Introduction into Biochemistry PDF Lecture Notes - Summer 2022-2023

Introduction into biochemistry PROF. NAFEZ ABU TARBOUSH NURSING SUMMER 2022 -2023 What is biochemistry? It is the chemistry of living organisms. It describes the structure, organization, and functions of living matter in molecular terms. It describes the flow of energy inside a living cell/organism...

Introduction into biochemistry PROF. NAFEZ ABU TARBOUSH NURSING SUMMER 2022 -2023 What is biochemistry? It is the chemistry of living organisms. It describes the structure, organization, and functions of living matter in molecular terms. It describes the flow of energy inside a living cell/organism.  Hydrogen Important terms Electronegativity e 3181at a tea Oxygen Covalent bonds I set a ta k Nitrogen Carbon ◦ Polar vs. non-polar covalent they havealmostthe bonds same electronegativity Non-covalent interactions Ii's ◦ Electrostatic interactions ◦ Hydrogen bonds (donor and no acceptor) 6.5 ◦ Hydrophobic interactions blol iamb x ◦ Hydrophobic versus hydrophilic molecules Carbon Why can carbon form complex molecules?  Properties of carbon (1) It can form four covalent bonds, which can be single, double, or triple bonds (example: fatty acids). Each bond is very stable. ◦ strength of bonds: triple > double > Single) They link C atoms together in chains and rings (example: sugars). ◦ These serve as a backbones Properties of carbon (2) Carbon bonds have angles giving molecules distinct three- dimensional structures. In a carbon backbone, some carbon atoms rotate around a single covalent bond producing molecules of different shapes. Properties of carbon (3) The electronegativity of carbon is between other atoms. ◦ It can form polar and non-polar molecules Pure carbon is not water soluble, but when carbon forms covalent bonds with other elements like O or N, the molecule that makes carbon compounds to be soluble. Nonpolar Properties of water (1) Water is a polar molecule as a whole because of ◦ the different electronegativity between Hydrogen and oxygen ◦ It is angular Water is highly cohesive. Water molecules produce a network. Properties of water (2) Water is an excellent solvent because It is small, and it weakens electrostatic forces and hydrogen bonding between polar molecules. Properties of water (3) It is reactive because it is a nucleophile. ◦ A nucleophile is an electron-rich molecule that is attracted to positively- charged or electron-deficient species (electrophiles) Properties of water (4) A water molecule is ionized to become a positively-charged hydronium ion (or proton), and a hydroxide ion: Acids versus bases - 1 Acid: a substance that produces H+ when dissolved in water (HCl, HNO3, CH3COOH, H2SO4, H3PO4) ◦ H+ Reacts with water producing hydronium ion (H3O+) Base: a substance that produces OH- when dissolved in water (NaOH, KOH) Acids and bases - 2 Acid: any substance (proton donor) able to give a hydrogen ion (H+-a proton) to another molecule ◦ Monoprotic acid: HCl, HNO3, CH3COOH ◦ Diprotic acid: H2SO4 ◦ Triprotic acid: H3PO4 Base: any substance that accepts a proton (H+) from an acid ◦ NH3 Water: acid or base? Both Products: hydronium ion (H3O+) and hydroxide  amphi- = “both” Amphoteric substances Substances that can act as an acid in one reaction and as a base in another are called amphoteric substances ◦ Example: water With ammonia (NH3), water acts as an acid because it donates a proton (hydrogen ion) to ammonia NH3 + H2O ↔ NH4+ + OH– With hydrochloric acid, water acts as a base water is acting as a bas HCl+ H2O → H3O+ + Cl- Acid/base strength Acids differ in their ability to release protons. ◦ Strong acids dissociate 100%. Bases differ in their ability to accept strong protons. acids ◦ Strong bases have strong affinity for save to fryu protons. For multi-protic acids (H2SO4, H3PO4), each proton is donated at a different strength. base  Rule The stronger the acid, the weaker the conjugate base Reactions favor