Module 3 Energy in Biochemistry S23 Students PDF

 When bonds are broken, catabolism chemical energy is released these chemical bonds have more energy than: these chemical bonds propane oxygen (O2) CO2 H2O where did the extra energy go?...

 When bonds are broken, catabolism chemical energy is released these chemical bonds have more energy than: these chemical bonds propane oxygen (O2) CO2 H2O where did the extra energy go? this is the central feature of our study of biochemistry…. Energy is neither created nor destroyed but can change forms in a series of biochemical reactions (chemical) (kinetic) either from a molecule to or from the surroundings to its surroundings… the molecule … measured in J (Joules) or in kJ (kilojoules) ∆H is always considered relative to the molecules that are reacting DH = Hfinal - Hinitial 8kJ 8kJ = -5kJ = +5kJ 3kJ 3kJ 8kJ 3kJ 3kJ 8kJ  (H = 20 kJ) (H = 5kJ) R eactants P roducts ∆H = HFinal – HInitial Initial H of molecule = 20kJ Final H of molecule = 5kJ ∆H = 5kJ – 20kJ ∆H = -15kJ environment oC  (H = 10 kJ) (H = 30kJ) R eactants P roducts ∆H = HFinal – HInitial initial H of molecule = 10kJ Final H of molecule = 30kJ ∆H = 30kJ – 10kJ ∆H = +20kJ environment oC (H = 10KJ) (H = 30KJ) X R → P environment the environment (the universe) does not want to give up energy All processes try to increase entropy (degree of disorder) in the universe Entropy is inversely proportional to energy increasing entropy, going from low entropy → high entropy, means that the reaction is moving from higher to lower energy → releasing energy, energy that helps a reaction occur spontaneously -ve DH, and +ve DS help a reaction be spontaneous  (less disorder) (more disorder) DS is measured in Joules/degree Kelvin* *degrees Kelvin is the SI unit of temperature: oK = oC + 273  entropy entropy energy energy (energy released to environment) (energy taken from environment) > liquid > sold  CH2OH CH2OH CH2OH CH2OH HO H H H OH HO H OH H H OH H OH 1 H H O 4 OH H H H OH H H + HO OH H H H OH H OH H OH H OH lactose galactose glucose entropy energy  entropy entropy  Reactants (R) Products (P) AB → A + B initial final 1. calculate ∆H: Hfinal – Hinitial = 14 J – 27 J = –13 J 2. calculate ∆S: Sfinal – Sinitial = 22 J/oK – 13 J/oK = +9 J/oK losing energy 3. from an enthalpy standpoint, the “system” is _____ gaining entropy, 4. from an entropy standpoint, the “system” is _______ losing energy which means that it is ______ losing energy 5. overall: the “system” is ______ 1. calculate ∆H: Hfinal – Hinitial = 25 J – 6 J = +19 J 2. calculate ∆S: Sfinal – Sinitial = 45 J/oK – 34 J/oK = +11 J/oK gaining energy 3. from an enthalpy standpoint, the “system” is _______ gaining entropy, 4. from an entropy standpoint, the “system” is _______ losing energy which means that it is ______ Gibbs free energy combines ????? energy enthalpy and entropy 5. overall: the “system” is ______  releases spontaneously requires non-spontaneously ∆G = ∆H – T∆S  ΔG = ΔH – TΔS DG, the Gibbs free energy of a reaction determines whether a reaction will proceed → the change in enthalpy (DH) and change in entropy (DS) If ΔH and ΔS then ΔG Result – + – Exergonic + – + Endergonic – –  Exergonic: ΔH < TΔS Endergonic: ΔH > TΔS + +  (note that -2 < -1 )  initial final Reactants Products initial final initial final DH = Hfinal – Hinitial = 8 kJ – 6 kJ = +2 kJ = +2000 J DS = Sfinal – Sinitial = 45 J/oK – 34 J/oK = +11 J/oK T = oC + 273 = 20 + 273 = 293 oK DG = DH – TDS = +2000 J – 293oK(11 J/oK) = +2000 J – 3233 J DG = –1233 J → Exergonic → spontaneous! Rule: ∆G and ∆Go’ for a given reaction usually correlate (but not always) This is what we need to determine meaning this value determines always +ve → is it +ve or –ve? whether we are adding a +ve or → spontaneous or not? -ve number to DGo’ constant variables  [C] = 0.3M [D] = 0.1M [A] = 0.2M [B] = 0.3M 0.3M x 0.1M 0.03 = = = 0.5 0.2M x 0.3M 0.06 ln ? ln(0.5) = -0.693 1. Unidirectional arrows: indicate irreversible reactions (i.e. only