Biochemistry 1 - Carbohydrates PDF

BIOCHEMISTRY LESSON I CARBOHYDRATES Macromolecules are composed of monomers which are linked together to create polymers. So monomers are represented as the building block of polymer which is defined as a long m...

BIOCHEMISTRY LESSON I CARBOHYDRATES Macromolecules are composed of monomers which are linked together to create polymers. So monomers are represented as the building block of polymer which is defined as a long molecule consisting of similar building blocks. We know that there are four different classes of biological molecules: 1) Lipids; 2) Carbodydrates; 3) Proteins; 4) Nucleic acids. But it is important to highlight that one of these does not derive from the polymerization of different monomers: the lipids. Lipids are just characterized by hydrophobicity which represents the common feature of lipids. Carbohydrates Literally means hydrates of carbon, but actually carbohydrates may be defined as polyhydroxy aldehydes or polyhydroxy ketones or compounds producing them by hydrolysis. The term sugars is utilized just for those carbohydrates soluble in water and sweet in taste. Carbohydrates are characterized by many functions: 1) Source of energy: for example, glucose is the principal fuel of metabolism, supplying energy to all living cells, both plants and animals; 2) Storage form of energy: Also it serves as the storage form of energy (starch in plants and glycogen in animals) to satisfy the immediate energy demands of the organism. 1  3) Structural component: The form structural components of many organisms such as: - The fibers of plants (cellulose); - Exoskeleton of some insects or crustaceans (chitin); - The cell wall of microorganism (bacteria). 4) RNA AND DNA COSTINTUENT: in the. Case of ribose and 2-deoxyribose; 5) Conjugated with proteins and lipids; 6) Involved in cell-cell interaction and Recognition events They are utilized also as raw materials for several industries (paper, plastic, ecc..). The basic composition of the simplest sugar, the monosaccharide, follows this formula. (CH20)n where n ≥ 3 so we can expect that the simplest sugar is made up of 3 carbon atoms. We can classify monosaccharides using different classification based on:  The number of carbon atoms (3, 4, 5, 6, 7)  On the function group - Aldoses= have an aldehyde group at C1; - Ketoses= have a ketone group at C2. We have to remember these structure and also to be able to write them down. They are the simplest sugars with 3 carbons. (trioses). If you want to make these molecules longer when we have to put the CH20H groups? Or if I start from a ketone structure, we can create the elongation starting from the third carbon creating for example a 6 atoms carbon sugars such as fructose. 2 Is it possible to easily distinguish the correct version, that is, the one used in our metabolism, of glucose and fructose? No because we have to take into account the isomers. In fact, we need to specify which type of stereoisomer, as enzymes in our metabolism work with specific forms of glucose or fructose. For example, the first enzyme of glycolysis, hexokinase, which transfers a phosphate group from ATP to glucose, forming glucose-6-phosphate, acts on alpha-D-glucose and not on a generic form of glucose." 3 As we can see in the picture above, D-Glyceraldehyde, D-Ribose, D-Glucose, D-Mannose and D- Galactose are the most common sugars in our metabolism but in any case, the aldose family is characterized by the D stereoisomers (in animal cells). Also in the case of the ketose family we have the phenomenon of D stereoisomerism except for dihydroxyacetone. We will see why later. When two monosaccharides are covalently linked we have the disaccharides: - Maltose -> glucose + glucose; - Sucrose -> glucose + fructose; - Lactose -> glucose + galactose. OLIGOSACCHARIDES instead are made up of 2 to 10-15 monosaccharidic residues covalently linked. POLYSACCHARIDES (glycans) are polymers consisting of long chains of monosaccharide (about 100 sugars monomers) or disaccharide units. They have high molecular weights (up to 106 dalton). They are generally tasteless (they are not sugars1) and form a colloidal solution with water. We can distinguish them into:  HOMOPOLYSACCHARIDES: (monosaccharides of a single type)  HETEROPOLYSACCHARIDES (monosaccharides of 2 or more different types, or their derivatives) 1 Disaccharides are sugars! 4  What is this? It is the D-glucose, in fact as we can see we have 6 carbon atoms with an aldehyde group and this is a sugar. Nomenclature of monosaccharides The nomenclature of monosaccharides is directly related to the CONFIGURATION OF EACH ASYMMETRIC CARBON. This is the first important concept. In fact, we have to remember something about organic chemistry in order to define which kind of stereoisomers we deal with: D or L. We know that carbon can form a maximum of 4 bonds while the minimum number of bonds is 2 but with double bonds (?). It depends on the hybridization of carbon. The asymmetric or chiral carbon is the carbon when attached to 4 groups of atoms. So in order to define the stereoisomer we have at first to identify the chiral carbon. Considering the Aldo triose glyceraldehyde, having just 1 asymmetric carbon (C*) in second position, we can have 2 configurations named as D & L. 5 In these Fischer projections we can see how in the second position we have a chiral carbon atom bonded to 4 different groups of atoms. This allows us to distinguish two different stereoisomers: D and L. So for each chiral carbon we have different stereoisomers. The simplest way to designate the different forms of stereoisomers is as follows:  if the hydroxyl group of the chiral carbon is on the right, we have a D-stereoisomer; conversely;  if it is on the left, we have an L-stereoisomer. In sugars with more than one chiral center there is a rule to the define the type of stereoisomer: D or L refers to the position of the hydroxyl group in the asymmetric C farthest from the aldehydic, or ketonic, group. (D = OH group is on the right) As we can see in the picture above, the fifth carbon of glucose and fructose is the one farthest from the functional group. It is important to highlight that most naturally occurring sugars are D isomers, in fact L-glucose is not involved in our metabolism. 6 The second important concept is related to the definition of stereoisomers. They are the compounds that have the same structural formulae but differ in their spatial arrangement. This is the case of hexoses that have the same chemical formula C6H12O6 and exist different stereoisomers having exclusive names: GLUCOSE, MANNOSE, GALACTOSE, etc. What is the difference between these hexoses? Only the D and L configuration? No! There are some rules to keep in our mind. In fact, we have a specific type of stereoisomers that is enantiomers. Enantiomers are stereoisomers that are not-overlapped mirror images of each other. If we look at our hands, they are mirror images but they cannot overlapped. (This is the typical example of enantiomers). There is a way to compute the number of enantiomers with this formula: 2n = total enantiomers where n = chiral centers. For example, when the aldoses have 4 asymmetric centers there are 2 4= 16 ENANTIOMERS (8 D- hexoses and 8 L-hexoses). So, according to convention, D & L sugars are mirror images of each other. They have the same name, D-glucose & L-glucose. When we decide to change name? What’s the difference between glucose and mannose? Epimers Epimers 2 are two sugars identical in configuration (so they have the farthest carbon from the functional group that is the same) the same except for only one chiral carbon. In any case we have the D-family, so the hydroxyl group of the chiral carbon is always on the right. As we can see in the picture below the D-Galactose differs from Glucose only in the position of the hydroxyl group of the fourth carbon even though the rest of the molecule is the same. 2 Epimers are not enantiomers. 7 LINEAR AND CYCLIC STRUCTURES Usually, monosaccharides are presented and reported using the linear molecular structure (Fischer projections) but the MOST STABLE FORM FOR SUGAR IN SOLUTION IS THE CYCLIC STRUCTURE We know that  An aldehyde can react with an alcohol to form a hemiacetal;  A ketone can react with an alcohol to form a hemiketal. What do these reactions lead to? We have something that does not appear among the reagents: a new chiral carbon, in fact if we look at the ketone and aldehyde we do not have a chiral carbon while we have one in hemiketal and hemiacetal. Since we have a chiral carbon we have new stereoisomers. The linear form of glucose (and other aldoses) undergoes to a similar INTRAMOLECULAR reaction forming a cyclic hemiacetal. This reaction occurs between the aldehydic functional group (C1) and the hydroxyl group of the last asymmetric carbon (C5) More specifically, the formation of the cyclic structure of D-glucose from its linear form occurs through a cyclization reaction. Specifically, the aldehyde group on the carbon chain of the linear form of glucose reacts with the hydroxyl group of the carbon in fifth position. We said that we will have the formation of a new chiral carbon which corresponds to the Carbon of the first position of the ring structure. Be careful, when we have the ring structure we have to count the groups clockwise and this means that on the right side the group are external to the rings while the groups on the left side are internal the rings. 8 When we have cyclic structures, we need to be able to recognize the molecule and distinguish it from others (for example, glucose from mannose). How? To be able to do this, we must consider how the hydroxyl groups are positioned. Remind that the cyclic structure of glucose is defined as glucopyranose. Since we have a chiral center at position 1, we will have new stereoisomers called anomers. So we can have two possible anomeric forms: α-D-glucopyranose and β-D-glucopyranose, which differ in the position of the hydroxyl group relative to the chiral carbon in the cyclic structure.  Alfa (α) -> OH below the ring;  Beta (β) -> OH above the ring. 9 As we can see in the above image, in alpha-D-glucose the hydroxyl group is positioned below the anomeric carbon at position 1, while in beta-D-glucose it is positioned above. To recap, a new chiral center resulting from a hemiacetal reaction leads to the formation of two new stereoisomers called anomers and C1 becomes the ANOMERIC CARBON. In similar manner, the linear form of D-fructose (and other ketoses) undergoes to a similar INTRAMOLECULAR reaction forming a cyclic hemiketal. This reaction occurs between the ketonic functional group (C2) and the hydroxyl group of the last asymmetric carbon (C5). So in this case the cyclic structure is defined as fructofuranose and it is possible to distinguish a new chiral center (anomeric carbon) which is the C2, so in second position. Also, in this case we have two different anomers:  Alfa -> OH below the ring;  Beta -> OH above the ring. It's important to emphasize a concept: when we talk about sugars, but generally for any molecule, we need to consider the molar concentration within a solution. The correct expression, for example for glucose in our bloodstream is '5 millimoles (which is the right value of our glycemia). It's important to remember that 80% of glucose is found in solution in the form of a cyclic structure, which is more stable, while the remaining 20% is present in linear form. In fact, pyranose sugars assume a "CHAIR" configuration since THIS is more stable as 3- dimensional spatial arrangement. (AXIAL AND EQUATORIAL DISPOSITION). So, the final situation is represented by the chair structure. 