. There is an increasing amount of inhibition with CMP (38%) and CTP (86%), indicating that inhibition is facilitated with an increasing number of phosphate groups. UTP could not inhibit the enzyme (only 8% inhibition), indicating that the amino group may be required for inhibition. ATP and dATP functioned as stimulators of the enzyme. CTP functions as an inhibitor because it is the product of the pathway. High concentrations of CTP signal the enzyme that CTP is no longer needed. ATP stimulates the pathway (inhibition expressed as -180%) because a high concentration of purines must be matched by a high concentration of pyrimidines for efficient DNA and RNA synthesis.

2. a. There are multiple aspartate binding sites on the ATCase enzyme, and the binding of aspartate to ATCase is cooperative–once one molecule of aspartate binds, subsequent aspartate molecules bind more easily. Thus, ATCase is an allosteric enzyme, as determined from the sigmoidal shape of the curve. The Michaelis-Menten equation does not describe the shapes of these curves. The enzyme has two conformations, a T (low activity) form and an R (high activity) form. b. In the presence of CTP, the KM value increases (from about 6 mM to about 9 mM) but the Vmax value stays the same. This might lead one to conclude that CTP is a competitive inhibitor (meaning that CTP and aspartate compete for the same binding site). This is unlikely because CTP and Asp have such different structures. Instead, CTP functions as an allosteric inhibitor and binds to a site distinct from the active site and stabilizes the T form of the enzyme, which has a lower affinity for substrate. The increased KM value indicates that a greater concentration of substrate is required to achieve half-maximal activity. Inhibition is overcome at high concentrations of aspartate, so Vmax is achieved. c. In the presence of ATP, the KM value decreases (from about 6 mM to about 4 mM) but the Vmax value stays the same. This indicates that ATP is an activator of ATCase. The Vmax is achieved, but at a lower substrate concentration. KM decreases because the apparent affinity of the substrate for the enzyme has increased in the presence of the allosteric activator. When ATP binds, the R form of the enzyme is stabilized.

3. CTP inhibits the enzyme 70% of its full activity. But ATP can overcome this inhibition. When ATP is present, the enzyme is inhibited only 17%. But since the ATP is present in the experiment at a higher concentration (2 mM) than the CTP (0.1 mM), CTP must be a more effective inhibitor than ATP is an activator. In the presence of 2 mM ATP, the “inhibition” of ATCase is listed in Table 16.2 as -180%. Adding just 0.1 mM CTP causes the inhibition to decrease to 83%. ATP and CTP must compete for a common binding site, and ATP must have a lower affinity for the site(s) than CTP. Otherwise, you’d expect ATP to occupy the site(s) since ATP is present at a higher concentration (and you’d expect to see stimulation, not inhibition).

1

4. ATCase catalyzes an early step in the synthesis of pyrimidines. It is an allosteric enzyme, consisting of multiple subunits and multiple binding sites for its substrate, aspartate. When the aspartate substrate binds to one subunit, binding of subsequent molecules of substrate occurs more readily.

The enzyme activity is influenced by two allosteric effectors, CTP and ATP. CTP is an inhibitor of ATCase. Binding of CTP stabilizes the T form of the enzyme, which has a low affinity for substrate.

CTP, a pyrimidine, exerts its effects via feedback inhibition–that is, the end product of a pathway regulates its own synthesis by inhibiting an enzyme early in the pathway. On the other hand, ATP, a purine, stimulates ATCase by stabilizing the R form, which has a high affinity for substrate. The inhibitor effect of CTP and the stimulatory effect of ATP regulate the purine/pyrimidine balance in the cell and assure that there are approximately equal amounts of each for DNA synthesis.

5. a. The combination that gives the most effective inhibition is that of CTP and UTP. About 95% of the ATCase activity can be inhibited when CTP and UTP are added together. In contrast, CTP alone inhibits only 70% of ATCase activity, and UTP alone inhibits only 5% of ATCase activity. b. UTP and CTP are both pyrimidines. Regulation of the ATCase in this manner makes metabolic sense because CTP is synthesized from UTP and thus is produced in larger quantities first. The CTP produced can then inhibit the ATCase enzyme, but not completely. Only when UTP levels rise do UTP and CTP act synergistically to completely inhibit the ATCase enzyme. c.