formation of a weak acid. ◦ Strong acids and bases are one-way reactions. HCl → H+ + Cl- NaOH → Na+ + OH- ◦ Weak acids and bases do not ionize completely. HC2H3O2 ↔ H+ + C2H3O2- NH3 + H2O ↔ NH4+ + OH-  Equilibrium constant Acid/base solutions are at constant equilibrium. We can write equilibrium constant (Keq) for such reactions: The value of the Keq indicates direction of reaction ◦ When Keq is greater than 1 the product side is favored The strength of O ◦ When Keq is less than 1 the reactants are favored an acid or base D is measure by Equilibrium constant for acids is Ka and for bases Kb. its Keq. ◦ Note: H3O+ = H+ HE is 41 I HE 344 84 1 g s y i saw 111 in Expression A solutions can be expressed in terms of its concentration or molarity (M). Acids and bases can also be expressed in terms of their equivalence (Eq). Molarity of solutions We know that moles of a solution are the amount in grams in relation to its molecular weight (MW). moles = grams / MW A molar solution is one in which 1 liter of solution contains the number of grams equal to its molecular weight. M = moles / volume Since (mol = grams / MW), you can calculate the grams of a chemical you need to dissolve in a known volume of water to obtain a certain concentration (M) using the following formula: grams = M x vol x MW Exercise How many grams do you need to make 5M NaCl solution in 100 ml (MW 58.4)? grams = 58.4 x 5 M x 0.1 liter = 29.29 g Equivalents My we omit owe When it comes to acids, bases and ions, it is useful to think of them as equivalents. An equivalent is the amount of moles of hydrogen ions that an acid can donate. ◦ or a base can accept. A 1 g-Eq of any ion is defined as the molar mass of the ion divided by the ionic charge. Examples For acids: 1 mole HCl = 1 mole [H+] = 1 equivalent 1 mole H2SO4 = 2 moles [H+] = 2 equivalents ◦ 1 eq of H2SO4 = ½ mol (because 1 mole gives two H+ ions) Remember: One equivalent For ions: of any acid neutralizes one One equivalent of Na+ = 23.1 g equivalent of any base. One equivalent of Cl- = 35.5 g One equivalent of Mg+2 = (24.3)/2 = 12.15 g  Molarity and equivalents Equivalents = n x M x volume (L) One equivalent of any acid neutralizes one equivalent of base. Based on the equation above, since x eq of an acid is neutralized by the same x eq of a base, then (n x M x vol) of an acid is neutralized by (n x M x vol) of a base. Problem 1 Note that each one produces 1 mole of H+ or OH-, so 1M of HCl is equal to 1M of NaOH, so 1M HCl produces 1M of H+, which is neutralized by 1M NaOH, which produces 1M OH- Eq of acid = Eq of base N x M1 x Vol1 = n x M2 x Vol2 1 x 0.12 x 22.4 = 1 x M2 x 12 M2 = (0.12 x 22.4) / 12 M2 = 0.224 M Problem 2 Note that 1 mole of HNO3 produces 1 mole of H+, but 1 mole of Ba(OH)2 produces 2 moles of OH-. In other words, the n is different. Eq of acid = Eq of base N x M1 x Vol1 = n x M2 x Vol2 1 x 0.085 x Vol = 2 x 0.12 x 15 Vol = (2 x 0.12 x 15) / 1 x 0.085 Vol = 42.35 mL Ionization of water Water dissociates into hydronium (H3O+) and hydroxyl (OH-) ions For simplicity, we refer to the hydronium ion as a hydrogen ion (H+) and write the reaction equilibrium as  Equilibrium constant The equilibrium constant Keq of the dissociation of water is The equilibrium constant for water ionization under standard conditions is 1.8 x 10-16 M Kw Since there are 55.6 moles of water in 1 liter, the product of the hydrogen and hydroxide ion concentrations results in a value of 1 x 10-14 for: This constant, Kw, is called the ion product for water [H+] and [OH-] For pure water, there are equal concentrations of [H+] and [OH-], each with a value of 1 x 10-7 M Since Kw is a fixed value, the concentrations of [H+] and [OH-] are inversely changing If the concentration of H+ is high, then the concentration of OH- must be low, and vice versa. For example, if [H+] = 10-2 M, then [OH-] = 10-12 M What is pH? Changes in [H+] have significant effects on many biochemical processes A logarithmic