[CHEM 160] Exam 1 reviewer.pdf

HMYD CHEMISTRY 160 REVIEWER Prokaryotic Cell & Its Parts 1st Exam Coverage...

HMYD CHEMISTRY 160 REVIEWER Prokaryotic Cell & Its Parts 1st Exam Coverage Prokaryotic Cell Lesson 1: The Cell Earliest and most primitive forms of life in structure Simple in structure ○ Are unicellular organisms Cell ○ No recognizable organelles The basic building block of life ○ Lacks a nucleus Smallest living unit of an organism Genetic material is concentrated on the nucleoid region of Capable of growth, reproduction, energy adaptation, and response to the the Cytoplasm environment Can be unicellular or multicellular: ○ Unicellular - an entire organism ○ Multicellular - one of the billions of cells that comprise an organism Cell and Biochemistry Cellular components are made up of biomolecules Biosynthesis and Degradation (Turnover) of biomolecules happens inside the cell Biochemical reactions take place inside the various cellular components Two Types of a Cell: Prokaryotic and Eukaryotic Cytoplasm The main difference between the eukaryotic and prokaryotic cells is the The area enclosed by the plasma membrane presence of a true nucleus. Eukaryotes have a nucleus, while prokaryotes only Contains cytosol, a thick intracellular fluid, which is composed of water, have a nucleoid region. The nucleus is where the genetic material can be seen, salts, and proteins while in prokaryotes, the genetic material is concentrated in the nucleoid region. Plasma Membrane Lipid bilayer with embedded proteins Encloses the cytoplasm and the nucleoid region Function: Controls the passage of molecules in and out of the cell Prepared by: Hans Dominguez To: Eizel Valenzuela Cell Wall Eukaryotic Cell Composed of peptidoglycan Located underneath the capsule and surrounds the plasma membrane Function: Eukaryotic Cell ○ Maintains the cell shape Have more complex structures and intracellular processes ○ Prevents mechanical injury Structural Features: ○ Prevents the cell from bursting when it takes up water ○ Compartmentalization - allows two incompatible reactions to occur in a single cell at the same time Capsule ○ 1000 times greater in volume than prokaryotes Not all bacteria have a capsule, it is present in some only ○ Has membrane bound organelles Is a gelatinous polysaccharide Pathogenic bacteria have thick capsules, making it hard to treat, e.g., Organelles Mycobacterium tuberculosis Specialized subunits within a eukaryotic cell Function: Has their own function/s ○ Protects the cell from defense of higher organism Suspended in cytosol ○ Help prokaryotes to cling to each other Usually separately enclosed within its own membrane ○ Prevents the cell from drying out Main Functions: ○ Synthesis and Degradation of Biomolecules Pili ○ Selective Transport of Biochemical Substances Hair-like structures on the surface ○ Metabolism Function: ○ Storage, Transmission, and Expression of Genetic Information ○ Conduit for DNA during sexual conjugation ○ Used for the attachment of the cell Two Types of a Eukaryotic Cell: Animal and Plant Cell Flagellum The main difference between the two is that there are unique organelles Present in some bacteria for each of them. Whip-like appendages Function: Locomotion Prepared by: Hans Dominguez To: Eizel Valenzuela  Central Vacuole Animal Cell: Unique Organelles Function: Reservoir of food, nutrients, ions, waste products, and specialized materials such as pigments Lysosomes Surrounded by the tonoplast Globules present in the cytoplasm Some animal cells also have a vacuole, however, the central vacuole is Function: Has a membrane containing hydrolytic digestive enzymes for bigger than the normal vacuole intracellular digestion Tonoplast Centrioles A semipermeable membrane Function: Formation of spindle apparatus during cell division Cell Wall Flagellum Composed of cellulose, unlike the prokaryotic cell, which is composed of Function: Locomotion peptidoglycan Function: Cellulosic inert supporting structures provide the plant cell Plant Cell: Unique Organelles rigidity Chloroplast Plasmodesmata Function: Site of Photosynthesis Function: A channel allowing neighboring cells to exchange small Contains Chlorophyll, the pigment responsible for harvesting light molecules Organelles Both Found in Animal & Plant Cells Nucleus Contains the genetic material of the cell Function: Site of DNA and RNA synthesis Ribosome Embedded in the cytoplasm or in the rough endoplasmic reticulum Function: Site of Protein Synthesis Has two important parts: ○ Large ribosomal subunit ○ Small ribosomal subunit Prepared by: Hans Dominguez To: Eizel Valenzuela Endoplasmic Reticulum Lesson 2: Overview of Biomolecules Network of membrane channels connected to the nuclear pore Function: Processes and Transports proteins and lipids from one site into Biologically Important Elements another site in the eukaryotic cell Has two types, the Rough ER and the Smooth ER: ○ Rough ER: Contains ribosomes ○ Smooth ER: Site of lipid synthesis Golgi Complex/Apparatus Function: Sorts and Transports molecules for export into the plasma membrane or other cells Peroxisomes Function: Has oxidative enzymes for intracellular digestion, unlike the lysosome which is a hydrolytic digestive enzyme which is also aimed for Bulk Elements intracellular digestion Serve as structural components of cells and tissues ○ Degrade hydrogen peroxide, a toxic compound produced during Required in the diet in grams daily metabolism H2O2 –peroxisome→ H2O + ½O2 Trace Elements Required in the diet in mg or fewer levels daily Mitochondrion The powerhouse of the cell C, H, N, O, P, and S Function: Site of cellular respiration Most important elements present in biomolecules Similar to the chloroplast, it has two membranes; however, instead of stroma, it has a mitochondrial matrix Prepared by: Hans Dominguez To: Eizel Valenzuela  Carbon is the most important atom for different biomolecules since a Lipids carbon atom is a tetravalent atom and can form a maximum of 4 bonds. It can Heterogenous form single, double, and