Kod/Nama Kursus : BCM ( + ). (Biokimia Asas) ... Salleh (ABS). Jabatan. :
Biokimia. Jadual Kuliah. ( Masa dan Tempat) : 2 ... Page 11 .... Cuti Pertengahan
Semester ..... important equation called the Henderson-Hasselbalch equation. pH
= pK a.
BCH 3000 PRINSIP BIOKIMIA FOR PJJ (Semester 1 -2013/14) 1
Kod/Nama Kursus :
BCM 3000 (4+0) (Biokimia Asas)
Nama Pensyarah
Prof. Dr. Mohd Arif Syed (MAS) -Penyelaras
:
Prof. Dato’ Dr. Abu Bakar Salleh (ABS) Jabatan
:
Biokimia
Jadual Kuliah ( Masa dan Tempat) :
2
Sinopsis (This course encompasses the main biomoleculer components in biochemistry. Metabolism involving the anabolism and catabolism of major biomolecules are also explained)
3
Learning Outcome 1. Distinguish the structure and function of biomolecules found in biological systems (C4) 2. State the various key metabolic processes (P2) 3. Describe the biochemical reactions (A3) 4. Solve problems related to the metabolism of biomolecules by using information from various sources (CTPS, LL)
4
Brief Lecture Contents 1. Introduction-Biochemistry? Contributions? Important life components 2. Carbohydrates – Classification – mono, di polysaccharides – Structure –configuration & stereochemistry; reactions – glucose and other sugars 3. Amino acid & protein – biological roles, structure, classification, reactions, analysis. Peptides – primary, secondary, tertiary and quaternary structures 4. Lipid –functions & distribution, characteristics of fatty acids-saturated & unsaturated f/acids. Structures & characteristics of triacylglycerols, phospholipids, sphingolipids, terpenes & steroids 5
Brief Lecture Contents 5. Nucleic acids – components – purines, pyrimidines. Structure, reactions & importance of nucleosides, nucleotides & polynucleotides. DNA, RNA – structure, functions & types 6. Enzymology – Classification, naming, active sites. Enzyme kinetics. Factors affecting enzyme activity- enzyme & substrate concentration, pH, temperature . Substrate specificity – single & multiple substrate. Enzyme inhibitors – competitive, con-competitive, uncompetitive. Control of enzyme reactions – product inhibition, Isoenzymes, multienzyme system and allosteric enzymes
6
Brief Lecture Contents 7. Carbohydrate metabolism – Metabolic energy cycle – Bioenergetics: ATP other high energy compounds. Storage & energy transfer. Glycolysis & fermentation. Electron transport system. Compartmentation & mitochondria. Phosphorylation & production of ATP. Anaplerotic reactions. Glyoxylate cycle. Gluconeogenesis. Pentose phosphate pathway. Integration and control. 8. Photosynthesis – Fixation of CO2 during photosynthesis. Chlorophyll, components of photosynthesis. Photosystem I & II. Photophosphorylation. Calvin cycle. Hatch-Slack cycle. 7
Brief Lecture Contents 9.
Lipid metabolism – Lipid oxidation- Enzymes involved, energy production. Oxidation of saturated & branched fatty acids. Formation of ketone bodies. Lipid biosynthesis –mitochondrial system and extramitochondrial. Cycle & enzymes involved. Synthesis of saturated & unsaturated fatty acids. Cholesterol synthesis & control.
10. Protein & amino acid metabolism – Degradation of amino acids- transamination, deamination, decarboxylation. Cycle involved- intermediates for the TCA cycle. Ammonia and urea metabolism. Biosynthesis of amino acids- role in the metabolism of porphyrin and nucleic acids. Nitrogen fixation. 8
Brief Lecture Contents 11. Nucleic acid metabolism – synthesis of mononucleotides – purines, pyrimidines – cycle and enzymes involved; control. Biosynthesis of ribo & deoxyribonucleotides. Characteristics of genetic materials – chromosomes. Genetic code, base sequence. DNA replication. DNA repair. Protein synthesis – ribosome, co-factor involved & phase of synthesis. Inhibition and control of synthesis. 12. Membrane Biochemistry – Modification & structure. Model for membrane structure. Transport mechanism across membrane – passive & active transport.
9
Brief Lecture Contents 13. Hormones- Introduction to plant & animal hormones. Reactions & control of endocrine hormones. Hormone reactions at the molecular level. 14. Integration & control of metabolism. Relationship between carbohydrate, lipid and protein metabolism.
10
Course Evaluation 1. Mid Term
= 35% (17% + 18%)
1. Final Exam
= 35%
2. SCL
= 30%
Total
= 100%
11
Course Evaluation 1. Mid Term 2. SCL i. Model and presentation ii. Assignment 3. Final Examination TOTAL
35% (17% + 18%) 30% 20% 10% 35% 100%
12
Course Evaluation Mid Term
= 35% - 2013
Topics covered in test 1. Introduction-Biochemistry 2. Carbohydrates 3. Amino acid & protein 4. Lipid 5. Nucleic 6. Enzymology
13
Course Evaluation Mid Term
= 35% -
Types of Questions 1. Duration – 2 hours 2. Multiple choice – 60 questions (1 mark each) 3. Short Answers - Choose 8 out of 10 questions ; 5 marks each
14
Course Evaluation Final Examination
= 35%
Topics covered in exam 7. Carbohydrate metabolism 8. Photosynthesis 9.
Lipid
10. Protein & amino acid metabolism 11. Nucleic acid metabolism 12. Membrane 13. Hormones 14. Integration & control of metabolism 15
Course Evaluation SCL (Student Centered Learning = 30% 1. MODEL AND PRESENTATION In this exercise, each student is required to produce a model of an oligopeptide using materials from the environment. The model should be able to demonstrate clearly the structural configuration of the oligopeptide . The student will be asked to present the model and explain the structural configuration
16
Course Evaluation SCL (Student Centered Learning = 20% Model Requirements 1. The student must design and produce a model of an oligopeptide 2. All amino acids must be different from one another and of different group 3. Materials used must be from the environment. No model kit will be allowed. This is also not computer modeling 4. The model should clearly show the structure of the amino acid 5. Student will be asked to explain their respective models 17
Course Evaluation SCL (Student Centered Learning = 25% PRESENTATION 1. Week 14 (The latest date. Can be arranged) 2. Place – Biotek 1 3. Date – please inform when you are available 4. Evaluation by a panel
18
19
BCH 3000 PRINSIP BIOKIMIA (Semester 1 -2012/13) 20
Kod/Nama Kursus :
BCM 3000 (4+0) (Biokimia Asas)
Nama Pensyarah
Prof. Dr. Mohd Arif Syed (MAS) -Penyelaras
:
Puan Zetty Jabatan
:
Biokimia
Jadual Kuliah ( Masa dan Tempat) : SK 10-12; DKBiotek 1.1
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Sinopsis (This course encompasses the main biomolecule components in biochemistry. Metabolism involving the anabolism and catabolism of major biomolecules are also explained)
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Learning Outcome 1. Membezakan struktur dan fungsi biomolekul yang terdapat dalam sistem biologi (C4) 2. Menyatakan pelbagai proses metabolisme yang utama (P2) 3. Menerangkan tindakbalas biokimia (A3) 4. Menyelesaikan masalah dalam metabolisme biomolekul dengan menggunakan maklumat dari pelbagai sumber (CTPS, LL)
23
Brief Lecture Contents 1. Introduction-Biochemistry? Contributions? Important life components 2. Carbohydrates – Classification – mono, di polysaccharides – Structure –configuration & stereochemistry; reactions – glucose and other sugars 3. Amino acid & protein – biological roles, structure, classification, reactions, analysis. Peptides – primary, secondary, tertiary and quaternary structures 4. Lipid –functions & distribution, characteristics of fatty acids-saturated & unsaturated f/acids. Structures & characteristics of triacylglycerols, phospholipids, sphingolipids, terpenes & steroids 24
Brief Lecture Contents 5. Nucleic acids – components – purines, pyrimidines. Structure, reactions & importance of nucleosides, nucleotides & polynucleotides. DNA, RNA – structure, functions & types 6. Enzymology – Classification, naming, active sites. Enzyme kinetics. Factors affecting enzyme activity- enzyme & substrate concentration, pH, temperature . Substrate specificity – single & multiple substrate. Enzyme inhibitors – competitive, con-competitive, uncompetitive. Control of enzyme reactions – product inhibition, Isoenzymes, multienzyme system and allosteric enzymes
25
Brief Lecture Contents 7. Carbohydrate metabolism – Metabolic energy cycle – Bioenergetics: ATP other high energy compounds. Storage & energy transfer. Glycolysis & fermentation. Electron transport system. Compartmentation & mitochondria. Phosphorylation & production of ATP. Anaplerotic reactions. Glyoxylate cycle. Gluconeogenesis. Pentose phosphate pathway. Integration and control. 8. Photosynthesis – Fixation of CO2 during photosynthesis. Chlorophyll, components of photosynthesis. Photosystem I & II. Photophosphorylation. Calvin cycle. Hatch-Slack cycle. 26
Brief Lecture Contents 9.
