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1984 - Introduction Notes

Chapter 21 - Reaction, Revolution, and Romanticism, 1815-1850

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<html><body>CHAPTER Biochemical Evolution 2

The authors have used evolution as the unifying theme of this book. The present

chapter is an ambitious attempt to illustrate how all aspects of cellular functioning can be better understood from an evolutionary perspective. Truly mastering

the breadth of information touched on here may be difficult but the reward will be a more basic, thorough, and intuitive grasp of the whole remainder of the book.

The origin of life is considered in four stages—generation of biomolecules, tran

sition to replicating systems, interconversion of light and chemical energy, and adaptability to change. This discussion is theoretical, since the origins are obscure and hard data is lacking about actual mechanisms.

Evolution requires three properties: a system must reproduce, there must be vari

ation, and there must be competition in a selective environment. Any system that satisfies these requirements will evolve, whether pure RNA in solution with a replicating enzyme, or a population of cells or higher plants and animals.

After touching on ribozymes as evidence that life passed through an “RNA

World” stage, the authors illustrate how duplication and variation led to the many features of modern cells including DNA genes, ATP, lipid membranes, ion pumps, energy transducers, receptors with second messengers, etc. Cells have to move, either with flagella (procaryotes) or by changing shape using microfilaments, microtubules, and molecular motors (eucaryotes). Multicellular organisms require cells to differentiate according to developmental programming and signals from neighboring cells. All life on Earth came from a single progenitor, so we can learn about human biochemistry by studying any species, even simple single-celled organisms. 9

CHAPTER 2 LEARNING OBJECTIVES

When you have mastered this chapter, you should be able to complete the following objectives. Key Organic Molecules Are Used by Living Systems (Text Section 2.1)

1. List the four stages leading from inert chemicals to modern living cells.

2. Explain the Urey-Miller experiment, and diagram the apparatus. Describe the major

products produced by this experiment. Evolution Requires Reproduction, Variation, and Selective Pressure (Text Section 2.2)

3. Identify the three principles necessary for evolution to occur.

4. Describe Spiegelman’s experiment with Qb RNA. Understand how the three principles

of evolution are included in this experiment.

5. Most enzymes are composed of protein. Explain how ribozymes differ from more nor

mal enzymes.

6. Describe what is meant by a “hammerhead ribozyme.”

7. Explain how RNA bases are derived from amino acids.

8. Explicate the advantages that polymers of amino acids have over nucleic acid polymers

in providing catalysis for the cell.

9. Describe the roles of mRNA, tRNA, and rRNA in protein synthesis. Know that three

mRNA bases are required to code for a single amino acid.

10. Ribosomal catalysis of peptide bond synthesis is mediated by regions of rRNA, and not

by protein. Understand the implications of this catalysis for the concept of an RNA World.

11. Recall the three principles necessary for evolution as defined in Section 2.2. With these in

mind, explain how the genetic code is ideally suited as a medium for evolutionary change.

12. Transfer RNAs all have very similar structures with minor variations that lead to signif

icant differences in function. This is a common phenomenon in biochemistry. Describe how this situation would arise.

13. Explain the advantages of DNA compared to RNA for long-term storage of information.

14. The building blocks of DNA are made directly from the building blocks of RNA.

Understand that this leads to the deduction that RNA must be older than DNA.

15. Define transcription and translation. Energy Transformations Are Necessary to Sustain Living Systems (Text Section 2.3)

16. Describe the similarities between ATP production and use, and the function of money

in society. You should appreciate the fact that this leads to the description of ATP as “energy currency” in the cell.

17. Describe the properties of a cell membrane that are responsible for keeping important

cellular constituents (enzymes, nucleic acids, ATP, etc.) inside.

18. Define osmosis, ion pump, and ion gradient.

BIOCHEMICAL EVOLUTION

19 Describe the process of photosynthesis in general terms. Understand why photosynthe

sis must be membrane-associated.

20. Write the equation for the oxidation of water to oxygen.

21. Understand why oxygen is described as “toxic.”

22. Know how many ATPs are produced per glucose consumed when using oxygen in glu

cose metabolism. Cells Can Respond to Changes in Their Environments (Text Section 2.4)

23. Describe how E. coli responds when arabinose is the only source of carbon.

24. Define second messenger and signal transduction. Name two second messengers.

25. Distinguish between flagella, microfilaments, and microtubules.

26. Identify what happens on a molecular level when cells change shape.

27. Define cell differentiation.

28. Describe how the slime mold Dictyostelium uses signaling and changes in cell differen

tiation to respond to varying conditions. Understand that cAMP acts as a messenger (not a second messenger) for Dictyostelium.

29. Give a general description of how development is controlled in C. elegans. Notice the total

number of cells in an adult human, and contrast that with the number of cells in C. elegans.

30 Know why understanding enzymes and processes in single-celled organisms like yeast

or E. coli help us understand how human cells work.

31. Examine the time line in Figure 2.27, and explain during what time frame single-celled

anaerobes would have dominated life on Earth. SELF-TEST Key Organic Molecules Are Used by Living Systems

1. A reducing atmosphere as described in this chapter would not contain significant

amounts of

a.

CH4

d.

H2O

b.

CO2

e.

H2

c.

NH3 Evolution Requires Reproduction, Variation, and Selective Pressure

2. What would happen in Spiegelman’s experiment with Qb RNA if no selective conditions

were imposed (inhibitors, limited time, etc.)? Would a variety of different RNAs still arise?

3. Does RNA self-replicate?

4. Which amino acid is not mentioned in textbook Figure 2.6 as a source for synthesis of

RNA bases?

a.

glutamine

d.

serine

b.

glycine

e.

none of the above

c.

aspartic acid

a.

uracil

d.

cytosine

b.

adenine

e.

guanine

c.

thymine Energy Transformations Are Necessary to Sustain Living Systems

7. Osmosis tends to equalize concentrations on both sides of a membrane. Any living cell

will have protein and nucleic acid inside, which “draws” water inward. To prevent bursting, concentration of something inside the cell has to be made lower than the concentration outside. Concentration of what? How is this adjustment made?

8. Would the structure of an ion-driven ATP synthase have to be different from that of an

ATP-driven ion pump?

9. What is the advantage to the use of oxygen in metabolism? Cells Can Respond to Changes in Their Environments

10. What signal causes aggregation of Dictyostelium slime mold amoebae into mobile slugs?

11. Actin is an important part of human muscle. It is equally important in other species

including amoebas and slime molds. Is it surprising to find the same protein in such diverse species? ANSWERS TO SELF-TEST

1. b. CO2. Interestingly, modern theories based on observations of atmospheres of other

planets, and observations of the geochemistry of early minerals, hold that there was much more carbon dioxide (CO2) than hydrogen (H2) in the Earth’s early atmosphere.

2. Yes. Variability should remain constant, but the variant RNAs would presumably re

main in low concentrations or disappear, and the original RNA would probably remain dominant.

3. No. Despite much work to find a self-replicating RNA, the replication always requires

the presence of protein. Recent work by David P. Bartel at MIT is showing some promise toward finding an RNA replicase ribozyme (Science 292[5520]:1319). The fact that an RNA replicase may be produced in the laboratory does not, of course, prove that the ribozyme existed in nature.

4. d. Serine. In modern cells, the glycine plus two of the other carbons of the purine ring

can originate as parts of the amino acid serine.

5. Ribozymes are easy to find in modern cells, and probably the most abundant one is the

ribosome where peptide bonds are formed. Several others exist including certain ribonuclease enzymes.

6. c. Thymine. All of the other building blocks are found in RNA. Uracil is only found in

RNA. Thymine replaces it in DNA.

BIOCHEMICAL EVOLUTION

7. Small ions including sodium and protons (H;) are routinely pumped out of the cell. This

allows the outward osmotic pressure generated by the ions to match the inward pressure generated by cellular macromolecules.

8. No. In fact, textbook Figures 2-16 and 2-17 depict the same system functioning inward

or outward. And in living cells the structures are the same or very similar.

9. While aerobic cells have to have protection against oxygen damage, the rewards for deal

ing with oxygen are great. As stated in the text, glucose metabolism using oxygen affords 15 times as much ATP as anaerobic glucose metabolism. Thus anaerobes have to ingest 15 times as much sugar to do the same work as aerobes. Use of other fuels also produces much more ATP in aerobic cells. It is also true that using oxygen as an electron acceptor can aid in maintenance of a proton gradient.

10. Cyclic AMP causes the cells to aggregate into a multicellular organism. cAMP is found

as a kind of “hunger signal” in many different organisms, from procaryotes to man.

11. Considering the “Unity of Biochemistry” perhaps it is more surprising that actin does not

appear to play the same role in procaryotic cells. But actin is found in essentially all eucaryotes in a similar role, often paired with myosin as the contractile apparatus. PROBLEMS

1. Stanley Miller’s experiments are called the “Primordial Soup Theory.” There are other

schools of thought not mentioned in the chapter, notably Günter Wächtershäuser’s Pyrite World. He suggests that early life might have lived in the hot sulfur-rich environment near deep volcanic vents, and that precellular reactions could have taken place on the surface of pyrite crystals. One disadvantage is the extreme heat and pressure—over 110ÚC—but that environment is rich in life today. Can you think of other advantages or disadvantages of the Pyrite Theory versus the Soup Theory?

2. The RNA from Phage Qb was shown to evolve in an artificial system with no membranes

or cells. Why is it so important that organisms should have had membranes for them to evolve efficiently? What is the difference?

3. The antibiotic peptide, gramicidin, is assembled (in modern cells) without the use of

RNA. Peptide bonds are formed after the amino acids are activated by attachment to sulfur on the enzyme surface. Does this suggest an alternative, or a precursor, to the RNA world described in the chapter?

4. RNA bases are built from amino acids. Thus amino acids (which are produced in the

Urey Miller experiment) are older than RNA building blocks (which are not produced in this experiment). Is it reasonable that the only use to which amino acids were put was synthesis of RNA building blocks?

5. DNA has a remarkable ability to preserve complex information perfectly intact for mil

lennia. Would it be a favorable situation if DNA could always be reproduced with absolutely no errors, and never had any mutations?

6. Theorists of the RNA World have debated whether the constituents of the cell arose in the

sequence RNA-DNA-Protein or RNA-Protein-DNA. The universal use of ribonucleotide reductase enzymes provided an answer to this question. Can you see why?

CHAPTER 2

7. If arabinose is the only source of carbon, E. coli cells utilize it for metabolism. The sys

tem described in this chapter apparently is driven only by the presence of arabinose. What if glucose and arabinose are present in equal concentrations? The arabinose would not be the “sole source.” Is there an implication that the cell also checks for the absence of glucose?

8. Scientists know that the Earth’s early oceans around three billion years ago were very

rich in dissolved iron salts, including ferrous chloride (FeCl2). Many ferric compounds, including ferric oxide (Fe2O3), are rather insoluble in water. Given these facts, what kind of hard evidence would you look for to prove that oxygen entered the atmosphere about two billion years ago, as shown in textbook Figure 2.27? ANSWERS TO PROBLEMS

1. One major advantage of the “hot deep” origin of life is the fact that at volcanic vents,

one finds metal sulfides that would be insoluble at cooler temperatures, and hydrogen sul fide gas (H2S). These can combine to form pyrite or “Fool’s Gold” (H

+

2S+FeSDFeS2

H2). Thus a deep-sea volcanic vent is a reducing environ

ment (with electrons from hydrogen [H2]), and spontaneous synthesis of both amino acids and peptides has been observed in laboratory simulations of this environment. Much of the work and theory has come from the collaboration of Claudia Huber and Günter Wächtershäuser (recent papers published in Science). It is especially important to identify a terrestrial system with reducing properties now that the Earth’s early atmosphere is thought to have been a CO2 greenhouse and not reducing at all. The obvious disadvantage is that organic compounds can be destroyed by the extremely hot environment. But the fact that there are abundant living organisms at the vents illustrates that this is a problem that life has solved. (Huber & Wächtershäuser. Science 281[5377]: 670.)

