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Questions 61 - 71 ELMA design. 61. Making a Buffer What would be the most practical way of making the Tris solution A.Weigh out 2.4228 g of Tris, dissolve.

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Presentation on theme: "Questions 61 - 71 ELMA design. 61. Making a Buffer What would be the most practical way of making the Tris solution A.Weigh out 2.4228 g of Tris, dissolve."— Presentation transcript:

1 Questions 61 - 71 ELMA design

2 61. Making a Buffer What would be the most practical way of making the Tris solution A.Weigh out 2.4228 g of Tris, dissolve in 20 ml of water, then adjust the pH to 8.1 by adding concentrated HCl B.Dissolve 2.4228 g of Tris in 20 ml water, then adjust the pH to 8.1 by adding dilute HCl C.Dissolve 2.4228 g of Tris in 10 ml water, then adjust the pH to 8.1 with HCl, then add water to 20 ml D.Dissolve about 2.5 g of Tris in about 10 ml water, adjust the pH to 8.1 with HCl, then add water to give a volume equal to 20 x (weight taken/2.4228) E.Accurately measure 20 ml water and adjust its pH to 8.1 with HCl, then add 2.4228 g of Tris too fiddly and a small vol. change Tris changes pH of solution 72 45 71 26 4 too fiddly correct too fiddly and a small vol. change

3 62. Calculate volume of stock To achieve the above, how much of the stock acetyl-CoA solution (as recommended by your assistant) should you add to the cuvette? A.120 µl B.8.2 µl C.1.2 µl D.100 µl E.2.5 µl C 1 V 1 (initial) = C 2 V 2 (final) 12.2mM x V 1 = 0.1mM x 1mL V 1 = 0.008196721311mL = 8.2  L B is correct 19 174 9 6 8 We want 1 ml at 0.1 μmol/ml which is 100 nmol We get that from the stock which is 12.2 nmol per μ

4 63. Calculate concentration from Abs If the absorbance change was 0.75 per min, how fast is the AMOUNT of yellow product increasing in the cuvette? A.75 µmol per min B.50 nmol per min C.5 nmol per min D.15 nmol per min E.15 µmol per min A=  c l  A=  c l 0.75/min = 15mM -1 cm -1 x  c x 1cm  c = 0.05mM / min 23 111 16 32 20 n= C x V = 0.05mM x 1mL = 50nmoles B is correct

5 64. Adding enzyme You decide that you need 50 mU of citrate synthase (CS) in the cuvette. What is the MOST PRACTICAL way of doing this? A.Adding 0.025 µl of the stock CS directly to the cuvette B.Adding 25 µl of the stock CS directly to the cuvette C.Mixing 1 µl of the stock CS with 999 ul Tris buffer and then adding 25 ul of this diluted CS to the cuvette D.Mixing 10 µl of stock CS with 990 ul Tris buffer and then adding 2.5 ul of this diluted CS to the cuvette E.Adding 2.5 µl of the stock CS directly to the cuvette Stock = 10mg/mL = 200U/mg x 10mg / mL = 2000U/mL 50mU ÷ 2000U/mL = 0.025  L 19 41 43 70 25 Correct amount but vol. too small Incorrect amount Correct amount but 1  L is a bit small to pipette Incorrect amount Correct amount, dilutions and vol. OK

6 65. What is NOT a plausible interpretation of these observations? A.Something in the citrate synthase stock has a lot of free thiol groups B.The acetyl CoA solution that you made up has partially degraded to give CoA C.You have added a lot more citrate synthase than you thought D.The DTNB spontaneously degrades at the pH of the Tris E.Your acetyl CoA solution contains some free thiol groups HSCoA+DTNB  intense yellow CoA + DTNB  slightly yellow (even before adding CS) If it is degraded then there should be no reaction at all HSCoA + DTNB  slightly yellow Enzyme should always be in excess. There is no oxaloacetate for CS to catalyse  no CoA 21 29 62 86 18