Lipid metabolism – Lipid oxidation- Enzymes involved, energy production. Oxidation of saturated & branched fatty acids. Formation of ketone bodies. Lipid biosynthesis –mitochondrial system and extramitochondrial. Cycle & enzymes involved. Synthesis of saturated & unsaturated fatty acids. Cholesterol synthesis & control.
10. Protein & amino acid metabolism – Degradation of amino acids- transamination, deamination, decarboxylation. Cycle involved- intermediates for the TCA cycle. Ammonia and urea metabolism. Biosynthesis of amino acids- role in the metabolism of porphyrin and nucleic acids. Nitrogen fixation. 27
Brief Lecture Contents 11. Nucleic acid metabolism – synthesis of mononucleotides – purines, pyrimidines – cycle and enzymes involved; control. Biosynthesis of ribo & deoxyribonucleotides. Characteristics of genetic materials – chromosomes. Genetic code, base sequence. DNA replication. DNA repair. Protein synthesis – ribosome, co-factor involved & phase of synthesis. Inhibition and control of synthesis. 12. Membrane Biochemistry – Modification & structure. Model for membrane structure. Transport mechanism across membrane – passive & active transport.
28
Brief Lecture Contents 13. Hormones- Introduction to plant & animal hormones. Reactions & control of endocrine hormones. Hormone reactions at the molecular level. 14. Integration & control o f metabolism. Relationship between carbohydrate, lipid and protein metabolism.
29
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Kuliah Cuti- Alexandria Cuti Pertengahan Semester Peperiksaan Akhir Cuti Umum
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Why study biochemistry? ● Part of curriculum ● Explain a lot of the controversies in the news at the moment.
Stem cell study
Cloning of the human being
Diseases (defect in metabolism)
GM food and organisms (genetically modified)
31
BIOCHEMISTRY: A PROLOGUE Biochemistry = the chemistry of life. ● ∴ bridges the gap between chemistry (the study of the structures and interactions of atoms and molecules) and biology (the study of the structures and interactions of cells and organisms). ● Since living things are composed of inanimate molecules, life, at its most basic level, is a biochemical phenomenon.
Inert; not living; not lively 32
Living organisms ●
diverse in their macroscopic properties
●
BUT remarkable similarity in their biochemistry that provides a unifying theme with which to study them. ◊ For example, hereditary information is encoded and expressed in an almost identical manner in all cellular life ◊ the series of biochemical reactions= metabolic pathways, as well as the structures of the enzymes that catalyze them are, for many basic processes, are nearly identical from organism to organism. 33
This strongly suggests that all known life forms are descended from a single primordial ancestor in which these biochemical features first developed
34
Although biochemistry is a highly diverse field, it is largely concerned with a limited number of interrelated issues. These are 1. What are the chemical and three-dimensional structures of biological molecules and assemblies, how do they form these structures, and how do their properties vary with them? 2. How do proteins work?- what are the molecular mechanisms of enzymatic catalysis, how do receptors recognize and bind specific molecules, and what are the intramolecular and intermolecular mechanisms by which receptors transmit information concerning their binding states? 35
3. How is genetic information expressed and how is it transmitted to future cell generations? 4. How are biological molecules and assemblies synthesized? 5. What are the control mechanisms that coordinate the myriads of biochemical reactions that take place in cells and in organisms? 6. How do cells and organisms grow, differentiate, and re-produce?
36
History of Biochemistry ● fairly new field of science -the 20th century ● first landmark of biochemistry - Friedrich Wohler (1828) synthesized the organic compound urea from the inorganic compound ammonium cyanate ● ∴building blocks of life were the same as those of nonliving things ● The role of enzymes as catalyst - Buchner showed that a process of biochemistry, catalysis, could occur independently from living cells (enzymes in yeast extracts and fermentation)
37
History of Biochemistry ● Fischer developed the lock and key model (enzyme as rigid lock, substrate as key) A modified version of this model (induced fit) is still used today ● The second part of the 20th century saw advances in structural biology especially the structure of proteins ● The first protein structures were determined by John C. Kendrew and Max Perutz in the 1950s and 1960s. ● Now have determined the structures of more than 1000 proteins.
38
39
History of Biochemistry ●
The role of nucleic acid as information molecules
In 1944 Oswald Avery et al extracted DNA from a toxic strain of a bacteria and when added to a nontoxic strain resulted in the bacteria being transformed into a virulent strain.
Watson and Crick (1950) deduced the 3D structure of DNA.
Crick predicted that information encoded in DNA is transcribed to ribonucleic acid and then translated to protein.
This unidirectional information flow is referred to as the central dogma of molecular biology 40
BIOLOGICAL STRUCTURES Living things are enormously complex. ● simple E. coli cell contains some 3 to 6 thousand different compounds, most of which are unique to E. coli ; Homo sapiens (human beings), may contain 100,000 different types of molecules, although only a minor fraction of them have been characterized. ● ∴ biochemical understanding of any organism would be a hopelessly difficult task ??
No !!!! - Why ???? 41
Living things have an underlying regularity that derives from their being constructed in a hierarchical manner.
Multicellular organisms Organizations of organs Tissues Cells Subcellular organelles Supramolecular assemblies of macromolecules
42
An example of hierarchical organization of biological structures 43
What makes a living thing?
44
The Chemical Elements of Life ● Only six nonmetallic elements make up 97% of the weight of most organisms – carbon, oxygen, hydrogen, nitrogen, phosphorous and sulfur. ● All form stable covalent bonds. ● here are also 5 common ions found in all organisms: - Calcium (Ca2+), Potassium (K+) Sodium (Na+), Magnesium (Mg2+), Chloride (Cl-) ● Water is a major component of cells. ● Altogether, a total of 29 different elements are commonly found in living organisms.
45
Brown – important elements purple – essential ions dark blue – more common trace elements light blue – less common trace elements 46
Organic compounds ● Most of the solid material of cell consists of carboncontaining compounds (∴ ∴organic compounds). ● The organic compounds of interest is shown
47
Functional Groups ●
These organic compounds have own specific functional groups
48
● The elements of life are assembled into molecules with common structures and patterns – how ?? –
via linkages (bonds)
49
The polymeric organization of proteins, nucleic acids an dpolysaccharides 50
BIOPOLYMERS ● they are formed from smaller molecules called monomers that are linked together in a sequential way to form long chains ● After being joined together, the individual monomers in a chain are referred to as residues ● There are various levels in the hierarchy of life i.e. atoms, molecules, biopolymers, organelles, cells, tissues, organs and whole organisms
51
PROTEINS An example of a biopolymer ● polymers formed from the condensation of individual monomers called amino acids ● Twenty amino acids are incorporated into proteins in all cells ● Each amino acid contains an amino group and a carboxylate group and a side chain (R group). ● The amino group from one amino acid reacts with the carboxylate group of the other to form an amide linkage that is referred to as a peptide bond
52
53
● Many amino acids joined in this manner is called a polypeptide ( have N and C terminal ● The amino acids are combined in a specific sequence to produce proteins consisting of hundreds or thousands of amino acid residues
54
● A functional protein consist of one polypeptide or several different polypeptides tightly bound together. ● Proteins function as either enzymes or structural components of cells and organisms. ● The function of a protein depends on the 3D structure or conformation.
55
BCM 3000 PRINSIP BIOKIMIA
56
POLYSACCHARIDES ● Composed basically of carbon, oxygen and hydrogen. ● Monosaccharides – simple sugars, polysaccharides – polymers. ● Fischer projection – linear molecule. ● Haworth projection – ring form (usual biochemical form). ● Ribose – approx. 20 conformations
57
● Glucose – o most abundant six-carbon sugar o Monomer for cellulose, glycogen and starch o Differ in bonding between C-1 of the monomer to the next
58
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Nucleic acids ●
Biopolymers of monomers called nucleotides.
●
Nucleotides – five-carbon sugar, heterocyclic nitrogen base and at least one phosphate. o
Five – carbon sugar (ribose and deoxyribose).
o
Base (purines and pyrimidines).
Purines – adenine (A) and guanine (G).
Pyrimidines (cytosine ( C), thymine (T) and Uracil (U)
60
Nucleic acids are polymers formed from monomers called nucleotides that are joined in a phosphodiester linkage Polynucleotideformed by linking phosphate group to C-3 oxygen atom of another nucleotide
61
Examples of nucleic acids 1. DNA ●
uses deoxyribose sugars
● Usually double stranded 2. RNA ● Uses ribose sugars ● Usually single stranded ● 4 types of RNA : i. mRNA ii. tRNA iii. rRNA iv. heterogenous RNA. 62
Lipids and Membranes ● Diverse class of compounds rich in carbon and hydrogen (low in oxygen). Fatty acids, glycerophospholipids, wax and beeswax. ● Most lipids not soluble in water. ● In membranes lipid are polar - hydrophilic water loving head and hydrophobic (water fearing) tail. ● Form bi-layer membrane system.