operate in direction shown) 2. Bidirectional arrows: indicate reversible reactions (i.e. operate in both directions)  ∆Go’ = +13.8 kJ/mol this is what happens in step 1 of Glycolysis (previous slide, minus the enzyme and without ATP cleavage) H + HPO42- + H2O Glucose Glucose-6-phosphate ∆Go’ is large and positive → therefore reaction is non-spontaneous energetically unfavourable problem: the reaction doesn’t occur on its own (but we need it to) Reaction 1 (from 2 slides ago): Glucose + HPO42– → Glucose-6-phosphate + H2O ∆Go’R1 = +13.8 kJ/mol Reaction 2 (ATP hydrolysis): ∆Go’R2 = -30.5 kJ/mol H2O + ATP → ADP + HPO42– Can use ATP to change overall ∆Go’ of reaction 1 from positive to negative ‘Coupling’ to ATP hydrolysis Reaction therefore becomes energetically favourable energy of overall “coupled” reaction: Overall ∆Go’ = ∆Go’R1 + ∆Go’R2 = +13.8 kJ/mol + (–30.5 kJ/mol) = –16.7 kJ/mol Reaction 1: ∆Go’R1 = ??? kJ/mol R-PO42– + H2O → R-OH + HPO42– Reaction 2: ∆Go’R2 = +30.5 kJ/mol ADP + HPO42– → H2O + ATP Overall reaction: ∆Go’overall = +11.1 kJ/mol R-PO42– + ADP → R-OH + ATP Overall ∆Go’ = ∆Go’R1 + ∆Go’R2 +11.1 kJ/mol = ∆Go’R1 + (+30.5 kJ/mol) +11.1 kJ/mol - 30.5 kJ/mol = ∆Go’R1 –19.4 kJ/mol = ∆Go’R1  [R-PO42–] = 0.13 M Overall reaction: R-PO42– + ADP → R-OH + ATP reactants { [ADP] = 0.09 M [R-OH] = 0.002 M ∆Go’overall = +11.1 kJ/mol products { [ATP] = 0.0012 M ∆G = ∆Go’+ RT lnKeq ∆G = +11.1 kJ/mol + RT lnKeq Keq = [0.002]x[0.0012] [0.13]x[0.09] ∆G = 11,100 J/mol + (8.314 J/moloK ))(310)lnKeq Keq = 0.0000024 ∆G = 11,100 J/mol + (2,577.3 J/mol)(-8.49) 0.0117 ∆G = 11,100 J/mol + (-21,881.6 J/mol) Keq = 0.00021 ∆G = -10,781.6 J/mol) lnKeq = -8.49 exergonic… spontaneous (in these conditions)  1000 calories = 1 Calories = 1 kilocalories (kcal) calories Joules 1 calorie = the energy needed to increase 1 Joule = the energy needed to lift 1lb to a the temperature of 1 gram (i.e. 1 ml) of height of 9 inches (application of 1 Newton water by 1oC of force over distance of 1 metre) Equivalent of heat Equivalent of heat Not an SI unit but is commonly used to measure food energy SI unit of energy 1 calorie = 4.184 J 1 J = 0.239 calories 1 Calorie (or 1 kcal) = 4184 J (or 4.184 kJ) 5 Cal = 5 kcal = 5000 calories = 20920 J = 20.9 kJ Food Cal/g (or kcal/g) carbohydrate 4 Converted to the protein 4 “universal energy currency”: ATP alcohol (ethanol) 7 lipid 9 Combustion of 1 calorie of energy = food sample energy needed to raise the T of 1g generates heat (1ml) of water by which raises 1oC temperature ex. our bomb calorimeter of water has 200ml of water, burning food sample raised water by 3oC Calories = ??  Starbucks Caramel Macchiato does NOT factor in energy calculation Total Calories = 304 Calories (304 kcal) Total kJ = 1274 kJ (304 Cal x 4.184 kJ/Cal*) Calories from fat, carbs, protein? 12 g Fat = 108 Calories (12g x 9 Cal/g) 37 g CHO = 148 Calories (37g x 4 Cal/g) 12 g Protein = 48 Calories (12g x 4 Cal/g) note that 108 + 148 + 48 = 304 Calories Fiber, which is listed under carbohydrates cannot be digested, therefore contributes zero Calories, and the g of Fiber must be subtracted from Total Carbohydrate  Calculate the % of total calories in this food sample from fat, CHO, and protein Starbucks Caramel Macchiato Total Calories = 304 Calories (304 kcal) 12 g Fat = 108 Calories (12g x 9 Cal/g) (108 Cal / 304 Calories x 100%) = 36% 37 g CHO = 148 Calories (37g x 4 Cal/g) (148 Cal / 304 Cal x 100%) = 49% 12 g Protein = 48 Calories (12g x 4 Cal/g) (48 Cal / 304 Cal x 100%) = 16% NOTE: Values on the nutrition labels are approximations, and labels contain rounded quantities and therefore will often contain small errors. values listed on nutrition labels are Useable, digestible, metabolizable energy (not the Gross energy in the food serving) (useable E) ME = GE – (FE + UE + GasE) indigestible energy (Unuseable) ME is used for growth, cellular/tissue maintenance, activity (movement), reproduction BUT… negligible growth and movement (adult hens) (caged) here, ME