酮体的生成和利用.doc

1、 酮体的生成和利用【实验目的】了解酮体的生成部位及掌握测定酮体生成与利用的方法。【实验原理】在肝脏线粒体中，脂肪酸经 -氧化生成的过量乙酰辅酶 A 缩合成酮体。酮体包括乙酰乙酸、 -羟丁酸和丙酮三种化合物。肝脏不能利用酮体，只有在肝外组织，尤其是心脏和骨骼肌中，酮体可以转变为乙酰辅酶 A 而被氧化利用。本实验以丁酸为基质，与肝匀浆一起保温，然后测定肝匀浆液中酮体的生成量。另外，在肝脏和肌肉组织共存的情况下，再测定酮体的生成量。在这两种不同条件下，由酮体含量的差别我们可以理解以上的理论。本实验主要测定的是丙酮的含量。酮体测定的原理：在碱性溶液中碘可将丙酮氧化成为碘仿。以硫代硫酸钠滴定剩余的碘，可

2、以计算所消耗的碘，由此也就可以计算出酮体（以丙酮为代表）的含量。反应式如下：CH3COCH3 十 3I2 十 4NaOH CHI3 十 CH3COONa 十 3NaI 十 3H2OI2 十 2Na2S2O3 Na2S4O6 十 2NaI【实验材料】1. 实验器材试管；移液管；锥形瓶；滴定管及架。2. 实验试剂(1) 0.1% 淀粉液。(2) 0.9% NaCl 溶液。(3) 15% 三氯乙酸。(4) 10 NaOH 溶液。(5) 10 HCl 溶液。(6) 0.5mol/L 丁酸溶液：取 5ml 丁酸溶于 100ml 0.5mol/L NaOH 中。(7) 0.1mol/L 碘液：I 2 12

3、.5g 和 KI 25g 加水溶解，稀释至刻度 1L，用 0.1mol/L Na2S2O3 标定。(8) 0.02mol/L Na2S2O3: 24.82g Na2S2O35H2O 和 400mg 无水 Na2CO3 溶于 1L 刚煮沸的水中，配成 0.1mol/L 溶液，用 0.1mol/L KIO3 标定。临用时将标定 Na2S2O3 溶液稀释成 0.02mol/L。【实验操作】1.标本的制备：将兔致死，取出肝脏，用 0.9% NaCl 洗去污血，放滤纸上，吸去表面的水分，称取肝组织 5g 置研钵中，加少许 0.9% NaCl 至总体积为 10ml，制成肝组织匀浆。另外再取后腿肌肉 5g，

4、按上述方法和比例，制成肌组织匀浆。2.保温和沉淀蛋白质：取试管 3 只，编号，按下表操作：管号试剂 A B C肝组织匀浆 2.0 ml 2.0 ml预先煮沸的肝组织匀浆 2.0 ml pH 7.6 的磷酸盐缓冲液 4.0 ml 4.0 ml 4.0 ml正丁酸 2.0 ml 2.0 ml 2.0 ml43水浴保温 60 分钟肌组织匀浆 4.0 ml 预先煮沸的肌组织匀浆 4.0 ml 4.0 ml43水浴保温 60 分钟15%三氯醋酸 3.0 ml 3.0 ml 3.0 ml摇匀后，用滤纸过滤，将滤液分别收集在 3 支试管中,为无蛋白滤液。3.酮体的测定取锥形瓶 3 只，按下述编号顺序操作：编

5、 号试 剂 1 2 3无蛋白滤液 5.0 ml 5.0 ml 5.0 ml0.1mol/L I2-KI 3.0 ml 3.0 ml 3.0 ml10% NaOH 3.0 ml 3.0 ml 3.0 ml摇匀，静置 10 分钟，向各管中加入 10% HCl 3ml,加 1%淀粉液 1 滴呈兰色，分别用0.02mol/L Na2S2O3 滴定至溶液呈亮绿色为止。【实验结果与计算】肝脏生成的酮体量（mmol/g）=(CA)Na 2S2O3 的摩尔数1/6肌肉利用的酮体量（mmol/g）=(CB)Na 2S2O3 的摩尔数 1/6A: 滴定样品 1 消耗的 Na2S2O3 ml 数。B: 滴定样品 2

6、 消耗的 Na2S2O3 ml 数。C: 滴定样品 3 消耗的 Na2S2O3 ml 数。【思考题】为什么只有在肝外组织，酮体才可以被氧化利用？Experiment 11 Production and Degradation of Ketone Bodies【Purpose】Understand the production organs and master the method used for production and degradation of ketone bodies measurement.【Principle】Within the mitochondria of live

