Understanding Fatty Acid Metabolism and Insulin Signaling
School
Boston University**We aren't endorsed by this school
Course
CAS CH 373
Subject
Chemistry
Date
Dec 9, 2024
Pages
10
Uploaded by BarristerMaskCobra24
CH 373 Discussion 10 Fall 2024Fatty Acid Metabolism & Insulin signalingFocus Questions:Q1. Oleic acid is a mono-unsaturated omega-9 fatty acid found in various animal and vegetablesources. What is the “net” ATP production from the breakdown of one molecule of oleic acid inaerobic respiration?First Oleic acid (18:1Δ9) needs to be activated to oleoyl CoA at the expense of equivalent to 2ATP (1 ATP turning into 1 AMP)β-oxidation of Oleoyl CoA(18:1Δ9):https://www.youtube.com/watch?app=desktop&v=u9Owmim67b4# of β-oxidation rounds: 8𝑛2()− 1()# of acetyl-CoA: 9 (equals the number of β-oxidation rounds+1)# of NADH: 8 (equals the number of β-oxidation rounds)# of FADH2: 7 (Equals β-oxidation rounds – every degree of unsaturation that originates fromodd carbon, in this case only the one from #9)# of NADPH (consumed): NONE (Equals to the # of double bonds originating from an evennumber of carbons- NONE in this case)TCA cycle:1 Acetyl CoA = 3NADH + 1 FADH2+1GTP9 Acetyl CoA = 27 NADH +9 FADH2+ 9 GTP = 67.5 ATP+ 13.5 ATP+ 9 ATP= 90 ATPNow ADD the NADH & FADH2from β-oxidation:8 NADH + 7 FADH2= 20 ATP+ 10.5 ATP= 30.5 ATPTOTAL: 120.5 ATP (Note if you are in a tissue that uses shuttles to bring NADH into ETC youneed to account for those here- I will not expect you to know which one they would be, but beprepared to account for those if asked)120.5-2 = 118.5 ATP (considering loss in activation)Q2. Why do unsaturated fatty acids produce less energy than saturated fatty acids? Provide adetailed rationale.Unsaturated fatty acids are already more oxidized than saturated fatty acids. They contain C=Cin their hydrocarbon tails. Due to this C=C already being present, the acyl-CoA Dehydrogenasereaction is skipped. This means 1 x FADH2molecule is not produced for every round ofβ-oxidation where a C=C is encountered. So, for every round of β-oxidation that encounters aC=C, there is 1.5 less ATP produced.
Background:Regulating glucose in the body is done autonomously and constantly throughout eachminute of the day to keep the body running smoothly. A continuous concentration of 60 to 100mg/dL of glucose in blood plasma is needed to supply the body's cells with the requiredimmediate energy. During exercise or stress, the body needs a higher concentration (up to 140mg/dL) because muscles require glucose for energy.Brain cells don’t require insulin to drive glucose into neurons; however, normal amountsmust still be available. Too little glucose, called hypoglycemia, starves cells, and too muchglucose (hyperglycemia) creates a sticky, paralyzing effect on cells. Euglycemia, or blood sugarwithin the normal range, is ideal for the body’s functions. A delicate balance between hormonesof the pancreas, intestines, brain, and even adrenals is required to maintain normal BG levels.Of the three fuels for the body, glucose is preferred because it produces both energy andwater through the citric acid cycle and aerobic metabolism. The body can also use protein andfat; however, their breakdown creates ketoacids, making the body acidic, which is not optimal.Excess of keto acids can produce metabolic acidosis.Functioning body tissues continuously absorb glucose from the bloodstream. For peoplewho do not have diabetes, a meal of carbohydrates replenishes the circulating blood glucoseabout 10 minutes after eating and continues until about two hours after eating. A first-phaseinsulin release occurs about 5 minutes after a meal, and a second phase begins at about 20minutes. Because the duration of insulin’s effect is only about two hours, taking a 2-hourpostprandial (after meal), BG shows how well insulin is released and used by the body. The foodis broken down into small components, including glucose, and then absorbed through theintestines into the bloodstream. Glucose (potential energy) that is not immediately used isstored by the body as glycogen in the muscles, liver, and fat.The glucose concentration in the blood is determined by the balance between the rate ofglucose entering and the rate of glucose leaving the circulation. These signals are deliveredthroughout the body by two pancreatic hormones, insulin and glucagon, which are part of anegative feedback loop. Optimal health requires that:●When blood glucose concentrations are low, the liver is signaled to add glucose to thecirculation.●When blood glucose concentrations are high, the liver and the skeletal muscles aresignaled to remove glucose from the circulation.Watch this video:https://www.youtube.com/watch?v=OlHez8gwMgwand read section 25.2 inyour textbook to review how insulin response works at a molecular level.Q3. Circle the correct option for each of the 11 boxes on the following figure. Note that theoptions are indicated with a “/”. Explain your choice making specific references to thebackground information you reviewed using relevant scientific terminology.