quantity, pH, was developed as a convenient scale for working with levels of [H+] The pH of a solution is a measure of its concentration of H+ (acidity) The pH is defined as Acidic and basic pH On the pH scale, pH of 7.0 is considered neutral pH values below 7.0 indicate acidic solutions pH values above 7.0 indicate basic (alkaline) solutions Strong acids A strong acid dissociates completely, while a strong base completely binds to a proton. HA  H+ + A- In a 0.1 M solution of hydrogen chloride (HCl) in water, [H+] = 0.1 M because all the HCl has dissociated into H+ and Cl- ions For this solution pH = log 0.1 = 1.0 Strong bases Similarly, a 0.01 M solution of NaOH will have an OH- concentration of 1 x 10-2 The pH is calculated as shown below: [H+] = 1 x 10-14 / 1 x 10-2 = 1 x 10-12 The pH will be 12 Otherwise, if [OH-] = 10-2, then [H+] = 10-12 and the pH = 12 Weak acids and bases Many acids, such as the amino acids and acetic acid, do not dissociate completely in water Such acids are called weak acids, and similarly, some bases do not completely dissociate and are called weak bases The ionization equilibrium of a weak acid is given as Acid dissociation constant Ka, the acid dissociation constant, is the equilibrium constant for the reaction of a weak acid (HA) converting into a proton and the conjugate base (A-). What is pKa? As with pH, a logarithmic scale is useful for working with Ka values. pKa is defined as: pKa: the tendency of an acid to dissociate into H+ and its conjugate base A- Each weak acid and base have their own fixed pKa values Note: if Ka is very large, pKa is very small The Henderson-Hasselbalch equation What is the relation between pH, pKa, and the ratio of acid to base? A useful expression can be derived É ‫درجة تفاعل الحمض ملًا يتحلل نصفه‬ a P ka da t PH w yo 4 Ht t.LI to F When pKa = pH At the point of the dissociation where the concentration of the conjugate base [A-] is equal to that of the acid [HA]: pH = pKa + log The log of 1 = 0 pKa = pH pKa is the pH where 50% of acid is dissociated into conjugate base. Maintenance of equilibrium Because a weak acid (or base) can exist in equilibrium with its conjugate base, the equilibrium can be maintained even when small amounts of [H+] or [OH-] are added to it. Example If a small amount of [H+] is added to a solution of weak acid, then the [H+] can combine with it to produce [HA], maintaining [H+] level If a small amount of [OH-] is added, then [OH-] will bind to [H+] producing H2O, [HA] will dissociate to [H+] and [A-] to compensate for the loss of [H+], maintaining [H+] level What is a buffer? I 4 294 typy concentration The shift in equilibrium will then maintain the pH of the solution. A solution with the ability to resist changes in pH is called a buffer. Titration w wa o f wi 0688 62 3 both weak The pKa values of weak acids are determined by titration. This involves adding small amounts of a strong acid or base to the solution and measuring the resulting changes in pH generating a plot of titration curve. Midpoint For example, titration of acetic acid gives the curve shown to the right. Note that the pKa value is the midpoint of the curve. At this point, pH is equal to the pKa since there are equal concentrations of HA and A-.  a w wa a Buffering capacity in The ability of a buffer to minimize changes in pH is known as its buffering capacity. The buffering capacity of weak acids and bases is one pH unit from their pKa values. In other words, a buffer is effective at resisting changes in pH at pH +/- 1 of the pKa. For example, since the pKa of acetic acid is 4.7, then the buffering capacity ranges 3.7-5.7.  