The Molecules of Life: Physical and Chemical Principles (Instructor Solution Manual, Solutions) [1 ed.] 0815341881, 9780815341888

Citation preview

The MOLECULES of LIFE Physical and Chemical Principles

Solutions Manual Prepared by James Fraser and Samuel Leachman

Chapter 1 From Genes to RNA and Proteins Problems and Solutions True/False and Multiple Choice 1. When two atoms approach each other closely, the energy goes up because the nuclei of the atoms repel each other. True/False 2. Ionic interactions are stronger in water than in vacuum because water forms strong hydrogen bonds with polar molecules. True/False 3. An N–H•••O=C hydrogen bond has optimal energy when it: a. is bent. b. is linear. c. has a donor–acceptor distance of 4 Å. d. has a donor–acceptor distance of 2 Å. 4. A by-product of forming a peptide bond from two amino acids is water. True/False 5. Circle all of the polar amino acids in the list below: a. Phenylalanine b. Valine c. Arginine d. Proline e. Leucine 6. Proteins fold with their hydrophobic amino acids on the surface and their hydrophilic amino acids in the core. True/False 7. Which of the following is not a unit of structure found in proteins? a. β sheets b. Loop regions c. α helices d. γ arches

8. The central dogma of molecular biology states that RNA is translated from proteins. True/False 9. Which of these types of molecules serve as a template for messenger RNA? a. Protein b. DNA c. Transfer RNA d. Carbohydrates e. Ribosomes 10. DNA primase synthesizes DNA molecules. True/False Fill in the Blank 11. When two atoms approach closer than ________________ the interaction energy goes up very sharply. Answer: their van der Waals radii 12. The van der Waals attraction arises due to _______ induced dipoles in atoms. Answer: mutually 13. At room temperature, the value of __________ is about 2.5 kJ•mol−1. Answer: thermal energy 14. ________ is an operational nucleic acid, whereas _________ is strictly an informational nucleic acid. Answer: RNA, DNA 15. Amino acids are zwitterions: molecules with charged groups but an overall ______ electrical charge. Answer: neutral 16. The ______ consists of two subunits that assemble around messenger RNA. Answer: ribosome

2

Chapter 1: From Genes to RNA and Proteins

Quantitative/Essay 17. Order the following elements from lowest electronegativity to highest: P, C, N, H, O, S. Use your ordering of the atoms to rank the following hydrogen bonds from weakest to strongest: a. SH----O b. NH----O c. OH----O Answer: H = P < C = S < N < O; SH----O < NH----O < OH----O 18. The stabilization energy of a bond or interatomic interaction is the change in energy upon breakage of a bond between two atoms (that is, the change in energy when the atoms are moved away from each other). We can classify bonds into the following categories, based on their dissociation energies: Strong: > 200 kJ•mol−1

b. Only iv could be broken readily by thermal fluctuation. c.

d. ii, iii, and iv. 19. The diagram below shows a representation of the structure of a peptide segment (that is, a short portion of a larger protein chain). Hydrogen atoms are not shown. Nitrogen and sulfur atoms are indicated by “N” and “S.” Sidechain oxygen atoms are indicated by “*.” a. Identify each of the amino acid residues in the peptides. b. Draw a linear chemical structure showing the covalent bonding, including the hydrogen atoms, of the entire Question 1.19peptide.

Medium: 20–200 kJ•mol−1 Weak: 5–20 kJ•mol−1 Very weak: 0–5 kJ•mol−1

i. Same. ii. Slightly stronger; the Van der Waals interactions do not change, but the net interaction becomes stronger because of the hydrophobic effect. iii. Weaker. iv. Weaker.

N

S

N

N

N

* N

N

N

N N

N N N

d. Which of these bonds could be broken readily by thermal in water? Questionfluctuations 1.18

Answer:

(i)

b. See Figure at the top of the next page.

(iii)

P

C

(ii)

O

OH

(iv) O

C

–

O

Answer:

H

N +

C

HH

Answer: a.

i. Strong. ii. Very weak. iii. Weak. iv. Weak.

a. Ala-Gly-Ser-Gly-Ser-Tyr-Gly-Arg-Val-Met. 20. With 64 possible codons and 21 options to code (20 amino acids + stop), (a) what is the average number of codons per amino acid/stop codon? (b) Which amino acids occur more often than expected in the actual codon table?

O C

*

≈

c. Next, consider what happens when these molecules are immersed in water (fully solvated). For each bond, indicate whether it becomes weaker, stronger, or stays the same in water.

N

≈

Consider the bonds highlighted in purple in the diagram below. a. First consider the bonds in molecules isolated from all other molecules (in a vacuum). Classify each of them into the four categories given above, based on your estimation of the bond strength. b. Which of these bonds could be broken readily by thermal fluctuations?

a. Average = 64 codons/21 amino acids = 3.05 codons per amino acid. b.

Leucine (6 codons), valine (4 codons), serine (6 codons), proline (4 codons), threonine (4 codons), alanine (4 codons), arginine (6 codons), glycine (4 codons).

21. There are approximately 3 billion nucleotides in the human genome. If all of the DNA in the entire genome were laid out in a single straight line, how

*



PROBLEMS and solutions

1Q19b

3

NH

+

C

NH

NH

OH

CH3

CH2 OH

C O

H

CH3

N

C

C

H

O

A

H

H

N

C

C

H

O

G

OH

H

CH2

N

C

C

H

O

S

H

H

N

C

C

H

O

CH2

N

C

C

H

O

S

long would the line be? Express your answer in meters. (3 ×

bp) × (3.4 ×

10–10

m•bp–1)

= 1.02 m

22. The human immunodeficiency virus (HIV) is a retrovirus with an RNA genome. a. Assume that each HIV contains two RNA genomes and 50 molecules of the reverse transcriptase enzyme. b. Assume that each reverse transcriptase molecule acts on each RNA genome 10 times to produce DNA. c. Assume that an integrase enzyme successfully integrates 1% of the available reverse transcribed HIV genomes into the genome of a human host cell. d. Assume that each integrated copy of the viral genome is transcribed 500 times/day.

H

H

N

C

C

H

O

Y

H

H

N

C

C

H

O

G

CH

H

CH2

N

C

C

H

O

R

S

CH3

CH2

H

H

N

C

C

H

O

V

H

CH2

N

C

C

H

O

M

24. What changes to the central dogma were necessary after the discovery of retroviruses? Answer:

Answer: 109

CH2

H

G

H3C

How many HIV RNA genomes are created per day from one infected cell? Answer: 50 reverse transcriptases × 10 replications × 2 genomes = 1000 DNA genomes created. 1000 genomes created × 0.01 successfully integrated by integrase = 10 integrated genomes. 10 copies of the genome integrated × 500 transcriptions per day = 5000 HIV genomes created from one infection per day.

23. What is a step in RNA processing that occurs in eukaryotes but not in prokaryotes? Answer: Eukaryotes have introns, which necessitate splicing. Additionally, eukaryotes add a poly-A tail on the 3ʹ end and a methyl-G cap on the 5ʹ end.

The unidirectional flow from DNA to RNA was no longer true after the discovery of retroviruses. Retroviruses use reverse transcription to create a DNA copy and then generate many RNA copies. 25. What chemical properties have led to DNA being selected through evolution as the information molecule for complex life forms instead of RNA? Answer: DNA is inherently more stable. The 2ʹ-OH group in an RNA nucleotide, which DNA lacks, can react to break the backbone just downstream by forming a cyclic 2ʹ-3ʹ phosphodiester bond and breaking the sugar– phosphate backbone. Furthermore, because DNA typically exists as a double-stranded helix, in the event of DNA on one strand being lost or damaged it can be replaced or repaired using the other strand as a template. RNA, typically single-stranded, does not have this capability. 26. How do size considerations forbid G-A base pairs in a double helix? Answer: Both G and A are purines and have two rings. The double helix only has enough space for a single purine (two rings) and pyrimidine (one ring), but not enough for four rings total. 27. Why are DNA chains synthesized only in the 5ʹ-to-3ʹ direction in the cell? Answer: Replication proceeds only in the 5ʹ → 3ʹ direction, and not 3ʹ → 5ʹ, because only nucleotide

4

Chapter 1: From Genes to RNA and Proteins

triphosphates with the triphosphate group attached to the 5ʹ carbon of the sugar are present in cells. This is the high-energy, reactive end of the molecule. 28. Chemists are able to synthesize modified oligonucleotides (that is, polymers of nucleotides) in Question 1.29 which the phosphate linkage is replaced by a neutral amide linkage, as shown in the diagram below.

phosphodiester DNA O

O

O O

P

O–

O

H

O base

O

O

O

H

H

N

base

H base

O

O

Answer: You could postulate a nucleotide triphosphate with the triphosphate attached to the 3ʹ sugar carbon, rather than the 5ʹ. The high energy 3ʹ triphosphate would be susceptible to attack by the 5ʹ-OH group in the growing strand, allowing the chain to grow in the 3ʹ → 5ʹ direction.

amide DNA O

base

29. Suppose you were told that the two strands of DNA could in fact be readily replicated in opposite directions (that is, one in the 5ʹ → 3ʹ direction, and one in the 3ʹ → 5ʹ direction). Would you have to postulate the existence of a new kind of nucleotide? If so, what would the chemical structure of this compound be?

H

(Adapted from M. Nina et al., and S. Wendeborn, J. Am. Chem. Soc. 127: 6027–6038, 2005. With permission from the American Chemical Society.) a. Such modified oligonucleotides are able to form double-helical structures similar to those seen for DNA and RNA. Often these double helices are more stable than the natural DNA and RNA double helices with the same sequence of bases. Explain why such helices can form, and why they can be more stable. b. Given the increased stability of such modified nucleotides, why has nature not used them to build the genetic material? Provide two different reasons that could explain why these molecules are not used. Answer: a. The amide DNA helices lack the unfavorable negative charge concentration present in phosphodiester DNA. b. If the amide DNA double helices were too stable, it wouldn’t be possible to unwind them for replication and transcription. Also, without a high-energy phosphate linkage, the amide DNA monomers can’t internally store the energy needed for synthesis.

The MOLECULES of LIFE Physical and Chemical Principles

Solutions Manual Prepared by James Fraser and Samuel Leachman

Chapter 2 Nucleic Acid Structure Problems and Solutions True/False and Multiple Choice

Fill in the Blank

1. The tertiary structure of functional RNA molecules is easily predicted.

7. The ______________ structure of a nucleic acid is the sequence of nucleotides in the DNA or RNA molecule. Answer: primary

True/False 2. Which of the following is not a stabilizing force for the structure and stability of double-stranded nucleic acids? a. b. c. d.

base stacking hydrogen bonding disulfide bonds electrostatic forces

3. H-bond acceptor, H-bond donor, H-bond acceptor, methyl group is the pattern of potential interactions at the edge of which Watson-Crick base pair? a. b. c. d. e.

A-T G-C A-A C-G T-A

4. Genomic DNA can become deformed from its normal B-form by DNA binding proteins, such as the histone proteins and the TATA-box binding protein. True/False 5. Which nonstandard base pair is most likely to form a wobble base pair? a. b. c. d. e.

U-U A-G A-A G-U G-G

6. Classify the following RNA structural elements as secondary or tertiary structure: a. b. c. d. e. f. g.

coaxial helices pseudoknot hairpin junction adenosine platform ribose zipper bulge

Answer: Secondary: c, d, g; tertiary a, b, e, f

8. The modified RNA base in which two methyl groups are added to guanine is __________. Answer: N,N-dimethylguanine. The corresponding nucleoside is N,N-dimethylguanosine. 9. B-form DNA has a C2ʹ ________ sugar pucker. Answer: endo 10. Hoogsteen base pairs, where the hydrogen-bonding interactions utilize the Watson-Crick base-pairing edge on one base and the major groove edge in the other base, can be utilized to form an RNA _______ helix. Answer: triple 11. Metal ions, such as K+, Na+, and Mg2+, typically interact with the __________ group on the backbone of nucleic acids. Answer: phosphate

Quantitative/Essay 12. What physical factors force RNA to adopt only the C3ʹ endo sugar configuration, but allow DNA to adopt either the C2ʹ endo or the C3ʹ endo sugar configurations? Answer: The C2ʹ endo configuration results in close contact (1.9 Å) between one of the oxygen atoms of the 3ʹ phosphate group and the hydrogen atom at the 2ʹ position of the ribose ring in DNA. Compared with DNA, RNA has the 2ʹ hydrogen replaced by hydroxyl group (OH) and the two oxygen atoms (2ʹ ribose and 3ʹ phosphate) repel each other strongly. Rather than distorting the structure away from the Watson-Crick model, it is energetically favorable for RNA to switch to the C3ʹ endo conformation.

2

Chapter 2: Nucleic Acid Structure

2Q14

Problem question 2) (na_57_v1) 13. WhatSet type(quantitative of RNA interaction is shown below?

Answer:

H



H

minor groove

G

N N

O

H

N

N

N N

H

H Label the major and minor grooves of the Watson-Crick base pair and the potential hydrogen bonds between all bases. Answer: This shows an A-minor motif in which the minor groove edges of an adenine base interact with the minor groove 2Q13 of a G-C base pair.

major groove

U

O

H

O

N

H

major groove

Interactions on minor groove: H-bond acceptor, H-bond donor, H-bond acceptor. Interactions on major groove: hydrogen atom, H-bond acceptor, H-bond acceptor, H-bond acceptor. The G-U base pair is not a standard Watson-Crick base pair. 15. List at least three physical features that distinguish A- and Z-form DNA. Answer: The main differences between A- and Z-form DNA are as follows: For A-form DNA, the repeating unit is one base; however, in Z-form DNA, the repeating unit is two bases (most often alternating pyrimidine and purines).

minor groove

A-form DNA is a right-handed double helix. Z-form DNA is a left-handed double helix. In A-form DNA, the sugar pucker of the bases is C3ʹ endo, whereas Z-form DNA alternates between C2ʹ endo and C3ʹ endo.

14. Label the atoms, the bases, the hydrogen bonds across bases, the major and minor grooves, and the interactions along the grooves for the following base pair (oxygen and nitrogen atoms are not identified explicitly): Problem set (quantitative question 2) (na_56_v1)

The minor groove of A-form DNA is wide and shallow, whereas the Z-form minor grove is narrow. The major groove of A-form DNA is deep and narrow, whereas the Z-form major groove is relatively shallow. In A-form DNA the base pairs are tilted to the helical axis, whereas the Z-form base pairs are nearly perpendicular to the helical axis. 16. In water, why is base stacking relatively more important than hydrogen bonding to forming a DNA double helix? Answer: Water can make H bonds with the DNA base H-bond donor and acceptors, so the energetic gain in interactions when those H bonds are satisfied by other nucleotides is not great. However, the many small polar interactions along the nucleotide rings that are formed upon base stacking are energetically more favorable than the interactions with water.



Is the base pair a standard Watson-Crick base pair?

17. How are the interactions between a nucleic acid and Mg2+ mediated by water?



PROBLEMS and solutions Answer: Water molecules can form a hydration shell around Mg2+, facilitating indirect interactions between it and the phosphate and functional groups in nucleic acids.

18. Why is the R (purine) in GNRA tetraloops required rather than a pyrimidine? Answer: The R position, adjacent to the A in sequence and space, is functionally significant because it provides a large surface area (the base has the two rings of a purine, as opposed to one ring for a pyrimidine) against which the A can base stack. 19. Why do most transcription factors interact with the major groove of B-form DNA rather than the minor groove?

3

22. The structure of the large ribosomal subunit from Haloarcula marismortui (PDB code: 1FFK) has been solved by x-ray crystallography. The 23S RNA contains 2922 nucleotides (758 A, 889 G, 739 C, and 536 U). a. Assuming a random distribution of nucleotides, how many four-mer sequences with the sequence G-N (any base)-R (purine)-A are possible? b. There are actually 21 GNRA tetraloops in the structure. What percentage of possible GNRA sequences actually formed tetraloops in the structure? Answer: a. Probability of G = total G/total nucleotides = 889/2922 Probability of N = 1. Probability of R = (total A + total G)/total nucleotides = (758 + 889)/2922.

Answer:

Probability of A = total A/total nucleotides = 758/2922.

The major groove has more distinguishing interactions than the minor groove. Each base pair has a unique pattern of molecular interactions along the major, but not the minor, groove. Since transcription factors need to bind specifically to sequences of DNA, the best potential for distinguishing different sequences is to interact with the major groove. Further, the minor groove is narrow compared with the major groove. The depth of the major groove is compatible with the interaction with protein structural elements such as an α helix.

Each four-mer has a GRNA probability = (probability of G) × (probability of N) × (probability of R) × (probability of A) = 0.0445.

20. Which of the following DNA sequences is most likely to adopt the Z-form? Why? a. b.

GCGCGCGCATATGCGCGCGCC AGAGAGCTCTCTCTCTAAAAT

Answer: Sequence a is more likely to adopt Z-form because it alternates purine and pyrimidines in a GC-rich sequence. This alternating pattern of 2ʹ endo and 3ʹ endo puckers yields the zig-zag pattern, where the smallest repeating unit is two base pairs, characteristic of Z-form DNA. 21. Consider a relaxed, closed-circular DNA plasmid that has 1040 base pairs with writhe = 0. An intercalator is added, such that there is one intercalator per 104 base pairs. The effect of the intercalator is to cause the twist between the base pairs that flank it to be reduced to zero. Will the resulting intercalator-bound DNA be positively or negatively supercoiled? Answer: Positively supercoiled. The relaxed plasmid contains 100 turns (1040 bp × 1 turn/10.4 bp) and therefore has linking number L = 100. Because it is not initially supercoiled, it has writhe W = 0. By the relation W = L – T, initial twist T is 100, with each base pair contributing ~35°. The cumulative effect of binding 10 intercalators is a reduction of the twist by a full turn (~35° × 10 ≈ 360°) to T = 99. As L is constant, positive supercoiling with W = 1 results.

Total four-mers = 2922 – 3 = 2919. Probable GRNA four-mers = probability × total fourmers = 0.0445 × 2919 = 130. b. If only 21 GNRA tetraloops formed out of a possible 130, then16% of the sequences form a stable tertiary GNRA tetraloop motif. 23. A bacterial DNA polymerase moves at approximately 1000 base pairs per second when replicating DNA. The polymerase holoenzyme is approximately 110 Å long. By what multiple of its length does the polymerase move forward along the axis of the DNA double helix in 3 seconds? Assume the DNA stays fixed and ignore the rotational component of the motion of the polymerase along the DNA. Answer: Each bp of DNA ~3.4 Å rise per bp. In 3 s, the polymerase moves 3000 bp. Length = 3000 bp × 3.4 Å•bp–1 = 10,200 Å. Length/polymerase length = 10,200 Å/110 Å = 92.7 times the total length of the polymerase in 3 sec.

The MOLECULES of LIFE Physical and Chemical Principles

Solutions Manual Prepared by James Fraser and Samuel Leachman

Chapter 3 Glycans and Lipids

Problems and Solutions True/False and Multiple Choice 6. An individual lectin-binding domain is likely to bind with high affinity to many diverse sugars.

1. In most carbohydrate monomers, the ratio of carbon:hydrogen:oxygen is a. b. c. d. e.

True/False

1:1:1 2:1:1 1:2:1 1:1:2 1:1.5:1

7. In plants, the cell wall is generally higher in percent polysaccharide content and lower in percent lipid content than the cell membrane. True/False

2. Which of the following is not a lipid? a. b. c. d. e.

cholesterol sphingolipid apolipoprotein vitamin A triglyceride

Fill in the Blank 8. Simple sugars can exist in both linear and ______ forms. Answer: cyclic 9. ________ is an acetylated polymer found in the exoskeletons of insects and the cell walls of fungi.

3. Many sugars are structural isomers of other sugars. True/False

Answer: Chitin

4. Most detergents will organize spontaneously into bilayers when placed in aqueous solutions.

10. Inositol is an unusual six-carbon sugar because it does not contain an _______ atom in its ring.

True/False

Answer: oxygen

5. Which of the following fatty acids are unsaturated?

11. A lipid _____ is a section of membrane with both composition and physical properties that are distinct from the surrounding membrane.

Problems Q5 (lipids_probQ5_v1) (A)

(B) O

O

(C)

O

(D) O

O O

(E)

O

(F) O

O O

O O

Answer: raft 12. In eukaryotic cells, intracellular compartments bounded by membranes are called ________. Answer: organelles

Quantitative/Essay 13. Why are the most stable conformations of sugars most often nonplanar with hydrogens in the equatorial positions?

Answer: A, B, and E

Answer: Carbons which make four single bonds should have bond angles near the value in a tetrahedron, 109°. To maintain bond angles near this value the atoms in closed ring structures cannot all lie in a plane. Instead the rings “pucker,” moving one or more atoms out of

2

Chapter 3: Glycans and Lipids the plane. There is less space for nonhydrogen atoms in the axial positions, so the conformational energy is lower when most of the hydroxyls are in equatorial positions.

14. What information can be gained more readily from the Fischer representation than from the Haworth representation? Answer: The Fischer projection is very convenient for comparing stereochemistry of different sugar isomers, but poor for seeing the three-dimensional relationship of atoms and the real geometry of the molecules. The format indicating three-dimensional structure provides a better sense of conformational features, but is less easy for seeing relationships between stereoisomers. 15. How do the homopolymers cellulose and glycogen differ at the level of: (a) the stereochemistry of the monomer, (b) physical properties, and (c) the intramolecular interactions that maintain those physical properties? Answer: Glycogen contains only α-glucose, whereas cellulose contains only β-glucose. Glycogen branches off spiral structures through 1→6 linkages that result in a soft, gel-like substance. The gradual helical structure of glycogen is not stabilized by any particular intramolecular interaction. Cellulose consists of linear chains, assembled into stacked sheets that form stiff fibers. These sheets are stabilized by intramolecular hydrogen bonds between neighboring residues. 16. A student performs a biochemical fractionation of a cell and isolates the membrane-containing fraction. An analysis of the glycoproteins in the fraction reveals only N-linked sugars. a. What type of amino acid residue are the sugars linked to? b. What is the enzyme responsible for transferring the first sugar to that residue? c. What organelle has the student likely isolated? Answer: a. N-linked glycosylation links sugars to Asn residues. b. The enzyme oligosaccharyl transferase transfers a 14 residue sugar to the Asn from a lipid anchor. c. The student has most likely isolated the endoplasmic reticulum. 17. The O blood type group is known as the universal donor. What property of the glycosylation pattern of O-type individuals allows for the blood to be transfused into individuals of any blood type without eliciting an immune response? Answer: The other three blood types (A, B, and AB) all contain the same tetrasaccharide that defines the O blood type. Since the immune system of an individual with any blood type must be tolerant of the tetrasaccharide in the context of “self” blood cells, no immune reaction is generated by the tetrasaccharide of donated O-type blood cells.

18. What molecular forces drive the formation of bilayers when phospholipids are immersed in water? Answer: The charged phospho- head groups point out to form a favorably charged interface with water. The hydrophobic alkyl chains point in and interact with each other laterally, packing against the hydrophobic chains of the other layer. These hydrophobic forces exclude water from the interior region. This structure satisfies the interaction preferences for both segments (charged head group and uncharged tail) of the lipid molecules. 19. How does the packing of cholesterol against other lipids alter the biophysical properties of a membrane? Answer: The rigid core of cholesterol packs against alkyl chains of neighboring lipid molecules, reducing their flexibility. The net effect is to stiffen the membrane bilayer against distortion, while retaining fluidity. A membrane containing cholesterol is less permeable to small molecules, but does not substantially reduce the rate of diffusion of molecules within the membrane. 20. Many proteins need to be in close proximity to the membrane in order to perform their biological functions. What are two strategies employed by the cell to ensure that certain proteins are anchored to the membrane? Answer: Proteins can be covalently modified at a cysteine residue or at the C-terminus with a lipid-containing molecular anchor such as a polyisoprene. These modified proteins then associate with the membrane via the insertion of the lipid anchor into the membrane. Alternately, proteins containing specific lipid binding domains can interact with the cytosolic portion of lipids that are inserted into the membrane. These interactions are generally noncovalent. A third mechanism is for the protein to have a stretch of hydrophobic amino acids that insert into the membrane ensuring that the soluble domains of the protein are localized near the membrane. 21. Without the aid of an enzyme, is a carbohydratemodified lipid more likely to (a) move in an undirected lateral motion for 500 Å or (b) transfer across the ~40-Å thick cell membrane from one leaflet to the other? Answer: (a) is more likely. Without a flippase enzyme, transfer of a lipid molecule from one leaflet to another requires moving the head group (in this case, partially polar carbohydrate) across the hydrophobic region, a process that is energetically unfavorable and very slow. In contrast, diffusion laterally within the membrane occurs quite easily. 22. Scientists* have proposed a model for how yeast cells determine the length of a class of fatty acids that are synthesized using a family of fatty acid synthase enzymes. In different variants of the enzyme, a key lysine residue can be positioned at various places along an α helix of the synthase to control the length of the fatty acid chain that is produced.



PROBLEMS and solutions



Imagine a similar enzyme in which moving the position of a catalytic lysine residue along a β sheet creates a synthase that catalyzes four additional carbons into the fatty acid chain. The lysine residue can only be moved in two-residue units along the β sheet and still face the substrate (that is, it can be moved from position 1 to positions 3 or 7, but never to positions 2 or 4). Thus, movement from position 1 to position 3 results in a four-carbon change in the length of the fatty acid produced.



There are 14 residues in the β sheet. The maximum length, a 40-carbon fatty acid, is created when the residue is placed at the end of the β sheet (position 1). What is the shortest possible fatty acid that the synthase can create?



*[Denic V & Weissman JS (2007) A molecular caliper mechanism for determining very long-chain fatty acid length. Cell 130, 663–677.] Answer: The catalytic residue can be moved to position 13—this is 6 residue units. 40 carbon fatty acid – (6 residue units × 4 carbons per residue unit) = 40 carbons – 24 carbons = 16 carbons. The smallest fatty acid chain that can be created by this enzyme is 16 carbons long.

