35. Either
Compare the kinetics of reversible and non-reversible enzyme-catalysed reactions. (10 marks)
Show how the kinetic equations for a reversible reaction may be developed, stating the assumptions that need to be made. (A full derivation is NOT required) (10 marks)
or
Compare the kinetics of inhibited reactions. (10 marks)
Show how the kinetic equations for inhibited reactions may be developed. (10 marks)
36. (a) Explain what is meant by the 'productivity' of an enzyme in a process explaining the factors that might limit it. (8 marks)
(b) Given the expression
(Vmax/VolS) .t = [S]0X - KmLn(1 - X)
determine the percentage hydrolysis of 5 litres of 2 M lactose in 15 hours. (Km = 2 mM; Vmax = 10 mmol/min). (12 marks)
This page was last updated by Martin Chaplin on 18 January 2002
Proteomics tools for biomedicine (2002): Wasinger, V. C.and Corthals, G. L. (2002). J Chromatography B: Analytical Technologies in the biomedical and life sciences: 771(1-2): 33-48.
This article deals with proteomics as a tool to measure gene expression, activity and interactions of biological events at the protein level. Proteins are the major catalysts and contain information that can delineate the actual rather than the potential functional implications as measured by the analysis of mRNA levels.
This is a review article containing 119 references and discusses different aspects of integrated knowledge of proteomics, genomics and other technologies.
C2005/F2402 '06 -- Key to Recitation Problems #1
1A. Hints:
(i) Where will the N atoms go? Will they replace old N atoms already in molecules? Be used to make new molecules?
(ii) If you are not sure if some of the complex molecules ever contain N, check out your handouts. (What is the "R" in phospholipids?)
Answers: Amino acid, phospholipids, polypeptides, and the cell membrane (in proteins and phospholipids there).
1B. HINT: How many bacteria will be present at the end of 5 hrs? See approach #1.
Answer: Approach #1 -- Figure out the number of cells at the start and the # at the end. 6 mg of N was consumed, which is enough to build, at 2 X 10-9 mg N/cell, 6/(2 X 10-9) = 3 X 109 cells. You started with 1 X 109, so you have 4 X 109 cells at the end.
Approach #2 -- Concentrate on the amount of N in bacteria at start and end, instead of the # of bacteria. You started with the 1 x 109 cells which corresponds to 2 mg of N. You used up 6 mg, so you had 8 mg of N in bacteria at the end.
There are now two ways to view the calculation of doubling time: 1) Simple way: You make 3 X 109 new cells (or use up 6 mg N), so a 4-fold increase overall: 1 x 109 -> 4 x 109 cells, or 2 mg N --> 8 mg N. This means 2 generations (2 mg --> 4 mg --> 8 mg, or 1 x 109 cells --> 2 x 109 --> 4 x 109) in 5 hours. So doubling time = 5/2= 2.5 hrs/gen, or tD = 2.5h. 2) Using equations: The initial number of cells was 109, and the N consumption added 3 X 109 cells to this initial number, so the total at the end of the 5 hours would be 1 + 3 = 4 X 109. Using N = No ekt , ln(N/No)= ln(4/1) = 1.39 = kt., and t = 5h, so k = 1.39/5 = 0.278, Now k = ln2/tD = 0.69/tD. So 0.69/tD = 0.278, and so tD = (0.69)/0.278 = 2.48
1C. HINT: How much N is left in the medium? Enough for the bacteria to keep doubling?
Answer: Stationary phase. There is only 4 mg of N left in the culture at the end of the first 5 h hour growth period, which is only enough for 2 x 109 cells, or a 50% increase over the 4 x 109 cells present at that time. This is only enough for about a 1/2 of a cell division, so the cells will run out of N after a short time, and so for most of the second 5-hour period they will be in stationary phase.
1D. HINT: What is the limiting factor? Time or raw materials?
Answer: 6 X 109 cells. There were 4 x 109 after the first 5-hour period of growth in part 1 above; to the 4 x 109 present at the start of the second 5-hour growth period must be added the 2 x 109 cells that could be built with the 4 remaining mg of N, so the total will be 4 + 2 = 6 X 109 cells. (Alternatively, calculate how many bacteria you can make if you use up all of the 10 mg N, and add that to the starting number of bacteria.)
C2005/F2401 '06 Answers to Recitation Problems #2 1. Hint: Consider which molecules (or parts thereof) are hydrophobic or hydrophilic.