2. Spiegelman’s RNA system with a replicase enzyme is very artificial; there is only one mol

ecule being reproduced. A living cell has many constituents, and part of the competition in evolution involves which cell has the best mixture of constituents. The whole organism must evolve, with all its parts. This cannot happen in a “soup”; it requires individuals surrounded by a barrier, hence a membrane.

3. Yes, it does. While the thioester method of peptide synthesis used in making gramicidin

is cumbersome compared to RNA-directed peptide synthesis, it does suggest that proteins might be able to self-replicate. Several prominent theoreticians including Graham Cairns-Smith, Freeman Dyson, Robert Shapiro, and the Nobel laureate Christian de Duve see a period before the “RNA World” in which proteins are the dominant cellular macromolecule, and many aspects of metabolism would resemble what is seen in modern cells. The use of ATP and other nucleotides as energy currency in a very primitive system would lead naturally to an environment where RNA synthesis could occur spontaneously. This is in contrast to the “Primordial Soup” where nucleotides would be unstable and probably quite rare.

4. Not really. A system rich in amino acids would have at least some peptides. And there

are many processes that are easily catalyzed by simple proteins but have never been demonstrated using RNA ribozymes. An example would be the sort of electron transfer mediated by iron sulfur clusters. Cellular synthesis of purines and pyrimidines must be very ancient, but it would seem likely that these are merely representatives of many other processes involving amino acids and peptides.

BIOCHEMICAL EVOLUTION

5. No, it would not be favorable. While some critical genes, such as those for the histone

proteins found in the nuclei of eucaryotes, appear to remain pristine and never change, in fact there must be variation that is ruthlessly trimmed by selection. A lack of variation, of mutation in the DNA, would lead to an end to evolution. We would be “stuck” with the species that lived millions of years ago, or more accurately, “we” would never have come into existence. Considering the fact that DNA must vary, it is quite interesting that some of the earliest microfossils found by J. William Schopf and others appear to be cyanobacteria, or blue-green algae, which are morphologically almost identical to pond-scum living today. This is despite the fact that one or two billion years separate the fossils from the living examples (e.g., Entophysalis, living today, and Eoentophysalis, 2.1 billion years old [see Cradle of Life, Schopf, Princeton 1999, p. 229]).

6. The mere fact that DNA building blocks are made from RNA building blocks shows that

DNA is newer than RNA. But the fact that, universally, DNA building blocks are produced by a protein enzyme proves that protein also came before DNA. Note that it does not resolve the question of whether the correct sequence is RNA-Protein-DNA or ProteinRNA-DNA. There are several other indications that DNA is a “recent” development, including RNA genomes in some viruses, and inconsistencies in DNA structure and usage. There are some eucaryotic species (dinoflagellates) which use 5-Hydroxymethyluracil instead of Thymine, for example. The fact that DNA is bound to histones as chromatin in most eucaryotes is very different from the way DNA is handled in procaryotes. Freeland, Knight, & Landweber. Science 286(5440): 690.

7. Yes, there is. If there were no mechanism to check for the presence of glucose, then ara

binose or other sugars would be utilized whenever they were present, and not only when they were the “sole source.” In fact a second messenger mentioned in this chapter, cyclic AMP, is generated when glucose is absent. This “hunger signal” then allows the use of other sugars.

8. The evidence is as hard as iron! Geologists know that the best iron ore is found in

“banded iron formations,” which are usually in layers that were part of the ocean floor around two billion years ago. These attractive red-layered formations represent the precipitation of most of the dissolved iron in the world’s oceans as hematite, magnetite, and other insoluble ferric salts. The age of these layers can be clearly established by isotopic dating. The emergence of more abundant oxygen is the only possible explanation for this worldwide chemical reaction. EXPANDED SOLUTIONS TO TEXT PROBLEMS

1. For alanine, the NH2 would come from NH3; CH3, CH, and the other carbon from CH4;

OH and the other oxygen from H2O. (Some hydrogens could also be replaced from H2 if lost in earlier oxidation reactions.)

H C

H

3

OH

H N

2

K

O

2. The lone fast-replicating molecule will complete three “generations” for every replication

of the 99 other molecules. After n “generations,” each of 15 minutes duration, therefore,

CHAPTER 2

one population will be (99)(2n), while the other population will be (1)(23n). The results will be: Generation # Slow # Fast % Slow % Fast

n

(99)(2n)

(1)(23n)

0

99

1

99.00%

1.00%

1

198

8

96.12%

3.88%

10

1.0¥105

1.1¥109

0.01%

99.99%

25

3.3¥109

3.8¥1022

0.00%

100.00%

3. The more tightly bound nucleotide monomers would be more available for RNA repli

cation and could therefore cause a faster rate of replication. This advantage would be most important if the monomers were in short supply, that is, present only in low concentrations in the solution (environment).

4. Chemical or physical equilibrium between two compartments would require the same

ion concentrations in both compartments (a state of high entropy). To establish a gradient with unequal ion concentrations in the two compartments would require work to impose more order on the system (and move the system to a state of higher energy and lower entropy). (Consider also a bag of 100 red marbles and another equivalent bag that has 100 green marbles. It requires less effort [energy] to allow the marbles to mix together in a single bag than it does to separate the mixture back into the original all-red and all-green compartments.)

5. If a “gate” is opened to allow protons to flow out of the cell, then energy will be released.

If some of this energy could be captured for useful work, then the energy could be used for pumping a second type of ion out of the cell? (E.g., a proton ATPase would couple the synthesis of a high-energy bond in adenosine triphosphate (ATP) to the release of a proton gradient; the chemical energy stored in the ATP could then be used for another purpose, such as pumping the second ion.)

6. Eight protons, because the generation of hydroxide ion on one side is equivalent to the

generation of a proton on the other side.

7. Very hydrophobic molecules could cross the cell membrane without the assistance of a trans

port protein. For these molecules, therefore, only a gene-control protein would be needed.

8. From the early part of the time scale in Figure 2.26, it appears that there are between

five and seven cycles of approximately synchronous division before the respective cell division rates diverge. CHAPTER Protein Structure and Function 3

Proteins are macromolecules that play central roles in all the processes of life.

Chapter 3 begins with a discussion of key properties of proteins and continues with a description of the chemical properties of amino acids—the building

blocks of proteins. It is essential that you learn the names, symbols, and properties of the 20 common amino acids at this point, as they will recur throughout the text in connection with protein structures, enzymatic mechanisms, metabolism, protein synthesis, and the regulation of gene expression. It is also important to review the behavior of weak acids and bases, either in the appendix to Chapter 3 or in an introductory chemistry text. Following the discussion of amino acids, the chapter turns to peptides and to the linear sequences of amino acid residues in proteins. Next, it describes the folding of these linear polymers into the specific three-dimensional structures of proteins. The primary structure (or sequence of amino acids) dictates the higher orders of structure including secondary (a, b, etc.), tertiary (often globular), and quaternary (with multiple chains). You should note that the majority of functional proteins exist in water and that their structures are stabilized by the forces and interactions you learned about in Chapter 1. This chapter concludes with a discussion of the theory of how proteins fold, including attempts to predict protein folding from amino acid sequences. 17

CHAPTER 3 LEARNING OBJECTIVES

When you have mastered this chapter, you should be able to complete the following objectives. Introduction

1. List the key properties of proteins.

2. Explain how proteins relate one-dimensional gene structure to three-dimensional struc

ture in the cell, and their complex interactions with each other and various substrates. Proteins Are Built from a Repertoire of 20 Amino Acids (Text Section 3.1)

3. Draw the structure of an amino acid and indicate the following features, which are com

mon to all amino acids: functional groups, side chains, ionic forms, and isomeric forms.

4. Classify each of the 20 amino acids according to the side chain on the a carbon as aliphatic, aromatic, sulfur-containing, aliphatic hydroxyl, basic, acidic, or amide derivative.

5. Give the name and one-letter and three-letter symbol of each amino acid. Describe each

amino acid in terms of size, charge, hydrogen-bonding capacity, chemical reactivity, and hydrophilic or hydrophobic nature.

6. Define pH and pKa. Use these concepts to predict the ionization state of any given amino

acid or its side chain in a protein.

7. State Beer’s Law. Understand how it can be used to estimate protein concentration. Primary Structure: Amino Acids Are Linked by Peptide Bonds to Form Polypeptide Chains (Text Section 3.2)

8. Draw a peptide bond and describe its conformation and its role in polypeptide sequences.

Indicate the N- and C-terminal residues in peptides.

9. Define main chain, side chains, and disulfide bonds in polypeptides. Give the range of molecular weights of proteins.

10. Explain the origin and significance of the unique amino acid sequences of proteins.

11. Understand why nearly all peptide bonds are trans.

12. Define the f and y angles used to describe a peptide bond, and be able to read a Ramachandran plot. Secondary Structure: Polypeptide Chains Can Fold into Regular Structures Such as the Alpha Helix, the Beta Sheet, and Turns and Loops (Text Section 3.3)

13. Differentiate between two major periodic structures of proteins: the a helix and the b pleated sheet. Describe the patterns of hydrogen bonding, the shapes, and the dimensions of these structures.

14. List the types of interactions among amino acid side chains that stabilize the three- dimensional structures of proteins. Give examples of hydrogen bond donors and acceptors.

15. Describe a-helical coiled coils in specialized proteins and the role of b turns or hairpin turns in the structure of common proteins.

PROTEIN STRUCTURE AND FUNCTION Tertiary Structure: Water-Soluble Proteins Fold into Compact Structures with Nonpolar Cores (Text Section 3.4)

16. Using myoglobin and porin as examples, describe the main characteristics of native folded

protein structures. Quaternary Structure: Polypeptide Chains Can Assemble into Multisubunit Structures (Text Section 3.5)

17. Describe the primary, secondary, tertiary, and quaternary structures of proteins. Describe domains. The Amino Acid Sequence of a Protein Determines Its Three-Dimensional Structure (Text Section 3.6)

18. Using ribonuclease as an example, describe the evidence for the hypothesis that all of the

information needed to specify the three-dimensional structure of a protein is contained in its amino acid sequence.

19. Rationalize the conformational preferences of different amino acids in proteins and

polypeptides.

20. Give evidence that protein folding appears to be a cooperative transition, and explain

why that means it is an “all or none” process.

21. Explain how protein folding proceeds through stabilization of intermediate states rather

than through a sampling of all possible conformations.

22. Discuss the methods and advances in the prediction of three-dimensional structures

of proteins.

23. List examples of the modification and cleavage of proteins that expand their functional roles. SELF-TEST Introduction Proteins Are Built from a Repertoire of 20 Amino Acids

1. (a) Examine the four amino acids given below:

COO

+

J

HNJJ CJH

+H NJ CJH

+H NJ CJH

3

J

J

J

H C

CH

2

J

CH

2

J 2

CH

J

2

CH

J J

2

H C

CH

J

3

CH J 2

J

CH

OH

2

J

NH +

3 A B C D

CHAPTER 3

Indicate which of these amino acids are associated with the following properties:

(a) aliphatic side chain (b) basic side chain (c) three ionizable groups (d) charge of ;1 at pH 7.0 (e) pK ~10 in proteins (f)

secondary amino group

(g) designated by the symbol K (h) in the same class as phenylalanine (i)

most hydrophobic of the four

(j)

side chain capable of forming hydrogen bonds

(b) Name the four amino acids. (c) Name the other amino acids of the same class as D.

