Ch1 - Ch 3 Flashcards
1) the foundations of biochemistry 2) water 3) amino acids, peptides, and proteins (83 cards)
1
Q
What sets the limitations of cell size?
A
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.
2
Q
What are stereoisomers?
What are the 3 different categories of stereoisomers?
A
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.
- Geometric isomers
- enantiomers
- Diastereomers
3
Q
Geometric Isomers
A
- cis vs trans
- different physical and chemical properties
4
Q
Enantiomers
A
- mirror images
- identical physical properties (except rotating plane-polarized light)
- react identically with achiral reagents
5
Q
Diastereomers
A
- non-mirror images
- Different physical and chemical properties
6
Q
Chiral molecules in living organisms.
A
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
7
Q
Keq
A
- 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
8
Q
The equilibrium constant, Keq, 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 [Pi] = 5 mM, is this reaction at equilibrium in living cells?
A
The mass-action ratio (Q) is far from the Keq for the reaction.
The reaction is very far from the equilibrium in cells. Therefore, tends to go strongly to the right.
9
Q
For the reaction catalyzed by the enzyme hexokinase:
Glucose + ATP → glucose 6-phosphate + ADP
the equilibrium constant, Keq, 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?
A
At equilibrium,
Keq = 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.
10
Q
DeltaG
A
DeltaG is a measure of how far a reaction is from equilibrium.
At equilibrium, deltaG=0, Ki=Keq and the standard deltaG = -RTlnKeq
standard deltaG is independent of concentration
When Keq >> 1, exergonic reaction (standard DeltaG << 0 )
When Keq << 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
11
Q
Given that the standard free-energy change for the reaction
glucose + Pi → glucose 6-phosphate is 13.8 kJ/mol,
and the standard free-energy change for the reaction
ATP → ADP + Pi is -30.5 kJ/mol,
what is the free-energy change for the reaction
glucose + ATP → glucose 6-phosphate + ADP?
A
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.
12
Q
If the equilibrium constant, Keq, for the reaction
ATP → ADP + Pi is 2.22 x 105 M,
calculate the standard free-energy change, ∆G°, for the synthesis of ATP from ADP and Pi at 25 °C.
A
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.
13
Q
What is ∆G° under physiological conditions (E. coli grows in the human gut, at 37 °C) for the following reaction? Keq= 7.8 x 102
Glucose + ATP → glucose 6-phosphate + ADP
A
∆G° = -(8.315 J/mol·K)(310 K)(ln 7.8 x 102) = -17 kJ/mol
14
Q
How can you speed up a chemical reaction?
A
- High temperatures
- add more concentrations of reactants
- change the reaction by coupling to a fast reaction (universally used by living organisms)
- lower activation barrier by catalysis (universally used by living organisms)
15
Q
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?
A
The two enantiomers have different interactions with a chiral biological “receptor” (a protein). In living organisms, proteins are mostly L isomers.
16
Q
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?
A
(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.
17
Q
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.
A
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.
18
Q
covalent bonds
A
- 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
19
Q
ionic bonds
A
- 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
20
Q
H-bonds
A
- 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)
21
Q
van der Waal’s interactions
A
- 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
- attractive force (London dispersion) depends on the polarizability
- there is a minimum energy distance (van der Waals contact distance)
22
Q
hydrophobic effect
A
- 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
23
Q
What is so important about water?
A
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
24
Q
H-bonding in Water
A
- 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
25
Proton Hopping
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)
26
What is the concentration of H+ in a solution of 0.1 M NaOH?
27
What is the concentration of OH– in a solution with
an H+ concentration of 1.3 x 10-4 M?
28
Why is distilled water that has been exposed to the atmosphere somewhat acidic?
absorbed CO2 from the atmosphere reacts with water to become carbonic acid
29
acid dissociation constant (Ka)
* 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 Ka
weaker acids have smaller Ka
30
pKa
expresses the relative strength of a weak acid or base
the stronger the acid, the smaller the pKa
the stronger the base, the larger the pKa
the pKa = pH at the midpoint of the titration curve
where the concentrations of the [HA] and [A-] are equal
plus or minus 1 pKa = buffer region
31
buffer
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=pKa
32
Henderson-Hasselbalch equation
the relationship between pH, pKa and buffer concentration can be determined by this equation
33
biological buffer
* 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
34
buffering blood
* 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
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
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%
36
What is the pH of a mixture of 0.042 M NaH2PO4 and 0.056 M Na2HPO4?
pKa = 6.86
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.
