Ch1 - Ch 3

Question 1

What sets the limitations of cell size?

Answer 1

Lower limit: the minimum number of each type of biomolecule required by the cell.

Upper limit: the rate of diffusion of solute molecules in aqueous systems. With increasing cell size, the surface-to-volume ratio decreases. We don’t want metabolism to consume O₂ faster than diffusion can supply it.

Question 2

What are stereoisomers?
What are the 3 different categories of stereoisomers?

Answer 2

Stereoisomers: molecules with the same chemical bonds and the same chemical formula but different configuration. They cannot be interconverted without breaking covalent bonds. Stereoisomers are important because the function of molecules strongly depends on 3D structure.

1. Geometric isomers
2. enantiomers
3. Diastereomers

Question 3

Geometric Isomers

Answer 3

• cis vs trans
• different physical and chemical properties

Question 4

Enantiomers

Answer 4

• mirror images
• identical physical properties (except rotating plane-polarized light)
• react identically with achiral reagents

Question 5

Diastereomers

Answer 5

• non-mirror images
• Different physical and chemical properties

Question 6

Chiral molecules in living organisms.

Answer 6

proteins: L isomers
glucose: D isomers

Living cells produce only one chiral form of a biomolecule because the enzymes that synthesize that molecule are also chiral

Question 7

K_eq

Answer 7

• Equilibrium constant: [product]/[reactant]
• shows the tendency of a chemical reaction to go to completion
• no unit of measurement
• a large value means that the reaction tends to proceed until the reactants are almost completely converted into the products

Question 8

The equilibrium constant, K_eq, for the following reaction is 2 x 10⁵ M (see attached image)

If the measured cellular concentrations are [ATP] = 5mM, [ADP] = 0.5 mM and [P_i] = 5 mM, is this reaction at equilibrium in living cells?

Answer 8

The mass-action ratio (Q) is far from the K_eq for the reaction.

The reaction is very far from the equilibrium in cells. Therefore, tends to go strongly to the right.

Question 9

For the reaction catalyzed by the enzyme hexokinase:

Glucose + ATP → glucose 6-phosphate + ADP

the equilibrium constant, K_eq, is 7.8 x 10²

In living E. coli cells, [ATP] = 5 mM, [ADP] = 0.5 mM, [glucose] = 2 mM, and [glucose 6-phosphate] = 1 mM.

Is the reaction at equilibrium in E. coli?

Answer 9

At equilibrium,

K_eq = 7.8 x 10² = [ADP][glucose 6-phosphate]/[ATP][glucose]

In living cells, [ADP][glucose 6-phosphate]/[ATP][glucose] = (0.5 mM)(1 mM)/(5 mM)(2 mM) = 0.05.

The reaction is therefore far from equilibrium.
The reaction tends strongly to go to the right.

Question 10

DeltaG

Answer 10

DeltaG is a measure of how far a reaction is from equilibrium.

At equilibrium, deltaG=0, K_i=K_eq and the standard deltaG = -RTlnK_eq
standard deltaG is independent of concentration

When K_eq >> 1, exergonic reaction (standard DeltaG << 0 )
When K_eq << 1, endergonic reaction (standard DeltaG >> 0)
This just tells you which way the reaction is going to go - not how fast.

T = Kelvin
R = 8.315 J/mol

Question 11

Given that the standard free-energy change for the reaction
glucose + P_i → glucose 6-phosphate is 13.8 kJ/mol,

and the standard free-energy change for the reaction
ATP → ADP + P_i is -30.5 kJ/mol,

what is the free-energy change for the reaction
glucose + ATP → glucose 6-phosphate + ADP?

Answer 11

The standard free-energy change for two reactions that sum to a third is simply the sum of the two individual reactions.

A negative value for ∆G° (-16.7 kJ/mol) indicates that the reaction will tend to occur spontaneously.

Question 12

If the equilibrium constant, K_eq, for the reaction
ATP → ADP + P_i is 2.22 x 10⁵ M,

calculate the standard free-energy change, ∆G°, for the synthesis of ATP from ADP and P_i at 25 °C.

