Midterm Exam Review F24 PDF

Midterm Exam Review  Midterm Exam Questions 25 multiple-choice questions 1 hour in duration Closed book Calculator allowed Derived from units 1-6 inclusive (up until the end of Lipids)  Topics that are important Essentiality Different components of the pr...

Midterm Exam Review  Midterm Exam Questions 25 multiple-choice questions 1 hour in duration Closed book Calculator allowed Derived from units 1-6 inclusive (up until the end of Lipids)  Topics that are important Essentiality Different components of the proximate analysis and errors distinguish Different digestive systems -> thedifferiments Energy terminology, concepts of BMR, calculation ↳i What's the difference between using RQ. e. diff. laws *General structures of carbohydrates and lipids * Digestion and transport of carbohydrates and lipids #Macronutrient metabolism including regulatory steps (hormones, enzymes) and overall purpose; calculation of ATP equivalents  Essential Nutrients? An essential nutrient is a chemical that is required for metabolism, but that cannot be synthesized or cannot be synthesized rapidly enough to meet the needs of an animal or human for one or more physiological functions. Nutrients are essential to the human diet if: removed - adverse outcomes deficient and added - start to see recovery Which of the following statements regarding essential nutrients is correct? When an essential nutrient is reintroduced after its removal from the diet, the symptoms associated with a nutrient deficiency can be reversed provided no permanent damage occurred.  FEED SAMPLE (wet weight) Air dry 1. MOISTURE DRY MATTER don't ese needto memorize ↑ Kjeldahl Boil in acid 2. ETHER Residue Filtrate 4. NITROGEN EXTRACT Boil in alkali Residue ASH + Crude Fibre Filtrate Ignite Ignite 3. ASH Ash 5. CRUDE FIBRE  % moisture = wet weight – dry weight × 100% wet weight % dry matter = 100 - % moisture Potential sources of error / limitations: Drying can remove other volatile compounds, such as short chain fatty acids (SCFAs) and some minerals This can cause a slight under-estimation of dry weight Differences between human and agricultural applications? Agricultural industry more interested in composition of dry matter Human food labelling is based on wet weight  % crude fat = weight of ether extract ×100% wet weight of sample Potential sources of error / Gas chromatography (newer method) limitations: Other things are soluble in ether extract: – e.g., chlorophyll, resins, waxes in plants (which are not nutrients) – This will over-estimate crude fat determination Changes in current food labeling – Newer and more sensitive methods now exist  3. Ash (Mineral Content) % Ash = weight of ash × 100% wet weight of sample ! underestimate Potential sources of error / limitations: ↑ Volatile minerals may be lost when burning the residue No information about individual minerals ***It is now mandatory for food labels to indicate sodium content  Kjeldahl Analysis Potential sources of error / limitations: 1. Assumes all proteins have 16% nitrogen – Actual range is 13-19% e.g., peanuts have ~18% nitrogen…therefore conversion factor = 5.5 e.g., milk has 15.7% nitrogen…therefore conversion factor = 6.38 2. Other sources of nitrogen – Any nitrates, nitrites, urea, nucleic acids, etc. in the food sample would therefore be part of the crude protein calculation *This would slightly over-estimate crude protein content  Crude Fibre “Crude fibre” is not the same thing as “dietary fibre” “Crude fibre” is mainly cellulose and lignin (i.e., what remains after processing in the proximate analysis) The term “dietary fibre” is used to describe all fibre (both soluble and insoluble fibres) in a food. – To better estimate dietary fibre content, additional analyses are necessary Potential sources of error / limitations: Unable to distinguish different fibre components Measuring crude fibre under-estimates the actual dietary fibre content of feed by up to 50%. Why? – Dietary fibre includes cellulose, hemicellulose, pectin, mucilages, gums, lignin, etc. Soluble fibres are lost during the proximate analysis.  Nitrogen Free Extract (NFE) = Digestible Carbohydrate (CHO) Estimates starch & sugar content Potential sources of error / limitations: NFE accumulates ALL of the ERRORs that exist for the other components The proximate analysis forms the basis of food composition analysis. However, the components that are calculated within the proximate analysis can lead to errors. Which of the following components is generally under-estimated within the proximate analysis? Crude fibre  1. Simple System w/o caecum e.g. human, pig, cat, dog Key features: Monogastric Non-functional caecum Suited for a nutrient dense, low fibre diet **Fun fact** Average human gut is ~16 feet long  2. Simple System w/ functional caecum e.g. horse, rabbit, hamster Foregut: stomach Hindgut: after the + small intestine small intestine Key Features ‘Pseudo-ruminant’ Hindgut fermenter Functional caecum ↳ little pouch connecting large+ small intestines All other regions of the gut function similar to the monogastric system Suited for a diet with large amounts of fodder and foraging **Fun Fact** Horse digestive tract ~100 feet  III. Multiple System: Ruminant e.g. cattle, sheep, goats Key Features: Large stomach divided into 4 regions 1. Reticulum 2. Rumen 3. Omasum 4. Abomasum System highly suited for animals that eat a high quantity of fodder and forage plant materials. All other regions of the gut function similar to the monogastric system Theory why there is a difference between cows and horses?  