3

The MOLECULES of LIFE Physical and Chemical Principles

Solutions Manual Prepared by James Fraser and Samuel Leachman

Chapter 4 Protein Structure Problems and Solutions True/False and Multiple Choice 1. Globular proteins are generally embedded in the interior of a lipid bilayer. True/False 2. The secondary structure of a protein refers to the extent and order of its α helices and β sheets. True/False 3. Which of the following statements regarding protein domains is NOT true? a. Secondary structural elements of a domain generally pack so that a hydrophobic core is formed. b. A protein domain normally contains 50–200 residues. c. Protein domains are units of tertiary protein structure. d. Proteins are only comprised of one protein domain. 4. Match the following proteins with their quaternary structure: a. hemoglobin b. RNA polymerase c. myoglobin

i. one subunit ii. many structurally similar subunits iii. many subunits of varied structure

Answer: a–ii, b–iii, c–i 5. Most protein conformational changes involve breaking and reforming several covalent bonds along the polypeptide chain. True/False 6. The only genetically encoded amino acid without a stereoisomer is: a. b. c. d. e.

alanine tryptophan glycine proline lysine

7. The least restricted ϕ and ψ angles are found in polypeptides in which class of secondary structure? a.

right-handed α helix

b.

β sheet

c.

left-handed α helix

d. loop 8. The active site in open twisted α/β domains is in a crevice outside the carboxyl ends of the β strands. True/False

Fill in the Blank 9. Soluble proteins have mostly _________ sidechains on the inside and mostly ________ sidechains on the outside. Answer: hydrophobic, hydrophilic 10. The organization of the protein subunits in multimeric proteins is known as the ___________ structure. Answer: quaternary 11. SH2 domains bind to _________ and SH3 domains bind to _________. Answer: phosphorylated tyrosine residues, proline containing peptides 12. _______ residues form cis peptide bonds in proteins with significant frequency. Answer: Proline 13. α helices that have one hydrophobic face and one hydrophilic face are known as _________ helices. Answer: amphipathic

Quantitative/Essay 14. A group of scientists isolate a novel strain of bacteria. Although the bacteria have normal nucleic acids and proteins, the strain has an abnormal membrane. In this strain, the lipid bilayer is twice as thick (70 Å) as

2

Chapter 4: Protein Structure that of a normal strain of bacteria. This extra thickness is entirely due to longer hydrophobic tails and not due to changes in the charged head groups of the phospholipids that comprise the membrane. If you isolated a single transmembrane helix from a protein from this strain, how long would you expect it to be?

do not gain stability from the hydrophobic effect. While backbone hydrogen bonds could contribute stabilizing energy to the folded state, the backbone –NH and –C=O groups of the unfolded polypeptide hydrogen bond to water. Upon folding, formation of secondary structure elements replaces hydrogen bonds to water with hydrogen bonds to other parts of the protein backbone. Thus the summed energy of hydrogen bonding does not change upon folding.

Answer: 70 Å membrane/1.5 Å per residue of alpha helix = at least 47 residues. 15. After sequencing the genome of the bacteria in Problem 14, the scientists want to predict all of the membrane proteins.

18. Draw an alanine-alanine dipeptide. Indicate the peptide bond. What factors restrict the rotation about the 4Q18 bond? peptide Answer:

a. Describe the hydrophobicity index procedure for predicting α helical membrane proteins.

O

b. How would the hydrophobicity index calculation differ for this new strain compared to normal bacteria?

H2N

Answer:

CH3

a. The hydrophobicity index at each residue represents the mean hydrophobicity value of the surrounding contiguous set (“window”) of residues centered on the residue in question. Clustered regions of negative hydrophobicity index indicate the presence of a transmembrane helix. b. For this strain of bacteria, since the membrane is thicker than normal, the “window” size should be increased. Rather than calculating the average hydrophobicity for 19 residues, we should calculate the average for at least 40 residues. This represents a minimum contiguous stretch of residues to stretch across the lipid bilayer. 16. Using the hydrophobicity scale in Figure 4.74, calculate the hydrophobicity index for the 19 contiguous residue window defined by the following sequence:
 Pro-Gly-Ala-Val-Val-Ile-Trp-Phe-Val-Val-Met-Ser-Ala-IleIle-Phe-Tyr-Ala-Thr

CH

C

OH

CH3

Answer: The simplest Ramachandran diagram shows which conformations are prevented due to interatomic collisions. The repulsive energy of such collisions is so high that it dominates the forbidden regions of the diagram. Taking the other forces into account modulates the favorability of some regions of the Ramachandran diagram, but the overall forbidden and allowed zones are not changed.

Pro-Gly-Ala-Val-Val-Ile-Trp-Phe-Val-Val-Met-Ser-Ala-IleIle-Phe-Tyr-Ala-Thr.

20. How are the large loop elements of scorpion toxin stabilized without participating in the hydrophobic core?

This segment could be part of a transmembrane helix as the hydrophobicity index is negative and there are no charged residues.

Answer: In addition to the hydrophobic core, this particular protein fold is held together by four disulfide bonds. In the loop regions, the disulfide bonds provide covalent linkages between different parts of the protein backbone. Since the energy required to break these covalent bonds is very large, these loop elements can be stabilized without significant contributions from the hydrophobic effect.

17. Why are isolated secondary structural elements not usually stable in isolation, even though all backbone hydrogen bonds are satisfied? Answer: The hydrophobic effect is the major force driving protein folding and governing the stability of globular proteins. Isolated secondary structural elements cannot bury hydrophobic residues away from water and thus

N

19. The simplest form of the Ramachandran diagram (shown in Figure 4.20) is calculated by ignoring hydrogen bonding, interactions with water, and the hydrophobic effect. Why is this simple form of the diagram still an effective predictor of protein backbone conformation?

Answer:

= –36.6; –36.6/19 = –1.9

C

O

Rotations are strongly hindered because the peptide bond has partial double bond character. Evidence for this double bond character includes the observation from high resolution crystal structures that the peptide bond length is shorter than normal single N–C bond lengths.

Could this segment be part of a transmembrane helix?

0.4 + 4.5 + 2.1 – 2.1 – 2.1 – 4.5 – 8.8 – 7.1 – 2.1 – 2.1 – 2.9 + 2.1 + 2.1 – 4.5 – 4.5 – 7.1 – 2.9 + 2.1 + 0.8

CH

Peptide bond

21. Draw a helical wheel for the following sequences: 


a.

Leu-Asp-Lys-Ile-Val-Arg-Phe-Leu-Gln-Ser-Tyr



PROBLEMS and solutions



b.

Leu-Asp-Leu-Lys-Arg-Ser-Glu-Leu-Asn-Tyr-Asn



For each, highlight the hydrophobic residues by marking them with an asterix (*). Do these sequences form amphipathic helices? Answer: a. Y F I*

K

S

L* R

L* D V*

23. Why are binding sites in proteins often located at interdomain boundaries? Answer: Mutations that confer novel binding are more likely to be tolerated by the protein if they do not interfere with proper folding, making surface residues at interdomain boundaries better candidates for binding sites. Also, interdomain regions possess deep invaginations within which small molecules can bind. 24. In contrast to globular proteins and regions exposed to the cytoplasm, the integral membrane portions of membrane proteins contain very few residues not ordered into secondary structure elements such as β sheets and α helices. Why is satisfying backbone hydrogen bonds by forming secondary structure elements more important for protein folding in the membrane than in solution? Answer:

Q

Yes, this forms an amphipathic helix. b. N E K

L*

Y

3

L* S

L* D R N

No, this does not form an amphipathic helix. 22. The ridges and grooves of an α helix form an angle of ~25° or ~45° to the helical axis. What are the spacings between the residues that line these two types of grooves? How do these geometric considerations constrain the packing of pairs of helices at relative angles of 50° and 20°? Answer: ~25° grooves are separated by four residues; ~45° grooves are separated by three residues. In order to pack the two helices using two ~25° grooves, one helix must be turned 180° and placed on top of the other helix. In the interface between the two helices the directions of the ridges and grooves must thus be inclined by an angle of about 50° (25° + 25°) in order for the ridges of one helix to fit into the grooves of the other and vice versa. To form an interface between ridges formed using a ~25° and a ~45° groove, one helix must be rotated 180°. The helices are then on the same side of the helical axis and make an angle of 20° (45° – 25°) when the ridges and grooves fit into each other.

In solution, residues in loop segments can form backbone–water hydrogen bonds and thus do not pay a large energetic penalty for failing to satisfy these interactions with other backbone atoms. In contrast, in the membrane the lipid tails do not provide hydrogen bond donor or acceptors. Additionally, there is no energetic gain from burying a hydrophobic core in the membrane since the hydrophobic environment negates the importance of the hydrophobic effect. Secondary structure formation is an efficient mechanism for completely satisfying the hydrogen bonding requirements of the peptide backbone. These hydrogen bonds provide the major energetic force driving folding in the membrane. 25. A scientist isolates a membrane protein that transports sodium ions from inside the bacterium (where it is at low concentration) into solution (where the sodium ion concentration is high). a.

Does this protein use active or passive transport?

b. What are two energy sources for accomplishing this transport? Answer: a. This protein uses active transport since it is moving an ion to a region of higher concentration from inside the cell where the ion is at a lower concentration. b. Two mechanisms for coupling energy to transport against a concentration gradient are to convert light energy to a conformational or chemical change (as in rhodopsin) and to use ATP hydrolysis to drive a conformational in the protein (as in the maltose transporter). 26. What is the conformational change that occurs when bacteriorhodopsin absorbs light? Answer: The conformational change occurs on the covalently bound cofactor retinal. When retinal absorbs a photon, the bond between carbon atom 13 and carbon atom

4

Chapter 4: Protein Structure 14 is isomerized from a trans conformation to a cis conformation.

27. Two mutated bacteriorhodopsin proteins have a Val or Glu at the normal Asp 85 position. a. Do you predict that either of these mutated proteins will transport protons? b. If not, at what stage of the proton transport “wire” will the proton transport be blocked compared to wild type? Answer:

a. It is possible that the Asp-to-Glu mutation will still transport protons because the Glu sidechain can pick up the proton from retinal (acting as a Schiff base). In contrast, Val does not have a charged sidechain and cannot remove the proton from retinal, thus it will likely be unable to transport protons. b. In the Val mutant, the proton removed from retinal (acting as a Schiff base) cannot be moved to the hydrogen bonded network of water in the protein transport wire. Without the removal of this proton, retinal will be perpetually charged and unable to absorb an additional proton from Asp 85 to eventually conduct to the water portion of the proton “wire.”

The MOLECULES of LIFE Physical and Chemical Principles

Solutions Manual Prepared by James Fraser and Samuel Leachman

Chapter 5 Evolutionary Variation in Proteins Problems and Solutions True/False and Multiple Choice 1. The BLOSUM scoring matrix gives a measure of how conservative a mutation is. For substitutions of aspartic acid (Asp, D), which of the following orderings correctly places the amino acids from most conservative to least conservative? a. b. c. d.

K,L,A,C,E,S E,S,K,A,C,L L,C,A,K,S,E A,C,E,K,L,S

2. An environment profile in the 3D-1D profile method compares: I. the stability of the amino acid in varying solvents II. the burial of each amino acid in the structure III. the hydrophobicity surrounding each amino acid IV. the type of secondary structure element containing each amino acid a. b. c. d.

I, II, III, and IV I and IV II, III, and IV II and IV

7. The core of a protein generally contains residues from the ________ class of amino acids. Answer: hydrophobic/nonpolar 8. Globin proteins bind the iron-containing ________ cofactor. Answer: heme 9. A disulfide bond links two _______ residues. Answer: cysteine 10. According to the BLOSUM substitution matrix, the most conservative mutation from tryptophan (W), other than to itself, is to ______, which has a score of ______. Answer: tyrosine, 2 11. Many soluble human proteins can be expressed in the E. coli bacteria or using an in vitro translation system. How can these proteins fold without the cellular machinery present in human cells? Answer: The “thermodynamic hypothesis” states that proteins adopt native structures that optimize thermodynamic properties. Since sequence determines structure and most proteins do not require posttranslational modifications, external templates, or specific molecular chaperones to fold, many proteins can fold after translation by a different organism or even a cell-free (in vitro) translation system.

3. Protein domains can be assembled together in many different ways, because surface sidechains can be mutated easily without losing protein stability. True/False 4. In contrast to ribonuclease, some proteins cannot fold without the assistance of proteins known as molecular chaperones. This means the thermodynamic hypothesis of protein folding does not apply to these proteins. True/False 5. Two proteins that share more than 50% sequence identity over a 100-residue stretch are likely to have the same three-dimensional fold. True/False

Fill in the Blank 6. Two common chemical denaturants of proteins are guanidinium and _______. Answer: urea

12. What level of activity (1, 10, or 100%) is predicted for ribonuclease-A when it is subject to each of the following stepwise procedures? a. i. denatured, then ii. reduced, then iii. exposed to oxygen, then iv. refolded by removing urea b. i. denatured, then ii. reduced, then iii. refolded by removing urea while exposed to oxygen c. i. denatured, then ii. refolded by removing urea

Rationalize the predictions based on the effect of denaturation and reduction–oxidation of the ribonuclease-A cysteine residues. Assume that all

2

Chapter 5: Evolutionary Variation in Proteins However, we must correct for the vertical ordering: it does not matter which lysine–glutamate pair is picked first, second, etc., just that the correct pairs are picked. This means we need to divide the total number by 6! (6 pairs to choose first × 5 pairs to chose second...etc). 518,400/(6 × 5 × 4 × 3 × 2 × 1) = 720. Off the 720 random structures populated during crosslinking in experiment (b) only one will be correct. Thus the activity expected is 1/720 = 0.0014, or 0.14%.

possible unfolded conformations are equally likely and that folding is faster than oxidation. Answer: a. Low activity (~1%). (i) The protein is first denatured, forming mostly random structures; however, all disulfide bonds remain intact. (ii) Reduction causes all disulfide bonds to be broken. (iii) Exposure to oxygen causes the disulfide bonds to reform. They reform with random pairings, since the protein is unfolded and populating many random structures. (iv) The protein refolds, but only those molecules in which the disulfides formed correctly are active. b. High activity (~100%). (i) The protein is first denatured forming mostly random structures; however, all disulfide bonds remain intact. (ii) Reduction causes all disulfide bonds to be broken. (iii) Exposure to oxygen causes the disulfide bonds to reform, but since folding is faster than oxidation, the disulfide bonds reform correctly. c. High activity (~100%). (i) The protein is first denatured forming mostly random structures; however, all disulfide bonds remain intact. (ii) The protein refolds and since the disulfide bonds are already intact, full activity is restored. 13. A folded protein structure contains six ion pairs between lysine and glutamate. There are no other possible ion pairs in the protein. A chemical crosslinker forms a covalent bond between ion-paired lysine and glutamate sidechains. By analogy to the Anfinsen experiment, the following experiments are done:

a. folded protein → unfold with urea → remove urea to refold → add cross-linker → remove excess crosslinker → measure activity



b. folded protein → unfold with urea → add crosslinker → remove excess cross-linker → remove urea to refold → measure activity



The cross-linker does not by itself alter the activity of the protein when the correct ion pairs are formed. A protein with incorrect ion pairs cross-linked would be inactive. The activity measured at the beginning and end of experiment (a) is 100%. What percentage of the activity do you expect to observe at the end of experiment (b)? Assume that all unfolded conformations are equally likely. Answer: There are six ways of picking the first lysine and six ways of picking the first glutamate (6 × 6 = 36). There are five ways of picking the second lysine and five ways of picking the second glutamate (5 × 5 = 25). There are four ways of picking the third lysine and four ways of picking the third glutamate (4 × 4 = 16). There are three ways of picking the fourth lysine and three ways of picking the fourth glutamate (3 × 3 = 9). There are two ways of picking the fifth lysine and two ways of picking the fifth glutamate (2 × 2 = 4). There is one way of picking the last lysine and one way of picking the last glutamate (1 × 1 = 1). The total number of ways to pick is: 36 × 25 × 16 × 9 × 4 × 1 = 518,400.

14. Use the BLOSUM substitution matrix (Figure 5.11) to compute the sum of the substitution scores (Sij) and the overall likelihood ratio (L) of the following short alignments: a. PADKTN PEEKSA b. KFLASV ATWDPE Answer: a. Sequence: P A D K T N Sequence: P E E K S A Score: (7)+(–1)+(2)+(5)+(1)+(2) Sum of Scores = 12 Likelihood = 2(Score/2) = 2(12/2) = 64 b. Sequence: K F L A S V Sequence: A T W D P E Score: (–1)+(–2)+(–2)+(–2)+(–1)+(–2) Sum of Scores = –10 Likelihood = 2(Score/2) = 2(–10/2) = 1/32 = 0.3125 15. Based on the BLOSUM matrix, how much more likely is it that: a. tryptophan is substituted by a tyrosine than a tryptophan is substituted by a cysteine? b. sequence (i) DPKRFL is related to sequence (ii) EPKRFI than sequence (i) is related to sequence (iii) KGKRYA?

To answer this question, you must calculate the ratio of the likelihood ratios for each case. Explain the significance of higher likelihood.
 Answer: a. The score for W → Y substitution = 2; The likelihood = 2(Score/2) = 2(2/2) = 2. The score for a W → C substitution = (–2); The likelihood = 2(Score/2) = 2(–2/2) = 0.5. Lij/Lik = 2/0.5 = 4 times higher likelihood. This means that a W → Y substitution is more conservative than a W → C subsitution. b. The sum of scores for (i) → (ii) = 27; The likelihood = 11,585.2. The sum of scores for (i) → (iii) = 9; The likelihood = 22.6. Lij/Lik = 11,585.2/22.6 = 512 times higher likelihood. This means that (i) is more likely to be related to (ii) than to (iii).

16. Proteins known as cyclophilins catalyze proline cis-trans isomerization. A catalytic arginine residue is invariant in all cyclophilins. All other positions change residue identity in different cyclophilins despite the fact the variant proteins have the same overall fold and general catalytic activity. Explain, given the relationship between protein sequence and structure, how catalytic activity is



PROBLEMS and solutions retained even though most residues can change.
 Answer: The relationship between sequence and structure is asymmetric. Many proteins can have the same structure despite different (degenerate) sequences. However, sequences fold into only one structure (following the thermodynamic hypothesis). Only residues that perform specific functions, such as the catalytic arginine in cyclophilins and the histidine in helix F in globins are invariant.

17. Why is the sequence similarity generally higher when comparing two globins from mammals than when comparing a globin from a mammal and a globin from a plant? Answer: It is likely that the both the mammalian and plant globins derived from a single ancestral globin and have expanded through gene duplication. The mammalian globins share a common ancestor with each other that is more recent than the mammalian globins share with plant globins. 18. How might the tolerated variation in the hydrophobic core of the lambda repressor change if the hydrophobic core of wild type lambda repressor were more tightly packed? Answer: The protein would likely be less tolerant of mutations. Proteins undergo fluctuations, even in the hydrophobic core. They are not perfectly packed as interlocking pieces of a puzzle. This property allows for accommodation of differently shaped sidechains and for toleration of mutation.

3

Explain why the distribution of protein sizes has the periodicity that is seen in the diagram and estimate a value for x. Answer: A reasonable value for x is approximately 100–150. This is the average size of a protein domain (which are normally 50–200 residues). The periodicity is observed because proteins are modular and are expanded by addition of different domains. The peaks at 1x, 2x, and 3x derive from proteins containing 1, 2, or 3 domains. 21. The number of distinct protein folds is limited. Why might this be so? Approximately how many folds are there (hundreds, thousands, millions, or billions)? Answer: There are likely to be only thousands of folds because natural selection will favor protein sequences that will fold into stable structures. In order to fold into a stable structure, it is necessary to be built up of secondary structure elements and have a hydrophobic core. It is probable that there are a finite number of orientations and combinations of secondary structure elements that will produce a hydrophobic core of acceptable geometry to enable protein folding. 22. How many folds are represented below? Describe what CATH class each fold belongs to and how the secondary structure is arranged in each fold.

Q5.12 (seq_struct_70_v1)

19. What characteristics define a protein domain?
 Answer: Domains have a distinct topology and a self-contained hydrophobic core. They generally contain 50–200 residues.

A

B

C

D

E

F

20. The diagram below shows the size distribution for globular proteins produced by the bacterium E. coli.

frequency of occurrence

Q5.10 (seq_struct_71_v1)

Answer: There are four folds. A and E have the same Rossmannlike mixed α/β fold. B has a unique mixed α/β fold. C has an all β sheet structure arranged in a barrel. D and F have a four helix bundle.

1x 2x 3x number of residues in the protein

23. A threading program is used and it gives two possible predicted folds for a particular sequence. They differ mainly in the placement of a single helix. In predicted fold (A), the helix is entirely within the hydrophobic core. In predicted fold (B), the residues of the helix are

4

Chapter 5: Evolutionary Variation in Proteins mostly exposed to solvent. The sequence of the helix is LIVFLAIL. Explain how the 3D-1D profile method could be used to distinguish between the two possible folds. Answer: All of the residues in the helix are hydrophobic and have positive scores as α helical buried positions. In contrast, these residues have negative scores for exposed α helical positions. Thus, fold A correctly predicts that the α helix should be part of the hydrophobic core of the protein.

24. What structural features of the Rossmann domain enable it to bind nucleotides? Answer: The negatively charged phosphate groups of the nucleotides interact with the P loop at the positive pole of the helix dipole generated by the first α helix of the Rossmann fold. 25. Both thioredoxin reductase and glutathione reductase use fused FAD- and NADPH-binding domains and dimerization to accomplish their cellular functions. Which structural feature likely evolved first, the fused domains or dimerization? Answer: The fused FAD and NADPH domain structure is conserved between both structures. However, the relationship between the individual subunits of the dimers is different for each protein (see Figure 5.43). This implies that domain structure evolved before the split of individual thioredoxin and glutathione reductase lineages and that dimerization evolved later.

The MOLECULES of LIFE Physical and Chemical Principles

Solutions Manual Prepared by James Fraser and Samuel Leachman

Chapter 6 Energy and Intermolecular Forces

Problems True/False and Multiple Choice 1. Which of the following properties are extensive (choose all that apply): temperature pressure amount of heat released density energy molarity number of moles mass volume

5. Energy is a good indicator of the direction of spontaneous change for macroscopic mechanical objects but not for molecular processes. True/False

True/False 3. A protein is negatively charged, but it binds a negatively charged small molecule faster than it binds a positively charged small molecule. The most reasonable explanation for this phenomenon is: a. Even though the protein is negatively charged overall, electrostatic focusing effects provide a pathway for a negatively charged molecule to enter the active site. b. The presence of water molecules screen the electrostatic effects. c. The negatively charged molecule makes stronger hydrogen bonds than the positively charged molecule. d. Charged proteins normally bind substrates with the same overall charge. 4. The enthalpy change for a process is equal to the heat transferred to the system under which of the following conditions?

a. The potential energy of an ideal gas is equal to its kinetic energy. b. The potential energy of an atom is the work done in moving the atom from infinity to its present position. c. Potential energy is always absolutely conserved. d. The potential energy of a system always increases. e. Potential energy is an intensive function. 7. The type of function that best describes the energy of a hydrogen bond as the distance between the hydrogen atom and the acceptor atom varies between 1.5 and 2.5 Å, is:

Problems, multiple choice, Q7 (forces_53_v1) (a) (b)

a. constant volume b. constant pressure

energy (increasing)

2. A piston containing an ideal gas expands isothermally from 7 atm pressure to 2 atm pressure. The energy of the system (that is, the contents of the piston) remains constant during this process.

6. Which of the following statements about potential energy is true?

energy (increasing) 1.5

(c)

distance (Å)

2.5

1.5

(d)

distance (Å)

2.5

ing)

The Molecules of Life by John Kuriyan, Boyana Konforti, and David Wemmer © Garland Science ing)

a. b. c. d. e. f. g. h. i.

c. constant temperature d. reversible expansion e. expansion at a constant pressure followed by reversible compression

energy (incr

energy (incr

2

∆U = 0 (all isothermal expansions) w = –nRT ln

Chapter 6: Energy and Intermolecular Forces 1.5

2.5

distance (Å)

distance (Å)

2.5

energy (increasing) 1.5

2.5

distance (Å)

= –nRT ln

1.5

distance (Å)

2.5

Answer: Graph shown in a.

Fill in the Blank 8. A reaction that releases energy is called an ________ reaction. Answer: exothermic 9. The first law of thermodynamics states that the total energy of the system and surroundings remains __________. Answer: constant 10. Electrostatic interactions are governed by __________’s law Answer: Coulomb 11. An ion pair with lysine can be formed by the sidechain of _________ or ____________. 12. The energy for van der Waals interactions approaches _____ as the distance between atoms goes to zero and approaches _____ at infinite distance.

Answer: q = 0 (isolated system) w = 0 (no external pressure) ΔU = 0 (all isothermal expansions) 16. One mole of a different ideal monatomic gas is initially at 300 K and 1010 J•L–1 pressure inside a cylinder with a frictionless piston . Calculate ΔU (the change in energy), q (the heat transferred to the system), and w (the work done by the system) when the expansion is isothermal and against a constant pressure of 101 J•L–1. Answer: For work against constant pressure, w = –PEXT∆V = –PEXT (V2 – V1)

13. The force arising from a noncovalent interaction between two atoms is always ______ at the energy minimum. Answer: zero

Quantitative/Essay 14. One mole of an ideal monatomic gas is initially at 300 K and 1010 J•L–1 pressure inside a cylinder with a frictionless piston. (Note that 1 J•L–1 = 103 Pa.) The gas expands until the pressure is 101 J•L–1. Calculate ΔU (the change in energy), q (the heat transferred to the system), and w (the work done by the system) when the expansion is isothermal and reversible. Answer:

1 mol × 8.31 J•K–1•mol–1 × 300 K – 101 J•L–1 1 mol × 8.31 J•K–1•mol–1 × 300 K 1010 J•L–1

= –2240 J w = –2.24 kJ q = –w = 2.24 kJ 17. The complete oxidation of one mole of a sugar produces carbon dioxide and water. 2000 kJ of heat is transferred from the system to the surroundings. The rearrangement of bonds as 0.5 moles of the sugar are oxidized generates heat in an open test tube (101 J•L–1 pressure and 300 K temperature). What is the change in internal energy of the system (ΔU)? What is the change in enthalpy of the system(ΔH)? Answer:

ΔU = q + w. There is no work done. Therefore: ΔU = q = –2000 kJ•mol–1 × 0.5 mol released by the system

ΔU = –1000 kJ ΔH = ΔU + PΔV. Since there is no volume change, ΔV = 0 and ΔH = ΔU = –1000 kJ.