What are the ratios of [A-]/(HA] of the acetate buffer at point X and point Y? At point X, it is the end of the buffering capacity where the pH = pKa – 1. pKa = 4.8. so, pH = 3.8. Use the Henderson-Hasselbalch equation to solve for [A-]/[HA]. Y pH = pKa + log [A-]/[HA] F 3.8 = 4.8 = [A-]/[HA] X log [A-]/[HA] = 3.8 – 4.8 X 8 log [A-]/[HA] = -1 [A-]/[HA] = 1/10 For point Y, it is the other end of the buffering capacity where the pH = 5.8. If you solve it, [A-]/[HA] = 10/1. Conjugate bases Acid Conjugate base CH3COOH CH3COONa (NaCH3COO) H3PO4 NaH2PO4 H2PO4- (or NaH2PO4) Na2HPO4 H2CO3 NaHCO3 Properties of buffers They all have the same function. Their pKa’s are different. Buffering capacities are same. We choose a buffer based on its pKa, which should be close to the desired pH. Problems and solutions the Henderson-Hasselbalch (HH) equation is useful for determining the pH of a solution if the molar proportion of A- to HA and the pKa of HA are known. Example: Consider a solution of 0.1 M acetic acid and 0.2 M acetate ion. The pKa of acetic acid is 4.8. Hence, the pH of the solution is given by Similarly, the pKa of an acid can be calculated if the molar proportion of A- to HA and the pH of the solution are known.  Exercise What is the pH of a buffer containing 0.1M HF and 0.1M NaF? (Ka = 3.5 x 10-4)? HF is a weak acid meaning that it does not dissociate into the ions H+ and F-. The concentration of the acid (HF) = 0.1. The concentration of the conjugate base (NaF) = 0.1. There are two ways to solve it. The first is to calculate [H+}, then calculate the pH The other way is to convert Ka to pKa, then use the [H3O+] = Ka x [HF]/[F-] Henderson-Hasselbalch equation to solve for pH. = 3.5 x 10-4 x (0.1/ 0.1) pKa = -log Ka = 3.5 x 10-4 pKa = -log 3.5 x 10-4 pKa = 3.46 pH = -log 3.5 x 10-4 pH = pKa + log (F-]/[HF] = 3.46 pH = 3.46 + (0.1/0.1) pH = 3.46 titty Polyprotic weak acids It He is monoca dix this Some weak acids (buffers) can donate more than one proton. An example is phosphoric acid (H3PO4). Phosphoric acid can donate up to three protons. Different pKa’s These protons do not dissociate at the same time. Each proton has a certain pKa. and… Each proton dissociates at a certain pH.  What is the pH of a lactate buffer that contains 75% lactic acid and 25% lactate? (pKa = 3.86) Problem 1 Remember, lactic acid is HA that has the proton to donate and lactate is the conjugate base that does not have a proton and carries a negative charge. Use the Henderson-Hasselbalch equation to solve for the pH: pH = pKa + log (lactate]/[lactic acid] pH = 3.86 + log 25/75 pH = 3.86 + (-0.48) pH = 3.38 You can guess the answer as well. Since you have more acid than the conjugate base and it Is not by more than 10 times (the ratio between conjugate base to acid is not less than 10 times), the pH should be lower than the pKa. Problem 2 What is the pKa of a dihydrogen phosphate buffer when pH of 7.2 is obtained when 100 ml of 0.1 M NaH2PO4 is mixed with 100 ml of 0.1 M Na2HPO4? NaH2PO4 is acid and Na2HPO4 is its conjugate base because NaH2PO4 has an extra proton. Use the Henderson-Hasselbalch equation to solve for the pH: pH = pKa + log (Na2HPO4]/[ NaH2PO4] 7.2 = pKa + (0.1/0.1) 7.2 = pKa Buffers in human body Carbonic acid-bicarbonate system (blood) Dihydrogen phosphate-monohydrogen phosphate system Proteins (via the amino acids)  5 more or less normed change Blood buffering 7.357745ft Blood pH must be maintained at around 7.4. Any dramatic change (up or down) can be dangerous. By far the most important buffer for maintaining acid-base balance in the blood is the carbonic-acid-bicarbonate buffer. How? During metabolism, cells produce CO2. This CO2 is dissolved in blood and is converted to carbonic acid, which dissociates into bicarbonate ion (conjugate base) and H+. If alungs a icalosis t d causes Dissolved CO2 is in equilibrium with pressure of CO2 (pCO2) in lungs. acidosis If blood pH changes, then the body changes the rate of breathing in order to balance blood pH. The equation Relationships of the bicarbonate buffer system to the lungs and the kidneys. Roles of lungs and kidneys Maintaining blood is balanced by the kidneys and the lungs Kidneys control blood HCO3- concentration ([HCO3-]) Lungs control the blood CO2 concentration (PCO2)