Ruminant Digestion Reticulum Honeycomb appearance in order to capture nutrients and trap foreign materials (wire, nails, etc) Rich in bacteria (fermentation vat) Rumen 2 make senses The largest section of the stomach tentation fore gat Rich in bacteria (fermentation vat) ~ Rumen papillae increases surface area for absorption (like microvilli in the human intestine) Food is mixed & partially broken down, and stored temporarily papillae 60-80% of total energy produced here as SCFA  Ruminant Digestion Cont. Omasum Resorption of water and some electrolytes Filters large particles Abomasum > - What makes it diff? Digestive enzymes secreted from gastric glands (HCl, mucin, pepsinogen, lipase, etc) The ‘true stomach’, similar to that of monogastric animals cacidic component) The ruminant digestive system contains a large stomach divided into 4 regions. Which of the following can be thought as the true stomach similar to that of monogastric animals? Abomasum # FRACTION INTAKE EXCRETION Dry matter (DM) 1000 g (100%) 100 g (100%) Protein (CP) 20% 7% Fat (EE) 25% 16% CHO (NFE) 50% 28% Fibre (CF) 4% 39% Chromic oxide (fecal marker) 1% 10% Indicator Method (the marker is used) A–B Apparent Digestibility Coefficient = A A = Ratio of Nutrient/Marker in Feed; B = Ratio of Nutrient/Marker in Feces 20% 7% - 1% 10% Protein = = 0.965 20% 1% Protein = 96.5% Digestible You are a beginner farmer interested in determining the apparent digestibility of protein in a new feed for your cows. You elect to use the marker technique (indicator method) using chromic oxide as a marker. Based on the obtained data (in grams), what is the apparent digestibility of protein in this new feed? Fraction Intake Excretion Protein 95 g 6g Fat 90 g 9g Carbohydrate 136 g 11 g Fibre 75 g 20 g Chromic oxide 4g 4g = 0 93.7%  Definitions Cellular source of energy = ATP – This cellular source of energy is supplied by the macronutrients in the diet – Sustains physical energy, anabolism, active transport, etc. What is the energy value of a food? – Calorie is a measure of heat to express the energy content of food 1000 Chemistry calories = 1 Food Calorie 1 Food Calorie = 1 kcal = 4.18 KJ (kilojoules) Energy required to raise the temperature of 1 kg (1L) of water by 1ºC 1 Food Calorie is equivalent to which of the following? All of the above (1 kcal, 4.18 kJ, 1000 chemistry calories)  * don't have to know this Bomb Calorimetry Dry and weigh sample (~1g), and place in enclosed chamber (the ‘bomb’) with oxygen Sample ignited Heat released is absorbed by water and measured Heat of combustion (gross energy) – Gross energy = maximum energy Potential Errors? – Overestimates energy We don’t digest food like a bomb calorimeter (e.g. fibre) – Doesn’t take into account the energy needed for digestion & absorption  Physiological Fuel Values Heat of Energy Lost Apparent Physiological Combustion in Urine Digestibility Fuel Value (Gross Energy) (a) (b) (c) (a-b) x c units kcal/g kcal/g % kcal/g CHO 4.15 -- 97 4 Fat 9.40 -- 95 9 Protein 5.65 1.25 92 4 Physiological fuel values also called: Takes into account incomplete digestion Atwater Values Available energy Metabolizable energy Fatty Acid Structure & Gross Energy Stearic Acid Factors that affect – 18:0 heat of combustion of – 9.53 kcal/g fatty acids? Oleic Acid – Chain length – 18:1 Longer chain length – 9.48 kcal/g releases more energy – Degree of unsaturation Linoleic Acid The more double – 18:2 bonds, the less energy released (for equivalent – 9.42 kcal/g chain length fatty acids) Fatty Acid X has 21 carbons and 2 double bonds. Which of the following fatty acids will have higher energy than Fatty Acid X? Fatty Acid A that has 21 carbons and 1 double bond You are a new undergraduate research project student and discovered fatty acid X and fatty acid Y. Through a series of calculations, you conclude that fatty acid X has higher gross energy than fatty acid Y. Which of the following statements would provide the most likely explanation why fatty acid X has higher gross energy than fatty acid Y? You are a new undergraduate research project student and discovered fatty acid B and fatty acid C. Through a series of calculations, you conclude that fatty acid B has lower gross energy than fatty acid C. Which of the following statements would provide the most likely explanation why fatty acid B has lower gross energy than fatty acid C? Fatty acid X has more carbons in its structure compared to fatty acid Y. Fatty acid B has less carbons in its structure compared to fatty acid C.  