2. Draw the structure of cysteine at pH 1.

3. Match the amino acids in the left column with the appropriate side chain types in the

right column.

(a) Lys

(1) nonpolar aliphatic

(b) Glu

(2) nonpolar aromatic

(3) basic

(d) Cys

(4) acidic

(e) Trp

(5) sulfur-containing

(f)

Ser

(6) hydroxyl-containing

4. Which of the following amino acids have side chains that are negatively charged under

physiologic conditions (i.e., near pH 7)?

(a) Asp

(d) Glu

(b) His

(e) Cys

5. Why does histidine act as a buffer at pH 6.0? What can you say about the buffering ca

pacity of histidine at pH 7.6? Primary Structure: Amino Acids Are Linked by Peptide Bonds to Form Polypeptide Chains

6. How many different dipeptides can be made from the 20 L amino acids? What are the

minimum and the maximum number of pK values for any dipeptide?

7. For the pentapeptide Glu-Met-Arg-Thr-Gly,

(a) name the carboxyl-terminal residue. (b) give the number of charged groups at pH 7. (c) give the net charge at pH 1. (d) write the sequence using one-letter symbols. (e) draw the peptide bond between the Thr and Gly residues, including both side chains.

8. If a polypeptide has 400 amino acid residues, what is its approximate mass?

(a) 11,000 daltons

(b) 22,000 daltons

(d) 88,000 daltons

9. Which amino acid can stabilize protein structures by forming covalent cross-links be

tween polypeptide chains?

(a) Met

(d) Gly

(b) Ser

(e) Cys

10. Discuss the significance of Ramachandran plots. Contrast the conformational states of Gly

and Pro in proteins compared with other amino acid residues. Secondary Structure: Polypeptide Chains Can Fold into Regular Structures Such as the Alpha Helix, the Beta Sheet, and Turns and Loops

11. Which of the following statements about the peptide bond are true?

(a) The peptide bond is planar because of the partial double-bond character of the bond

between the carbonyl carbon and the nitrogen.

(b) There is relative freedom of rotation of the bond between the carbonyl carbon and

the nitrogen.

bonyl group.

(d) There is no freedom of rotation around the bond between the a carbon and the car

bonyl carbon.

12. Which of the following statements about the a helix structure of proteins is correct?

(a) It is maintained by hydrogen bonding between amino acid side chains. (b) It makes up about the same percentage of all proteins. (c) It can serve a mechanical role by forming stiff bundles of fibers in some proteins. (d) It is stabilized by hydrogen bonds between amide hydrogens and amide oxygens in

polypeptide chains.

(e) It includes all 20 amino acids at equal frequencies.

13. Which of the following properties are common to a-helical and b pleated sheet struc

tures in proteins?

(a) rod shape (b) hydrogen bonds between main-chain CO and NH groups (c) axial distance between adjacent amino acids of 3.5 Å (d) variable numbers of participating amino acid residues

14. Explain why a helix and b pleated sheet structures are often found in the interior of

water-soluble proteins. Tertiary Structure: Water-Soluble Proteins Fold into Compact Structures with Nonpolar Cores

15. Which of the following amino acid residues are likely to be found on the inside of a

water-soluble protein?

(a) Val

(d) Arg

(b) His

(e) Asp

16. Which of the following statements about the structures of water-soluble proteins, ex

emplified by myoglobin, are not true?

(a) They contain tightly packed amino acids in their interior. (b) Most of their nonpolar residues face the aqueous solvent. (c) The main-chain NH and CO groups are often involved in H-bonded secondary

structures in the interior of these proteins.

(d) Polar residues such as His may be found in the interior of these proteins if the

residues have specific functional roles.

(e) All of these proteins contain b sheet structural motifs. Quaternary Structure: Polypeptide Chains Can Assemble into Multisubunit Structures

17. Match the levels of protein structures in the left column with the appropriate descrip

tions in the right column.

(a) primary

(1) association of protein subunits

(b) secondary

(2) overall folding of a single chain, can

include a-helical and b sheet structures

(d) quaternary

(3) linear amino acid sequence (4) repetitive arrangement of amino acids

that are near each other in the linear sequence The Amino Acid Sequence of a Protein Determines Its Three-Dimensional Structure

18. Which of the following statements are true?

(a) Ribonuclease (RNase) can be treated with urea and reducing agents to produce a

random coil.

(b) If one oxidizes random-coil RNase in urea, it quickly regains its enzymatic activity. (c) If one removes the urea and oxidizes RNase slowly, it will renature and regain its

enzymatic activity.

(d) Although renatured RNase has enzymatic activity, it can be readily distinguished

from native RNase.

19. When most proteins are exposed to acidic pH (e.g., pH 2), they lose biological activity.

Explain why.

20. Which one of the following amino acids may alter the direction of polypeptide chains

and interrupt a helices?

(a) Phe

(d) His

(b) Cys

(e) Pro

21. If we know that a solution of protein is half-folded, what will we find in solution?

(a) 100% half-folded protein (b) 50% fully folded, 50% unfolded (c) 33% fully folded, 34% half-folded, and 33% unfolded

22. Several amino acids can be modified after the synthesis of a polypeptide chain to en

hance the functional capabilities of the protein. Match the type of modifying group in the left column with the appropriate amino acid residues in the right column.

(a) phosphate

(1) Glu

(b) hydroxyl

(2) Thr

(3) Pro

(d) acetyl

(4) Ser (5) N-terminal (6) Tyr

23. How can a protein be modified to make it more hydrophobic? ANSWERS TO SELF-TEST

1. (a) (a) C (b) D (c) B, D (d) D (e) B, D (f) A (g) D (h) B (i) C (j) B, D (k) D.

(b) A is proline, B is tyrosine, C is leucine, and D is lysine. (c) Histidine and arginine (basic amino acids).

2. See the structure of cysteine. At pH 1, all the ionizable groups are protonated.

COOH J

+H NJ CJH

3

J

CH J 2

SH Cysteine

3. (a) 3 (b) 4 (c) 1 (d) 5 (e) 2 (f) 6

4. a, d

5. Histidine acts as a buffer at pH 6.0 because this is the pK of the imidazole group. At pH

7.6, histidine is a poor buffer because no one ionizing group is partially protonated and therefore capable of donating or accepting protons without markedly changing the pH.

6. The 20 L amino acids can form 20¥20=400 dipeptides. The minimum number of

pK values for any dipeptide is two; the maximum is four.

7. (a) glycine

(b) 4, namely the 2 carboxyl groups of glutamate, the R group of arginine, and the alpha

amino group of glycine.

(c) ;2, contributed by the N-terminal amino group and the arginine residue (d) E-M-R-T-G (e) See the structure of the peptide bond below.

H

O

H

J K

NJ CJCJNJCJCJO

J

JJ

HJ CJOH H H

J

CH

3 Peptide bond

8. c

9. e

10. a. A Ramachandran plot gives the possible f and y angles for the main polypeptide chain

containing different amino acid residues. The fact that glycine lacks an R group means that it is much less constrained than other residues. In Figure 3.31, the left-handed helix region, which occurs rarely, generally includes several Gly residues. In contrast to glycine, proline is more highly constrained than most residues because the R group is tied to the amino group. This fixes f at about :65Ú. In Figure 3.26, the rare cis form of the peptide bond is shown as occurring about half of the time in X-Pro peptide bonds.

11. a, c

12. c, d

13. b, d

14. In both a-helical and b sheet structures, the polar peptide bonds of the main chain are

involved in internal hydrogen bonding, thereby eliminating potential hydrogen bond formation with water. Overall the secondary structures are less polar than the corresponding linear amino acid sequences.

15. a, c. Specific charged and polar amino acid residues may be found inside some proteins,

in active sites, but most polar and charged residues are located on the surface of proteins.

16. b, e. Statement (b) is incorrect because globular, water-soluble proteins have most of

their nonpolar residues buried in the interior of the protein. Statement (e) is incorrect because not all water-soluble proteins contain b sheet secondary structures. For example, myoglobin is mostly a-helical and lacks b sheet structures.

17. (a) 3 (b) 4 (c) 2 (d) 1

18. a, c

19. A low pH (pH 2) will cause the protonation of all ionizable side chains and will change

the charge distribution on the protein; furthermore, it will impart a large net positive charge to the protein. The resulting repulsion of adjacent positive charges and the disruption of salt bridges often cause unfolding of the protein and loss of biological activity.

20. e

21. b

22. (a) 2, 4, 6 (b) 3 (c) 1 (d) 5

23. The attachment of a fatty acid chain to a protein can increase its hydrophobicity and

promote binding to lipid membranes. PROBLEMS

1. The net charge of a polypeptide at a particular pH can be determined by considering the

pK value for each ionizable group in the protein. For a linear polypeptide composed of 10 amino acids, how many a-carboxyl and a-amino groups must be considered?

2. For the formation of a polypeptide composed of 20 amino acids, how many water mol

ecules must be removed when the peptide bonds are formed? Although the hydrolysis of a peptide bond is energetically favored, the bond is very stable in solution. Why?

PROTEIN STRUCTURE AND FUNCTION

3. Where stereoisomers of biomolecules are possible, only one is usually found in most or

ganisms; for example, only the L amino acids occur in proteins. What problems would occur if, for example, the amino acids in the body proteins of herbivores were in the L isomer form, whereas the amino acids in a large number of the plants they fed upon were in the D isomer form?

4. Many types of proteins can be isolated only in quantities that are too small for the direct

determination of a primary amino acid sequence. Recent advances in gene cloning and amplification allow for relatively easy analysis of the gene coding for a particular protein. Why would an analysis of the gene provide information about the protein’s primary sequences? Suppose that two research groups, one in New York and the other in Los Angeles, are both analyzing the same protein from the same type of human cell. Why would you not be surprised if they publish exactly the same primary amino acid sequence for the protein?

5. Each amino acid in a run of several amino acid residues in a polypeptide chain has f

values of approximately :140Ú and y values of approximately ;147. What kind of structure is it likely to be?

6. A survey of the location of reverse turns in soluble proteins shows that most reverse turns

are located at the surface of the protein, rather than within the hydrophobic core of the folded protein. Can you suggest a reason for this observation?

7. Wool and hair are elastic; both are a-keratins, which contain long polypeptide chains

composed of a helices twisted about each other to form cablelike assemblies with crosslinks involving Cys residues. Silk, on the other hand, is rigid and resists stretching; it is composed primarily of antiparallel b pleated sheets, which are often stacked and interlocked. Briefly explain these observations in terms of the characteristics of the secondary structures of these proteins.

8. In a particular enzyme, an alanine residue is located in a cleft where the substrate binds.

A mutation that changes this residue to a glycine has no effect on activity; however, another mutation, which changes the alanine to a glutamate residue, leads to a complete loss of activity. Provide a brief explanation for these observations.

9. Glycophorin A is a glycoprotein that extends across the red blood cell membrane. The

portion of the polypeptide that extends across the membrane bilayer contains 19 amino acid residues and is folded into an a helix. What is the width of the bilayer that could be spanned by this helix? The interior of the bilayer includes long acyl chains that are nonpolar. Which of the 20 L amino acids would you expect to find among those in the portion of the polypeptide that traverses the bilayer?

10. Before Anfinsen carried out his work on refolding in ribonuclease, some scientists ar

gued that directions for folding are given to the protein during its biosynthesis. How did Anfinsen’s experiments contradict that argument?

11. Early experiments on the problem of protein folding suggested that the native three-di

mensional structure of a protein was an automatic consequence of its primary structure.

(a) Cite experimental evidence that shows that this is the case.

