Metabolism I,II,III (tbc)

All chemical reactions that maintain the living state of cells and organisms.

Assimilation of molecules and complex structures from the building blocks of life. This also requires energy.

Breakdown of molecules to obtain the anabolic 'building blocks' of life and substrates for energy. It is a precursor for the anabolic process.

<ol> <li>Oxidation through aerobic glycolysis yields pyruvate - efficient</li> <li>Fermentation by anaerobic glycolysis yields lactate - rapid, inefficient</li> </ol> Signup and view all the answers

<p>Oxidation through pentose phosphate pathway → ribose-5-phosphate This acts a precursor for nucleotide synthesis and DNA repair → growth</p> Signup and view all the answers

<p>Via Na+/glucose symporters</p> <ul> <li>Via passive facilitated diffusion glucose transporters – GLUTs (1-5) <ul> <li>Different tissues have different GLUTs – different KM, regulated differently</li> </ul> </li> </ul> Signup and view all the answers

<p>The initial pathway for the conversion of glucose to pyruvate. There is a net gain of 2 ATP.</p> Signup and view all the answers

<ol> <li>Phosphorylation of glucose to give fructose-1,6-bisphospate – requires phosphofructokinase</li> <li>Two interconvertible three-carbon molecules are formed- 2 triose phosphate</li> <li>Through the oxidation of the 3C molecules generates 4 ATP and 2 NADH</li> <li>This produces pyruvate. (2 x pyruvate)</li> </ol> Signup and view all the answers

<p>Control points refer to specific enzymatic reactions within the pathway that play a key regulatory role in controlling the overall rate of glycolysis.</p> Signup and view all the answers

<p>The main control points in glycolysis are associated with enzymes that catalyze reactions involving significant changes in free energy. These control points allow the cell to regulate the flux through the glycolytic pathway based on various factors such as the availability of substrates, the energy status of the cell, and hormonal signals.</p> Signup and view all the answers

<ol> <li>Hexokinase- substrate entry</li> <li>Phosphofructokinase- rate of flow</li> <li>Pyruvate kinase- product exit</li> </ol> Signup and view all the answers

<p>Fructokinase is a key enzyme that controls the rate of substrate movement in glycolysis (rate-limiting)</p> Signup and view all the answers

<ol> <li>Activated by AMP – increases glycolysis when energy is needed</li> <li>Inhibited by ATP – decreases glycolysis when energy is abundant</li> </ol> Signup and view all the answers

<p>Controls conversion of phosphoenolpyruvate to pyruvate (product exit)</p> Signup and view all the answers

<p>Hexokinase catalyzes the phosphorylation of glucose, the rate-limiting first step of glycolysis</p> Signup and view all the answers

<p>It catalyzes the phosphorylation of fructose 6-phosphate to fructose 1,6-bisphosphate using adenosine triphosphate. It is a key regulatory enzyme that controls the rate of substrate movement in glycolysis (rate-limiting).</p> Signup and view all the answers

<p>Pyruvate provide the carbon to fuel the Krebs cycle (TCA cycle) in the mitochondria.</p> Signup and view all the answers

<p>Electron transport chain and ATP synthesis.</p> Signup and view all the answers

<p>In the absence of oxygen, pyruvate can act as a hydrogen acceptor, taking hydrogen ions from NADH. Pyruvate is converted into lactate and NAD is regenerated. It results in the production of lactic acid.</p> Signup and view all the answers

<p>The Warburg Effect is an observation of an increase in the rate of glucose uptake and preferential production of lactate, even in the presence of oxygen.</p> Signup and view all the answers

<p>Cancer cells have a low Km isoform of hexokinase. It supports rapid cell growth (proliferation) - energy production and supports pathways for nucleotide synthesis. - Produces H+ and lactate as end products - Inefficient ATP synthesis with high glucose demand</p> Signup and view all the answers

<p>The lower Km in cancer cells allows hexokinase to effectively phosphorylate glucose even when glucose concentrations are relatively low. This characteristic is thought to confer a metabolic advantage to cancer cells in a few ways:</p> <ol> <li>Enhanced glucose uptake- glucose from surrounding cells</li> <li>Maintaining glycolytic flux- Warburg effect and anaerobic glycolysis</li> <li>Reduced sensitivity to changes in glucose levels- adaptable in varying glucose availability environments.</li> </ol> Signup and view all the answers