How can we use the Atwater Values? Fat 8 g × 9 kcal/g = 72 kcal Carbohydrate 22 g × 4 kcal/g = 88 kcal Protein 25 g × 4 kcal/g = 100 kcal Fibre?  terminology + relationship Use of Metabolizable Energy Heat Increment of Feeding (HIF) Also called the thermic effect of food Energy used for the digestion, absorption, distribution & storage of nutrients Comprises 5-30% of daily energy usage Used to determine Net Energy (Net Energy = Metabolizable Energy – HIF) Snow much do at the end ? we get Net Energy: supports basal metabolism, physical activity, growth, pregnancy, etc.  Calculating BMR BMR = A×[M0.75] kcal/day don't need > - to do calculation Based on ‘metabolic weight’ – Metabolically active tissue (A) i.e., fat free mass (muscle and bone) Value for humans = 70; every species has its own value – Body weight (M) in kilograms – 0.75 (Kleiber’s Law) – a constant used for all vertebrates, invertebrates and even unicellular organisms e.g., What is the BMR for a 65 kg person? BMR = (70)×[(65 kg)0.75] = 1602 kcal  Harris-Benedict Equation Determining your daily caloric needs refines energy - calculation sex weight height age BMR = 66 + (13.7 × W) + (5 × H) - (6.8 × Age) = Daily calories required BMR = 665 + (9.6 × W) + (1.8 × H) - (4.7 × Age) = Daily calories required physical activity Sedentary (little or no exercise): BMR × 1.2 Lightly active (light exercise/sports 1-3 days/week): BMR × 1.375 Moderately active (moderate exercise/sports 3-5 days/week): BMR × 1.55 Very active (hard exercise/sports 6-7 days a week): BMR × 1.725 Extra active (very hard exercise/sports & physical job): BMR × 1.9 Basal Metabolic Rate (BMR) vs. Resting Metabolic Rate (RMR). Same thing? http://www.bmi-calculator.net/bmr-calculator/harris-benedict-equation/ Can you use body fat % to calculate BMR? If you can measure body fat %, you can have a more accurate measure of BMR. I indicated that the Harris-Benedict equation is commonly used to calculate BMR…which is not untrue…but there is another equation that can be used that takes into account fat % to calculate BMR. The Katch-Mcardle BMR equation – Same formula for men and women If a 70kg human finds out they have 20% body fat, this means they have 80% fat-free mass (FFM). 80% × 70 kg = 56 kg FFM BMR (using FFM) = 1579.6 kcal/d BMR (HB equation) = 1619 kcal/d There are several equations that exist to calculate one's basal metabolic rate (BMR). Which of the following statements is correct? The Harris-Benedict equation considers weight, height, age, biological sex and physical activity when calculating BMR.  Respiratory Quotient (RQ) Provides information about: Energy expenditure Biological substrate being oxidized Ratio of metabolic gas exchange RQ = CO2 produced O2 consumed (Non-protein RQ because protein contributes little to energy metabolism)  For each non-protein RQ value there is a caloric value for each L of O2 consumed or CO2 produced Table also tells you how much CHO and fat contribute to energy (This table will be given to you on exams)  Using RQ to determine energy expenditure… RQ = CO2 / O2 12.0L / 15.7L = 0.764 Kcal produced per L of O2 consumed or L of CO2 produced: (Use the RQ table) 4.751 kcal for 1 L O2 and 6.253 kcal for 1 L CO2  Using RQ to determine energy expenditure… From the previous table, caloric equivalent – 4.751 kcal for 1L O2 and 6.253 kcal for 1L CO2 Energy expenditure per hour: 15.7L O2 × 4.751 kcal/L = 74.6 kcal ~ 75 kcal/h or… 12.0L CO2 × 6.253 kcal/L = 75 kcal/h If you want to determine the basal metabolic rate (BMR): – 75 kcal/h × 24 h = 1800 kcal per day Assumptions Made? 1. Only CHO and fat are metabolized. 2. No synthesis is taking place at the same time as breakdown. 3. Amount of CO2 exhaled = amount of CO2 produced by tissues A person consumed 25.0 L O2/h and exhaled 20.0 L CO2/h under standard conditions. What is this person’s basal metabolic rate (BMR) in kcal per day? RQ =2 = 0. 8 02 = 4 801. CO2 = 6 001. 2x24n 25 01x 4 801 = 120 025 Kcal/h 2880. = 6 Keal/day... 2,880 kcal per day Under standard conditions, Annie consumed 20.0 L O2/h and exhaled 15.0 L CO2/h whereas Brandy consumed 21.5 L O2/h and exhaled 20.0 L CO2/h. Which of the following statements is correct? Please use the provided RQ table. Under standard conditions, Carol consumed 12.5 L O2/h and exhaled 12.0 L CO2/h whereas Annie consumed 20.0 L O2/h and exhaled 15.0 L CO2/h. Which of the following statements is correct? Please use the provided RQ table. Under standard conditions, Brandy consumed 21.5 L O2/h and exhaled 20.0 L CO2/h whereas Annie consumed 20.0 L O2/h and exhaled 15.0 L CO2/h. Which of the following statements is correct? Please use the provided RQ table. Annie has lower energy expenditure than Brandy. For Carol, carbohydrate was the primary source of energy compared to fat whereas for Annie, fat was the primary source of energy compared to carbohydrate. For Brandy, carbohydrate was the primary source of energy compared to fat whereas for Annie, fat was the primary source of energy compared to carbohydrate.  