Later, the discovery that proteins are synthesized directionally on ribosomes, from the amino to the carboxy terminus, complicated the earlier view of protein folding.

(b) Explain what the complicating circumstance might be.

The discovery of chaperone proteins allows both earlier views to be reconciled.

CHAPTER 3

12. Suppose you are studying the conformation of a monomeric protein that has an unusu

ally high proportion of aromatic amino acid residues throughout the length of the polypeptide chain. Compared with a monomeric protein containing many aliphatic residues, what might you observe for the relative a-helical content for each of the two types of proteins? Would you expect to find aromatic residues on the outside or the inside of a globular protein? What about aliphatic residues?

13. As more and more protein sequences and three-dimensional structures become known,

there is a proliferation of computer algorithms for the prediction of folding based on sequence. How might it be possible to winnow through the possibilities and find the best computer programs? Bear in mind that if the sequence and the structure are available, it is too easy to “reverse engineer” a routine that will produce the correct answer.

14. In its discussion of protein modification and cleavage, the text refers to the synthesis and

cleavage of a large polyprotein precursor of virus proteins, as well as to the synthesis of multiple polypeptide hormones from a single polypeptide chain. Is there an advantage to synthesizing a large precursor chain and then cleaving it to create a number of products?

15. What is the molarity of pure water? Show that a change in the concentration of water by

ionization does not appreciably affect the molarity of the solution.

16. When sufficient H+ is added to lower the pH by one unit, what is the corresponding

increase in hydrogen ion concentration?

17. You have a solution of HCl that has a pH of 2.1. What is the concentration of HCl needed

to make this solution?

18. The charged form of the imidazole ring of histidine is believed to participate in a reac

tion catalyzed by an enzyme. At pH 7.0, what is the probability that the imidazole ring will be charged?

19. Calculate the pH at which a solution of cysteine would have no net charge. ANSWERS TO PROBLEMS

1. Only the N-terminal a-amino group and the C-terminal a-carboxyl group will undergo

ionization. The internal groups will be joined by peptide bonds and are not ionizable.

2. For a peptide of n residues, n-1 water molecules must be removed. A significant acti

vation energy barrier makes peptide bonds kinetically stable.

3. All metabolic reactions in an organism are catalyzed by enzymes that are generally spe

cific for either the D or the L isomeric form of a substrate. If an animal (an herbivore in this case) is to be able to digest the protein from a plant and build its own protein from the resulting amino acids, both the animal and the plant must make their proteins from amino acids having the same configuration.

4. Because the sequence of DNA specifies, through a complementary sequence of RNA, the

amino acid sequence of a protein, knowledge about any one of the three types of sequences yields information about the other two. One would also expect the coding sequence for a particular protein to be the same among members of the same species, allowing for an occasional rare mutation. For that reason, the published primary amino acid sequences are likely to be the same.

5. From the Ramachandran plot in Figure 3.35 of the text, we see that b conformation is

accommodated by f values of approximately :140Ú and y values of approximately

;147Ú. The structure is most likely a b sheet. In fact, the “low” numbers here imply that it is an antiparallel beta sheet. The parallel b sheet would have higher numbers, more like f=:160Ú and y=;160Ú.

PROTEIN STRUCTURE AND FUNCTION

6. Figure 3.42 in the text shows that in a reverse turn the CO group of residue 1 is hydro

gen-bonded to the NH group of residue 4. However, there are no adjacent amino acid residues available to form intrachain hydrogen bonds with the CO and NH groups of residues 2 and 3. These groups cannot form hydrogen bonds in the hydrophobic environment found in the interior portion of a folded protein. They are more likely to hydrogen bond with water on the surface of the protein.

7. When the a helices in wool are stretched, intrahelix hydrogen bonds are broken as are

some of the interhelix disulfide bridges; maximum stretching yields an extended b sheet structure. The Cys cross-links provide some resistance to stretch and help pull the a helices back to their original positions. In silk, the b sheets are already maximally stretched to form hydrogen bonds. Each b pleated sheet resists stretching, but since the contacts between the sheets primarily involve van der Waals forces, the sheets are somewhat flexible.

8. Both alanine and glycine are neutral nonpolar residues with small side chains, whereas

the side chain of glutamate is acidic and bulkier than that of alanine. Either feature of the glutamate R group could lead to the loss of activity by altering the protein conformation or by interfering with the binding of the substrate.

9. Since each residue in the a helix is 1.5 Å from its neighbor, the length of the chain that

spans the membrane bilayer is 19¥1.5 Å=28.5 Å, which is also the width of the membrane. One would expect to find nonpolar amino acid residues in the polypeptide portion associated with the membrane bilayer. These would include Ala, Val, Leu, Ile, Met, and Phe (FILMV+A). The actual sequence of the buried chain is

I–T–L–I–I–F–G–V–M–A–G–V–I–G–T–I–L–L–I.

10. The fact that ribonuclease folded in vitro to yield full activity indicated that the biosyn

thetic machinery is not required to direct the folding process for this protein.

11. (a) The experiment by Anfinsen on ribonuclease, described in Section 3.6 of the text,

is the classic observation. When native ribonuclease is treated with mercaptoethanol to disrupt disulfide bonds and with urea as a denaturant, it unfolds, as indicated by the fact that it becomes enzymically inactive. When urea is removed by dialysis and disulfide bonds reform by oxidation, it regains enzymic activity, suggesting that its native structure has been restored.

(b) The discovery that proteins are synthesized directionally on ribosomes beginning

at the amino terminus complicates matters somewhat because folding of the amino end of the polypeptide chain could begin before the carboxyl end had been synthesized. Such folding could represent the most stable conformation over a short range, but there would be no guarantee that it would be part of the energy minimum for the entire molecule.

it from undergoing final folding until the entire molecule was synthesized.

12. The higher the proportion of aromatic side chains (such as those of phenylalanine) in

the protein, the more likely that steric hindrance among closely located residues could interfere with the establishment of the regular repeating structure of the a helix. Smaller aliphatic side chains like those of leucine, isoleucine, and valine would be less likely to interfere. Structural studies on many proteins reveal that the number of aromatic residues in a-helical segments is relatively low, while the content of aliphatic side chains in such segments is unremarkable, compared to that of other nonhelical regions of a folded protein. Both aliphatic and aromatic side chains (especially that of phenylalanine) are hydrophobic, so that many of them are buried inside a globular protein, away from water molecules.

CHAPTER 3

13. Protein scientists have devised a competition called CASP, or Critical Assessment of

Techniques for Protein Structure Prediction, which is held every other year. Laboratories that are working on determination of three-dimensional structure by x-ray crystallography (or nmr) announce that they expect to release the structure in a few months. They give a description of the sequence of the protein and its use in the cell, and withhold the actual structural coordinates until a certain date. In the meantime, laboratories with predictive algorithms publicly post the structure they think the protein will have. The success or failure of the prediction takes place in a public arena, and the better predictors have bragging rights. CASP-4 in 2000 showed that there are several effective programs available, notably ROSETTA, used by David Baker of the University of Washington. Results of the competition are published in the journal Protein and online (in technical language) at the website http://predictioncenter.llnl.gov/.

14. The primary advantage of precursor chain synthesis is that the production of related pro

teins can be coordinated. This could be important in viral infection, and it may also be important for coordinated synthesis of hormones with related activities. It is worth noting that there are other reasons for the synthesis of polyprotein precursors. For example, the genome of the poliovirus consists of a single RNA molecule that acts as a messenger on entering the cytoplasm of the host. In eukaryotic cells a messenger RNA molecule can be translated into only one polypeptide chain. Therefore the poliovirus can reproduce only by synthesizing its proteins by sequential cleavages.

15. The molarity of water equals the number of moles of water per liter. A liter of water

weighs 1000 grams, and its molecular weight is 18, so the molarity of water is

1000

M =

= 55 6

.

18

At 25ÚC, Kw is 1.0¥10:14; at neutrality, the concentration of both hydrogen and hydroxyl ions is each equal to 10:7 M. Thus, the actual concentration of H2O is (55.6-0.0000001) M; the difference is so small that it can be disregarded.

16. Because pH values are based on a logarithmic scale, every unit change in pH means a

tenfold change in hydrogen ion concentration. When pH=2.0, [H;]=10:2 M; when pH=3.0, [H;]=10:3 M.

17. Assume that HCl in solution is completely ionized to H; and Cl:. Then find the con

centration of H;, which equals the concentration of Cl:.

pH = log H

[ +]= .21

H

[ +]= .

10 2 1

=

.

100 9

10 3

= .

7 94 10 3 M

Thus, H+

Cl

HCl

.

M

[ ]=[ ]=[ ]=794 10 3

18. Use the Henderson-Hasselbalch equation to calculate the concentration of histidine,

whose imidazole ring is ionized at neutral pH. The value of pK for the ring is 6.0 for a histidine residue in a protein (see Table 3.1).

PROTEIN STRUCTURE AND FUNCTION

His

[ ]

pH = pK + log His

[ +]

His

[ ]

.

7 0 = .

6 0 + log His

[ +]

His

[ ]

log

.

1 0

His

[ +] =

His

[ ] 10

His

[ +] =

At pH 7.0, the ratio of uncharged histidine to charged histidine is 10:1, making the probability that the side chain is charged only 9%.

19. To see which form of cysteine has no net charge, examine all the possible forms, begin

ning with the one that is most protonated:

COOH

COO

+OH

J

+OH

+OH

J

+H NJ CJH

+H NJ CJH

H NJ CJH

3

G

J

+H NJ CJH

3

G

3

G

2

J

J

CH

CH

2

2

2

J

J 2

J

SH

Net

charge

+1

0

-1

-2

The pH of the cysteine solution at which the amino acid has no net charge will be that point at which there are equal amounts of the compound with a single positive charge and a single negative charge. This is, in effect, the average of the two corresponding pK values (see the Appendix to Chapter 3), one for the a-carboxyl group and the other for the side chain sulfhydryl group. Thus, (1.8+8.3)/2=5.05. This value is also known as the isoelectric point. EXPANDED SOLUTIONS TO TEXT PROBLEMS

1. (a) Since tropomyosin is double-stranded, each strand will have a mass of 35 kd. If the

average residue has a mass of 110 d, there are 318 residues per strand (35,000/110), and the length is 477Å (1.5 Å/residue¥318).

(b) Since 2 of the 40 residues formed the hairpin turn, 38 residues formed the an

tiparallel b pleated sheet which is 19 residues long (38/2). In b pleated sheets, the axial distance between adjacent amino acids is 3.5 Å. Hence, the length of this segment is 66.5 Å (3.5¥19).

2. Branching at the b carbon of the side chain (isoleucine), in contrast to branching at the

g carbon (leucine), sterically hinders the formation of a helix. This fact can be shown with molecular models.

CHAPTER 3

3. Changing alanine to valine results in a bulkier side chain, which prevents the correct in

terior packing of the protein. Changing a nearby, bulky, isoleucine side chain to glycine apparently alleviates the space problem and allows the correct conformation to take place.

4. The amino acid sequence of insulin does not determine its three-dimensional structure.

By catalyzing a disulfide-sulfhydryl exchange, this enzyme speeds up the activation of scrambled ribonuclease because the native form is the most thermodynamically stable. In contrast, the structure of active insulin is not the most thermodynamically stable form. The three-dimensional structure of insulin is determined by the folding of preproinsulin, which is later processed to mature insulin.

5. Appropriate hydrogen-bonding sites on the protease might induce formation of an in

termolecular b pleated sheet with a portion of the target protein. This process would effectively fully extend a helices and other folded portions of the target molecule.

6. Being the smallest amino acid, glycine can fit into spaces too small to accommodate other

amino acids. Thus, if sharp turns or limited spaces for amino acids occur in a functionally active conformation of a protein, glycine is required; no substitute will suffice. In view of this, it is not surprising that glycine is highly conserved.