triple bonds with other carbon atoms, and single and Exhibit low solubility/insoluble in water and soluble in organic solvents multiple bonds with other atoms in the figure above. Has very large hydrophobic portions; non-polar Function/s: See PPT for commonly found classes of organic compounds, linkages, and ○ Fuel for metabolic processes, e.g., fats and oils chemical reactions (Important; Lumabas sa Final Exam). ○ Building blocks and vitamins, e.g., phospholipids ○ Insulation, e.g., fats and oils ○ Protective Covering, e.g., suberin Types of Biomolecules Nucleic Acids Protein Chemical carriers of genetic information Most abundant molecule in the cell at around 15% Biopolymers of nucleotides linked by phosphodiester bonds Most varied and diverse in terms of structure and function Has two forms: DNA and RNA Function/s: ○ 2’-deoxyribonucleic acid ○ Establishment and maintenance of structure, e.g., collagen ○ Ribonucleic acid ○ Transport, e.g., hemoglobin Function/s: ○ Control and Regulation, e.g., hormones ○ Energy currency in metabolic processes, e.g., ATP and ADP ○ Defense and Immunity, e.g., immunoglobulin ○ Structural components of cofactors and coenzymes, e.g., NADH and ○ Catalysis, e.g., enzymes FADH2 ○ Movement, e.g., myosin (muscle tissue) ○ Building blocks for molecular repositories of genetic information, ○ Storage, e.g., ferritin (storage of iron) e.g., DNA and RNA Carbohydrates Most abundant molecule in Earth Either a polyhydroxy aldehyde or a polyhydroxy ketone Function/s: ○ Energy source and storage, e.g., starch ○ Structural integrity, e.g., cellulose ○ Conjugates to other macromolecules, e.g., glycoproteins ○ Components of nucleic acids, e.g., DNA and RNA Prepared by: Hans Dominguez To: Eizel Valenzuela  Together, the Cohesion and Adhesion is important for the transport of Lesson 3: Water water from the roots to other parts of the plant. The direction of water movement is against the pull of gravity. Adhesion of the water to cell walls Major component of the Earth’s surface (made of cellulose, has lots of -OH groups) via H bonding helps resist gravity Most abundant compound in living organism around 70% or more by while Cohesion between the water molecules via H bonding helps hold together weight of the organism the columns of water within the cells. Structure of Water High Surface Tension Molecular geometry: Bent or V-shaped Surface Tension - a measure of the resistance of a substance to increase ○ The bent structure is important in hydrogen bonding its surface area Has two lone pairs of electrons and two hydrogen atoms; four electron ○ Due to high cohesion of water molecules groups in tetrahedral arrangement The electronegative O and the 2 H atoms forms a net dipole moment, making the water a polar molecule Water can serve as both as a H-donor and acceptor Has the potential to form 4 H bonds: ○ One each for the hydrogen atom bonding with the oxygen atom of other water molecules ○ Lone pairs of the oxygen atom bonding to the hydrogen atoms of other water molecules Unique Properties of Water In the molecular perspective, water has two types of molecules: inner and surface. The inner molecule is attracted to the molecules around it via H bonding. Meanwhile, the surface molecule is attracted to other surface The unique properties of water can be attributed to its ability to form molecules via H bonding as well. It can only attract those molecules in its left, hydrogen bonds right, downwards, and not above it, resulting in fewer molecules interacting with each other, making H bonding stronger there, forming a layer of strongly High Cohesion bonded water molecules. Cohesion - act of sticking together; H2O to H2O Moreover, the inner molecules have no net force since the forces exerted by neighboring molecules cancel out. On the other hand, the surface molecules High Adhesion have a net inward force due to the absence of an attractive force above, causing Adhesion - act of sticking to something; H2O to Fxn group them to contract and resist being stretched. ○ Due to the ability of water molecules to form H-bonds Prepared by: Hans Dominguez To: Eizel Valenzuela  This allows things or animals denser than water to float when usually they This is important in the sweating process and the evaporation of water at would sink, e.g., water striders. This concept is also involved in the formation of the surface of the leaves. water droplets. Liquid at Room Temperature High Specific Heat Specific Heat - the amount of heat per unit mass required to raise the temperature by 1OC ○ 1 cal/gOC Temperature can be moderated by water since it can absorb heat from warmer air and release that heat to cooler air. This is possible because of water’s high specific heat. In liquid water, water molecules can form an average of 3.4 H bonds with An increase in temperature means that the molecules are moving faster, each other, while in Ice, it can form a maximum of four H-bonds, creating a due to an increase in kinetic energy (KE). This causes the breaking of H bonds regular crystal lattice. Ice crystals are spacious as they have fewer molecules which absorbs heat. compared to an equal volume of liquid water. Solid ice has a higher volume Meanwhile, a decrease in temperature means the molecules are moving than liquid water. 