Answer 12

to synthesize 1 mol of ATP under standard conditions (25 °C (298 K), 1 M concentrations of ATP, ADP, and Pi), at least 30.5 kJ of energy must be supplied

Hint: get the ∆Gº for the ATP breakdown reaction first.

Question 13

What is ∆G° under physiological conditions (E. coli grows in the human gut, at 37 °C) for the following reaction? K_eq= 7.8 x 10²

Glucose + ATP → glucose 6-phosphate + ADP

Answer 13

∆G° = -(8.315 J/mol·K)(310 K)(ln 7.8 x 10²) = -17 kJ/mol

Question 14

How can you speed up a chemical reaction?

Answer 14

1. High temperatures
2. add more concentrations of reactants
3. change the reaction by coupling to a fast reaction (universally used by living organisms)
4. lower activation barrier by catalysis (universally used by living organisms)

Question 15

The quantitative differences in biological activity between the two enantiomers of a compound are sometimes quite large. For example, the D isomer of the drug isoproterenol used to treat mild asthma is 50 to 80 times more effective as a bronchodilator than the L isomer.

Identify the chiral center in L isoproterenol.
Why do the two enantiomers have such radically different bioactivity?

Answer 15

The two enantiomers have different interactions with a chiral biological “receptor” (a protein). In living organisms, proteins are mostly L isomers.

Question 16

In studying a particular biomolecule (a protein, nucleic acid, carbohydrate, or lipid) in the laboratory, the biochemist first needs to separate it from other biomolecules in the sample—that is, to purify it. Specific purification techniques are described later in the book. However, by looking at the monomeric subunits of a biomolecule, how would you separate (a) amino acids from fatty acids and (b) nucleotides from glucose?

Answer 16

(a) Only the amino acids have amino groups; separation could be based on the charge or binding affinity of these groups. Fatty acids are less soluble in water than amino acids and much longer and bigger (solubility, size, and shape)
(b) Glucose is a smaller molecule than a nucleotide. The nitrogenous base and phosphate group are not in glucose. Size, solubility, and charge can be used for separation from glucose.

Question 17

Some years ago, two drug companies marketed a drug under the trade names Dexedrine and Benzedrine. The structure of the drug is shown below.

The physical properties (C, H, and N analysis, melting point, solubility, etc.) of Dexedrine and Benzedrine were identical.

The recommended oral dosage of Dexedrine (which is still available) was 5 mg/day, but the recommended dosage of Benzedrine (no longer available) was twice that. Apparently it required considerably more Benzedrine than Dexedrine to yield the same physiological response.

Explain this apparent contradiction.

Answer 17

There is a chiral center in the structure of the drug molecule. Dexedrine and Benzedrine are enantiomers. Enantiomers have the same chemical and physical properties (except for their interactions with plane-polarized light).

This drug needs to bind to a receptor (protein) to induce activity. Because proteins in living organisms have the L isomer, the Dexedrine is probably the D isomer. Benzedrine is probably a racemic mixture where both enantiomers are present. That is why the effectiveness is halved and a double dosage is required to get the same effect as 5 mg of Dexedrine.

Question 18

covalent bonds

Answer 18

• strong interaction
• bond dissociation energy super high (100kJ/mol to over 1,000 kJ/mol)
• formed between atoms that can share their unpaired electrons
  
  • double and triple bonds are where atoms share more than one electron pair
• the stronger the bond, the closer the two atoms are in space

Question 19

ionic bonds

Answer 19

• noncovalent, weak interaction
• do not share a pair of electrons
• involve the transfer of electrons between atoms so one becomes negatively charged and the other positively charged
• electrostatic interactions between two atoms with a large difference in electronegativity

Question 20

H-bonds

Answer 20

• noncovalent, weak interaction
• strong dipole-dipole (uncharged, but polar molecules) or charge-dipole interaction that arises between an acid (proton donor) and a base (proton acceptor)
• strongest when the bonded molecules are oriented to maximize electrostatic interaction (ideally the three atoms involved are in a line)