-

HCO3 + Glutamine + ATP

+ AT Ca se

Carbamoyl phosphate + Aspartate

N-carbamoylaspartate

-

UMP

UTP

CTP

d. The graph shows a hypothetical curve for the effect of CTP and UTP combined on ATCase activity. The KM value is increased, indicating that a higher concentration of aspartate is required to achieve half-maximal activity.

2

3

Case 17 A Possible Mechanism for Blindness Associated with Diabetes Sodium-Dependent Glucose Uptake by Retinal Cells 1. Glucose uptake by pericytes nearly doubles when sodium is present. In contrast, in endothelial cells, the uptake of glucose is constant, regardless of whether sodium is present or absent.

2. Glucose uptake increases as sodium concentration increases in pericytes. In endothelial cells, glucose uptake is constant regardless of sodium ion concentration. The shape of the curve for the pericytes indicates that a protein transporter is involved–glucose uptake increases linearly as sodium ion concentration increases, and then reaches a plateau at high sodium ion concentrations, indicating that the transporter is saturated and is operating at its maximal capacity.

3.

KM

Vmax

With sodium ions

1 KM = − −1 − 0.25 mM KM= 4 mM

Without sodium ions

1 K = − − 0.25 mM−1 M = 4 mM

1 / 0.0 2 8mg /hr nmol Vmax = 36 nmol/ hr/ mg 1 / Vmax = 0.0 6 3mghr/nmol / / Vmax =16 nmol hr mg Vmax =

K The KM values for glucose transport in the presence and absence of sodium ions are the same, indicating that the transporter has equal affinity for glucose in the presence and absence of sodium M ions. However, the maximal velocity is much greater in the presence of sodium ions than in the absence of sodium ions. Because the KM values are the same, sodium ions probably don’t increase the transporter’s affinity for glucose. Instead, it is more likely that the sodium ions cause some kind of conformational change which allows the transporter to transport glucose more effectively.

The investigators have convincingly demonstrated that an SGLT exists in pericytes. Glucose transport increases as the Na + ion concentration increases. The transport curve is hyperbolic, indicating the existence of a protein transporter with specific binding sites for glucose in pericytes.

There is probably not a SGLT in endothelial cells, since glucose transport is independent of Na + concentration.

4. a. Galactose and 2-deoxyglucose inhibited glucose transport only slightly in the presence and absence of NaCl. b. Phlorizin inhibits glucose transport in the presence of sodium ions, but does not inhibit glucose transport in the absence of sodium ions.

1

5. When glucose concentration increases, cell density increases, which is due to increased collagen synthesis in pericytes. Thus, increased glucose transport leads to increased collagen synthesis.

6. a. In the absence of phlorizin, glucose transport increases with increasing concentration of glucose.

But in the presence of phlorizin, glucose transport decreases. b. When glucose concentration increases, the cell density increases in the absence of phlorizin. But when phlorizin is present, glucose transport decreases, as shown in Figure 17.6 (left). In the presence of phlorizin, both Type IV and Type VI collagen synthesis decreases also. Thus, a link has been established between glucose transport and collagen synthesis: When glucose concentration increases, transport into the cell increases, which stimulates collagen production.

7. In the untreated diabetic, plasma glucose concentrations will be high because insulin is not present to stimulate the uptake of glucose into the cells. Pericytes, however, do not rely on insulin for glucose uptake; instead glucose uptake is stimulated by sodium ions. The experimental data show that when glucose concentration is increased, glucose uptake via the SGLT transporter also increases. The results also show that increased glucose uptake leads to increased collagen synthesis, which damages the pericytes and leads to retinopathy.

In order to prevent this from occurring, diabetics should carefully control their plasma glucose concentration through exogenous insulin and diet. In addition, drugs that act in the same manner as phlorizin might be helpful, since the results show that in the presence of phlorizin, glucose uptake by the SGLT decreases, even in the presence of high glucose concentration.

2

Case 19 Purification of Rat Kidney Sphingosine Kinase 1. The sphingosine kinase enzyme elutes last from the DEAE-cellulose (anion exchange) column.

Since anion exchange beads are positively charged, the sphingosine kinase must be negatively charged at pH 7.4. The negatively charged sphingosine kinase was attracted to the positively charged beads and did not elute until the salt concentration of the eluting buffer was raised so as to interfere with the ionic attractive forces. Other proteins in the rat kidney are less negatively charged than sphingosine kinase. If positively charged, the proteins would be repelled by the beads and would be eluted first. If neutral, the proteins would be neither attracted nor repelled and would be eluted next.