<ol> <li>2 deoxyglucose- competitive inhibitor. Blocks further metabolism of G6P.</li> <li>3 bromopyruvate- competitive inhibitor. Blocks production of 1,3 bisphosphoglycerate</li> <li>Dichloracetate- promotes conversion of lactic acid to pyruvate. By reengaging mitochondrial metabolism, it slows glycolysis rate. Cells can't sustain nucleotide synthesis and cannot grow.</li> </ol> Signup and view all the answers

<p>Only limited amounts of NAD+ is present in cells. Glycolysis reduces NAD+ to NADH and H+ ions. So NADH must be deoxidised to let the glycolysis continue. NAD+ is regenerated through the oxidative metabolism of pyruvate.</p> Signup and view all the answers

<p>The Krebs cycle occurs in the mitochondria. The inner membrane contains proteins for electron transport chain, ATP synthase and transport proteins. The matrix contains enzymes for the TCA cycle.</p> Signup and view all the answers

<ol> <li>H+ gradient from cytosol to matrix</li> <li>Pyruvate transporter: H+/pyruvate symport by facilitated diffusion</li> <li>A similar process regulates ADP, ATP &amp; inorganic phosphate (Pi) movement into and out of mitochondria</li> </ol> Signup and view all the answers

<ol> <li>The pyruvate dehydrogenase complex (PDC) catalyses the oxidative decarboxylation of pyruvate to acetyl-CoA.</li> <li>PDC consists of 3 enzymes and is allosterically regulated by phosphorylation.</li> <li>PDC activity determines glucose oxidation in well oxygenated tissues.</li> <li>The reaction is irreversible. Acetyl-CoA cannot be converted back to pyruvate.</li> </ol> Signup and view all the answers

<ol> <li>C2 (acetyl-CoA) condenses with C4 (oxeloacetate) → C6 (citrate)</li> <li>C6 is decarboxylated twice, yielding 2x CO2</li> <li>Four oxidation reactions yield NADH+, H+ and FADH2</li> <li>One GTP formed</li> <li>C4 recreated</li> </ol> Signup and view all the answers

<p>Citrate synthase</p> Signup and view all the answers

<p>It forms reducing equivalents and not necessarily the energy through GTP, as it is only a small energy yield. It is necessary in processes like beta oxidation of lipids and metabolism of amino acids.</p> Signup and view all the answers

<ol> <li>3 pairs of electrons transferred in the conversion of NAD+ to NADH and H+</li> <li>1 pair of electrons needed to reduce FAD to FADH2</li> </ol> <p>From each acetyl-CoA, the TCA cycle generates:</p> <ul> <li>3 NADH + H+</li> <li>1 FADH2</li> <li>1 GTP</li> <li>2 CO2</li> </ul> Signup and view all the answers

<p>All enzymes involved in the Krebs cycle are located in the matrix, except the succinct dehydrogenase which is integrated into the inner mitochondrial membrane.</p> Signup and view all the answers

<ol> <li>Citrate synthase.</li> <li>Isocitrate dehydrogenase.</li> <li>α-ketoglutarate dehydrogenase</li> </ol> Signup and view all the answers

<ol> <li>First step of the cycle</li> <li>The acetyl-CoA is combined with oxaloacetic acid to form citrate.</li> <li>It is inhibited by high concentrations of ATP, acetyl-CoA, and NADH</li> </ol> Signup and view all the answers

<ol> <li>Important catalyst in the third step of the reaction.</li> <li>It regulates the speed at which the isocitrate forms α-ketoglutarate.</li> <li>The coenzyme NADH is a product of the reaction and, at high levels, acts as an inhibitor by directly displacing the NAD+ molecules it is formed from.</li> </ol> Signup and view all the answers

<ol> <li>Important catalyst in the fourth step of the cycle</li> <li>α-ketoglutarate forms succinyl CoA.</li> <li>The two products of the reaction, succinyl CoA and NADH, both work as inhibitors at large concentrations.</li> </ol> Signup and view all the answers