Question 21

van der Waal’s interactions

Answer 21

• noncovalent, weak interaction
• between all atoms (universal), regardless of the polarity (always attractive)
  
  • stronger in polarizable molecules
• two components
  
  • attractive force (London dispersion) depends on the polarizability
    
    • dominates at longer distances
    • polarization due to fluctuating charge distributions
  • repulsive force (steric repulsion) depends on the size of the atoms
    
    • dominates at shorter distances
• there is a minimum energy distance (van der Waals contact distance)

Question 22

hydrophobic effect

Answer 22

• noncovalent, weak interaction
• association/folding of nonpolar molecules in the aqueous solution
• does not arise because of an attractive force between two nonpolar molecules
• water is highly ordered around nonpolar substances (low entropy, thermodynamically unfavorable, low solubility)
• when nonpolar molecules are clustered, only the edge of the cluster forces the ordering of water. Fewer water molecules are ordered, increasing entropy
• binding sites in enzymes and receptors are often hydrophobic

Question 23

What is so important about water?

Answer 23

Water is the medium for life

• Life evolved in water (UV protection)
• organisms typically contain 70-90% water
• chemical reactions occur in aqueous milieu
• water is a critical determinant of the structure and functions of proteins, nucleic acids, and membranes

Question 24

H-bonding in Water

Answer 24

• water can serve as both H-donor and H-acceptor
• up to 4 H-bonds per water molecule gives water its
  
  • high boiling point
  • high melting point
  • unusually large surface tension
• H-bonding in water is cooperative
• salts in water increase entropy by breaking the highly ordered crystal lattic

Question 25

Proton Hopping

Answer 25

No individual proton moves very far through the bulk solution, but a series of proton hops between hydrogen-bonded water molecules causes the net movement of a proton over a long distance in a remarkably short time. As a result of the high ionic mobility of H+, acid-base reactions in aqueous solutions are exceptionally fast. (faster than true diffusion)

Question 26

What is the concentration of H⁺ in a solution of 0.1 M NaOH?

Answer 26

Question 27

What is the concentration of OH^– in a solution with
an H⁺ concentration of 1.3 x 10^-4 M?

Answer 27

Question 28

Why is distilled water that has been exposed to the atmosphere somewhat acidic?

Answer 28

absorbed CO₂ from the atmosphere reacts with water to become carbonic acid

Question 29

acid dissociation constant (K_a)

Answer 29

• equilibrium constants for ionization reactions
  
  • shows the tendency of a chemical reaction to go to completion
  • no unit of measurement
  • a large value means that the reaction tends to proceed until the reactants are almost dissociated

stronger acids have larger K_a

weaker acids have smaller K_a

Question 30

pK_a

Answer 30

expresses the relative strength of a weak acid or base

the stronger the acid, the smaller the pK_a
the stronger the base, the larger the pK_a

the pK_a = pH at the midpoint of the titration curve
where the concentrations of the [HA] and [A-] are equal

plus or minus 1 pK_a = buffer region

Question 31

buffer

Answer 31

an aqueous system that tends to resist changes in pH when small amounts of acid or base are added

at the midpoint of the buffering region (relatively flat zone of the titration curve)

• [HA] = [A-]
• the buffering power of the system is maximal (the pH changes least on the addition of H+ or OH^–)
• pH=pK_a

Question 32

Henderson-Hasselbalch equation

Answer 32

the relationship between pH, pKa and buffer concentration can be determined by this equation

Question 33

biological buffer

Answer 33

• vital to all cells
  
  • enzyme-catalyzed reactions have optimal pH
  • the solubility of polar molecules depends on H-bond donors and acceptors
  • equilibrium between CO₂ gas and dissolved HCO₃^– depends on pH
• mainly based on
  
  • phosphate: acts in the cytoplasm of all cells (in millimolar range)
  • bicarbonate: important for blood plasma
  • histidine: efficient at neutral pH