Let’s review CHO Nomenclature Anomeric vs. chiral carbon – Anomeric carbon is the carbon with the carbonyl group – Chiral carbon has 4 different atoms / groups attached to it right left a D vs. L configuration – D confirmation is nutritionally more important down u Alpha or beta configuration of hemiacetal or hemiketal group – More than configuration in nature – has -OH group on anomeric carbon pointing up Fischer Projection vs. Haworth Model – Linear vs. cyclic  Monosaccharides TRIOSE METABOLITES OF GLUCOSE PENTOSE COMPONENTS OF DNA AND RNA HEXOSE NUTRITIONALLY THE MOST IMPORTANT Functional carbonyl group * * * * D-glucose D-fructose (an aldose) (a ketose) (an aldohexose) (a ketohexose) *Anomeric carbon (location of carbonyl group)  Stereoisomerism Fischer Projection Chiral carbon – Counting begins at the anomeric carbon for an aldose Exist in two forms: D vs. L Determined by the -OH group on the highest chiral carbon -OH on the right = D -OH on the left = L The number of stereoisomers for a molecule = 2n (where n = # chiral carbons) D-monosaccharides are nutritionally important because digestive enzymes are stereospecific for D sugars  Haworth Model HEMIKETAL From Fischer projection to Haworth Model OH Rules for what’s up and what’s down? OH Same rules apply for hemiacetal and hemiketal sugars From biochemistry… LAB vs. RBA (left/above/beta vs. right/below/alpha) For the given diagram, how many chiral carbons are there? Y anomeric carbon 4 = : not chiral The given diagram shows "Monosaccharide X" in the Haworth model whereby carbon #1 is indicated as the new chiral centre. Based on this diagram, which of the following correctly describes the structure? The given diagram shows "Monosaccharide Y" in the Haworth model whereby carbon #1 is indicated as the new chiral centre. Based on this diagram, which of the following correctly describes the structure? Hexose, alpha-configuration Hexose, beta-configuration  Polysaccharides Starch (Amylose, Amylopectin) is rich in plants – Both forms are polymers of D-glucose Glycogen is rich in animal tissue Is there an advantage for branching? – Yes. This provides a larger number of ends from which to cleave glucose when energy is needed. straight Chain but branched branchedriveseanimal tissue You discovered a new form of carbohydrate called “Carbohydrate X” that is branched and contains more than 20 glucose units joined together by glycosidic bonds. You conclude that Carbohydrate X resembles which of the following? Amylopectin  Characteristics of Dietary Fibre Characteristics of solubility? – Water-holding ability Ability of a fibre to hold water, becoming a viscous solution – Adsorptive ability Ability of a fibre to bind enzymes and nutrients Insoluble (cellulose, lignin, some hemicelluloses) Remain intact throughout the digestive system Reduce transit time (i.e., things move quickly through the gut) Increases fecal bulk Soluble (pectins, gums, -glucans, some hemicelluloses) Forms a gel Delays gastric emptying, increases transit time Slows down the rate of nutrient absorption  Health Benefits of Fibre Maintains function & health of the gut constipation (insoluble fibre) > - reducing transit time Stimulates muscle contraction to break down waste Decreases risk of bacterial infections satiety (soluble fibre) Delays gastric emptying Slows down nutrient uptake  Soluble Fibre and Disease Risk Decrease cardiovascular disease risk by lowering blood cholesterol ***Can also lower the risk of type II diabetes by binding some glucose in the digestive tract Psyllium seed husks Soluble fibre and cholesterol from foods reach the stomach and travel to the small intestine. Soluble fibre forms a gel which binds some cholesterol in the small intestine and carries it out of the body (i.e., in feces). Which of the following characteristics associated with soluble fibres is correct? Soluble fibres can form a gel.  Carbohydrate Digestion Mouth – -amylase (salivary) breaks down -1,4-glycosidic bonds – Produces only a few monosaccharides – Cellulose and lactose are resistant, as are -1,6-bonds – Stomach – -amylase digestion continues until pH drops, then enzyme is inactivated (distinctionbetween lipid digestion) carb/ – At this point, the pool of dietary CHO consists of small polysaccharides and maltose – Small Intestine – -amylase (pancreas) – Active at a neutral pH – -1,6 bonds are resistant and eventually produce isomaltose Which of the following statements regarding carbohydrate digestion is correct? Pancreatic alpha-amylase is active in the small intestine, which the pH of the small intestine is neutral. Salivary alpha-amylase is inactivated by the low pH of the stomach.  