10 Modification of sugars A variety of chemical and enzymatic reactions produces DERIVATIVES of the simple sugars (monosaccharides). THESE MODIFICATIONS GENERATE A VARIED COLLECTION OF SACCHARIDE DERIVATIVES The most common modifications are:  Phosphorylation: is a biochemical process where a ortho-phosphate group (P0—33) is added to a molecule, for example β-D-Glucose with the formation of β-D-Glucose-6-phosphate. We need to identify the position to attach the phosphate. Carbon at position 6 is the most reactive to favor this reaction and thus attaches the phosphate. So our cells expend a lot of energy to ensure that this reaction occurs. We must imagine that glucose alone is very stable. What is then the purpouse of this phosphorylation reaction? The aim is to trap glucose inside the cell, and to achieve this, we need to create a negative charge; indeed, free glucose is capable of crossing the membrane, although not easily, while glucose-6-phosphate cannot do so and therefore remains trapped inside the cell. Therefore, the objective is to store glucose inside the cell. If we consider, for example, blood glucose levels (i.e., the concentration of glucose in the blood), an increase or decrease in blood sugar causes a change in phosphorylation because we need to allow glucose to exit the cell or, conversely, we need to create more phosphorylated forms to trap glucose inside the cells 4. 3 The presence of the two minus signs is the result of the loss of two hydrogens. 4 Why does glucose utilize phosphate and not other negative charges? When we study metabolism, we will see that phosphate transfer potential is the most common form for exchanging energy between macromolecules 11 The preferred position for phosphorylation is the sixth position, but it can occur at the first position The same is for fructose. We can have Fructose-6-phopshate or Fructose 1-6-phosphate. In this last case, we have a double phosphorylation, and this is an important strategy not to keep fructose inside the cells but to prepare the cells for a breakdown. When there are two phosphates close to each other, it means that we are preparing for an enzymatic attack to 12  break the molecule. This is typical of the first reaction of glycolysis, where glucose is broken down.  Not very common generally, but common regarding polysaccharides and disaccharides is amination which refers to the process of introducing an amino group (-NH2) into a compound or molecule. In this case, we obtain β-D-glucosamine. In this case, the preferred position to add the amine is carbon 2. This reaction is very important because it represents the beginning of another modification, which consists of adding an acetic acid to the amino group at position 2. This leads to the formation of the N-Acetyl-B-D-Glucosamine. This is a very typical modification. 13 So, amination represents the starting point for further modifications. The amino group can be acetylated by adding an acid molecule. For example, if we add a molecule of acetic acid we get the N-acetyl-B-D-glucosamine (component of bacteria cell walls). Another example is the case of the formation of muramic acid. In this case, we have the addition of lactic acid molecule at C3 (with loss of OH) of glucosamine. This leads to the formation N-Acetylmuramic acid (another component of bacteria cell walls). The same amination can occur in the cases of galactose and mannose. Again, at position 2, we have the addition of the amino group resulting in the formation of galactosamine and mannosamine 14   Deoxygenation: it is a process that leads to the formation of Deoxy sugars which are monosaccharides with one or more hydroxyl groups replaced by hydrogens. For example, the transformation from β-D-ribose to β-D-deoxyribose involves the removal of a hydroxyl group (-OH) from carbon 2 of ribose. This process is known as deoxygenation and can be considered a form of reduction, as it involves the loss of oxygen. Specifically, β-D- deoxyribose lacks a hydroxyl group on carbon 2, which is present in β-D-ribose.  Oxidation: In glucose, especially, there are some preferred positions where oxidation occurs. We can have different situations depend on the position in which the oxidation occurs: o As we can observe in the picture, at position 6, we have a carboxyl group in β-D- glucuronate5 acid, which is an organic compound derived from the oxidation of the primary alcohol group (the OH bonded to C-6) of D-glucose to a carboxyl group. So if the oxidation occurs at position 6 we have an uronic acid as in the case of B-D- glucuronate. o If we have the oxidation of the carbon in first position (C1) we will get an aldonic acid for example the D-Gluconic. The first position is now in an unstable state. The reaction between the carboxylic group at position 1 and the adjacent hydroxyl group 5 Remember that when we have the suffix “uro” means that the oxidation is in position 6. 15  in position 6 is an esterification reaction that leads to the formation of an intramolecular ester called D-glucono-δ-lactone. These are the most famous oxidation examples. o In the oxidation occurs at position C1 and C6 (so both position) we will get an aldaric acid as in the case of D-glucaric acid. Recap  In neurons is very typical to find a very strong modification of glucose, in fact we can find the sialic acid. Look at the structure: in first position we have the carboxyl group again while in 4 position we have an ammino group (NH), also in this case the amination is linked to the acetyl group. The result in this case is a very negative charge in the molecule. On the right side we can see the glycerol structure (three carbons linked to 3 hydroxyl group). PARTE ESTREMAMENTE CONFUSA, CHIARIRE DAL LIBRO. 16 DISACCHARIDES They represent the union between two monosaccharides through a glycosidic bond in which the anomeric hydroxyl group of a monosaccharide and a hydroxyl of another sugar can join together, splitting out a water molecule to form a GLYCOSIDIC BOND. When talking about disaccharides, two features must be remembered (La prof.ssa spiega veramente troppo male, non sono riuscito a capire ciò che ha detto, ma si riferisce a qualcosa sulla priorità). Maltose It is the simplest disaccharide, and it derives from the union of two molecules of glucose. More specifically, is a disaccharide with an α (1 α 4) glycosidic bond between –OH-C1 (α anomer) of the first sugar and –OH-C4 of second glucose (β anomer)6. So we have α -1,4 glycosidic bond 6 In other words, the alfa-anomeric carbon atom 1 of one molecule is joined through an oxygen atom to carbon atom 4 of the second molecule. To form maltose, the two OH (Hydroxides) groups on these carbon atoms react condensing out H2O and leaving the O atom bridge. 17 Cellobiose It is a product of the cellulose breakdown, is the equivalent β anomer of maltose. The β (1 -> 4) glycosidic bond is represented as an oblique link. Sucrose Sucrose is formed by a glycosidic (1’-2’) bond between the C1 atom of the alpha anomer of glucose with the C2 atom of the beta anomer of fructose. So α- β 1,2 glycosidic bond (ATTENZIONE: LA PROF.SSA DICE CHE LA FORMA ANOMERICA DEL FRUTTOSIO E’ ALFA, MA E’ SBAGLIATO). As we can see in the image below, fructose is inverted in such a way as to expose carbon C2 for glycosidic bond. 18 The complete name of sucrose is α-D-glucopyranosyl-(1->2)-D-fructopyranose. Lactose It is milk sugar, is composed by galactose + glucose, with β (1->4) glycosidic bond (it utilizes the anomeric OH of galactose). The B-anomeric carbon of the galactose unit (C1) forms a linkage with the Oxygen at carbon 4 of the glucose unit (C4). Its complete name is β -D-galactopyranosyl-(1->4)-D-glucopyranose POLYSACCHARIDES Previously we said that polysaccharides can be distinguished into:  HOMOPOLYSACCHARIDES: (monosaccharides of a single type)  HETEROPOLYSACCHARIDES (monosaccharides of 2 or more different types, or their derivatives) In turn homo and heteropolysaccharides can be branched or unbranched. 1. Homopolysaccharides: o Unbranched: These polysaccharides are composed of a single type of monosaccharide repeating unit. Examples include glycogen and starch, both of which consist of glucose units linked together. In unbranched homopolysaccharides, the repeating units are typically linked in a linear fashion without branching points. o Branched: Some homopolysaccharides may exhibit branching, where additional monosaccharide units are attached to the main chain at specific positions. For example, glycogen, which serves as a storage form of glucose in animals, has a highly branched structure. 2. Heteropolysaccharides: o Unbranched: These polysaccharides are composed of two or more types of monosaccharide units. They often serve structural roles in cells and extracellular matrices. For example, hyaluronic acid, a component of the extracellular matrix, is composed of repeating disaccharide units of glucuronic acid and N- acetylglucosamine. o Branched: Heteropolysaccharides may also exhibit branching, where additional monosaccharide units or other molecules are attached to the main chain. For instance, glycosaminoglycans (GAGs), which are components of connective tissues, can be highly branched heteropolysaccharides. 19 STORAGE HOMOPOLYSACCARIDES: STARCH AND GLYCOGEN Glycogen Remember that the most famous polysaccharide in our body is Glycogen! Glycogen is the reserve polysaccharide of animals. It is found in all the cells, especially in the liver and muscles as cytoplasmatic granules. ATTENTION: liver has an higher concentration of glycogen but we take into account the total concentration of glycogen this is higher in muscles cells (this is due to the fact that our body is composed mostly of muscles). It presents alfa 1-4 glycosidic bond but also 1-6 bonds. The structure is like that of amylopectin but more densely ramified, with ramifications every 8-14 glucose residues. Glycogen is degraded by glycogen phosphorylase that breaks down alpha (1-4) bonds from the non- reducing ends of the molecule. The highly ramified structure allows the quick mobilization of glucose in moments of metabolic necessity. 