395G Exam 2 - Fall 2006 1. Refer to figures 11-1 in the textbook. a. Show the conversion of the Fisher projection of D-idose to the Haworth projection of β-D-idose. (4 points) b. Raffinose, also called melitose, is a trisaccharide that is found in beans, cabbage, brussels sprouts, and broccoli. Humans cannot digest this saccharide and it is fermented in the large intestine by gas-producing bacteria. Give the systematic name for this sugar. (5 points) α-D-galactopyranosyl-(1→6)-α-D-glucopyranosyl-(1→2)-β-D-fructofuranoside c. Proteins called lectins reversibly bind to specific carbohydrate residues in glycoproteins. Peanut lectin binds specifically to the disaccharide composed of galactose and N-acetylgalactosamine. The two monosaccharides are linked by a β(1-3) glycosidic bond. Draw the structure of this disaccharide. Hint: the non-reducing end is galactose. (4 points)2. a. Given the following changes to a lipid membrane, would you expect the melting temperature of the membrane to increase or decrease? Explain each answer. (3 points each) i. The lengths of the acyl tails are increased. Increase. More free energy (and a higher temperature) is required to disrupt the more extensive van der Waals interactions in longer acyl chains. ii. Further sites of unsaturation in the hydrocarbon tails are introduced. Decrease. Cis double bond produces a kink in the acyl chain and so it is less able to pack efficiently against it neighbors and disrupts van der Waals interactions. b. In a membrane, would you expect a cis double bond or a trans double bond to cause a greater change in the membrane transition temperature? Explain your answer. (4 points) A cis double bond would cause a greater change in membrane transition temperature. A trans double bond does not produce a kink in the molecule as cis double bonds do, therefore, its geometry more closely resembles that of a single bond. c. The distance between the Cα atoms in a β sheet is approximately 3.5 angstroms. Can a single 9-residue protein segment with a β-conformation reasonably serve as the transmembrane portion of an integral membrane protein? Explain your answer. (4 points) No. Although the 9-residue β-strand could theoretically span the membrane, a single strand would be unstable because its backbone could not form the hydrogen bonds it would form with water in an aqueous environment. d. Does the phosphatidyl glycerol “head group” of cardiolipin (see Table 12-2) project out of a lipid bilayer like other glycerophospholipid head groups? Explain your reasoning. (3 points) No. The two acyl chains of the head group, being hydrophobic, would bury themselves in the lipid bilayer interior, leaving the head group of diphosphoglycerol. e. What properties of triacylglycerols make them unsuitable to be major components of lipid bilayers, such as those found in membranes? (3 points) Triacylglycerols lack polar head groups so they do not orient themselves in a bilayer with their acyl chains inward and their glycerol moiety toward the surface.3. a. A kinetics experiment was performed with a particular enzyme and substrate A or substrate B. A Lineweaver-Burk plot was constructed from the data collected to give a graph similar to the following: A B 1/V0 1/[S] i. Which substrate has a higher Vmax value? Explain. (3 points) B The value of the y-intercept is equal to 1/Vmax for each substrate. Due to this inverse relationship, the substrate corresponding to the line with the lower 1/vo value, has the higher Vmax. In this case, substrate B has the higher Vmax value. ii. Which substrate has a higher Km value? Explain. (3 points) A The value of the x-intercept is equal to 1/KM for each substrate. Due to this inverse relationship, the substrate corresponding to the line with the lower 1/[S] value, has the higher KM. In this case, substrate A has the higher KM value. iii. Assuming the same amount of enzyme was used in both experiments, with which substrate does the enzyme attain the highest catalytic efficiency? Explain your answer. (5 points) B catalytic efficiency ≡ kcat/KM = Vmax/(KM·[E]T), so the substrate with the higher Vmax value and the lower KM value represents the substrate with which the enzyme has the highest catalytic efficiency. b. For an enzyme that follows simple Michaelis-Menten kinetics, what is the value of Vmax if vo is equal to 1 µmole/min at 1/10 Km? (4 points) 11 µmoles/min3. c. Protein phosphatase 1 (PP1) catalyzes a reaction which yields products that are important in regulating cell division. Consequently, PP1 is a possible drug target to treat certain types of cancers. The PP1 enzyme acts to hydrolyze a phosphate group from a specific substrate. One of PP1’s substrates is myelin basic protein (MBP). The reaction is shown below: PP1 MBP-phosphate MBP + Pi The activity of PP1 was measured in the presence and absence of the inhibitor phosphatidic acid (PA). The concentration of PA was 300 nM. [MBP] (mg/mL) vo without PA (nmol Pi released·mL-1min-1) vo with PA (nmol Pi released·mL-1min-1) 0.010 0.0209 0.00381 0.015 0.0335 0.00620 0.025 0.0419 0.00931 0.050 0.0838 0.0140 i. Use the data provided to construct a Lineweaver-Burk plot on the attached sheet of graph paper. (5 points) What kind of inhibitor is PA? Explain your answer. (3 points) 1/vo (min·mL-1·nmol-1) 1/[MBP] (mL/mg) with inhibitor y = 2.38x + 15.64 without inhibitor y = 0.43x + 4.19 Graphs should resemble the graph above. PA is a mixed inhibitor – it has a different x- and y-intercept, corresponding to a change in Vmax and KM. ii. What are the Km and Vmax values for PP1 in the presence and the apparent Km and apparent Vmax values in the absence of the inhibitor? (8 points) Without inhibitor: KM = 0.102 mg/mL; Vmax = 0.239 nmol·mL-1·min-1 With inhibitor: KM = 0.152 mg/mL; Vmax = 0.064 nmol·mL-1·min-1d. For a one-substrate enzyme catalyzed reaction, double-reciprocal plots were deteremined for 3 different enzyme concentrations. Which of the following three families of curve would you expect result? Explain. (5 points) 1/vo