Brush Border enzyme activity question on this a Also called sucrase sucrose glucose + fructose Also called isomaltase isomaltose 2 glucose lactose glucose + Maltose 2 glucose galactose See Diagrams There are several enzymes that function to break down different disaccharides. Which of the following is a correct statement? Maltase breaks down maltose into 2 glucose molecules. Lactose Intolerance  how to absorb glucose into bloodstream Monosaccharide absorption Enterocytes are polarized cells (in other words, they have an Very efficient “up” and a “down) Nearly all monosaccharides are taken up by enterocytes – What happens to the glucose then? Small amounts leak back out into the lumen from the enterocyte Small amounts diffuse into blood APICAL SGLT1 through the basolateral membrane Majority is transported into blood through GLUT2 Transport of glucose and galactose from BASOLATERAL 7 lumen into blood is dependent on basolateral Na-K ATPase activity / facilitated 3 Na+ 2K+ transport * want to In contrast, fructose taken up by maintain favourable facilitated transport gradient for – GLUT5 on apical surface glucose Glucose, galactose and fructose all SGLT1 sodium glucose transport 1 enter blood via basolateral GLUT2 Which of the following statements regarding monosaccharide absorption is correct? Fructose is taken up by enterocytes by facilitated transport via GLUT5 on the apical surface.  know difference between Glycogenesis hexo and glucokinase Insulin Insulin +ve +ve Glycogen Synthase -ve Hexokinase (muscle) want to store it ASAP to g prepare Glucokinase (liver) +ve Stimulates Insulin  Glycogenesis, Cont. helps to prime glycogenesis Glycogenin is an enzyme that serves as a scaffold on which to attach glucose molecules to build glycogen. – Think of this enzyme as a “primer”. It initially attaches glucose molecules to itself before glycogen synthase takes over and adds glucose to the growing glycogen store – 30,000+ glucose molecules can be contained in a single glycogen store Glycogenesis refers to the pathway which glucose is converted to its storage form glycogen. Which of the following statements is correct? Glycogen synthase is stimulated by insulin and facilitates the formation of glycogen.  Glycogenolysis ↳ breakdown of glycogen couring fasting Glucagon +ve STIMULATES Glycogen Phosphorylase (breaks -1,4- glycosidic bond) Glucose-6- phosphatase (liver only) = GLYCOLYSIS GALACTOSE GLUCOSE Phosphofructokinase is the first committed step (irreversible) in FRUCTOSE Glucokinase / Hexokinase glycolysis Phosphofructokinase Fructose-1,6- bisphosphate -ve -ve ATP Glucagon 2 × Glyceraldehyde- (in the liver) 3 phosphate Red blood cells have no mitochondria Net Energy Yield from 1 Glucose 2 × Pyruvate In red blood cells, [2 NADH + 2 ATP (equivalent to ~8 ATP)] glycolysis is the only way these cells can Aerobic* Anaerobic* generate ATP *Metabolic fate of pyruvate depends on the cell’s Kreb’s Cycle Lactic Acid oxygen status  Anaerobic metabolism of glucose Lactic acid production – Occurs in muscle during prolonged exercise and in red blood cells – Pyruvate is converted into lactic acid in the cell’s cytosol – Regenerates NAD+, which allows glycolysis to continue – A net of 2 ATP is produced when glucose is converted to lactic acid Making ethanol Conly applicable for yeast) – Doesn’t happen in the body – Yeast breaks down pyruvate into CO2 and ethanol – This is the basis of fermentation when you make wine and beer – Regenerates NAD+, which allows glucose to continue being broken down in glycolysis  Cori Cycle The Cori Cycle occurs in times where oxygen is unavailable (anaerobic state) in the muscle, leading to the production of lactate. Lactate is transported back to the liver, where gluconeogenesis allows for the conversion of pyruvate back to glucose. For 2 molecules of lactate to form glucose the cell consumes 6 ATP molecules. rid of it gets i. e. What is relationship between tissues This is not sustainable because more energy is consumed than produced. need 6 , produce 2  alternate Hexose Monophosphate Shunt ↳ to build things ! Chucleotides , FAS pathway to glycolysis Important for NADPH production and ribose synthesis Ifatty acid production Occurs in the cytoplasm of a cell BLOSYNTHESIS OF PRECURSORS 6PGL 6PG G6P = glucose 6-phosphate; 6PGL = 6-phosphogluconolactone; 6PG = 6-phosphogluconate F6P = fructose 6-phosphate The Hexose Monophosphate Shunt is used to generate NADPH and precursors for nucleotide synthesis Which of the following processes forms biogenic molecules for fatty acid synthesis and nucleotide synthesis? Hexose monophosphate shunt  Pyruvate Dehydrogenase: The “gatekeeper” to