Example only; numbers will differ for each person! Procedural Table Question: Name: 1. Fill in the below table with the appropriate amounts of H2O (water) such that there will be 2.5 ml of solution in total after adding the enzyme, but before adding the DNPH. Note that the enzyme amount is in ul (microliters), while the other amounts (including the H2O) are in ml. Tube letter: | A | B | C | D | E | Leucine-PPi (ml): | 0.4 | 0.4 | 0.4 | 0.4 | 0.4 | FAD (ml): | 0.25 | 0.25 | 0.25 | 0.25 | 0.25 | H2O (ml): | | | | | | Warm at 37 deg | | | | | | Enzyme (ul) | 250 | 100 | 50 | 25 | 0 | Incubate 10 min | | | | | | DNPH (ml): | 0.5 | 0.5 | 0.5 | 0.5 | 0.5 | 2. Fill in the below table with the right amounts of 0.2 M PPi to keep the concentration of PPi in each tube (after adding enzyme) the same as it is in the last tube (tube F). The total volume in each tube after adding 0.1 ml of enzyme will be 1; in addition to 0.2 M PPi, water will be used to make up the volume to this (I am not providing the volumes of water except for tube F because that would tell you how much PPi to use). You do not need to fill in the amount of water. Note that the molarity of the PPi in the Valine-PPi (Val-PPi) is 0.1 M, not 0.2 M. Tube letter: | A | B | C | D | E | F | FAD (ml): | 0.1 | 0.1 | 0.1 | 0.1 | 0.1 | 0.1 | H2O (ml): | ? | ? | ? | ? | ? | 0.15 | Val + 0.1 M PPi (ml): | 0 | 0.13 | 0.26 | 0.39 | 0.52 | 0.65 | 0.2 M PPi (ml): | | | | | | 0 | Warm at 37 deg | | | | | | | Enzyme (ml) | 0.1 | 0.1 | 0.1 | 0.1 | 0.1 | 0.1 | Purification Table Question: In the below table, one value is left blank on each of the last 4 lines. Fill in the number. Note that the Yield is as per a normal purification table - as in vs the maximum - not as in question 1 in the enzyme lab (for which Yield was equal to the units at the present step divided by units at the previous step). Key to abbreviations: Vol = Volume Prot = Protein Conc = Concentration Spec = Specific Activ = Activity uni/mg = Units/mg | | Units | | Prot | Spec | | Total | | Vol | per | Total | Conc | Activ | Yield | Prot | | (ml) | ml | Units | (mg | (uni | % | (mg) | | | | | /ml) | /mg) | | | | ---- | ---- | ---- | ---- | ---- | ---- | ---- | Crude extract: | 1000 | 1.5 | 1500 | 7.5 | 0.2 | 100 | 7500 | | | | | | | | | Before dialysis: | 102 | 12.8 | | 13.5 | 0.95 | 87.1 | 1377 | | | | | | | | | After dialysis: | 108 | 10.5 | 1134 | 12.7 | 0.83 | 75.6 | | | | | | | | | | Hydroxylapatite: | 60 | 16.5 | 990 | 4.5 | 3.67 | | 270 | | | | | | | | | Final enzyme: | 6.2 | 139 | 862 | 8 | | 57.5 | 49.6 | | | | | | | | | ---- ---- ---- ---- ---- ---- ---- Correct answer: 1. Fill in the below table with the appropriate amounts of H2O (water) such that there will be 2.5 ml of solution in total after adding the enzyme, but before adding the DNPH. Note that the enzyme amount is in ul (microliters), while the other amounts (including the H2O) are in ml. Tube letter: | A | B | C | D | E | Leucine-PPi (ml): | 0.4 | 0.4 | 0.4 | 0.4 | 0.4 | FAD (ml): | 0.25 | 0.25 | 0.25 | 0.25 | 0.25 | H2O (ml) | 1.6 | 1.75 | 1.8 | 1.825 | 1.85 | Warm at 37 deg | | | | | | Enzyme (ul) | 250 | 100 | 50 | 25 | 0 | Incubate 10 min | | | | | | DNPH (ml): | 0.5 | 0.5 | 0.5 | 0.5 | 0.5 | 2. Fill in the below table with the right amounts of 0.2 M PPi to keep the concentration of PPi in each tube (after adding enzyme) the same as it is in the last tube (tube F). The total volume in each tube after adding 0.1 ml of enzyme will be 1; in addition to 0.2 M PPi, water will be used to make up the volume to this (I am not providing the volumes of water except for tube F because that would tell you how much PPi to use). You do not need to fill in the amount of water. Note that the molarity of the PPi in the Valine-PPi (Val-PPi) is 0.1 M, not 0.2 M. Tube letter: | A | B | C | D | E | F | FAD (ml): | 0.1 | 0.1 | 0.1 | 0.1 | 0.1 | 0.1 | H2O (ml): | ? | ? | ? | ? | ? | 0.15 | Val + 0.1 M PPi (ml): | 0 | 0.13 | 0.26 | 0.39 | 0.52 | 0.65 | 0.2 M PPi (ml): | 0.325 | 0.26 | 0.195 | 0.13 | 0.065 | 0 | Warm at 37 deg | | | | | | | Enzyme (ml) | 0.1 | 0.1 | 0.1 | 0.1 | 0.1 | 0.1 | In the below table, one value is left blank on each of the last 4 lines. Fill in the number. Note that the Yield is as per a normal purification table - as in vs the maximum - not as in question 1 in the enzyme lab (for which Yield was equal to the units at the present step divided by units at the previous step). Key to abbreviations: Vol = Volume Prot = Protein Conc = Concentration Spec = Specific Activ = Activity uni/mg = Units/mg | | Units | | Prot | Spec | | Total | | Vol | per | Total | Conc | Activ | Yield | Prot | | (ml) | ml | Units | (mg | (uni | % | (mg) | | | | | /ml) | /mg) | | | | ---- | ---- | ---- | ---- | ---- | ---- | ---- | Crude extract: | 1000 | 1.5 | 1500 | 7.5 | 0.2 | 100 | 7500 | | ---- | ---- | ---- | ---- | ---- | ---- | ---- | Before dialysis: | 102 | 12.8 | *1306 | 13.5 | 0.95 | 87.1 | 1377 | | ---- | ---- | ---- | ---- | ---- | ---- | ---- | After dialysis: | 108 | 10.5 | 1134 | 12.7 | 0.83 | 75.6 | *1372 | | ---- | ---- | ---- | ---- | ---- | ---- | ---- | Hydroxylapatite: | 60 | 16.5 | 990 | 4.5 | 3.67 | *66 | 270 | | ---- | ---- | ---- | ---- | ---- | ---- | ---- | Final enzyme: | 6.2 | 139 | 862 | 8 | *17.4 | 57.5 | 49.6 | ---- ---- ---- ---- ---- ---- ----