𝑚 slower due to a decrease in KE. This causes the formation of H bonds which 𝑝= 𝑉 release heat. Since ice has a higher volume compared to liquid water, assuming equal This concept is important because it can regulate the rate of change in mass, it is expected that ice has a lower density than liquid water. This allows ice air’s temperature, especially near oceans; the change is gradual instead of a to float on top of water, forming a layer of insulation on top of lakes and seas, sudden change. Also, it is important in the regulation of organisms’ internal which helps life below water. If ice were to be denser than liquid water, the body temperature. plankton under the sea could be destroyed due to freezing in the seafloor, destroying the ecosystem and leading to the extinction of marine organisms. High Heat of Vaporization Heat of Vaporization - the amount of heat needed to turn 1 g of liquid Excellent Solvent water into a vapor without a rise in temperature of the liquid As a polar solvent, water readily dissolves polar compounds and other ○ 540 cal/g at 25OC salts via the formation of electrostatic interaction due to the high dielectric constant of water Evaporative cooling is the process at which as liquid evaporates, the liquid Dielectric constant - the measure of the ability of a solvent to keep surface that remains behind cools down. This is due to the water’s high heat of opposite charges apart vaporization since the presence of a hydrogen bonding network between the water molecules, requires a high input energy to transform 1 gram of liquid NaCl –H2O→ Na+ + Cl- water into water vapor. Prepared by: Hans Dominguez To: Eizel Valenzuela  These ions would undergo hydration, which is when they are surrounded Water as a Medium for Biochemical Processes by water molecules such that the charges would be attenuated (coulombic forces). Weak interactions in the aqueous system ○ These happen inside our living cells which can influence the structure, function, and specificity of biomolecules ○ Hydrogen bonding, Ionic Interactions (presence of salts), Hydrophobic Interactions, and Van der Waals forces are present These are individually weak but collectively has a significant influence on the 3D structure and chemical properties of biomolecules Water and its ionization products (H3O+ and OH-) ○ Since it is an amphiprotic molecule, it can undergo self-ionization and form H3O+ and OH-, these can influence the structure, Hydrophobic Exclusion/Effect self-assembly, and properties of all cellular components Can be seen in amphiphiles (e.g., soap). Water tends to hydrate its hydrophilic portion and exclude its hydrophobic portion. Weak acid-base equilibria/physiological buffers It's usually due to: (1) the presence of amphiphiles disrupt the H-bonding ○ These can exist in our living cells which are important to maintain between water molecules. Here, the water molecules surround each amphiphilic the optimum pH for biochemical reactions molecule; however, this is not favored because of the resulting highly-ordered structure that decreases the entropy of the system. The state should be See PPT for the different H-bonds found in biological systems. disoriented for it to be good. (2) To increase the entropy, amphiphiles will cluster together to avoid the formation of a highly ordered structure. Here, fewer water molecules are needed to surround the amphiphiles. (3) The perfect clustering is called Micelles, wherein all hydrophobic groups are sequestered from water, and the ordered shell of H2O molecules is minimized, and entropy is further increased. Polar heads cannot interact with the water molecules since they are covalently bonded to the hydrophobic tail. Thus, the hydrophobic effect is the tendency of the amphiphiles to self-associate in water rather than to dissolve individually. Prepared by: Hans Dominguez To: Eizel Valenzuela  −7 Lesson 4: Buffers 𝑝𝐻 =− 𝑙𝑜𝑔(1. 0 × 10 𝑀) = 7 To interpret pH: Autoionization of Water pH = 7.0, neutral pH < 7.0, acidic Since water is an amphiprotic molecule. One water molecule can serve as pH > 7.0, basic a proton donor, while the other can serve as a proton acceptor. pKa of Weak Acids: Determining the Predominant Species Consider Acetic Acid, CH3COOH (Theoretical Value): − + 𝐶𝐻3𝐶𝑂𝑂𝐻(𝑎𝑞) + 𝐻2𝑂(𝑙) ⇋ 𝐶𝐻3𝐶𝑂𝑂(𝑎𝑞) + 𝐻3𝑂(𝑎𝑞) y56$\ Equilibrium Constant Expression: − + [𝐶𝐻3𝐶𝑂𝑂(𝑎𝑞)][𝐻3𝑂(𝑎𝑞)] −5 + − [𝐻3𝑂 ]𝑎𝑞[𝑂𝐻 ]𝑎𝑞 𝐾𝑎 = [𝐶𝐻3𝐶𝑂𝑂𝐻(𝑎𝑞)] = 1. 8 × 10 𝐾𝑒𝑞 = 2 [𝐻2𝑂]𝑙 2 + − 𝐾𝑒𝑞[𝐻2𝑂]𝑙 = [𝐻3𝑂 ]𝑎𝑞[𝑂𝐻 ]𝑎𝑞 𝑝𝐾𝑎 =− 𝑙𝑜𝑔𝐾𝑎 + − −14 𝑜 −5 𝐾𝑤 = [𝐻3𝑂 ]𝑎𝑞[𝑂𝐻 ]𝑎𝑞 = 1. 0 × 10 (25 𝐶) 𝑝𝐾𝑎 =− 𝑙𝑜𝑔(1. 8 × 10 ) = 4. 74 + − Since [𝐻3𝑂 ]𝑎𝑞 = [𝑂𝐻 ]𝑎𝑞, let either be x: 2 Pattern pHsoln < pKa pHsoln = pKa pHsoln > pKa 𝐾𝑤 = 𝑥 −14 2 Predominant Both [HA] and Conjugate Base 1. 0 × 10 = 𝑥 Weak Acid [HA] Species [A-] [A-] −7 𝑥 = 1. 0 × 10 𝑀 + − −7 [𝐻3𝑂 ]𝑎𝑞 = [𝑂𝐻 ]𝑎𝑞 = 1. 0 × 10 𝑀 Example using Acetic Acid, CH3COOH pHsoln < 4.74 pHsoln = 4.74 pHsoln > 4.74 pH Scale Predominant CH3COOH and CH3COOH CH3COO- Species CH3COO- + 𝑝𝐻 =− 𝑙𝑜𝑔[𝐻3𝑂 ]𝑎𝑞 For pure water at 25OC: Prepared by: Hans Dominguez To: Eizel Valenzuela Sample Problem 1, Succinic Acid ○ Experimentally, pKa of acetic acid is 4.76, which is slightly different to its theoretical value of 4.74 5.6 3.0 7.0 4.21 5.0 pH = pKa2 pH < pKa1 pH > pKa2 pH = pKa1 pKa1 < pH > pKa2 B and C A C A and B B Pattern pHsoln < pHIP pHsoln = pHIP pHsoln > pHIP Sample Problem 2, Phosphoric Acid Predominant Both [HA] and Conjugate Base Weak Acid [HA] Species [A-] [A-] Buffers 7.21 5.0 10 2.20 Solutions that resist drastic change in pH upon addition of small amounts of acid or base pH = pKa2 pKa1 < pH > pKa2 pKa2 < pH > pKa3 pH = pKa1 ○ Has buffering action, which is the ability to resist drastic pH changes upon addition of small amounts of an acid or base [H2PO4-] = [HPO42-] H2PO4- HPO42- [H3PO4] = [H2PO4-] Can be made up of either of the following: ○ Weak Acid + Salt of its Conjugate Base Titration of Weak Acids: Determining the Predominant Species ○ Weak Base + Salt of its Conjugate Acid Used in the study of biomolecules to mimic the natural conditions inside pKa for certain weak acids such as acetic acid can be determined the cell experimentally through titration ○ Physiological pH range = 6.4-7.6 pKa corresponds to the inflection point of the resulting titration curve The two chemical species involved in the buffer system must be in appreciable concentrations Prepared by: Hans Dominguez To: Eizel Valenzuela Dissociation Equation b. Give the dissociation equation involved Buffer made up of Weak Acid and its Conjugate Base: c. Give the buffering action when the buffer system is added with a small − + 𝐻𝐴(𝑎𝑞) + 𝐻2𝑂(𝑙) ⇋ 𝐴(𝑎𝑞) + 𝐻3𝑂(𝑎𝑞) amount of acid and a small amount of base Buffer made up of Weak Base and its Conjugate Acid: + − 𝐵(𝑎𝑞) + 𝐻2𝑂(𝑙) ⇋ 𝐵𝐻(𝑎𝑞) + 