20 SECOND PART OF THE LESSON Starch Plants store glucose as starch, which is formed of two different components: amylose and amylopectin. Starch is stored in plant cells in granules in the stroma of plastids, particularly in seeds and tubers. It is the most abundant dietary carbohydrate, and we find it in rice, potatoes, pasta. AMYLOSE is a homopolymer of D-glucopyranose with (14) linkages meaning that the position of the anomeric hydroxyl group OH is the position 1. Amylose: forms the 15 – 20% of the total amount of starch; has a molecular weight from several thousands to million of Daltons, so it is very huge; has a non-branched linear structure. Although poorly soluble in water, it forms micelles in which amylose assumes a helical conformation, because of the distorted form (chair) of each glucose unit. If we want to write down the liner structure of amylose, it would be like this: In reality, due to the presence of the presence of with (14) linkages and the typical chair form of the glucose in nature, amylose looks like a left-handed spiral: AMYLOPECTIN is a glucose homopolymer with mainly (14) linkages, but it also has branches formed by (16) linkages. Branches (1 every 20 residues) are chains generally composed by 24- 30 glucose units. Amylopectin forms the 80-85% of the total amount of starch. The ramifications: are an important advantage for the molecule itself from, for example, the energetic point of view (having more bonds means having more energy) or the practical point of view (having branches means to have more monomers available in less time); give rise to a compact and stable structure; provide multiple chain ends at which enzymatic cleavage can occur. 21 CELLULOSE We know that plants do not synthesize structural fibrous proteins as animals, then their function is replaced by the structural polysaccharides. The most common and present structural polysaccharide in plants is cellulose, which is is the major constituent of plant cell walls. Even if it is a carbohydrate, cellulose forms extended structures similar to the beta sheet of proteins because of its role: impart strength and rigidity to plant cell walls, in order to prevent the osmotic swelling. Humans are not able to digest cellulose, because of our lack in the enzyme that decompose cellulose itself, the cellulase. Cellulase catalyses the cleavage of the (1->4) linkages that are present in the molecule of cellulose between each monomer of glucose. In fact, this is the structure of the cellulose: Even if starch and cellulose are both chains of glucose, the difference stays in their structure. As we know, starch stores energy, while cellulose has a structural function. What is the difference in the structures then? The difference is: the linkages. In starch, as we said, we find (14) linkages, and in cellulose we find (1->4) linkages. This means that we have 2 different anomers in the 2 molecules. 22 This is the structure of 2 monomers of glucose in the cellulose molecule. We have a β because of the position of the OH linked to the first carbon of the first monomer, which is the same side of the sixth carbon. The second carbon contributes to the linkage with is fourth carbon. For this reason, there is a distortion of the first and the second sugar, which is not visible while writing down the linear formula of cellulose, but it is in reality. This is how cellulose actually looks like: We can see a rotation of 180° between the first carbon of the first monomer and the fourth carbon of the second monomer, giving to the cellulose molecule a filiform structure. Why does the structure of the molecules of starch and cellulose give rise to different functions in plants? Starting from the cellulose, since the molecules are filiform, they can overlap easily and use more and more hydrogen bonds and not so strong other type of bonds that are anyways numerous, to compact the whole structure. The starch, as we said, has an easy-to-break structure, so it is less compact but more likely to give glucose in order to obtain ATP. GLYCOGEN Glycogen represents the form of glucose storage in animals. It is similar in structure to amylopectin and present in granules inside liver muscle and cells. Glycogen is composed by glucose monomers linked through (14) linkages for the linear situation, but we can also find (16) branches (1 every 8 residues, instead of 1 every 20 residues like the amylopectin. So in glycogen we have more branches than in starch). The highly branched structure permits rapid glucose release from glycogen stores, e.g., in muscle during exercise. The ability and possibility to quickly mobilize glucose is very important in animals, because of the energy demand of metabolic pathways. But since animals need a lot of free monomers of glucose, why do we store it as glycogen? 23 The first reason is the regulation. Our cells work in a constant equilibrium of break-building molecules. The energy released by the breaking down of the bonds is very useful for the organism. The second one is that storage of low molecular weight metabolites in polymeric form avoids the very high osmolarity that would result from storing them as individual monomers. If the glucose in liver glycogen were monomeric, the glucose concentration in liver would be so high that cells would swell and lyse from the entry of water by osmosis. The reason of this is the Van't Hoff coefficient in the osmotic pressure π formula (𝜋 = 𝑖 × 𝑀 × 𝑅 × 𝑇). If I have thousands of monomers polymerized in one single molecule, the Van’t Hoff coefficient is going to be 1; if, on the other hand, I have the same exact number of free monomers this time, the Van’t Hoff coefficient is going to be really high, leading to an increase of the osmotic pressure as well. In particular, in liver: The concentration of glycogen is equal to 0,01 µM; The concentration of glucose monomers is equal to 0,4 M. The difference between these two concentrations is huge. In terms of osmotic pressure this means that such a high concentration of monomers results in an increase in osmotic pressure with an increase in intracellular water. This is a truly dangerous situation. That's why nature has ensured that monomers are compacted into polymers, thereby avoiding variations in osmotic pressure. Furthermore, in the blood circulation we need to keep a very stable concentration of 5 mM. We cannot change this value. GLYCOSAMINOGLYCANS AND PROTEOGLYCANS Among all heteropolysaccharides, glycosaminoglycans and proteoglycans are the most important for our organism. With these heteropolysaccharides we have to mix some of the information we learnt about the modification of glucose (amino modifications, the sulphuration, phosphorylation) Most mammalian cells are located in tissues where they are surrounded by a complex gellike material, extracellular matrix (ECM). It is now widely demonstrated that ECM is the substrate on which all tissue cells can adhere, migrate, proliferate and differentiate, and which also influence its survival, shape and function. ECM can cover lots of actions like regression, compression, differentiation, and so on. 24 In this picture we can observe the position of the ECM in the epithelial tissue. ECM covers lots of functions, such as: binding and packing of tissues (connective tissue); it is a connector, an anchor and a support for the organs, consequently for the whole body; transport of metabolites between capillaries and tissues; defence against infection; repair of injury (via cell proliferation and fibbers formation). The ECM contains three major classes of biomolecules: structural proteins: collagen, elastin; specialized proteins such as fibrillin, fibronectin and laminin; heteropolysaccharides such as glycosaminoglycans and proteoglycans. This picture shows the transversal section of the plasmatic membrane, with some membrane annexes, such as heteropolysaccharides. We can notice:  Location  here are heteropolysaccharides free in the ECM but also connected to the cellular membrane.  Structure  there are linear and branched heteropolysaccharides.  Dimension  there are long and short heteropolysaccharides. GLYCOSAMINOGLYCANS (GAGs) These molecules are long linear (unbranched) polysaccharides containing repeated disaccharide units. We find them not only in the ECM, but they are also connected with the extracellular side of the plasma membrane. 25 The disaccharide units contains: One amino sugar (monomer). This can be either N-acetylglucosamine (GlcNAc) or N- acetylgalactosamine (GalNAc); One acidic sugar which is the uronic acid with the oxidation of the sixth carbon (C6), becoming glucuronic acid a negatively charged molecule. Due to the presence of glucuronic acid the GAGs are negatively charged compounds with the following characteristics: High viscosity; Low compressibility; Structural integrity. This means that these molecules are particularly suitable to constitute the lubricating fluids, for example, those present in the joints (synovial fluid). In some GAGs, the amino sugar shows one (or more) esterification by sulfate leading to the highest concentration of negative charges, improving all the previous characteristics of the GAGs themselves. But why does the negative charge lead to these characteristics of the GAGs? The reason is that these molecules, since they have a charge, they are polar, leading to an higher affinity with water. Some of the most clinically relevant GAGs are: 26  Hyaluronic acid It is the sum of D-glucuronic acid + N-acetyl-D- glucosamine Hyaluronic acid is the only GAGs that does not contain any sulphate group. It is a non-covalently linked component of complexes with proteoglycans of the ECM. The glycosidic linkages are β (13) between the two monomers of the disaccharide and β (14) between the two disaccharides. It may consist of 25000 disaccharide units, with a molecular weight up to 10 7 and it is the only GAG present both in animals and bacteria. It is present in: o Synovial fluid because it acts as lubricator in articular joints, and has a function in wound repair; o Vitreous humor of the eye; o ECM of loose connective tissue because hyaluronic acid may be important in allowing tumour cells to migrate through the ECM. Tumour cells can induce fibroblasts to synthesize greatly increased amounts of this GAG, to facilitate their own diffusion; o Umbilical cord and other embryonic tissues because it plays an important role in permitting cell migration during morphogenesis; o Cartilage. Heparin (it is the most negative one) D-iduronate-2-sulfate + N-sulfo-glucosamine-6-sulfate The D-iduronate-2-sulfate is the oxidate form of idose, a rare isomer of glucose. In nature we cannot find the idrose (?) but just the structure with the carboxylic group. 