∆U = 0 (all isothermal expansions)

= –nRT ln

nRT nRT – P1 P2

= –101 J•L–1

Answer: infinity kcal•mol–1, 0 kcal•mol–1

w = –nRT ln

P2

15. One mole of an ideal monatomic gas is initially at 300 K and 1010 J•L–1 pressure inside a cylinder with a frictionless piston. The cylinder is an isolated (adiabatic) system. The gas expands against zero external pressure. When the volume expands to 24.7 L, a peg stops the piston. Calculate ΔU (the change in energy), q (the heat transferred to the system), and w (the work done by the system) when the expansion is isothermal and reversible.

= –PEXT

Answer: aspartic acid, glutamic acid



V1 P1

10 = –1 mol × 8.31 J•K–1•mol–1 × 300 K × ln 1 = –5740 J w = –5.74 kJ q = –w = 5.74 kJ

(d)

energy (increasing)

(c)

1.5

V2

V2

V1 P1 P2 –1

–1

10

18. A system is maintained at thermal equilibrium (at the same temperature) with its surroundings and has an

= –1 mol × 8.31 J•KKuriyan, •molBoyana × 300 K × lnand David Wemmer © Garland Science The Molecules of Life by John Konforti, 1 = –5740 J w = –5.74 kJ



PROBLEMS and solutions enthalpy of 50 kJ. It has 100 kJ of heat transferred to it, which causes it to expand against a constant pressure of 1 atm. It is then compressed back to its initial volume. All steps are at constant temperature. What is its final enthalpy? Answer: 50 kJ. Because all steps are at constant temperature, ΔU = 0, and because the system undergoes no volume change, ΔV = 0 and ΔH = ΔU. The heat initially put into the system to change its state is balanced by the work put in to return it to its initial state. The enthalpy doesn’t change from its initial value because enthalpy is a state function.

21. Consider a protein with different energy levels for electronic transitions and molecular vibrations. What are the approximate energy differences between the levels for each type? To what extent do the populations of levels within each type explain why the heat capacity of a protein is highest at the unfolding point? Answer: Electronic energy levels are far apart in energy (103–104 kJ•mol–1). These are not generally involved in changes in heat capacity when a protein is unfolding, since the energy absorbed is not enough to populate the higher energy levels. Translational energy level spacings are very small (10–14 kJ•mol–1). The translational movement of a protein does not greatly change depending on whether it is folded or not. These levels are present during unfolding, but do not contribute the increased heat capacity since both folded and unfolded proteins can readily access many translational energy levels.

19. How much kinetic energy does a system containing 3 moles of an ideal gas at 300 K possess? What is the heat capacity at constant volume? How much heat would need to be transferred to the system to raise the temperature by 15°C? Answer: 3 K.E. =  nRT 2 3 = × 3 mol × 8.31 J•K–1•mol–1 × 300 K 2 = 11,218.5 J = 11.2 kJ 3 CV =  nR 2 3 = × 3 mol × 8.31 J•K–1•mol–1 2 = 37.4 J•K–1 q = CV × ΔT = 37.4 J•K–1 × 15 K = 561 J

Molecular vibration energy levels are moderate (1–10 kJ•mol–1). A folded protein that is heated will adjust its arrangements and absorb energy through fluctuations breaking many weak interactions. These rearrangements are highest during unfolding, explaining why the maximum heat capacity occurs at the unfolding temperature. Access to more molecular vibration energy levels in the unfolded state explains the higher heat capacity of unfolded proteins relative to folded proteins.

20. Below are two graphs of heat capacity changes in a differential scanning calorimeter experiment. In the experiment, protein molecules in solution are unfolded by increasing the temperature. In both graphs, temperature increases from left to right. Which graph correctly characterizes the heat capacity changes of a protein as it is heated? Which characteristics in the Problems, long Q7 (forces_54_v1) graph lead to thisanswer, conclusion? 9.0

22. The only two accessible conformations of a protein differ by 2 kJ•mol–1. What percentage of protein molecules will be in the higher energy state: a. at 350 K? b. at 270 K? Answer: a. –(Uhigh–Ulow) Phigh = e RT = e Plow Phigh

9.0 CP (mJ•°C–1)

CP (mJ•°C–1)

Phigh = 6.0

3.0

0

temperature

6.0

b.

3.0

Phigh Plow

0

3

Answer: The graph on the left is correct. Both graphs correctly indicate that the heat capacity at low temperatures, when the protein is folded, is lower than the heat capacity at high temperatures, when the protein is unfolded. However, the heat capacity of a protein peaks at its melting temperature because relatively more energy goes into breaking intramolecular interactions. It then decreases at higher temperatures when fewer unbroken interactions persist. Only the left graph indicates this correctly.

8.31 J•K–1•mol–1 × 350 K

= 0.503

Plow = 33.5 % Phigh +1 Plow –(Uhigh–Ulow) RT

=e

Phigh =

temperature

–2000 J•mol–1

=e

–2000 J•mol–1 8.31 J•K–1•mol–1 × 270 K

= 0.410

Phigh Plow Phigh Plow

= 29.1 %

+1

23. Consider a ribosome translating a codon to an amino acid. Compared to a perfect pairing, imagine that a codon–anticodon pairing with one incorrect base pair is 10 kJ•mol–1 higher in energy, two incorrect base pairs is 20 kJ•mol–1 higher, and three incorrect base pairs is 30 kJ•mol–1 higher.

How many ways are there of making a one-, two-, or three-base-pair mismatch? What is the partition

The Molecules of Life by John Kuriyan, Boyana Konforti, and David Wemmer © Garland Science

4

Chapter 6: Energy and Intermolecular Forces function of the codon–anticodon system at 300 K? If energy difference is the only consideration, what percent of decoding events occur without an error (no mismatches) at 300 K?





Answer: a. 1 qiqj U= ε rijÅ

Hint: There are three incorrect single-base mismatches in each of the three positions (A, B, and C)(3 × 3 = 9); 3 × 3 = 9 incorrect two-base mismatches in three possible combinations (AB, AC, and BC) of positions (9 × 3 = 27); and 3 × 3 × 3 = 27 incorrect three-base mismatches.

b.

Answer: Q = ∑e

–Ui kBT

–Uzeromismatches

=e

–Utwomismatches

kBT

kBT

+9×e

kBT

i

+ 27 × e

–Uthreemismatches

+ 27 × e =

–Uonemismatch

–Ui e kBT +

–10 kJ–1•mol–1

9×e

kBT

–20 kJ–1•mol–1

+ 27 × e

kBT

–30 kJ–1•mol–1

+ 27 × e

kBT

–Uo

Ntotal e kBT Q

=

100 e

–0 kBT

1.1724

=

100 1.1724

1 –1 80 3.5

× 1390 kJ•mol–1 = –4.96 kJ•mol–1

U=

1 qiqj ε rijÅ

× 1390 kJ•mol–1

U=

1 –1 2 3.5

× 1390 kJ•mol–1 = –199 kJ•mol–1

26. A particular sidechain has a dihedral angle for which the energy is governed by the following equation: Udihedral = 2 × cos (3 × angle) kJ•mol–1

= 85.3 %

Answer:

a. What is the energy of a sidechain with an angle of 60°? b. With an angle of 90°? c. What is the relative population of sidechains with an angle of 60° versus 90° at 300 K? Answer:

Therefore, approximately 85.3% of all amino acids are translated faithfully. The redundancy built into the genetic code and other energetic and kinetic considerations raise this number considerably. 24. The energy of a covalent bond as a function of interatomic distance, r, is described by the following equation: U (r ) = 2000 ( r – 1.5 ) 2 In this equation, energy is expressed in units of kJ•mol−1 and distance in Å. What is the equation that describes the force with respect to position? What is the energy when the interatomic distance is 2 Å? At the same value of the interatomic distance, what is the force and is the force pushing the bonded atoms closer together or further apart?

a.

Udihedral = 2 × cos(3 × 60) kJ•mol–1 = –2 kJ•mol–1

b.

Udihedral = 2 × cos(3 × 90) kJ•mol–1 = 0 kJ•mol–1

c.

P60

RT

8.31 J•K–1•mol–1 × 300 K

=e

= 2.23 times as many

 rmin   r 

2

U (r ) = ε  



where ε = 2 kJ•mol−1. Calculate the value of rmin if the energy at 3 Å is 1.5 kJ•mol−1.

U(s) = 2000 kJ•mol–1× Å–2 (2 Å – 1.5 Å)2 = 2000 kJ•mol–1× Å–2 (0.25 Å2) = 500 kJ•mol–1

U(r) = ε

25. A salt bridge between an arginine and a glutamic acid has the two bridging atoms 3.5 Å apart. Assume that these bridging atoms have elementary charges of +1 and –1, respectively.

–(–2000 J•mol–1 – 0 J•mol–1)

27. For two atoms, separated by distances between 0 and 4 Å, the van der Waals energy is well described by the following equation:

Answer:

The force is attractive. If one atom were held fixed, the force on the second atom would accelerate it in the negative direction, towards the first atom. Also, notice that the ideal bond length is 1.5 Å, so atoms spaced 2 Å apart would be attracted.

–(U60–U90)

=e

P90

F(r) = –dU/dr = –4000 kJ•mol–1 × Å (r – 1.5 Å)

F(2) = –4000 kJ•mol–1 × Å–2 (2 Å – 1.5 Å) = 2000 kJ•mol–1× Å

× 1390 kJ•mol–1

U=

kBT

= 1 + 9 (0.018) +27 (0.0003) + 27 (0.000006) = 1 + (0.162) + (0.0081) + (0.0002) = 1.1724 N0 =

What is the energy if the residues are a) on the protein surface (ε = 80) and b) in the core of the protein (ε = 2)?

rmin

2

r

1.5 kJ•mol–1 = 2 kJ•mol–1 0.75 =

rmin

2

3Å

rmin2 9 Å2

6.75 Å2 = rmin2 rmin = 2.6 Å 28. A calorimeter is used to measure the heat capacity of two molecules, A and B. Molecule A is a simple small molecule, whereas molecule B is a complex polymer (like DNA). Define heat capacity and explain why molecule B is likely to have a higher heat capacity.

The Molecules of Life by John Kuriyan, Boyana Konforti, and David Wemmer © Garland Science



PROBLEMS and solutions

5

Answer: The heat capacity is the amount of heat required to raise the temperature of the system one Kelvin. Molecule B likely has a higher heat capacity because it has more internal energy modes (vibrational and conformational modes) than the simpler molecule. These low energy transitions will allow it absorb more heat per change in temperature. 29. Water forms hydrogen bonds with proteins. How might these hydrogen bonds alter the ability of a protein to undergo conformational changes in water versus in the gas phase? Answer: Water weakens the effective strength of hydrogen bonds in proteins by competing for hydrogen bond donors or acceptors. With stronger hydrogen bonds, as in the gas phase, certain conformational changes will require more energy and rarely occur spontaneously. In water, however, the relative strength of protein hydrogen bonds is closer to kBT and conformational changes are more likely to occur.

The Molecules of Life by John Kuriyan, Boyana Konforti, and David Wemmer © Garland Science

The MOLECULES of LIFE Physical and Chemical Principles

Solutions Manual Prepared by James Fraser and Samuel Leachman

Chapter 7 Entropy

Problems True/False and Multiple Choice 1. It is easier to predict the bulk behavior of a small number of molecules than a large number of molecules. True/False 2. Two sets of molecules are mixed into a system at time zero. After 10 seconds, equilibrium has been reached. At what time was the entropy of the system maximal? a. 10 seconds (equilibrium) b. 5 seconds (half the time it takes to reach equilibrium) c. immediately upon mixing of the two sets of molecules d. prior to mixing of the two sets of molecules 3. Consider a coin with two sides (H = heads; T = tails). The probability of observing HHHHHTTTTT is equal to the probability of observing HTHTHTHTHT. True/False 4. On a 10-sided die, with a side for each number from 1 to 10, the probability of rolling a 5 is: a. b. c. d. e.

5/10 1/10 1/9 5/51 1/6

statements correctly characterizes the system? a. It contains only blue molecules on the left side. b. It contains only red molecules on the left side. c. It contains a mixture of red and blue molecules throughout the system. d. It contains only blue molecules at the bottom of the system. e. It has an equal number of red and blue molecules on each side. 7. A state corresponds to many different microstates. True/False

Fill in the Blank 8. The work done in a near-equilibrium expansion of an ideal gas is _____________ than for a nonequilibrium expansion. Answer: greater 9. When the volume of a system increases, its multiplicity ________. Answer: increases

5. An isolated molecular system exists in two states of equal energy. State A has high multiplicity, whereas State B has low multiplicity. Without any external agents, a system in State B will spontaneously convert to State A. True/False

10. The log of the multiplicity of the system is an __________________ property of the system. Answer: additive/extensive 11. The combined entropy of the system and surroundings always _______________ for a spontaneous process. Answer: increases

6. A system is divided into two halves separated by a removable divider. Initially, with the divider in place, the left half has only red molecules and the right half has only blue molecules. The divider is removed and equilibrium is reached. Which of the following

12. A drop of dye is added to a container of solvent. When equilibrium is reached the concentration of dye will be ____________ within the container. Answer: uniform

The Molecules of Life by John Kuriyan, Boyana Konforti, and David Wemmer © Garland Science

2

Chapter 7: Entropy

Quantitative/Essay 13. A coin is weighted deliberately so that the probability of tossing heads is twice the probability of tossing tails. What is the probability of tossing three heads in a row? What is the probability of tossing three tails in a row? What is the relative probability of tossing three heads versus three tails?

16. Calculate the entropy of the system depicted below. There are four types of molecules (that is, X, O, Y, and Z) that can be arranged in any way in the available boxes. Problems, long answer, (Hint: Use Equation 7.2.2.) Q3 (entropy_28_v1)

X

Answer: increases 2/3 chance of heads, 1/3 of tails. 3 heads in a row = (2/3)3 = 8/27 = 0.296. 3 tails in a row = (1/3)3 = 1/27 = 0.037. Relative probability = (8/27)/(1/27) = 8 times more likely.

Y Z

14. Shown below is a portion of Pascal’s triangle, with the tenth row filled in. 1

10

45

120 210 252 210 120 45

10

a. 1Fill 10 in the eleventh row. 45values 120 for 210the 252 210 120 45What 10 simple 1 rule can you use to fill in these values?



b. Using Pascal’s triangle, calculate how much more likely1 it is10to get and 210 six tails a series 45 five 120 heads 210 252 120 in45 10 of 1 11 coin tosses than getting four heads and seven tails. 1

11

1 1

55

10 11

165 330 462 462 330 165

45 55

120 210 252 210 120

55

45

165 330 462 462 330 165

11

10 55

1

1 11

1

a. Simple rule: The numbers at the edges are always 1. All the other numbers are the sums of the two numbers diagonally above. b. According to Pascal’s triangle, the multiplicity for getting five heads is 462, and for four heads it is 330. So, the ratio of the probabilities is: 462 330

Y

O

Y

O

O

Z

O

1



Answer:

X

Answer: The multiplicity of the system (W) is given by M!/(X!O!Y!Z!(M – X – O – Y – Z)!). W = 25!/(2!3!4!2!14!) = 308,897,820,000 The entropy is given by kB lnW = kB ln(308,897,820,000) = 26.46 kB. 17. Consider the systems below consisting of different numbers of identical molecules (indicated by the symbol X) and equal-sized grid boxes that are either empty or occupied by a molecule. Which system has a higher entropy? What is the difference in entropy Problems, longB?answer, Q4 (entropy_29_v1) between A and

A.

X X X

= 1.4

X

15. Consider the following three cases of grid boxes. Molecule X can move between and occupy any box. Calculate the multiplicity for each case. Problems, long answer,and Q2entropy (entropy_27_v1)

a.

X

b.

X

c.

X

X

X

B.

X

X

X X

X X

X

X X

Answer: For each let X be the number of X molecules and M be the total number of grid spaces. For each the multiplicity can be calculated by M!/(X!(M – X)!). a. W = 6!/(4!(6 – 4)!) = 15; S = kB ln(W) = 2.71 kB b. W = 6!/(2!(6 – 2)!) = 15; S = 2.71 kB c. W = 6!/(3!(6 – 3)!) = 20; S = 3.00 kB

Answer: 16! WA = 12!4! WB =

16! 6!10!

5×6 The Molecules of Life by John Kuriyan, Boyana Konforti, and David Wemmer © Garland WAScience WB

=

12 × 11

= 0.227

X

WA =



WB = WA WB

=

16! 12!4!

PROBLEMS and solutions

16! 6!10! 5×6 12 × 11

= 0.227

The multiplicity of the top (A) is lower; therefore, (B) has a higher entropy. The difference in entropy is WA kB ln = kB ln (0.227) = –1.48 kB WB 18. A bag contains 20 coins, each marked by the oneletter code for an amino acid. What is the probability of drawing each of the three large amino acids (Y, W, and F) once, in any order, without returning the coins to the bag after each draw? What is the probability if coins are returned to the bag after each draw?

What is the probability of drawing the three letters out in exactly the order of hydrophobicity (a unique ordering) without returning any coins to the bag? Answer: Large amino acids: Drawing each without replacement = 3/20 × 2/19 × 1/18 = 0.00088. With replacement = 3/20 × 2/20 × 1/20 = 0.00075. Hydrophobicity: Probability of drawing the first amino acid = 1/20, second = 1/19, third = 1/18. 1/20 × 1/19 × 1/18 = 1/6840 = 1.46 × 10–4.

Problems, long answer, Q6 following (entropy_30_v1) 19. What is the multiplicity of the system?

3

X

X

O

O

X O

X

21. In a near-equilibrium expansion, 15 kJ of work are done by a system at a constant temperature of 300 K. To reach the same state, 5 kJ of work are done in a nonequilibrium process. What is the entropy change for the two processes? Answer: Entropy is a state function, so the change will be the same in both cases: ∆S = qrev/T = 15 kJ/300 K = 50 J•K–1 22. Considering the following system at two time points, A Problems, long answer, Q9 (entropy_31_v1) and B.

A.

X

X

B.

X

X

X X

X

X

X

X

The system is divided by a movable partition. The molecules (X) cannot move across the partition, but they can move between the grid boxes on the same side of the partition. At which time point does the system have the higher multiplicity? At which time point is the system closer to equilibrium? Answer: At time point A: The multiplicity of the left side is: M!/(X!(M – X)!) = 4!/3! = 4. The multiplicity of the right side is: M!/(X!(M – X)!) = 6!/(2!4!) = 15. The total multiplicity = mulitpicityright × multiplicityleft = 4 × 15 = 60. At time point B: The multiplicity of the left side is: M!/(X!(M – X)!) = 5!/(3!2!) = 10. The multiplicity of the right side is: M!/(X!(M – X)!) = 5!/(2!3!) = 10. The total multiplicity = mulitpicityright × multiplicityleft = 10 × 10 = 100. The multiplicity of time point B is higher. Since the multiplicity of time point B is higher, it is closer to equilibrium than time point A.

O

23. Consider a coin that is weighted so that it is twice as likely to come up heads as tails. Answer the following questions using the Gaussian distribution description for the results of a series of coin tosses:

O

If an additional five empty grid boxes are added, does the multiplicity of the system increase or decrease? Answer: The multiplicity is M!/(X!O!(M – O – X)!) = 25!/(4!5!16!) = 257,414,850. Adding additional volume always increases the multiplicity of the system. 20. System A has a multiplicity of 15, whereas System B has a multiplicity of 12. What is the total entropy of Systems A and B? Answer: The entropies of the two parts will be additive. Therefore we can calculate them separately. SA = kB ln(15) = 2.71 kB; SB = kB ln(12) = 2.48 kB. Total entropy = 5.19 kB.

a. What is the most likely outcome of a series of 2000 coin tosses using this biased coin? b. What is the relative probability of getting 1500 heads versus 1000 heads in a series of 2000 coin tosses using this coin? Express your answer as a power of 10. Answer: a. The mean value (μ, most likely outcome) is given by: μ = Mp where M is the number of events and p is the probability of a favorable outcome. The probability of obtaining heads is twice as large as obtaining tails, so we can write: p 2

p+

=1

3 The Molecules of Life by John Konforti, and David Wemmer © Garland Science p =Boyana 1 ⇒ Kuriyan, 2

2

4

Chapter 7: Entropy p p+ =1 2 This change would not occur spontaneously because the entropy is lower in the final state.

3 p=1 2 2 ⇒ p= = 0.6667 3 Hence, the mean value is: 0.6667 × 2000 = 1333

⇒

b. To calculate the relative probabilities we need to use the Gaussian distribution:

 –(x – µ)2  P(x) = exp  1/2  2σ  2σ2  1

The standard deviation, σ, is given by: 2000 × 0.6667 ×0.3333 = 21.1

M × p × (1 – p) =

The ratio of probabilities is given by:

– (1500 – 1333.3)2 + (1000 – 1333.3)2 P(1500) = exp   P(1000) 2 × 21.082   – 27778 + 111111  = exp   889   = exp(93.75) = 1040 24. A system consists of molecules that convert between two colors (green and yellow). There are spaces for 10,000 molecules, but there are only 6000 molecules in the system. The system starts in a state with 2000 green molecules and 4000 yellow molecules. What is the entropy of the system? (Hint: Use Stirling’s approximation.) Answer: W=

Spaces!

=

Green!Yellow!Empty!

S = kB ln

Answer: V2 V1 = –293 K × 8.31 J•K•mol–1 × ln (0.250)

w = –nRT ln

where x is the number of favorable outcomes.

σ=

26. How much work is done in compressing one mole of an ideal gas from a starting volume of 1 L to a final volume of 250 mL at a constant temperature of 293 K? What is the change in entropy? Assume that the process occurs near-equilibrium (reversibly).

10000! 2000!4000!4000!

= 3375.3 J –w –3375.3 J = 11.5 J•K–1 ∆S = = T 293 K 27. If the increase in entropy indicates the direction of spontaneous change, how can a system ever undergo a process that results in a decrease in the entropy of the system? Answer: The only way for a system to decrease its entropy spontaneously is to couple to an external agent that is increasing its entropy. The external agent must provide an external entropy increase that ensures that the summed entropy of the system plus the external increases. 28. In our calculations, why do we work with the natural log of the multiplicity? Answer: ln(W) is an additive property of the system, and is more convenient to work with than W because its numerical value remains manageable as the number of molecules in the system increases.

10000! 2000!4000!4000!

= kB ln((10000 ln10000 – 10000) – (2000 ln 2000 – 2000 + 4000 ln 4000 – 4000 + 4000 ln 4000 – 4000) = 10549.2 kB 25. The system in Problem 24 converts to a final state with 6000 yellow molecules and no green molecules. Assuming no energy difference between yellow and green molecules, would this conversion occur spontaneously? Answer: W=

Spaces! Yellow!Empty!

S = kB ln

=

10000! 6000!4000!

10000! 6000!4000!

= kB ln((10000 ln10000 – 10000) – (6000 ln 6000 – 6000 + 4000 ln 4000 – 4000) = 6730.1 kB The Molecules of Life by John Kuriyan, Boyana Konforti, and David Wemmer © Garland Science

The MOLECULES of LIFE Physical and Chemical Principles

Solutions Manual Prepared by James Fraser and Samuel Leachman

Chapter 8 Linking Energy and Entropy: The Boltzmann Distribution

Problems True/False and Multiple Choice 1. For a system converting from state 1 to state 2: (kBlnW2 − kBlnW1) = qrev/T. True/False 2. If the second-lowest energy level is separated from the ground state by 0.5kBT, then the second-lowest energy level will not be occupied appreciably. True/False 3. There are many equivalent microstates corresponding to a particular energy distribution. True/False 4. Which of the following definitions of entropy are equivalent for a large system: a. b. c. d.

probabilistic definition thermodynamic definition statistical definition all of the above

8. The Boltzmann distribution describes the energy of molecules at _________________. Answer: equilibrium 9. The Boltzmann constant is the gas constant (R) divided by ____________ . Answer: Avogadro’s number 10. Kinetic energy is due to the __________ of atoms. Potential energy is due to the ___________ of atoms. Answer: motion, configuration/relative positions 11. The direction of spontaneous change is governed by an increase in ____________, thus explaining how energy can be transferred “uphill,” that is from a system of lower total energy to one with higher total energy. Answer: entropy/multiplicity

5. The probabilistic definition of entropy is more accurate than the statistical definition for small systems. True/False 6. After spontaneous heat transfer between systems, the overall multiplicity is: a. b. c. d.

Fill in the Blank

lower than it was before the transfer of heat zero maximized minimized

7. The partition function (Q) is needed to describe: a. The volume of two gases on either side of a partition. b. The extent to which many different energy levels are occupied. c. How the statistical definition of entropy is equivalent to the probabilistic definition of entropy. d. The relationship between work and pressure.