2. Both of the species in the phosphate buffer, H 2PO4- and HPO4 are negatively charged and would be attracted to the DEAE-cellulose beads. If phosphate species are bound to the beads, binding of the sphingosine kinase would be prevented. The conjugate base of the Tris buffer is neutral, and the weak acid is positively charged. Thus, the species in the Tris buffer would not be attracted to the DEAE-cellulose beads and would not prevent the sphingosine kinase from binding.

3. a. The molecular weight of sphingosine kinase is about 49 kilodaltons. b. Since treatment with β-mercaptoethanol yielded the same results as the gel run in the absence of the reducing agent, it’s possible that sphingosine kinase consists of a single polypeptide chain with no disulfide bonds. Alternatively, there may be intramolecular disulfide bond(s) within a single chain. The results of this gel do not distinguish between these two possibilities.

4. a. Calculations for the KM and Vmax values for both substrates are shown below. The enzyme has a greater affinity for sphingosine than for ATP, as indicated by the KM values. sphingosine

ATP

1 K = − − 0.14 M−1 M = 7.1 M

1 K = − −16.6 M − 1 M = 0.0 6 0mM=60M

K Vmax =

K Vmax =

1 0.0081mg / min/mol M Vmax = 123molmg / /min

1

0.0078 mg / min/mol M Vmax = 128molmg / /min

b. Estimated KM and Vmax values are shown below. The threo-dihydrosphingosine is a competitive inhibitor since the Vmax has remained relatively unchanged but the KM has increased four-fold.

Competitive inhibitors usually have molecular structures that are similar to the substrate, and a stereoisomer of sphingosine like threo-dihydrosphingosine has a structure similar to sphingosine.

1

1 1 Vmax = K = − 0.0 1 1mg / min/mol − 0.0 4 1mM−1 M = 24.4 mM Vmax = 91molmg / /min K M

2

Case 20 NAD+-dependent Glyceraldehyde-3-Phosphate Dehydrogenase fromT. tenax 1. Hexokinase, phosphofructokinase-1, and pyruvate kinase.

2. In the GAPDH reaction in E. coli, NAD+ is reduced to NADH and H+ and glyceraldehyde-3phosphate is phosphorylated to 1,3-bisphosphoglycerate. Thus this reaction couples oxidation and phosphorylation. This reaction is slightly endergonic, so it is coupled to the next step, which is sufficiently exergonic to drive both reactions. In the next step, ATP will be generated when 1,3bisphosphoglycerate donates one of its phosphates to ADP (first substrate level phosphorylation of glycolysis). Subsequent reactions (production of lactate, ethanol, or oxidative phosphorylation) must re-oxidize the NADH to NAD+ so that the glycolytic pathway can continue.

3. This reaction produces 3-phosphoglycerate instead of 1,3-BPG, a high-energy intermediate. It is irreversible and regulated (see subsequent questions on regulation). The next step will not involve substrate level phosphorylation, i.e. production of ATP, because 3-phosphoglycerate is not a high energy intermediate. In the reaction in E. coli, as described in question 2, the next step involves the production of ATP. The reaction is reversible and not one of the regulated steps of glycolysis (see Question 1).

4. a. NADP+ is an inhibitor of GAPDH because its KM value is greater than the control, indicating that GAPDH has a low affinity for substrate in the presence of the inhibitor and is in the T form.

Glucose-1-phosphate and AMP are activators of GAPDH because the KM values are less than the control, indicating a greater affinity (R form) for substrate in the presence of these activators.

Glucose-1-phosphate is a particular potent activator. b and c NADH and ATP are inhibitors of GAPDH because NADH is a product of the reaction and ATP is a product of the glycolytic pathway as a whole. Increased concentrations of NADH and ATP indicate a high energy charge of the cell. Glycolysis is inhibited if NADH and ATP are not needed. In the same manner, AMP and ADP are activators of GAPDH because AMP and ADP indicate that the energy charge of the cell is low and ATP is needed and thus glycolysis is stimulated. Glucose-1-phosphate is an activator because glucose-1-phosphate is the product of glycogen breakdown. Glucose-1-phosphate is converted to glucose-6-phosphate, and then the glucose-6-phosphate enters glycolysis. Thus, the glucose-1-phosphate acts via feed-forward stimulation and stimulates the activity of a “downstream” enzyme so that additional substrate can be accommodated in the glycolytic pathway. The inhibition by NADP+ might be explained by noting that the other GAPDH isozyme is dependent upon NADP+. Inhibition of the NAD+dependent isozyme may ensure that both enzymes are not simultaneously active.