27  Heparin has an extended helical conformation and the repulsion by the tons of negatively charged groups may contribute to the conformation itself. As regards the linkages, both in between the monomers and between the disaccharides, we find α (14) linkages. Heparin is an intracellular GAG with the highest negative charge compared to the other GAGs, it is a component of intracellular granules of mast cells lining the arteries of the lungs, liver and skin and it is an important anticoagulant (it binds with factors of the clotting process). Chondroitin sulphate D-glucuronic acid + N-acetyl-D-galactosamine-4 (or 6)-sulphate One of the most abundant GAG, found in: o cartilage (bind collagen and hold the fibres strongly); o tendons, ligaments, heart valves. D-glucuronic acid + N-acetyl-D-galactosamine-4-sulphate D-glucuronic acid + N-acetyl-D-galactosamine-4-sulphate Dermatan sulphate Iduronic acid + N-acetylgalactosamine-4 sulfate Heparan sulphate D-iduronate-2-sulfate + N-acetylated-glucosamine-6-sulfated Keratan sulphate D-galactose + N-acetylglucosamine-6-sulfate 28 PROTEOGLYCANS The GAGs (except for hyaluronic acid and heparin) in the body are linked to core proteins forming proteoglycans (also called mucopolysaccharides). Mammalian cells can produce 40 types of proteoglycans. These molecules act as tissue organizers, and they influence various cellular activities, such as growth factor adhesion and activation. The basic proteoglycan unit consists of a “core protein” to which glycosaminoglycans are covalently attached. The GAGs extend perpendicularly from the protein core in a brush-like structure. We can actually find for example hyaluronic acid in this kind of compounds, but only if we are talking about proteoglycan aggregates. In this case the bond between the core protein and the acid is an ionic bond. 29 The connection of GAGs with the protein core involves the N-α-acetylgalactosamine associated to the -OH group of serine or threonine (or also hydroxylysine) residues in the protein. This is defined as O-glycosidic bond. GAGs are linked to the protein core also through a N- glycosidic bond, through the amide nitrogen of an asparagine residue. Aggrecan is the main proteoglycan of articular cartilage. This molecule is important for the proper functioning of articular cartilage because it provides a hydrated gelatinous structure. When a trauma occurs, the aggrecan absorbs the tension by immediate release of water and gradually it re- accumulates water by releasing the tension gradually as well. This need to displace the water this way is the reason behind the tons on negative charges in this kind of molecules. GLYCOPROTEINS They are branched glycans. They can have the function of: structural proteins; enzymes; membrane receptors; 30  transport proteins; immunoglobulins. Glycolipids and Lipopolysaccharides In glycolipids and lipopolysaccharides, complex oligosaccharide chains are attached to membrane- anchored lipidic structures. Other glycosphingolipids, the globosides, contain oligosaccharide chains that serve as the blood group antigens. In glycolipids and lipopolysaccharides, complex oligosaccharide chains are attached to membrane-anchored lipidic structures. The distinction between blood groups (A, B, 0 and AB) depends on the oligosaccharidic portion of glycolipids on the erythrocyte surface. Blood groups In all blood types the oligosaccharidic portion contains L-fucose. o In blood group antigen A: at the non-reducing end, sugar is N-acetylgalactosamine; o In blood group antigen B: at the non-reducing end, sugar is galactose; In blood group antigen 0: at the non-reducing end are absent both types of saccharides; o In blood group antigen AB: at the non-reducing end are present both types of saccharides. During transfusions, we must pay a lot of attention to the blood groups, because the erythrolysis can happen, due to the presence of antigens and anti-bodies. 31 MUCOPOLYSACCHARIDOSIS Several genetically inherited diseases, for example the lysosomal storage diseases, result from defects in the lysosomal enzymes responsible for the metabolism of complex membraneassociated GAGs. These specific diseases, defined mucopolysaccharidoses (MPS) lead to an increase of GAGs concentration within lysosomes of affected cells. There are at least 14 known types of lysosomal storage diseases that affect GAGs catabolism, characterized by mental retardation and/or structural defects of the body. Some of them are: MPS Type I or Hurler’s syndrome: It results from a deficiency of α-L-iduronidase. Heparan sulphate and dermatan sulphate accumulate. There is growth and mental retardation with characteristic facial changes. MPS Type II or Hunters syndrome: It is similar to Hurler’s syndrome, but the enzyme deficiency is for iduronate sulfatase, and the inheritance is X-linked. MPS Type III or Sanfilipo’s syndrome: It is caused by a deficiency of one of four enzymes of which three are hydrolases and one is an N-acetyltransferase. There is severe mental retardation but only mild structural features. Other MPS Types are IV, VI and VII. All are autosomal recessive disorders, except Hunters syndrome which is X-linked. Specific lab tests: 1. urine; 2. enzymes assay; 3. tissue biopsy; 4. DNA testing; 5. prenatal diagnosis. AMMINO ACIDS 32 Amino acids are the monomers of proteins. There are 20 of them that actually forms proteins and others that DO NOT form proteins, all different from each other, and it is possible to obtain an infinite number of proteins depending on the type, number and sequence order with which the different amino acids are linked. We can have:  standard AA: those that form proteins. They are 20. They can be: o essential  it is NOT possible to biosynthesize them and therefore they must be introduced through the diet. They include valine, leucine, isoleucine, phenylalanine, threonine, tryptophan, methionine and lysine. o non-essential  it is possible to biosynthesize them and therefore they must NOT be introduced through the diet. non-standard AA: modified, those that rarely form proteins. Nutritionally, amino acids are of three types: Essential; Non-essential; Semi-essential: these are growth promoting factors since they are not synthesised in sufficient quantity during growth. They include arginine and histidine. They become essential in growing children, pregnancy and lactating women. AAs have some chemical properties in common: They are all α-amino acids and the amino group and the carboxyl group are bonded to the same carbon atom at physiological pH they are found in the form of zwitterions (i.e. molecules that carry the two opposite charges at the same time, maintaining neutrality. 33  They are called α because the central carbon is α, due to the fact that in carboxylic acids’ nomenclature we call α the second carbon of the chain, the one bonded to the carbon one, which is the one from the carboxylic group. This does not mean, however, that in nature we cannot find β-amino acid. We find them as the product of the catalysis of nucleotides; They are therefore amphipathic compounds, meaning that they can behave as both acids and bases based on the pH of the medium in which they are found; They have optical activity and are all found in the L form (with the exception of glycine). This is because they are made by enzymes and enzymes grab the chemical precursors on one way only. Like other ionic compounds they are more soluble in polar solvents than in non-polar ones. The 20 standard amino acids can be divided into groups depending on the charge and polarity of their side chains: Nonpolar neutral side chains: o Glycine → a single hydrogen atom; o Alanine - Valine - Leucine - Isoleucine → aliphatic hydrocarbon side chains (dimensions ranging from those of a methyl group for alanine to those of an isomeric butyl group for leucine and isoleucine); o Methionine → side chain containing a THIOL ETHER; o Proline → side chain made up of a CYCLIC PYRROLIDINE; o Phenylalanine → aromatic side chain (PHENYL GROUP); o Tryptophan → aromatic side chain (INDOLE GROUP). Polar neutral side chains: o Serine - Threonine → hydroxyl R groups with different sizes; o Asparagine - Glutamine → side chains with amide groups; o Tyrosine → side chain with phenolic group 34  o Cysteine → side chain with thiol group. Basic charged side chains: o Lysine → butylamine as side chain; o Arginine → side chain with guanidine group; o Histidine → side chain with imidazole group; o Aspartic acid - Glutamic acid → side chain with carboxyl group (ionized state: aspartate and glutamate). Acidic charged side chains: aspartate, glutamate. 35  BIOCHEMISTRY LESSON II AMINO ACIDS During the last lesson, we had an introduction about amino acids. Remember the structure and the features of amino acids: each amino acid has an amino group (- NH2) and a carboxyl group (-COOH) linked to the central carbon atom. This central C atom is a chiral carbon (except for the case in which the R’ group is an H atom: in particular this is the case of glycine, in which the central C is bonded to two H atoms). We discussed about α (alpha) amino acids. Why are they called α? It is related to the nomenclature of the carboxylic acids:  If we use the Greek letters, we name the carbon after the carboxylic carbon with the first letter of the Greek alphabet which is alpha, so it’s the: α carbon.  if we want to use numbers, we can say that: the α carbon is the carbon number 2, after the carboxyl C which is the number 1. So, that’s why, specifically speaking, we can talk about alpha amino acids: because the -NH2 and the -COOH groups are linked to the C atom named alpha (which can also be called C number 2). R chains7 Now, we’re going to focus on the R groups. Proteins (that derive from the polymerization of amino acids that are linked together) have a multitude of very different functions; the “secret” of this differentiation (of the many functions of proteins) depends on the specific amino acids: their charge, their R group, and how they are linked together. N.B. the following part is very important. The 20 standard amino acids can be divided into 4 groups depending on the charge and polarity of their side chains:  Nonpolar neutral side chains: alanine, phenylalanine, glycine, isoleucine, leucine, methionine, proline, tryptophan, valine.  Polar neutral side chains: asparagine, cysteine, glutamine, serine, tyrosine, threonine.  Acidic charged side chains: aspartate, glutamate. (Remember the definition of an acidic compound, that is a compound that tend to give H+ to the solution). 7 Let’s delve in what kind of side chains these amino acids have. N.B. lei ha detto letteralmente: “Try to write just few of these amino acids. But look at the category in general”. N.B. in the images there are amino acids marked with stars ( ): these are the essential amino acids. We have to assume them from the diet, because we’re not able to synthesize them. 36  Basic charged side chains: arginine, histidine, lysine. (Remember the definition of a basic compound, that is a compound that tend to attract H+ from the solution).  Nonpolar neutral side chains: Obviously, from a “non-polar neutral side chain” we can expect hydrocarbon components (as you can see from the image, we have many methyl groups).  Glycine → the R group is a single hydrogen atom; it’s the only amino acid without the central chiral carbon. It’s the simplest amino acid.  Alanine - Valine - Leucine - Isoleucine → aliphatic hydrocarbon side chains; Their dimensions range from those of a methyl group for alanine to those of an isomeric butyl group for leucine and isoleucine; in particular: - alanine has a methyl group, - valine has an isopropyl group (lei dice semplicemente: it has 3 methyl groups), - leucine has an isobutyl group, (anche qui dice semplicemente: it has 4 methyl groups) - isoleucine present an isomeric form of the side chain of leucin.  