𝑂𝐻(𝑎𝑞) Buffering Action Upon addition of a small amount of strong acid: ○ Increases [H3O+] in the solution Counteracted by a neutralization reaction with the basic component of the buffer system + − 𝐻3𝑂(𝑎𝑞) + 𝐴(𝑎𝑞) → 𝐻2𝑂(𝑙) + 𝐻𝐴(𝑎𝑞) or + + + 𝐻3𝑂(𝑎𝑞) + 𝐵(𝑎𝑞) → 𝐻2𝑂(𝑙) + 𝐵𝐻(𝑎𝑞) Upon addition of a small amount of strong base: ○ Increases [OH-] in the solution Counteracted by a neutralization reaction with the acidic component of the buffer system − − 𝑂𝐻(𝑎𝑞) + 𝐻𝐴(𝑎𝑞) → 𝐻2𝑂(𝑙) + 𝐴(𝑎𝑞) or − + 𝑂𝐻(𝑎𝑞) + 𝐵𝐻(𝑎𝑞) → 𝐻2𝑂(𝑙) + 𝐵(𝑎𝑞) Sample Problem For the following buffer systems: 1. CH3COOH - NaCH3COO buffer system 2. NH3 - NH4Cl buffer system a. Identify the buffer components Prepared by: Hans Dominguez To: Eizel Valenzuela Try this Problem acid and conjugate base, wherein, at this stage, the weak acid is exactly half-neutralized. For the following buffer systems: 𝑝𝐻 = 𝑝𝐾𝑎 + 𝑙𝑜𝑔 − [𝐴 ] [𝐻𝐴] NaHCO3 - Na2CO3 buffer system NaH2PO4 - Na2HPO4 buffer system 𝑝𝐻 = 𝑝𝐾𝑎 + 𝑙𝑜𝑔(1) 𝑝𝐻 = 𝑝𝐾𝑎 a. Identify the buffer components b. Give the dissociation equation involved Buffering Region c. Give the buffering action when the buffer system is added with a small amount of acid and a small amount of base pH range in which the buffers is effectively resisting drastic pH changes Buffering Region = pKa 土 1 Henderson-Hasselbach Equation − [𝐴 ] 𝑝𝐻 = 𝑝𝐾𝑎 + 𝑙𝑜𝑔 [𝐻𝐴] Derived from the acid dissociation constant expression for the hypothetical acid HA: − + 𝐻𝐴(𝑎𝑞) + 𝐻2𝑂(𝑙) ⇋ 𝐴(𝑎𝑞) + 𝐻3𝑂(𝑎𝑞) − + [𝐴 ][𝐻3𝑂 ] 𝐾𝑎 = [𝐻𝐴] − [𝐴 ][𝐻3𝑂 ] + Note: Always choose a buffer range that covers the given pH value. At this 𝑙𝑜𝑔 𝐾𝑎 = 𝑙𝑜𝑔 ( [𝐻𝐴] ) range, the buffer’s capacity to resist change is at maximum. − + [𝐴 ] 𝑙𝑜𝑔 𝐾𝑎 = 𝑙𝑜𝑔[𝐻3𝑂 ] + 𝑙𝑜𝑔( [𝐻𝐴] ) − Sample Problem + [𝐴 ] − 𝑙𝑜𝑔[𝐻3𝑂 ] =− 𝑙𝑜𝑔 𝐾𝑎 + 𝑙𝑜𝑔( [𝐻𝐴] ) − [𝐴 ] Which of the following is/are can be used to prepare a buffer solution with 𝑝𝐻 = 𝑝𝐾𝑎 + 𝑙𝑜𝑔 [𝐻𝐴] a pH of 6.5? a. Acetate buffer (pKa = 4.74) With this equation, you may prove that the maximum buffering capacity b. Phosphate buffer (pKa1 = 2.2; pKa2 = 7.21; pKa3 = 1.27) happens when pH = pKa. It will also happen if you have equal amounts of weak Prepared by: Hans Dominguez To: Eizel Valenzuela  Answer: The target pH is between 5.5 to 7.5, so the pKa2 of the phosphate buffer. The components of the buffer system are H2PO4- and HPO42-. Sample Problems on the Determination of pH Steps: 1. Identify the components of the buffer system Sample Problem 2 ○ If it has multiple pKas, it is best to write the complete dissociation What is the pH of a buffer prepared by mixing 100 mL 0.050 mM NaH2PO4 equation of such polyprotic acids and 25 mL 0.075 mM Na2HPO4? (pKa1 = 2.2; pKa2 = 7.21; pKa3 = 12.7) 2. Determine the final concentration of each component of the buffer system ○ A dilution formula may be used when mixtures are mixed, both final concentrations of the buffer components must be determined. i. CiVi = CfVf 3. Use the Henderson-Hasselbalch Equation to compute the pH ○ Sometimes, the pKa is not given, and other variables are given, such as Kb. Remember that Kw = KaKb, as such, to get Ka, manipulate the formula into Ka = Kw/Kb. Note also that pKa = -logKa. Sample Problem 1 Calculate the pH of a buffer solution that is 0.5 M CH3COOH and 0.8 M NaCH3COO. The pKa for acetic acid is 4.74. Prepared by: Hans Dominguez To: Eizel Valenzuela Sample Problem 3 From Solid or Liquid Reagents Calculate the pH of a buffer solution that is 0.25 M NH3 and 0.10 M NH4Cl. Mass of Solid Reagent The Kb for ammonia is 1.8 x 10-5. 𝑀𝑎𝑠𝑠 = 𝐶 × 𝑉 × 𝑓 × 𝑀𝑀 ○ C = Concentration of the Buffer Solution (in M) ○ V = Volume of the Buffer Solution (in L) ○ f = Fraction of the Component in the Buffer System ○ MM = Molar Mass (in g/mol) Volume of Liquid Reagent 𝑀𝑎𝑠𝑠 𝐶 ×𝑉×𝑓×𝑀𝑀 𝑉𝑜𝑙𝑢𝑚𝑒 = ρ = ρ ○ p = Density (g/mL) Prepare 500 mL of 100 mM acetate buffer, pH 5.0 using NaCH3COO (MM = 82 g/mol) and CH3COOH (MM = 60 g/mol, p = 1.46 g/mL). The pKa for acetic acid is 4.74. Sample Problems on Buffer Preparation Steps: 1. Identify the components of the buffer system 2. Calculate the ratio of the conjugate base over the ratio of weak acid ○ The working equation is derived from the Henderson-Hasselbalch Equation: − [𝐴 ] (𝑝𝐻−𝑝𝐾𝑎) [𝐻𝐴] = 10 3. Use the appropriate equation to compute for the relative amounts of each component Prepared by: Hans Dominguez To: Eizel Valenzuela In Words In Words To prepare the buffer solution, transfer 2.65 g NaCH3COO and 0.73 mL Measure 1.39 L of 60 mM NaH2PO4 stock solution and 0.22 L of 75 mM CH3COOH in a 500 mL volumetric flask. Dissolve the solids with distilled water Na2HPO4 stock solution and transfer it to a 2L volumetric flask. Dilute the and dilute the resulting solution up to the 500 mL mark with distilled water. resulting solution to the 2L mark with distilled water. Shake the flask properly. Shake the flask properly. From Stock Buffer Solution of the Same pH From Stock Solutions of Buffer Components Use the simple dilution formula Volume of Component 𝐶𝑖𝑉𝑖 = 𝐶𝑓𝑉𝑓 𝐶𝑏𝑢𝑓𝑓𝑒𝑟𝑉𝑏𝑢𝑓𝑓𝑒𝑟 𝑉𝑜𝑙𝑢𝑚𝑒 = ( 𝐶𝑠𝑡𝑜𝑐𝑘 )×𝑓 Prepare 2L of 50 mM phosphate buffer pH 6.5 using 1 M phosphate buffer Prepare 2L of 50 mM phosphate buffer pH 6.5 using 60 mM NaH2PO4 and pH 6.5. 