7 66. Limiting reagent I Over the first 30 seconds, you find that the absorbance rises by about 0.35. Which statement is the MOST LIKELY prediction of what will happen next? A.The rate of increase in absorbance will remain constant for at least 10 minutes B.The absorbance will plateau when the acetyl-CoA runs out C.The rate will fall after about 5 minutes as the citrate synthase becomes degraded D.The absorbance will fall as the all the DTNB gets used up E.In less than five minutes all the oxaloacetate will run out Oxaloacetate (0.05mM) runs out before acetyl-CoA (0.1mM) Abs will not ‘fall’, can only ‘plateau’ Not likely Oxaloacetate (0.05mM) is the limiting reagent Enzyme rate will be less than Vmax as [substrate] decrease 21 128 9 45

8 67. Limiting reagent II To make this system measure acetyl-CoA, you replace the stock acetyl-CoA with a volume of an ‘unkonwn’ sample (eg, a tissue extract)What else do you have to do to make this system useful as an assay for acetyl ‑ CoA? A.Ensure that oxaloacetate is never in excess B.Ensure that DTNB is in excess C.Ensure that citrate synthase does not drive the reaction to completion in less than 5 minutes D.Ensure that the absorbance of the system plateaus above 2 E.Ensure that the unknown acetyl-CoA is added in excess Only acetyl-CoA should be limiting Faster rate means you don’t have to wait as long! The spec is not very good above 2!! Acetyl-CoA should be limiting Everything should be in excess except acetyl-CoA 19 137 38 4 17

9 68. Function of reagent How does the Triton X-100 and freezing/thawing help you get the most accurate results from the tissue samples? A.It activates citrate synthase B.It increases the release of citrate synthase from the mitochondrial matrix C.It causes the hydrolysis of acetyl-CoA D.It increases the release of metabolites from the cell E.It increases the turbidity of the extract CS can work without adding tissue sample Reaction depends on acetyl CoA Hydrolysis  no acetyl-CoA to start reaction It does not increase turbidity & turbidity is not desirable  accurately reflect true [acetyl-CoA] in tissue 14 31 19 125 21

10 69. Troubleshooting When you use the real tissue extract, you anticipate problems will occur. Which is NOT likely to be a cause of such problems?. A.Tissue extracts can contain pigments that absorb in the visible range B.Tissue extracts may contain CoA C.Tissue extracts contain citrate synthase D.The acetyl CoA concentration in the tissue extracts may be too low E.Tissue extracts can be cloudy Tisse extract could be yellowish CoA reacts with DTNB spontaneously increase Abs reading Low [acetyl-CoA]  Abs not detectable (<<0.1) CS is not limiting, more CS should not affect the assay 25 53 65 35 33

11 70. Calculate original conc. If the acetyl-CoA concentration of a sample of heart homogenate is found to be 50 µM, what is the concentration of acetyl-CoA in the tissue? A.1000 nmol g-1 B.50 nmol g-1 C.10 nmol g-1 D.5000 nmol g-1 E.100 nmol g-1 10%(w/v) = 10g tissue in 100mL homogenate = 1g tissue in 10mL homogenate number of moles in 1g tissue = number of moles in 10mL homogenate = C x V = 50  M x 10mL =500nmoles 16 59 7 83 24 No answer is correct!!

12 71. Reliability What would give you the most confidence that your acetyl-CoA assay was working well? A.Each tissue extract gives a different final result for [acetyl-CoA] B.Different volumes of the same tissue extract plateau at the same absorbance C.The final result for [acetyl-CoA] for each tissue extract is identical D.For a particular tissue extract, the plateau absorbance is proportional to the amount of extract added E.The absorbance of each assay plateaus at the same value [acetyl-CoA] alone does not prove assay worked Results reproducible for replicates (plateau abs = completion) Only proves duplicates work (not replicates!) Different vol. should give different plateau abs 8 22 42 122 15 [acetyl-CoA] alone does not prove assay worked

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