63
The Energetics of Life ● Life requires input of energy – ultimately from the sun ● Metabolism Describes the numerous reactions in which organic compounds are synthesized and degraded and useful energy is extracted, stored and used. Bioenergetics o The study of the changes in metabolic energy 64
● ∆G – the free-energy change for a reaction is the difference between the free energy of the product and the free energy of the reactants.
∆G = ∆H - T ∆S where ∆H is the change in heat content (enthalpy change) ∆S is the change in randomness (entropy change) T is the temperature in Kelvin.
∆G < 0 reaction is spontaneous ∆G > 0 reaction requires input of energy ∆G = 0 reaction is at equilibrium. 65
Metabolic Processes ●
Many many types of chemical reactions simultaneously occur in any living cell.
●
Yet, these reactions follow a pattern that organizes them into the coherent process we refer to as life. o For instance, most biological reactions are members of a metabolic pathway - sequence of reactions that produce one or more specific products. o Moreover, the rates of its reactions are so tightly regulated that there is rarely an unsatisfied need for a reactant in a metabolic pathway or an unnecessary buildup of some product. 66
Metabolism
- traditionally divided into two major
categories: 1. Catabolism or degradation, in which nutrients and cell constituents are broken down so as to salvage their com-ponents and/or to generate energy. 2. Anabolism or biosynthesis, in which biomolecules are synthesized from simpler components
The energy required by anabolic processes is provided by catabolic processes largely in the form of adenosine triphosphate (ATP). 67
68
Expression and Transmission of Genetic Information ● Deoxyribonucleic acid (DNA) is the cell's master repository of genetic information ● Genetic information is encoded in the sequence of these bases. ● The division of a cell must be accompanied by the replication of its DNA. ◊ each DNA strand acts as a template for the formation of its complementary strand ◊ ∴ every progeny cell contains a complete DNA molecule each of which consists of one parental strand and one daughter strand. Mutations arise when, through rare copying errors or damage
69
repository is a place where data or specimens are stored and maintained for future retrieval. Encoded - prearranged; programmed, determined, set,preset Replication - duplication,copying, reproduction
70
The cell is the basic unit of life Cells are classified as eukaryotes or prokaryotes 1. Prokaryotes (bacteria) Ubiquitous, no nucleus, only approx 1000 genes
2. Eukaryotes (plants, animals, fungi and protists)
Membrane bound nucleus, 1000 fold greater in volume.
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An animal cell
72
A plant cell
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Water Properties ● All living cells depend on water for their existence. ● Metabolic machinery of cells has to operate in an aqueous environment ● Water molecule – polar ● Water molecule is V-shaped with an angle of 104.5O 75
76
Oxygen atom ● has 8 electrons with 6 in the outer shell ● The outer shell can accommodate 4 sp3 orbitals. 2 lone pairs containing 2 electrons each and 2 orbitals which can share their electron with the electron in hydrogen an oxygen atom covalently bonded to two hydrogen atoms It has a tetrahedral bond structure that results from 4 sp3 hybridized orbitals ● The oxygen nucleus attracts electrons more strongly than the single proton in hydrogen nucleus 77
78
Tetrahedral bond structure of water 79
● There is a small negative charge on oxygen (δ δ-) and a small positive charge on the hydrogen (δ δ +) ● Water forms hydrogen bonding (up to 4 hydrogen bondings)
80
Noncovalent Interactions in Biomolecules There are 4 types of noncovalent interactions.
1. Charge-charge interactions ● Electrostatic interactions between two charged particles e.g. NaCl ● The strongest noncovalent forces ● Also responsible for mutual repulsion of similarly charged ionic groups
81
2. Hydrogen bonds ● A type of electrostatic interaction which occurs in many macromolecules ● Among the strongest noncovalent forces in biological systems ● Strong enough to confer structural stability but weak enough to be readily broken. ● Can form when a hydrogen covalently bonded to a strong electronegative atom, such as nitrogen, oxygen and sulfur
82
83
Hydrogen bonding 84
3. Van der Waals Forces ● Weak intermolecular forces produced between all neutral atoms by transient electrostatic interactions ● Have attractive and repulsive components ● forces that exist between MOLECULES of the same substance
85
Van der Waals Forces 86
4. Hydrophobic interactions ● The association of a relatively nonpolar molecule or group in aqueous solution with other nonpolar molecules rather than with water ● The non-polar groups mutually repel water and other polar groups and results in a net attraction of the nonpolar groups for each other. Hydrocarbon alkyl groups on ala, val, leu, and ile interact in this way
87
Hydrophobic interactions 88
Ionization of Water ● ● ● ●
Important property of water – slight tendency to ionize. Pure water consist of H2O and small quantity of hydronium ions (H3O+) and hydroxide ions (OH-) Important property of water – slight tendency to ionize. Pure water consist of H2O and small quantity of hydronium ions (H3O+) and hydroxide ions (OH-)
89
The pH Scale A logarithmic scale to measure the concentration of H+.
90
91
AIR – H2O Medium for biological system ∴ need to know the role in the degradation/dissociation of ions from biomolecules Water = neutral molecule, but it can ionise –
92
AIR – H2O From the dissociation/ionization of water equation
pH = -log [H+] ie. pH = negative log of hydrogen ion (H+) concentration pOH
= negative log of hydroxyl ion (OH-) concentration
93
ACID = substance that donates proton a substance which produces hydrogen ions (H+) by dissociation For example
HCl H+ + ClHydrogen ions (H+) associate with water to form H3O+ (Hydronium ion) 94
= Substances that receive proton
BASE
= can associate with H+ (e.g. NH3) =
i.
may produce hydroxide ions (OH- ) either by direct dissociation or subsequent to reaction with water
KOH K+ + OHNH3 + H2O NH4OH NH4+ + OH-
95
STRONG ACIDS are compounds that dissociate (ionize) almost 100% in an aqueous solution.
HCl H+ + Cl-
96
The common acids that are almost one hundred percent ionized are: HNO3 - nitric acid HCl - hydrochloric acid H2SO4 - sulfuric acid HClO4 - perchloric acid HBr - hydrobromic acid HI - hydroiodic acid 97
WEAK ACIDS = Substances that dissociate partially
HAc
H+ + Ac-
98
Examples of strong bases LiOH - lithium hydroxide NaOH - sodium hydroxide KOH - potassium hydroxide RbOH - rubidium hydroxide CsOH - cesium hydroxide Mg(OH)2 - magnesium hydroxide Ca(OH)2 - calcium hydroxide Sr(OH)2 - strontium hydroxide Ba(OH)2 - barium hydroxide 99
When a weak acid or base dissociates in an aqeous solution, will get an equilibrium between the acid and its conjugate base. This equilibrium is called
Equilibrium constant = Ka Using this equilibrium and taking into consideration the equation for the dissociation of weak acid, will get an important equation called the Henderson-Hasselbalch equation
pH = pKa + log [A-] [HA] 100
IONIZATION OF WATER H2O + H2O H2O
H+
H3O + + OH+ OH ………… (1)
Law of Mass Action to get equilibrium point (the rate of a chemical reaction is directly proportional to the molecular concentrations of the reacting substances ) Keq = [H+] [OH-] ..................... (2) [H2O] Keq = Equlibrium constant [ ] = concentration in mole/liter (M) 101
If we know the value of Keq pure water can determine the values of [H+] dan [OH-] Keq for pure water = 1.8 x 10-16M [Water] = 55.5M ∴ 1.8 x 10-16 = [H+] [OH-] 55.5M [H+] [OH-]
=
1.0 x 10-14M .................... (3)
[H+] = [OH-] – equation (1) ∴[H+] = [OH-] =
1.0 x 10-14M = 1.0 x 10-7M 102
In exponential form - (1.0 x 10-7M) too small Simpler method – use - Svenson – pH dan pOH
pH = -log [H+] pOH = -log [OH-]
Equation (3) becomes log [H+] + log [OH-] = -14 -log [H+] - log [OH-] = 14 i.e. pH + pOH
= 14 103
Because
[H+] = [OH-] pH = pOH = 7
pH = pOH = 7, solution is neutral pH = pOH < 7, solution is acidic pH = pOH > 7, solution is basic
104
HENDERSON-HASSELBALCH EQUATION For weak acid eg. CH3COOH (HA) H+ + A-
HA
In equilibrium conditions Ka =
[H+][A-]
Ka = equilibrium constant
[HA] [H+] = Ka [HA] [A-] 105
HENDERSON-HASSELBALCH EQUATION
log [H+] = log Ka + log [HA] [A-] - log [H+] = - log Ka + log [A-] [HA] - log [H+] = pH ; -log Ka = pKa
pH = pKa + log
[A-] [HA] 106
pH = pKa + log [A-] [HA] pH = pKa + log [conjugate base] [conjugate acid]
pH = pKa + log [proton acceptor] [proton donor]
107
This equation is called the HENDERSONHASSELBALCH Equation This equation is useful to determine 1. the amount of acid 2. the amount of base or 3. the amount of salt required to prepare a buffer with the pH that we want
108
109
BCH 3000 PRINCIPLES OF BIOCHEMISTRY (Semester 1 -2012/13) 110
CARBOHYDRATE •
Carbohydrates as the main component of life
•
Classification
•
Structure
•
Chemical reactions and biochemical functions of carbohydrate
111
CARBOHYDRATE
112
113
CARBOHYDRATES ● 'hydrate of carbon', with structural formula (CH2O)n ● Carbohydrate = saccharides = biological molecules ● The basic building blocks of carbohydrates are the monosaccharides –i.e. simplest = monosaccharides, eg. glucose ● These are linked together in longer chains to form oligosaccharides (2 - 20 residues - eg. Maltose) and polysaccharides (> 20 residues eg. starch, cellulose , glicogen )
114
CARBOHYDRATES ●
The root sacchar- comes from the Latin saccharum, "sugar".