75 mM Na2HPO4 solutions (pKa1 = 2.2, pKa2 = 7.21, pKa3 = 12.7) In Words Measure 0.10 L of 1 M phosphate buffer pH 6.5 and transfer it to a 2 L volumetric flask. Dilute the resulting solution up to the 2 L mark with distilled water. Shake the flask properly. Prepared by: Hans Dominguez To: Eizel Valenzuela From a Solid or Liquid Reagent and a Strong Acid or Base Physiological Buffer Systems Prepare 2 L of 50 mM phosphate buffer pH 6.5 using Na2HPO4 ⬤ 6 H2O (MM = 250 g/mol) and concentrated HCl solution (37% w/w, sp. gr. 1.18, MM = 36.458 g/mol). (pKa1 = 2.2, pKa2 = 7.21, pKa3 = 12.7) Phosphate Buffer System Buffers the extracellular fluids and cytoplasmic components (pH ≈ 6.9 to 7.4) The dissociation equation involved: H2PO4-(aq) + H2O(l) ⇋ HPO42-(aq) + H3O+(aq) pKa = 6.86 Buffering Region = 5.9 to 7.9 Bicarbonate Buffer System Partially buffers human blood plasma (pH ≈ 7.4) The dissociation equation involved: H2CO3(aq) + H2O(l) ⇋ HCO3-(aq) + H3O+(aq) When blood pH falls to 6.8 or below, irreparable cell damage and death may occur ○ Found in severely uncontrolled diabetes where metabolic acids are largely produced When one undergoes strenuous activities, lactic acid concentration in our blood increases, due to this, the concentration of H+ increases in the blood, decreasing the pH of our blood. To avoid the sudden decrease in pH, the bicarbonate buffer system’s bicarbonate ion will react to excess H+ in blood to produce carbonic acid which promotes the dissociation reaction to form water and carbon dioxide, wherein it can enter the lungs and be exhaled Prepared by: Hans Dominguez To: Eizel Valenzuela  Lesson 5: Amino Acids Amino Acids Building block of proteins General Structure At neutral pH, predominantly in dipolar or zwitterionic form ○ Zwitterion - molecules that bear charges of opposite polarity; net charge is zero Zwitterionic Form ○ Explains the physical properties of amino acids similar with ionic compounds Crystalline solid Why is it called an ɑ-amino acid? Very high melting point In naming organic compounds, the most oxidized carbon atom is given High solubility in water the highest priority Have an asymmetric carbon except for glycine ○ In amino acids, the carboxyl group is the most oxidized form of the ○ Four different groups of atoms attached to the ⍺-carbon carbon atom ○ Chiral molecule ○ Adjacent carbon atom to the carboxyl group is referred to as the ○ Optically active alpha carbon Stereochemistry of Amino Acids ○ Amino group is attached to the alpha-carbon ○ Exists as a pair of enantiomers D- and L- are relative configurations around the chiral carbon D-form – amino group is on the right L-form – amino group is on the left ○ Amino acid residues in naturally occurring protein molecules are exclusively L-isomers ○ D-amino acids are found only in a few peptides E.g., bacterial cell wall and some antibiotics Prepared by: Hans Dominguez To: Eizel Valenzuela  Note that in drawing amino acids that they are always in the L-form Classification of Amino Acids: Based on Structure unless specified and are in their zwitterionic form. Aliphatic/Alkyl Amino Acids Classification of Amino Acids: Based on Nutritional Requirement Stabilize proteins through hydrophobic interaction Proline limits structural flexibility because of its rigid ring Essential Amino Acid Gly-ciers in Ala-ska Val-iantly Leu-cated in iso-iles’ Pro-ducts ○ cannot be synthesized by the human body ○ Glycine, Gly, G – H root; non-chiral carbon ○ tryptophan, isoleucine, phenylalanine, arginine, valine, leucine, ○ Alanine, Ala, A – CH3 root methionine, threonine, lysine, histidine ○ Valine, Val, V – CH root splits into two CH3s Arginine is only essential for infants ○ Leucine, Leu, L – Additional CH2 on the beginning of the root To remember, use this phrase, Here lies the most famous ○ Isoleucine, Ile, I – “Conformational Isomer” of Leu; H-C-CH3 at the versatile woman I’ve known, right? beginning ○ Proline, Pro, P – Non-aromatic Ring; 5:NH3 → NH2, 3 CH2s Non-essential Amino Acid ○ Can be synthesized by the body Note that the following forms of amino acids are in their zwitterionic form. Prepared by: Hans Dominguez To: Eizel Valenzuela  Aromatic Amino Acids Alcohols Amino Acids Stabilize protein through hydrophobic interaction Stabilize protein and interact with the aqueous environment through tyr (—OH) can participate in H-bonding hydrogen bonding trp and tyr account for absorbance of light exhibited by most Alcohol is a Ser-ious Thre-at proteins at 280 nm ○ Serine, Ser, S – CH2 + OH root ○ Absorbance at 280 nm is used for spectrophotometric ○ Threonine, Thr, T – Ser group (=) + methyl determination of protein concentration The aroma of Fine Pine and Yellow Tyr-es are Worth the Try-p ○ Phenylalanine, Phe, F – benzyl root ○ Tyrosine, Tyr, Y – benzyl root + hydroxyl ○ Tryptophan, Trp, W – 5/6: Tabinging House NH2 ang roof, phenyl Prepared by: Hans Dominguez To: Eizel Valenzuela  Basic Amino Acids Sulfur-Containing Amino Acids Have —NH2 side chains which are protonated at pH 7; positively charged Methionine is very hydrophobic Basically, His Lys-ed Kid Arg-ued Relentlessly Cysteine is responsible for disulfide bridges in proteins ○ Histidine, His, H - CH2 + tabinging bahay where the two ends of the Met-hodically Cys-se the path for sulfur roof are NH and and N, doon sa walang H may extra protection ○ Methionine, Met, M - 2 CH2 + S + CH3 (double bond) and yung lapag din double bond ○ Cysteine, Cys, C - CH2 - SH ○ Lysine, Lys, K - 4CH2 + N+H3 root ○ Arginine, Arg, R - Lys root pero nagaway yung NH3 at CH2 kaya nagswitch places sila pero naiwan yung 2H ni NH3 sumama sa nanay nila na si CH2. Lumaki na na yung 2H naging NH2 sila parehas kaso yung isa mas angat sa buhay kaya may extra bond at positive charge Prepared by: Hans Dominguez To: Eizel Valenzuela  Acidic Amino Acids Amide Amino Acids Have —COOH side chains which are ionized at pH 7 Amide counterparts of acidic amino acids Negatively charged Can stabilize proteins by H-bonding A-ci-D G-lov-E N-ice Asp-aragus, Q-uite Glu-m ○ Aspartate, Asp, D - CH2 + COO- ○ Asparagine, Asn, N - CH2 + C + NH2 = O ○ Glutamate, Glu, E - CH2 + Asp ○ Glutamine, Gln, Q - CH2 + Asn Prepared by: Hans Dominguez To: Eizel Valenzuela  Uncommon Amino Acids Protonic Equilibria 4-hydroxyproline In aqueous solutions, amino acids can exist in various forms, depending ○ Found in plant cell wall proteins on the pH of the medium ○ Found in collagen Amino acids are amphoteric 5-hydroxylysine ○ Can act as an acid or a base ○ Found in collagen ○ Mainly due to carboxyl and amino groups present on its structure 6-N-methyllysine ○ Found in myosin Ionizable Groups in Amino Acids Y-carboxyglutamate Carboxyl Group ○ Found in prothrombin, a blood clotting protein Desmosine Amino Group Sulfhydryl Group of Cys ○ Found in elastin Phenolic Group of Tyr Ornithine and Citrulline ○ Metabolites in the biosynthesis of arginine and in the urea cycle