12. If two systems of different temperatures are brought together, there will be a net transfer of heat from the system of __________ temperature to the system of _________ temperature. Answer: higher, lower

Quantitative/Essay Assume kB = 1 J•K–1 for all problems. 13. A system starts with a multiplicity of 2000. Two kJ of heat are transferred into the system reversibly at 298 K. What is the multiplicity now? Answer: Initial entropy = kB ln W = 1 × ln 2000 = 7.6 J•K–1 Change in entropy = 2000 J/298 K = 6.7 J•K–1 Final entropy = 14.3 J•K–1 Final multiplicity = eS/kB = 1.64 × 106

The Molecules of Life by John Kuriyan, Boyana Konforti, and David Wemmer © Garland Science

2

Chapter 8: Linking Energy and Entropy: The Boltzmann Distribution

14. A system starts with a multiplicity of 1099. Heat is transferred out of the system reversibly at 298 K such that the final multiplicity is 105. How much heat was lost? Answer:

∆S = kB ln

Wfinal Winitial

= kB ln

105 1099

qrev = T∆S = (298 K)(–216

= –216

J•K–1)

J•K–1

= –64.5 kJ

64.5 kJ of heat are lost. 15. A system of noninteracting atoms has a positional multiplicity of 120 and an energetic multiplicity of 5. What is the total multiplicity? What are the positional, energetic, and total entropies of the system? Answer: Total multiplicity = positional multiplicity × energetic multiplicity = 120 × 5 = 600 Positional entropy = kB ln 5 = 1.6 J•K–1; energetic = kB ln 120 = 4.8 J•K–1 Total entropy = positional + energetic entropy = 1.6 + 4.8 = 6.4 J•K–1 16. A system with 1 mole of an ideal gas changes its volume from 10 L to 1 L by isothermal compression. What is the change in entropy of the system? What is the change in the entropy associated with the energy distribution? (Use R = 8.31 J•K–1•mol–1.) Answer: The configurational entropy change is given by: ∆S = nR ln (V2/V1) = 1 mol × 8.31 J•K–1•mol–1 × ln (1 L/10 L) = –19.1 J•K–1 For the isothermal expansion of an ideal gas there is no change in energy. Thus, the energy distribution is unchanged by the process, and the associated entropy is also unchanged. 17. Consider a system with five molecules and three energy levels. The energy levels are such that a molecule at energy level 1 contributes 1 J of energy to UTOTAL, a molecule at energy level 2 contributes 2 J of energy to UTOTAL, and a molecule in energy level 3 contributes 3 J of energy to UTOTAL. Which has more internal energy, problems, Q17 (eng_ent_34_v1) state Amultiple or state choice, B?

state A

state B 3 2 1

3 2 1

Answer: For State A UTOTAL = ΣNiUi = 3 × N3 + 2 × N2 + 1 × N1 = 3 × 1 + 2 × 3 + 1 × 1 = 10 For State B UTOTAL = ΣNiUi = 3 × N3 + 2 × N2 + 1 × N1 =3×0+2×2+1×3=7 Therefore, system A has greater energy.

18. Consider the system and the two states from Problem 17. Which state has greater multiplicity? Answer: For State A: W = 5!/(1!3!1!) = 20 For State B: W = 5!/(2!3!) = 10 Therefore, system A has greater multiplicity. 19. A system has 100,000 molecules at energy level 1, 10,000 molecules at energy level 2, and 1000 molecules at energy level 3. What is the entropy of the system? (Hint: Use the probabilistic definition.) Answer: Total molecules = Nt =111,000 Probability of level 1 = N1/Nt = 100,000/111,000 = 0.90 Probability of level 2 = N2/Nt = 10,000/111,000 = 0.09 Probability of level 3 = N3/Nt = 1000/111,000 = 0.009 Entropy = –NtkB Σpi ln pi = –111,000 J•K–1 × (0.90 ln 0.90 + 0.09 ln 0.09 + 0.009 ln 0.009) = –111,000 J•K–1 × (–0.094 – 0.217 – 0.042) = –111 kJ•K–1 × (–0.353) = 39 kJ•K–1 20. A system with 100,000 molecules has two energy levels (A and B). At first, the two energy levels are populated equally. After a reversible process, energy level A is populated by 65% of the molecules and the system is at 293K. a. What is the difference in energy between the two levels? b. How much heat was added or removed from the system? c. What is the change in entropy? Answer: a. P(B) P(A) 0.35 0.65

 –∆G    RT 

= exp 



= exp 

–∆G

 

 (293 K)(8.31 J•K–1•mol–1)

⇒ ∆G = 1.5 kJ•mol–1 b. Initial entropy = –NtkB Σpi ln pi = –100,000 J•K–1 × (2 × 0.5 ln 0.5) = 69.3 kJ•K–1 Final entropy = –100,000J•K–1 × (0.65 ln 0.65+0.35 ln 0.35) = 64.7 kJ•K–1 The change in entropy (∆S) is –4.57 kJ•K–1 ∆S = qrev/T; qrev = ∆S × T = –4.57 kJ•K–1 × 293 K = –1339 kJ c. See above. As a check, we can verify that heat is removed from the system because the entropy of the system decreases during this process. 21. Two systems, A and B, are placed in thermal contact and isolated from the rest of the world. Shown next are three possible microstates for the combined system. Two of these are valid microstates for the combined system, and one is not. Identify the inconsistent microstate and explain why it does not belong with the other two. (Hint: Consider Figure 8.15.)

The Molecules of Life by John Kuriyan, Boyana Konforti, and David Wemmer © Garland Science



8Q22 problems, calculations, Q21 (eng_ent_35_v1)

PROBLEMS and solutions

max

5 4 3 2 1 0

log (W )

energy

microstate 1 system A system B

energy

microstate 2 system A system B

energy

5 4 3 2 1 0

23. What is the entropy of the following energy distribution? What is the multiplicity? N3 = 200 molecules N2 = 1100 molecules N1 = 10,500 molecules Answer: We first compute the entropy using the probabilistic definition. Initial entropy = –NtkB  Σpi ln pi = –11,800 J•K–1 × (0.017 ln 0.017 + 0.093 ln 0.093 + 0.89 ln 0.89) = –11,800 J•K–1 × (–0.394) = 4651 J•K–1 S = kB ln W; W = eS/kB = e4651 ~ 102020.

microstate 3 system A system B energy

3

5 4 3 2 1 0

Answer: Microstate 3 is inconsistent with the other two. Because the two systems are isolated from the rest of the world, the total energy must be observed. Microstates 1 and 2 have the same total energy, but microstate 3 does not. 22. A system has two types of molecules (black and gray) that interconvert. The log of the multiplicity of energy of the two types of molecules individually versus the total system energy is plotted below. Plot the log of the total multiplicity of energy and indicate where the maximum problems, Q10 (eng_ent_36_v1) value forcalculations, the combined multiplicity occurs.

24. A system has a partition function of 1.3 at 293 K. At that temperature, an energy level is populated by 10% of the molecules. What is the energy of that level? Answer: e–u/(kBT) = p × Q; U = –kB × T × ln (p × Q) = –1 J•K–1 × 293 K × ln (0.1 × 1.3) = 598 J relative to ground state. 25. Explain how energy can move spontaneously from a system of low total energy to a system of higher total energy. Answer: Spontaneous processes are governed by an increase in universal entropy. If moving energy to the system of higher total energy increases its entropy more than it decreases the entropy of the system with low total energy, the change will be spontaneous.

log (W )

26. Why is it more common to think of temperature as being related to the rate of change of energy with respect to entropy than it is to think of it as being related to the rate of change of entropy with respect to energy?

energy Answer: Because the log of the multiplicity is additive, the log of the total multiplicity is the sum of the black and gray lines. The total maximum is the same as the maximum value for the gray molecule, because after that point, the log(W) of the gray molecule decreases faster than the log(W) of the black molecule increases.

Answer: While both definitions are viable, it is much more intuitive to use the former definition because it makes the direction of heat flow from higher to lower temperatures, and because it makes all typical temperatures much greater than 1, rather than fractions much less than 1. For example, it makes more sense to say 37° is hotter than 25° than it does to say 0.003224 is hotter than 0.003354.

The Molecules of Life by John Kuriyan, Boyana Konforti, and David Wemmer © Garland Science

The MOLECULES of LIFE Physical and Chemical Principles

Solutions Manual Prepared by James Fraser and Samuel Leachman

Chapter 9 Free Energy

Problems True/False and Multiple Choice 1. For a system at equilibrium, Gibbs free energy is maximized. True/False 2. The work done by biological systems is most commonly which type of work? a. b. c. d. e.

pressure chemical home volume heat

a. A reaction always favors the formation of as much product as possible. b. Reactants will only react upon the addition of external heat or pressure. c. The system will adjust, if more reactants are added, and a new equilibrium will be established to balance the change. d. The Gibbs free energy is always greater than the amount of work done plus the Helmholtz free energy. 4. The value of

for elemental oxygen is:

a. –15 kJ•mol–1. b. 0 kJ•mol–1. c. Much larger than the free energy of elemental hydrogen and much less than the free energy of elemental uranium. d. Equal to the free energy of elemental tungsten. e. Both b and d. 5. The cell generates ATP from ADP and Pi by coupling the synthesis to an energetically unfavorable process.

Answer: pressure and temperature, volume and temperature 8. The standard state has a pressure of ___________. Answer: 1 atm 9. The change in free energy of a process is equal to the ________________ amount of work extracted from that process. Answer: maximum 10. In biochemistry, the standard state for water is ____ M. Excepting H+, for all other solutions, the standard state is ______ M. Answer: 55, 1

Quantitative/Essay Problems 11. A system is at constant volume and 29°C. The internal energy is –40 kJ and the entropy is 31 J•K–1. What is the value of the Helmholz free energy? Answer:

True/False 6. At equilibrium: a.

Fill in the Blank 7. Gibbs free energy is used to describe systems with constant ___________, while Helmholtz free energy is used to describe systems with constant _______________.

3. According to Le Châtelier’s principle:

∆f Go

b. The free energy of the system is minimized. c. The combined entropy of the system and surroundings is maximized. d. Only b and c. e. a, b, and c.

There is no change in temperature over time.

First convert 29°C to K by adding 273. A = U – TS = –40 kJ – (302 K × 31 J•K–1) = –40 kJ – 9.4 kJ = –49.4 kJ

The Molecules of Life by John Kuriyan, Boyana Konforti, and David Wemmer © Garland Science

2

Chapter 9: Free Energy

12. Consider the distribution of molecules in energy levels in the diagram below, where each “X” represents one molecule. A molecule in energy level 1 contributes one unit of energy to the total internal energy, a molecule in energy level 2 contributes two units of energy to the total internal energy, etc. The system is at 273 K.

16. What is the expression for the equilibrium constant for the reaction A B + C? Answer: [B][C]/[A] 17. Consider the reaction A + 2B → C. Values of ∆f Go for A, B, and C are as follows:

Energy level 4

X

Energy level 3

XX

Energy level 2

XXXXX X

C = –112 kJ•mol–1

Energy level 1

XXXXX XXXXX XXXXX

What is the value of ∆Go for the reaction?

A = –34 kJ•mol–1 B = 84 kJ•mol–1

Answer:



What are the values of the (a) internal energy (U), (b) entropy (S), and (c) Helmholtz free energy (A) of the system? Assume that kB = 1 energy unit•K–1. Answer: a. Utotal = ΣNiUi = 1 × 4 + 2 × 3 + 6 × 2 + 15 × 1 = 4 + 6 + 12 + 15 = 37 units b. W = (Total)!/(Energy level 4)! × (Energy level 3)! ... = 24!/(1!2!6!15!) = 329,491,008 S = kB ln W = 19.6 units•K–1

∆G = ∆f Gproducts – ∆f Greactants = –112 kJ•mol–1 –

(–34 kJ•mol–1 + 2 × 84 kJ•mol–1) = –246 kJ•mol–1 18. Consider the following reactions in which A, B, C, and D are elements in their most stable states:
 A+B→Y C+D→Z
 Y+Z→J

c. A = U – TS = 37 units – 273 K × 19.6 units•K–1 = –5317 units 13. A system at 275 K in state A has an enthalpy of –25 kJ and an entropy of 2 J•K–1. In state B, it has an enthalpy of –20 kJ and an entropy of 10 J•K–1. Will state A convert spontaneously to state B? Answer: GA = H – TS = –25 kJ – 275 K × 2 J•K–1 = –25.55 kJ GB = H – TS = –20 kJ – 275 K × 10 J•K–1 = –22.75 kJ Since the free energy is lower in state A, it will not convert spontaneously to state B at 275 K. 14. Assume that entropy and enthalpy changes are independent of temperature. A system in state A has an enthalpy of –22 kJ and an entropy of 7 J•K–1. In state B, it has an enthalpy of –12 kJ and an entropy of 15 J•K–1. At what temperatures will state B be favored? Answer: The state B will be favored when GB < GA Therefore: HB – TSB < HA – TSA (HB – HA)/(SB – SA) < T (because SB – SA > 0) (– 12,000 J + 22,000 J)/(15 J•K–1 – 7 J•K–1) = 1250 K 1250 K < T. The temperature must be greater than 1250 K. 15. A reaction has an enthalpy change of 200 kJ and an entropy change of 250 J•K–1. Assume that entropy and enthalpy changes are independent of temperature. At what temperature will the Gibbs free energy be zero? Answer: G = H – TS, G = 0. H = TS 200 kJ = T(250 J•K–1) T = 200 kJ/(250 J•K–1) = 800 K

Using the following table of ∆Go values, what is the value of ∆f Go for the formation for J? Reaction

∆Go (kJ•mol–1)

A+B→Y

–70

C+D→Z

23

Y+Z→J

15

Answer:

∆f G(Y) – (∆f G(A) + ∆f G(B)) = –70 kJ•mol–1 → ∆f G(Y) =

–70 kJ•mol–1 ∆f G(Z) – (∆f G(C) + ∆f G(D)) = 23 kJ•mol–1 → ∆f G(Z) = 23 kJ•mol–1 ∆f G(J) – (∆f G(Y) + ∆f G(Z)) = 15 kJ•mol–1

∆f G(J) = 15 kJ•mol–1 – 70 kJ•mol–1 + 23 kJ•mol–1 = –32 kJ•mol–1

19. Consider the reaction, 2I + J → K, and the following table of enthalpies and entropies of formation at 298 K:

∆f Ho

(kJ•mol–1)

∆f So (J•K–1•mol–1)

I

J

K

0

–30

–300

120

200

60

What is the standard free-energy change (∆Go) for the reaction? Answer:

∆G = ∆f G(K) – [2 × ∆f G(I) + ∆f G(J)]

= –300 kJ – 298 K × 60 J•K–1 – [2 × (0 – 298 K × 120 J•K–1) + (–30 kJ – 298 K × 200 J•K–1)] = –317.9 kJ – (–71.5 kJ – 89.6 kJ) = –156.8 kJ 20. Consider the same reaction as in Problem 19. Assume that enthalpy and entropy do not vary with temperature.

The Molecules of Life by John Kuriyan, Boyana Konforti, and David Wemmer © Garland Science



PROBLEMS and solutions At what temperature will the reaction begin to proceed spontaneously in the opposite direction (that is, at what temperature will ∆Goreactants < ∆Goproducts)? Answer: At 298 K, the reaction proceeds to products in Problem 19. To proceed towards reactants: ∆Greactants < ∆Gproducts 0 < ∆Gproducts – ∆Greactants 0 < ∆f Hproducts – T∆f Sproducts – (∆f Hreactants – T∆f Sreactants) 0 < –300,000 J – T × 60 J•K–1 + 240 J•K–1 × T + 30,000 J + 200 J•K–1 × T 0 < –270,000 J + 380 J•K–1 × T 710.5 K < T Therefore the temperature must be at least 710.5 K.

21. The value of ∆Go for forming water from elemental oxygen and hydrogen is –237 kJ•mol–1. 2H2 +O2 → 2H2O

Calculate the standard enthalpy of formation of water using the entropies below. (Hint: Elemental oxygen and hydrogen have the same enthalpy of formation.)

So (J•K–1•mol–1)

O2

H2

H2O

205

130

70

Answer: Elemental oxygen and hydrogen both have an enthalpy of formation = 0. ∆f G(H2O) = ∆f H(H2O) – T∆f S(H2O) ∆f H(H2O) = ∆f G(H2O) – T(∆f S(H2O)) ∆f H(H2O) = ∆f G(H2O) – T(So(H2O) – So(H2) – ½ × So(O2)) ∆f H(H2O) = –237 kJ•mol–1 + 298 K × (70 J•K–1•mol–1 – 130 J•K–1•mol–1 – 205/2 J•K–1•mol–1) ∆f H(H2O) = –285 kJ•mol–1

3

23. A molecular motor moves along a microtubule track in steps of 100 Å displacements. The motor hydrolyzes one molecule of ATP per step. The motor operates under conditions where the free-energy change for ATP hydrolysis is –60 kJ•mol–1. What is the maximum resistive force against which the motor can move cargo? Answer: F=

w x

=

60000 J•mol–1 100 × 10–10 m

×

1 mol 6.022 × 1023

= 9.96 × 10–12 N = 10 pN

24. Explain why the absolute value of the work done by a process at constant pressure can never be greater than the absolute value of the Gibbs free-energy change for the process. Answer: The maximum amount of work that can be performed by a process at constant temperature and pressure is the Gibbs free energy. It incorporates the minimum energy that must necessarily be lost as heat. When the temperature isn’t held constant, additional energy is lost as heat, which cannot be used as work. 25. The equilibrium constant for a chemical system is 1. More reactants are added to the system. Explain what will happen to the system as it reestablishes equilibrium. Answer: If the equilibrium constant is 1, then there is always an even ratio of products to reactants. According to Le Chatalier’s principle, as more reactants are added to the system, more product will form to equalize the ratio.

22. A chemist wants to develop a fuel by converting water back to elemental hydrogen and oxygen using coupled ATP hydrolysis to drive the reaction. Given that the value of ∆ f Go for water is –237 kJ•mol–1 and that one mole of ATP hydrolyzed to ADP yields –30 kJ•mol–1, how much ATP is needed to yield three moles of H2 gas? Answer: Reaction 1: 2H2+O2 → 2H2O The ATP reaction: 2H2O + X ATP → 2H2 + O2 + ADP + phosphate The molar ratio of water to hydrogen gas is 1:1, therefore 3 moles of H2O will need to be converted. ∆Greaction1 = –711 kJ must be balanced by ATP hydrolysis. ∆Greaction1 > ∆GreactionATP –711 kJ > –30 kJ × (moles of ATP) –711 kJ/–30 kJ < moles of ATP 23.7 < moles of ATP Therefore at least 23.7 moles of ATP must be hydrolyzed to yield 3 moles of hydrogen gas. This also demonstrates how ATP hydrolysis can be used to drive an otherwise unfavorable process. The Molecules of Life by John Kuriyan, Boyana Konforti, and David Wemmer © Garland Science

The MOLECULES of LIFE Physical and Chemical Principles

Solutions Manual Prepared by James Fraser and Samuel Leachman

Chapter 10 Chemical Potential and the Drive to Equilibrium

Problems True/False and Multiple Choice 1. Molecules move spontaneously from regions of low chemical potential to regions of high concentration. True/False 2. The difference in chemical potential for a region with 500 mM of molecule B and a region with 1 M of molecule B is equal to: a. kBT ln(0.5) b. 0 c. 1 d. kBT ln(500) e. kBT ln(1) 3. The proton concentration in pure water at standard state (298 K) is: a. b. c. d. e.

equal to 14 always less than –7 the square root of the ion product 107 81 kJ•mol–1

7. During protein folding, the entropy of water: a. increases b. decreases c. is equal to the protein entropy change d. is zero

Fill in the Blank 8. A region with a high chemical potential for molecule A has a _______ concentration of molecule A than a region with low chemical potential. Answer: greater/higher 9. To make a buffer, add a weak acid to its conjugate _______. Answer: base 10. _____, ______, and ______ sidechains generally have a pKa less than 7.0. _______, ________, and ______ sidechains have a pKa greater than 9.

4. The melting temperature (TM) is the temperature at which 100% of the protein molecules are unfolded. True/False 5. Which of the following must be independent of temperature when properly applying the van’t Hoff equation? a. Keq b. ∆So c. 1/T d. pH 6. The pKa of a protein sidechain depends only on the chemical identity of the sidechain, not on the surrounding environment. True/False

Answer: His, Asp, Glu; and Tyr, Lys, Arg 11. The integral of the melting curve of heat capacity versus temperature yields the _____________ change of protein unfolding. Answer: enthalpy change 12. The _______________ change for a reaction determines the direction of spontaneous change. Answer: free energy or ∆G

Quantitative/Essay 13. Two regions of an ideal dilute solution have a difference in concentration of potassium ions (K+). At 293 K, what is the difference in chemical potential between region

The Molecules of Life by John Kuriyan, Boyana Konforti, and David Wemmer © Garland Science

2

Chapter 10: Chemical Potential and the Drive to Equilibrium 1, with a concentration of 0.5 M K+, and region 2, which has a concentration of 2 mM? Answer: C2

∆µ = RT ln

C1

= 8.31 J•K–1•mol–1 × 293 K × ln

18. The arginine-tRNA synthetase enzyme catalyzes the reaction that charges a tRNA with the amino acid arginine:

0.002 M 0.5 M

ATP + arginine + tRNA

= –13.5 kJ•mol–1

Answer:

0.2 M x

0.2 M x

C2 C1

0.2 M x

= 2.05

= 7.79

15. A cell with an internal calcium ion (Ca2+) concentration of 20 µM is placed in media with a Ca2+ concentration of 70 mM. What is the difference in chemical potential for Ca2+ ions between the inside and outside of the cell at 310 K? Answer:

∆µ = µin – µout = RT ln

Cin Cout

∆µ = 8.31 J•K–1•mol–1 × 310 K × ln

20 × 10–6 M 70 × 10–3 M

∆µ = 21.0 kJ•mol–1 16. At equilibrium, in a test tube, the concentration of GDP is 1 M, of GTP is 20 µM, and of Pi is 1 M. What is the equilibrium constant of the reaction, GTP GDP + Pi? Answer: [1][1] [20 × 10–6]

= 50000

17. The reaction, A + 2B C, has an equilibrium constant of 2000. During a reaction, the concentration of A is 0.01 M, of B is 0.2 M, and of C is 0.5 M. a. b.

Answer: [500 × 10–6] [500 × 10–6] [RtRNA*] [2 × 10–6] [500 × 10–3] [10 × 10–6]

[RtRNA*] = 4.52 × 10–5 = 45.2 µM 19. The pKa of a weak acid is 5. What is the pH when the concentration of the acid form is 0.5 M and the concentration of the conjugate base form is 0.05 M? Answer: pH = pKa + log10([base]/[acid]) = 5 + log10(0.05/0.5) = 5 + log10(0.1) = 4

x = 25.7 mM

K=

At equilibrium, following an in vitro reaction, the concentration of ATP is 2 µM, of arginine is 500 mM, and of arginyl-tRNA is 10 µM. The concentrations of AMP and of PPi are 500 µM. What is the concentration of arginyl-tRNA?

K = 1.13 =

5000 J•mol–1 = 8.31 J•K–1•mol–1 × 293 K × ln

ln

AMP + PPi + arginyl-tRNA

The value of the equilibrium constant is 1.13.

14. The difference in chemical potential for a particular molecule between two regions of an ideal dilute solution is 5 kJ•mol–1. The region with the higher chemical potential has a concentration of 200 mM. What is the concentration of the molecule in the other region at 293 K?

∆µ = RT ln

Answer: a. Q = [C]/([A][B]2) = 0.5/(0.01 × 0.22) = 1250 b. The reaction will proceed towards the right since Q < Keq.

What is the reaction quotient (Q)? In what direction will the reaction proceed?

20. The pH of a 0.15 M propionic acid/0.1 M sodium propionate buffer is 4.71. What is the pKa of propionic acid? Answer: pKa = pH – log([base]/[acid]) = 4.71 – log(0.1/0.15) = 4.71 + 0.18 = 4.89 21. Consider a protein with a surface-exposed histidine residue in a pH 4 solution. What is the fraction of protein molecules in which this histidine residue is charged? (Assume that the pKa is 6.0.) Answer: [His]/[His+] = 10(pH – pKa) = 10(4 – 6) = 10–2 = 0.01 Fraction[His] = [His]/[His+] / ([His]/[His+] + 1) = 0.9901 Therefore, 99% of the sidechains will be charged. 22. For a protein with a surface-exposed aspartic acid, at what pH will this residue be neutral in 75% of the protein molecules? (Assume that the pKa is 4.0.) Answer: [Asp–]/[Asp] = 0.25/0.75 = 1/3 pH = pKa + log(1/3) pH = 4 – 0.48 pH = 3.52 23. A histidine is involved in an interaction with a glutamic acid that stabilizes the charged form of the histidine, such that the value of ∆Go for deprotonation is 15 kJ•mol–1 at pH 7.0 and 293 K (calculated using the biochemical standard state). What is the pKa of this histidine?

The Molecules of Life by John Kuriyan, Boyana Konforti, and David Wemmer © Garland Science

Answer: [His]/[His+] = e(–∆G/RT) = e(–15,000/(293 × 8.31)) = 0.002118 pKa = pH – log([His]/[His+]) = 7 – log10(0.002118) = 9.67 24. At the TM of a protein (55°C), the value of ∆Hounfolding is 15 kJ•mol–1. What is the value of ∆Sounfolding? Answer:

∆Sounfolding = ∆Hounfolding/TM = 15 kJ•mol–1/328 K =

45.7 J•mol–1•K–1

25. A protein has a ∆Hunfolding value of 140 kJ•mol–1 at 25°C. The value of ∆CP is 7.5 kJ•K–1•mol–1. The value of ∆Hounfolding at the TM is 230 kJ•mol–1. What is the value of TM? Answer: ∆CP (298 K – TM) = ∆Hounfolding @ 25 – ∆Hounfolding @ TM 298 K – TM = (∆Hounfolding @ 25 – ∆Hounfolding @ TM)/∆CP TM = –[(∆Hounfolding @ 25 – ∆Hounfolding @ TM)/∆CP – 298 K] TM = 310 K or 37°C 26. A lysine sidechain has four torsion angles, each of which can take on three different values (60°, –60°, and 180°). Each unique combination of angles is called a rotamer. For example, a lysine residue where the first, second, third, and fourth torsion angles are all 60° is one unique rotamer, whereas a residue where the first, second, and third torsion angles are 60° and the fourth torsion angle is 180° is a second rotamer. In contrast, a serine sidechain has only one torsion angle, which can take on three different values. Assume that all possible dihedral angles are allowed at each angle for residues at this surface-exposed position. a. What is the difference in molar entropy between a protein with a surface-exposed lysine and an otherwise identical protein with a serine mutation at that position? b. Why might the simplification that each lysine torsion angle is able to adopt any of the three staggered positions, independent of the conformation at other torsion angles, lead to an overestimate of the number of low-energy conformations that lysine can adopt? Answer: a.