●
Why saccharide called carbohydrate ? Most have general formula (CH2O)n
●
Many saccharides a. Modified b. Contains amino groups, sulphates, phosphates, etc.
115
Functions of carbohydrates ● important source of energy for the body - 1g of carbohydrate provides 4.2 kcal of energy o Almost all of the cells use glucose to distribute energy -brain cells ; erythrocytes (red blood cells) are completely dependant on glucose as an energy source. ● Act as energy storage - glycogen stores act as a readily available energy reserve. A person weighing 70kg has a glycogen reserve of about 350 - 400g, which is about 1.500 kcal.
116
Functions of carbohydrates ● Carbohydrates find many uses as structural elements – eg. Cellulose – cell walls in plants, bacteria, exoskeleton of insects, ● Carbohydrates are utilized as raw materials for several industries. For e.g., paper, plastics, textiles etc. ● Marker molecules for cell recognition – Blood types – A,B,O ● Found in biological molecules e g coenzymes and nucleic acids
117
Classification of Carbohydrates
1. Monosaccharide- one sugar residue. Most well known is glucose, C6H12O6
2. Oligosaccharide- a few (2-9) sugar residues . Most well known is cane sugar or sucrose, C12H22O11.
3. Polysaccharide - many sugar residues. Most common are glycogen, starch and cellulose, from animals, plants and plants.
118
Monosaccharides 1. Two classes- aldoses (aldehyde) and ketoses (keto) 2. In our formula, (CH2O)n, n is 3 or more 3. Simplest are dihydroxycetone, a ketose where n=3, and glyceraldehyde, an aldose where n=3 - Triose 4. Simple sugars – monomeric 5. can have different number of carbon atoms 6. can be combined to form disaccharides and polysaccharides 7. some can have a linear or ring structure
119
Monosaccharides
● Aldose = an aldehyde with two or more hydroxyl groups. ● Ketose = a ketone with two or more hydroxyl groups ● Both are trioses = simplest monosaccharides; threecarbon sugars 120
Aldehyde
Ketone
back
121
Aldoses and Ketoses 122
Monosaccharides ● Both have the same compositions = TAUTOMERS = (isomer) ● Tautomers are organic compounds that are interconvertible by a chemical reaction called tautomerization. ● Can change from one form to the other but takes a very long time ● Catalysts can speed up the change
Why tautomer ?? Change from one form (aldehyde) to another (ketone) 123
Isomers = are molecules ●
with the same chemical formula
●
and often with the same kinds of bonds between atoms,
●
but in which the atoms are arranged differently.
●
Many isomers share similar if not identical properties in most chemical contexts
124
ENANTIOMER (Optical isomers) ● In glycerldehyde – chiral carbon = 4 different groups ● get 2 isomers = enantiomer mirror images which are not superimposable - D and L isomer
125
126
127
128
Enantiomers -if they are – ● mirror images of each other- one is the mirror image of the other, ● two stereoisomers are enantiomers if they are different but each can be superimposed on the mirror image of the other. 129
Stereoisomers ● means that the two molecules differ in their three-dimensional shapes only but that they have the same structural formulas. ● This means they have the same exact groups attached in the same way. ● Only the three-dimensional orientation of these groups are different. ● ∴ enantiomers are stereoisomers
130
Diastereomers (or diastereoisomers) ● are stereoisomers that are not enantiomers (nonsuperimposable mirror images of each other). ● can have different physical properties and different reactivity. ● pairs of isomers that have opposite configurations at one or more of the chiral centers but are not mirror images of each other ● cis-trans isomerism is a form of diastereomerism 131
Diastereomers
132
133
QUESTION Which is L isomer? Which is D isomer? Fischer Projections ● Glyceraldehyde is actually the basis for the L and D nomenclature ● Solution of D-glyceraldehyde rotates polarized light to the right (dextrorotatory) and L-glyceraldehyde rotates light to the left (levorotatory) 134
Plane Polarized Light 135
136
137
138
Fischer Projections ● Monosaccharides can have multiple chiral centers ● ∴ need some conventions for drawing their structures. ● For linear chains, the stereochemistry is often represented as a Fischer Projection:
139
● In a Fischer projection the carbon chain is oriented in the vertical direction, with a conformation that projects the carbon bonds onto a flat plane, and with all horizontal bonds projecting out, in front of the plane ● When the molecule is oriented with the C1 aldehyde at the top, pointing away from the viewer, this defines a convention where the C2 hydroxyl group will be on the left for L-glyceraldehyde, and on the right for Dglyceraldehyde 140
Stereochemistry of Longer Monosaccharides i.
For longer monosaccharides, the assignment of the L and D configuration is determined by the configuration of the chiral carbon farthest away from the C1 carbonyl (ie. Highest chiral number)
ii. Eg. - glucose, a 6-carbon sugar, the C5 carbon is used. If the C5 hydroxyl group is on the left, the molecule is Lglucose. If the hydroxyl group is on the right, it is Dglucose iii. In a carbon chain with 2 possible configurations for each chiral center, there are a total of 2n stereoisomers for a compound with n chiral carbons 141
142
Some DAldose Isomers
143
Some DD-Ketose Isomers
144
Under natural conditions, only one enantiomer predominates – the D-isomer cf. amino acid – L-isomer e.g. monosaccharide – monosaccharide - = Dmonosaccharide
145
Diastereoisomer : Diastereomers are stereoisomers that are not enantiomers or mirror images of each other. Diastereomers can have different physical properties and different reactivity.
146
147
Chiral carbons
Tetrose Are they enantiomers?
148
149
PENTOSE Contains 5 carbon atoms 3 chiral carbons = 23 stereoisomers - 4 pairs of enantiomers - untuk aldose = aldopentose Stereochemistry of Longer Monosaccharides
BUT ketopentose – only 2 chiral carbons 4 isomers only
150
Some DAldose Isomers
151
Aldopentose
Ketopentose 152
HEXOSE ● Monasaccharide with 6 carbon atoms ● Number of isomers – high ● Common hexosee = glucose & fructose, ● mannose & galactose also abundant – all play important biological roles
153
154
Cyclic Structures Cyclic structures • are the common form of monosaccharides with 5 or 6 carbon atoms. O
O
• form when the hydroxyl group on C-5 reacts with the – aldehyde group or – ketone group. 155
Drawing the Cyclic Structure for Glucose STEP 1 : Number the carbon chain and turn clockwise to form a linear open chain.