Imidazole Group of His Prepared by: Hans Dominguez To: Eizel Valenzuela  Guanidino Group of Arg Sample Problems on Protonic Equilibria and IpH Steps: 1. Determine the total number of possible ionic forms of an amino acid, using the formula below: Total Ionic Forms = # of pKa Values + 1 2. When drawing the protonic equilibria, start with the most protonated Amino Group of Proline form going to the most deprotonated form ○ Amino acids undergo ionization such that the functional group with the lowest pKa is ionized first while those with the highest pKa value is ionized last 3. To compute for the net charge, use the equation below: Net Charge = # of Positive Groups - # of Negative Groups 4. To compute for the IpH of an amino acid ○ Look for the zwitterionic form of the amino acid on the protonic Isoelectric point/pH (IpH) equilibria and determine the pKa values on its left and right side pH wherein the net charge of the molecule is zero then get their average Has the formula: 𝐼𝑝𝐻 = 𝑝𝐾𝑎1+𝑝𝐾𝑎2 Sample Problem 2 Draw the complete protonic equilibria of glycine (pKa ɑ-COOH = 2.34; pKa ɑ-NH3+ = 9.60). Indicate the net charge of each form and calculate its IpH. Prepared by: Hans Dominguez To: Eizel Valenzuela Sample Problem Titration of Amino Acids Draw the complete protonic equilibria of glutamic acid (pKa ɑ-COOH = 2.19; pKa ɑ-NH3+ = 9.67; pKa side chain = 4.25). Indicate the net charge of each form and calculate its IpH. Several important pieces of information that can be derived from titration curves of amino acids: ○ Quantitative measure of pKa values of the ionizable groups ○ Relationship between its net electric charge and the pH of the solution ○ pH ranges where the amino acids can act as buffers Titration Curve of Glycine Point A: Cationic Form Point B: Equal amounts of Cationic and Zwitterionic Form Point C: Dipolar/Zwitterionic Form Point D: Equal amounts of Zwitterionic and Anionic Form Beyond D: Anionic Form Prepared by: Hans Dominguez To: Eizel Valenzuela Titration Curve of Acidic Amino Acid (Aspartate) Titration Curve of Basic Amino Acid (Lysine) 1. Which point (A, B, C, or D) corresponds to a. pKa of the alpha-carboxyl group? A 1. Which point (A, B, C, or D) corresponds to b. pKa of the alpha-amino group? D a. pKa of the alpha-carboxyl group? A c. pKa of the R-group? C b. pKa of the alpha-amino group? B d. IpH of the amino acid? B c. pKa of the R-group? D 2. Which is/are the predominant form(s) of amino acid(s) at these points: d. IpH of the amino acid? C a. Point A: 1 and 2 2. Which is/are the predominant form(s) of amino acid(s) at these points: b. Point B: 2 a. Point A: 1 and 2 c. Point C: 2 and 3 b. Point B: 2 and 3 d. Point D: 3 and 4 c. Point C: 3 d. Point D: 3 and 4 Prepared by: Hans Dominguez To: Eizel Valenzuela  Paper Electrophoresis Lesson 6: Peptides Peptides Can be used to separate amino acids based on net electric charge Spots are visualized using ninhydrin Polymers of amino acids ○ From colorless to purple spots Formed by condensation reaction of amino acids resulting to the formation of a peptide bond During paper electrophoresis, Pattern pHBuffer < IpHAA pHBuffer = IpHAA pHBuffer > IpHAA Some Important Terminologies Peptides/Oligopeptides Predominant + NC AA; Toward 0 NC AA; Does - NC AA; Toward Species Cathode not move Anode ○ Continuous chains of few amino acids usually not more than 50 residues Polypeptides Sample Problem ○ Amino acid chains with MW < 10,000 Da Glycine and glutamic acid are subjected to paper electrophoresis using a Proteins buffer solution at pH 7.0. Predict which electrodes each amino acid will migrate. ○ Amino acid chains with MW ≥ 10,000 Da Glycine = 5.97 < 7; - NC AA, Toward the Anode Note that all proteins are polypeptides but not all polypeptides are Glutamate = 3.22 < 7; - NC AA, Toward the Anode proteins. Structure of a Peptide Parts N-terminal – with free alpha amino group C-terminal – with free alpha carboxyl group ○ Drawn starting with the N-terminal ending with the C-terminal ○ Note that not all polypeptides have N- and C- terminal residues E.g., cyclic peptides Prepared by: Hans Dominguez To: Eizel Valenzuela Peptide Bond Nomenclature of Peptides Is a semi-rigid bond ○ Exhibits a partial double bond character due to resonance ○ Chemically stable – inert and not easily cleaved Steps for Naming: Experimentally 1. Start with the N-terminal and end with the C-terminal ○ Peptide bond (C—N) length is 0.133nm 2. Replace -ine with -yl for all residues except for the C-terminal, and the Shorter than the adjacent C—N bond (0.145nm) following: Longer than the C=O bond length of 0.123nm ○ Aspartate = Aspartyl ○ Support peptide bond has a partial double bond character ○ Glutamate = Glutamyl Consequences ○ Glutamine = Glutaminyl 1. Restricts rotation around the peptide bond ○ Tryptophan = Tryptophanyl Allows rotation only around the ɸ angle and the ψ angle 3. Write the whole name as one word ○ ɸ angle - angle formed around the C⍺-peptide N bond ○ ψ angle - angle formed around the C⍺-carbonyl carbon Sample Problem bond Write the name of the pentapeptide wherein the N-terminal residue is Ramachandran Plot serine, and followed by glycine, phenylalanine, alanine, and leucine ○ A plot of ψ versus ɸ angle used to serylglycylphenylalanylalanylleucine determine what values are sterically permitted in a Steps for Determining the Sequence: polypeptide chain 1. The three-letter or one-letter designation may be used ○ Permissible angles are shown as 2. Start with the N-terminal and end with the C-terminal colored region in the plot 3. For three-letter designation, use a hyphen (-) while for one-letter designations, a hyphen is not needed 2. Six atoms composing the peptide bond tend to be coplanar Sample Problem Forms the amide plane of the polypeptide backbone Give the one- and three-letter designation of serylglycylphenylalanylalanylleucine 1: SGFAL 3: ser-gly-phe-ala-leu Prepared by: Hans Dominguez To: Eizel Valenzuela Steps for Drawing the Peptide: Acid-Base Properties of Peptides 1. Start with the N-terminal 