The reasons for knowing these points relate to the way purines and pyrimidines interact in nucleic acids, which we'll cover shortly. You need to know which are purines and which are pyrimidines, and whether it is the purines or the pyrimidines that have one ring. There are four dominant bases here are three of them:īe aware that uracil and thymine are very similar they differ only by a methyl group. The presence or absence of the 2' -OH has structural significance that will be discussed later. Deoxyribose, which lacks a 2' -OH (in deoxyribonucleotides).Nucleotides polymerize to yield nucleic acids. Let's now turn to nucleotides and nucleic acids. If the branch ends are the reactive sites, more branches provide more reactive sites per molecule. Since most monosaccharides have more than one hydroxyl, branches are possible, and are common.īranches result in a more compact molecule. A reducing end (with a free anomeric carbon).If the anomeric hydroxyl reacts with a non-anomeric hydroxyl of another sugar, the product has ends with different properties. If two anomeric hydroxyl groups react (head to head condensation) the product has no reducing end (no free anomeric carbon). Monosaccharides can polymerize by elimination of the elements of water between the anomeric hydroxyl and a hydroxyl of another sugar. The point is, a monosaccharide can therefore be thought of as having polarity, with one end consisting of the anomeric carbon, and the other end consisting of the rest of the molecule. The rest of the carbohydrate consists of ordinary carbons and ordinary -OH groups. Sugars with free anomeric carbons are therefore called reducing sugars. They can reduce alkaline solutions of cupric salts. (note: it's easy to pick out because it is the only carbon with TWO oxygens - ring and hydroxyl - attached.)įree anomeric carbons have the chemical reactivity of carbonyl carbons because they spend part of their time in the open chain form. The anomeric carbon (the carbon to which this -OH is attached) differs significantly from the other carbons. The ring can close in either of two ways, giving rise to anomeric forms, ( 5-OH adds across the carbonyl oxygen double bond.) Glucose exists mostly in ring structures. This is what you need to know about glucose, not its detailed structure. It has two important types of functional group:Ī carbonyl group (an aldehyde in glucose, some other sugars have a ketone group instead.) Monosaccharides polymerize to yield polysaccharides. Let's look at the three major classes of macromolecules to see how this works, and let's begin with carbohydrates. These macromolecules are polar because they are formed by head to tail condensation of polar monomers. WHEN THEY POLYMERIZE IN A HEAD-TO-TAIL FASHION, THE RESULTING POLYMERS ALSO HAVE HEADS AND TAILS. The main point of the first segment of this material is this: THE MONOMER UNITS OF BIOLOGICAL MACROMOLECULES HAVE HEADS AND TAILS. We will conclude this section of the course with a consideration of denaturation and renaturation - the forces involved in loss of a macromolecule's native structure (that is, its normal 3-dimensional structure), and how that structure, once lost, can be regained. So you'll need to learn only one pattern, then apply that pattern to the other systems. The stories for proteins, monosaccharides and nucleotides are just variations on the same theme. We will investigate macromolecular interactions and how structural complementarity plays a role in them. The three-dimensional structure of each type of macromolecule will then be considered at several levels of organization. We will then look at the monomers in each major type of macromolecule to see what specific structural contributions come from each. We will describe the features of representative monomers, and see how the monomers join to form a polymer. In fact, the principles governing the organization of three-dimensional structure are common to all of them, so we will consider them together. Often they are treated separately in different segments of a course. There are three major types of biological macromolecules in mammalian systems. In the example given, Q 1 = +1(1.6022 × 10 −19 C) and Q 2 = −1(1.This text is divided into five major sections:Ĭhemistry of the bonds in biological macromolecules Helicity in macromolecules Macromolecular folding Macromolecular interactions Denaturation Introduction In this case, the proportionality constant, k, equals 8.999 × 109 J The equation can also be written using the charge of each ion, expressed in coulombs (C), incorporated in the constant. This value of k includes the charge of a single electron (1.6022 × 10 −19 C) for each ion. The proportionality constant k is equal to 2.31 × 10 −28 J Where each ion’s charge is represented by the symbol Q.
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