H O 1C
H OH H
H
O
2
H C OH 3
HO C H 4
H C OH
HOCH 2 C C C C C 6
5
4
3
2
1
H
OH OH H OH
5
H C OH 6 CH
2OH 156
H O 1C
H
H OH H
O
2
H C OH 3
HO C H 4
H C OH
HOCH 2 C C C C C 6
5
4
3
2
1
H
OH OH H OH
5
H C OH 6 CH
2OH
157
Drawing the Cyclic Structure for Glucose STEP 2: Fold into a hexagon. • Bond the C5 –O– to C1. 6 CH2OH • Place the C6 group above the 5 O ring. • Write the –OH groups on C2 4 1 and C4 below the ring. OH OH OH • Write the –OH group on C3 3 2 above the ring. OH • Write a new –OH on C1. 158
Drawing the Cyclic Structure for Glucose STEP 3 : Write the new –OH on C1 •
•
down for the α form. up for the β form. CH2OH
CH2OH
O
OH
OH
O OH
α
OH
α-D-Glucose
OH
β OH
OH OH
β-D-Glucose 159
Summary of the Formation of Cyclic Glucose
160
α-D-Glucose and β-D-Glucose in Solution When placed in solution, • cyclic structures open and close. • α-D-glucose converts to β-D-glucose and vice versa. • at any time, only a small amount of open chain forms. CH2OH CH2OH CH2OH H O O O OH O
OH
OH
OH
OH OH
α-D-glucose (36%)
C
OH OH
D-glucose (open) (trace)
H
OH
OH OH
β-D-glucose (64%) 161
Cyclic Structure of Fructose Fructose • is a ketohexose. • forms a cyclic structure. • reacts the —OH on C-5 with the C=O on C-2. CH2OH C O
CH2OH
HO C H
CH2OH
CH2OH
OH
H C OH H C OH
O
O
OH OH
OH
CH2OH
OH
OH
α--D-fructose
β-D-fructose
CH2OH
D-fructose
162
CYCLICAL STRUCTURE CHO with 5 & 6 carbon atoms normally exist as ring structures Cyclization = interactions between functional groups at ● C-1 & C-5 hemiacetal (in aldohexose) ● Or between C-2 dan C-5 hemiketal (in ketohexose) carbonyl carbon becomes new chiral centre = anomeric carbon
Cyclic sugars – 2 different forms - α dan β =
Anomers 163
164
Hemiacetal & Hemiketal
165
Drawing the Cyclic Structure for Glucose New Chiral centre
Glucose 166
Drawing the Cyclic Structure for Fructose New Chiral centre 1 2 6 3 4 5
2 5 4
3
1
6
Fructose 167
168
169
170
CYCLICAL STRUCTURES ● CHO 5 carbon = furanose – furan ● CHO 6 carbon = pyranose – Pyran ● Normally CHO > 5 carbon – in cyclic form
o Free carbonyl group can form α @ β anomer o Can change from one form to another 171
172
173
Furanose
174
pyranose
175
Which isomer used for reaction? ● Certain reactions – any isomer ● Others – only one anomer eg. RNA & DNA – requires α-D-ribose & β D-deoksiribose • Fischer Projection – usefull to explain `stereochemsitry’ sugars, But does not give true picture of overall shape accurately. ∴ use HAWORTH PROJECTION FORMULA 176
Haworth Projection Formula Fischer Projection Formula
α or β? 177
178
179
Important Simple Monosaccharides
1. Glucose 2. Mannose 3. Galactose 4. Fructose 5. Ribose
180
RNA - only ribofuranose Dinding sel (polysaccharide) - pyranose KETOPENTOSE – almost all in cyclic form, but only furanose eg. a -D-ribulose
181
182
REACTIONS OF MONOSACCHARIDES 1. Mutarotation. 2. Oxidation to CO2 + H2O Reactions due to aldehyde group 3. Reducing sugars. 4. Reduction to polyols. Reactions due to alcohol group 5. Esterification. 6. Formation of acetals, also called glycosides
183
184
REACTIONS OF MONOSACCHARIDES 1. Oxidation & Reduction ● Important in biochemistry – provides energy when CHO completely oxidised ● Photosynthesis – reversible process – when CO2 & H2O reduced ● Oxidation reaction can be used to detect the presence of carbohydrates eg. Aldehyde [O] carboxyl – basis for test for aldose When aldehyde is oxidised, the oxidising agent is reduced 185
Because of his property, they are called reducing agents. Ketose is also a reducing agent – Y? Ketoses can also be reducing sugars because they can isomerise (a tautomerisation) to aldoses via an enediol:
186
REDUCING SUGARS ● Sugars that contain aldehyde groups that are oxidised to carboxylic acids are classified as reducing sugars. ● They are classified as reducing sugars since they reduce the Cu2+ to Cu+ which forms as a red precipitate, copper (I) oxide.
187
● Common test reagents are : Benedicts reagent (CuSO4 / citrate) Fehlings reagent (CuSO4 / tartrate) hemi-acetal ● Remember that aldehydes (and hence aldoses) are readily oxidised ● In order for oxidation to occur, the cyclic form must first ring-open to give the reactive aldehyde (?) ● So any sugar that contains a hemi-acetal will be a reducing sugar. ● But glycosides which are acetals are not reducing sugars. 188
189
Another example of reagent for reducing sugars – Tollen’s Reagent Use silver ammonium complex Ag (NH3)2+ as oxidising agent mirror precipitate on the walls of test tube RCHO + 2Ag(NH3)2+ + OH- RCOO- + 2 Ag+ + 3NH3 + NH4+ + H2O
Other methods – use enzyme – glucose oxidase – to detect glucose 190
ESTERIFICATION REACTION ● Hydroxyl group (OH) in CHO reacts with acids ester ● E.g. Phosphate ester – intermediate in the breakdown of CHO energy ● Phosphate ester is normally formed when the phosphate from ATP reacts with sugars phosphorylated sugar – important in the metabolism of CHO
191
MONASACCHARIDE DERIVATIVES Monosaccharides have several OH groups – can bind/exchange with other groups modify the original structures
A. Phosphate esters Phosphate esters – found in many metabolsic pathways - ATP, ADP, etc
B. Acid & lactones Oxidation of monosaccharides produces 1. Aldonic acids – e.g. aldose reacting with alkaline solution of Cu+ 2. Lactone & uronic acids - [O] with enzim 192
Mannose
193
194
C. Alditols Reduction of C=O polyhydroxy compounds = eg Dmannitol & D-glucitol
Ribitol or Adonitol is a crystalline pentose alcohol (C5H12O5) formed by the reduction of ribose. It occurs naturally in the plant Adonis vernalis
195
196
● Mannitol or 1,2,3,4,5,6-hexanehexol (C6H8(OH)6) is a vasodilator which is used mainly to reduce pressure in the cranium, ●
Chemically, mannitol is an alcohol and a sugar, or a polyol; it is similar to xylitol or sorbitol.
● Mannitol is also used as a sweetener for people with diabetes.
197
198
D. Amino sugars ● 2 derivatives of amino sugars – in polysaccharides – glucosamine & galactosamine ● Exchange of OH with NH2
199
200
E. Glycoside ●
glycosides are certain molecules in which a sugar part is bound to some other part
●
Bond = glycosidic bond
●
Formed when H2O is removed from OH – of saccharide & other compounds containing OH-
●
Found in plants/animals
●
e.g. - Salicin, a glycoside related to Aspirin
NB: OH- must be attached to anomeric carbon 201
Salicin, a glycoside related to Aspirin
202
203
● OH- from sugar with another OH- ether bond ● OH- - must be anomeric ● bond= glycosidic bond ● product = glycoside furanose = furanoside, pyranose = pyranoside.
204
OLIGOSACCHARIDES Glycosidic bonds between monosaccharides = basis for the formation of oligosaccharides and polysaccharides Bonds = between a anomer or b anomer and another OH- of another sugar lots of combinations - OH- must be numbered to differentiate notation for glycosidic bond – which anomeric atom involved e.g.. α(1 → 4), α(1 → 6), β (1 → 1) 205
206
OLIGOSACCHARIDES Formed when two monosaccharides are joined together by glycosidic bond Plays important roles in living organisms Simplest and most important oligosaccharide = disaccharide Examples of disaccharide = sucrose, lactose, maltose
207
4 important characteristics to differentiate disaccharides 1. monomer found and their configuration 2. which carbon involved in the bond 3. the arrangement of the monomers 4. the anomeric configuration of the hydroxyl group
208
Chemical characteristics of poly- & oligosaccharides formed will depend on 1. Chemical characteristics of monosaccahrides 2. The type of glycosidic bonds formed (i.e. which anonmer; which carbon atom etc ) e.g. The differences between celluloses and starch is because of the difference in the glycosidic bond formed between glucose Because of the variation in the glycosidic bond, can get various types of polymers – branched or linear 209
210
211
Bond: β (1 → 4)
212
213
214
215
216
217
Example. ● Arrangement – starts with non-reducing end –left – anomeric & enantiomeric – with prefixes ● Cyclic configuration with suffix ● Atoms between glycosidic bonds – the number inside bracket between residues Not all oligosaccharides are dimeric; also possible to have trimers, tetramer and bigger
218
219
Sucrose ● 2 Monosaccharides = α−D-Glucose & β-D- fructose ● Glucose = aldohexose = pyranose ; Fructosee = ketohexose = furanose ● α-C-1glucose attached to fructose ● Not reducing sugar - why? – because both anomeric groups are involved in glycosidic bond ● But free glucose and fructose are reducing sugars. When sucrose is digested, hydrolysed glucose & frctose energy
220
221
Lactose ● A disaccharide formed from β-D- galactose & βD-glucose ● Galctose = epimer of glucose - ie. Reverse position at C-4 ●
Glycosidic bond = β(1 4) between anomeric C1 ( β form) of galactose and C-4 carbon of glucose
222
Because anomeric carbon of glucose is NOT involved in the bond formation, can be in either α &β Lactose = reducing sugar because the groups at the anomeric carbon atom (glucose) is not involved in glycosidic bond formation; hence can react with oxidising agents
223
224
LACTOSE INTOLERANCE Lactose= milk sugar Human can be allergic to milk or milk products – why??? • lack of lactase (breaks lactose to galactose and glucose) lactose will accumulate •
Lactose will be acted on by lactase bacteria produce Hydrogen gas, CO2 & organic acids – problems with digestion – bloating and diarrhea
Although lactose can be degraded to galactose, galactose has to be isomerised to glucose before being absorbed can accumulate GALACTOSEMIA – mental retardation 225
226
Maltose a disaccharide – hydrolytic product of starch 2 molecules of D-glucose -α α-D-glucose & β-D-glucose joined by α(1 4) bond Different from celobiose o
hydrolysis of cellulose
o
different glycosidic bond - D-glucose attached through β(1 4)bond
o
maltose can be digested by humans, cellobiose cannot 227
228
Epimers are diastereomers that differ in configuration of only one stereogenic center Diastereomers are a class of stereoisomers that are nonsuperposable, non-mirror images of one another
229
Epimers 230
POLYSACCAHARIDE various functions sequence of monmer determines the primary structure – normally simple monomers ●
1 type of monomer = homopolysaccharide
●
2 @ more = heteropolysaccharide
●
normally not complex – not more than 2 residues
●
Cf. Protein & nucleic acids - well defined length’ – polysaccharide chain - random length -
231
Storage Polysaccharide 1.