Krebs Cycle Pyruvate Dehydrogenase (PDH) Complex Several enzymes requiring several cofactors, 2× including 4 vitamins: 1) Thiamine, 2) Niacin, 3) Riboflavin, and 4) Pantothenic acid. COA COMMITTED STEP acetyl Net Energy Yield (~6 ATP) = - pyruvate 2 × [1 NADH / pyruvate]  Krebs Cycle (TCA or citric acid cycle) Pyruvate need to know total Pyruvate count of ATP equivalents Carboxylase ATP ADP Takes place in mitochondrial matrix NAD+ NAD+ NAD+ GDP FAD Energy Yield from 1 Acetyl-CoA [3 NADH, 1 FADH2, 1 GTP (equivalent to ~12 ATP)] What is the total number of ATP equivalents produced from 1 molecule of glucose in the Kreb’s cycle alone? ↳ 2 acetyl Coa : 12x2 24  How much energy do you get from 1 molecule of glucose? ATP yields from the complete oxidation of 1 molecule of glucose Discrepancy in the literature about ATP yields from glucose: – In class we said: Glycolysis net energy = 2 ATP + 2 NADH (~8 ATP) Pyruvate dehydrogenation net energy = 1 NADH (~3 ATP) – X 2 because two pyruvates (total = 6 ATP) Kreb’s cycle net energy = 3 NADH, 1 FADH2 & 1 GTP (~12 ATP) – X 2 because 2 acetyl CoA (total = 24 ATP) TOTAL ENERGY = 38 ATP THIS IS THE NUMBER WE WILL CONSIDER IN NUTR*3210! – However, some textbooks account for the fact that 2 ATPs are needed to import the 2 NADH produced from glycolysis into the mitochondria. THUS SOME BOOKS SAY TOTAL ENERGY = 38 ATP – 2 ATP = 36 ATP You discovered a new organism that is capable of completely oxidizing 1 molecule of glucose through Process W that generates 12 NADH, 2 FADH2 and 2 GTP. Using the conversion factors we learned in class for converting energy molecules to ATP equivalents, how many total ATP equivalents can this organism produce through Process W for 1 molecule of glucose? You discovered a new organism that is capable of completely oxidizing 1 molecule of glucose through Process X that generates 15 NADH, 5 FADH2 and 2 GTP. Using the conversion factors we learned in class for converting energy molecules to ATP equivalents, how many total ATP equivalents can this organism produce through Process X for 1 molecule of glucose? You discovered a new organism that is capable of completely oxidizing 1 molecule of glucose through Process Y that generates 20 NADH, 5 FADH2 and 6 GTP. Using the conversion factors we learned in class for converting energy molecules to ATP equivalents, how many total ATP equivalents can this organism produce through Process Y for 1 molecule of glucose? You discovered a new organism that is capable of completely oxidizing 1 molecule of glucose through Process Z that generates 25 NADH, 4 FADH2 and 2 GTP. Using the conversion factors we learned in class for converting energy molecules to ATP equivalents, how many total ATP equivalents can this organism produce through Process Z for 1 molecule of glucose? 42 ATP equivalents 57 ATP equivalents 76 ATP equivalents 85 ATP equivalents GLUCONEOGENESIS Glucokinase / Glucose-6- Hexokinase phosphatase Phosphofructokinase Fructose-1,6- bisphosphatase GLUCONEOGENESIS GLYCOLYSIS Enzymes to “bypass” irreversible steps and allow gluconeogenesis (enzymes expressed in liver, but not muscle or adipose) Pyruvate carboxylase Pyruvate kinase & Phosphoenolpyruvate (PEP) carboxykinase (PEPCK) In anaerobic states, muscle lactate returns to the liver for gluconeogenesis (Cori cycle)  Fatty Acids > HOW - TO NAME THEM Saturated – Maximum # of H HYDROPHOBIC HYDROPHILIC atoms – Only single bonds Unsaturated – “Missing” H atoms – Double bonds Monounsaturated Monounsaturated fatty acid (Oleate) Polyunsaturated – cis or trans configuration Carboxylic acid group  Fatty Acid Nomenclature Delta System ( ) Omega System ( ) – Numbering starts from – Numbering starts from carboxyl end of fatty acid methyl end of fatty acid CH3-(CH2)5 18:2 9,12 18:2 n-6 or 18:2 -6 Location of 1st # Carbons Double Bond # Double Bonds Position of Double Bonds # Carbons # Double Bonds Which of the following statements is CORRECT when describing the given structure?  Essential Fatty Acids (EFA) Linoleic Acid (18:2 n-6) Alpha Linolenic Acid (18:3 n-3) Nuts, seeds, vegetable oils, etc Fatty fish, canola oil, almonds, etc Why are these fatty acids essential? Humans lack the enzymes necessary to insert double bonds beyond the delta-9 position of a fatty acid The delta-12 and delta-15 fatty acids are expressed in plants  Signs of Essential Fatty Acid Deficiency Are ↳ Whodeficiencies common? know the is susceptible to NO. a deficiency? distinction Infants and hospitalized patients n-6 deficient n-3 deficient Skin dermatitis Ok Growth Ok Reproductive Ok maturity CNS Ok IQ development Retinal Ok Visual acuity development 2-3% of energy in diet 1% of energy in diet Patient A has n-6 fatty acid deficiency. Which of the following would you expect this patient to display? Dermatitis  EFA Desaturation and Elongation Eicosanoid Desaturation insert double bond, remove 2 H