Methionine → side chain containing a thiol ether (we can see from the image that there is a S atom between two methyl groups).  Proline → side chain made up of a cyclic pyrrolidine: in proline, the nitrogen of the usual amino group is inside the ring. It's a typical feature of this amino acid that will be very important during the formation of the secondary structure of proteins.  Phenylalanine → aromatic side chain: it’s a phenyl group.  Tryptophan → aromatic side chain: it’s an indole group: there are two rings together, a six-membered ring and a five-membered ring. : as we said before these are the essential amino acids.  Polar neutral side chains:  Serine - Threonine → hydroxyl R groups with different sizes; in particular: 37  - in serine the -OH group is linked to a methyl group. - in threonine there are a -OH group and two methyl groups.  Tyrosine → side chain with phenolic group: the benzenic ring is attached to a -OH group.  Cysteine → side chain with thiol group unlike methionine, cysteine's S atom is not between two methyl groups but is free. It means that this S atom is available for another link with another -SH group: the reaction between a thiol and a thiol results in the formation of a disulfide bridge (S-S bridge). It's likely that we observe a disulfide bridge in the structure of insulin, resulting from the oxidation of cysteine.  Asparagine - Glutamine → side chains with amide groups. N.B. it's essential to remember asparagine and glutamine because they’re very important in the amino acid metabolism. In amino acid metabolism: the release of - NH2 groups is crucial, as the accumulation of free -NH2 groups can be toxic to the body. In fact, a thing is when the amino group is inside the structure of the amino acid and another thing is when it's free. Hence, when the catabolism of amino acids occurs, it's fundamental to quickly remove free -NH2 groups from the solution to prevent toxicity. In fact, as humans, unlike plants, we lack the ability to recycle -NH2 groups for energy and other processes. Consequently, we must excrete all excess free NH2 groups through urine to eliminate them from our body. So: free -NH2 group is very dangerous! So, when we have molecules like asparagine and glutamine, we have a concentration of -NH2 groups, meaning that they’re amino acids that recover a lot of NH2- groups, which are readily available to “go away” in the metabolic processes of urea cycle (we’ll see it next year). From the image we can see that also Threonine is an essential amino acid.  Basic and acidic charged side chains: (look at the image below): “charged” means that in solution these amino acids undergo a strong ionization. Ultimately, the amino acid will carry a net positive or negative charge, depending on its basic or acidic properties, respectively. Acidic:  Aspartic acid - Glutamic acid (it’s important to remember these two amino acids for some reasons we’ll discuss later) → side chain with carboxyl group (ionized state: aspartate and glutamate). - Aspartic acid has one methyl group and one carboxyl group; aspartic acid at physiological PH of 7.4 carries a negative charge 38  - Glutamic acid has two methyl groups and one carboxyl group. So, it’s longer than aspartic acid for one methyl group. Basic:  Lysine → butylamine as side chain: it has 4 methyl group and one amino group; it carries a positive charge, since it is basic: it attracts protons from the solution;  Arginine → side chain with guanidine group; also arginine is a positively charged amino acid. Arginine is the amino acid with the richest content of amino groups: three amino groups. This abundance makes it famous for recovering all the amino groups present in a solution  therefore, arginine collects and carry out amino groups from solutions! Therefore, its role is important in the context of the urea cycle. So, arginine is the “chief” of this process, to array all the amino groups from solution to the external environment.  Histidine → also histidine is a positively charged amino acid. Its side chain is with an imidazole ring. and this is important because in every situation of the solution (meaning at any value of the PH) HISTIDINE IS THE BEST BUFFER. From the image we can see that Lysine is an essential amino acid (in fact it’s indicated by the blue star). And we can also notice that Arg and His are denoted by two yellow stars ( ). Why? Because they’re not essential as Lys but they’re SEMI-ESSENTIAL. What does “semi-essential” mean?  they are not essential throughout our entire lives; at the beginning they’re not essential, but become essential during specific periods: for instance, during phases like childhood growth and pregnancy. So, in these periods, we have to assume them from the diet. ENANTIOMERS Yesterday we also discussed about L-stereoisomers: why L? Chemically, AA are organic molecules that have both the amino (-NH2) and carboxyl (-COOH) functional groups. With the exception of glycine, AAs are chiral molecules (there is a stereo-centre: chiral carbon atom bonded to four different functional groups), so for each AA there are two enantiomers with DIFFERENT biochemical properties: L configuration (NH3 on the left) and D configuration (NH3 on the right). 39 So, what’s the rule to establish the L or D form? As we’ve seen for sugars (in which the D- or L- enantiomer depended on the position of the -OH group linked to the C farthest from the carbonyl group), also in the context of amino acids we can distinguish two enantiomers: D and L. In particular, it is determined by the amino group. When we depict an amino acid in a two-dimensional representation, we can distinguish its L or D form by observing the position of the amino group relative to the central carbon atom:  If the amino group is on the right side, it indicated the D form.  Conversely, if it's on the left side, it indicates the L form. What is the biologically active form for our cells? In biochemistry we speak of L-α-amino acids  our enzymes recognize and interact with the L enantiomers. (Remember that in sugars, it's the D form that's biologically active!). N.B. you should remember that in the construction of bacterial cells’ wall, there are also D-amino acids. Thus, our cells are “blind” to the D-amino acids present in bacterial walls: that’s one of the reasons why bacteria are enemies for us! (a questo punto qualcuno ha fatto una domanda su una curiosità, ma non si sente benissimo.) Very recent researches say that in our neural cells we can synthesize also D-amino acids. Coming back to the concept of enantiomers, we can see the example of tyrosine: Now, let’s talk about the concept of “charge” in the context of amino acids. 40 N.B. Il seguente paragrafo è preso direttamente dalle slides ed è un sunto della spiegazione. Subito dopo di esso si trova la spiegazione un po’ più ampia della prof In general: due to the presence of the amino and carboxylic groups, the amino acids show themselves as zwitterions, i.e. molecules that simultaneously bear the two opposite charges, maintaining neutrality. They can therefore behave as both acids and bases depending on the pH of the medium in which they are found. In addition, each AA has a specific side group (R group) that can be acidic, basic, hydrophilic (or polar) and hydrophobic (or apolar). The bulk of the various R groups and the interactions between similar groups contribute to model the spatial conformation of the protein, a conformation on which the biological activity of the protein itself essentially depends. Now, we’re going to analyse these situations in a more detailed way. First, we have to recall some important information from the inorganic chemistry, particularly regarding the strength of dissociation. This is the “secret” to determine the charge of an amino acid in a given solution at given pH values. (In questa parte ha fatto le seguenti domande per introdurre l’argomento, ma ha parlato in maniera molto confusionaria, senza dare risposte chiare).  Now, let's consider the following question: given a charged or polar amino acid with the amino group, the carboxyl group and also their specific R group, what will be the overall charge at a particular pH value of the solution?  Why do we have to answer this question? Because when we create a protein, we have to deal with various interactions: polar interactions, ionic interactions, hydrophobic interactions, etc. These interactions contribute to the diverse characteristics of proteins. So, we need to understand why and what is the total charge of amino acids in specific pH environments.  When we mention that the physiological pH is around 7.4, is this applicable to all body fluids? Or do different bodily regions have distinct pH levels? The answer is yes: for example, during digestion: the stomach produces HCl to aid in the breakdown of food that we introduce. This results in an increase in the H+ concentration and in a decrease in pH, making the environment more acidic. What’s the behaviour of amino acids in this case, for example? Look at the following table, in which we have the pKa. PKa represents the negative logarithm (in base 10) of the acid dissociation constant (Ka). So, ka is a measure of the strength of dissociation and the values of ka are generally very small; that’s why chemists introduced pka, because it’s an easier operator to use. So, these amino acids are characterised by two pKa: pKa1 and pKa2 for the carboxylic acid and the amine respectively. (Remember their tendency to release and acquire protons). 41 N.B. The empty rows indicate the amino acids that have no charged R-side chain: so, we only have the pk for the “usual” carboxyl group and amino group in the common structure of the amino acids. The pKa values of the two acidic groups, COOH and NH3+, are around 2.34 and 9.69 respectively.  These values are typical of pKa’s of the carboxylic and amine groups on the α carbon. This means that in an environment with a pH of 7.4, amino acids tend to be in a neutral situation. This neutrality arises from the balance between the negative and the positive charges. So: in a physiological solution, we have a neutral situation (N.B.: it’s not uncharged, but we have charges that balance each other: the negative charge of the carboxylic group and the positive charge of the amino group; so together they are equal to zero). The problem is when we have a R-side chain that carry a charge, a strong charge (negative or positive). In such case, predicting the overall charge becomes crucial. Will it be negative, positive, or neutral? This depends on the specific characteristics of the charged groups and their interactions. N.B. Then she mentioned the following slide just to remind us of the concept of pka: 42 If the Ka value is higher, the higher is the value of the dissociation, so we can find in the solution more H+. Instead, if we consider the pka the correlation is inverted: the higher the value of pka, the lower the dissociation level (just for a mathematical reason). Therefore, in a solution with a higher pKa value, we would expect to find a lower concentration of H+ ions. Now, we’ll analyse this example to understand better the concept: We try to get a titration: so, we try to understand how the charge of the amino acid changes as we change the PH of the solution. So: 43 Qui sopra ho inserito le 3 slide di cui lei ha parlato in questa parte della lezione; ha detto che contengono tutte le informazioni importanti. Nel paragrafo successivo si trova la sua spiegazione.  