7. To answer this question one needs to know some of the characteristics of the guani

dinium group of the side chain of arginine and of the other functional groups in proteins. Most of the needed information is presented in Figure 3-42 in your textbook; note that the guanidinium group has a positive charge at pH 7 and contains several hydrogen bond donor groups. The positive charge can form salt bridges with the negatively charged groups of proteins (glutamate and asparatate side chains and the terminal carboxylate). As a hydrogen bond donor, the guanidinium group can react with the various hydrogen bond acceptors shown in Figure 3-42 (glutamine, asparagine, aspartate, and the main chain carbonyl). It can also hydrogen bond with the hydroxyl group of serine and threonine (not shown in Figure 3-42). Hydroxyl groups accept hydrogen bonds much like water does.

8. The keratin of hair is essentially a bundle of long protein strands joined together by disul

fide bonds. If these bonds are broken (reduced) by the addition of a thiol and the hair curled, the keratin chains slip past each other into a new configuration. When an oxidizing agent is added, new disulfide bonds are formed, thus stabilizing the new “curled” state.

9. There is a considerable energy cost for burying charged groups of non-hydrogen-bonded

polar groups inside a hydrophobic membrane. Therefore, an a-helix with hydrophobic side chains is particularly suited to span a membrane. The backbone hydrogen-bonding requirements are all satisfied by intramolecular interactions within the a-helix. Good candidate amino acids with hydrophobic side chains would include Ala, Ile, Leu, Met, Phe, and Val. (Pro is also hydrophobic but will cause a bend in the helix.) Additionally, the aromatic (and amphipathic) amino acids Trp and Tyr are often found toward the ends of membrane-spanning helices, near the phospholipid head groups in the membrane/water interface region.

10. The protein is not at equilibrium, but is in a state where the peptide bond is “kinetically

stable” against hydrolysis. This situation is due to the large activation energy for hydrolyzing a peptide bond.

PROTEIN STRUCTURE AND FUNCTION

11. One can effectively apply the Henderson-Hasselbalch equation successively to the amino

group and to the carboxyl group and multiply the results to arrive at a ratio of 10:5.

So, considering first the amino group, pH=pK+log [NH

;

2]/[NH3 ]. With pH=7

and pK=8, one has 7=8+log [NH

;

2]/[NH3 ] or [NH2]/[NH3 ]=10:1. Considering

now the carboxyl group with pK of 3, one has 7=3+log [COO:]/[COOH], or [COO:]/[COOH]=10;4 or [COOH]/[COO:]=10:4. Then to consider the two simultaneous ionizations that relate the zwitterionic form to the neutral form of an amino acid such as alanine, one needs to multiply the ratio of [NH

;

2]/[NH3 ] by the ratio of

[COOH]/[COO:], i.e., (10:1)¥(10:4)=(10:5).

12. The presence of the larger sulfur atom (next to the beta carbon of Cys) alters the rela

tive priorities of the groups attached to the a carbon. The stereochemical arrangement of the b carbon with respect to the a hydrogen does not change, but the convention for assigning the R configuration changes when the Cb-sulfur is present. (With methionine, the sulfur is too far removed for Cb to influence the group priority.)

13. SAVE ME I'M TRAPPED IN A GENE.

14. No. Unlike the Pro nitrogen in X-Pro, the nitrogen of X in the peptide bond of Pro-X is not bonded between two tetrahedral carbon atoms. Therefore, the steric preference for the trans conformation will be similar to that of other (non-proline) peptide bonds.

15. Model A shows the reference structure for extended polypeptide chain with f=180Ú and

y=180Ú, so the answer is c. Models C and E have one torsion angle identical to model A and the other angle changed to 0Ú. In model C f is changed to 0Ú (answer d), and in model E y is changed to 0Ú (answer b). Comparing model B with a reference for which

f=0Ú (model C), we see a 60Ú counterclockwise rotation of f, when viewed from Ca, so answer e is correct for model B. Finally, comparing model D with the f=0Ú reference in model C, we see a 120Ú clockwise rotation of f, when viewed from Ca (answer a).

16. One should use Beer’s Law and remember that each mole of protein contains 3 moles of

tryptophan. Then for the protein, A=3ecl, where e is the molar extinction coefficient for tryptophan at 280 nm. With A=0.1, e=3400 M:1 cm:1 and l=1.0 cm, one has c=A / (3el). Therefore c=(0.1)/((3)(3400 M:1))=9.8¥10:6 M. For the concentration in grams per liter, one multiplies 9.8¥10:6 moles/liter by 100,000 grams per mole to arrive at 0.98 g/liter, or 0.98 mg/mL. CHAPTER 4 Exploring Proteins

Chapter 4 extends Chapter 3 by introducing the most important methods used

to investigate proteins. Many of these were essential in discovering the principles of protein structure and function presented in the preceding chapter.

These methods also constitute the essentials of the armamentarium of modern biochemical research and underlie current developments in biotechnology. First, the authors define the concept of the proteome, the sum of functioning proteins in the cell and their interactions. Then they outline methodological principles for the analysis and purification of proteins. Next they describe methods of sequencing the amino acids in proteins, and explain why the knowledge revealed by these techniques is so important. They continue with a discussion of antibodies as highly specific analytical reagents, followed by a discussion of the uses of peptides of defined sequence and how they are chemically synthesized. To close the chapter, there is a discussion of the use of x-ray crystallography and nuclear magnetic resonance spectroscopy to determine the three-dimensional structures of proteins. 33

CHAPTER 4 LEARNING OBJECTIVES

When you have mastered this chapter, you should be able to complete the following objectives. The Proteome Is the Functional Representation of the Genome (Text Section 4.0.1)

1. Distinguish between the genome and the proteome, and define both terms. The Purification of Proteins Is an Essential First Step in Understanding Their Function (Text Section 4.1)

2. Describe how a quantitative enzyme assay can be used to calculate the specific activity

during protein purification.

3. Define differential centrifugation, and describe how it would be used to produce a protein

mixture from a cell homogenate.

4. List the properties of proteins that can be used to accomplish their separation and purifi- cation, and correlate them with the appropriate methods: gel-filtration chromatography, dialysis, salting out, ion-exchange chromatography, and affinity chromatography. Describe the basic principles of each of these methods.

5. Describe the principle of electrophoresis and its application in the separation of proteins.

6. Explain the determination of protein mass by SDS-PAGE: sodium dodecyl sulfate-polyacrylamide gel electrophoresis.

7. Define the isoelectric point (pI) of a protein and describe isoelectric focusing as a separation

method.

8. Explain the quantitative evaluation of a protein purification scheme.

9. Define the sedimentation coefficient S, and give its common name. Note the range of S values

for biomolecules and cells.

10. Describe zonal centrifugation and sedimentation equilibrium and explain their applications

to the study of proteins.

11. Outline the application of mass spectrometry to the analysis of proteins and compare the

merits of the various methods of determining the molecular weights of proteins. Amino Acid Sequences Can Be Determined by Automated Edman Degradation (Text Section 4.2)

12. Outline the steps in the determination of the amino acid composition and the amino-ter- minal residue of a peptide.

13. Describe the sequential Edman degradation method and the automated determination of

the amino acid sequences of peptides.

14. List the most common reagents used for the specific cleavage of proteins. Explain the ap

plication of overlap peptides to protein sequencing.

15. Describe the additional steps that must be used for sequencing disulfide-linked polypep- tides and oligomeric proteins.

EXPLORING PROTEINS

16. Give examples of the important information that amino acid sequences reveal.

17. Explain, in general terms, how recombinant DNA technology is used to determine the

amino acid sequences of nascent proteins. Note the differences between a nascent protein and one that has undergone posttranslational modifications. Immunology Provides Important Techniques with Which to Investigate Proteins (Text Section 4.3)

18. Define the terms antibody, antigen, antigenic determinant (epitope), and immunoglobulin G.

19. Contrast polyclonal antibodies and monoclonal antibodies and describe their preparation.

20. Outline methods that use specific antibodies in the analysis or localization of proteins.

21. Describe how the use of fluorescent markers allows direct observation of changes within

living cells. Peptides Can Be Synthesized by Automated Solid-Phase Methods (Text Section 4.4)

22. List the most important uses of synthetic peptides.

23. Outline the steps of the solid-phase method for the synthesis of peptides. Three-Dimensional Protein Structure Can Be Determined by NMR Spectroscopy and X-Ray Crystallography (Text Section 4.5)

24. Describe the fundamentals of the method and basic physical principles underlying nu- clear magnetic resonance spectrometry as applied to protein structure determination.

25. Provide a similar description of the x-ray crystallographic analysis of a protein, and give

the basic physical principles underlying this technique.

26. Compare the relative advantages and disadvantages of x-ray crystallography and NMR

spectroscopy for protein structure determination. SELF-TEST The Proteome is the Functional Representation of the Genome

1. The genome sequence tells us all of the proteins an organism can make. Are all of these

proteins expressed? The Purification of Proteins Is an Essential First Step in Understanding Their Function

2. The following five proteins, which are listed with their molecular weights and isoelec

tric points, were separated by SDS–polyacrylamide gel electrophoresis. Give the order of their migration from the top (the point of sample application) to the bottom of the gel.

CHAPTER 4 Molecular weight (daltons) pI

(a) a-antitrypsin

45,000

5.4

(b) cytochrome c

13,400

10.6

17,000

7.0

(d) serum albumin

69,000

4.8

(e) transferrin

90,000

5.9

Top —————————————— Bottom

3. If the five proteins in question 2 were separated in an isoelectric-focusing experiment,

what would be their distribution between the positive (+) and negative (−) ends of the gel? Indicate the high and low pH ends.

Cathode (−)——————-(+) Anode

4. Which of the following statements are NOT true?

(a) The pI is the pH value at which a protein has no charges. (b) At a pH value equal to its pI, a protein will not move in the electric field of an elec

trophoresis experiment.

(c) An acidic protein will have a pI greater than 7. (d) A basic protein will have a pI greater than 7.

5. SDS–polyacrylamide gel electrophoresis and the isoelectric-focusing method for the sep

aration of proteins have which of the following characteristics in common? Both

(a) separate native proteins. (b) make use of an electrical field. (c) separate proteins according to their mass. (d) require a pH gradient. (e) are carried out on supporting gel matrices.

6. Before high-performance liquid chromatography (HPLC) methods were devised for the

separation and analysis of small peptides, electrophoresis on a paper support was frequently used. Separation was effected on the basis of the charge on a peptide at different pH values. Predict the direction of migration for the following peptides at the given pH values. Use C for migration toward the cathode, the negative pole; A for migration toward the anode, the positive pole; and O if the peptide remains stationary. pH 2.0 4.0 6.0 11.0

(a) Lys-Gly-Ala-Gly

__________________________

(b) Lys-Gly-Ala-Glu

__________________________

(d) Glu-Gly-Ala-Glu

__________________________

(e) Gln-Gly-Ala-Lys

__________________________

7. How would the time required for the separation described in question 6 be changed if

all the solutions that were used during the electrophoresis contained 100 g/L of table sugar (sucrose)?

8. Examine Table 4.1 in the text, evaluating a protein purification scheme. Does “total ac

tivity” go up or down as the protein is purified? Would it have been a good idea to try affinity chromatography at an earlier stage of purification?

EXPLORING PROTEINS

9. The molecular weight of a protein can be determined by SDS–polyacrylamide gel elec

trophoresis or by sedimentation equilibrium. Which method would you use to determine the molecular weight of a protein containing four subunits, each consisting of two polypeptide chains cross-linked by two disulfide bridges? Explain your answer.