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?
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
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?
The NaOH dissociates completely into Na+ and OH–, giving [OH–] = 0.01 mol/L.
pOH = –log[OH-] = 2
pH = 14-2 = 12
39
Why does intravenous administration of a bicarbonate solution raise the plasma pH?
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.
40
Draw a titration of a monoprotic molecule
41
Draw the "generic" amino acid and glycine
Glycine R group is a hydrogen atom
42
Amino acids: atom naming
start from a-carbon and go down the R-group
43
R group classification
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
44
Ionization of Amino Acids
* 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
45
amino acids as buffers
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–]
46
How do you get the pI when the side chain is ionizable?
1. Identify the form that carries a net zero charge
2. Identify the pKa value that defines the acid strength of this zwitterion (pK2)
3. Identify pKa that defines the base strength of this zwitterion (pK1)
4. Take the average of these two pKa values
47
the pKa of any functional group is affected by its chemical environment
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)
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
a) 4.8
b) 9.2
c) 4
d) 4.8
49
What is the H+ concentration of a solution with pH of
a) 3.82
b) 6.52
c) 11.11
a) 1.51 x 10-4 M
b) 3.02 x 10-7 M
c) 7.76 x 10-12 M
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 Ka than the weak acid?
c) Does the strong acid have a higher or lower pKa than the weak acid?
a) greater tendency
b) higher Ka
c) lower pKa
51
Which is the conjugate base in each of the pairs below?
a) RCOOH, RCOO–
b) H2PO4–, H3PO4
c) RNH2, RNH3+
d) H2CO3, HCO3–
a) RCOO–
b) H2PO4–
c) RNH2
d) HCO3–
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)
a) HCl
b) NaOH
c) NaOH
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 pKa of the amino group?
a) pH 8.6 – 10.6
b) 4/5
c) 10 mL
d) pH = pKa – 2
54
A buffer contains 0.010 mol of lactic acid (pKa = 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?
a) 4.6
b) 0.1 pH unit
c) 4 pH units
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?
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.
56
Draw a peptide bond
57
Polarity of proteins
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
58
oligopeptide vs polypeptide vs proteins
* 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
59
ionization of peptides
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
60
What are the physical and chemical properties used to purify proteins?
* charge
* size
* affinity for a ligand
* solubility
* hydrophobicity
* thermal stability
61
What are the general steps of a protein purification process?
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
62
Column chromatography separates proteins by utilizing which properties?
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
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?
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.
64
SDS PAGE: Molecular Weight
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
65
IEF: isoelectric focusing
used to determine the pI of a protein
66
2D Gel Electrophoresis
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
67
electrophoresis
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
68
What is Specific Activity?
How is it different from Relative Activity?
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
69
Why does specific activity increase even as total activity falls with each purification step?
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
70
Direct protein sequencing (Sanger)
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
71
Edman degradation
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.
72
Breaking disulfide bonds in proteins
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
73
Proteases for protein sequencing
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
74
electrospray ionization mass spectrometry (ESI MS)
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
75
tandem MS (MS/MS)
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.
76
How do amino acid sequences provide clues to evolutionary relationships?
* 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
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
a) 2
b) 4
c)
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.
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.
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; pKR = 4.25
Histidine: pKR = 6.00
Arginine: pKR = 12.48
Glycine: pK1 (–OCOOH) = 2.34
a) at pH 3 = +2; at pH 8 = 0; at pH 11 = –1
b) pI = 7.8
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
a) (Glu)20
b) (Lys-Ala)3
c) (Asn-Ser-His)5
d) (Asn-Ser-His)5
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?
```
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
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%
Peptide C elutes first, Peptide B second, Peptide A last.
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.
88%, 97%
all residues released in the first cycle are correct, even though the efficiency of cleavage is not perfect.