Question 34

buffering blood

Answer 34

• blood plasma is buffered by the bicarbonate system, consisting of carbonic acid (H₂CO₃) as proton donor and bicarbonate (HCO₃^–) as proton acceptor.
• carbonic acid is formed from dissolved carbon dioxide CO₂(d) and water
• CO₂ is produced as a by-product of energy metabolism in our cells
• the rate of respiration – the rate of inhaling and exhaling – can quickly adjust these equilibria to keep the blood pH nearly constant.
• the rate of respiration is controlled by the brain stem, where detection of an increased blood pCO₂ or decreased blood pH causes an increase in breathing and removal of the excess CO₂ through our lungs

Question 35

Calculate the fraction of histidine that has its imidazole side chain protonated at pH 7.3. The pKa values for histidine are pK1 = 1.8, pK2 (imidazole) = 6.0, and pK3 = 9.2

Answer 35

5% of histidine has its imidazole side chain protonated at pH 7.3

Need to convert the ratio of unprotonated form to a fraction. That fraction is 20/21 (20 parts His per 1 part HisH+, in a total of 21 parts histidine in either form), or about 95.2%

Question 36

What is the pH of a mixture of 0.042 M NaH₂PO₄ and 0.056 M Na₂HPO₄?

pK_a = 6.86

Answer 36

When more conjugate base [A-] than acid [HA] is present, the acid is more than 50% titrated and thus the pH is above the pKa (6.86), where the acid is exactly 50% titrated.

Question 37

If 1.0 mL of 10.0 M NaOH is added to a liter of the buffer
with a mixture of 0.042 M NaH₂PO₄ and 0.058 M Na₂HPO₄ (pKa = 6.86),
how much will the pH change?

Answer 37

A liter of the buffer contains 0.042 mol of NaH₂PO₄. Adding 1.0mL of 10.0 M NaOH (0.010 mol) would titrate an equivalent amount (0.010 mol) of NaH₂PO₄ to Na₂HPO₄, resulting in 0.032 mol of NaH₂PO₄ and 0.068 mol of Na₂HPO₄. The new pH is 7.2

Question 38

If 1.0 mL of 10.0 M NaOH is added to a liter of pure water at pH 7.0, what is the final pH?

Answer 38

The NaOH dissociates completely into Na+ and OH^–, giving [OH^–] = 0.01 mol/L.

pOH = –log[OH^-] = 2

pH = 14-2 = 12

Question 39

Why does intravenous administration of a bicarbonate solution raise the plasma pH?

Answer 39

The ratio of HCO₃^– to [CO₂(d)] determines the pH of the bicarbonate buffer, according to the equation pH = 6.1 + log [HCO₃^–]/[H₂CO₃]

where [H₂CO₃] is directly related to pCO₂, the partial pressure of CO₂. So if [HCO₃^–] is increased with no change in pCO₂, the pH will rise.

Question 40

Draw a titration of a monoprotic molecule

Answer 40

Question 41

Draw the “generic” amino acid and glycine

Answer 41

Glycine R group is a hydrogen atom

Question 42

Amino acids: atom naming

Answer 42

start from a-carbon and go down the R-group

Question 43

R group classification

Answer 43

1. nonpolar, aliphatic: glycine, alanine, valine, leucine, isoleucine, proline, methionine
2. aromatic: tryptophan, phenylalanine, tyrosine
3. polar, uncharged: serine, threonine, cysteine, asparagine, glutamine
4. positively charged: histidine, lysine, arginine
5. negatively charged: aspartate, glutamate

Question 44

Ionization of Amino Acids

Answer 44

• cationic form: at acidic pH, the carboxyl group is protonated
• zwitterions: at neutral pH, the carboxyl group is deprotonated and the amino group is protonated; the net charge is zero
  
  • can act as a base or an acid
• anionic form: at alkaline pH, the amino group is neutral –NH₂

Question 45

amino acids as buffers

Answer 45

amino acids with uncharged side chains have two pKa values and can act as a buffer in two pH regimes

for amino acids without ionizable side chains, the zwitterion form is at the isoelectric point (pI). At this point, the amino acid is least soluble in water and does not migrate in an electric field (charge = 0)

pI = (pK₁ + pK₂) / 2

pK is when [HA] = [A^–]

Question 46

How do you get the pI when the side chain is ionizable?

