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1 Chapter 16 Carbohydrate Biosynthesis 第 16 章碳水化合物的生物合成 1. Role of Oxidative Phosphorylation in Gluconeogenesis( 氧化磷酸化在糖异生中的作用 ) Is it possible to obtain a net synthesis of glucose from pyruvate if the citric acid cycle and oxidative phosphorylation are totally inhibited? 如果三羧酸循环和氧化磷酸化被完全抑制, 是否可能从丙酮酸净合成葡萄糖? 2. Pathway of Atoms in Gluconeogenesis( 糖异生中原子的路径 )A liver extract capable of carrying out all the normal metabolic reactions of the liver is briefly incubated in separate experiments with the following 14 C-labeled precursors: 肝脏提取物可以进行肝脏中所有的正常代谢反应, 用下面 14 C- 标记的前体在隔离的实验中简单培养 : Trace the pathway of each precursor through gluconeogenesis. Indicate the location of 14 C in all intermediates and in the product, glucose. 追踪糖酵解中各个前体的路径指出在所有中间产物和产物中 14 C 的定位 3. Pathway of CO 2 in Gluconeogenesis( 糖异生中 CO 2 的路径 ) In the first bypass step of gluconeogenesis, the conversion of pyruvate to phosphoenolpyruvate. pyruvate is carboxylated by pyruvate carboxylase to oxaloacetate, which is subsequently decarboxylated by PEP carboxykinase to yield phosphoenolpyruvate. The observation that the addition of CO 2 is directly followed by the loss of CO 2 suggests that 14 C of 14 CO 2 would not be incorporated into PEP. glucose, or any intermediates in gluconeogenesis. However, when a rat liver preparation synthesizes glucose in the presence of 14 CO 2, 14 C slowly appears in PEP and eventually at C-3 and C-4 of glucose. How does the 14 C label get into PEP and glucose? (Hint: During gluconeogenesis in the presence of 14 CO 2, several of the four-carbon citric acid cycle intermediates also become labeled.) 在糖异生的旁路的第一步中, 丙酮酸转化成磷酸烯醇式丙酮酸丙酮酸被丙酮酸羧化酶羧化成草酰乙酸, 然后又被 PEP 羧化激酶去羧基变成磷酸烯醇式丙酮酸观测到 CO2

2 的增加直接又伴随着 CO2 的丢失, 暗示着 14 CO 2 中的 14 C 不会进入到 PEP, 葡萄糖, 或其它任何糖异生的中间产物中去然而, 当鼠的肝脏制备物在有 14 CO 2 存在时 14 C 会慢慢地出现在 PEP 中并最终存在于葡萄糖的 C-3 和 C-4 上标记的 14 C 如何进入到 PEP 以及葡萄糖中?( 提示 : 当糖异生中有 14 CO 2 时, 好几种 4 碳的三羧酸循环的中间产物也被标记上了 ) 4. Energy Cost of a Cycle of Glycolysis and Gluconeogenesis( 一个循环的糖酵解和糖异生的能量消耗 ) What is the cost (in ATP equivalents) of transforming glucose to pyruvate via glycolysis and back again to glucose via gluconeogenesis? 通过糖酵解将葡萄糖转化为丙酮酸, 然后又经过糖异生回到葡萄糖的过程需要消耗多少能量 ( 以 ATP 的形式 )? 5. Regulation of Glycolysis and Gluconeogenesis( 糖酵解和糖异生的调节 ) Because both glycolysis and gluconeogenesis are irreversible, there is no thermodynamic barrier to their simultaneous operation. What would the result be if both pathways were operating at the same time and the same rate? What prevents such simultaneous operation in cells? What considerations govern which pathway is operating at any time? 由于糖酵解和糖异生都是不可逆的, 他们的同时进行不存在热力学障碍如果这两个过程同时以相同的速率进行会有什么结果? 在细胞内什么阻止了它们同时发生? 什么利益使得哪个过程可以随时进行? 6. Regulation of Fructose 1,6-Bisphosphatase and Phosphofructokinase-1(1,6- 二磷酸果糖和磷酸果糖激酶的调节 ) What are the effects of increasing concentrations of ATP and AMP on the catalytic activities of fructose 1,6-bisphosphatase and phosphofructokinase-1? What are the consequences of these effects on the relative flow of metabolites through gluconeogenesis and glycolysis? ATP 和 AMP 浓度的增加对 1,6- 二磷酸果糖和磷酸果糖激酶的催化活性有什么影响? 这种影响对糖异生和糖酵解中代谢物的流动会造成什么后果? 7. Glucogenic Substrates( 生糖的底物 ) A common procedure for determining the effectiveness of compounds as precursors of glucose in mammals is to starve the animal until the liver glycogen stores are depleted and then administer the compound in question, A substrate that leads to a net increase in liver glycogen is termed glucogenic because it must first be converted to glucose 6-phosphate. Show by means