Answer: Phospholipid: the glycolipid has the same structure as a phospholipid except that it has a different hydrophilic (polar) group. It has 3 groups attached to glycerol as in a P-lipid. The polar group attached to the glycerol is (galactose) instead of the phosphate or phosphate ester group of the phospholipid; the nonpolar groups attached to the glycerol (2 fatty acids) are the same as in a phospholipid.
2. Hints: A. What differences in properties are exploited by each of these methods? B. How do the properties of asp (aspartic acid) and gln (glutamine) compare? Which method is most likely to take advantage of these differences? C. Where does the carbon in aspartate (aspartic acid)* come from (in minimal medium)? * The terms 'aspartate' and 'aspartic acid' are used interchangeably by most biochemists. Strictly speaking, 'aspartate' is the ionized form and 'aspartic acid' is the un-ionized form. In solution, at most pH values, you have a mixture of the two.
Answers: 2A. Method of Fingerprinting (meaning a 2D separation, using chromatography in one direction and electrophoresis in the other). This is the best bet because it combines the separation power of both paper chromatography and paper electrophoresis. Next best: paper chromatography, since it separates on the basis of small differences among the amino acids in their solubility in organic solvents. Note: The term "fingerprinting" usually refers to the 2D separation of peptide fragments, that is, the separation of the products of partial hydrolysis of a protein. "Fingerprinting" usually does not refer to the separation of amino acids. Why? If you hydrolyze proteins completely to amino acids, and separate the amino acids, almost every protein will give the same pattern or "fingerprint" -- almost every protein will give the same 20 spots corresponding to the same 20 amino acids. On the other hand, if you partially hydrolyze a protein and separate the peptide products, each protein will usually give a unique pattern of spots corresponding to a unique mixture of peptides. What is meant by "method of fingerprinting," in this case, is the application of the fingerprinting procedure (that is, a 2D separation) to a mixture of amino acids.
2B. Paper electrophoresis at pH 12, since asp will have a net charge of -2 while gln will have a net charge of -1, OR paper chromatography using a different pH and or organic solvent than the first paper chromatographic separation. Paper electrophoresis at pH 2 would not work because both molecules would have a net charge of +1.
2C. 20. All the carbons of aspartate are derived from glucose (the only source of C in the minimal medium). The glucose contained 30 units per mole, and had 6 carbons, all equally labeled, so there are 30/6 = 5 radioactivity units per carbon atom (per mole of each carbon atom) in glucose. Aspartate has 4 carbon atoms, or 4 moles of carbon atoms per mole of aspartate. Each carbon derived from glucose has 5 units of radioactivity. Four carbons, at 5 units each, yields 5 X 4 = 20 radioactivity units per mole of aspartate.
3. Hints: A. Is the glucose L or D? If the picture shows an alpha glucose, what does a beta glucose look like? B. Would the sugar-sugar connects be straight across or at an angle? C. What determines the overall shape of a polysaccharide chain? D. What are the products of complete hydrolysis? Will any of them isomerize?
Answers: 3A. Equatorial to equatorial. C1 is equatorial in beta glucose (axial in alpha). All the other OH’s are equatorial in glucose, as can be seen in the structure of D glucose, the natural isomer.
3B. A helix but not like starch . The
1.3 connection, while allowing the connected monomer sugar chairs to
occupy the same plane, nevertheless forces the polymer to turn a
corner at each connection, since carbons 3 ,1 , 3 (or 1, 3, and 1) do
not lie in a straight line. The polymer will thus curve and
turn on itself, forming a helix. The alpha 1,4 turn in starch
is more
extreme, not allowing the sugars to exist in the same
plane, and starch has the 1,6 branches, which make it even more
compact. The inability to form straight chains that can line up
with each other to form a bundle of polymers makes curdlan unlike
cellulose. Thus it has an intermediate physical character.
3C.
Carbon #5. The C5 OH is positioned so
that when it swings around and approached C1, it can form a bond to
the C1 carbon. The C1 double bond opens up and a hydroxyl is
formed on the C1 O.
Another way to reason it without
having to remember that it is the C5 OH that attacks:
Inspection of the ring structure
shows that the ring O lies
between carbons 1 and 5, so the O must have come from either C1 or
C5. The anomeric carbon is number 1, which has a carbonyl (C=O)
in the straight chain form. It is this double bond that is
attacked by a hydroxyl to form the ring. After this attack, the
carbonyl O becomes a hydroxyl, so that is not the one in the ring, so
it must be the C5 OH that becomes the ring O.