Answer 46

1. Identify the form that carries a net zero charge
2. Identify the pK_a value that defines the acid strength of this zwitterion (pK₂)
3. Identify pK_a that defines the base strength of this zwitterion (pK₁)
4. Take the average of these two pK_a values

Question 47

the pKa of any functional group is affected by its chemical environment

Answer 47

due to intramolecular interactions

• a-carboxy group is much more acidic than in carboxylic acids because of the amino group on the a-carbon (electronegative)
• a-amino group is slightly less basic than in amines because of the electronegative oxygen in the carboxyl group

similar effects can be caused by chemical groups that happen to be positioned nearby (e.g. in the active site of an enzyme)

Question 48

What is the pH of a solution that has an H+ concentration of

a) 1.75 x 10^–5 mol/L
b) 6.50 x 10^-10 mol/L
c) 1.0 x 10^-4 mol/L
d) 1.5 x 10^-5 mol/L

Answer 48

a) 4.8
b) 9.2
c) 4
d) 4.8

Question 49

What is the H+ concentration of a solution with pH of

a) 3.82
b) 6.52
c) 11.11

Answer 49

a) 1.51 x 10^-4 M
b) 3.02 x 10^-7 M
c) 7.76 x 10^-12 M

Question 50

a) Does a strong acid have a greater or lesser tendency to lose its proton than a weak acid?
b) Does the strong acid have a higher or lower K_a than the weak acid?
c) Does the strong acid have a higher or lower pK_a than the weak acid?

Answer 50

a) greater tendency
b) higher Ka
c) lower pKa

Question 51

Which is the conjugate base in each of the pairs below?

a) RCOOH, RCOO^–
b) H₂PO₄^–, H₃PO₄
c) RNH₂, RNH₃⁺
d) H₂CO₃, HCO₃^–

Answer 51

a) RCOO^–
b) H₂PO₄^–
c) RNH₂
d) HCO₃^–

Question 52

Are the following compounds more soluble in an aqueous solution of 0.1 M NaOH or 0.1 M HCL? (the dissociable protons are shown in red)

Answer 52

a) HCl
b) NaOH
c) NaOH

Question 53

The amino group of glycine has pKa = 9.6

a) In what pH range can glycine be used as an effective buffer due to its amino group?
b) In a 0.1 M solution of glycine at pH 9.0, what fraction of glycine has its amino group in the –NH₃⁺ form?
c) How much 5 M KOH must be added to 1.0 L of 0.1 M glycine at pH 9.0 to bring its pH to exactly 10.0?
d) When 99% of the glycine is in its –NH₃⁺ form, what is the numerical relation between the pH of the solution and the pK_a of the amino group?

Answer 53

a) pH 8.6 – 10.6
b) 4/5
c) 10 mL
d) pH = pKa – 2

Question 54

A buffer contains 0.010 mol of lactic acid (pK_a = 3.86) and 0.050 mol of sodium lactate per liter.

a) Calculate the pH of the buffer.
b) Calculate the change in pH when 5 mL of 0.5 M HCl is added to 1 L of the buffer.
c) What pH change would you expect if you added the same quantity of HCl to 1 L of pure water?

Answer 54

a) 4.6
b) 0.1 pH unit
c) 4 pH units

Question 55

a) Explain the effect of holding one’s breath, hypoventilation, or breathing in and out of a paper bag on the blood pH.
b) Explain why blood pH increases with hyperventilation.
c) During a short-distance run, the muscles produce a large amount of lactic acid (K₂ = 1.38 x 10^–4 M) from their glucose stores. Why might it be useful for short-distance runners to hyperventilate before a dash?

Answer 55

Blood pH is controlled by the carbon dioxide–bicarbonate buffer system

a) During hypoventilation, [CO₂] increases in the lungs and arterial blood, driving the equilibrium to the right, raising [H⁺] and lowering pH.
b) During hyperventilation, [CO₂] decreases in the lungs and arterial blood, driving the equilibrium to the left, reducing [H⁺] and increasing pH above the normal 7.4 value
c) lactate is a moderately strong acid, completely dissociating under physiological conditions and thus lowering the pH of blood and muscle tissue. Hyperventilation removes H⁺, raising the pH of blood and tissues in anticipation of the acid buildup.