3 of known enzymatic reactions which of the following substances are glucogenic. 在动物中决定某种化合物作为葡萄糖前体的效果的一般程序就是使动物饥饿直到储存的肝糖原耗尽, 然后给予该化合物, 一种导致肝糖原净增的底物适合糖异生因为它首先必须被转化为 6- 磷酸果糖用已知的酶反应说明下面给出的物质那些是生糖的? 8. Blood Lactate Levels during Vigorous Exercise.( 剧烈运动时的血液乳酸水平 )The concentration of lactate in blood plasma before, during, and after a 400 M sprint are shown in the graph. 在一个 400 米的疾跑之前, 之中以及之后的血浆中乳酸浓度如下表 : (a) What causes the rapid rise in lactate concentration? 什么引起了乳酸浓度的迅速增加? (b) What causes the decline in lactate concentration after completion of the sprint? Why does the decline occur more slowly than the increase? 什么引起了疾跑结束后乳酸浓度的降低? 为什么? (c) Why is the concentration of lactate not zero during the resting state? 为什么静止状态时乳酸的浓度不是零? 9. Relationship between Fructose 1,6-Bisphosphatase and Blood Lactate Levels.

4 ( 1,6- 二磷酸果糖和血液乳酸水平的关系 ) A congenital defect in the liver enzyme fructose 1,6-bisphosphatase results in abnormally high levels of lactate in the blood plasma. Explain. 肝脏 1,6- 二磷酸果糖酶的先天缺陷会导致血浆中乳酸水平不正常地偏高请解释 10. Ethanol Affects Blood Glucose Levels( 酒精对血糖水平的影响 ) The consumption of alcohol (ethanol), especially after periods of strenuous activity or after not eating for several hours, results in a deficiency of glucose in the blood, a condition known as hypoglycemia. The first step in the metabolism of ethanol by the liver is oxidation to acetaldehyde, catalyzed by liver alcohol dehydrogenase: 酒精的消费, 尤其是在经过激烈运动或者好几个小时没有进食之后, 会导致血糖缺乏, 即所知的血糖过低酒精代谢的第一步就是被肝脏氧化成乙醛, 由肝醇脱氢酶催化 : Explain how this reaction inhibits the transformation of lactate to pyruvate. Why does this lead to hypoglycemia? 解释这个反应如何抑制了乳酸向丙酮酸的转化为什么导致了血糖过低 11. Effect of Phloridzin on Carbohydrate Metabolism( 根皮素对碳水化合物代谢的影响 ) Phloridzin, a toxic glycoside from the bark of the pear tree, blocks the normal reabsorption of glucose from the kidney tubule, thus causing blood glucose to be almost completely excreted in the urine. Rats fed phloridzin and sodium succinate excreted about 0.5 moles of glucose (made by gluconeogenesis) for every mole of sodium succinate ingested. How is the succinate transformed to glucose? Explain the stoichiometry. 根皮素是一种有毒的糖苷, 来自于梨树的树皮, 会阻断肾小管中正常的葡萄糖的再吸收因而引起血糖几乎完全从尿液中排出喂食根皮素和琥珀酸钠的兔子每吸收 1 摩尔的琥珀酸钠会排泄大概 0.5 摩尔的葡萄糖 ( 糖异生产生 ) 琥珀酸如何转化成葡萄糖? 用化学算法解释 12. Excess O 2 Uptake during Gluconeogenesis.( 糖异生时过量 O 2 的吸收 ) Lactate absorbed by the liver is converted to glucose, with the input of 6 mol of ATP for every mole of glucose produced. The extent of this process in a rat liver

5 preparation can be monitored by administering [ 14 C] lactate and measuring the amount of [ 14 C]glucose produced. Because the stoichiometry of O 2 consumption and ATP production is known, we can predict the extra O 2 consumption above the normal rate when a given amount of lactate is administered. However, when the extra O 2 used in the synthesis of glucose from lactate is actually measured, it is always higher than predicted by known stoichiometric relationships. Suggest a possible explanation for this observation 肝脏吸收的乳酸被转变成葡萄糖, 每产生 1 摩尔葡萄糖需要 6 摩尔的 ATP 这个过程在鼠肝脏中的全程可以用[ 14 C] 乳酸和测量 [ 14 C] 葡萄糖的产量来监测因为 O 2 的消耗和 ATP 的产生的化学计量是已知的, 我们可以推断在给予一个剂量的乳酸时正常速率下额外的 O 2 的消耗, 它总是比通过已知化学计量关系推算出来的要高对这个观测提出一个可能的解释 13. At What Point Is Glycogen Synthesis Regulated? ( 糖原的合成在哪一点受到调节 )Explain how the following observations identify the point of regulation in the synthesis of glycogen in skeletal muscle. 解释下面哪种观测证实了骨骼肌中肝糖合成的调节点 (a) The measured activity of glycogen synthase in resting muscle, expressed in micromoles of UDP-glucose used per gram per minute, is lower than the activity of phosphoglucomutase or UDP-glucose pyrophosphorylase, each measured in terms of micromoles of substrate transformed per gram per minute. 在静止肌肉中测得的肝糖合成的活性 ( 用 UDP- 葡萄糖每克每分钟的微摩尔数来表示 ), 比葡萄糖磷酸变位酶或 UDP- 葡萄糖焦磷酸化酶的活性 ( 都根据每克每分钟底物转变的微摩尔数来测量 ) 要低 (b) Stimulation of glycogen synthesis leads to a small decrease in the concentrations of glucose 6-phosphate and glucose 1-phosphate, a large decrease in the concentration of UDP-glucose, but a substantial increase in the concentration of UDP. 肝糖合成的刺激导致 6- 磷酸葡萄糖和 1- 磷酸葡萄糖浓度的少量降低,UDP- 葡萄糖浓度的大量降低, 但是 UDP 的浓度大幅度增加 14. What Is the Cost of Storing Glucose as Glycogen? ( 以肝糖形式储存葡萄糖的代价是什么?)Write the sequence of steps and the net reaction required to calculate the cost in ATP molecules of converting a molecule of cytosolic glucose