Important examples - amylose & amylopectin - starch in plants and & glycogen in mammals and bacterial cells
2.
amylose, amylopectin & glycogen = homopolysaccharide (glucan) - deposited in the liver, otot = polymer α-D-glucopyranoseDifference – bond between residues
232
● amylose - linear, α(1 4) ● Amylopectin & glycogen α(1 4) + α(1 6), branched polymer ● Glycogen - more branched; if not they are very similar
233
234
•
Regular and simple structure regular secondary structure
•
α(1 4) bond , each residue leans slightly compared to the previous resisue helical conformation
•
However, the helix is not stable- e.g. amylose form random coils
•
Iodine – can stabilise helix – because it can fit the core of the helix
•
Complex – blue in colour
235
236
Glycogen & amylopectin – cannot get blue colour ??? ● Branches – inhibit formation of helix ● To form a helix – need 12 residue for ever turn ● amylopectin - 10 -12 residues, glycogen - 8 residue – not enough
237
238
239
STRUCTURAL POLYSACCHARIDE Plants – do not use/synthesise structural protein
use specialIsed polysaccharide
animal – use both
Cellulose main polymer in plants - woody/fibrous Polymer linear - D-glucose ( glucan) Joined through β(1 4)bond 240
241
• Animals- can digest starch – can cleave bond • Cellulose - cannot – requires symbiotic bacteria - produce enzyme cellulase
• Ruminant - OK- cellulase is present • White ants – protozoa – can digest cellulse • Fungi - eg. mushroom – live on rottting wood, etc Other polysaccharides also found. e.g. xylans = polymer β(1→ 4) - linked Dxylopyranose Glucomanan = hemicellulose
242
● Cellulose – Not confined to plants only ● Marine Invetebrata - eg. Tunicates - cellulose in the mantle ● in connective tissue - human
243
Tunicates 244
Tunicates
245
246
CHITIN
•
Similar to cellulose
•
smalll difference - homopolymer N-asetil-Dglucosamine - minor constituent in fungi and algae- replace cellulose
•
Role in invertebrate - exoskeleton of arthropod & mollusks
247
248
249
250
GLYCOSAMINOGLYCAN In vertebrata - previously known as mucopolysacharide ● chondroitin sulphate ● keratan sulphate connective tissue ● dermatan sulphate – skin ● Hyaluronic acid
1. All are polymers = repeating units of disaccharides 2. Sugar (CHO) = N-asetylgalactosamine @ Nasetylglucosemine
251
252
253
Main functions of glycosaminoglycan ● formationn of matrix that bind protein components with connective tissue eg. Proteoglycan – inside cartilage - filemental structure synthesised from hyaluronic acid with protein core ● Protein - keratin sulphate chain and chondroitin sulphate chains are attached Structure - collagen fibers becomes compact and strong ● Bonds – electrostatic between sulphate and basic collagen side chains
254
255
Non Structural function of glycosaminoglycan Hyaluronic acid ● very soluble in water- found in-synovial fluid- lubricating agent for joints ● vitreous humor – eyes – agent for increasing fluidity
Heparin - anticoagulant – in body tissues – bound strongly to blood proteins (prothrombin III) – prevets enziymes in the coagulation of blood
256
Polysaccharides in bacterial cell walls ● Gram + & Gram – based on cell wall ● Gram + = peptidoglycan = polysaccharide – peptide complex which are multilayered and crosslinked ● Gram - = single layer of peptidoglycan covered with membrane layer
257
Gram + bacteria 258
Gram - bacteria 259
Importance of other peptidoglycas ● A few antibiotics inhibit bacterial growth by preventing peptidoglycan layers ● Lysozymes can dissolve cell walls cell lysis bacterial death ● Also found in bacteriophage, white of egg, eyedrops of humans
260
GLYCOPROTEIN ●
Many proteins are bound to saccharide = glycoprotein
●
Different functions
●
Saccharide chain (= glycan) bound to protein -2 ways ◊ Bound to N of the amino group of asparagine - (NIinked Glycans) ◊ Through N-acetylglycosmine @ N-acetylgalactosamine
Differet structures – complex branched structures Different functions – protein indicators – old proteins that need to be destroyed
261
ABO blood group system, the classification of human blood based on the inherited properties of •
red blood cells (erythrocytes) as determined by the presence or absence of the antigens A and B, which are carried on the surface of the red cells.
•
These antigens may be proteins, carbohydrates, glycoproteins, or glycolipids, depending on the blood group system
•
Persons may thus have type A, type B, type O, or type AB blood
262
263
eg. immunoglobin – sialic acid residues will be cleaved slowly, then receptor will recognise and bind to the protein, engulf the protein
• O-linked Glycans Different functions – e.g.. ● Antartic fish – have glycoprotein which acts as an “anti freez’ – fluids do not freeze although temperature below freezing point ● Mucins - glycoprotein – in salive – increase viscosity of fluid 264
Humans can produce antibodies against the A & B, but NOT O; i.e. O is antigenic antigenic Normally, antibodies react against other antigens. eg. Type A carries antibodies against B - therefore when receiving blood type B - will clot & precipitate Blood type O carries antibodies against A & B ∴cannot receive blood type A and B; but CAN donate to both Bllod Type AB – carries A & B antigens; therefore no antibodies against A @ B ; only donatieto AB only
265
Oligosaccharides as CELL MARKERS Blood group antigens – a cell recognition phenomenon Cells need to be marked –(on the surface) so that they can interact with other cells – can recognize own cells from other external cells In animals – there a layer of saccharides bound to a protein or lupid in the membranes – e.g. glycocalyx - can interact with bacteria in the intestine; collagen To act as signal, need to have a specific protein bound specifically - imunoglobulin Other examples – lectin – interact between cells and proteins in the intercellulr matrix – to maintain the structure of tissues and organs 266
267
Amino acid & Protein -1
Amino acids and proteins • Draw a general amino acid and identify the two functional groups common to all. • Classify each amino acid according to the chemical nature of its R group. • Define the meaning of an essential amino acid. • Draw the reaction that joins two amino acids to form a peptide bond. • Describe and differentiate primary, secondary, tertiary, and quaternary protein structures. • Describe and differentiate co-enzymes and prosthetic groups. • List and discuss four forces that stabilize globular protein structure.