3D. A carbonyl, beta-D-glucose, alpha-D-glucose, carbon with tetrahedral bonds, monomers. The free D-glucose would be comprised of an equilibrium mixture of the straight chain form (containing an aldehyde, which has a carbonyl), and the alpha and beta anomer ring forms. No L-glucose would form: one would have to break and reform the bonds between the hydroxyls and carbons at positions 2, 3 and 4 for that transformation to occur. No dimers or glycosidic bonds would be left as these are what is destroyed by the hydrolysis. The D-glucose monomers would be the products.
C2005/F2401 ’06– Recitation Problems #3 -- Answers
1A. Hint: What's being held together here?
1A. Answer: Quaternary structure is defined as the association of two or more polypeptides in a protein.
1B. Hint: What R groups are involved?
1B. Answer: Hydrophobic forces and van der Waals bonds. The leucine side chains consist of hydrocarbon, are non-polar and thus hydrophobic. Once aggregated due to hydrophobic forces, they can form van der Waals bonds due to fluctuating induced dipoles where two leucine molecules approach each other closely.
1C Hint: Try to draw a picture of the structure.
1C. Answer: The alpha-helix is formed by hydrogen bonding interactions between the backbone atoms of a polypeptide; the side chains protrude from the surface of this cylinder in all directions. For a series of side chains of two alpha-helices to interact with each other, they must all protrude from the same side of the cylinder; leucines pointing away from other leucines cannot contribute to the interaction. Since the alpha-helix has 3.6 amino acids per turn, leucines that occur about every 3-4 amino acids will be positioned on the same face of the helix, so one answer is 36/3.6 = 10 leucines. A better answer for the minimum number of leucines would be half this number, or 5. A space between leucines on one helix would allow the leucines from the partner helix to interdigitate between them. If the glycines and alanines, with their small side chains, alternate between these protruding leucines, there would be room for the leucine from the other chain to fit in. Such an arrangement would maximize the hydrophobic interactions along the length of the leucine hydrocarbon side chains.
1-D-1. Hint: What is protonated or ionized at pH 1? At pH 7?
1-D-1. Answer: At pH 1, all basic side chains will be charged, but the glutamates and aspartates will not be charged (their carboxyls will be protonated). The alpha amino group at the N-terminus will also be charged at pH 1, so there must be 6 basic side chains present to give a total charge of +7. At pH 7, all ionizable side chains will be charged, and the N and C-terminus charges will cancel each other out. At pH 7, the negatively charged glutamate + aspartates must just equal the number of basic amino acids with positively charged side chains to produce a zero net charge, so glu + asp must also equal 6, the number of basic side chains.
1-D-2. Hint: What amino acids contain sulfur? 1-D-2. Answer: Sulfur >2; sulfate 0, sulfhydryls 2, disulfides 1 or 0 (depending on how you read the Q; see below). Sulfur could be present in methionine residues (in addition to the one cysteine/subunit): so max # is >2. There's no sulfate group in any of the amino acids, so max # is 0. Sulfhydryls are found only in the amino acid cysteine, and 2 could be present if the 2 cysteines (1 per subunit) are not hooked up in a disulfide. One inter-chain disulfide could be formed between the 2 cysteines. However the statement in the beginning of the question says that "these (leucine) interactions are the only ones responsible for holding the two polypeptides together." This statement precludes disulfides from playing a role in the 4o structure. So the max. # of disulfides is 1 (if you forgot about the statement at the beginning) or 0, if you take it into account.
2. Hint: Is V vs S linear?
2. Answer: (> 1/2 the Vmax but < the Vmax).
In words:
As [S] increases, rate of increase of V increases more slowly than value of [S].
Value of V starts to plateau, so adding more S doesn’t increase V very much at large values of S.
Doubling [S] won’t double V.
Graphical solution:
Show a line rising from 2X Km to the V curve, and then left to the y-axis to intersect it between 1/2 Vmax and Vmax.
3A. Hint: Can you calculate the Vmax? (What value of [S] was used here? Does that help you get the Vmax?) How do you get from Vmax to the turnover number?
3A. Answer: 33.3 per second or 2000 per min.
How to get the Vmax: Since the substrate concentration of lactose used was equal to the Km of the enzyme, the velocity in that experiment must have been one-half the Vmax. So the Vmax was 2 X 10-5M/minute.
How to get the T. O. #: Turnover number is k3 = Vmax/[E], or 2 X 10-5 M/minute / 10-8M = 2000 per minute, or = 2000/60 = 33.3 per second.
Note: This answer may appear to contradict the answer to Q1. It doesn’t. If you know the Km, and the V at [S] = Km, you can calculate Vmax (as in this question); but you can’t say how much [S] is needed to get to Vmax (see question 1). That’s because V approaches Vmax as [S] increases, but V only reaches Vmax at infinite [S].