2. Form a peptide bond between the alpha-carboxyl group of the N-terminal residue and the alpha-amino group of the succeeding amino acid, and so All ⍺-NH2 and ⍺-COOH involved in peptide (amide) bond formation are not on anymore basic or acidic 3. End the drawing of structure by forming a peptide bond between the Ionizable groups in a peptide include: ⍺-carboxyl group of growing polypeptide chain and the ⍺-amino group of ○ Amino group of the N-terminal residue the C-terminal residue ○ Carboxyl group of the C-terminal residue ○ Side chains of the acidic and basic amino acids, cys and tyr Note that to always assume that the pH is neutral (unless specified). Steps in Drawing the Complete Protonic Equilibria of a Peptide: Sample Problem 1. Determine the total number of possible ionic forms of the peptide Draw the structure of lys-gly-cys-asp ○ In counting the number of pKa values, remember that the ⍺-NH2 and ⍺-COOH are not counted anymore except for the N-terminal amino group and the C-terminal carboxyl group Total Number of Ionic Forms = # of pKa values + 1 2. Start drawing the most protonated form of the peptide going to the most deprotonated form ○ Peptides undergo ionization such that the functional group with the lowest pKa is ionized first while those with the highest pKa is last 3. Compute for the net charge Net Charge = # of Positive Groups - # of Negative Groups 4. Compute for the IpH of the peptide ○ Look for the zwitterionic form of the peptide on the protonic Note that ionization should still be followed. equilibria ○ Determine the pKa values on its left and right sides then get their average Prepared by: Hans Dominguez To: Eizel Valenzuela Sample Problem Biological Activities of Some Peptides Draw the complete protonic equilibria of ser-gly-asp. Indicate the net charge in each form and the IpH. Their pKa values are the following As HORMONES Chemical messengers of the body Insulin ○ Responsible for the absorption of glucose from the bloodstream Oxytocin ○ A nonapeptide that stimulates uterine contraction ○ Love hormone Vasopressin ○ A nonapeptide that prevents urination at night Aspartame ○ A dipeptide aspartylphenyalanyl methyl ester used as artificial sweetener Phenylketonuria Person cannot metabolize phenylalanine due to low levels of phenylalanine hydroxylase; accumulation of phe leads to mental disorders ○ phenylalanine hydroxylase – converts phe to tyr Prepared by: Hans Dominguez To: Eizel Valenzuela  Lesson 7: Proteins Classification of Proteins Proteins According to COMPOSITION Simple Proteins Greek word “proteios” = “first of rank” ○ Yield only amino acids upon hydrolysis Most abundant molecule in the cell Conjugated Proteins Made up of one or more polypeptide chains folded in a specific ○ Yield simple proteins and non-protein substances conformation ○ Some important conjugated proteins: MW ≥ 10,000 Daltons (Da) ○ Common unit used to express the molecular weight of proteins; a Type of Conjugated Protein Non-Protein Portion unit of mass nearly equal to a hydrogen atom nucleoprotein nucleic acid ○ Some common proteins seen in the laboratory: glycoprotein carbohydrate lipoprotein lipid phosphoprotein Phosphate group hemoprotein heme flavoprotein flavin nucleotide metalloprotein metal ion According to BIOLOGICAL FUNCTION Enzymes ○ Catalyze biological reactions E.g., invertase – hydrolyze sucrose into its monosaccharide units: glucose and fructose Transport Proteins ○ Bind or Carry specific molecules or ions from one organ to another E.g., hemoglobin – transfer oxygen in the human blood Prepared by: Hans Dominguez To: Eizel Valenzuela  Nutrient and Storage Proteins According to SOLUBILITY ○ Mobilized by the body to drive building blocks or energy Albumins E.g., ferritin, casein, and ovalbumin ○ Soluble in water and dilute aqueous solutions Defense Proteins E.g., egg albumin and lactalbumin of milk ○ Defend organisms against invasion by foreign substances Globulins E.g., immunoglobulin ○ Soluble in dilute salt solutions but are insoluble and sparingly Contractile or Motile Functions soluble in water ○ Allow cells and organisms to contract, change shape, or move E.g., ovoglobulin of egg and myosin of muscle about Glutelins E.g., actin and myosin ○ Soluble in dilute solutions of acids and bases Structural Proteins E.g., glutelin of wheat and oryzenin of rice ○ As supporting filaments, cables, or sheets for strength and Albuminoids protection of biological structures ○ Insoluble in most ordinary solvents like water, salt solutions, dilute E.g., fibroin – protein found in the webs of Golden Orb acids, alkalis, and alcohols Spiders E.g., keratins of hair and collagen of bones Regulatory Proteins Prolamins ○ Regulate physical or cellular activities ○ Soluble in 70%-80% alcohols (others in 50%-90%), insoluble in E.g., insulin – major hormone that regulates blood sugar level water, neutral solvents or absolute alcohol E.g., zein of corn According to SHAPE Globular Levels of Protein Architecture ○ Polypeptide chain coiled into a compact spherical shape Being compact makes it surface area lower and more spherical, as a consequence, it is more soluble in water and more mobile within the cells E.g., amylase and hemoglobin Fibrous ○ Polypeptide chains are arranged side-by-side in long filaments Due to its bigger surface area, it is expected to be insoluble in water ○ Mechanically strong and have a structural and protective function E.g., collagen Prepared by: Hans Dominguez To: Eizel Valenzuela Primary Structure Amino group is part of the rigid ring, and the φ and ψ angles Amino acids stabilized by peptide bonds are fixed or restricted Linear order of amino acids from the N-terminus to the C-terminus The peptide N cannot participate in H- bonding Dictates the structure and function of the protein Introduces a destabilizing kink in the helix Secondary Structure Instead of being completely coiled, some parts are not Spatial arrangement of amino acid residues that are near one another in coiled the linear sequence of your peptide ○ E.g., Presence