1

Metabolite

Apparent KM, mM

Inhibitor or activator?

None

3.1

NADP+

4.5

inhibitor

Glucose-1-phosphate

0.4

activator

AMP

1.3

activator

NADH

inhibitor

ATP

inhibitor

ADP

activator

5. The NAD+-dependent GAPDH is composed of a tetramer and it is quite possible that there are multiple binding sites for its substrates. Figure 20.2 shows that the NADP+ inhibition is represented by a sigmoidally shaped curve and an increased value of its Hill coefficient (n >1), indicating that the NAD+ substrate binds cooperatively to the enzyme in the presence of the NADP+ allosteric modulator.

6. The ATP yield for the pathway is 2 ATP. T. tenax stores energy in the form of glycogen, which is degraded to glucose-1-phosphate. The glucose-1-phosphate is then converted to glucose-6phosphate, an isomerization reaction that does not require an investment of ATP. Glucose-6phosphate is then converted to fructose-6-phosphate. The next step, in which F6P is converted to F16BP via the PFK reaction, also does not require an investment of ATP, since the phosphofructokinase reaction in T. tenax is reversible, and is dependent upon pyrophosphate rather than ATP. Since the GAPDH reaction does not produce 1,3BPG, there is no ATP produced until the last step, in which PEP is converted to pyruvate, with the subsequent generation of 2 moles ATP per mole of glucose.

2

Case 21 Characterization of Pyruvate Carboxylase from M. thermoautotrophicum 1.

COOCH3 C

O

+ ATP + HCO3

PYC biotin

-

CH2

+ ADP + Pi C

O

COOCOO-

pyruvate

oxaloacetate

2.

COOCH2 C

O

Pi

+ HCO 3

PPC

-

CH2

+P i

C

O

COOCOO-

phosphoenolpyruvate

oxaloacetate

3. citrate

isocitrate

succinate

isocitrate lyase g lyo xyla t e + cycle, acetyl CoA condenses with oxaloacetate to form 4. In the first step of the tricarboxylic acid acetate

acetyl CoA

malate malate synthase

OAA malate dehydrogenase

citrate. The net result at the end of the cycle is that the two acetyl CoA carbons are oxidized to CO2, reducing equivalents in the form of NADH and FADH2 are generated and the oxaloacetate is regenerated so that the cycle can begin again. Without OAA, the tricarboxylic acid cycle cannot take 1

place. The cycle provides the last step for the oxidation of fuel molecules and also produces intermediates for biosynthetic reactions, so the proper functioning of the cycle is important to the health of the organism.

5. Biotin is a cofactor for pyruvate carboxylase. It serves as a carrier of the carboxylate group. The investigators reasoned that pyruvate carboxylase activity, if present, might increase in the presence of high concentrations of this required cofactor.

6. Two bands at approximately 52 and 75 kD on the denaturing gel indicate that the PYC enzyme has two distinct polypeptide chains. These chains can be termed α and β. These two chains come together to form a complex of about 135 kD as shown on the non-denaturing gel. This is approximately a multiple of four of the enzyme complex, as indicated by the gel filtration results.

Thus, the enzyme structure is α4β4.

7. a. Avidin binds tightly to biotin. Since biotin is a cofactor for pyruvate carboxylase, the presence of avidin will remove biotin from the assay solution and it will be unavailable to the pyruvate carboxylase. Thus the PYC enzyme is inactive. b. The enzyme requires ATP for full activity. No other nucleotide can substitute. If ATP is present, additional nucleotides have no effect on the activity with the exception of ADP and ITP which are mildly inhibitory. c. Acetyl CoA and α-ketoglutarate are mildly inhibitory. Aspartate and glutamate have very little effect on the activity of the enzyme. d. Mg2+ ions are required for full activity. Co2+ and Mn2+ can substitute, but the activity of the enzyme is decreased. Zn2+ cannot substitute for the Mg2+ ion.