Proteins Reverse transcription
DNA
Translation
RNA
protein
Transcription
Genotype Genome
Phenotype Proteome
(similar in all cells)
(unique to all cells)
Amino Acid Structure R H
+
Cα
H N H H
C O O
Common Amino Acids in Proteins
Common Amino Acids in Proteins
Amino Acid R-groups Polar Charged
Uncharged
Arginine (+) Glutamic acid (-) Aspartic Acid (-) Lysine (+) Histidine (+)
Cysteine Proline Serine Glutamine Asparagine
Non-Polar Hydrophobi c Tryptophan Phenylalanine Isoleucine Tyrosine Leucine Valine Ambivalent Methionine Glycine Threonine Alanine
Hydrophobic Indexes • • • • • •
Arginine Arg [R] -11.2 Glutamic Acid Glu [E] -9.9 Aspartic Acid Asp [D] -7.4 Lysine Lys [K] -4.2 Histidine His [H] -3.3 Cysteine Cys [C] -2.8
• • • •
Proline Serine Glutamine Asparagine
Pro [P] Ser [S] Gln [Q] Asn [N]
-0.5 -0.3 -0.3 -0.2
• Glycine • Threonine • Alanine • • • • • • •
Gly [G] 0 Thr [T] 0.4 Ala [A] 0.5
Methionine Met [M] 1.3 Valine Val [V] 1.5 Leucine Leu [L] 1.8 Tyrosine Tyr [Y] 2.3 Isoleucine Ile [I] 2.5 Phenylalanine Phe [F] 2.5 Tryptophan Trp [W] 3.4
Essential amino acids • Definition - Those amino acids that cannot be synthesized in the body in sufficient quantities for anabolic needs. • In humans, Isoleucine Tryptophan Lysine Phenylalanine
Leucine Methionine
Valine
Threonine
Histidine
Zwitterion Form of Amino Acids H
+
H3N
CO2
C R
“zwitterion” = hybrid ion form that exists at pH = 7
Recall from Acid-Base Equilibria: Henderson-Hasselbalch Equation -
[X ] pH = pKa + log [HX] or
[base] pH = pKa + log [acid]
Amino acid & Protein -2
Recall: Amino Acids are Chiral*
*except glycine - R group is another H
A zwitterion is formed when a proton (a hydrogen nucleus) moves from the carboxylic acid group to the amine group H
O
O C
-O
O C
proto R C H R C H + n, H , transf + H N H N ers to H H the H amin e group The zwitterion ion forms when an amino acid is dissolved in water – and this is how they are usually found in nature
Alanine is an amino acid. Its molecular formula is C3H7O2N Draw the structural formula for alanine. Click to see if you got it right Make a molecular model of alanine and use it to H O O C H 3C
C
H
2. show how its zwitterion is formed
N H
1. investigate the shape of the molecule and explain why can it exist in two forms
H
Click to continu e
Click to continue
Look at the picture of the zwitterion.
Replay Close window
1. where is the – NH
+
group?
The shapes of some amino acid molecules O
O
O H3C OH
H2N OH
NH2
HS
OH NH2
glycin alanini cystei e ne ne Click for a useful website that allows you to compare the shapes of amino acids
Formation of the peptide bond O
O
R1
R2
OH
OH
NH2
NH2
O
The molecules must be orientated so that the carboxylic acid group of one can react with the amine group of the other
R2
R1 OH
H2N O
NH2
Two amino acid molecules; the nature of the R group (R1 and R2) determines the amino acid
HO
O R2
R1
H2O
NH O
NH2 HO
The peptide bond forms with the elimination of a water molecule; molecule; it is another example of a condensation reaction
Hydrolysis of the peptide bond
The peptide bond holds two amino acid ‘residues’ together. It is a flat, rigid group A water molecule reacts with this group
The two amino acids form or, if the peptide bond is somewhere in a long peptide chain, two smaller peptide molecules are formed
O R2
R1
H2O
NH O
NH2 HO
O R2
R1 OH
H2N O
NH2 HO
Bonding between peptide chains Bonding within a peptide chain (intramolecular) and between one chain and another (intermolecular) C
Hydrogen bonds These form in all proteins. The hydrogen atom of the peptide link is attracted to the oxygen of another peptide link.
C H
O
H
O
N
N peptide chains
Covalent bonds In a very small number of proteins, sulfursulfur-sulfur covalent bonds (also called cystine bonds or disulfide bridges) bridges) are present.
- CH2 – S – S – CH2 -
Ionic bonds If some of the amino acids in the proteins have carboxylic acid or amine side groups, an ionic bond can form.
- COO-
H 3N+ -
Amino acid & Protein -3
Types of Proteins • • • • • • •
Type Examples Structural tendons, cartilage, hair, nails Contractile muscles Transport hemoglobin Storage milk Hormonal insulin, growth hormone Enzyme catalyzes reactions in cells Protection immune response 314
Amino Acids • • • •
Building blocks of proteins Carboxylic acid group Amino group Side group R gives unique characteristics R
side chain
I H2H—C —COOH I H
315
Examples of Amino Acids H I H2N—C —COOH I H I
glycine
CH3
H2N—C —COOH I H
alanine
316
Types of Amino Acids Nonpolar
R = H, CH3, alkyl groups, aromatic O Polar ll R = –CH2OH, –CH2SH, –CH2C–NH2, (polar groups with –O-, -SH, -N-) Polar/Acidic R = –CH2COOH, or -COOH Polar/ Basic R = –CH2CH2NH2 317
L-Form Amino Acid Structure
Carboxylic group
COO
-
Amino group
+ H3 N R group
α
H H = Glycine CH3 = Alanine Juang RH (2004) BCbasics
Mirror Images of Amino Acid
α
Mirror image
α
Same chemical properties Stereo isomers Juang RH (2004) BCbasics
Northwest line Chung-San line -C-C-C-N-C-N
=
Aromatic Trp W -C-
N+
Basic Arg R
This is NOT a metabolic pathway
N
-C-
-OH
Tyr Y
Lys K +
-C-C-CONH2
Asn N
Gln Q Amide
Asp D
Glu E Acidic
-C-COOH
-C-C-COOH
-C-
-C-C C N N+
His H
Central line Gly G
Phe F
Ala A A
Val V
Ile I
-CH3
C C -C
C -C-C-C
-H -C-OH
Ser S
Cys C
-C-SH
Circular line
South line Non-polar Polar
Nan-Kan line
-C-CONH2
-C-C OH
Thr T
Met M
Hydroxy
Sulfur
-C-C-S-C
Aliphatic Leu L C -C-C-C C
C C HN C-COOH α
Pro P
Imino, Circular
Juang RH (2004) BCbasics
-C-C-C-C-NH3
Amino Acid Subway Map
Classification of Amino Acids by Polarity
NONPOLAR
POLAR
Acidic
Neutral
Basic
Asp
Asn Ser Arg Cys Tyr His Gln Thr Lys Glu Gly Ala Ile Phe Trp Val Leu Met Pro
Polar or non-polar, it is the bases of the amino acid properties. Juang RH (2003) Biochemistry
Formation of Peptide Bonds by Dehydration Amino acids are connected head to tail NH2
1
COOH
NH2
2
COOH
Dehydration
Carbodiimide
-H2O
O NH2
1
C N
2
COOH
H Juang RH (2004) BCbasics
Peptide Bond Is Rigid and Planar
C
H
C O
N C
Juang RH (2004) BCbasics
Protein Structure -4
Proteins Reverse transcription
DNA
Translation
RNA
protein
Transcription
Genotype Genome
Phenotype Proteome
(similar in all cells)
(unique to all cells)
Peptide bond formation O Aspartate C H H
C
O
H Alanine H
H
Cα
C Cα
H
H
+ - H N H N H C O H C O H H O O +
condensation H2O
Peptide bond formation O
O C H H
+
C
H
C
H
Cα
H N H C H O
H Cα
H
N H C O O
Peptide bond
Primary Structure
H
Dipeptide Peptide bond resonance O
O C
H H
C Cα
H
H
C
H
H Cα
H
H N H C N H C O + H O O+
Peptide bond
α-helixes
Intrachain H-bonds
Secondary Structure
β-strands
Interchain H-bonds
Secondary Structure
Tertiary structure
Hb monomer (or myoglobin)
Quaternary structure
Hb α2β2 tetramer
Protein Structure Primary structure is the amino acid sequence. Secondary structure is how the amino acids in sequence fold up locally. Examples are αhelixes and β-strands and loops. Tertiary structure is the 3-dimensional folding of the secondary structural elements and connecting loops in space. Quaternary structure is the association of multiple subunits, each with a tertiary structure and each a unique gene product.
Stabilization of Protein Structure
Electrostatic interactions involve the interaction of (+) and (-) charged side groups. Hydrogen bonds involve sharing of a hydrogen atom between two eletronegative atoms (e.g., O, N). Van der Waal’s forces are weak forces based on optimal overlap of adjacent electronic orbitals. Can be repulsive. Hydrophobic interactions are, by far, the most powerful force stabilizing protein structure. Basis of force is entropy gain realized by burying hydrophobic residues.
Hierarchy of Protein Structure • 20 different amino acids: many combinations
Primary Structure The order of amino acids: Protein sequence
Secondary Structure Conformation varies depending on sequence
Tertiary Structure Overall structure of the chain in full 3D
Give the name and structure of at least 2 examples of each of the following: a. heterocyclic amino acid b. aromatic amino acid c. neutral amino acid d. acidic amino acid e. basic amino acid f. sulfur containing amino acid
Secondary Structure Local structure of consecutive amino acids Common regular secondary structures α Helix β Sheet β turn Amino acids have a greater propensity to form some secondary structures versus others Chemical properties Spatial constraints
Secondary StructureStructure- α Helix
H-bond
Secondary StructureStructure- β Sheet
Oxygen
Nitrogen
Hydrogen
Carbonyl C
Carbon α
R Group
H Bond
What Causes 2º, 3º, 4º Structure? • intermolecular forces peptide hydrogen bonds side-chain hydrogen bonds salt bridges London dispersion forces • metal ion coordination • disulfide bridges (covalent interaction)
Forces in Proteins
Some Ways to Denature Proteins • heat • pH changes • chemical reagents urea urea DMSO HMPA 2-mercaptoethanol
Give the name and structure of at least 2 examples of each of the following: a. heterocyclic amino acid b. aromatic amino acid c. neutral amino acid d. acidic amino acid e. basic amino acid f. sulfur containing amino acid
Show me a chemical structure of a tripeptide with 1 nonpolar,1 polar, and 1 charge amino acids? Ensure that the charged amino acid is at the C-terminal and the polar amino acid is at the N-terminal.