WK = 34 WS = 31

∆S = R ln

WK WS

= 8.31 J–1•K–1•mol–1 × ln

34

27. In the hydrophobic core of a folded protein, there are three alanine and five phenylalanine residues that are buried, and do not interact with water. Assume: • In solution, waters can take on seven energetically equal states. • Two waters are ordered around each alanine in the unfolded state. • Six waters are ordered around each phenylalanine in the unfolded state. • In the unfolded state, waters are ordered around alanine or phenylalanine residues and can take on only two energetically equal states. What is the difference in the entropy of the water due to the burying of these residues as this protein folds? Answer: Total water molecules = 2 × number of Ala + 6 × number of Phe =2×3+6×5 = 36 waters Sfolded = R ln736 = 8.31 J•K–1•mol–1 × 36 × ln7 = 582 J•K–1•mol–1 Sunfolded = R ln236 = 8.31 J•K–1•mol–1 × 36 × ln2 = 207 J•K–1•mol–1 ∆So = Sfolded – Sunfolded = 375 J•K–1•mol–1 28. Why do proteins denature at cold temperatures? Answer: While the physical mechanism behind cold denaturation is not yet understood, the phenomenon can be predicted from the curvature of protein stability curves. The constant curvature, arising from the difference in heat capacity between the folded and unfolded states, means that because ∆Go = 0 at the TM, it must also equal zero at some other point. 29. How do hydrophobic interactions provide favorable entropy for protein folding? Answer: Since water molecules cannot hydrogen bond with hydrophobic groups on the protein, the rotation of water around these groups is restricted. When the hydrophobic groups collapse into the core of the protein, water no longer surrounds them. Thus, the water that previously was restricted around the hydrophobic groups is released to the bulk solvent and free to move between many configurations (increasing the entropy of the system).

31

∆S = 27.4 J–1•K–1•mol–1 b. Some rotamers might be disallowed because they sterically clash with other residues, so not all independent conformations of torsion angles might be possible.

The Molecules of Life by John Kuriyan, Boyana Konforti, and David Wemmer © Garland Science

4

Chapter 10: Chemical Potential and the Drive to Equilibrium

The Molecules of Life by John Kuriyan, Boyana Konforti, and David Wemmer © Garland Science

The MOLECULES of LIFE Physical and Chemical Principles

Solutions Manual Prepared by James Fraser and Samuel Leachman

Chapter 11 Voltages and Free Energy

Problems True/False and Multiple Choice 1. An oxidative half-reaction involves the release of electrons from a molecule. True/False 2. Which of the following is a common consequence of a protein coordinating a metal: a. The free energy of the reduced and oxidized form of the metal are altered. b. The metal is protected from some chemical reactions. c. The protein undergoes a conformational change. d. It enables the transfer of electrons over long distances. e. All of the above. 3. Standard reduction potentials in biochemistry correspond to solutions with [H+] = 1 M. True/False 4. Which of the following conformational changes is not thought to occur as a K+ channel closes due to a voltage change: a. b. c. d. e.

Helix S6 becomes straighter. Helix S4 moves downward. Charged residues on helix S4 form new interactions. The selectivity filter changes size. The C-terminus of helix S4 forms a 310 helix.

5. Most ion channels have little selectivity because ions are small relative to the size of the channel pore. True/False 6. Which feature allows mitochondria and chloroplasts, unique from other organelles, to maintain large proton gradients: a. b. c. d.

They contain membrane spanning proteins. They have a second internal membrane. There are only one of each per cell. Their membranes consists of a lipid bilayer.

7. Which of the following statements about the passive spread of a voltage perturbation in a neuron are true: i The voltage spike is not regenerated. ii The potassium conductance of the membrane changes locally. iii The sodium conductance of the membrane remains constant. iv It only occurs in non-myelinated axons. a. b. c. d. e.

None of the above. All of the above. (i) and (iii). (i), (ii) and (iii). (ii) and (iv).

Fill in the Blank 8. The oxidized and reduced forms of a molecule are known as a ____________. Answer: redox couple 9. The energy from _______ is used to produce reduced compounds in photosynthesis. Answer: light 10. Sodium and potassium pumps hydrolyze ________ to move ions in and out of the cell. Answer: ATP 11. By decreasing the capacitance of the surface of the axon, ______________ greatly facilitates the transmission of action potentials. Answer: myelination 12. A _________________ can build up charge on two conducting surfaces with an insulator sandwiched between them. Answer: capacitor

The Molecules of Life by John Kuriyan, Boyana Konforti, and David Wemmer © Garland Science

2

Chapter 11: Voltages and Free Energy

Quantitative/Essay 13. How does phosphorylation contribute to increasing the diversity of redox active molecules in the cell? Answer: Organic compounds that undergo redox reactions can be phosphorylated on parts of the molecule that are not important for the redox reaction. For example, NAD+/NADH can be converted to NADP+/NADPH. The phosphorylated and unphosphorylated forms of the compound bind specifically to different subsets of cellular proteins, which increase the diversity of potentially active redox molecules in the cell. 14. A new microorganism is isolated from a lake and is placed into a solution of KCl. The voltage difference across its membrane is measured at 120 mV. How much energy is required to move a proton from the negative side of the membrane to the positive side? Answer: ∆U = q∆E = 1.602 × 10–19 C × 0.12 J•C–1 = 1.922 × 10–20 J 15. An electrochemical cell couples manganese and copper, each in the presence of its 2+ ion. a. b.

What is the combined redox reaction? What is the the value of ΔGo for the reaction?

Answer: a. Mn → Mn2+ + 2e–, E = 1.18 V (positive, the standard reaction is reversed) 2e– + Cu2+ → Cu, E = 0.340 V Cu2+ + Mn → Mn2+ + Cu, E = 1.52 V b. ΔG = –υFΔE = –2 × 96,500 C•mol–1 × 1.52 J•C–1 = –293,360 J = –293.4 kJ•mol–1 16. A researcher assembles an electrochemical cell with silver and an unknown metal in a 1 M solution of its ion. She measures the free energy of the combined reaction under standard conditions to be –150 kJ•mol−1. What is the potential of the oxidative half reaction if one electron is transferred by the unknown metal to silver under standard conditions? Answer: ΔE = –ΔG/υF = 150,000 J/96,500 C•mol–1 × 1 = 1.55 V Eunknown = ΔE – EAg = 1.55 V – 0.8 V = 0.75 V 17. A new ATP-producing protein is discovered that couples ATP production to the oxidation of NADPH by oxidative phosphorylation. Assume that the value of ΔGo for ATP synthesis is 30 kJ•mol−1. If this protein only produces 1 molecule of ATP per reaction that consumes one NADPH: a. How much free energy is wasted, under standard conditions? b. How many more ATP molecules could be created by a perfectly efficient electron transport chain from one NADPH?

Answer: a. The reaction is NADPH → NADP+ + H+ + 2e–, E = 0.32 1/2 O2 + 2H+ + 2e– → H2O, E = 0.815 NADPH + H+ + 1/2 O2 → H2O + NADP+, E = 1.135

ΔG = –υFΔE = –2 × 96,500 C•mol–1 × 1.135 J•C–1 = 219 kJ•mol–1 ΔGwasted = ΔG – ΔGATP = 219 kJ – 30 kJ = 189 kJ

b. 189/30 = 6.3 more ATPs could be produced. 18. Consider the structure of the potassium channel and answer the following questions. a. Given that the four K+ binding sites are roughly isoenergetic, why are usually only two of these sites occupied at any given time? b. Based on your answer to (a), how does the “knockon” model explain the very high conductance of potassium channels? c. What establishes the directionality of K+ ion flow through a potassium channel? Answer: a. The binding sites are close enough such that electrostatic repulsion of K+ ions in adjacent sites is highly unfavorable. In contrast, occupancy of alternating binding sites (1 and 3, or 2 and 4) is favorable. b. Two ions cannot simultaneously occupy adjacent binding sites within the channel, due to electrostatic repulsion. In the knockon model for ion movement, when an ion enters a vacant binding site within the channel, it causes the ion in the next binding site (if occupied) to hop into the next available site. For example, if ions are present initially in sites 2 and 4, the entry of an ion into site 1 at one end of the channel will cause the site 2 ion to move to site 3, and the site 4 ion to exit the channel at the other end. c. The directionality is established by the electrochemical gradient. There is no directionality preference for the channel itself. 19. A scientist measures the potential across the membrane of a cell. At room temperature, the pH outside the cell is 7.4 and the pH inside the cell is 7.1. What is the membrane potential for protons across the bilayer? Answer: EH = RT/F ln ([H+]out/[H+]in) = 59mV × (–7.4 + 7.1) = –17.7 mV 20. A squid axon is immersed in seawater during a laboratory experiment, and the resting potential across the axonal membrane is −76 mV, at room temperature. The concentrations of Na+ and K+ in the seawater and inside the squid axon are given in the table below: [K+] (mM)

[Na+] (mM)

inside squid axon

150

30

seawater

5

200

The Molecules of Life by John Kuriyan, Boyana Konforti, and David Wemmer © Garland Science



PROBLEMS and solutions a. What are the membrane potentials for Na+ and K+ under these conditions? b. Do you expect the resting membrane potential to be equal to the sum of the membrane potentials for Na+ and K+? Explain your answer. Answer: a. EK = 0.0257 V × ln ([K+]out/[K+]in) = 25.7mV × –3.4 = –87.4 mV ENa = 0.0257 V × ln ([Na+]out/[Na+]in) = 25.7mV × 1.9 = 48.8 mV b. The resting potential is influenced by the relative conductance of the two ions across the membrane. Since the membrane is more permeable to K+ than Na+, the actual potential is weighted more heavily to the K+ potential (see also equation 11.76).

21. A scientist grows a tiny synthetic nerve cell with a surface area of 3.1 × 10−7 cm2. In its culture medium, the cell has a resting membrane potential of 75 mV. During an action potential that depolarizes the cell to 0 mV, 6.2 × 10−12 C of charge move across the membrane. What is the capacitance of the cell per unit area of the cell? Answer: q = CV C = q/V = 6.2 pC/75 mV = (6.2 × 10–12 C)/(0.075 V) = 8.267 × 10–11 F C = CAA CA = C/A = 8.267 × 10–11 F/3.1 × 10–7 cm2 = 2.67 × 10–4 F•cm–1 = 267 μF•cm–2 22. A cell has the intracellular and extracellular concentrations of K+ and Na+ as shown below, at room temperature. If the resting conductance of its K+ channels is 30 times greater than its Na+ channels, what is the resting potential of the cell (assuming no other ions contribute significantly)? [K+] (mM)

[Na+] (mM)

intracellular

200

15

extracellular

12

360

3

RK = C/t = 1 μF/cm2/0.12 sec = 120,000 ohm•cm–2 Next, find the current: RK = EM/IM IM = EM/RK = 0.1 V/120,000 ohm•cm–2 = 8.33 × 10–7 C•sec–1•cm–2 Finally, solve for flow: IM = flow × F × number of channels/NA Flow = IM × NA/F × number of channels = (8.33 × 10–7 C•sec–1•cm–2 × 6.022 × 1023)/(96,500 × 25,000) = 2.08 × 108 C•sec–1 24. Immediately prior to and subsequent to an action potential, there is little conductance through sodium channels; however, conductance is blocked through different mechanisms before and after the action potential passes. Explain the structural mechanisms underlying the closed states of the channel before and after the action potential passes. Answer: Prior to the action potential, the voltage sensor is in the closed conformation, which is established by interactions of the positively charged residues on the voltage sensor and the negative electric field on the inside of the membrane. This causes the helices that comprise the channel to be closed. After the action potential, the electric field is not as negative, so the voltage sensor is in the open conformation, but flexible cytosolic inactivation domains bind to the base of the channel and prevent ion conductance.

Answer: E = (gNa/(gNa + gK)) × ENa + (gK/(gNa + gK)) × EK = (1/(1 + 30)) × 0.0257 V × ln (360/15) + (30/(1+30)) × 0.0257 V × ln (200/12) = 2.6 mV – 70.0 mV = –67.3 mV 23. A region of an axon has a time constant of 120 msec. The capacitance of the membrane is 1 μF•cm−2 and there are 25,000 potassium channels per cm2 of surface area of the membrane. How many ions does a typical potassium channel conduct per second when the membrane potential is 100 mV? Answer: First solve the resistance: t = RK × C The Molecules of Life by John Kuriyan, Boyana Konforti, and David Wemmer © Garland Science

The MOLECULES of LIFE Physical and Chemical Principles

Solutions Manual Prepared by James Fraser and Samuel Leachman

CHAPTER 12 Molecular Recognition: The Thermodynamics of Binding

Problems True/False and Multiple Choice 1.

Which of the following is not a process governed by molecular recognition? a. b. c. d. e.

2.

5.

The fidelity of DNA replication. Passive diffusion. Active transport. Translation by the ribosome. Transcription by RNA polymerase.

True/False

Noncovalent interactions, such as ionic and hydrophobic interactions, are generally much weaker than covalent interactions.

6.

A protein binds to the DNA sequence AAAAA with a 10-nM affinity (that is, the value of the dissociation constant is 10 nM). The same protein binds to the RNA sequence AUAAUA with a 15-nM affinity and to a lipid with a 10-mM affinity. Which of the following statements is true: a. The protein binds to RNA and DNA with high affinity, but low specificity. b. The protein has high specificity for lipids over RNA and DNA. c. The protein binds with low affinity for RNA and DNA, but high specificity. d. The protein binds with low affinity and low specificity, for all three targets.

4.

7.

Which of the following would lead to a decrease in binding affinity? a. Releasing more protein-bound water molecules upon binding. b. Decreasing the number of rotational degrees of freedom of the ligand upon binding. c. Adding more hydrophobic interactions to the ligand–protein interface. d. Increasing the number of hydrogen bonds between the ligand and protein.

Fill in the Blank 8.

The value of KD corresponds to: a. The reciprocal of KA. b. The concentration of ligand at half saturation of receptor. ∆ Go c. e RT. E
 e. All of the above.

Some inhibitors of HIV reverse transcriptase are nucleotide analogs that displace the natural nucleotide substrates in the active site. True/False

True/False 3.

A linear ligand binding curve indicates a simple binding equilibrium, whereas a hyperbolic binding curve indicates that the system contains a receptor that is allosteric or contains multiple binding sites with different affinities.

A _______ is a small molecule that binds to a macromolecule receptor. Answer: ligand

9.

In general, higher affinity is achieved by increasing the __________ of a ligand, while increasing specificity relies on increasing _______________. Answer: hydrophobicity, hydrogen bonds

The Molecules of Life by John Kuriyan, Boyana Konforti, and David Wemmer © Garland Science

2

CHAPTER 12: Molecular Recognition: The Thermodynamics of Binding

10. A typical drug has a dissociation constant for its receptor in the __________ to __________ range.

16. At 298 K a variant of the protein streptavidin binds to biotin with a dissociation constant of 150 pM. What are o the values of KA and ∆Gbind ?

Answer: nanomolar, picomolar 11. Because they bind to the same site as ATP, many kinase inhibitors are __________ with respect to ATP. Answer: competitive 12. Without ___________, some drugs cannot enter or exit the active site of a kinase domain. Answer: breathing motions/conformational changes

17. Enough estrogen is dissolved into a solution containing the estrogen receptor that the total concentration of estrogen is 3 nM. The value of KD for the interaction between estrogen and its receptor is 5 nM. What fraction of the protein is bound to estrogen? Assume that the concentration of estrogen is much greater than that of the protein.

Quantitative/Essay 13. The binding between the drug cyclosporin and the protein cyclophilin is measured, giving a KD value of o 1.5 nM. What is the the value of ∆Gbind at 298 K? Answer:

∆Gobinding = RT ln(KD)



+t,–1tNPM–1 × 298 K × ln(1.5 × 10–9) 
o
L+tNPM–1 14. Shown below is a binding isotherm for an enzyme called a phosphatase and a lead compound known as JF99. There is a tryptophan near the binding site, and the fluorescence from this residue increases upon binding the compound (shown on the y axis). At the highest concentration of JF99 tested (1 μM), the fluorescence signal is 490 units (not shown on the graph). Estimate the value of KD.

Answer: f = [L]/([L] + KD) = 3/(3 + 5) = 0.375 37.5 percent of the protein will be bound to estrogen. 18. A binding assay is performed with a DNA-binding protein and a small DNA duplex. When the concentration of the free DNA duplex is 10 nM, 8% of the protein is bound to DNA. Assume the protein concentration is much lower than that of DNA. What is the value of KD for the interaction? What are the units of KD? Explain your answer. Answer: KD = [L]/f – [L] = 10 nM/0.08 – 10 nM = 125 nM – 10 nM = 115 nM

400 350 300 250

19. Using a purified sample of the protein cyclophilin, the Scatchard plot shown below is obtained when a form of cyclosporin is added. What is the value of KD?

200 150 100 50 0

0

10

20

30

40

50

[JF99] (nM)

Answer: At the KD the protein will have the half-maximal fluorescent signal. Since the maximal signal is ~500, we look for the ligand concentration that gives a signal of ~250. The estimated KD is ~20 nM. 15. What is the value of ∆Gbind for the interaction between the phosphatase and the inhibitor, based on the data shown in previous question, at 298 K? o

Answer: If the KD is ~20 nM. ∆Gobinding = RT ln(KD) 

+t,–1tNPM–1 × 298 K × ln(2 × 10–8) 
o
L+tNPM–1

60

[bound cyclosporin]/[free cyclosporin]

fluorescence signal (arbitrary units)

Answer: KA = 1/KD = 1/(1.5 × 10–10) = 6.7 × 109 ∆Gobinding = RT ln(KD) 

+t,–1tNPM–1 × 298 K × ln(1.5 × 10–10) 
o
L+tNPM–1

0.6 0.5 0.4 0.3 0.2 0.1 0

0

0.01

0.02 0.03 0.04 0.05 [bound cyclosporin] (nM)

Answer: KD = –1/slope = –1/((0.5 – 0.1)/(0.01 – 0.05 nM)) = 100 pM

The Molecules of Life by John Kuriyan, Boyana Konforti, and David Wemmer © Garland Science

0.06

PROBLEMS AND SOLUTIONS 20. The drug jafrasitor (molecular weight 540 daltons) binds the histone deacetylase enzyme Sir2 with a dissociation constant of 0.1 nM. What mass of jafrasitor should be administered to a patient with a blood volume of 5.5 L such that Sir2 is at least 91% inhibited? Answer: From the “universal” binding curve, we can see that approximately 91% saturation occurs at 10 × KD, which is 1 nM. Mass = concentration × volume × molar mass 

ONPMt-–1
¨

-
¨

HtNPM–1 = 2.97 × 10–6 g = 2.97 μg 21. How do the conformational changes during kinase activation enable imatinib to distinguish between Abl and the hundreds of other related kinases in the cell? Answer: When active, kinases are relatively similar structurally especially in the ATP binding pocket. However, inactive kinases are quite different from each other. Imatinib binds to the inactive conformation of Abl, which is distinct from the inactive conformations of most other kinases.

[bound rapamycin]/[free rapamycin]

22. Using 1 mL of partially purified cell lysate, which contains 1 mg of total protein, the following Scatchard plot is obtained for the binding of the protein FKBP (molecular weight 20,000 daltons) and the drug rapamycin. What is the purity of the lysate with respect to FKBP?

3

23. The IC50 value for a drug called razundib for a Src kinase is 15 nM. The Src kinase is known to bind ATP with a dissociation constant of 0.05 mM. If the concentration of ATP in the cell is 0.75 mM, what is the value of KI for razundib with respect to the Src kinase? Answer: KI = IC50 × (KD/(KD + [L])) = 15 nM × (0.05/(0.8)) = 0.94 nM 24. An NMR analysis of the the drug fluvastatin indicates that it loses rotational and translational degrees of freedom upon binding to HMG-CoA reductase. Additionally, many internal bond rotations are restricted. Yet, an isothermal titration calorimetry experiment indicates that the binding reaction has a favorable change in entropy. What is the likely source of the favorable binding entropy change? Answer: The binding surface for fluvastatin is a large hydrophobic cavity that traps many water molecules when it is not occupied by fluvastatin. Fluvastatin displaces these water molecules and the increases in the rotational and translational degrees of freedom for the water molecules. This increase in entropy more than offsets the entropy loss of binding fluvastatin.

0.6 0.5 0.4 0.3 0.2 0.1 0

0

0.002 0.004 0.006 0.008 0.01 0.012 0.014 0.016 0.018 0.02 [bound rapamycin] (nM)

Answer: First calculate the concentration of FKBP. KD = –1/Slope = –1/((0.5 – 0.1)/(0.01 – 0.018)) = 20 pM When [L]bound = 0.01 nM, [L]bound/[L]bound = 0.5, 0.5 = –1/(0.020 nM) × 0.01 nM + [P]/20 pM [P] = 1 × 20 pM = 20 pM 
QNPMt-–1
¨
 
HtNPM–1 = 4 × 10–7
Ht-–1, which is 4 × 10–7
PG

NHtN-–1

The Molecules of Life by John Kuriyan, Boyana Konforti, and David Wemmer © Garland Science

The MOLECULES of LIFE Physical and Chemical Principles

Solutions Manual Prepared by James Fraser and Samuel Leachman

Chapter 13 Specificity of Macromolecular Recognition

Problems True/False and Multiple Choice 1. The affinity of a macromolecular interaction reflects the strength of an interaction for one receptor relative to all other possible receptors.

6. Interface residues that do not contribute greatly to binding affinity are also generally unimportant for specificity. True/False

True/False 2. Which of the following is an attribute of the FGF–FGFR family of interactions?

7. Which of the following is not an important aspect of protein–nucleic acid recognition?

a. Each FGF ligand has high specificity for an FGF receptor. b. Because there are 18 FGF and 4 FGFR genes, there are 72 potential interactions. c. Signaling specificity is enhanced through selective expression of only a few receptors per tissue type. d. FGFRs are soluble serine/threonine kinases. e. FGF proteins have highly diverse folds. 3. Specificity depends on the concentration of ligand, whereas affinity is independent of ligand concentration. True/False 4. Most of the binding energy for SH2 domain–peptide interactions is contributed by: a. b. c. d.

Amino–aromatic interactions. The phosphorylation of the peptide tyrosine. The amino acid in the peptide + 1 position. The amino acid in the peptide + 3 position.

5. Which of the following statements regarding interfacial waters is true? a. They contribute favorably to the entropy of protein– protein recognition. b. They help to optimize the packing at the contact surface. c. They contribute only to affinity, but not to the specificity of protein–protein recognition. d. There are typically fewer water-mediated hydrogen bonds than direct hydrogen bonds at protein–protein interfaces.

a. Insertion of arginine sidechains into the minor groove of the nucleic acid. b. Conformational changes of the protein and nucleic acid. c. Stacking interactions between serine sidechains and nucleic acid bases. d. Stacking interactions between tyrosine sidechains and nucleic acid bases. e. Water-mediated hydrogen bonds between protein and nucleic acid.

Fill in the Blank 8.

_____________ residues, identified by mutating residues to alanine, contribute a larger than expected energy to the interaction affinity of human growth hormone with human growth hormone receptor. Answer: Hot spot

9. The kinase Zap-70 contains two _______ domains that bind to two __________ residues in its targets. Answer: SH2, phosphotyrosine 10. Proteins distinguish double-helical RNA and DNA by differences in _______ shape. Answer: groove 11. Protein–protein interfaces typically bury approximately _____________ in surface area. Answer: ~1000 Å

The Molecules of Life by John Kuriyan, Boyana Konforti, and David Wemmer © Garland Science

2

Chapter 13: Specificity of Macromolecular Recognition

12. Examining complexes of nucleic acids and their binding proteins reveals a high _____________ in both shape and charge. Answer: complementarity

Quantitative/Essay (Assume T = 300 K and RT = 2.5 kJ•mol–1 for all questions.) 13. A single-molecule microscopy experiment is performed on a slide containing 1000 molecules each of green, red, cyan, and yellow fluorescent proteins (GFP, RFP, CFP, and YFP, respectively). The experiment measures the number of antibodies bound to each type of protein, finding that 900 molecules of GFP, 10 of RFP, 1 of CFP, and 3 of YFP are bound at 1 nM concentration of antibody. a. b.

What is the specificity of the antibody for GFP? What is the KD of the GFP–antibody interaction?

Answer: a. α = [RGFP•L]/([RRFP•L] + [RCFP•L] + [RGFP•L]) = 900/(10 + 1 + 3) = 64.3 b. KD = [RGFP] × [L]/[RGFP•L] = 100 × 10–9 M/900 = 111 pM 14. A scientist wants to engineer an antibody to distinguish between two proteins (Cyclophilin A and Cyclophilin B) with a specificity of 500 at 1 nM concentration for each protein. Her starting material is an antibody that binds with 10 nM KD to both proteins. She finds that she can easily make mutations that decrease the affinity for Cyclophilin B without affecting the affinity for Cyclophilin A. When she achieves the desired specificity, what is the KD for Cyclophilin B? Answer:

α = (1/(1 + KD,CypA/[L]))/(1/(1 + KD,CypB/[L])) KD,CypB = α × ([L] + KD,CypA) – [L] KD,CypB = 5.5 μM 15. A tetracycline repressor (TetR) protein, which is present in E. coli at 10–8 M concentration, binds to the tetO site with a KD of 10–10 in the absence of the drug tetracycline and a KD of 10–6 in the presence of tetracycline. There is one tetO site in the E. coli genome, giving a concentration of 10–9 M per cell. There are 2 × 105 nonmatching sites per cell (a concentration of 5 × 10–2 M), to which TetR binds nonspecifically with a KD of 10–5.

b. α = ([tetO]/(1 + (KD,TetO/[L]))/([nonspecific]/(1 + (KD,nonspecific/[L]))) = ([1 × 10–9/(1 + (1 × 10–6/1 × 10–8)))/(5 × 10–2/(1 + (1 × 10–5/1 × 10–8))) = 1.98 × 10–7 c. The low specificity values are tuned through selection to ensure that a switch in transcription can be achieved. If the equilibrium constants were used without considering competing nonspecific sites, then the tetO sites would almost always be bound. In essence the nonspecific sites titrate the TetR away from the specific site allowing switch-like behavior despite the low specificity values. 16. A zinc finger protein is isolated from a yeast cell. The value of KD for its binding site is 3 µM. In the presence of glucose, the protein dimerizes and recognizes an inverted repeat binding site. a. What is the expected value of KD if the binding is additive? b. The dimeric KD is measured at 5 nM. Why does this value deviate from the expected KD? Answer: a. KD,dimer = (KD,monomer)2 = (3 × 10–6) = 9 × 10–12 = 9 pM b. The simple additivity of free energy is probably reduced because energy is used to induce conformational changes in the DNA and protein to stabilize the dimer. 17. A tryptophan residue near the periphery of a protein– protein interface is mutated to alanine and changes the KD of binding from 1 nM to 40 µM at 300 K. a. How much binding energy was contributed by that residue? b. Explain whether or not the tryptophan residue is likely a hot spot residue. Answer: a. ∆∆G = ∆GTrp – ∆GAla = RT ln(KD,Trp) – RT ln(KD,Ala) = RT ln(KD,Trp/KD,Ala) = RT ln(10–9/4 × 10–5) = –27 kJ•mol–1 b. Non-hotspot residues typically contribute 0–5 kJ•mol–1 of binding energy, whereas hotspot residues often contribute more than 20 kJ•mol–1. Since the Trp contributes 27 kJ•mol–1 it is likely a hotspot residue.

a. What is the specificity for the tetO site over the nonmatching sites when no tetracycline is present? b. What is the specificity for the tetO site over the nonmatching sites when tetracycline is present? c. How can TetR repressors switch transcription with such low specificity values?