Imagine placing a general amino acid into a solution with a very low pH: a pH close to zero, although it's not exactly zero (because it wouldn’t be possible, since it’s not a life-compatible situation).  Now, we try to get the titration.  So, starting from a solution with a pH of zero we can imagine that: the solution is full of hydrogen ions (H+). Under these conditions, we are sure that the amino acid is completely 44  “covered” by these H+. So, the total charge of the amino acid will be +1 (without considering the R group!).  Now, let's consider what happens as the pH increases. At this point, the presence of H+ is weaker. So, we can consider that the final charge of the amino acid reaches a neutral balance. Why? Because as the concentration of hydroxide ions (OH-) increases, they start attracting H+ away from the amino acid. So, to sum up what we’ve said in these few points: initially, there was an abundance of H+. Now, as the concentration of H+ decreases and the concentration of OH- increases, the OH- try to withdraw the H+ from the amino acid. In this situation, there's a “battle/fight” between the pH and the pKa of the carboxyl group. The H- donor has the tendency to donate H+ due to its pKa. So, there's a fight between the molecule's tendency and the situation of the solution driven by the PH of the solution. Until now we’ve seen the “normal” situation at a physiological PH of 7.4.  in this physiological situation, the competition between the pka and the PH is exactly this one: the amino group carries a positive charge, while the carboxyl group carries a negative charge, resulting in a neutral overall situation 8. Then, when the pH increases, there is a higher concentration of OH- compared to H+: this means that more H+ are withdrawn from the amino acid, leading to a total charge of -1. So, according to this titration process, we can imagine the final charge for the total amino acid without considering the R group.  We can say all amino acids a neutral in a physiological PH. 8 (*): Domanda da parte di qualcuno: about the fight between the PH and the pKa; this was the professor’s answer: In the context of the solution, there's a conflict between the chemical pka and PH. Specifically, the donation and acceptance of H+ are managed by on the Ka in the case of the molecule and the pH in the case of the solution. This sets up a competition. So, which is the best donor or the best acceptor of H+ ions? It depends on the strength, whether it's the Ka of the molecule or the force of the solution given by the PH. But when they balance together, we have an equilibrium. 45  However, how does the presence of R-charged chains change the situation? Can we imagine the total charge for an acidic or a basic amino acid? Are we sure that we can find them in a neutral form at a physiological pH? So, can we imagine the total charge for an acidic or a basic amino acid at a physiological PH?  N.B. Remember the competition between the pKa and the PH.  So, due to this competition between the pKa and the PH, in the case of an acidic amino acid (with an R-group that tends to release H+ in the solution, for example an amino acid that has a -COOH as R-side-chain), we’re sure that at a physiological PH the acidic amino acid will have a total charge equal to -1. So, now, we don’t have a condition of neutrality anymore, but we have a -1 charge, because the R group carries this negative charge.  Conversely, for a basic amino acid (for example an amino acid that has a -NH2 as R-side- chain), at a physiological PH this basic amino acid would carry a +1 charge. Why? Because we have an additional aminic group that receives H+ from the solution. 46 (a questo punto è stata fatta una domanda sulla “competition between pka and PH”; ho inserito la risposta della prof nello stesso paragrafo in cui ne aveva già parlato prima, trovate un (*) affianco) IN SLIDES this concept is summed up as: At a very acidic pH the amino acid will have an overall +e charge and at very basic pH the amino acid will have an overall -e charge. The isoelectronic point is the pH value at which the molecule is neutral. It is the average of, these two pKas, i.e. pI = 1/2 (pKa1 + pKa2). For the simplest amino acid, glycine, pKa1= 2.34 and pKa2 = 9.6, pI = 5.97. N.B.: she didn’t mention the table above. Non-standard Amino Acids So, let's discuss nonstandard amino acids, which are amino acids that are very useful but do not participate in protein formation (therefore, they do not contribute to the peptide bonds). Although they are not part of proteins, nonstandard amino acids are very “famous” and important in our cells. Let’s see some of them:  Beta-alanine: It’s a beta-amino acid; in fact, as we said in the previous lesson, while most amino acids are alpha amino acids, they exist also the beta amino acids. Beta-alanine is a product of the catabolism of pyrimidine and is a component of Coenzyme A (CoA), a famous 47  well-known coenzyme that functions alongside ATP. Specifically, CoA is the molecule that links acidic groups and transport fatty acids in our cells.  Taurine: Taurine is nowadays famous because it can be found in some energy drinks. However, Taurine is mainly found in bile acids, where it is attached to cholesterol esters to form bile acids. This highlights its importance beyond just energy drinks.  Ornithine and citrulline: They are intermediates in urea cycle. The (previously mentioned) urea cycle serves to eliminate excess amino groups from our bodies (animal cells). It's named the urea cycle because the primary product in animal cells is urea, not uric acid. So, in our urine we have urea from amino acids. Uric acid is the result of another metabolic pathway.  Thyroxine (T4) and Tri-iodo Thyronine (T3): Thyroid hormones synthesised from tyrosine. so, tyrosine is the beginning of the cascade of the pathway to obtain the thyroid hormones.  γ-aminobutyric acid (GABA): A neurotransmitter produced from glutamic acid.  β-amino isobutyric acid: These are end product of pyrimidine metabolism.  δ-aminolaevulinic acid (δ-ALA): These are intermediate in haem synthesis.  S-adenosyl methionine (SAM): These are methyl donor formed from L-methionine  3, 4-dihydroxy phenyl alanine (DOPA): A precursor of melanine pigment. D-AMINO ACIDS These are non-standard amino acids. Amino acids normally isolated from animal and plants are L-amino acids. But certain D-amino acids are found in bacteria and antibiotics and in brain tissues of animals. In fact, as mentioned early, we now know that also neurons can synthesize amino acids in the D form. N.B.: lei non ha menzionato i seguenti esempi, che sono invece presenti nelle slides:  D-glutamic acid and D-Alanine are constituents of bacterial cell walls.  D-amino acids are found in certain antibiotics, e.g. gramicidin-S, Actinomycin-D. Animal tissues contain L-amino acids which are deaminated by L-amino acid oxidase. But there is also present D-amino acid oxidase the function of which was not known. Now D-amino acids like D- aspartate and D-serine have been found in brain tissue. This explains the existence of D-amino acid oxidase. AA derivatives As we said for sugars, also in the case of amino acids we can have covalent modifications. Some proteins also contain other types of NON-standard amino acids. In almost all cases, these unusual amino acids derive from specific modifications of an amino acid residue (“residue” after they have been incorporated into a protein (called a “post-translational modification”)) that include the simple addition of a small chemical group to the side chain, thus determining:  Hydroxylation  Methylation  Acetylation  Carboxylation 48   Phosphorylation FUNCTIONS OF AMINO ACIDS (non si è soffermata sulle seguenti slides, anche perché alcune cose erano già state dette): Polymerization of AAs Amino acids polymerize by PEPTIDIC BINDING ---CO—NH-. 49 The link between two amino acids may seem “easy” at first, but it's actually quite complex. Looking at the process, it is formed through a condensation reaction, involving the release of water molecules into the surrounding medium. This reaction occurs between the carboxyl group of the first amino acid and the amino group of the second one. This process results in the formation of a peptide bond, formed by this condensation reaction. It's important to note that while it may seem like a single bond, actually it’s not! Researches have measured the length of this bond: its length was different from that of a typical single and of a typical double bond  its length is intermediate between the one of a single bond and the one of a double bond. What’s the secret of this bond? This bond's “secret” lies in the concept of the resonance, due to the movement of the lone pair of electrons on the N atom, and it also lies in the three different values of electronegativity for the three atoms N, C, O (so, between these 3 atoms there’s a scale of electronegativity: the strongest value is the one of O, then N and finally C). As a result of resonance and of this difference in the electronegativity, there’s a flow of electrons directed first of all by oxygen: O recalls the available electrons (the lone pair of electrons on the N atom), so electrons flow starting from the N, passing to the C and then reaching the O. So, in the final situation, we have this resonance form (an extreme form of the same situation): we have O that has a negative charge and N that at the end will have a +1 charge. This movement of electrons leads to a bond length that is shorter than a typical single bond but longer than a double bond. So, to sum up: the secret is in the movement of electrons between N, C and O. So, the peptide bond is neither a typical single nor a double bond. Instead, it involves the movement of electrons between these 3 atoms. As it’s possible to see from the slide, for convenience, chemists refer to the beginning of the peptide as the amino group  N terminus. Conversely, the end of the chain is known as the carboxyl group  C terminus. Coming back to the particularity of the peptide bond: look at the movement of electrons we discussed earlier. 