10. After isolating and purifying to homogeneity a small enzyme (110 amino acids long)

from a culture of bacteria, you are confused as to whether you grew wild-type bacteria or a mutant strain that produced the enzyme with a valine residue at position 66 instead of the glycine found in the wild-type strain. How could you quickly determine which protein you had? Amino Acid Sequences Can Be Determined by Automated Edman Degradation

11. Which of the following statements concerning the Edman degradation method are true?

(a) Phenyl isothiocyanate is coupled to the amino-terminal residue. (b) Under mildly acidic conditions, the modified peptide is cleaved into a cyclic deriva

tive of the terminal amino acids and a shortened peptide (minus the first amino acid).

sequential degradation can begin.

(d) If a protein has a blocked amino-terminal residue (as does N-formyl methionine,

for example), it cannot react with phenyl isothiocyanate.

12. Which of the following are useful in identifying the amino-terminal residue of a protein?

(a) cyanogen bromide

(d) dabsyl chloride

(b) fluorodinitrobenzene

(e) phenyl isothiocyanate

13. When sequencing proteins, one tries to generate overlapping peptides by using cleav

ages at specific sites. Which of the following statements about the cleavages caused by particular chemicals or enzymes are true?

(a) Cyanogen bromide cleaves at the carboxyl side of threonine. (b) Trypsin cleaves at the carboxyl side of Lys and Arg. (c) Chymotrypsin cleaves at the carboxyl side of aromatic and bulky amino acids. (d) 2-Nitro-5-thiocyanobenzoate cleaves on the amino side of cysteine residues. (e) Chymotrypsin cleaves at the carboxyl side of aspartate and glutamate.

14. What treatments could you apply to the following hemoglobin fragment to determine

the amino-terminal residue and to obtain two sets of peptides with overlaps so that the complete amino acid sequence can be established? Give the sequences of the peptides obtained.

Val-Leu-Ser-Pro-Ala-Lys-Thr-Asn-Val-Lys-Ala-Ala-Trp-Gly-Lys-Val-Gly-Ala-His-Ala-GlyGlu-Tyr-Gly-Ala-Glu-Ala-Thr-Glu

15. Which of the following techniques are used to locate disulfide bonds in a protein?

(a) The protein is first reduced and carboxymethylated. (b) The protein is cleaved by acid hydrolysis. (c) The protein is specifically cleaved under conditions that keep the disulfide bonds

intact.

(d) The peptides are separated by SDS–polyacrylamide gel electrophoresis. (e) The peptides are separated by two-dimensional electrophoresis with an intervening

performic acid treatment.

CHAPTER 4

16. Which of the following are important reasons for determining the amino acid sequences

of proteins?

(a) Knowledge of amino acid sequences helps elucidate the molecular basis of biolog

ical activity.

(b) Alteration of an amino acid sequence may cause abnormal functioning and disease. (c) Amino acid sequences provide insights into evolutionary pathways and protein

structures.

(d) The three-dimensional structure of a protein can be predicted from its amino acid

sequence.

(e) Amino acid sequences provide information about the destination and processing of

some proteins.

(f)

Amino acid sequences allow prediction of the DNA sequences encoding them and thereby facilitate the preparation of DNA probes specific for the regions of their genes.

17. In spite of the convenience of using recombinant DNA techniques for determining the

amino acid sequences of proteins, chemical analyses of amino acid sequences are frequently required. Explain why. Immunology Provides Important Techniques with Which to Investigate Proteins

18. Match the terms in the left column with the appropriate item or items from the right

column.

(a) antigens

(1) immunoglobulins

(b) antigenic determinants

(2) foreign proteins, polysaccharides,

or nucleic acids

(d) monoclonal antibodies

(3) antibodies produced by hybridoma

cells

(4) groups recognized by antibodies (5) heterogeneous antibodies (6) homogeneous antibodies (7) antibodies produced by injecting an

animal with a foreign substance

(8) epitopes

19. The methods used to localize a specific protein in an intact cell are

(a) Western blotting.

(d) immunoelectron microscopy.

(b) solid-phase immunoassay.

(e) fluorescence microscopy.

20. Explain why immunoassays are especially useful for detecting and quantifying small

amounts of a substance in a complex mixture. Peptides Can Be Synthesized by Automated Solid-Phase Methods

21. The amino acid sequence of a protein is known and strong antigenic determinants have

been predicted from the sequence; however, you do not have enough of the pure protein to prepare antibodies. How could you circumvent this problem, using knowledge of peptide synthesis?

EXPLORING PROTEINS

22. Which of the following is commonly used as a protecting group during peptide synthesis?

(a) tert-butyloxycarbonyl (b) dicyclohexylcarbodiimide (c) dicyclohexylurea (d) hydrogen fluoride (e) phenyl isothiocyanate

23. The following reagents are often used in protein chemistry:

(1) CNBr (2) urea (3) b-mercaptoethanol (4) trypsin (5) dicyclohexylcarbodiimide (6) dabsyl chloride (7) 6 N HCl (8) fluorescamine (9) phenyl isothiocyanate (10) chymotrypsin (11) dilute F3CCOOH Which of these reagents are best suited for the following tasks?

(a) determination of the amino acid sequence of a small peptide (b) identification of the amino-terminal residue of a peptide (of which you have less

than 10−7 grams)

(c) reversible denaturation of a protein devoid of disulfide bonds (d) hydrolysis of peptide bonds on the carboxyl side of aromatic residues (e) cleavage of peptide bonds on the carboxyl side of methionine (f)

hydrolysis of peptide bonds on the carboxyl side of lysine and arginine residues

(g) reversible denaturation of a protein that contains disulfide bonds (two reagents are

needed)

(h) activation of carboxyl groups during peptide synthesis (i)

determination of the amino acid composition of a small peptide

(j)

removal of t-Boc protecting group during peptide synthesis Three-Dimensional Protein Structure Can Be Determined by NMR Spectroscopy and X-Ray Crystallography

24. Must we be able to crystallize a protein in order to learn the three-dimensional structure?

25. Which of the following statements concerning x-ray crystallography is NOT true?

(a) Only crystallized proteins can be analyzed. (b) The x-ray beam is scattered by the protein sample. (c) All atoms scatter x-rays equally. (d) The basic experimental data are relative intensities and positions of scattered electrons. (e) The electron-density maps are obtained by applying the Fourier transform to the

scattered electron intensities.

(f)

The resolution limit for proteins is about 2 Å.

26. How can x-ray crystallography provide information about the interaction of an enzyme

with its substrate?

CHAPTER 4 ANSWERS TO SELF-TEST

1. In any given cell, at any given time, it is quite likely that many genes capable of pro

ducing protein are not being expressed. Single-celled organisms tend to respond to their environment, producing enzymes to deal with the nutrients and conditions in the area. Multicellular organisms need different proteins for different parts of the body. Humans have very different needs in the retina, the liver, and muscle cells. This is why the proteome is a useful concept, it is a description of the proteins actually present in a functioning cell.

e d a c b

2. Top ———— Bottom

b c e a d

3. High pH (−) ———— Low pH (+) 4. a, c. Regardless of the pH, a protein is never devoid of charges; at the pI, the sum of all

the charges is zero.

5. b, e

6.

pH

2.0

4.0

6.0 11.0

(a) C

C

(b) C

C

O

C

A

(d) C

O

A

(e) C

C

For example, peptide b carries a net charge of +1.5 at pH 2.0 (Lys side chain, +1; a-amino group, +1; Glu side chain, 0; and terminal carboxyl, −0.5, since the pH coincides with its pK value). At pH 4.0, the net charge is +0.5; the Glu side chain is half ionized (−0.5), but the terminal carboxyl is almost completely ionized (−1). At pH 6.0, the net charge is 0 due to a +2 charge contributed by the Lys residue and a −2 charge contributed by the Glu residue. At pH 11.0, the a-amino group is deprotonated (charge of 0) and the Lys side chain is half-protonated (charge of +0.5); thus, the net charge is −1.5. The same answer for peptide b can be given graphically (see Figure 4.1): FIGURE 4.1

Net charge

(COO:)

pH 2.0: + H N J LysJGlyJAlaJGluJCOOH ;1.5

3

J

NH ;

COOH

3

pH 4.0: + H N J LysJGlyJAlaJGluJCOO: ;0.5

3

J

NH ;

COOH

3

(COO:)

pH 6.0: + H N J LysJGlyJAlaJGluJCOO: 0

3

J

;

COO:

NH3

pH 11.0: H N J LysJGlyJAlaJGluJCOO: -1.5

2

J

COO:

NH2 (NH ;)

3

EXPLORING PROTEINS

7. More time would be required for the separation, since the velocity of movement of the

compounds would be slowed because of the increased viscosity of the solution. The velocity at which a molecule moves during electrophoresis is inversely dependent on the frictional coefficient, which is, itself, directly proportional to the viscosity of the solution.

8. Total activity drops as material is lost in each step of purification. In a good purification,

total protein drops much faster, so that specific activity goes up dramatically. Not all proteins can be purified with affinity chromatography. When a protein has a unique substrate and works this well with affinity chromatography, it may be a good idea to leave out the ion exchange and molecular exclusion steps, and go straight to the “home run” technique. A standard source on protein purification states that “One-step purifications of 1,000-fold with nearly 100% recovery have been reported” with this technique. (R. K. Scopes. [1994]. Protein purification, principles and practice. [3rd ed.]. New York: Springer-Verlag.)

9. Determinations of mass by SDS–polyacrylamide gel electrophoresis are carried out on

proteins that have been denatured by the detergent in a reducing medium; the reducing agent in the medium disrupts disulfide bonds. Therefore, to determine the molecular weight of a native protein containing subunits with disulfide bridges, you must use the sedimentation equilibrium method. Another nondenaturing method, gel-filtration chromatography, can be used to obtain approximate native molecular weights.

10. Mass spectrometry—Electrospray or MALDI-TOF—could easily distinguish a protein of

this approximate mass (101 amino acids × approximately 110 d per amino acid = 12 kd) that contained an extra 41 atomic mass units as a result of the substitution of a valine for a glycine residue.

11. a, b, c, d

12. b, d, e

13. b, c, d

14. The amino-terminal residue of the hemoglobin fragment can be determined by labeling

it with fluorodinitrobenzene or dabsyl chloride or by analyzing the intact fragment by the Edman degradation method; this shows that the amino-terminal residue is Val. Trypsin digestion, separation of peptides, and Edman degradation give

Val-Leu-Ser-Pro-Ala-Lys

Thr-Asn-Val-Lys

Ala-Ala-Trp-Gly-Lys

Val-Gly-Ala-His-Ala-Gly-Glu-Tyr-Gly-Ala-Glu-Ala-Thr-Glu

Chymotrypsin digestion, separation of peptides, and Edman degradation give

Val-Leu-Ser-Pro-Ala-Lys-Thr-Asn-Val-Lys-Ala-Ala-Trp

Gly-Lys-Val-Gly-Ala-His-Ala-Gly-Glu-Tyr

Gly-Ala-Glu-Ala-Thr-Glu

15. c, e. The performic acid oxidizes the disulfide bonds to SO3− groups and releases new

peptides.

16. a, b, c, e, f. Answer (d) may be correct in some cases where homologous proteins are

compared in terms of amino acid sequences and known three-dimensional structures.

CHAPTER 4

17. The amino acid sequence derived from the DNA sequence is that of the nascent polypep

tide chain before any posttranslational modifications. Since the function of a protein depends on its mature structure, it is often necessary to analyze the protein itself to determine if any changes have occurred after translation.

18. (a) 2 (b) 4, 8 (c) 1, 5, 7 (d) 1, 3, 6

19. d, e. Immunoelectron microscopy provides more precise localization than does fluores

cence microscopy.