18. A protein–protein interface has a 10 nM affinity at 300 K. A series of mutants are made in which each residue at the interface is replaced by alanine. A lysine residue at the center of the interface is mutated, and found to contribute 4 kJ•mol–1 to the binding free energy.

Answer: a. α = ([tetO]/(1 + (KD,TetO/[L]))/([nonspecific]/(1 + (KD,nonspecific/[L]))) = (1 × 10–9/(1 + (1 × 10–10/1 × 10–8)))/(5 × 10–2 /(1 + (1 × 10–5/1 × 10–8))) = 1.98 × 10–5

Answer: a. ∆∆G = –4 kJ•mol–1 = ∆GLys – ∆GAla = RT ln(KD,Lys) – RT ln(KD,Ala)

a. What is the new KD? b. Explain whether or not the lysine residue is a hot spot residue.

The Molecules of Life by John Kuriyan, Boyana Konforti, and David Wemmer © Garland Science



PROBLEMS and solutions RT ln(KD,Ala) = RT ln(KD,Lys) – 4 kJ•mol–1 = RT ln(10–8) + 4 kJ•mol–1 = –42 kJ•mol–1 KD,Ala = e(–16.8) = 5.0 × 10–8 = 50 nM b. Since the Lysine residue contributes only 4 kJ•mol–1 of binding energy, it is likely not a hotspot residue.

19. The transcription factor FraJ binds a poly-A DNA sequence with a 10 nM KD and a poly-G DNA sequence with a 27 µM KD. Mutation of a critical Phe residue to Ala results in a loss of 20 kJ•mol–1 in binding free energy for the poly-A sequence, but only a loss of 4 kJ•mol–1 on binding to the poly-G sequence. What is the change in specificity for the poly-A sequence over the poly-G sequence at 10–8 M concentration of FraJ at 300 K? Answer: For poly-A: ∆∆G = –20 kJ•mol–1 = ∆GPhe – ∆GAla = RT ln(KD,Phe) – RT ln(KD,Ala) –20/2.5 = ln(10–8) – ln(KD,Ala) –8 = –18.4 – ln(KD,Ala) (KD,Ala) = e(∆G/RT) KD,Ala = e(–10.4) = 3.0 × 10–5 For poly-G: ∆∆G = –4 kJ•mol–1 = ∆GPhe – ∆GAla = RT ln(KD,Phe) – RT ln(KD,Ala) –4/2.5 = ln(2.7 × 10–5) – ln(KD,Ala) –1.6 = –10.5 – ln(KD,Ala) (KD,Ala) = e(∆G/RT) KD,Ala = e(–8.9) = 1.3 × 10–4

αPhe = (1/(1 + KD,poly-A/[L]))/(1/(1 + KD,poly-G/[L])) = (1/(1 + 10–8/10–8))/(1/(1 + 2.7 × 10–5/10–8)) = 1350.5 αAla = (1/(1 + KD,poly-A/[L]))/(1/(1 + KD,poly-G/[L]))

= (1/(1 + 3 × 10–4/10–8))/(1/(1 + 1.4–4/10–8)) = 4.49 The change in specificity is 1346.

20. A protein–protein interface comprises 22 residues at the contact surface. From structures of the isolated proteins, it is expected that completely burying these residues would cause a surface area reduction of ~2000 Å2. However, a structure of the interface reveals that only 1200 Å2 of surface area is buried. Why is there a discrepancy between the expected and measured surface area reductions? Answer: Because the shape complementarity is not perfect, many residues are not completely buried in the bound complex. Additionally, residues that interfacial waters, which are important for providing hydrogen bonding networks at interfaces, are not normally counted as part of the buried surface area. 21. Consider the dsRB domain and its potential for interacting with DNA and RNA (see Figure 13.32). What is the predicted effect on the specificity and affinity of recognition for the two types of nucleic acids of: 
 a. An Arg to Ala mutation at the binding interface? 


3

b. Insertion of loop residues that change the relative spacing of helix A and helix B? Answer: a. The binding affinity of either complex would likely be reduced by mutating an Arg to an Ala. The positively charged Arg residue can make ionic contacts to the negatively charged phosphate backbone. This mutation would likely have a negligible effect on specificity between RNA and DNA. b. The dsRB domain recognizes the distinct groove spacing between RNA and DNA. A mutation that changed that spacing would definitely have a detrimental effect on RNA recognition. If the mutation changed the spacing so that it was more complementary to DNA grooves, it would switch the specificity dramatically. 22. A DNA-binding domain binds the sequence GATCGCAATATCGATCGATC with a 25 nM affinity. A mutation of an Arg to Ala in the protein or a mutation of the underlined “T” to “G” in the DNA sequence both result in a 9 kJ•mol–1 loss of binding free energy. Simultaneous mutation of both the protein and the DNA also results in a 9 kJ•mol–1 loss of binding free energy. a. What is the effect on the KD of any of these mutations? b. What does the double mutant result suggest about the structural basis for the protein–DNA interaction? Answer: a. ∆∆G = –9 kJ•mol–1 = ∆GArg/G/Arg+G – ∆GAla = RT ln(KD,Arg/G/Arg+G) – RT ln(KD,Ala) –9/2.5 = ln(2.5 × 10–8) – ln(KD,Ala) –3.6 = –17.5 – ln(KD,Ala) KD,Ala = 9.1 × 10–7 b. The lack of additivity suggests that the Arg and T might interact directly, as removing either side of the interaction has the same effect as removing both sides. Given that the T occurs in an AT-rich stretch of DNA and that Arg residues can recognize such stretches through a narrowed minor groove, it is likely that this mechanism contributes to the specific recognition of this stretch of DNA. 23. Each subunit of a homodimeric transcription factor can individually recognize a DNA half-site with a 5 µM KD. The dimeric form of the transcription factor recognizes the full inverted repeat DNA site with a 50 nM KD. How much free energy is used to induce the conformational changes of the protein and DNA during the binding of the dimeric transcription factor? Answer: We expect KD,dimer = (KD,monomer)2 = (5 × 10–6) = 2.5 × 10–11 ∆∆Gconf = RT ln(KD, dimer measured) – RT ln(KD, dimer expected) = RT ln(50 × 10–8) – RT ln(2.5 × 10–11) = 19.0 kJ•mol–1 24. A complex of seven transcription factors binds a DNA enhancer element. The binding is cooperative. What

The Molecules of Life by John Kuriyan, Boyana Konforti, and David Wemmer © Garland Science

4

Chapter 13: Specificity of Macromolecular Recognition are two molecular mechanisms that the transcription factors might use to achieve this cooperativity? Answer: The transcription factors may make contact with each other facilitating the assembly of the context. These protein–protein interactions would increase the apparent cooperativity of the protein–DNA interactions. Second, some of the transcription factors may recognize a distorted DNA structure. If one transcription factor stabilizes this distorted structure (and in essence pays the energetic penalty for distorting it away from ideal geometry) then subsequent binding events can rely on the distorted DNA for specific interactions without contributing energy to induce a conformational change in the DNA.

The Molecules of Life by John Kuriyan, Boyana Konforti, and David Wemmer © Garland Science

The MOLECULES of LIFE Physical and Chemical Principles

Solutions Manual Prepared by James Fraser and Samuel Leachman

Chapter 14 Allostery

Problems True/False and Multiple Choice 1. Which of the following can result in an allosteric modulation of activity? a. Covalent modification such as phosphorylation or acetylation. b. Oligomerization. c. Binding of a ligand. d. Stabilizing an alternative conformation. e. All of the above. 2. The transcription of a gene is controlled by a transcription factor binding either glucose or lactose. When there is 0.2 mM of either glucose or lactose in the cell, the gene is transcribed at about 10% of the maximum. When there is more than 2 mM glucose in the cell, the gene is fully induced. However, when there is 2 mM lactose in the cell, the amount of transcription is approximately half of the maximum. At 40 mM of either glucose or lactose in the cell, the gene is fully induced. The transcriptional regulation is likely: a. Ultrasensitive with respect to both glucose and lactose. b. Ultrasensitive with respect to lactose but not glucose. c. Graded with respect to glucose. d. Ultrasensitive with respect to glucose but not lactose. e. Hyperbolic with respect to glucose but not lactose. 3. In allosteric proteins, ligand binding can only result in positive cooperativity. True/False 4. Which of the following is not a known allosteric effector of hemoglobin? a. b.

oxygen bisphosphoglycerate

c. low pH d. myoglobin e. CO2 5. Bacterial chemotaxis involves only random, Brownian movement. True/False 6. Because of the importance of the F helix in oxygen binding, the allosteric hemoglobin proteins of all organisms are tetrameric assemblies. True/False 7. Which of the following statements about the Hill coefficient are true? i. It is the steepness of the log-log binding isotherm at the half saturation point. ii. Allosteric systems have a Hill coefficient of exactly 1. iii. The maximum value for a dimeric protein is 2. a. b. c. d.

Only (i) is true. Both (i) and (iii) are true. Both (ii) and (iii) are true. All of the statements are true.

Fill in the Blank 8. When a system is ultrasensitive, the response to an input is _______ than expected from the graded, hyperbolic response. Answer: sharper 9. Oxygen biases the equilibrium of hemoglobin towards the _____ state. Answer: R or relaxed

The Molecules of Life by John Kuriyan, Boyana Konforti, and David Wemmer © Garland Science

2

Chapter 14: Allostery

10. The extent of flagellar clockwise rotation is governed by the ___________________ of the protein CheY. Answer: phosphorylation 11. The __________ histidine residue of hemoglobin senses the effective size of the heme iron atom. Answer: proximal 12. An effect of colocalization is to greatly increase the local __________ of two proteins. Answer: concentration

Quantitative/Essay 13. A dimeric enzyme, glucokinase, has a binding site for glucose in each subunit. The KD for the first binding event is 1 mM and the KD for the second event is 10 µM. a. What is the Hill coefficient? b. Is this protein positively or negatively cooperative with respect to glucose binding? Answer: a. nH = 2/(1 + √(KD2/KD1)) = 2/(1 + √(0.01/1)) = 1.8 b. nH > 1, therefore it is positively cooperative. 14. A dimeric hemoglobin is isolated from a fish. Each subunit contains a binding site for a xenon gas atom. The KD for the first binding event is measured to be 23 nM. The KD for the second binding event is measured to be 3.5 µM. a. What is the Hill coefficient? b. Is this protein positively or negatively cooperative with respect to xenon binding?

a. What fraction of the protein has ligand bound at 10 nM concentration of free ligand? b. A single point mutation abolishes all cooperativity in the protein such that the protein binds its ligand with an apparent KD equal to 25 nM. What is the fraction of the protein that has ligand bound at 10 nM concentration of free ligand? Answer: a. KD2 = KD1 × (2/nH – 1)2 = 25 × (2/1.9 – 1)2 = 1.56 nM f/(1 – f) = ([L]/KD1 + ([L]/KD1) × ([L]/KD2))/(1 + [L]/KD1) = (10/25 + (10/25) × (10/1.56))/(1 + 10/25) = 2.11 f = 2.11 – 2.11f 3.11 = 2.11 f = 2.11/3.11 f = 0.68 b. f/(1 – f) = [L]/KD = 10/25 = 0.4 f = 0.4/1.4 = 0.29 18. The first and second binding sites of a positively cooperative allosteric dimeric protein have KD values of 100 mM and 10 µM, respectively. a. Sketch the binding isotherms as log[f/(1 – f )] versus log([L]). b. What is the value of the Hill coefficient?

Q14.18a Answer: a.

Answer: a. nH = 2/(1 + √(KD2/KD1)) = 2/(1 + √(3.5/0.023)) = 0.15 b. nH < 1, therefore it is negatively cooperative. 15. A dimeric enzyme with two identical binding sites has a Hill coefficent of 1.3. If the KD of the first binding site is 200 nM, what is the KD of the second binding site?

8 7 6 5 4 3 2 1

KD1 KD2 isotherm

–7

–6

–5

–4

–3

–2

–1 –1

Answer: f/(1 – f) = ([L]/KD1 + ([L]/KD1) × ([L]/KD2))/(1 + [L]/KD1) = (15/20 + (15/20) × (15/2))/(1 + 15/20) = 3.64 17. A dimeric allosteric protein is isolated. A scientist determines that the value of KD for the first binding site is 25 nM and that the Hill coefficient is 1.6.

2

3

–2 –3 –4 –5 –6

Answer: KD2 = KD1 × (2/nH – 1)2 = 200 × (2/1.3 – 1)2 = 58 nM 16. What is the ratio of bound to unbound receptor [f/(1 – f )] for a dimeric protein with two binding sites (KD1 = 20 nM; KD2 = 2 nM) at 15 nM concentration of ligand?

1

b. nH = 2/(1 + √(KD2/KD1)) = 2/(1 + √(10–5/10–1)) = 1.98 19. The first and second binding sites of a negatively cooperative allosteric dimeric protein have KD values of 100 µM and 10 mM, respectively. a. Sketch the binding isotherms as log[f/(1 – f )] versus log([L]). b. What is the value of the Hill coefficient?

The Molecules of Life by John Kuriyan, Boyana Konforti, and David Wemmer © Garland Science



PROBLEMS and solutions

Q14.19a

Answer: a.

–7

–6

–5

AAC displays positive cooperativity at low temperatures and negative cooperativity at high temperatures. At 10°C, AAC has KD1 = 330 µM and KD2 = 130 µM. At 37°C, AAC has KD1 = 200 µM and KD2 = 230 µM.

7 6 5 4 3 2 1

KD1 KD2 isotherm

–4

–3

–2

–1 –1

3

a. What is the Hill coefficient at each temperature? b. What is the change in fraction bound at 50 µM ligand between 10°C and 37°C? c. Describe a reasonable mechanism underlying this unusual behavior. 1

2

3

–2 –3 –4 –5

b. nH = 2/(1 + √(KD2/KD1)) = 2/(1 + √(10–2/10–4)) = 0.18 20. Two homologs of a protein are isolated. Homolog A is a monomer that binds glucose with a KD of 4 mM. Homolog B is a positively cooperative dimer. The KD of the first binding site of homolog B is measured at 4 mM. At 1 mM concentration of glucose, it is found that homolog B binds twice as much glucose as homolog A. What is the Hill coefficient of homolog B? Answer: For homolog A: f/(1 – f) = [L]/KD = 1/4 = 0.25 f = 0.25/(1 + 0.25) = 0.2 Since homolog B binds twice as much glucose, f = 0.4; f/(1 – f) = 0.667 f/(1 – f) = 0.667 = ([L]/KD1 + ([L]/KD1) × ([L]/KD2))/(1 + [L]/ KD1) 0.667 = (1/4 + (1/4) × (1/KD2))/(1 + 1/4) = (0.25 + 0.25 × (1/KD2))/(1.25) KD2 = 0.429 mM nH = 2/(1 + √(0.429/4)) = 1.51 21. A cyclist is interested in cheating in a race by delivering more oxygen to his muscles. The cyclist reasons that since bisphosphoglycerate (BPG) stabilizes the “T” state of hemoglobin, which reduces its affinity for oxygen, that reducing the BPG concentration in his blood cells should be good for his performance. How might removing BPG have a detrimental effect on the delivery of oxygen to his muscles? Answer: BPG acts as an allosteric effector for hemoglobin. Removal of BPG will result in a sharper switch between saturated and empty hemoglobin. Without this effect, the binding isotherm will not match the difference in oxygen concentration between the arterial and venous blood and likely lead to suboptimal delivery of oxygen to his muscles. 22. Due to a fascinating coupling between folding and ligand binding, the dimeric acetyl transferase enzyme

Answer: a. For 10°C nH = 2/(1 + √(KD2/KD1)) = 2/(1 + √(130/330)) =1.23 For 37°C nH = 2/(1 + √(KD2/KD1)) nH = 2/(1 + √(230/200)) = 0.97 b. For 10°C f/(1 – f) = ([L]/KD1 + ([L]/KD1) × ([L]/KD2))/(1 + [L]/KD1) = (50/330 + (50/330) × (50/130))/(1 + 50/330) = 0.18 f = 0.18/1.18 = 0.15 For 37°C f/(1 – f) = ([L]/KD1 + ([L]/KD1) × ([L]/KD2))/(1 + [L]/KD1) = (50/200 + (50/200) × (50/230))/(1 + 50/200) = 0.243 f = 0.243/1.243 = 0.20 Thus, 4% more enzymes are bound at the higher temperature. Even though AAC is negatively cooperative at higher temperatures, more of the protein is bound to ligand. This is driven by the favorable change in KD1 at higher temperatures. c. One possibility is that binding to the first subunit destabilizes the second subunit by disrupting the interface between them. At high temperatures this effect is enough to partially unfold the second subunit, overriding the intrinsic cooperativity of the native dimer. Hint: See LA Freiburger et al. and AK Mittermaier (2011) Nat. Struct. Mol. Biol. 18, 288–294. 23. How does CO2 directly and indirectly stabilize the “T” state of hemoglobin in venous blood? Answer: CO2 directly stabilizes the “T” state by reacting with the N-terminal amino groups of the protein. This alters the interfacial interactions between the α and β subunits, stabilizing the “T” state relative to the “R” state. CO2 indirectly stabilizes the “T” state by reducing the pH within red blood cells. The reduction in pH leads to the Bohr effect where β subunit His 146 becomes protonated. This disrupts interactions centered around the ion pair with β Asp 94 and an interaction to α Lys 40. Without these interactions, the “T” state is stabilized. 24. Two proteins are modified by myristoylation, which targets them to the plasma membrane in a cell at 25°C.

The Molecules of Life by John Kuriyan, Boyana Konforti, and David Wemmer © Garland Science

4

Chapter 14: Allostery This changes their effective local concentration from 10 nM to 1 µM. Assume that any favorable mutation would decrease the value of ΔGo for binding by 4 kJ•mol–1. How many favorable mutations would have to occur in the absence of colocalization to result in an equivalent effective affinity as observed when the two proteins are colocalized? Answer: ∆∆G = RT ln(10–8 × 10–8)/(10–6 × 10–6) = 2.5 × ln (0.001) = –23 kJ•mol–1 If each mutation added 4 kJ• mol–1, the at least six mutations would be required to result in as high an effective affinity.

25. A scientist finds two pathways, both of which depend on distinct kinase activities, which activate a stress response in yeast. One pathway responds to elevated levels of salt and the other responds to elevated levels of caffeine. Both pathways result in transcription of the chaperone Hsp90. Below is a table listing the relative concentrations of salt or caffeine and the measured transcription of Hsp90. Explain which pathway likely contains a single kinase and which pathway likely contains a kinase cascade similar to the MAP kinase pathway. [NaCl] (mM)

Hsp90 transcription

[Caffeine] (mM)

Hsp90 transcription

1

0

1

0

2

1

2

0

3

1

3

1

4

40

4

1.1

5

100

5

2

10

100

10

10

100

100

100

50

1000

100

1000

100

Answer: Only the salt response is ultrasensitive (in that it undergoes a complete activation in a ~2-fold rather than ~100-fold concentration change). Ultrasensitivity resulting from the coupling of multiple kinases converts a graded input signal into a sharp switch. This situation is observed only for the salt response, which likely has a kinase cascade, and not for the caffeine response, which is more hyperbolic.

The Molecules of Life by John Kuriyan, Boyana Konforti, and David Wemmer © Garland Science

The MOLECULES of LIFE Physical and Chemical Principles

Solutions Manual Prepared by James Fraser and Samuel Leachman

Chapter 15 The Rates of Molecular Processes

Problems True/False and Multiple Choice 1. What is the order of this elementary reaction: A + 2B → 1C a. 2 b. 7 c. 4 d. 3 e. 8 2. The addition of a catalyst increases the rate of the reaction but not the equilibrium constant. True/False 3. For a reaction with a larger ΔG compared to a reaction with a smaller ΔG, a. The reaction with the larger ΔG is always the faster reaction. b. The reaction with the larger ΔG is always the slower reaction. c. The two rates are equal. d. It is impossible to decide which reaction is faster. 4. For elementary reactions of zero and second order with a rate constant of 1 (with appropriate units), the second-order reaction half-life is a shorter period of time. True/False 5. The most basic step used to describe a reaction process is: a. The transition state. b. An elementary reaction. c. A unimolecular reaction. d. The steady state. e. The equilibrium rate. 6. The probability of a bimolecular reaction occurring is related to the rate of collisions between the two species of molecules. True/False

7. After excitation, the intensity of light emitted by a sample of fluorescent molecules drops exponentially with time. True/False

Fill in the Blank 8. The symbol “‡” is used to represent the ______________ of the reaction. Answer: transition state 9. The rate law gives the relationship between the rate of a reaction and the ________ of the chemical species involved in the reaction. Answer: concentrations 10. The rate constant for a _____-order elementary reaction is M–1•sec–1. Answer: second 11. The rate of product formation at the end of a pathway is determined by the rate of the ______ step. Answer: slowest 12. Catalysts can work by changing the reaction mechanism, ________, or _________. Answer: decreasing the activation energy, increasing the pre-exponential factor/[increasing the rate of collisions with favorable orientation]

Quantitative/Essay 13. How do the kinetics of hydrolysis of ATP to ADP and phosphate by water make it a good energy reservoir for the cell?

The Molecules of Life by John Kuriyan, Boyana Konforti, and David Wemmer © Garland Science

2

Chapter 15: The Rates of Molecular Processes Answer: Hydrolysis of ATP yields a large amount of energy (~30 kJ•mol–1) but the forward rate is very slow. Unless an enzyme catalyst is present, water does not spontaneously react frequently with ATP. Due to this slow rate of hydrolysis, very little energy is lost unless a catalyst initiates the reaction.

14. Problem Below is a15.14 two-dimensional energy versus reaction coordinate diagram for the reaction AB + C → A + BC.

r1



r2 Circle the approximate area that represents the transition state and justify your answer. If r1 corresponds to the A-B distance, and r2 to the B-C distance, indicate what molecules would be present in the regions corresponding to the left hand and right hand sides of the reaction above. Answer: The transition state is the highest point of energy on the 15Q14 trajectory between reactants and products.