7、r, the excess acetyl-CoA produced during fatty acid -oxidation is converted to acetoacetate, -hydroxybutyrate, and acetone, this group of molecules is called the ketone bodies. Liver can not use ketone bodies as an energy source. Only in several tissues out of liver, most notably cardiac and skeleta

8、l muscle, ketone bodies are converted to acetyl-CoA, the acetyl-CoA is then oxidated to generate energy .We use butyric acid as initial stuff in this experiment, the butyric acid is heated with the liver plasm, and then measure the content of ketone bodies in liver plasm. Moreover, measure the conte

9、nt of ketone bodies under the condition of coexistence of liver plasm and skeletal muscle in reaction system. We can comprehend the above theories from the difference of the ketone bodies content under the two different conditions. We determine the content of acetone in this experiment.The ketone bo

10、dies measurement principle is shown below: In alkaline aqua the iodine can oxidize acetone to become iodoform , titrate the remainder iodine in the reaction system with hyposulphite, we can calculate the consumption of iodine according to the result of hyposulphite titration, we can also calculate t

11、he content of ketone bodies (take acetone as to represent) according to the titration result.The equation of Reaction is as follows:CH3COCH3 十 3I2 十 4NaOH CHI3 十 CH3COONa 十 3NaI 十 3H2OI2 十 2Na2S2O3 Na2S4O6 十 2NaI【Materials】1. Apparatus:Tubes, Pipets, Flasks Burettes and burette support.2. Reagents:(

12、1) 0.1% aqua of starch.(2) 0.9% aqua of NaCl.(3) 15% trichlorine acetic acid.(4) 10% aqua of NaOH.(5) 10% aqua of HCl.(6) 0.5mol/L butyric acid: Dissolve 5ml butyric acid in 100ml 0.5mol/L NaOH.7 0.1mol/L aqua of iodine: Dissolve I2 12.5g and KI 25g in distilled water, dilute the solution to 1L, dem

13、arcate the solution with 0.1mol/L Na2S2O3.8 0.02 mol/L Na2S2O3: Dissolve 24.82 g Na2S2O3 5 H2O and 400 mg anhydrous Na2CO3 in 1L fresh boiled water to get 0.1 mol/L solution, demarcate the solution with 0.1 mol/L KIO3. Dilute the solution to 0.02 mol/L just before using.【Procedures 】1. Preparation o

14、f the specimen:Execute the rabbit, take out the liver, scour off the blood with 0.9% NaCl, put the liver on filter paper to suck away the surface humidity, weigh 5 g of the liver organize, place it into the mortar, add a few 0.9% NaCl to total volume 10 ml. Then take 5 g muscle of rear leg, make it

15、into the liver organization plasm according to above- mentioned method and comparisons.2. Heat preservation and precipitation of the protein:Take three tubes, number the tubes and operate as the table followed：Tube No.Reagent A B CThe liver organization plasm 2.0 ml 2.0 mlThe liver organizationn pla

16、sm that boiled in advance 2.0 ml Phosphoric acid salt buffer liquid of pH7.6 4.0 ml 4.0 ml 4.0 mlorthobutyric acid 2.0 ml 2.0 ml 2.0 mlHeat preservation at 43 water bath for 60 minutes.The muscle organization plasm 4.0 ml The muscle organization plasm that boiled in advance 4.0 ml 4.0 mlHeat preserv

17、ation at 43 water bath for 60 minutes.15% trichlorine acetic acid 3.0 ml 3.0 ml 3.0 mlFilter with the filter paper after shaking evenly, collect the filtrate respectively in 3 tubes, then we get the filtrate without protein.3. Determination of the ketone bodies Take 3 flasks, operate as below-mentio

18、ned serial number in proper order: Flask NoReagent 1 2 3Filtrate without protein 5.0 ml 5.0 ml 5.0 ml0.1mol/L I2-KI 3.0 ml 3.0 ml 3.0 ml10% NaOH 3.0 ml 3.0 ml 3.0 mlPlace Statically for 10 minutes after shaking evenly, add 10% HCl 3ml to each tube, and add one drop of 1% starch liquid to each tube,

19、then we can see the solution presents the color of orchid, titrate with 0.02mol/L Na2S2O3 until the solution presents the color of bright green respectively.【Result and calculation】Ketone bodies produced in liver( mmol/g)=(CA)Moore content of the Na2S2O3 1/6Ketone bodies consumed in muscle ( mmol/g)

20、=(CB)Moore content of the Na 2S2O3 1/6A: volume of the Na2S2O3 (ml) consumed during titration of sample 1 B: volume of the Na2S2O3 (ml) consumed during titration of sample 2C: volume of the Na2S2O3 (ml) consumed during titration of sample【Advisement after experiment】Ketone bodies are oxidated to generate energy only in several tissues out of liver. Why?