ANSWER:Insulin is a peptide hormone made in the beta cells of the pancreas that is central toregulating carbohydrate metabolism in the body.As a meal containing carbohydrates is eatenand digested, rising BG levels stimulate synthesis and release of insulin from the beta cells ofthe pancreas and turn off glucagon production. When insulin reaches insulin-sensitivecells—liver cells, fat cells, and striated muscle—it stimulates them to take up and metabolizeglucose by activating several enzymes that convert the glucose to chains of glycogen—so longas both insulin and glucose remain plentiful. Glycogen provides an energy reserve that can bequickly mobilized to meet a sudden need for glucose. In this postprandial or “fed” state, the livertakes in more glucose from the blood than it releases. Insulin has various effects that can becategorized as anabolic or growth-promoting. During fasting states, the liver is directed to breakdown chains of glycogen to supply energy in the form of glucose for cells, while during a fedstate, insulin is released to supply the liver and skeletal muscles with the uptake of glucose toprovide for energy. A delicate balance is required among cells.In-Class QuestionsQ1. Based on the given structure on the right:a.Write down the systematic name and lipid designationfor Linoleic acid.9,12-Octadecadienoic acid, 18:2 Δ9,12 (all cis)b.What kind of omega fatty acid is it and what is itsshorthand lipid designation based on its omega name?Omega-6 (ω-6), 18:2 (n-6)
c.Identify the end products when linoleic acid undergoes complete β-oxidation.This is an even (18) numbered fatty acid with 2 degrees ofunsaturation, so:# of β-oxidation rounds: 8𝑛2()− 1()# of acetyl-CoA: 9 (equals the number of β-oxidation rounds +1for even-numbered fatty acids)# of NADH: 8 (equals the number of β-oxidation rounds)# of FADH2: 7 (Equals β-oxidation rounds – every degree ofunsaturation that originates from odd carbon, in this case onlythe one from #9)# of NADPH: 1 consumed (Equals to the # of double bondsoriginating from an even number of carbon- in this case, only theone from position #12)Solutions: https://www.youtube.com/watch?v=zCLC9PV5ZUoFigure:https://pharmaxchange.info/2013/10/oxidation-of-fatty-acids/Q2. a.Saturated fatty acids with an odd number of carbons are common in the lipids of manyplants and some marine organisms. Identify the end products of pentadecanoic acid at the endof beta-oxidation, which is the major dietary source of butterfat in cow's milk.# of β-oxidation rounds: 6then round down𝑛2()− 1()# of acetyl-CoA: 6 (equals the number of β-oxidation rounds)# of NADH: 6 (equals the number of β-oxidation rounds)# of FADH: 6 (equals the number of β-oxidation rounds)1 Propionyl CoAb. What is propionyl-CoA converted into, and where does it enter into one of the metabolicpathways you studied to be completely oxidized to produce energy?
