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The Tyrosine Hydroxylase assay You are investigating the up-regulation of the enzyme, Tyrosine Hydroxylase (TH) in adrenal gland cell culture samples after certain stimuli. You have an assay to measure its activity, using its substrate tyrosine which is converted to the product, Dopa. The rate of appearance of Dopa is monitored spectrophotometrically at 265 nm (  = 5 mM -1 cm -1 ) over a 10 min time period. The assay is carried out at 37 o C in 0.1 M Tris HCl, pH 7.8 containing 5 mM MgCl 2 and 1 mM biopterin, a cofactor required for the hydroxylation. Tyrosine was added to a final concentration of 0.2 mM.

The reaction

The Question You have carried out the TH assay in a 3 mL final volume, using a quartz cuvette. You have obtained a change in absorbance of 0.1 at 265 nm over a 10 min time period. HOW MUCH substrate is left after the reaction has been running for 10 min?

The working The question asks for how much substrate not the concentration. What do you know? The initial [S] = 0.2 mM How much substrate is this? 0.2 mmol/L  0.2 umol/mL  200 nmol/mL = 600 nmol in 3 mL (the assay volume)

The working How much substrate has been converted to product? We know the change in absorbance is 0.1 over 10 min and the  is 5 mM -1 cm /5 = 0.02 mM = 0.02 umol/mL = 20 nmol/mL = 60 nmol/3 mL So 60 nmol of product is produced in 10 min, this means 600 – 60 = 540 nmol of substrate remains

The Answer A.60 nmoles B.0.02 mM C.540 nmoles D.0.18 mM E.0.02 mmoles Answer C

The Question If you performed the TH assay at 0.4 mM tyrosine (instead of 0.2 mM) which of the following measurements would double? A.The reaction rate (nmoles/min) B.The specific activity (U/mg) C.Absorbance at 265 nm after 2 min D.All of the above E.None of the above

The Answer Let’s consider what is happening in the assay. The assay is designed to estimate the amount of enzyme in the sample by measuring the rate of the reaction i.e. the more enzyme the faster the rate  it is enzyme dependent. This means it should be operating at ~Vmax

The Answer If it is operating around Vmax, adding more substrate will have no effect on the rate of the reaction; it is already at a maximum rate. So the reaction rate, the specific activity and the absorbance change at 265 nm will not change. So the answer is E.

The Question If the Kcat for Tyrosine Hydroxylase is 50 sec -1 how many  moles of product would a nmole of enzyme produce in one minute? A.3 B.30 C.50 D.50,000 E.3,000

The Answer If the Kcat is 50 sec -1 then this means that 50 molecules product are produced per sec per molecule enzyme. So in 1 min 1 molecule of enzyme would produce molecules of product. 1 nmole of enzyme would produce nmol Product in 1 min or 3 umoles. Answer A

ELMA The results of incubations that contain old GOD that has lost much of its activity? The results of incubations that contain too little AAP? The results of incubations using a glucose standard with significant impurities in it? The results of incubations that probably contain too much precious GOD? Which set of data (A – E) that would give the WORST standard curve?

The results of incubations that contain old GOD that has lost much of its activity? While you can’t “run out” of enzyme you may not have enough to complete the reaction is a certain time such as 40 min. Therefore the answer is D where the graphs don’t plateau.

The results of incubations that contain too little AAP? In this case you have a limited amount of one of the reactants so those reactions with the most glucose in them will give a limited response. Answer B, the graphs where the higher [glucose] gave the same response

The results of incubations using a glucose standard with significant impurities in it? This will mean the standard glucose will have less in it than expected. The response will be lower but will plateau. Answer C where the maximum response only reaches A500 of 0.1 but the graphs plateau.

The results of incubations that probably contain too much precious GOD? You are looking for a response where the maximum plateau value is reached quickest. Answer E. Note that the response doesn’t go higher than 1, it just gets there quicker.

Which set of data (A – E) that would give the WORST standard curve? Answer would be B. This won’t be a linear response and you couldn’t use it above the second concentration.

Which of the graphs (A – E) could represent the blood glucose and insulin levels of: A patient given an oral glucose tolerance test (i.e. 50 g glucose as a 20% w/v solution) whose pancreatic beta cells have been destroyed by an autoimmune disease? Someone given an equivalent molar amount of inulin (a carbohydrate polymer that cannot be metabolised i.e. degraded to monomers) rather than glucose?

A patient given an oral glucose tolerance test (i.e. 50 g glucose as a 20% w/v solution) whose pancreatic beta cells have been destroyed by an autoimmune disease? This is a description of type 1 diabetes where the beta cells cannot produce insulin and the blood glucose levels are uncontrolled. Answer C

Someone given an equivalent molar amount of inulin (a carbohydrate polymer that cannot be metabolised i.e. degraded to monomers) rather than glucose? In this case there would be no change in the blood glucose, it would remain at 5 mM and the insulin would be very low and invariant. There is no reason to stimulate the secretion of insulin. Answer D