1. How many peptide bonds are there 2. What are amino acids? 3. How many α-carbon can you find? 4. How many chiral centers? 5. In a solution of these oligopeptides, What sort of interactions can develop?
Protein Structure - 5
Hierarchy of Protein Structure • 20 different amino acids: many combinations
Primary Structure The order of amino acids: Protein sequence
Secondary Structure Conformation varies depending on sequence
Tertiary Structure Overall structure of the chain in full 3D
Protein Sequences Amino terminus
Carboxyl terminus
N Residue number 1
C 2
3
4
• 20 different amino acids: many combinations
A protein of n residues 20n possible sequences! 100 residue protein has 10020 possibilities 1.3 X 10130! The latest estimates indicate on 40,000 sequences in the human genome −> THERE MUST BE RULES!
Protein Sequences Amino terminus
Carboxyl terminus
N Residue number 1
C 2
3
4
• 20 different amino acids: many combinations
A protein of n residues 20n possible sequences! 100 residue protein has 10020 possibilities 1.3 X 10130! The latest estimates indicate on 40,000 sequences in the human genome −> THERE MUST BE RULES!
Protein Sequences *Length is generally 100-1000 residues* Minimum length for performing a function ~40 Molecular machines not perfect- errors Function requires specific amino acid properties Not all amino acids are equally useful Abundant: Leu, Ala, Gly, Ser, Val, Glu Rare: Trp, Cys, Met, His Post-translational modifications Addition of co-factors- metals, hemes Chemical modification- phosphate, glycos.
Protein Sequences The pattern of amino acid side chains determines the secondary (and tertiary!!) structure
*Pattern is more important than exact sequence* Reporting/Comparing Protein Sequences h-CaM A T V R L L E W E D L b-CaM A T V R L L E Y K D L 5
10
conservative non-conservative
Secondary Structure Local structure of consecutive amino acids Common regular secondary structures α Helix β Sheet β turn Amino acids have a greater propensity to form some secondary structures versus others Chemical properties Spatial constraints
The Peptide Bond Peptide plane is flat ω angle ~180º Partial double-bond:
Peptide bond
-
H -C = NO-
-
=
-
H -C - NO
Resonance structures
Βackbone Conformation φ
R
H
ψ
Cα
Peptide planes ω angle is fixed, φ and ψ angles vary Many φ/ψ ψ combinations cause atoms to collide
Βackbone Conformation φ
R
H
ψ
Cα Cα H
R
Cα H
R
Side chains collision also limit φ/ψ ψ combinations Backbone restricted → Secondary structure limited
Specific Secondary Structures Three acceptable backbone φ/ψ φ/ψ combinations 1. Right-hand helix: α-helix (-40°, -60°) 2. Extended: antiparallel β-sheet (140°, -140°) 3. Left-hand helix (rare rare): α-helix (45°, 45°) Glycine: special because it has no side chain! Side chains positioned to minimize collisions → amino acids prefer specific secondary structures Hydrogen bonds between backbone atoms provide unique stability to secondary structures
Secondary StructureStructure- β Sheet
Oxygen
Nitrogen
Hydrogen
Carbonyl C
Carbon α
R Group
H Bond
Secondary StructureStructure- α Helix
H-bond
Secondary StructureStructure- β Turn 3
4
2
1
Reverses direction of the chain
Titration of Amino Acids -6
Titration curves Of amino acids and weak acids(acetic acid)
Titration • Titration curves are produced by monitoring the pH of given volume of a sample solution after successive addition of acid or alkali • The curves are usually plots of pH against the volume of titrant added or more correctly against the number of equivalents added per mole of the sample
Titration of acetic acid • At the starting point the acid form predominates (CH3COOH). • As strong base is added (e.g. NaOH), the acid is converted to its conjugate base. • At the mid point of the titration, where pH=pK, the concentrations of the acid and the conjugate base are equal. • At the end point(equivalence point), the conjugate base predominates, and the total amount of OH added is equivalent to the amount of acid that was present in the starting point.
Titration
Titration Determination of pKa values: pKa values can be obtained from the titration data by the following methods: 1. The pH at the point of inflection is the pKa value and this may be read directly 2. By definition the pKa value is equal to the pH at which the acid is half titrated. The pKa can therefore be obtained from the knowledge of the end point of the titration.
Titration of amino acids • Titration of glycine • Titration of arginine
Titration • When an amino acid is dissolved in water it exists predominantly in the isoelectric form. • Upon titration with acid, it acts as a base, and upon titration with base, it acts as an acid( a compound that can act as either an acid or a base is known as an amphoteric compound).
• +H3N-CH2-COO- + HCl +H3N-CH2-COOH + Cl(base) (acid) (1) +H N-CH -COO3 2
+ NaOH H2N-CH2-COO- + Na+ +H2O
(acid) (base) (2) In this experiment, the amino acid represents either the A- or the HA form in the Henderson-Hasselbalch equation, depending on the titration.
Acid–base properties • All of the amino acids have an acidic group (COOH) and a basic group (NH2) attached to the α carbon. • Two of the amino acids have acidic side chains: aspartate and glutamate. • Three of the amino acids have basic side chains: arginine, histidine, and lysine.
• All amino acids contain ionizable groups that act as weak acids or bases, giving off or taking on protons when the pH is altered. These ionizations follow the HendersonHasselbalch equation: pH=pKa+log [unprotonated form(base)] [protonated form (acid) ]
• When the conc of the unprotonated form equals that of the unprotonated form, the ratio of their concentrations equals 1, and log 1=0. • Hence, pKa can be defined as the pH at which the concentrations of the protonated and unprotonated forms of a particular ionizable species are equal. • The pKa also equals the pH at which the ionizable group is at its best buffering capacity; that is the pH at which the solution resists changes in pH most effectively.
• Consider applying the Henderson-Hasselbalch equation to the titration of glycine with acid and base. • Glycine has two ionizable groups: a corboxyl group and an amino group, with pKa values of 2.4 and 9.6 respectively. • In water at pH 6, glycine exists as a dipolar ion, or zwitterion, in which the carboxyl group is unprotonated(-COO- ) and the amino group is protonated to give the substituted ammonium ion(-NH3+).
• Addition of acid to the solution lowers the pH rapidly at first and then more slowly as the buffering action of the carboxyl is exerted. • At pH 2.4 the pKa is reached, one-half the acid has been consumed, and the carboxyl group is half ionized and is most effective as a buffer. • Titration of the amino group with base follows a similar curve into the alkaline region. • The intersection between the titration of the carboxyl group and the titration of the amino group describes in this case the point at which glycine has no net charge, and is called the isoelectric point (pI).
The isoelectric point (pI) • the isoelectric point, pI, is the pH of an aqueous solution of an amino acid at which the molecules have no net charge. In other words, the positively charged groups are exactly balanced by the negatively charged groups. • For simple amino acids such as alanine, the pI is an average of the pKa's of the carboxyl (2.34) and ammonium (9.69) groups. Thus, the pI for alanine is calculated to be: (2.34 + 9.69)/2 = 6.02. • If additional acidic or basic groups are present as side-chain functions, the pI is the average of the pKa's of the two most similar acids.
Cont.. (pI) • In the case of aspartic acid, the similar acids are the alpha-carboxyl function (pKa = 2.1) and the side-chain carboxyl function (pKa = 3.9), so pI = (2.1 + 3.9)/2 = 3.0. • For arginine, the similar acids are the guanidinium species on the side-chain (pKa = 12.5) and the alpha-ammonium function (pKa = 9.0), so the calculated pI = (12.5 + 9.0)/2 = 10.75.
• Most amino acids contain carboxyl and amino groups having pKa values similar to those of glycine. • In addition to these groups, many amino acids contain other ionizable groups, which introduce other “steps” or pKa values into their titration curves.
Titration curves • The pK is the pH at the midpoint of the buffering region (where the pH changes only slightly upon addition of either acid or base). • The pK is the pH corresponding to the inflection point in the titration curve. • The end point of a titration curve represents the observed end of the titration. • The isoelectric point (isoelectric pH; pI) is the pH at which the amino acid has a net zero charge. For a simple diprotic amino acid, the pI falls halfway between the two pK values. For acidic amino acids, the pI is given by ½(pK1 + pK2) and for basic amino acids it’s given by ½(pK2 + pK3)
Nucleic acids -1 Purines
Purine metabolism
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