16. A reaction is half complete after 20 minutes. After 40 minutes the reaction is two-thirds complete. When will the reaction be 90% complete? Answer: First we must determine the order of the reaction. The reaction cannot be first order because the half-life is not constant (it halves in 20 minutes, and then in the next 20 minutes only decreases to one-third). It cannot be a zeroth order reaction, where we would expect the half life to decrease as the reaction proceeded, since as the reactants decrease the half life increases. We can test whether the reaction is second order: t1/2 = 1/k[A0] 20 min = 1/k[1] k = 1/20 min = 0.05 min–1 Test for second order kinetics when 2/3 complete (A40 = 1/3): 1/[A0] – 1/[A40] = –k × t 1 – 1/(1/3) = –0.05 min–1 × t –2 = –0.05 min–1 × t t = 40 min, which agrees with the statement, confirming second-order kinetics When the reaction is 90% complete, [A] = 1/10 1/[A0] – 1/[A] = –k × t 1 – 10 = –0.05 × t –9/–0.05 min–1 = t t = 180 minutes 17. A protein (P) can either fold properly into the native state (N) or aggregate into a misfolded form (A). Both processes obey first-order kinetics. The branching ratio ([N]/[A]) is 9 and the effective rate constant, keff , is 15 sec–1. What is the rate constant for native state folding?

r1

Right side A + BC

Transition state

Apply the rate constant, k, the final mass (1.5 μg) [A], and starting mass (15 μg) [A0] to the first order rate equation: [A] = [A0]e–kt t = –ln([A]/[A0])/k t = –ln(1.5 μg/15 μg)/0.053 hour–1 t = –ln(0.1)/0.053 hour–1 t = 43 hours

Left side AB + C

r2 15. Iodine-123 is important for medical imaging studies and follows first-order decay kinetics. A 15-µg sample of I-123 has decayed to 7.5 µg after 13 hours. After how much time will it decay to only 1.5 µg? Answer: It has decayed to half the original mass after 13 hours, which therefore corresponds to the half-life (t1/2). To find the rate constant: k = ln(2)/t1/2 = 0.693/13 hour = 0.053 hour–1

Answer: From the branching ratio: kN/kA = 9 kA = kN/9 keff = (kN + kA) keff = (kN + kN/9) keff = 10kN/9 (15 sec–1 × 9)/10 = kN kN = 13.5 sec–1 18. Upon excitation, a modified green fluorescent protein emits photons that yield an initial intensity of 10,000 units in the fluorimeter. After 2 nanoseconds, the signal has decayed to 300 units. If the rate constant for fluorescent production of light (kf) is 0.1 nsec–1, what are the values of the rate constant for heat production (kh) and the fluorescence lifetime (τf)? Answer: I2nsec/I0nsec = kf[F*]0e–(kf+kh)t/kf[F*]0e–0 = e–(kf+kh)(2 nsec) 300/10000 = e–(0.1 nsec–1 + kh)(2 nsec)

The Molecules of Life by John Kuriyan, Boyana Konforti, and David Wemmer © Garland Science



PROBLEMS and solutions 0.03 = e–(0.1 nsec–1 + kh)(2 nsec) ln(0.03) = –(0.1 nsec–1 + kh)(2 nsec) 0.1 nsec–1 + kh = 1.75 nsec kh = 1.65 nsec–1 τf = 1/(kf + kr) = 1/(0.1 nsec–1 + 1.65 nsec–1) = 0.570 nsec

19. An experiment is performed to measure the affinity of the human proline isomerase hCypA to the inhibitor cyclosporin, analogous to the experiments described for imatinib in this chapter. At 25 µM cyclosporin, the value of kobs is measured to be 12.55 sec–1. Given that the offrate is 0.01 sec–1, what is the on-rate? Answer: kobs = kf[L] + kr 12.55 sec–1 = kf[25 μM] + 0.01 sec–1 12.54 sec–1/25 μM = kf kf = 0.501 μM–1•sec–1 kf = 0.501 × 10–6 M–1•sec–1 approximate values from M.A. Wear et al. and M.D. Walkinshaw, Anal. Biochem. 345, 214–226, 2005. 20. A homologous proline isomerase, vCypA, is isolated from a deadly virus. The kinetics of cyclosporin binding to vCypA were measured. The off-rate is 0.1 sec–1 and the on-rate is 0.05 µM–1•sec–1. What concentration of cyclosporin was used to yield kobs = 120 sec–1? Answer: kobs = kf[L] + kr 120 sec–1 = 0.05 μM–1•sec–1[L] + 0.1 sec–1 [L] = 119.9 sec–1 / 0.05 μM–1•sec–1 [L] = 2398 μM = 2.398 mM 21. Several experiments indicate that the vCypA is essential for viral replication. Use the parameters calculated in Problems 19 and 20, and assume that the concentration of cyclosporin would be much larger than the concentration of either vCypA or hCypA. For each protein, calculate the value of KD for cyclosporin. Given the ratio of the two KD values, explain whether cyclosporin could be an effective treatment for this virus. Answer: For hCypA: KD = kr/kf = 0.01 sec–1/0.501 × 10–6 M–1•sec–1 = 2 × 10–8 M–1•sec–1 For vCypA: KD = kr/kf = 0.1 sec–1/0.05 × 10–6 M–1•sec–1 = 2 × 10–6 M–1•sec–1 The ratio of KDs: KD-hCypA / KD-vCypA = 2 × 10–8 M–1•sec–1/2 × 10–6 M–1•sec–1 = 0.01 The ratio of cyclosporine-bound to unbound hCypA is 100 times greater than for vCypA. This suggests that the drug might not be a very effective treatment for the virus, since it will preferentially bind to the human protein rather than the viral one. 22. The activation energy for proline isomerization of a peptide depends on the identity of the preceding residue and obeys Arrhenius rate behavior. Experiments

3

are conducted on the isomerization of an alanineproline peptide. At 25°C (298 K) the observed rate constant is 0.05 sec–1 and the value of EA is calculated to be 60 kJ•mol–1. What is the value of the preexponential factor (A)? Similar measurements are performed on a phenylalanine-proline peptide at 25°C, with a measured rate constant of 0.005 sec–1. Assuming an identical preexponential factor as the alanine-proline peptide, what is the activation energy for this peptide? Answer: ln (k) = –EA/RT + ln(A) ln(k) + EA/RT = ln(A) ln(0.05) + 60,000/(8.314 × 298) = ln(A) A = e21.22 = 1.64 × 109 sec–1 Assuming that A = 1.64 × 109 sec–1 for the phenylalanine–proline peptide EA = –(ln(k) – ln(A)) × RT = –(ln(0.005) – ln(1.64 × 109)) × 8.314 × 298 = 65.7 kJ•mol–1 23. A relaxation experiment probes the homodimerization of an oligonucleotide. At 10-nM oligonucleotide, the apparent rate constant (kapparent) is 2.06 sec–1. At 100-nM oligonucleotide, the apparent rate constant (kapparent) is 6.34 sec–1. What are the association and dissociation rate constants? Answer: kapp2 = kr2 + 8kfkr[N] At 10 nM: 4.24 = kr2 + 80kfkr (4.24 – kr2)/80 = kfkr Substitute into 100 nM 40.2 = kr2 + 8 × (4.24 – kr2)/80 × [100] 40.2 = kr2 + (4.24 – kr2) × 10 40.2 = kr2 + 42.4 – 10kr2 9 kr2 = 2.24 kr2 = 0.249 kr = 0.500 sec–1 Substitute this value to solve for kf 4.24 = 0.25 + 80kf × 0.5 4 = 40kf kf = 0.100 sec–1•nM–1 24. A protein folding reaction has two intermediate states, each of which individually obeys Arrheniustype behavior. At low temperatures, forming the first intermediate is rate-limiting. At high temperatures, forming the second intermediate is rate-limiting. a. Does the protein folding reaction obey Arrheniustype behavior over all temperatures? b. Forming which intermediate has a higher activation energy? Answer: a. No. The rate-determining step changes as the temperature changes, so the apparent rate constant will change as well. Above the temperature where forming the second intermediate is rate-determining, the slope of the temperature dependence will be dominated by the activation energy of forming the second intermediate. Below that temperature, the slope will be dominated by the activation energy of

The Molecules of Life by John Kuriyan, Boyana Konforti, and David Wemmer © Garland Science

4

Chapter 15: The Rates of Molecular Processes forming the first intermediate. This generates curvature in the Arrhenius plot that is uncharacteristic of linear Arrhenius-type behavior. b. Formation of the first intermediate has a higher activation energy. At low temperatures, forming the first intermediate is slower than forming the second. As the temperature is increased, the rate of forming the first intermediate increases faster than the rate of forming the second, such that at high temperatures, formation of the second intermediate is slower. Because forming the first intermediate has greater temperature dependence, it must necessarily have a higher activation energy.

The Molecules of Life by John Kuriyan, Boyana Konforti, and David Wemmer © Garland Science

The MOLECULES of LIFE Physical and Chemical Principles

Solutions Manual Prepared by James Fraser and Samuel Leachman

Chapter 16 Principles of Enzyme Catalysis

Problems True/False and Multiple Choice 1. The initial reaction velocity for an enzyme reaction reaches a maximum at high substrate concentration because the free enzyme can no longer regenerate at the end of each reaction cycle. True/False

6. An enzyme inhibitor is observed to alter the KM but not the Vmax of a reaction. This inhibitor is most likely: a. b. c. d. e.

A noncompetitive inhibitor. A competitive inhibitor. An allosteric inhibitor. A substrate-dependent noncompetitive inhibitor. A covalent inhibitor.

2. The turnover number for an enzyme obeying Michaelis– Menten kinetics is: a. k2. b. kcat/KM. c. k1/k–1. d. (k1 + k2). e. ΔG‡.

7. Due to its extremely slow dissociation kinetics, the protein bovine pancreatic trypsin inhibitor (BPTI) has broad specificity and inhibits more proteases than protease inhibitors that are small molecules.

3. Catalytic antibodies are generally less efficient than natural enzymes that catalyze the same reactions.

Fill in the Blank

True/False 4. A metabolic enzyme generates the amino acid methionine. For a given substrate concentration, an experiment conducted in the presence of high initial concentrations of methionine generates less new methionine than an experiment conducted with no initial methionine present. This is likely an example of: a. A ping-pong mechanism of substrate binding. b. A proximity effect. c. Substrate strain. d. Product inhibition. e. A reaction intermediate. 5. Which of the following is not a commonly observed feature of proteases? a. The catalytic triad in the active site. b. Exclusively hydrophobic residues in the active site. c. A cysteine residue in the active site. d. Metal ions coordinated in the active site. e. A pair of acidic residues in the active site.

True/False

8. In the schemes for the catalyzed reactions considered in this chapter, S, E, and P refer to _______________, ______________, and ______________, respectively. Answer: substrate, enyzme, product 9. The specificity constant or catalytic efficiency is the ratio between __________ and _________. Answer: kcat, KM 10. In a plot of initial velocity versus substrate concentration, an allosteric enzyme displays a ____________ curve, whereas a non-allosteric enzyme that obeys Michaelis–Menten kinetics displays a ____________ curve. Answer: sigmoidal, hyperbolic 11. The geometry of competitive inhibitors commonly mimics the _____________ of the reaction that the enzyme normally catalyzes. Answer: transition state

The Molecules of Life by John Kuriyan, Boyana Konforti, and David Wemmer © Garland Science

Figure Q16A 2

Chapter 16: Principles of Enzyme Catalysis

12. Proteins are not the only polymers that act as catalysts. Catalytic ______ molecules are also essential for cells, including playing an essential role in protein synthesis.

a.

Answer: RNA/ribozyme

14. Show that the equations plotted in Lineweaver–Burk and Eadie–Hofstee plots are equivalent. Answer: Start with the Lineweaver–Burk equation: 1/v0 = 1/Vmax + KM/Vmax[S] Multiply all by Vmax, Vmax/v0 = 1 + KM/[S] Multiply all by v0, Vmax = v0 + v0KM/[S] Rearrange for v0, v0 = –v0KM/[S] + Vmax (which is the Eadie–Hofstee equation).   → E•⋅ S → E + P , calculate 15. In a reaction, E + S ←  k k2

−1



the value of KM if the forward rate constant (k1) for E•S formation is 4.3 × 106 sec–1•M–1, the reverse rate constant (k–1) for E•S dissociation is 2.4 × 102 sec–1, and the turnover number (k2) is 1.2 × 103 sec–1. Answer: KM = (k–1 + k2)/k1 = (2.4 × 102 sec–1 + 1.2 × 103 sec–1)/4.3 × 106 sec–1•M–1 = 3.3 × 10–4 M

16. Presented below are Lineweaver–Burk plots for enzymatic reactions with (red) and without (blue) inhibitor. What type of inhibition is occurring in each case?

1.0

Figure Q16B0

0

0.2

0.4

0.6 0.8 –1 1/[S] (M )

1.0

1.2

0

0.2

0.4

0.6 0.8 –1 1/[S] (M )

1.0

1.2

0

0.2

0.4

1.0

1.2

b.

0.7 0.6 1/v (M–1•sec)

Answer: o‡cat ‒ ΔGo‡uncat)/RT) kcat/kuncat = e(‒(ΔG o‡cat ‒ ΔGo‡uncat)/8.314 × 298) 105 = e(‒(ΔG 5 o‡ ln(10 ) = (‒(ΔG cat ‒ ΔGo‡uncat)/2.5) ΔGo‡cat ‒ ΔGo‡uncat = –28.5 kJ•mol–1

1.5

0.5

0.5 0.4 0.3 0.2 0.1

Figure Q16C 0

c.

4.5 4.0 3.5

1/v (M–1•sec)

13. At 25°C, an enzyme accelerates a reaction by a factor of 105 over the uncatalyzed reaction in water. If the effect of the enzyme is solely to reduce the energy of the transition state, by what amount does it reduce the energy of the transition state (EA)?

1/v (M–1•sec)

2.0

Quantitative/Essay

k1

2.5

3.0 2.5 2.0 1.5 1.0 0.5 0

The Molecules of Life by John Kuriyan, Boyana Konforti, and David Wemmer © Garland Science

0.6 0.8 1/[S] (M–1)



PROBLEMS and solutions Answer: a. Competitive inhibition. b. Substrate-dependent noncompetitive inhibition. c. Noncompetitive inhibition.

Substitute slope into: 1/v = slope × 1/[S] + 1/Vmax 0.5 = 0.8 × 0.5 + 1/Vmax 0.1 = 1/Vmax Vmax = 10 mM•sec–1 KM/Vmax = 0.8, K* M = 8 mM

17. The table below lists initial velocities measured for an enzymatic reaction at different substrate concentrations in the presence and absence of an inhibitor. The enzyme concentration is identical in both reactions.

v, no inhibitor 
 (mM•sec–1)

v, with inhibitor 
 (mM•sec–1)

1

2.50

1.11

2

4.00

2.00

5

6.25

3.85

10

7.69

5.56

20

8.70

7.14

[S] (mM)

a. Graph a Lineweaver–Burk plot. b. What are the apparent values of Vmax and KM for each experiment? c. What is the inhibition mechanism? d. If the concentration of inhibitor is 0.5 mM, what is the value of KI? Answer: Q16.17a a. Red line is with inhibitor, blue line is with no inhibitor.

1/v (mM–1•sec)

1.0

no inhibitor with inhibitor

0.8 0.6

3

c. The slope changes but the y intercept does not. This is reflected in the apparent KM increase upon addition of inhibitor and the constant Vmax between the two reactions. These observations are indicative of a competitive inhibitor. d. For a competitive inhibitor, the apparent KM with inhibitor is related to the KM without inhibitor: K* M = KM(1 + [I]/KI) 8 mM = 3 mM (1 + 0.5 mM/KI) 8 mM = 3 mM + 1.5 mM/KI 5 mM = 1.5 mM/KI KI = 1.5 mM/5 mM KI = 0.3 mM 18. The table below lists initial velocities measured for an enzymatic reaction at different substrate concentrations in the presence and absence of an inhibitor. The enzyme concentration is identical in both reactions. [S] (mM)

v, no inhibitor (mM•sec–1)

v, with inhibitor 
 (mM•sec–1)

1

1.000

0.923

5

1.154

1.053

10

1.176

1.071

50

1.195

1.087

100

1.198

1.089

a. Graph a Lineweaver–Burk plot for each set of data. b. What are the apparent values of Vmax and KM for each experiment? c. What is the inhibition mechanism? d. If the concentration of inhibitor is 10 nM, what is the value of KI?

0.4 0.2 0.0 0.0

0.2

0.4

1/[S]

0.6

0.8 –1 (mM )

1.0

1.2

Answer: a. Red line is with inhibitor, blue line is with no Q16.18a inhibitor. 1.2 1.0

1/v (mM–1•sec)

b. First calculate the slope of the lines—the data are nearly perfectly linear, so we will just use the first two data points to calculate the relevant parameters. For no inhibitor: Slope = KM/Vmax = (y1 – y2)/(x1 – x2) = (0.4 – 0.25)/ (1 – 0.5) = 0.3 Substitute slope into: 1/v = slope × 1/[S] + 1/Vmax 0.25 = 0.3 × 0.5 + 1/Vmax 0.1 = 1/Vmax Vmax = 10 mM•sec–1 KM/Vmax = 0.3, KM = 3 mM For with inhibitor: Slope = KM/Vmax = (y1 – y2)/(x1 – x2) = (0.9 – 0.5)/(1 – 0.5) = 0.8

0.8 0.6 0.4 no inhibitor with inhibitor

0.2 0.0 0.0

0.2

0.4

0.6

0.8

1/[S] (mM–1)

1.0

1.2

The Molecules of Life by John Kuriyan, Boyana Konforti, and David Wemmer © Garland Science

4

Chapter 16: Principles of Enzyme Catalysis

c. The y intercept changes, but the slope does not. This is reflected in the apparent KM decrease upon addition of inhibitorand the apparent Vmax decrease, but the maintenance of a constant ratio between the two constants. These observations are indicative of a substrate-dependent noncompetitive inhibitor. d. 10 nM is 0.00001 mM, K*M = KM/(1 + [I]/KI) K*M = KM/(1 + [I]/KI) 0.182 mM = 0.2 mM/(1 + 0.00001 mM/KI) 0.00001 mM/KI = 0.2 mM /0.182 mM – 1 KI = 0.1 μM or 100 nM 19. The table below lists initial velocities measured for an enzymatic reaction at different substrate concentrations in the presence and absence of an inhibitor. The enzyme concentration is identical in both reactions. [S] (µM)

v, no inhibitor 
 (µM•sec–1)

v, with inhibitor 
 (µM•sec–1)

10

8.93

6.94

20

10.42

8.93

30

11.03

9.87

100

12.02

11.57

200

12.25

12.02

a. Graph a Lineweaver–Burk plot for each set of data. b. What are the values of Vmax and KM for each experiment? c. What is the inhibition mechanism ? d. If the concentration of inhibitor is 100 nM, what is the value of KI?

Q16.19a Answer: a. 0.16 0.14

1/v (µM–1•sec)

b. For no inhibitor: Slope = KM/Vmax = (y1 – y2)/(x1 – x2) = (1 – 0.85)/(1 – 0.1) = 0.167 Substitute slope into: 1/v = slope × 1/[S] + 1/Vmax 1 = 0.167 × 1 + 1/Vmax Vmax = 1.2 mM•sec–1 KM/Vmax = 0.167, KM = 0.167 × 1.2 = 0.2 mM With inhibitor: Slope = KM/Vmax = (y1 – y2)/(x1 – x2) = (0.95 – 0.92)/(0.2 – 0.02) = 0.167 Substitute slope into: 1/v = slope × 1/[S] + 1/Vmax 0.95 = 0.167 × 0.2 + 1/Vmax Vmax = 1.09 mM•sec–1 KM/Vmax = 0.167, KM* = 0.167 × 1.09 = 0.182 mM

0.12 0.10 0.08 0.06 0.04 no inhibitor with inhibitor

0.02 000 0.0

0.02

0.04

0.06

1/[S]

0.08 –1 (µ M )

0.10

0.12

b. First calculate the slope of the lines—the data are nearly perfectly linear, so we will just use the first two data points to calculate the relevant parameters. For no inhibitor: Slope = KM/Vmax = (y1 – y2)/(x1 – x2) = (0.112 – 0.096)/ (0.1 – 0.05) = 0.32 Substitute slope into: 1/v = slope × 1/[S] + 1/Vmax 0.112 = 0.32 × 0.1 + 1/Vmax 0.08 = 1/Vmax Vmax = 12.5 μM•sec–1 KM/Vmax = 0.32, KM = 4 μM For no inhibitor: Slope = KM/Vmax = (y1 – y2)/(x1 – x2) = (0.144 – 0.112)/ (0.1 – 0.05) = 0.64 Substitute slope into: 1/v = slope × 1/[S] + 1/Vmax 0.112 = 0.64 × 0.05 + 1/Vmax 0.08 = 1/Vmax Vmax = 12.5 KM/Vmax = 0.64, K* M = 8 μM c. The slope changes but the y intercept does not. This is reflected in the apparent KM increase upon addition of inhibitor and the constant Vmax between the two reactions. These observations are indicative of a competitive inhibitor. d. For a competitive inhibitor, the apparent KM with inhibitor is related to the KM without inhibitor: K* M = KM (1 + [I]/KI) 8 = 4(1 + 0.1/KI) 8 = 4 + 0.4/KI 4 = 0.4/KI KI = 0.4/4 KI = 0.1 μM or 100 nM 20. An experiment with 10 nM of an enzyme obeying Michaelis-Menten kinetics yields a Vmax of 7 × 10–3 M•sec–1. a. What is the turnover number (k2)? b. The experiment is repeated in the presence of a noncompetitive inhibitor and the Vmax is reduced to

The Molecules of Life by John Kuriyan, Boyana Konforti, and David Wemmer © Garland Science



PROBLEMS and solutions 5 × 10–4 M•sec–1. What fraction of the enzyme is bound to the inhibitor?

Answer: Recognize v0 = Vmax/(1 + KM/[S]), so a concentration near KM will show half the velocity of the highest velocity observed, assuming it is close to Vmax. a. ~3–5 × 10–3 M b. ~4 × 10–9 M c. ~200 × 10–3 M

Answer: a. Vmax = k2[E] 7 × 10–3 M•sec–1 = (k2)(1 × 10–8 M) k2 = 7 × 105 sec–1 b. Vmax = k2(1 – f)[E] (1 – f) = 5 × 10–4 M–1•sec–1/(7 × 105 sec–1 × 1 × 10–8 M) 1 – f = 0.07 f = 0.93 Therefore 93% of the enzymes are bound to the inhibitor. 21. Given the following three data tables of substrate concentrations and initial velocities for enzymes that obey Michaelis–Menten kinetics, estimate KM for each enzyme in molar units.

22. When the bi-substrate analog PALA is added to the enzyme ATCase at low concentration it increases the rate of reaction of aspartate and carbamylphosphate. However, at higher concentrations it decreases the reaction rate. How can PALA act as both an activator and inhibitor of ATCase? Answer: PALA mimics the two substrates of ATCase. Its affinity is greater than the substrate, and therefore outcompetes substrate for the active site. At moderate concentrations (where not every site on ATCase is occupied by PALA), the equilibrium of ATCase is shifted to favor the active conformation. Since the active conformation has a higher affinity for substrate, binding sites that are not occupied by PALA are able to increase the reaction velocity. At high concentrations of PALA all binding sites become occupied by inhibitor and turnover decreases.

a. [S] (mM)

v0 (mM•sec–1)

1

266.7

3

553.8

5

705.9

50

1121.5

500

1191.7

5000

1199.2

b. [S] (nM)

v0 (mM•min–1)

4

123.5

5

137.4

6

148.5

10

177.3

100

240.2

1000

249.0

[S] (mM)

v0 
(M•hour–1)

c.

1

0.00

10

0.01

100

0.07

200

0.10

1000

0.17

5000

0.19

5

23. In the search for the catalytic mechanism of an enzyme, three mutations of charged residues to alanine (which is uncharged) are made and compared with the wild type (WT) enzyme. At otherwise identical conditions and concentrations of enzymes, the following initial velocities (µM•sec–1) are measured as a function of pH.

pH

WT

Arg55Ala

Glu63Ala

Lys113Ala

4

1.3 × 103

1.3 × 103

1.3 × 102

1.5 × 103

5

8.7 × 105

8.7 × 105

1.3 × 102

9.5 × 105

6

8.1 × 104

8.1 × 104

1.3 × 102

8.1 × 104

7

5.3 × 103

5.3 × 103

1.3 × 102

5.2 × 103

a. Explain which residue likely acts as a general acid/ base during catalysis? b. What is a possible mechanism for the slightly increased reaction velocities observed in the Lys113Ala mutants at lower pH? Answer: a. Glu63 likely participates directly in catalysis. All other mutations have much higher catalytic efficiency (comparable to wild type). Additionally, Glu63Ala no longer displays any pH dependency when mutated to an uncharged residue. b. Having a Lys near the catalytic Glu will shift the pKa down and make it less likely to donate a hydrogen to the reaction. Removing this charge with the Lys113Ala mutant shifts the pKa slightly and leads to increased reactivity and increased reaction velocities. The Molecules of Life by John Kuriyan, Boyana Konforti, and David Wemmer © Garland Science

6

Chapter 16: Principles of Enzyme Catalysis

24. Why is triose phosphate isomerase considered to be an example of a “perfect enzyme”? Answer: Triose phosphate isomerase is considered a “perfect” enzyme because it catalyzes its reaction so fast that it is diffusion controlled.

The Molecules of Life by John Kuriyan, Boyana Konforti, and David Wemmer © Garland Science

The MOLECULES of LIFE Physical and Chemical Principles

Solutions Manual Prepared by James Fraser and Samuel Leachman

Chapter 17 Diffusion and Transport

Problems True/False and Multiple Choice 1. Bacterial movement towards nutrients resembles a biased random walk. True/False

c. d.

solute flux is higher in acetone. it is easier to concentrate water than it is acetone.

2. The concentration gradient of bicoid protein in Drosophila embryos is established by: a. passive diffusion of protein molecules. b. an mRNA gradient. c. spatially biased maternal deposition of mRNA. d. all of the above.


Fill in the Blank

3. The distance moved in two-dimensional diffusion has a square-root dependence on time ( t ). In three dimensions, the distance covered through diffusion has a cubed-root dependence on time ( 3 t ). True/False

9. The width of the Gaussian distribution describing a one-dimensional random walk ____________ with an increasing number of time steps.

4. In passive transport with no barriers, if the concentration of molecules is high in one area but low in another area, then a. the molecules will have a net flux to the area of low concentration. b. the concentration will tend to be equal in both areas after a long period of time. c. the molecules will not move unless ATP is added. d. both a and b. e. none of the above.
 5. A spherical virus with a radius of 30 nm will move more slowly by diffusion than a protein dimer with a radius of 5.5 nm. True/False 6. The rate of diffusion increases as friction increases. True/False 7. Water has a higher viscosity than acetone because a. it has a greater molecular mass. b. hydrogen bonds allow for the transfer of momentum.

8. ________ transport moves molecules through diffusion, while __________ transport drives movement through the coupling to chemical energy. Answer: Passive, active

Answer: increases or broadens 10. Viscosity is a measure of a fluid’s ___________ to particles moving through it. Answer: resistance 11. An oblate or prolate object will have increased friction relative to a _______ object that occupies the same volume. Answer: spherical 12. The velocity at which friction balances applied force is called the _____________. Answer: terminal velocity

Quantitative/Essay 13. The protein cyclophilin is a monomer with a diffusion constant of 1.2 × 10–8 cm2•sec–1 in water at 25°C. Cyclophilin binds HIV capsid and the complex has a diffusion constant of 7 × 10–9 cm2•sec–1. What is the difference in the typical (r.m.s.) distance that cyclophilin alone will travel versus the cyclophilin-capsid complex in 10 seconds?