50 Moreover, it appears as a covalent bond, but it’s not really covalent: in fact, the O attracts the lone pair of electrons that are on N, so the N has a positive charge, and the O has a negative charge. So, now we can write the peptide bond in two different “ways”, in two different forms: the charged form and the neutral form (that we saw before) these are the two extreme forms of the same molecule (in fact, the molecule has in both cases a neutral overall charge: the essence remains unchanged, only the depiction varies). Thus, to sum up, it’s possible to write the peptide bond in these 2 ways, that are two extremes of the same situation:  In the charged form (with the 2 positive and negative charges): just to show the movement of electrons.  In the neutral form. Remember that movement of electrons is continuous, so we cannot establish a real “strict” and fixed structure of the molecule. As we said, the consequence is that: the length of this bond is between that of a single and a double bond.  so, it’s a very special bond. The planar trans configuration of the peptide bond 51 What’s the consequence of the resonance? At the level of the peptide bond (look at the yellow part of the image above) there’s a structural consequence that we can notice. The peptide bond assumes a planar (resembling a flat sheet) and rigid configuration: it’s impossible to rotate at the level of the peptide bond!  Everything is blocked at the level of this connection between the three atoms (N, C, O). If we wanted to perform a rotation, we would have to spend a lot amount of energy (a lot of Kilojoule/mole), so we can assume that there is no free rotation at this level. So, the answer to the question about the consequence of the resonance is that: due to resonance peptide bonds have a planar and rigid configuration. 52 However, another question: we just said that this rotation is precluded within the peptide bond; so, how are proteins able to organize and wrap in the three-dimensional space? Where can the movement occur? The “secret” is in the “neighbours”: while the peptide bond remains fixed, neighbours can rotate in space. So, rotation can occur on the two sides of the peptide bond: this rotation is possible around the single bonds between the α-carbon and the N (=ready for a new peptide bond) on one side, and between the α-carbon and the carbonyl C on the other side. (look at the image). (A questo punto ha aggiunto una frase, ma non si capisce bene dall’audio) So, around these links, rotation can occur, and this gives rise to several different conformations in the space for the protein. In the following image, it’s possible to look at a “longer situation” (since we have more amino acids linked together). So, again: the peptide bond is BLOCKED (like a planar sheet), but the NEIGHBOURS are free to rotate. 53 Another important feature is the configuration of this bond: the most stable conformation in solution is the trans configuration. Why trans configuration? It’s trans, because the R groups are positioned on the opposite sides of the peptide bond: this is the most stable configuration, because there’s no conflict between big R groups. The names of the angles involved in the rotation are phi (Φ) and psi (Ψ), signifying the degree of rotation around the peptide bond. Obviously, there’s a limit when we consider the combinations between Φ and Ψ, because they could not be feasible due to clashes between R groups. So theoretically, the rotation could be of 360°, but in reality, there are some limitations.  so, there are some permitted and some non-permitted combinations. THE RAMACHANDRAN PLOT There was a famous researcher, Ramachandran, that studied all the permitted combinations and created a diagram to elucidate these permitted and these non-permitted combinations of these rotations; so. to predict the conformation of proteins in space we can use this diagram. So, he studied all the combinations in order to have a possible secondary structure. In a polypeptide the main chain N-C alpha and C alpha-C bonds are free to rotate. These rotations are represented by the torsion angles phi and psi, respectively. G N Ramachandran used computer models of small polypeptides to systematically vary phi and psi with the objective of finding stable conformations. For each conformation, the structure was examined for close contacts between atoms. Phi and psi angles which cause spheres (=atoms) to collide correspond to sterically disallowed conformations of the polypeptide backbone. 54 POLYMERIZATION OF AA: PEPTIDES Polymers containing:  two amino acids: dipeptides  few (3 to 10) amino acids: oligopeptides, in particular those with 3 amino acids are referred to as tripeptides.  many amino acids: polypeptides But are very often simply called "peptides". After being incorporated into a peptide, individual amino acids are called amino acid residues. So, amino acids residues (but we can also talk about carbohydrate residues) are the monomers that create the polypeptide; they are “residues” because they “sacrifice” a group to create the link and participate in the formation of peptide bonds; so, the amino acid enters the polypeptide chain not in its entire structure. That’s why we talk about an aminoacidic residue and not about a “total amino acid”  in fact, you have to remember that in the peptide bond’s formation we have the loss of a H2O molecule (one - OH from one side and one -H from the other): so, at the end we have no longer a complete amino acid but a residue of the original molecule. 55 By convention, as we said before, the amino terminus is considered the beginning of the polypeptide chain. Variations in the length and amino acid sequence of polypeptides contribute to determining the differences in shape and biological function that characterize the various proteins. The polipeptide chain is not linear and rigid… CONFORMATION: Spatial arrangement of atoms depending on bonds and bond rotations, and it is particularly stable and has the lowest level of free energy.  THE AMINO ACID SEQUENCE DETERMINES THE FINAL PROTEIN SHAPE (CONFORMATION). Our aim is to reach the tertiary structure to explain the function of proteins: fibrous proteins (like collagen) or haemoglobin, etc.; and to explain the function of enzymes. STRUCTURAL ORGANISATION OF PROTEINS PROTEIN STRUCTURE IS NORMALLY DESCRIBED AT FOUR LEVELS OF ORGANISATION.  now, we’re going to analyse these hierarchical levels from the primary, to the secondary, tertiary and quaternary structures. These different levels of complexity are dependent between each other: each level builds upon the preceding one; secondary structures emerge from primary ones, and tertiary structures are built upon secondary elements. PRIMARY STRUCTURE Primary structure is the linear sequence of amino acids held together by peptide bonds in its peptide chain. So, “Gly, Gly, Tyr” and “Gly, Tyr, Gly” are two different polypeptides. Primary structure serves as the foundation for the subsequent structural levels, as secondary and tertiary structures. The peptide bonds form the backbone and side chains of amino acid residues project outside the peptide backbone. The free -NH2 group of the terminal amino acid is called as N-terminal-end and 56 the free -COOH end is called as C-terminal-end. It is a tradition to number the amino acids from N- terminal-end as No. 1 towards the C-terminal end. N.B. Presence of specific amino acids at a specific number is very significant for a particular function of a protein. Any change in the sequence is abnormal and may affect the function and properties of protein. So, the primary structure:  Linear amino acid sequence of the protein  It is present either in PEPTIDES ( function”. It means that each different structural organization possessed by a protein is associated with a specific biochemical function. From this point of view, it can be stated that the classification of proteins is related to: 1. Functional properties:  Extended or fibrous proteins perform generally biochemical functions, offering a valid defence against the outside world.  Globular proteins are involved in specific and multiple biological functions: enzymes, respiratory pigments, many hormones, toxins, antibodies. 2. Shape and size:  A protein is fibrous when the ration between length and width of the protein is more than 10, so the length is 10 times bigger than the width. They are linear proteins. For example: α-keratin from hair, collagen.  A protein is globular when the width of a protein molecule is less than 10. They are not linear, but are very compacted. For examples: Myoglobin, haemoglobin, ribonuclease, etc 3. Solubility and physical properties:  Simple proteins: proteins who walk alone, both fibrous and globular. -Fibrous are generally insoluble in aqueous solvents and, sometimes, unattackable by proteolytic enzymes. For example: collagen (essential constituent of connective tissue); elastin (main component of elastic fibres and vessel walls); keratin (essential component of the epidermis) -Globular are soluble and crystallizable. An example are histones.  Conjugated proteins (consisting of a simple protein and a non-protein prosthetic group) such as hemoglobin (globin + heme group). Biomedical importance of proteins:  Proteins are the main structural components of the cytoskeleton.  Biochemical catalysts known as enzymes are proteins.  Proteins known as immunoglobulins serve as the first line of defence against bacterial and viral infections. 64   Several hormones are protein in nature.  Structural proteins furnish mechanical support and some of them, like actin and myosin, are contractile proteins and help in the movement of muscle fibre, microvilli, etc.  Some proteins present in cell membrane, cytoplasm and nucleus of the cell act as receptors.  The transport proteins carry out the function of transporting specific substances either across the membrane or in the body fluids.  Storage proteins bind with specific substances and store them, e.g. iron is stored as ferritin.  Few proteins are constituents of respiratory pigments and occur in electron transport chain or respiratory chain, e.g. cytochromes, hemoglobin, myoglobin.  Under certain conditions proteins can be catabolised to supply energy.  Proteins by means of exerting osmotic pressure help in maintenance of electrolyte and water balance in body. Fibrous proteins The main fibrous proteins are keratin, collagen and elastin. There is a big different between keratin and the others two proteins because collagen and elastin belong to an extracellular environment. In fact, they belong to extracellular matrix and, in addition, they are basic structural elements and have special mechanical properties. They are found as components of skin, connective tissue, blood vessels, sclera and cornea of eye. On the contrary, keratin is inside the cell, is a component of cytoskeleton, belong to the family of intermediate fibres, it is component of hairs, horns and nails. Keratin The main features of keratin are insolubility, usually high-sulcus content filament forming proteins. These keratinous materials are formed by cells filled with keratin and are considered “dead tissues”. Nevertheless, they are among the toughest biological materials, serving as a wide variety of interesting functions, e.g. scales to armour body, nails and claws to increase prehension, hair and fur to protect against the environment. Structure of keratin Molecule formed by 2 right-handed helices coiled around each other (coiled-coil shape) and organized in protofilaments and protofibrils. The starting point is the dimer coiled coil formed by 2 keratins polypeptides, then there is the formation of protofilaments composed of two staggered rows of associated coiled coil, finally protofilaments dimerize to form a protofibrils, four of which form a microfibrils that in the cytoskeleton are called intermediate filaments. 