8. Biotin attaches to the protein via a lysine residue. A highly conserved lysine is therefore a good candidate for the biotin attachment site. Biotin binds to the bicarbonate; if biotin were not present, the carboxylation reaction could not occur.

2

Case 21 Characterization of Pyruvate Carboxylase from M. thermoautotrophicum 1.

COOCH3 C

O

+ ATP + HCO3

PYC biotin

-

CH2

+ ADP + Pi C

O

COOCOO-

pyruvate

oxaloacetate

2.

COOCH2 C

O

Pi

+ HCO 3

PPC

-

CH2

+P i

C

O

COOCOO-

phosphoenolpyruvate

oxaloacetate

3. citrate

isocitrate

succinate

isocitrate lyase g lyo xyla t e + cycle, acetyl CoA condenses with oxaloacetate to form 4. In the first step of the tricarboxylic acid acetate

acetyl CoA

malate malate synthase

OAA malate dehydrogenase

citrate. The net result at the end of the cycle is that the two acetyl CoA carbons are oxidized to CO2, reducing equivalents in the form of NADH and FADH2 are generated and the oxaloacetate is regenerated so that the cycle can begin again. Without OAA, the tricarboxylic acid cycle cannot take 1

place. The cycle provides the last step for the oxidation of fuel molecules and also produces intermediates for biosynthetic reactions, so the proper functioning of the cycle is important to the health of the organism.

5. Biotin is a cofactor for pyruvate carboxylase. It serves as a carrier of the carboxylate group. The investigators reasoned that pyruvate carboxylase activity, if present, might increase in the presence of high concentrations of this required cofactor.

6. Two bands at approximately 52 and 75 kD on the denaturing gel indicate that the PYC enzyme has two distinct polypeptide chains. These chains can be termed α and β. These two chains come together to form a complex of about 135 kD as shown on the non-denaturing gel. This is approximately a multiple of four of the enzyme complex, as indicated by the gel filtration results.

Thus, the enzyme structure is α4β4.

7. a. Avidin binds tightly to biotin. Since biotin is a cofactor for pyruvate carboxylase, the presence of avidin will remove biotin from the assay solution and it will be unavailable to the pyruvate carboxylase. Thus the PYC enzyme is inactive. b. The enzyme requires ATP for full activity. No other nucleotide can substitute. If ATP is present, additional nucleotides have no effect on the activity with the exception of ADP and ITP which are mildly inhibitory. c. Acetyl CoA and α-ketoglutarate are mildly inhibitory. Aspartate and glutamate have very little effect on the activity of the enzyme. d. Mg2+ ions are required for full activity. Co2+ and Mn2+ can substitute, but the activity of the enzyme is decreased. Zn2+ cannot substitute for the Mg2+ ion.

8. Biotin attaches to the protein via a lysine residue. A highly conserved lysine is therefore a good candidate for the biotin attachment site. Biotin binds to the bicarbonate; if biotin were not present, the carboxylation reaction could not occur.

2

Case 32 Glucose-6-phosphate dehydrogenase activity and cell growth 1. a. Glucose-6-phosphate dehydrogenase catalyzes the conversion of glucose-6-phosphate to 6phosphogluconolactone. The G6P is oxidized, concomitantly, NADP+ is reduced to NADPH. This is the first step in the oxidative branch of the PPP, and is the committing reaction of this branch of the pathway. The activity of this enzyme is regulated according to the cell’s need for NADPH. b. The oxidative branch of the PPP continues with the conversion of 6-phosphogluconolactose to 6phosphoglucoconate via the lactonase reaction, which opens the ring. Next, 6-phosphogluconate dehydrogenase enzyme converts 6-phosphogluconate to ribulose-5-phosphate. This is also a redox reaction, which involves the release of CO2 and the reduction of a second molecule of NADPH. Ribulose-5-phosphate may be isomerized to ribose-5-phosphate, which can be used for DNA and coenzyme biosynthesis. NADPH is important as a reducing agent in many biochemical pathways (such as fatty acid biosynthesis, for example) and is important in ridding the cell of damaging oxidizing agents.