is mostly used for repair/maintenance (body weight) and reproduction (laying eggs) THE HEN EXPERIMENT Control Low Energy Very Low Diet Diet Energy Diet Gross Energy (GE) 3.48 3.11 2.43 (Cal/g diet) Metabolizable 2.90 2.60 2.00 Energy (ME) (Cal/g diet) This is a good illustration of the amount of useable energy in a mammalian diet (here ~83% of GE) THE HEN EXPERIMENT Two main uses of energy in these hens (i.e. Energy OUT) → Egg production (reproduction) → Body weight (body repair/maintenance) Total Eggs This graph show how 14.00 12.00 lowering dietary 10.00 12.10 10.57 energy causes: 8.00 # of eggs 9.21 6.00 a reduced ability to 4.00 2.00 reproduce 0.00 Control Low Energy Very Low Energy Diet THE HEN EXPERIMENT Two main uses of energy in these hens (i.e. Energy OUT) → Egg production (reproduction) → Body weight (body repair/maintenance) Weight Change This graphs show how 0.00 lowering dietary Control Low Energy Very Low Energy -20.00 energy causes: Weight (g) -40.00 -51.72 -55.48 -60.00 -65.67 a reduced ability to -80.00 maintain/repair tissues -100.00 Diet The main finding: There is an association between ME and bodily functions → Energy expenditure (measured in Calories over 24 hrs) Energy expenditure (in 24 h) = BMR + activity + thermic effect of food 11 2 3 1 at rest, awake temperature neutral environment “post-absorptive” state (fasting) → Energy expenditure (measured in Calories over 24 hrs) Energy expenditure (in 24 h) = BMR + activity + thermic effect of food 1 2 3 2 Activity level Description Energy expended “Inactive” Little/no exercise, desk job BMR x 20% “Light” Light exercise 1–3 d/wk BMR x 30% “Moderate” Moderate exercise, 3–5 d/wk BMR x 40% “High” Intense exercise, 6–7 d/wk BMR x 50% → Energy expenditure (measured in Calories over 24 hrs) Energy expenditure (in 24 h) = BMR + activity + thermic effect of food 1 2 33 3 → Energy expenditure (measured in Calories over 24 hrs) Energy expenditure (in 24 h) = BMR + activity + thermic effect of food 1 2 3 1 2 3  70 lb x 10 Cal/lb = 700 Cal 700 Cal x 40% = 280 Cal Susie 1050 Cal x 10% = 105 Cal Total energy expended (in 24h) = 700 + 280 + 105 = 1085 Cal Question: Is Susie in energy balance? yes…. remember that these numbers are all rough approximations  people have a range of daily caloric expenditures due to: their weight (BMR) their activity level accounted for in formula how much food they are eating (TEF) age (older people tend to have lower BMR) sex (males tend to have higher BMR than females) amount of body fat (higher % body fat → lower BMR, higher muscle mass, higher BMR) genetics (can be high or low based on inherited genetics) body temperature (higher body temps (as in fever or sauna) increase BMR) disease (can increase or decrease) meanwhile, an obesity epidemic is occurring leading to: coronary disease, strokes, diabetes, colon and other cancers, osteoarthritis, sleep apnea, and even reproductive problems. How do we define Obesity? Body mass index (BMI) = weight (in kg)/height (metres)2 BMI example Monty: 180 lb → 180/2.2 = 81.8 kg 5’10” → 70in x.025 = 1.75 m Monty’s BMI = (81.8 kg)/(1.75 m)2 = (81.8 kg)/(3.06 m2) = 26.7 kg/m2 Weight: lb /2.2 = kg height: inches x.025 = metres  45.0 40.0 39% 36% 35.0 % Canadians 30.0 25.0 23% 20.0 15.0 10.0 5.0 2% 0.0 BMI Range: 30 Data Source: 2004 Canadian Community Health Survey: Nutrition (test subject BMI = 26.7 kg/m2) → overweight! Question: why are Obesity rates increasing?  *USDA’s Center for Health Policy & Promotion Why did food production per capita begin to rise in 1970?  + grains +++ added fats +++ added sugars *USDA’s Center for Health Policy & Promotion Question: where is obesity observed? Any thoughts why there? Question: why this distribution of obesity? Bowman et al, PEDIATRICS Vol. 113 No. 1 January 2004 → could be one reason why obesity is linked to high consumption of added sugars Portion sizes increase energy intake intake of fast food intake of added sugar (sweetened beverages) large portion sizes