Question 56

Draw a peptide bond

Answer 56

Question 57

Polarity of proteins

Answer 57

peptide ends are not the same: N-terminus and C-terminus

numbering and lettering starts from the amino terminus (left) and ends at the carboxyl terminus (right)

Ser-Gly-Tyr-Ala-Leu

Question 58

oligopeptide vs polypeptide vs proteins

Answer 58

• oligopeptide: when a few amino acids are joined by peptide bonds
• polypeptide: molecular weight below 10,000
• protein: thousands of amino acid residues; have higher molecular weights

Question 59

ionization of peptides

Answer 59

The acid-base behavior of a peptide can be predicted from its free a-amino and a-carboxyl groups combined with the nature and number of its ionizable R groups

like amino acids, peptides have characteristic titration curves and a characteristic isoelectric pH (pI) at which they do not move in an electric field

Question 60

What are the physical and chemical properties used to purify proteins?

Answer 60

• charge
• size
• affinity for a ligand
• solubility
• hydrophobicity
• thermal stability

Question 61

What are the general steps of a protein purification process?

Answer 61

1. grow cells, acquire tissue sample, harvest plants, harvest milk
2. concentrate sample and wash away non-specific contaminating materials
3. resuspend and lyse cells
4. non-specific purification steps
5. purification steps with high specificity for the desired protein

Question 62

Column chromatography separates proteins by utilizing which properties?

Answer 62

1. charge: proteins move through the column at rates determined by their net charge at the pH being used. with a column containing cation exchangers, proteins with a more negative net charge move faster and elute earlier
2. size: larger molecules pass more freely, appearing in the earlier fractions
3. affinity: column contains a polymer-bound ligand specific for the protein of interest; solution of a ligand is added to the column to wash the ligand-protein complexes off the column

Question 63

A biochemist wants to separate two peptides by ion-exchange chromatography. At the pH of the mobile phase to be used on the column, one peptide (A) has a net charge of −3 due to the presence of more Glu and Asp residues than Arg, Lys, and His residues. Peptide B has a net charge of +1.

Which peptide would elute first from a cation-exchange resin? Which would elute first from an anion-exchange resin?

Answer 63

A cation-exchange resin has negative charges and binds positively charged molecules. Peptide B, with its net positive charge, will interact more strongly than peptide A, and thus peptide A will elute first.

On the anion-exchange resin, peptide B will elute first.

Question 64

SDS PAGE: Molecular Weight

Answer 64

SDS micelles bind to and unfold all the proteins

• gives all proteins a uniformly negative charge
• the native shape of proteins does not matter
• rate of movement will only depend on size (mass): small proteins will move faster

can be used to calculate the molecular weight of a protein

Question 65

IEF: isoelectric focusing

Answer 65

used to determine the pI of a protein

Question 66

2D Gel Electrophoresis

Answer 66

Isoelectric focusing and SDS-PAGE are combined in 2D electrophoresis

1. separate proteins in the first dimension on gel strip with isoelectric focus (pI)
2. separate proteins in the second dimension on SDS-PAGE (molecular weight)

separates proteins of identical MW that differ in pI, or proteins with same pI but different mass

Question 67

electrophoresis

Answer 67

a process that separates proteins based on the migration of charged proteins in an electric field

often adversely affect the structure and thus the function of proteins

permits a quick estimation of the number of different proteins in a mixture or the degree of purity of a particular protein preparation

determines crucial properties of a protein like pI and approximate MW

Question 68

What is Specific Activity?

How is it different from Relative Activity?

Answer 68

Used to assess protein purity; specific activity increases during purification of an enzyme and becomes maximal and constant when the enzyme is pure

specific activity = enzyme activity/total protein (mg)
relative activity = activity/total solution (mL)

one can resuspend their sample in any arbitrary volume, but that would not change the specific activity

Question 69

Why does specific activity increase even as total activity falls with each purification step?