Propionyl-CoA is converted into succinyl-CoA and succinyl-CoA enters into the TCA cycle at theSuccinyl-CoA Synthetase reaction.So at the expense of an ATP you can create a succinyl CoA that can go into the citric acid cycleat the seventh step.Q3. Triacylglycerol (TAG) in adipose tissue is the major energy storage form in highereukaryotes. It needs to be converted to Acyl-CoA and go through β-oxidation to turn into ATP.Below is a flowchart for the first two stages of this process..a.Describe the conversion of TAGs stored in the adipose tissue to metabolic intermediatesby filling boxes 1-4 in the figure above.b.Describe the mechanism for Acyl CoA production by filling boxes 5-8 in the figure aboveand explain the involvement of pyrophosphatase in this series of reactions.The hydrolysis of PPi couples the cleavage of a second high-energy bond to theformation of thioester bond to make its formation exergonic (thermodynamicallyfavorable). This in turn, makes the FA activation step irreversible. In effect, two ATPs areused to make one acyl CoA.(Answer must include both an explanation for the involvement of pyrophosphatase andfilled-in boxes 5-8)c. Place the following incomplete list of reactions or locations during the β-oxidation of fattyacids in the proper order
(a) Reaction with carnitine(b) Fatty acid in cytosol(c) Activation of fatty acid by joining to CoA(d) Hydration(e) NAD+-linked oxidation(f)Thiolysis(g) Acyl CoA in mitochondria(h) FAD-linked oxidationFirst - b, c, a, g, h, d, e, f - Lastd. Explain the involvement of carnitine in the β-oxidation of fatty acids.Acyl CoA is formed in the cytosol, and the enzymes of the β-oxidation pathway are in the matrixof the mitochondrion. The mitochondrial inner membrane is impermeable to CoA and its acylderivatives. However, a translocase protein can shuttle carnitine and its acyl derivatives acrossthe inner mitochondrial membrane. The acyl group is transferred to carnitine on the cytosol sideof the inner membrane and back to CoA on the matrix side. Thus, carnitine acts as atransmembrane carrier of acyl groups.e. Describe the regulatory strategy for the use of lipids as fuel.The fatty acids are used as fuel by many tissues. Triacylglycerols are stored inside a fat cell asa lipid droplet. The liver processes glycerol by either the glycolytic or gluconeogenic pathway,depending on the liver’s metabolic circumstances. The triacylglycerol storage form must behydrolyzed before fats can be used as fuels to yield isolated fatty acids. Triacylglycerols can bemobilized by the hydrolytic action of lipases under the control of glucagon and epinephrine,while insulin inhibits lipolysis.Q4. When Acetyl CoA produced by β-oxidation exceeds the capacity of the citric acid cycle,ketone bodies are produced. Although Acetyl CoA is not toxic, mitochondria must divertacetyl CoA to ketone bodies to keep functioning. Explain why? What would happen ifketone bodies were not generated?There could be a shortage of free coenzyme A. Eventually, all of the CoA would be in the formof acetyl-CoA, no acceptor for additional carbons from fatty acids, and no energy could beproduced. Moreover, the high concentrations of acetyl-CoA would eventually inhibit fatty acidoxidation (not to mention glucose oxidation). Reducing the acetyl-CoA (by generating ketonebodies) solves these problems. It allows for generating high-energy electrons by fatty acidoxidation, not to mention providing fuel for other tissues.Q5. Please complete the following steps for fatty acid degradation and synthesis.I.Fill in the missing reactants, products, or cofactors in each box.II.In the boxes next to the numbers, add the type of reaction
III.In the boxes at the bottom, write the name of the carrier protein for each pathway (R’ andR’’).IV.For each chemical structure, draw a box around the part of the molecule being changedin that step.V.Outline the similarities and differences between fatty acid synthesis and degradation.Acetyl-CoA, the end product of fatty acid degradation, is the precursor for fatty acidsynthesis. In that sense, fatty acid synthesis reverses fatty acid degradation. However,there are differences in mechanisms and in strategies for regulating the pathways. Somekey differences are as follows: (i) Fatty acid degradation occurs in the mitochondrialmatrix, whereas synthesis occurs in the cytoplasm. (ii) The chemical intermediates arelinked to sulfhydryl groups in both cases, but the sulfhydryl carriers differ (being acylcarrier protein for synthesis as opposed to coenzyme A for degradation). (iii) Althoughthe degradative enzymes are separate molecules, the enzymes of fatty acid synthesisare joined in a single polypeptide chain (in higher organisms). (iv) Fatty acids are brokendown by the release of two-carbon units of acetyl coenzyme A, whereas the activateddonor of two-carbon units for synthesis is malonyl-ACP so that the chain elongationreaction can be driven by the release of CO2(v) Different reductants are used—NADPHbeing used for fatty acid synthesis as opposed to NADH and FADH2being producedduring fatty acid degradation.