Production Elongation add 2 carbons (from malonyl CoA) (Pro-inflammatory) Linoleic Acid -Linolenic Acid Dihomo- -Linolenic Acid Arachidonic Acid (LA) (18:2 n-6) (GLA) (18:3 n-6) (DGLA) (20:3 n-6) (AA) (20:4 n-6) Desaturation Elongation Desaturation Delta-6 Desaturase Elongase 5 Delta-5 Desaturase -Linolenic Acid Stearidonic Acid Eicosatetraenoic Acid Eicosapentaenoic Acid* (ALA) (18:3 n-3) (SDA) (18:4 n-3) (ETA) (20:4 n-3) (EPA) (20:5 n-3) Conversion Efficiency < 8% Eicosanoid Production (Anti-inflammatory) *EPA can be further converted into DHA Merino et al, Lipids in Health Dis, 2010 With respect to essential fatty acids, which of the following statements is correct? The 2 essential fatty acids are linoleic acid (18:2 n-6) and alpha-linolenic acid (18:3 n-3), both of which have 18 carbons in their structures.  Eicosanoids Metabolites of 20-carbon fatty acids (primarily AA and EPA) PGD2 prostaglandins Produced by most cells in the body Hormone-like, function TXA2 locally thromboxanes Role in inflammation, platelet aggregation, blood pressure, etc. Implications for disease LTE4 leukotrienes  Triglycerides (TAG) Main dietary lipid and storage lipid Key in several pathways: Lipogenesis Lipolysis Transport in lipoproteins Structures Monoacylglycerol (MG, MAG) Diacylglycerol (DG, DAG) 3 Triglyceride / Triacylglycerol (TG, 2 TAG) 1 Fatty acid composition determines sn-1, sn-2, sn-3 positions physicochemical properties Ester bonds -form between glycerol backbone and fatty acid chains  Phospholipids (PL) Structural features More polar than TAGs Hydrophilic phosphate head group 1 2 Primary functions 3 Components of membranes Source of physiologically active fatty acids for eicosanoid synthesis Anchors membrane proteins Intracellular signaling  Sterols > - FOUR RING Steroid alcohols – Monohydroxy alcohols Structural features – Free or esterified with a fatty acid Cholesterol ester (CE) Sources of cholesterol: – Diet: meat & eggs (~40%) – Endogenous production (~60%) Primary functions – Essential components of membranes – Precursor for : Bile acid production Steroid sex hormone production Vitamin D synthesis Which of the following statements correctly describes phospholipids? Phospholipids are more polar than triglycerides.  Lipid Digestion > - need more players Mouth – Lingual lipase – Continuously secreted Stomach – Gastric lipase – Continuously secreted – Lipases stable at low pH Liver Gallbladder – Storage of bile acids – Release of bile triggered by hormones Small Intestine – Pancreatic enzymes include pancreatic lipase, cholesterol - esterase Lingual and gastric lipases are important enzymes in the initial stages of lipid digestion. Which of the following correctly characterizes these lipases? Lingual and gastric lipases are stable at low pH of the stomach environment.  Mixed Micelles > - structural components important for lipid digestion ↳ bring helps lipids into enterocytes Digested lipids are emulsified by bile acids Small, spherical complexes containing lipid digestion products plus bile acids (also called bile salts) Can access the spaces between microvilli in the intestine Originally thought that digested lipids were delivered into intestinal enterocyte cells by passive diffusion, but carrier- mediated transporters have now been identified Bile acids are reabsorbed in our digestive tract BS = bile salt CL = cholesterol PL = phospholipid FA = fatty acid Source: Shi & Burns, Nat Rev Drug Disc 2004  Enterohepatic Circulation a reduces ***Soluble fibres cholesterol Bile acids reduce the efficiency made in Cholesterol of enterohepatic liver circulation by holding Bile acids on to bile acids, ~95% bile which are then acids secreted in feces reabsorbed Isforcing body to utilize and recycled endogenous Cholesterol by making bile acid back to the liver Bile acids VERY EFFICIENT stored in gallbladder ~5% bile acids lost in feces Source: http://zavantag.com/docs/1632/index-2974.html?page=5 Which of the following statements regarding bile acids in enterohepatic circulation correct? Soluble fibres reduce the efficiency of enterohepatic circulation by holding onto bile acids for elimination in feces.  Lipid Absorption - wantto get- Brush border enzymes: Pancreatic lipase Mixed Micelle Cholesterol esterase Phospholipase BROKEN DOWN ↑ Lipoprotein reassemble  Lipid Transport Lipoprotein classification determined by: 1. Ratio of Lipid-to- Protein (which affects density) LARGEST 2. Specific apolipoprotein (Apo) content (which affects receptor interactions) ↑ low density due to high lipid content tends to deposit things in places you don't want High Lipid High Lipid ‘BAD CHOLESTEROL’ ‘GOOD CHOLESTEROL’ Low Protein Low Protein Relative to VLDL, protein Low Lipid ApoB-48 ApoB-100 to lipid ratio increases High Protein ApoC, ApoE ApoC, ApoE ApoB-100, ApoC, ApoE ApoA family Source: www.medscape.com  Chylomicrons Chylomicrons increase in