The Molecules of Life by John Kuriyan, Boyana Konforti, and David Wemmer © Garland Science

2

Chapter 17: Diffusion and Transport Answer: rcyclophilin – rcyclophilin–capsid = √(6Dcyclophilint) – √(6Dcyclophilin–capsidt) = √(6 × 1.2 × 10–8 cm2•sec–1 × 10 sec) – √(6 × 7 × 10–9 cm2•sec–1 × 10 sec) = 8.5 × 10–4 cm – 6.5 × 10–4 cm = 2.0 × 10–4 cm

14. There are approximately 250,000,000 hemoglobin molecules per red blood cell. How many collisions will an oxygen molecule (O2; mass of 32 amu = 5.3 × 10–26 kg) of radius 0.21 nm have with hemoglobin of radius 3 nm in a blood cell of volume 10–16 m3 in 1 sec at 25°C? Assume both molecules are spheres, and that the viscosity is that of pure water. Answer: collision rate = 4πr0 (DO2 + Dmyo) CO2 + Cmyo [Equation 17.46] r0 = collision radius = rO2 + rmyo = (0.21 + 3.0) × 10–7 cm D = kBT/6πρr collision rate = 4 × 3.21 × 10–2 cm × (1.38 × 10–16 gcm2 sec–2 K–1 × 298 K)/6 × 10–2 g cm–1 sec –1) × [(1/0.21 × 10–7 cm) × (1/3.0 × 10–7 cm)] × 1/10–8 cm3 × (2.5 × 108/10–8 cm3) = 1.1 × 1010 sec–1 cm–3 so in the cell volume of 10–8 cm3 there will be Problem 17.18 110 collisions

is capable of making weak non-sequence-specific contacts with genomic DNA. Thus rather than searching the entire nucleus (in three dimensions) a DNA binding protein can slide along the DNA (in one dimension) through these weak interactions until the recognition sequence is located. Additionally, the oligomeric structure of the lac repressor enables weak contacts to be made across DNA strands. In eukaryotic cells, nucleosome structure brings many DNA segments nearby further reducing the three-dimensional space that needs to be searched. 18. Below are traces from dynamic light scattering experiments over the same length of time using purified wild-type and mutant kinase protein. The wild-type protein exists primarily as a dimer, whereas the mutant protein is primarily a monomer. Explain which trace Problem 17.18 came from which protein.

15. What is the typical (r.m.s.) distance traveled by a spherical HIV virus (radius 120 nm) in 1 hour through the bloodstream (viscosity = 3 cP) at 37°C?

16. Why is the diffusion-limited rate of collisions essentially constant at 1010 M–1•sec–1—that is, why is it essentially independent of the size of the molecules involved? Answer: The diffusion limited collision rate is given by k = 4π(Da + Db)(ra + rb). The molecular size and the diffusion coefficient have a reciprocal relationship (D ~ 1/r). Thus any increase in molecular size slows the diffusion rate (Da) by a factor (r) which is canceled by the increase in the radius term (ra). 17. What physical and chemical properties of DNA and the lac repressor account for the observation that the association rate is faster than the three-dimensional diffusion limit? Answer: The association rate can be faster because of restrictive diffusion, which effectively generates a one-dimensional movement in a three-dimensional search. Most DNA binding proteins have a positively charged face that

Answer: Because it is larger, the dimer will cause slower fluctuations in the intensity of light as it diffuses slowly into or out of the scattering volume. The smaller monomer will cause faster fluctuations in the intensity of light. Therefore the top DLS trace is the wild-type protein, and the bottom trace is the mutant protein.

Problem 17.19 19. The same proteins from the previous problem are analyzed by equilibrium ultracentrifugation. The logarithm of the absorbance is plotted versus squared distance from the top of the measurement cell at 10,000 rpm for each r2 (cm2) protein. a. Explain which sample is the mutant and which is the wild type. b. Is there monomer–dimer exchange in either the wild-type or mutant kinase? log(A280)

Answer: DHIV = kBT/(6πηr) = 1.38 × 10–23 J•K–1 × 310 K/(6π × 3 cP × 120 nm) = 4 × 10–21 kg•m2•sec–2/(0.57 g•cm–1•sec–1 × 1.2 × 10–5 cm) = 4 × 10–21 m2•sec–1/(6.84 × 10–9) = 6.3 × 10–13 m2•sec–1 rrms = √(6 × 6.3 × 10–13 × 60 sec/min × 60 min/hour) = 0.117 mm

Answer: a. Equation 17.69 indicates that the slope of ln(c) with respect to (r2) is proportional to ω2(1 – ρsolve/ ρprotein)/2RT × M, but everything except M (mass) is

The Molecules of Life by John Kuriyan, Boyana Konforti, and David Wemmer © Garland Science



PROBLEMS and solutions constant between the two experiments. Therefore, the steeper slope is the larger M, which is from the dimeric protein. The wild-type kinase is the red line, while the mutant kinase is the blue line. b. Both lines are roughly linear suggesting that only one molecular weight species is populated at that concentration. Thus neither the wild-type nor the mutant kinase exchange between monomer and dimer to any measureable extent.

3

24. How much resistive force does a 50-nm vesicle experience if it is transported by dynein at 1 µm•sec–1 in the cytoplasm (η = 0.2 g•cm–1•sec–1)? Answer: f = 6πηrv = 6 × 3.14 × 0.2 g•cm–1•sec–1 × 5 × 10–6 cm × 1 × 10–4 cm•sec–1 = 1.9 × 10–9 g•cm•sec–2 = 19 fN

20. Sedimentation coefficients (in Svedberg units) are often non-additive for macromolecular complexes. For example, the assembled ribosome and proteosome each have lower total sedimentation coefficients than one might expect given the constituents. How might a macromolecular assembly have a higher sedimentation coefficient than the sum of its subunits? Answer: The sedimentation coefficient is related to shape. If the constituent macromolecules were mostly spherical, but the assembled complex was highly nonspherical, the sedimentation coefficient would be non-additive and increase above expectation. For example the assembly of mostly spherical proteins into a filamentlike structure would increase the relative sedimentation coefficient. 21. Why is it necessary to add agarose or polyacrylamide for electrophoresis experiments? Answer: Agarose and polyacrylamide form a gel that prevents convection and increases the friction factor, allowing for a more robust measurement of relative electrophoretic mobility. 22. Why do proteins, but not nucleic acids, need to be covered in SDS to estimate the mass by gel electrophoresis? Answer: Proteins have heterogeneous charge and therefore would not migrate strictly in relation to mass. The number of SDS molecules associated with the protein is roughly proportional to the number of residues, ensuring an even charge:mass ratio. In contrast, nucleic acids are dominated by the charge of the phosphate backbone and have a relatively constant charge:mass ratio independent of sequence. 23. Dynein is a cytoplasmic motor similar to kinesin, but it travels along microtubules in the opposite direction. A single dynein transports a vesicle 0.6 µm along an axon in 5 sec. Dynein steps use one cycle of ATP hydrolysis that move it 80 Å along a microtubule filament. Assuming all steps are forward along one filament, what is the ATP hydrolysis rate of dynein? Answer: Steps = 0.6 × 10–6 m / 80 × 10–10 m = 75 steps or 75 ATPs hydrolyzed. ATP hydrolysis rate = 75/5 sec = 15 ATPs per second per dyenin. The Molecules of Life by John Kuriyan, Boyana Konforti, and David Wemmer © Garland Science

The MOLECULES of LIFE Physical and Chemical Principles

Solutions Manual Prepared by James Fraser and Samuel Leachman

Chapter 18 Folding

Problems True/False and Multiple Choice 1. Unlike protein folding, RNA folding is generally hierarchical. True/False 2. Which of the following effects favors RNA folding? a. Interactions between phosphate groups. b. Kinetic traps of alternative structures. c. Neutralization of backbone charge by counterions. d. The entropy of the native state compared to the unfolded state. 3. Which of the following is least likely to lead to RNA unfolding? a. b. c. d.

Increasing [Mg2+] by 1 mM. Increasing the temperature to 75°C. Adding 1 M EDTA. Adding 6 M urea.

d. They recognize exposed hydrophobic segments. e. They can bind and release proteins several times during a folding cycle. 7. Amyloid fibrils are formed by zipper-like hydrogen bonds within and across β sheets. True/False

Fill in the Blank 8. Counterions for nucleic acids carry a ___________ charge. Answer: positive 9. In hydroxyl radical footprinting, regions with ________ solvent accessibility are likely to be cleaved. Answer: high

4. Which of the following statements is true at Tm for a monomeric protein or RNA? a. ΔGfolding equals 0. b. Equal concentrations of folded and unfolded protein are present. c. Kfolding equals 1. d. The unfolding curve is at its midpoint. e. All of the above. 5. A protein with two-state folding populates two intermediates as folding occurs. True/False 6. Which of the following statements about molecular chaperones is not true? a. They accelerate the folding of client proteins. b. They prevent protein aggregation. c. They are required for in vitro, but not in vivo, folding.

10. When local secondary structure elements form quickly, protein folding is said to follow a ______________ mechanism. In contrast, when distant contacts are quickly formed without secondary structure formation, folding is said to follow a _____________ mechanism. Answer: diffusion collision, nucleation condensation 11. A mutation that has similar interactions in the folding transition state as in the folded state has a __________ of 1. Answer: ϕ-value 12. Folding can be visualized as movement on a multidimensional surface known as a _______________. Answer: free-energy landscape

Quantitative/Essay (Assume T = 300 K and RT = 2.5 kJ•mol–1 for all questions.) 13. The concept of hierarchy is important for understanding

The Molecules of Life by John Kuriyan, Boyana Konforti, and David Wemmer © Garland Science

e

2

Chapter 18: Folding

Answer: Although proteins and RNA have hierarchical structure, protein secondary structure elements are not generally stable in isolation. These elements are only stable when incorporated into the tertiary structure of a folded protein. In contrast, RNA secondary structure elements are generally stable in solution both in isolation and as part of the folded structure. Protein folding is generally non-hierarchical as formation of secondary and tertiary structure can occur simultaneously. RNA folding is kinetically hierarchical in that secondary structure elements tend to form in a distinct phase (or in distinct phases) prior to the formation of tertiary contacts. 14. Below are reaction coordinate diagrams for two RNA molecules (A and B). For each molecule of RNA, we present a reaction coordinate for a misfolded conformation (left) and a correctly folded conformation (right). Which of the RNA folding reactions is more likely to be caught in a misfolded kinetic trap? Explain your reasoning. Figure 18.Q13

unfolded

unfolded misfolded

folded

reaction coordinate

free energy, G

a. Which region becomes protected first? b. Which region becomes more solvent accessible Figure 18.Q14 during the folding reaction?

time

5

10 12

Answer: a. Bases 7–9 become protected first, then base 12. b. Base 4 becomes more solvent accessible during the reaction.

Answer: Both the rugged and smooth energy landscape may contain many local minima; however, only in the rugged landscape do large free-energy barriers separate unfolded unfolded local minima and slow the rate of interconversion misfoldedbetween wells. The slowing of interconversion between wells leads tofolded falling into kinetic traps in the rugged landscape, which is not a factor during folding on the reaction smooth coordinate landscape.

(B)

folded

15. Below is a hydroxyl radical footprint, similar to data shown in Figure 18.51 for a 12-nucleotide segment of RNA.

16. How do rugged and smooth energy landscapes differ with regard to the number of local minima, interconversions between minima, and kinetic traps?

(B) free energy, G

free energy, G

(A)

be caught in this misfolding trap, even though the free energy of the folded state is lower.

nucleotides

the differences between RNA and protein folding. How do the two processes differ in structural stability hierarchy and folding kinetics hierarchy?

17. A mutation of core residues in the E. coli protein Rop speeds up the rate of protein folding from 0.013 sec–1 to 1.3 sec–1.

unfolded

unfolded misfolded

folded

reaction coordinate Answer: RNA molecule A is more likely to be misfolded in a kinetic trap. Although the free energies of the folded states are similar for the two RNA molecules, the barrier height of the folding reaction for A is much higher than for B. In addition, the barrier for folding is higher than for misfolding in A. Thus, A will form the misfolded state faster than the folded state and can

a. What is the effect of the mutation on the free energy of the transition state relative to the unfolded state? b. Given that a simple mutation can increase the folding rate, what does this result suggest about the relationship between the protein folding rate and natural selection? (Data based on M. Munson, K.S. Anderson, and L. Regan, Structure: Folding and Design, 2: 77–87, 1997.) Answer: a. ΔΔGU→TS = RT ln (kfmut/kfWT) = –2.5 × ln (1.3/0.013) = –11.5 kJ•mol–1 b. Proteins must satisfy many functional constraints, so a fast folding rate is not necessarily a trait experiencing strong selection. It appears that Rop has experienced stronger selection for other aspects of function and that its slow wild-type folding rate is sufficient to support those functions.

The Molecules of Life by John Kuriyan, Boyana Konforti, and David Wemmer © Garland Science



PROBLEMS and solutions

18. Shown below are two equally populated alternative conformations of a simplified protein with three β strands. Which conformation is likely to fold and Figure 18.Q17 unfold more rapidly?

(A)

3

slower for the mutant than for the wild type, what is the unfolding rate of F4S? Answer:

ϕF4S = ΔΔGU→RTS/(ΔΔGU→RTS – ΔΔGF→RTS) –ΔΔGF→RTS = ΔΔGU→RTS/ϕF4S – ΔΔGU→RTS

RT ln (kumut/kuWT) = (–RT ln (kfmut/kfWT))/0.21 + RT ln (kfmut/kfWT)) 2.5 × ln (kumut/0.5) = (–2.5 × ln (0.5))/0.21 + 2.5 × ln (0.5) = 6.52 kumut/0.5 = 13.57 kumut = 6.78 sec–1

(B)

21. Why is the rate of protein folding limited to 106 sec–1 , even in the absence of any specific freeenergy barriers? Answer: Conformation A has a lower contact order as the central strand is contiguous in sequence. A lower contact order is highly correlated with faster protein folding. Therefore, it is likely that conformation A folds and unfolds more rapidly. 19. The following data were obtained from a mutational analysis on the folding kinetics of the Fyn SH3 domain:

wild type I28A V55A



30.2 1.24 29.1

ku (sec–1)

22. Shown below is a plot of relative fluorescence (reporting on fraction folded) versus time for a protein in the absence (solid line) and presence (dashed line) of a molecular chaperone. a. How many folding intermediates occur when the protein folds without assistance? Figure 18.Q21 b. What is the effect of the molecular chaperone?

0.5 1.8 18.1

Use these data to answer the following questions: a. What are the ϕ values of the two mutants? b. Which residue forms more native interactions at the transition state? (Data from J.G. Northey, A.A. Di Nardo, and A.R. Davidson, Nat. Struct. Biol. 9: 389–402, 2002.) Answer: a. ϕI28A = ΔΔGU→RTS/(ΔΔGU→RTS – ΔΔGF→RTS) = (–RT ln (kfmut/kfWT))/(–RT ln (kfmut/kfWT) + RT ln (kumut/ kuWT)) = (–2.5 × ln (1.24/30.2))/(–2.5 × ln (1.24/30.2) + 2.5 × ln (1.8/0.5)) = 0.71 ϕV55A = ΔΔGU→RTS/(ΔΔGU→RTS – ΔΔGF→RTS) = (–RT ln (kfmut/kfWT))/(–RT ln (kfmut/kfWT) + RT ln (kumut/ kuWT)) = (–2.5 × ln (29.1/30.2))/(–2.5 × ln (29.1/30.2) + 2.5 × ln (18.1/0.5)) = 0.01 b. A ϕ-value close to 1 suggests that native contacts have formed in the transition state, whereas a value close to 0 suggests that mostly unfolded (non-native) contacts are formed. Since I28 has a higher ϕ-value, it likely has formed more native interactions than V55.

20. The ϕ-value analysis of the previous problem is continued for an additional mutant, F4S. This mutation has a ϕ value of 0.21. If the folding rate is two times

1 fraction folded

kf (sec–1)

Answer: Even in the absence of any kinetic barriers, the speed of protein folding is limited by the rate of collisions causing small rotations around bonds that change the relative positions of different segments of the protein. These collisions are diffusive in character, so the speed limit of protein folding is set by diffusion.

0

time

Answer: a. The native protein has two well-populated folding intermediates as indicated by the plateaus on the graph. b. The effect of the chaperone is to accelerate (or bypass) the folding through the first intermediate; however, it appears to have no effect on the second intermediate. 23. How might ATP binding and hydrolysis contribute to the active and passive mechanisms of folding mediated by GroEL-GroES? Answer: In an actively folding mechanism, GroEL-GroES might energetically couple client protein unfolding to ATP hydrolysis prior to protein folding in the “Anfinsen cage”.

The Molecules of Life by John Kuriyan, Boyana Konforti, and David Wemmer © Garland Science

4

Chapter 18: Folding In contrast, in a passive mechanism, the nucleotide state of GroEL–GroES triggers conformational changes that gate client protein entry to, and exit from, the “Anfinsen cage,” where protein folding can occur.

24. A solution x-ray scattering experiment for a protein is performed in increasing amounts of urea. The data show a difference in the radius of gyration at 6 M urea and 0.1 M urea. a. Why is the radius of gyration different between an unfolded and folded protein even though both contain the same number of residues? b. How will the measured radius of gyration depend on the concentration of urea? Answer: a. Radius of gyration (Rg) is related to the distance of an average atom from the center of mass of the protein. Because folded proteins are compact and well packed, they tend to have a significantly smaller Rg than unfolded proteins, which populate an ensemble of random coil conformations. b. Since urea is a denaturant, increasing concentrations should result in unfolding of the protein. The measured Rg will then tend to increase monotonically with the concentration of urea until the protein is completely unfolded, such that the Rg at 6 M urea is significantly higher than at 0.1 M urea.

The Molecules of Life by John Kuriyan, Boyana Konforti, and David Wemmer © Garland Science

The MOLECULES of LIFE Physical and Chemical Principles

Solutions Manual Prepared by James Fraser and Samuel Leachman

Chapter 19 Fidelity in DNA and Protein Synthesis

Problems True/False and Multiple Choice 1. DNA polymerases from both prokaryotes and eukaryotes resemble a right hand made up of fingers, palm, and thumb subdomains. True/False 2. Which of the following is not a factor that favors DNA synthesis? a. The release of pyrophosphate. b. The base-stacking interactions. c. The change of entropy when the free dNTP is added to the polynucleotide chain. d. The formation of hydrogen bonds. e. The hydrolysis of pyrophosphate that is released. 3. The 5ʹ base in the codon is less important for codon– anticodon recognition and is therefore known as the “wobble base.” True/False 4. The error rate of DNA replication in an E. coli strain that lacked the mismatch repair system would increase by what amount? a. b. c. d. e.

d. The large subunit dissociates from the small subunit. e. EF-Tu dissociates from the ribosome. 7. The large ribsomal subunit is comprised of proteins and RNA, whereas the small subunit is comprised only of RNA. True/False

Fill in the Blank 8. Mismatches can be removed by a 3ʹ to 5ʹ _________ enzymatic activity. Answer: exonuclease 9. Conserved ______ residues bind divalent metal ions in the palm domain. Answer: acidic/negatively charged 10. DNA polymerase can add a nucleotide to the free ____ hydroxyl group of DNA or RNA. Answer: 3ʹ

~109 ~107–108 ~106 ~102 ~104–105

11. The “________” of the “finger-palm-thumb” subdomain structure for DNA polymerase I inserts into the minor groove of DNA.

5. Mismatched base pairs are symmetric about the glycosyl bonds. True/False 6. Which of the following statements is not an effect of codon–anticodon recognition in the ribosome? a. The 30S subunit changes from an open to closed conformation. b. The aminoacyl-tRNA structure is distorted. c. EF-Tu hydrolyzes GTP.

Answer: thumb 12. The movement of mRNA, as the tRNA shifts between sites on the ribosome, is called ______________. Answer: translocation

Quantitative/Essay (Assume T = 300 K and RT = 2.5 kJ•mol–1 for all questions unless otherwise stated.)

The Molecules of Life by John Kuriyan, Boyana Konforti, and David Wemmer © Garland Science

2

Chapter 19: Fidelity in DNA and Protein Synthesis

13. A scientist compares a bacterial polymerase (without a proofreading domain) with a yeast polymerase (also without a proofreading domain). The bacterial enzyme has lower fidelity. Surprisingly, the two enzymes have approximately the same value of KM for correctly matched base pairs. Explain which enzyme, the bacterial or yeast, would likely have a higher KM for mismatched base pairs? Answer: The yeast enzyme makes fewer errors than the bacterial enzyme. The similar KM for correct matches suggests that this difference is due to a difference solely in the affinity for incorrect matches. Since the bacterial polymerase makes more errors it likely has a greater affinity for incorrect base pairs. Thus, the yeast enzyme likely has the higher KM value (indicating a lower affinity for incorrect bases). 14. How would inactivating a mutation of pyrophosphatase affect the rate of DNA synthesis in the cell? Answer: Without pyrophosphatase, the rate of DNA synthesis would likely decrease. This is because pyrophosphatase breaks down pyrophosphate near the active site that could otherwise drive a reverse synthesis reaction where a base would be removed by addition of pyrophosphate to the chain. 15. What significant free-energy barriers affect the rate of DNA synthesis? Answer: The largest barrier in polymerase-catalyzed synthesis has been shown to be the slow conformational change that closes the polymerase after binding of the dNTP to be incorporated. Other smaller but significant barriers include the conformational change prior to pyrophosphate release and the release of pyrophosphate itself. 16. In the “two-metal” reaction mechanism for nucleic acid phosphoryl transfer, what roles are played by a) only metal ion A, b) only metal ion B, and c) both metal ions? Answer: a. Metal ion A activates the 3ʹ sugar hydroxyl group by facilitating deprotonation. b. Metal ion B promotes the bond breakage step by directly ligating the pyrophosphate leaving group and neutralizing its developing negative charge. c. Both metals orient the phosphate group in the polymerase active site and stabilize the negative charge in the transition state by ligating the nonbridging oxygen of the phosphate being transferred. 17. Why does EDTA quench DNA polymerase reactions? Answer: EDTA is a metal-chelating molecule that has higher affinity for Mg2+ than DNA polymerase does. EDTA outcompetes DNA polymerase for Mg2+ and thus binds all the Mg2+ in the reaction buffer. Since Mg2+ is required by DNA polymerase for catalysis, DNA synthesis cannot occur.

18. a. Why are fluorinated bases readily incorporated during DNA synthesis? b. Draw a fluorinated thymidine analog-adenine base pair and label the C1ʹ–C1ʹ distance and angles between the C1ʹ–C1ʹ vector and the glycosyl bonds with predicted values. Answer: a. Shape complementarity and base-stacking interactions are more important than the hydrogen bonds in determining which bases are incorporated. Since the fluorination only alters the hydrogen bonding and not the steric interactions, analogs are readily incorporated. b. The illustration is similar to Figure 19.33A with Q19.18b base from Figure 19.34. The C1ʹ–C1ʹ distance should be ~11 Å, eta angles should be ~50°.

dFTP H 3C

H F

H N

N

N N C1′ 50º

F

H

A

N

C1′ 51º

11.1 Å

19. How does the interaction between the O helix and the incoming nucleotide change between the “open” and “closed” forms of DNA polymerase? Answer: In the open form, conserved residues in the O helix help to bind the incoming nucleotide. When the polymerase closes, the O helix packs against the face of the new base pair, helping to seal the new nucleotide completely into the active site of the polymerase. 20. A scientist probes the thermodynamics of forming an RNA hairpin (H) from the unfolded state (R). She measures ∆ SoH→R = 908 J•K–1•mol–1 and ∆ HoH→R = 295 kJ•mol–1. What are the values of a) the equilibrium constant KH→R and b) ∆GoH→R? Answer: a. –RT lnKH→RR = ΔHH→RR – TΔSH→RR = 295 kJ•mol–1 – 300 K × 0.908 kJ•K–1•mol–1 = 22.6 kJ•mol–1 lnKH→RU = –9.0 KH→RU = 1.2 × 10–4 b. ΔG = ΔH – TΔS = 295 kJ•mol–1 – 300 K × 0.908 kJ•K–1•mol–1 = 22.6 kJ•mol–1 21. The Tm of a DNA duplex is measured to be 55°C at 10 μm. When the concentration is 1000 times higher, the Tm changes to 60°C. What are the ΔS and ΔH values for formation of this duplex? Answer:

ΔHH→RR = –R ln (C1/C2)/(1/T1 – 1/T2) =

–8.314 J•K–1•mol–1 × ln(0.001)/(1/328 – 1/333) =

The Molecules of Life by John Kuriyan, Boyana Konforti, and David Wemmer © Garland Science



PROBLEMS and solutions

3

1250 kJ•mol–1 Since ΔG = –RT ln(KD) = –RT ln(Ctot/4), at 55°C, –RT ln(10 μM/4) = ΔHH→RR – TΔSH→RR; ΔSH→RR = ΔHH→RR/T + R ln(10 μM/4) = 3.72 kJ•K–1•mol–1 22. Measurements on several variants of a DNA double helix are performed. Although the free energy of double helix formation does not vary significantly, the enthalpy and entropy measurements show a large range. A scientist observes that mutations that are favorable with respect to binding enthalpy are entropically disfavored. What effect has the scientist observed and why is this phenomenon common in biological molecules? Answer: The scientist is observing enthalpy–entropy compensation. Many biological molecules, including DNA, are highly flexible when not bound to other molecules. As tighter, more stable interactions are formed between two binding partners, their breathing motions decrease. This reduction in the degrees of freedom is unfavorable entropically. These offsetting enthalpic and entropic effects lead to small variations in the free energy of binding. 23. A DNA polymerase is isolated and found to have an error rate of 1 in 106. a. Suppose that the error rate is determined solely by the relative stabilities of incorrect and correct base pairs. What would the difference in free energy between correct and incorrect nucleotides incorporated by the polymerase have to be in order to explain the error rate? 
 b. Solution studies of isolated oligonucleotides indicate that the energetic difference is actually –1.2 kJ•mol–1. What is the equilibrium constant of correct–incorrect base pair discrimination based on these solution studies? c. What other enzymatic activity, in addition to nucleotide insertion, contributes to the increased fidelity of DNA polymerase? Answer: a. ΔG = –RT ln (Kex) = –2.5 × ln (10–6) = –36 kJ•mol–1 b. Kex = e–ΔG/RT = e1.2/2.5 = 1.62 c. The exonuclease activity contributes to the observed fidelity by excising incorrect bases, leading to a higher observed fidelity. 24. Both thermodynamics and kinetics play an important role in achieving fidelity. How is kinetic control exercised by a) DNA polymerase and b) the ribosome? Answer: In DNA polymerase, the rate of extension after an error is 200–2500 times slower, which gives the exonuclease domain a chance to act on mismatched nucleotides. In the ribosome, kinetic proofreading occurs through the combination of two steps (EF-Tu GTP hydrolysis and tRNA dissociation) that change the rate of correct codon–anticodon recognition without altering the equilibrium constants. The Molecules of Life by John Kuriyan, Boyana Konforti, and David Wemmer © Garland Science