*** Step 4 in the synthesis is a reduction with NADPH→NADP+Metabolism Integration Question:How many ATP equivalents are produced from the total β-oxidation of one stearic acid (18:0)?How does that compare to glucose? Show how you got your numbers.120 net ATP for stearate and 30 net ATP for glucoseStearic Acid have 18 carbon; therefore, we will get 8 rounds of b-oxidation. However, we get 9Acetyl-CoA
For each b-oxidation, we get1 NADH1 FADH21 Acety-CoA →TCA3 NADH1 FADH21 GTP35 NADH (2.5) = 87.5 ATPS17 FADH2(1.5) = 25.5 ATPS9 GTP (1) = 9 ATP122 ATP- 2 ATP(activation)=120 ATPWe know that we need 8 beta oxidation rounds to break down stearic acid into 9 Acetyl CoAunits completely. Thus, we make32ATPs.For the Citric Acid Cycle, we are sending 9 units to that pathway.Following the amount of ATP equivalents made in the table above from the citric acid cyclethese 9 Acetyl CoAs would produce 90 ATPs.Add everything together:90+ 32= 122, but we needed 2 ATPs to get started so that the net yield would be 120 ATPs.For Glucose: (You worked this out in Discussion 9 FQ1)ATP equivalents from the complete oxidation of glucose: Glycolysis: 1 glucose = 2 NADH = 5 ATP = 2 ATP = 2 ATP = 2 pyruvate Pyruvate Dehydrogenase: 2 pyruvate = 2 NADH = 5 ATP = 2 Acetyl-CoA Citric Acid Cycle: 2 Acetyl-CoA = 6 NADH = 15 ATP = 2 FADH2 = 3 ATP = 2 GTP = 2 ATP Total: = 32 ATP Note that your book removes 2 ATP for transporting the two NADH from glycolysis via glycerol3-phosphate shuttle. Also, be aware that different sources will use different ATP equivalents forthe electron carriers.REFLECTION
Q. After watching the following video(https://www.youtube.com/watch?v=oxX2fq2DBBo) onmembrane transport:a.Describe the three different types of ion transport through membranes in your ownwords. Make sure that your descriptions clearly outline their similarities and differencesand provide one example of each kind of transport.Type oftransportDefinitionExamplesSimilarities toothersDifferences toothers(Passive)-SimpleCan passthrough themembrane freelyCO2Similar tofacilitateddiffusion in thatthere is noenergy usedDifferent toactive transportin that nonenergy needed(Passive)-Facilitated diffusionA channel ortransporterprotein that doesnot use ATPIon channels,GLUT2 receptorSimilar to activetransport in thatthere is a proteinor channelneededSimilair to simplediffusion in thatno ATP is useddifferent toactive transportin that no ATP isused(Active)-primaryUses ATP fortransportSodiumpotassium pumpSimilar tofacilitateddiffusion in thetransportmachineryneededDifferent tosimple diffusionin that ATP isneeded(Active)-secondaryUses storedelectrochemicalgradients fortransportSodium-glucosetransporterSimilair toprimary activetgrasnport in thatenergy is usedDifferent toprimary activetransport in thatstored energy isused rather thandirect energyb.Indicate what type of transport is carried out by ATP synthase and justify your choice.Facilitated transport. If the inner mitochondrial membrane were open to diffusion by thehydrogen ions, the ions would spontaneously diffuse back across into the matrix, drivenby their electrochemical gradient. However, many ions, including H+, cannot diffusethrough the nonpolar regions of phospholipid membranes without ion channels. ATPsynthase is the ion channel that allows protons to flow back into the matrix down theirelectrochemical gradient, an example of facilitated transport. In chemiosmosis, theformation of an ion gradient leads to the generation ofpotential energythat is sufficientto drive the process without any need for ATP input.