circulation after a meal Enter circulation at a slow rate Peaks between 30min-3hr after eating Since chylomicrons enter the lymphatic system before entering the blood, dietary lipids are available to adipose and muscle before arriving at the liver. diff recognizes ↑ fingerprints Lipoprotein lipase (LPL) is located on the surface of endothelial cells lining small blood vessels and capillaries LPL not expressed in liver, but is expressed by adipose tissue and muscle LPL is activated by ApoC in chylomicrons LPL hydrolyzes the TAG in chylomicrons When chylomicrons become TAG-depleted, they are now referred to as a “chylomicron remnant” (CR) CR are removed from circulation through ApoE-mediated interactions with a receptor in the liver  Lipoproteins Cont. insulin after meal ApoC + lots of lipids body needs to clear Chylomicron LPL ApoC Intestine ApoE ApoE CR (Dietary TAG) ApoB48 ApoB48 TAG hydrolyzed ApoC VLDL LPL ApoC Liver ApoE ApoE (Liver LDL TAG) ApoB100 ApoB100 HDL TAG hydrolyzed Which of the following statements correctly describes lipoprotein lipase (LPL)? LPL hydrolyzes triglycerides.  Low-Density Lipoprotein (LDL) ApoC VLDL is the main ApoE VLDL transporter of ApoB100 newly synthesized TAG hepatic TAG either stored or LPL LDL taken up by the liver used for via LDL-receptors (i.e., receptor mediated energy endocytosis) *60% of blood cholesterol in LDL In a fasting subject ApoC Brown & Goldstein ApoE Nobel Prize in Medicine in 1985 -goes through LDL main transport endocytosis mechanism ApoB100  High-Density Lipoprotein (HDL) ApoA Cholesterol obtained from plasma membranes and then esterified directly Formation HDL on HDL Most blood cholesterol is esterified of LCAT with fatty acids cholesterol esters Reverse Cholesterol Transport REVERSE CHOLESTEROL TRANSPORT brings lipid molecules back when HDL picks up cholesterol around the body and delivers it to the liver. Liver to liver LCAT = lecithin-cholesterol acyltransferase (esterifies a fatty acid to cholesterol) esterifying SR-BI = scavenger receptor class B1 (HDL receptor in the liver) transfers brougholiver CETP = cholesterol ester transfer protein (transfers cholesterol ester from HDL to VLDL and/or LDL) In comparing low-density lipoprotein (LDL) versus high- density lipoprotein (HDL), which of the following statements is correct? Only HDL has surface ApoA that activates lecithin- cholesterol acyltransferase (LCAT), which forms cholesterol esters.  Dietary Cholesterol Plant sterol For healthy people, limiting Cholesterol onlyplanterare dietary cholesterol does not pumped aa absorbe > can't be change blood cholesterol much - LUMEN But for those with high blood cholesterol, dietary cholesterol Plant sterols compete with cholesterol for ABCG5 uptake by NPC1L1, but NPC1L1 plant sterols are then ABCG8 pumped back into the lumen by ABCG5 / G8 apical transporters. chylomicron NPC1L1: Niemann-Pick C1-Like 1 ABCG5: ATP-binding cassette sub-family G 5 LYMPH ABCG8: ATP-binding cassette sub-family G 8 chylomicron  Trans Fatty Acids ↳ banned Trans fats: Unsaturated fatty acids with at least one double bond in the trans configuration. There are both industrial and natural trans fats. Industrial trans fats produced during the hydrogenation of vegetable oils. – Companies do this to increase stability during cooking, longer shelf-life, and for palatability – Hydrogen atoms are added catalytically across double bonds Partial hydrogenation results in double bonds being converted from cis to trans Complete hydrogenation results in all double bonds becoming full saturated – Industrial trans fats are completely banned in Canada as of 2018. Trans fats are also found naturally in ruminant fat – Milk fat contains 4-8% trans fat (best known is conjugated linoleic acid – CLA) – Natural trans fats are made in the rumen through bacterial fermentation. – Health affects linked with natural trans fats are equivocal. Lipolysis and Gluconeogenesis Lipases hydrolyze ester linkages (lipolysis) Triglyceride to break needed In adipose tissue a down TAG – HSL (hormone sensitive lipase) -ve insulin – Cleaves a fatty acid from the HSL glycerol backbone The complete breakdown of a TAG molecule releases 1 glycerol and 3 fatty acids -oxidation generates energy Glycerol can enter into glycolysis or gluconeogenesis (depends on cell needs) Fatty acids can undergo - oxidation and used be to generate energy  -oxidation and Kreb’s Cycle STEPS Kreb’s cycle net energy = 3 NADH, 1 FADH2 & 1 GTP 1. Dehydrogenation FADH2 (~12 ATP) for 1 acetyl CoA 2. Hydration ETC 3. Oxidation NADH 4. Thiolysis *Each round of B-oxidation Acetyl CoA removes 2 carbons (acetyl CoA), and produces Kreb’s Cycle 1 NADH + 1 FADH2  1 cut (SATP) a You have 2 molecules of 4-carbon fatty acid. How many total ATP equivalents are produced by the complete oxidation of these 2 molecules of 4-carbon fatty acid? 24 + 5 = 29 29x2 = 38 58 ATP equivalents