20. Because the interaction of an antibody with its antigen is highly specific, recognition and

binding can occur in the presence of many other substances. If the antibody is coupled to a radioactive or fluorescent group or an enzyme whose activity can be detected in situ, then a sensitive method is available for the detection and quantitation of the antigen-antibody complex.

21. You could synthesize peptides containing the putative antigenic determinants, couple

these peptides to an antigenic macromolecule, and prepare antibodies against the synthetic peptides. If the same antigenic determinants are present in the protein of interest and are not occluded by the structure of the protein, then the antibodies prepared against the synthetic peptides should also react with the protein.

22. a

23. (a) 9 (b) 6, 7 (c) 2 (d) 10 (e) 1 (f) 4 (g) 2, 3 (h) 5 (i) 7, 8 (j) 11

24. No. Crystals are necessary for x-ray crystallography, but not for NMR spectroscopy. If a

highly concentrated solution (1 mM) of a pure protein can be obtained, then significant information can be derived about the three-dimensional shape.

25. c, d, e. In x-ray crystallography, x-rays, not electrons, are scattered and detected.

26. If an enzyme can be crystallized with and without its substrate and the three-dimen

sional structures of both are obtained using x-ray crystallography, the difference between the two structures should reveal how the substrate fits in its binding site and which atoms and what kind of bonds are involved in the interaction. PROBLEMS

1. How can a protein be assayed if it is not an enzyme?

2. Many of the methods described in Chapter 4 are used to purify enzymes in their native

state. Why would the use of SDS–polyacrylamide gel electrophoresis be unlikely to lead to the successful purification of an active enzyme? What experiments would you conduct to determine whether salting out with ammonium sulfate would be useful in enzyme purification?

3. Of the techniques for analyzing proteins discussed in Chapter 4 of the text, which one

would be the easiest to use for accurately determining the molecular weight of a small monomeric protein? Comment on the standards you would wish to use in this technique. What types of proteins might not be analyzed accurately by your suggested method?

4. Mass spectrometry is often used for the sequence analysis of peptides from 2 to 20 amino

acids in length. The procedure requires only microgram quantities of protein and is very sensitive; cationic fragments are identified by their charge-to-mass ratio. In one procedure, peptides are treated with triethylamine and then with acetic anhydride. What will such a procedure do to amino groups? Next, the modified peptide is incubated with a

strong base and then with methyl iodide. What groups will be methylated? Which two amino acids cannot be distinguished by mass spectrometry?

5. A glutamine residue that is the amino-terminal residue of a peptide often undergoes

spontaneous cyclization to form a heterocyclic ring; the cyclization is accompanied by the release of ammonium ion. Diagram the structure of the ring, showing it linked to an adjacent amino acid residue. How would the formation of the ring affect attempts to use the Edman procedure for sequence analysis? A similar heterocyclic ring is formed during the biosynthesis of proline; in this case, glutamate is the precursor. Can you propose a pathway for the synthesis of proline from glutamate?

6. A peptide composed of 12 amino acids does not react with dabsyl chloride or with

phenyl isothiocyanate. Cleavage with cyanogen bromide yields a peptide with a carboxylterminal homoserine lactone residue, which is readily hydrolyzed, in turn yielding a peptide whose sequence is determined by the Edman procedure to be

E-H-F-W-D-D-G-G-A-V-L

On the other hand, cleavage with staphylococcal protease (see Table 4.3 in text) yields an equivalent of aspartate and two peptides. Use of the Edman procedure gives the following sequences for these peptides:

G-G-A-V-L-M-E and H-F-W-D

Why does the untreated peptide fail to react with dabsyl chloride or phenyl isothiocyanate?

7. A hexapeptide that is part of a mouse polypeptide hormone is analyzed by a number of

chemical and enzymatic methods. When the hexapeptide is hydrolyzed and analyzed by ion-exchange chromatography, the following amino acids are detected:

Tyr Cys Glu

Ile Lys Met

Two cycles of Edman degradation of the intact hexapeptide released the following PTH–amino acids (see Figure 4.2): FIGURE 4.2

J

OK

K

N

S

O

N

S

J

J

J

J

K

C

C

C

J

J

J

J

HCJ

J NH

HCJ

J NH

J

J

CH

HJCJ

CH

2

3

J

J

CH

CH

2

J

S

CH

3

J

CH

3

Cleavage of the intact protein with cyanogen bromide yields methionine and a pentapeptide. Treating the intact hexapeptide with trypsin yields a dipeptide, which contains tyrosine and glutamate, and a tetrapeptide. When the intact hexapeptide is treated with carboxypeptidase A, a tyrosine residue and a pentapeptide are produced. Bearing in mind that the hexapeptide is isolated from a mouse, write its amino acid sequence, using both three-letter and one-letter abbreviations.

CHAPTER 4

8. The production of a small acidic protein, hCG or human chorionic gonadotropin dur

ing pregnancy is the basis of most pregnancy test kits. What method makes the most sense for detecting a known protein like this?

9. A laboratory group wishes to prepare a monoclonal antibody that can be used to react

with a specific viral coat protein in a Western blotting procedure. Why would it be a good idea to treat the viral coat protein with SDS before attempting to elicit monoclonal antibodies?

10. A map of the electron density is necessary for the determination of the three-dimensional

structure of a protein, but other information is also needed. Hydrogen atoms have one electron and cannot be visualized by x-ray analyses of proteins. Bearing this in mind, compare the structures of amino acids like valine, threonine, and isoleucine and then describe what additional information would be needed along with an electron-density map.

11. Some crystalline forms of enzymes are catalytically active, and thus are able to carry out

the same chemical reactions as they can in solution. Why are these observations reassuring to those who are concerned about whether crystallographic determinations reveal the normal structure of a protein?

12. In some ways it is easier to obtain protein structures by NMR spectroscopy. Proteins need

not be crystallized, just purified and dissolved. Would it be desirable to use NMR to learn about the structure of the active site of an enzyme, to design an inhibitor? Why or why not? ANSWERS TO PROBLEMS

1. This is a rather serious problem. Some proteins have a slight catalytic activity that can

be utilized in an assay although they are not enzymes. Or perhaps the protein will serve as a substrate for a reaction. If a protein has no enzyme activity, but the molecular weight and/or pI is known, it can be detected by gel electrophoresis by looking for protein concentration at the right spot on the gel. Some proteins actually fluoresce, like the GFP (green fluorescent protein) produced by jellyfish. In these cases, the intensity of the fluorescence could serve as the basis for an assay.

If the gene is known, then there are ways to “fish” out the protein in very high yield

by modifying the sequence. This bypasses the need for an assay. One common procedure is “his-tagging” in which six tandem histidines are added to the sequence of the gene. The protein expressed can then be purified in one step on a nickel-containing column, and eluted with imidazole. (A good review of various “Affinity Fusion Strategies” [Nilsson et al., Prot. Exp. Purif. 11(1997):1].)

2. SDS disrupts nearly all noncovalent interactions in a native protein, so the renaturation

of a purified protein, which is necessary to restore enzyme activity, could be difficult or impossible. You should therefore conduct small-scale pilot tests to determine whether enzyme activity would be lost upon SDS denaturation. Similarly, when salting out with ammonium sulfate is considered, pilot experiments should be conducted. In many instances, concentrations of ammonium sulfate can be chosen such that the active enzyme remains in solution while other proteins are precipitated, thereby affording easy and rapid purification.

3. SDS–polyacrylamide gel electrophoresis, a sensitive and rapid technique which takes

only a few hours and which has a high degree of resolution, is probably the easiest and

most rapid method for providing an estimate of molecular weight. Small samples (as low as 0.02 mg) can be detected on the gel. For standards or markers on the gel, you should use two or more proteins whose molecular weights are higher than that of the protein to be analyzed, as well as two or more whose molecular weights are lower. The relative mobilities of these markers on the gel can then be plotted against the logarithms of their respective molecular weights (see Figure 4.10 in the text), providing a straight line that can be used to establish the molecular weight of the protein to be analyzed. Proteins that have carbohydrate molecules covalently attached, or those that are embedded in membranes, often do not migrate according to the logarithm of their mass. The reasons for these anomalies are not clear; in the case of glycoproteins, or those with carbohydrate residues, the large heterocyclic rings of the carbohydrates may retard the movement of the proteins through the polyacrylamide gel. Membrane proteins often contain a high proportion of hydrophobic amino acid residues and may not be fully soluble in the gel system.

4. Treatment with triethylamine and then with acetic anhydride will yield acetylated amino

groups. The strong base removes protons from amino, carboxyl, and hydroxyl groups. These groups would then be methylated with methyl iodide. Leucine and isoleucine have identical molecular weights, so they cannot be distinguished by mass spectrometry.

5. Glutamine cyclizes to form pyrrolidine carboxylic acid (shown in Figure 4.3): FIGURE 4.3

R

J

CK

O

R

J

;H NJCJ

H

CK

O

3

J

CH

;

NH +

HNJ

J

J

CH

2

4

J

H;

J

J

CH

OKC

CH

2

J

2

J

C

CK

O

H

J

2

NH N-terminal Pyrrolidone carboxylate glutamine residue residue

The Edman procedure begins with the reaction of phenyl isothiocyanate with the terminal a-amino group of the peptide. In the pyrrolidine ring, that group is not available. Therefore, the cyclized residue must be removed enzymatically before the Edman procedure can be used. During the biosynthesis of proline, glutamate undergoes reduction to form glutamate g-semialdehyde; this compound cyclizes, with the loss of water, to form D1-pyrroline-5-carboxylate, which is then reduced to form proline (see Figure 4.4). FIGURE 4.4

COO:

COO:

COO:

COO:

+

J

;H NJCJ

;

H

H NJCJ

H O

HNJ

J

J

CH

2 H

2

H NJ

J CH

3

2

K

J

J

J

J

CH

CH

HC

CH

H C

CH

2

J

2

2

J

2

J

J

C

C

CH

CH

2

H

J

2

2

J

COO:

C

J

K

O

H Glutamate Glutamic

D1-Pyrroline- Proline

g-semialdehyde 5-carboxylate

CHAPTER 4

6. The sequences of the peptides produced by the two cleavage methods are circular per

mutations of each other. Thus, the peptide is circular, so it has no free a-amino group that can react with dabsyl chloride or with phenyl isothiocyanate (see Figure 4.5). FIGURE 4.5

M

L

E

V

H

A

F

G

W

G

D

D

Staphylococcal

Cyanogen

protease

bromide

E-H-F-W-D-D-G-G-A-V-L

G-G-A-V-L-M-E

+

M (as homoserine lactone)

H-F-W-D

+

D (as free aspartate)

7. The sequence of the mouse hexapeptide is Met-Ile-Cys-Lys-Glu-Tyr, or MICKEY.

Cyanogen bromide treatment cleaves methionine from one end, and the PTH-Met derivative places Met at the N-terminal end, with Ile next in the sequence. Trypsin treatment cleaves on the carboxyl side of Lys, so that Lys is on the C-terminal end of the tetrapeptide, next to Cys. Tyrosine is located on the C-terminal end, as shown by the observation that it is released as a single amino acid when the intact hexapeptide is treated with carboxypeptidase A. Glutamate must therefore be located between Lys and Tyr.

8. Most blood proteins do not show up in the urine, but hCG does. And it is produced very

soon after the egg is fertilized, and then in increasing amounts as the pregnancy progresses. Sandwich ELISA (see Figure 4.35 in the text) is the ideal method for complex biological fluids, and it is relatively easy to produce two different monoclonal antibodies to epitopes on opposite sides of the protein. All home pregnancy test kits are based on variations of this method.