6 6-phosphate to glycogen and back to glucose 6-phosphate. What fraction of the maximum number of ATP molecules available from complete catabolism of glucose 6-phosphate to CO 2 and H 2 O does this cost represent? 写出一个胞液中 6- 磷酸葡萄糖分子转变成肝糖然后又回到 6- 磷酸葡萄糖的步骤和净反应式以及需要的 ATP 分子数 ATP 分子总量的哪部分是 6- 磷酸葡萄糖完全代谢成 CO 2 和 H 2 O 的消耗的表现? 15. Identification of a Defective Enzyme in Carbohydrate Metabolism( 碳水化合物代谢中缺陷酶的鉴定 ) A sample of liver tissue was obtained postmortem from the body of a patient believed to be genetically deficient in one of the enzymes of carbohydrate metabolism. A homogenate of the liver sample had the following characteristics: (1) it degraded glycogen to glucose 6-phosphate, (2) it did not synthesize glycogen from any sugar nor did it use galactose as an energy source, and (3) it synthesized glucose 6-phosphate from lactate. Which of the following enzymes was deficient? Give reasonsyou re your choice. 从一个被确信是某种糖代谢酶的基因缺陷的病人的尸体检验中获得了一份肝脏组织的样本这份肝脏样本的匀浆有着以下特征 :(1) 肝糖向 6- 磷酸葡萄糖的转化退化,(2) 不能从任何糖类合成肝糖, 也不能利用半乳糖作为能源,(3) 从乳酸合成 6- 磷酸葡萄糖这是缺乏下面哪种酶? 给出你选择的理由 (a) Glycogen phosphorylase 肝糖磷酸化酶 (b) Fructose 1,6-bisphosphatase 1,6- 二磷酸果糖酶 (c) UDP-glucose pyrophosphorylase UDP- 葡萄糖焦磷酸化酶 16. Ketosis in Sheep( 羊的酮症 )After the birth of a lamb, the udder of a ewe uses almost 80% of the total glucose synthesized by the animal. The glucose is used for milk production, principally in the synthesis of lactose and of glycerol 3-phosphate used in the formation of milk triacylglycerols. During the winter when food quality is poor, milk production decreases and the ewes sometimes develop ketosis: increased levels of plasma ketone bodies. Why do these changes occur? A standard treatment for sheep ketosis is the administration of large doses of propionate (readily converted to succinyl-coa in ruminants). How does this treatment work?: 在小羊羔出生后,

7 母羊的乳房几乎利用了葡萄糖总量的 80% 来生产乳汁, 主要是合成乳酸和 3- 磷酸甘油用于乳汁三酰甘油的形成在冬季食物质量较差时, 乳汁产量减少并且母羊有时会表现出酮症 : 血浆中酮体的水平升高为什么会发生这些变化? 治疗羊的酮症的标准做法就是给予大剂量的丙酸 ( 在反刍动物中易于转化成琥珀酰辅酶 A) 这种治疗方法怎么起作用的?

脂肪酸胰脏分泌的脂肪酶原要在小肠中激活 ) 2 磷脂酶 A 2 ( 水解磷脂, 产生溶血磷酸和脂肪酸 ) 3 胆固醇脂酶 ( 水解胆固醇脂, 产生胆固醇和脂肪酸 ) 4 辅脂酶 (Colipase)

脂肪酸胰脏分泌的脂肪酶原要在小肠中激活 ) 2 磷脂酶 A 2 ( 水解磷脂, 产生溶血磷酸和脂肪酸 ) 3 胆固醇脂酶 ( 水解胆固醇脂, 产生胆固醇和脂肪酸 ) 4 辅脂酶 (Colipase) Chapter 8 Lipid Metabolism (7h) 教学目的通过本章教学, 使学生掌握脂肪酸的 β- 氧化及其氧化过程中能量的计算 ; 酮体的生成和利用熟悉脂肪酸的合成及其与脂肪酸分解的区别 ; 甘油磷脂的降解和合成了解脂类的分类和生理功