Answer 69

Activity and total protein generally decrease with each step.

Activity decreases because there is always some loss due to inactivation or nonideal interactions with chromatographic materials or other molecules in the solution.

Total protein decreases because the objective is to remove as much unwanted or nonspecific protein as possible.

in a successful step, the loss of nonspecific protein is much greater than the loss of activity; therefore, specific activitiy increases even as total activity falls

Question 70

Direct protein sequencing (Sanger)

Answer 70

1. react protein with FDNB (Sanger’s reagent) to label the N-terminal and then identify the amino acid
2. Select cleavage reagent or protease
3. Cleave into smaller polypeptides (with a different protease or cleavage reagent)
4. sequence each polypeptide
5. Determine the order of polypeptides in protein
6. use overlaps with sequences obtained by cleaving the protein with different reagents and proteases

Question 71

Edman degradation

Answer 71

in 2 steps, labels and removes only the amino-terminal residue from a peptide, leaving all other peptide bonds intact

1. the peptide is labeled under mildly alkaline conditions
2. the peptide bond is cleaved in an acidic condition
3. the derivatized amino acid is extracted with organic solvents, converted to the more stable form, and then identified

the process is repeated until as many as 40 sequential amino acid residues are identified. this is now automated.

Question 72

Breaking disulfide bonds in proteins

Answer 72

Two approaches to irreversibly break disulfide bonds in proteins.

1. oxidation of cystine residue to produce 2 cysteic acid residues
2. reduction of cystine residue and modification of the cysteine residues to produce carboxymethylated cysteine residues

*alklyation with N-ethyl maleimide

Question 73

Proteases for protein sequencing

Answer 73

some proteases and a few chemical reagents cleave only the peptide bond adjacent to particular amino acid residues

trypsin: hydrolyze peptide bonds in which the carbonyl group is contributed by either a Lys or an Arg residue

Question 74

electrospray ionization mass spectrometry (ESI MS)

Answer 74

used for determining the molecular mass of a protein

1. a protein solution is dispersed into highly charged droplets by passage through a needle under the influence of a high voltage electric field
2. as the protein is injected into the gas phase, it acquires a variable number of protons, and thus positive charges, from the solvent
3. the ions enter the mass spectrometer for m/z measurement
4. the variable addition of these charges creates a spectrum of species with different mass-to-charge ratios
5. each successive peak on the spectrum(from right to left) corresponds to a charged species with both mass and charge increased by 1

Question 75

tandem MS (MS/MS)

Answer 75

used to sequence proteins

1. a solution containing the protein under investigation is first treated with a protease or chemical reagent to hydrolyze it to a mixture of shorter peptides
2. the peptide mixture is sorted in the first mass spectrometer so that only one of the several types of peptides produced by cleavage emerges
3. the sample then travels through a vacuum chamber (collision cell) where the peptide is further fragmented by high-energy impact with a “collision gas” like He or Argon
   
   • each individual peptide is broken in only one place, forming two ions; although the breaks are not hydrolytic, most occur at the peptide bonds
4. the ions enter the second mass spectrometer for m/z measurement
5. one or more sets of peaks are generated (one set with y ions and one with b ions); the peaks are labeled with the molecular weight of the amino acid ion
6. each successive peak has one less amino acid than the peak before; the difference in mass from peak to peak identifies the amino acid that was lost in each case (thus revealing the sequence of the peptide)

7.

Question 76

How do amino acid sequences provide clues to evolutionary relationships?

Answer 76

• sequences of homologous proteins (proteins with similar sequence and function) from a wide range of species can be aligned and analyzed for differences
• differences indicate evolutionary divergences
• analysis of multiple protein families can indicate evolutionary relationships between organisms, ultimately the history of life on Earth

Question 77

Naming the Stereoisomers of Isoleucine

a) How many chiral centers does it have?
b) How many optical isomers?
c) Draw perspective formulas for all the optical isomers of isoleucine

Answer 77

a) 2
b) 4

c)

Question 78

A protein has a molecular mass of 400 kDa when measured by size-exclusion chromatography. When subjected to gel electrophoresis in the presence of sodium dodecyl sulfate (SDS), the protein gives three bands with molecular masses of 180, 160, and 60 kDa. When electrophoresis is carried out in the presence of SDS and dithiothreitol, three bands are again formed, this time with molecular masses of 160, 90, and 60 kDa. Determine the subunit composition of the protein.