For a better understanding of the use of ELISA in home pregnancy tests, view the

Animated Technique: Elisa Method for Detecting HCG at www.whfreeman.com/biochem5.

9. Samples for assay by Western blotting are separated by electrophoresis in SDS before

blotting and antibody staining, so the reacting proteins are denatured. The use of an SDSdenatured antigen to generate the monoclonal antibody response to the viral coat protein could assure that similar specificities are achieved in the test.

10. Even though individual atoms can be delineated at a resolution of 1.5 Å, the structure

of individual side chains that are similar in shape and size cannot be clearly established. The primary structure of the polypeptide chain must be available. The path of the polypeptide backbone can be traced and the positions of the side chains established. Those that are similar in size and shape can be distinguished by using the primary structure as a guide as they are fitted by eye to the electron-density map.

11. As later chapters will demonstrate, the chemical reactions carried out by proteins like

hemoglobin and lysozyme depend on the precise orientation of atoms involved in binding and acting on substrates. The fact that catalysis can occur in enzyme crystals argues that these same orientations are preserved and that the structure of the enzyme must be the same as that found in solution.

EXPLORING PROTEINS

12. The great advantage of x-ray crystallography is the high resolution that can be obtained.

While it is true that NMR is an easier method, the structures obtained are always approximate. Look at Figure 4.48 in the text. This is a family of approximations of a single structure—not a group of related structures. NMR generally gives “fuzzy” results like this. So determining the dimensions of an enzyme’s active site would not work well with NMR; x-ray crystallography would be preferred. NMR is excellent for obtaining approximate structures of small proteins. EXPANDED SOLUTIONS TO TEXT PROBLEMS

1. (a) The Edman method is best because it can be used repeatedly on the same peptide;

hence, phenyl isothiocyanate.

(b) Since you have a very small amount of sample, sensitivity is important. Hence, dab

syl chloride or dansyl chloride is the reagent of choice over FDNB (Sanger’s reagent).

bonds are present, they must first be reduced with b-mercaptoethanol to obtain a random coil by urea treatment. The known cleavage specificities of chymotrypsin (d), CNBr (e), and trypsin (f) provide easy answers.

2. Whereas the hydrolysis of peptides yields amino acids, hydrazinolysis yields hydrazides

O K

(JCJNHNH )

2

of all amino acids except the carboxyl-terminal residue. The latter can be separated from the hydrazides by the use of anion exchange resin. (The hydrazides of aspartic and glutamic acids would also be picked up by the anion exchange resin; thus, a further purification step might be necessary.)

3. Ethyleneimine reacts with cysteine to form S-aminoethylcysteine, which has the follow

ing structure:

NH ;

3

(;H NJCH JCH JSJCH JCJCOO:)

3

2

J

H

Note that the cysteine side chain has increased in length and has added a plus charge. It closely resembles lysine in both size and charge, and therefore, its carboxyl peptide bonds are susceptible to hydrolysis by trypsin.

4. A 1-mg/ml solution of myoglobin (17.8 kd) is 5.62 × 10−5 M (1/17,800). The ab

sorbance is 0.843 (15,000 × 1 × 5.62 × 10−5). Since this is the log of I0/I, the ratio is 6.96 (the antilog of 0.843). Hence, 14.4% (1/6.96) of the incident light is transmitted. Note that when we say the ratio is 6.96, we are really saying 6.96/1. Inverting this ratio gives the 1/6.96.

The above assumes a myoglobin mol. wt. of 17.8 kd. Most myoglobins have mol. wts.

of about 16.8–17.8 kd.

5. Rod-shaped molecules have larger frictional coefficients than do spherical molecules.

Because of this, the rod-shaped tropomyosin has a smaller (slower) sedimentation coefficient than does the spherical hemoglobin, even though it has a higher molecular weight.

CHAPTER 4

Imagine a metal pellet and a nail of equal weight and density sinking in a syrup. The pellet will sink in an almost straight line, whereas the nail will twist and turn and sink more slowly.

6. From equation 2 on page 83 and the equation on page 88 of your text, we can derive

the expression s ∝ m2/3, where s is the sedimentation coefficient and m is the mass, if we assume that the buoyancy (1-nr) and viscosity (h) factors are constant. Then the mass (m) of a sphere is proportional to its volume (v). Since n = 4/3pr3, m is ∝ r3 and r ∝ m1/3. Also, if ƒ = 6phr (equation 2, p. 83), ƒ ∝ r ∝ m1/3.

When the buoyancy factor is constant, s is proportional to m/ƒ (equation on page 88)

∝ m/m1/3 ∝ m2/3. Note that this says that the sedimentation coefficient is proportional to the two-thirds root of the mass.

Therefore, s(80 kd)/s(40 kd) = 802/3/402/3 = 22/3 = 1.59.

7. Electrophoretic mobilities are usually proportional to the log of the molecular weight

(see textbook Figure 4.10, p. 85). Note in Figure 4.10 that

D log MW

log 92000 log 30000

log x

log 30000

Slope =

=

−

=

−

D mobility

.

0 41 − .

0 80

.

0 62 − .

0 80

where x is the molecular weight of the unknown. Solving,

( .

4 964

.

4

)

476 (

.

0 1 )

8

L

og x =

−

+ .

4 476 = .

4 701

− .

0 39

and x = antilog of 4.701 = 50 kd

8. Compare the diagonal electrophoresis (textbook, p. 96) patterns obtained with the normal

and the mutant proteins. If these patterns are essentially identical, the disulfide pairing is the same in both proteins; if they are not, the new cysteine residue is probably involved in a new disulfide pairing in this mutant.

9. Assuming that cells with receptors that bind bacterial degradation products may also

bind fluorescent derivatives of those products, you could synthesize a fluorescent-labeled derivative of a degradation product (perhaps some peptide of interest) and use this derivative to detect cells having receptors for this peptide.

10. (a) The digestion products will be AVGWR, VK, and S. The products have slightly

different sizes (though all are quite small on a macromolecular scale) and somewhat different isoelectric points. These isoelectric points are approximately 1⁄2 (3.1 + 8.0) = 5.5 for serine (the pK values for ionizable groups from Table 3.1), approximately 1⁄2 (8.0 + 10.0) = 9.0 for VK, and approximately 1⁄2(8.0 + 12.0) = 10.0 for AVGWR. If high-pressure (high resolution) liquid chromatography is used, either an ion-exchange or a molecular-exclusion approach should work.

(b) The digestion products will be AV, GW, RV, and KS. Because all of the products are

dipeptides (similar in size), an ion-exchange column should be used.

11. A probable explanation is that an inhibitor of the enzyme was removed during a partic

ular purification step. When the inhibitor is absent, the apparent activity will increase. (Several other scenarios may be possible.)

EXPLORING PROTEINS

12. The specific activity is (total activity) divided by (total protein). The purification level is

(specific activity) divided by the (initial specific activity). The yield is (total activity) divided by (initial total activity), multiplied by 100%. Answers are given in the table below. Total Specific Purification Total activity activity Purification Yield procedure protein (mg) (units) (units mg−1) level (%)

Crude extract

20,000

4,000,000

200

1.0

100

(NH4)2SO4 precipitation

5,000

3,000,000

600

3.0

75

DEAE-cellulose chromatography

1,500

1,000,000

667

3.3

25

Size-exclusion chromatography

500

750,000

1,500

7.5

19

Affinity chromatography

45

675,000

15,000

75.0

17

13. Two types of 15-kD subunits are present, one type beginning with N-terminal Ala and

the other with N-terminal Leu. Pairs of these small 15-kD subunits are linked by covalent disulfide bonds that are broken only by the mercaptoethanol. The disulfide-linked subunits comprise the 30-kD species. (There is insufficient information to discern whether the 30-kD species consists of two different homodimers in which identical subunits are linked by disulfide bonds, or a unique heterodimer in which one Ala-initiated subunit is linked precisely to one Leu-initiated subunit.) Finally, two of the 30-kD species associate noncovalently to form the 60-kD particle that contains two copies each of two different 15-kD subunits. The 60-kD particle constitutes the native protein that is observed by molecular exclusion chromatography. (Urea disrupts the noncovalent subunit association, without breaking the disulfide bonds.)

The final quaternary structure may be described as either (A-A)(B-B), or (A-B)2, where

a hyphen (-) indicates a disulfide bond, “A” designates a subunit beginning with Ala, and “B” designates a subunit that begins with Leu.

14. The key question is whether the 30-kD units are two different homodimers or a unique

heterodimer? (Either population would give a 50/50 mixture of dabsyl-Ala and dabsylLeu upon N-terminal analysis.) Therefore, one needs a method that will separate the (putative) A-B dimers from A-A and B-B dimers before the disulfide bonds are broken. While no method is absolutely (100%) certain to accomplish this, methods based on nativegel electrophoresis or high-resolution ion-exchange chromatography will provide opportunities for a favorable outcome. A good choice would be to use two-dimensional electrophoresis consisting of isoelectric focusing of 30-kD units (in the presence of 6 M urea), followed by mercaptoethanol/SDS-PAGE in the second direction. The possible outcomes are diagrammed below. The A-B unit would travel as a single entity in the first direction (isoelectric focusing), whereas the A-A and B-B units may be separable under high-resolution isoelectric focusing.

A-B subunit

A-A and B-B subunits

E

G

A

B

A

B

A

SDS-P

Isoelectric focusing

Isoelectric focusing

CHAPTER 4

15. Light was used to direct the synthesis of these peptides. Each amino acid added to the

solid support contained a photolabile protecting group instead of a t-Boc protecting group at its a-amino group. Illumination of selected regions of the solid support led to the release of the protecting group, thus exposing the amino groups in these sites and making them reactive. The pattern of masks used in these illuminations and the sequence of reactants define the ultimate products and their locations. (See S. P. A. Fodor, J. L. Read, M. C. Pirrung, L. Stryer, A. T. Lu, & D. Solas. Science 251[1991]:767, for an account of light-activated, spatially addressable–parallel-chemical synthesis.)

16. The peptide is AVRYSR.

Trypsin cleaves after R. The other R is at the C-terminal, and carboxypeptidase will not cleave the C-terminal R. Chymotrypsin cleaves after Y.

17. The full peptide is: S-Y-G-K-L-S-I-F-T-M-S-W-S-L.

The peptide will give these digestion patterns:

Carboxypeptidase cleaves the C-terminal L.

Cyanogen bromide: S-Y-G-K-L-S-I-F-T-M* and S-W-S-L (*cleavage after M, with conversion of the M to homoserine).

Chymotrypsin: S-Y, G-K-L, S-I-F, T-M, S-W, and S-L (cleavage after Y, L, F, M, and W).

Trypsin: S-Y-G-K and L-S-I-F-T-M-S-W-S-L (cleavage after K).

18.

CH

H

N=C=S

J

3

J

O

J

H N

+

J

2

(1)

NH

2

H

J

J

N

S

N

NH

S

K

2

H;

KO

+NH ;

K

N

4

(2)

N

H

O

H

J

J

H

CH

3

3

K

J H

H O;

CH3

3

J

N

(PTH-alanine)

(3)

K

N

J

H

S

In step (1), the free a-amino group of alanine-amide reacts with phenylisothiocyanate to form the phenylthiocarbamyl derivative. Step (2) involves the use of anhydrous acid for a concerted cyclization of the phenylthiocarbamyl-peptide, cleavage of the peptide bond, and release of ammonia along with alanine-phenylthiazoline. (For the case of a real peptide, the shortened peptide of length (n-1) would be released instead of the ammonia.) Finally, in step (3), the phenylthiazolinone is converted to the corresponding phenylthiohydantoin (PTH) derivative using aqueous acid. </body></html>

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