65 Keratin is able to have this special compact structure thanks to α-helices. Usually α-helices are secondary structure proteins, so we can expect N-terminal head and C-terminal tail. Two of these proteins are coiled in a very special way called coiled-coil robe (very specific for keratin), this way to wrap the alfa-helices is not found in other proteins. Starting from the dimer (so two keratin polypeptides) that is joint in a special way, so tail to head and head to tail, and so on, creating a linear composition. Two of these filaments joint together also in a more special way that is not an exact overlapping, but in a stable row. There is a gap in the overlapping (there isn’t a linearity) that’s way? Because I have to shift the overlapping of these two dimers. What about the alpha-helices? Why are they able to form coiled-coil shape? Tandem repetition of 7 residue segments, in which the first and the fourth are hydrophobics, and the fifth and the seventh are polars. Since a 7 aa segment represents two complete turns of an alpha-helix (n=3.6 amino acids and each of the α-helix is about 5 angstrom), the apolar residues form an apolar edge along one side of the helix. This apolar edge interacts with polypeptide apolar edges of other chains to form the coiled coil superhelical structure. The polar edge interacts with water on the outside of the superhelix and stabilizes the superhelix. So, one side is hydrophobic (1/4) and the other side is polar (5/7). The hydrophobic residues are in the facing parts of the helices. The twisting of the helices is used to hide the segments with hydrophobic residues from the water. It is a loss of energy offset by the exclusion of water. 66 Additional, to create a stable structure, its possible to create disulphide bonds (cysteine-cysteine bridge) that can hold chains rigidly together. Collagen Collagen is a protein that belongs to the extra cellular matrix. Fibres of collagen outside the cell drives the movement of the cell into channel, collagen is able to bring the nutrients to the cells. It is a very important connection between cells. Collagens are the most abundant proteins in mammals (30% of total protein mass). The collagen superfamily comprises 28 members numbered with roman numerals in vertebrates (I–XXVIII). The common structural feature of collagens is the presence of a triple helix that can range from most of their structure (96% for collagen I) to less than 10% (collagen XII). Structure: Collagens consist of three polypeptide chains, called α-chains. Beyond the existence of 28 collagen types, further diversity occurs in the collagen family because of the existence of several molecular isoforms for the same collagen type (e.g., collagens IV and VI) and of hybrid isoforms comprised of α chains belonging to two different collagen types (type V/XI molecules). Functions, collagen playes structural roles and contribute to:  Mechanical properties, organization, and shape of tissues.  Found in cartilage, bones, blood vessels, skin and other connective tissues.  They interact with cells via several receptors families and regulate their proliferation, migration, and differentiation.  Some collagens have a restricted tissue distribution and hence specific biological functions. 67 Stick-shaped molecule (300.000 Da), length 3.000 angstroms, thickness 15 angstroms, it’s a very big triple helix. Collagen chains have a unique amino acid composition that caused to structure no fold canonically, but in the much more extensive helical form of the alpha-helix. It is composed by three alpha-helices wrapped around one another in a rope like fashion. Collagen is a glycoprotein containing galactose and glucose as carbohydrate contents. Collagen has a particular primary structure (in each helix), where the repeated sequence occurs glycine, proline (80%) and hydroxyproline or hydroxylysine. G – X – Y. Why glycine and proline? Both are important in the formation of the triple-stranded helix. Proline facilitates the formation of the helical conformation of each α-chain, because its ring structure causes “kinks” in the peptide chain (the amino group in proline is inside the dense ring). Glycine, the smallest amino acid, is found in every third position of the polypeptide chain. It fits into the restricted spaces where the three chains of the helix come together, so it stabilizes the structure. It’s a strategy to compact the structure. Why hydroxyproline and hydroxylysine? First of all, the formation of these two molecules is due to the addition of OH group to proline and lysine. We can have 4-hydroxyproline of 3-hydroxyproline (according to the position of OH group, that is changed by the enzymes proline or lysine hydroxylases). All these changes in the structure of polypeptide chains occur after the translational codification. In a second modification, hydroxyl group of hydroxylysines of collagen are enzymatically glycosylated by adding galactose and glucose as the carbohydrate content. The enzymes involved are glucose-transferase and galactosyl-transferase. Most commonly, glucose and galactose are sequentially attached to the polypeptide chain prior to triple–helix formation. Formation of collagen After the translation, the protein undergoes hydroxylation, that is the first modification in the ER. When it moves from ER to citosol there is the addition of the sugar. Then three of α-helix combine together, but there is unstable state, called pro collagen, that regret from cytosol to final destination, that is extracellular space. In the extracellular space the N-end and C-end are cut, so 68 there is the formation of tropocollagen. Finally occurs the crosslink formation with the consequent formation of collagen fibres. Along with hydrogen bonds forming between the three chains, there are also covalent bonds present. Covalent bonds also form cross-links between R groups of amino acids in interacting triple helices when they are arranged parallel to each other. The cross-links hold the collagen molecules together to form fibrils. The collagen molecules are positioned in the fibrils so that there are staggered ends (this gives the striated effect seen in electron micrographs). When many fibrils are arranged together, they form collagen fibres. Collagen fibres are positioned so that they are lined up with the forces they are withstanding. Of course, at the beginning, three alpha-helix come together creating hydrogen bonds. Relations between vitamin C and collagen There is a great connection between collagen and vitamin C. Some of the enzymes mentioned before needs the help of vitamins C to walk, for example the hydroxylases (enzyme that create the bond between proline or lysine and OH group) walks with the help of both vitamin C and iron. If we don’t have these two factors we can’t arrive to the formation of collagen because it’s impossible to do the hydroxylation. So, it’s necessary the presence of co-factors (vitamin C, iron, oxygen for hydroxylation). The clinical manifestation of ascorbic acid deficiency is the morpho-functional alteration of the connective tissue with the appearance of wrinkles, slowing of wound healing time, loss of tissue elasticity, disorganisation of macromolecules, loss of extracellular matrix. In the humans, collagen deficiency produces a disease called Scurvy. In scurvy, failure of conversion of protocollagen to collagen, due to the failure of hydroxylation, may lead to a rapid destruction of the collagen intermediates. 69 In addition, it is useful the absorption of acids to absorb iron. Ascorbic acid in food helps in the absorption of iron by converting the inorganic ferric iron to the ferrous form, forming water-soluble Fe-ascorbate chelate. It also helps in mobilization of iron from its storage form of ‘Ferritin’. In fact, disturbances of these functions may contribute to the development of hypochromic microcytic anaemia in SCURVY. Absorption of Fe both in normal and Fe-deficient patients is increased by over 10 per cent after administration of vitamin C. - The main defect is a failure to deposit intercellular cement substance, but also capillaries are fragile and there is tendency to haemorrhages: petechial, subcutaneous, subperiosteal and even internal haemorrhages can occur. - Wound healing is delayed due to deficient formation of collagen. - Poor dentine formation in children, leading to poor teeth formation. - Gums are swollen and becomes spongy and bleeds on slightest pressure–Hyperaemia, swelling, sponginess and bleeding of gums are seen. In severe scurvy, may lead to secondary infection and loosening and falling of teeth. 70 HEMOGLOBIN AND MYOGLOBIN The globin is the real protein part, the heme is the prosthetic group. Why the structure of hemoglobin is in this way? What are the features of a good transporter? The linkage and releasing of the molecule that the transport protein is transporting. Hemoglobin is able to joint these two different actions thanks to the quaternary structure. In quaternary structure there are more then two polypeptide chain joints together and, in this case, there are four polypeptide chains. Heme-containing proteins are characteristic of the aerobic organisms, but why? The solubility of oxygen. Oxygen is a gas, and when a gas match with another gas (or goes to solution), this two also match with different environments. There is a gap of affinity, oxygen is insoluble to melt in a solution, so there is a big problem to connect oxygen and water solution. For this reason, is necessary something that carries it, the transport protein. The normal concentration of Hb in an adult male varies from 14.0 to 16.0 g% and there are about 750 g of Hb in the total circulating blood of a 70 kg man, (very huge quantity). The percentage is related to the total volume that we have in our circulatory system that is 5 litres. Putting 14 to 16 g% and 5 litre (converted in kg) in an equation/proportion we arrive to 750g. 71 About every 3-month, hemoglobin is not destroyed but occurs the separation of heme part (in which iron is re-used for other reactions) and globin part. This last part is cut in the amino-acids component and used for other functions. This turnover is a hard work for our body. In fact, approximately 6.25 g (90 mg/kg) of Hb are produced and destroyed in the body each day. General structure: The structure of Hem remains the same in Hb from any animal source. The basic protein globin varies from species to species in its amino acid composition and sequence becaus