2. In the long term, G6PD can be regulated simply by the concentration of the enzyme protein available in the cell. This can be accomplished through regulation of transcription and translation of the G6PD gene. In the short term, the activity of the enzyme can be controlled by altering the activity of the enzyme once the protein is made, either through the use of allosteric inhibitors or activators, or through sequestration of the enzyme, then rapid release of the enzyme when it is needed.

3. When serum-starved cells were incubated with serum for 1 hour (2), there was about a 50% increase in enzyme activity. This activity increase is more dramatic for cells grown in serum all along (3)— the G6PD activity is four-fold greater in cells grown in the presence of serum than in serum-starved cells (1). This increase in activity is due in part to an increase in expression of the protein, as shown by the Western blot, in which there is a heavier band for the serum-grown cells. But the dramatic increase in activity indicates that the increase in activity cannot be attributed solely to an increase in protein (which is less dramatic)—it is likely that existing G6PD enzyme activity has been stimulated as well. There was no effect on PGD, the other enzyme in the oxidative branch of the PPP that produces NADPH as one of its products. PGD is included as a control in this experiment.

4. Increasing concentrations of DHEA caused a corresponding decrease in the activity of G6PD. A concentration of 100 μM DHEA caused the G6PD activity to decrease by half. In contrast, DHEA had no effect on the activity of PGD. DHEA also caused a corresponding decrease in the incorporation of [3H]-thymidine in cells incubated in the presence of growth factor, indicating that DHEA inhibited cellular growth. In other words, DHEA abolished the stimulation of cell growth that occurs when PGDF is added to cells. The decrease in cellar growth is closely correlated with the decrease in G6PD activity. Whatever the mechanism for this effect, it doesn’t involve changes in G6PD gene expression, since enzyme protein levels did not change when either PGDF or DHEA was added. Taken together, these data imply that DHEA inhibits cell growth via its inhibition on G6PD, although DHEA’s ability to inhibit cell growth via a mechanism independent of the regulation of G6PD cannot be ruled out.

1

5. a. Since serum withdrawal decreases G6PD activity, serum withdrawal should decrease the NADPH/NADP+ ratio. Since NADPH is a product of the G6PD reaction and NADP+ is a reactant, a decreased amount of the enzyme should lead to an increase in the oxidized form of the coenzyme and a decrease in the reduced form. b. In the presence of DHEA, the NADPH/NADP+ ratio should decrease, since DHEA inhibits G6PD activity, for the reasons outlined in (a) c. In the absence of any other factor, adding H2O2 should not affect the ratio, since G6PDH can increase its activity to produce more NADPH to react with the H2O2. d. In the presence of the inhibitor, G6PD cannot handle the increased load of H2O2, and the ratio would decrease.

6. The COS cells overexpress the G6PD enzyme, and as a consequence, cellular growth is stimulated, as noted by the increase (nearly 2-fold) in [3H]-thymidine incorporation in the absence of the DHEA inhibitor. Can the increase in cellular growth be attributed to the increase in G6PD activity? One way to strengthen this correlation is to see if DHEA can inhibit this increased growth. As shown in the figure, DHEA, an inhibitor of G6PD activity, does in fact decrease cell growth in a dosedependent manner. In the presence of 100 μM DHEA, cellular growth of the COS cells is about equal to the growth of non-transfected control cells. This experiment, along with the others presented earlier, strengthens the correlation between G6PD activity and cell growth.

7. Tyrosine phosphorylation has been shown to be an important part of the cascade of reactions that stimulate cellular growth. The authors of this study report that DHEA-treated cells show a decrease in the level of intracellular phosphorylated tyrosines (data not presented in the case study). PGDF binds to receptors in the plasma membrane, causing dimerization and autophosphorylation. The phosphate groups provide docking sites for the enzyme phospholipase-C-γ, an enzyme that hydrolyzes PI 4,5-bisphosphate to diacylglycerol and inositol 1,4,5-triphosphate (IP3). At this point, it would be difficult to say what happens next without more experimental evidence, but since IP3 acts on the ER to release Ca2+, it’s possible that G6PD could be activated by some sort of Ca2+dependent mechanism. Alternatively, diacylglycerol could activate protein kinase C and then the G6PD could be activated. If PI-3-kinase is involved, its activation by PGDF could lead to the phosphorylation of inositol phospholipids. These phosphorylated inositol phospholipids could serve as docking sites for cytosolic proteins, which may result in G6PD activation by a mechanism that has yet to be deciphered.

2