of glycine residue Common regular folding patterns of the polypeptide backbone Resulting α-helix structure has more conformational Main Interaction: Hydrogen bonds between peptide bonds flexibility ○ Between carbonyl oxygen and amide nitrogen The polymers take up coiled structures quite different from Three Types: α-helix, β-pleated sheets, and collagen helix the alpha-helix ○ E.g., Presence of lots of acidic and basic amino acids α-helix It will form electrostatic interactions or electrostatically repel polypeptide backbone is tightly wound each other, resulting in disrupted α-helix structure around an imaginary axis that is drawn ○ E.g., Presence of lots of amino acid having bulky or branched side longitudinally in the middle of the helix chains For each turn: Interfere in the formation of α-helix structure, if present in ○ rises by 5.4 Å / turn large numbers ○ each turn has ~3.6 amino acid residues Sample Problem ○ Height Difference: ~1.5 Å rise At what pH value does an alpha-helix with lots of glutamic acid residues between amino acid residues undergo destabilization? Why? Can be right-handed (counterclockwise) or At a very high pH, since it is acidic, the glutamic acid’s side chain will left-handed (clockwise) deprotonate at higher pH, undergoing destabilization, getting a negative charge Stabilized by H-bonds between C=O and which will have a tendency to repulse each other N—H in the polypeptide backbone that are 3-4 residues apart. β-pleated sheet Usually found in globular proteins Backbone of the polypeptide chain is extended into a zigzag rather than a helix Amino acids can affect the resulting α-helix structure structure Zigzag polypeptide chains are lying side by ○ E.g., Presence of a proline residue side and held together by hydrogen bonds “pleated” due to alternating position of α-carbon Prepared by: Hans Dominguez To: Eizel Valenzuela  Antiparallel Tertiary Structure ○ Neighboring strands extends in opposite Overall 3D arrangements of all atoms in the protein directions Results from folding into specific 3D shapes and unique binding sites ○ contain 2-15 strands, with an average of 6 Highest level of protein architecture common to all biologically active strands which is approximately 25 Å wide proteins ○ Each strand can contain up to 15 amino acid ○ Since not all protein structures have a quaternary structure residues, with an average of 6 Stabilized by the Following Interactions: Parallel 1. Hydrophobic Interactions ○ Strands runs in the same direction Overall tendency of side chains of non-polar residues to ○ Rarely contain fewer than 5 strands collect in the interior of the protein; seen in the interior part ○ Less stable than antiparallel of the protein structure Due to the offset of H-bonding which 2. Hydrogen Bonding weakens the interaction between the Between side groups or between peptide bonds; amino strands acids that can form H-bonding 3. Electrostatic Interaction Sample Problem Charged side chains What type of a beta-pleated sheet can be formed with one polypeptide ○ E.g., –COO- of aspartate and glutamate, and ε-NH3+ of strand? Why? Assume there is enough number of amino acid residue present in lysine the polypeptide. 4. Disulfide Bonds/Bridges Antiparallel beta-pleated sheet Formed by two cysteine residues ○ An actual covalent bond Collagen Helix Three separate polypeptide chains that are super-twisted about each Structural Motifs (Supersecondary Structures) other Regular 3D conformation or structural element of mostly 2° or some 3° ~3 amino acid residues per turn structure common to many different proteins Superhelical twisting is right-handed Indicative of a particular 3D architecture that are associated with specific Repetitive amino acid sequence: function ○ gly-pro-hyp (hydroxyproline) Non-essential amino acid Prepared by: Hans Dominguez To: Eizel Valenzuela Some Important Structural Motifs Quaternary Structure 1. Leucine Zipper (Coiled Coil) Describes the organization of subunits (polypeptide chains with 3° ○ made up of mostly hydrophobic amino acids structure) in a protein with multiple subunits (multimeric) ○ two helices can associate through hydrophobic interactions Subunits are held together by non-covalent and covalent interactions ○ important in DNA binding to proteins similar in tertiary structures 2. HELIX-turn-HELIX Can have homo-multimers or hetero-multimers ○ Two alpha helices have the same orientation relative to each other ○ Homo-multimers – same subunits and they are connected by a loop region ○ Hetero-multimers – at least one subunit is different ○ Motif used in DNA recognition and in calcium-binding proteins Multimeric or Oligomeric proteins are more stable than dissociated 3. Zinc Finger subunits ○ Consisting of an alpha helix coordinated by Zn2+ thru two Cys and two His residues and two beta strands Protein Folding ○ Has characteristic sequence motif: X- Folding of the protein gives its three-dimensional shape which is critical (Tyr/Phe)-X-Cys-X2,4-Cys-X3-Phe-X5-Leu-X2-His- X3,4-His-X4 in the function of that protein (e.g., formation of the binding site) where X is an unspecified amino acid; other hydrophobic ○ Manner by which proteins fold is dictated by its amino acid residues can be utilized aside from F, Y and L sequence or primary structure 4. β—α—β Loop Proteins fold into a specific conformation that results in a minimized free ○ With two parallel beta-strands having a long crossover between the energy of the molecule end of first strand and the beginning of the second strand made ○ Once minimized, the molecule becomes more stable usually by a helix In aqueous environment, the polypeptide chain folds in such a way ○ With ligand binding functions wherein hydrophobic side chains are buried in the interior and polar side ○ Found in ion channels chains are on the surface 5. β Barrel Folding of proteins can be stable or must the need the aid of a molecular ○ Beta-sheet that twists and coils to form a closed structure in which chaperone the first strand is hydrogen bonded to the last