Answer 78

The protein has four subunits, with molecular masses of 160, 90, 90, and 60 kDa. The two 90 kDa subunits (possibly identical) are linked by one or more disulfide bonds.

Question 79

a) What is the net charge of this peptide at pH 3, 8, and 11?
b) Estimate the pI for this peptide

Glutamate: pK₂ (–NH₃⁺) = 9.67; pK_R = 4.25

Histidine: pK_R = 6.00

Arginine: pK_R = 12.48

Glycine: pK₁ (–OCOOH) = 2.34

Answer 79

a) at pH 3 = +2; at pH 8 = 0; at pH 11 = –1
b) pI = 7.8

Question 80

Which of each pair of polypeptides that follow is more soluble at the indicated pH?

a) (Gly)₂₀ or (Glu)₂₀ at pH 7.0
b) (Lys-Ala)₃ or (Phe-Met)₃ at pH 7.0
c) (Ala-Ser-Gly)₅ or (Asn-Ser-His)₅ at pH 6.0
d) (Ala-Asp-Gly)₅ or (Asn-Ser-His)₅ at pH 3.0

Answer 80

a) (Glu)₂₀
b) (Lys-Ala)₃
c) (Asn-Ser-His)₅
d) (Asn-Ser-His)₅

Question 81

A biochemist discovers and purifies a new enzyme, generating the purification table below.

a) From the information given in the table, calculate the specific activity of the enzyme after each purification procedure
b) Which of the purification procedures used for this enzyme is most effective (i.e. gives the greatest relative increase in purity?)
c) Which of the purification procedures is least effective?
d) Is there any indication based on the results shown in the table that the enzyme after step 6 is now pure? What else could be done to estimate the purity of the enzyme preparation?

Answer 81

a) step 1 (crude extract): 200 units/mg
step 2 (salt): 600 units/mg
step 3 (pH): 250 units/mg
step 4 (ion): 4,000 units/mg
step 5 (affinity): 15,000 units/mg
step 6 (size): 15,000 units/mg

b) step 4 (ion)
c) step 3 (pH)
d) yes; specific activity did not increase in step 6; SDS PAGE electrophoresis to see if there’s a single band

Question 82

At pH 7.0 in what order would the following three peptides (described by their amino acid composition) be eluted from a column filled with a cation-exchange polymer?

Peptide A: Ala 10%, Glu 5%, Ser 5%, Leu 10%, Arg 10%,, His 5%, Ile 10%, Phe 5%, Tyr 5%, Lys 10%, Gly 10%, Pro 5%, and Trp 10%

Peptide B: Ala 5%, Val 5%, Gly 10%, Asp 5%, Leu 5%, Arg 5%, Ile 5%, Phe 5%, Tyr 5%, Lys 5%, Trp 5%, Ser 5%, Thr 5%, Glu 5%, Asn 5%, Pro 10%, Met 5%, and Cys 5%

Peptide C: Ala 10%, Glu 10%, Gly 5%, Leu 5%, Asp 10%, Arg 5%, Met 5%, Cys 5%, Tyr 5%, Phe 5%, His 5%, Val 5%, Pro 5%, thr 5%, Ser 5%, Asn 5%, and Gln 5%

Answer 82

Peptide C elutes first, Peptide B second, Peptide A last.

Question 83

A peptide with the primary structure Lys-Arg-Pro-Leu-Ile-Asp-Gly-Ala is sequenced by the Edman procedure. If each Edman cycle is 96% efficient, what percentage of the amino acids liberated in the fourth cycle will be leucine?

Do the calculation a second time, but assume a 99% efficiency for each cycle.

Answer 83

88%, 97%

all residues released in the first cycle are correct, even though the efficiency of cleavage is not perfect.