Why the nitrogenous base in DNA is planar

Nucleic Acid Chemistry

Transcript

1 Chemistry of nucleic acids Note: You will find the CD "The Nature of Genes" in the studio. The most important topics in molecular biology are conveyed in an easily understandable manner using tutorials and exercises. Synthesis and breakdown of the pyrimidine bases Note: The synthesis and breakdown of the pyrimidine and purine bases are subject of the 2nd year of study. They are therefore only presented here in summary form. 1

2 The biosynthesis of the pyrimidines takes place in 3 phases: Carbamoyl phosphate is formed by the cytoplasmic carbamoyl phosphate synthetase II: 2 ATP + glutamine + CO2 + H2O -> 2 ADP + glutamate + carbamoyl-P Carbamoyl phosphate, which has a high group transfer potential, reacts with aspartate to carbamoyl aspartate, which cyclizes to dihydroorotate in an equilibrium reaction (enzyme: aspartate transcarbamoylase). This is dehydrated to orotate (enzyme: dihydroorotase). Orotate reacts with 5-phosphoribosyl-1-diphosphate (PRPP) to form orotidine-5'-phosphate which is decarboxylated to uridine-5'-phosphate (UMP). Phosphorylation to UTP by kinases The other nucleotides (CTP and TTP) or their deoxyribose derivatives dctp and dttp for DNA are derived from uridine derivatives! Degradation: the ring is partially hydrogenated and hydrolytically opened between N-3 and C-4. However, since CO2 was split off between orotic acid and uridine during the synthesis, the breakdown product is not aspartic acid but alanine. Released NH3 is further metabolized to urea in the urea cycle and thus excreted. Synthesis and degradation of the purine bases The biosynthesis of the purines is much more complicated than that of the pyrimidines. The synthesis begins at the ribose-5-phosphate part of the nucleotide: Ribose-5-phosphate + ATP -> 5-phosphoribosyl-1-pyrophosphate (PRPP) The synthesis of the purinric system begins with an NH3 transfer: PRPP + glutamine + H2O - > 5-phosphoribosyl-1-amine + glutamate + PP This reaction is limiting for the entire purine synthesis and is inhibited by AMP and GMP (end product inhibition). Phosphoribosylamine is then converted in several steps into inosine monophosphate and this into anenosine and guanosine monophosphate. Degradation: Degradation is divided into two steps: first phosphate is split off from the nucleotides (by a phosphatase) and then the ribose residue is split off by a nucleoside phosphorylase. Two enzymes now act on the free purines and nucleosides: deaminases, which convert amino-purines into oxy-purines, and xanthine oxidases, which convert hypoxanthine via xanthine into uric acid, which can be excreted. In most mammals, uric acid is broken down further into allantoin by uricase. However, this step is not possible in humans because the uricase enzyme is missing. 2

3 Chemistry of bases Four bases are involved in the structure of DNA, namely adenine (A), guanine (G) [purine derivatives], cytosine (C) and thymine (T) [pyrimidine derivatives]. RNA contains uracil (U) instead of thymine. The purine and pyrimidine bases are planar and hydrophobic. They consist of heterocyclic ring compounds which contain nitrogen and carbon and carry peripheral non-polar hydrocarbon and hydrogen nitrogen groups. They are called bases because they add H + at physiological pH. C, T and U are simple derivatives of a pyrimidine six-membered ring, A and G are purine compounds (a five-membered ring condensed with the six-membered ring). The absorption maximum of these bases is 260 nm, i.e. in the UV range. Tautomeric Forms of the Bases One of the many reasons for a mutation is the occurrence of tautomeric forms of the bases. This is understood to mean a change in the bases that occurs spontaneously but with low frequency. This tautomeric form now pairs with another base, which leads to a mutation, e.g .: 3

4 Normal form A (amino) paired with TC (amino) paired with GG (keto) paired with CT (keto) paired with A Rare tautomeric form A (imino) paired with CC (imino) paired with AG (enol) paired with TT (Enol) paired with G Tautomers are structural isomers that are in equilibrium. The difference in the structures is due to the displacement of a proton. The equilibrium is strongly on the side of the amines and ketones, but there is also a small proportion of imines and enols. The H-bonds between bases need an H + donor and an H + acceptor. If there is a different tautomeric form because a proton has shifted within a compound, then the acceptor or donor changes depending on the base, and the same H + bridges can no longer be built up. 4th

5 Other ways of change offer two spontaneous chemical reactions that can cause serious DNA damage in cells. It is the deamination and the depurination of the bases, in which either another base is created through the elimination of ammonia (deamination) or the base is completely split off from the sugar (depurination). Metabolism of the sugars The sugars for nucleoside synthesis can be provided in two ways, namely in an oxidative and a non-oxidative way: oxidative way (hexose monophosphate way): 5

6 second way (sugar conversions): Note: The metabolism of sugar is treated as a matter of biochemistry and is therefore only presented here in summary form. Sugar chemistry A distinction is made between deoxyribonucleic acid (DNA) with 2-deoxyribose and ribonucleic acid (RNA) with ribose as sugar. The sugars form the "backbone" of the double helix because they are on the outside and are connected to each other via phosphate towards the "top" and "bottom". The 2'-OH group of ribose is responsible for the chemical reactivity of the RNA. If this OH group is attacked by a base (e.g. OH-), an alcoholate (O-) is formed. This attacks the neighboring phosphorus atom and thus splits the RNA chain (nucleophilic attack). Towards the center of the helix, the sugars are N-glycosidically attached to a purine resp. Pyrimidine base bound. If you think of the bases and the sugars as planar slices, then these form a right angle: 6

7 Chemistry of nucleic acids The center of the DNA is formed by the bases and is therefore hydrophobic. The basic rings that protrude inwards are comparable to the steps of a spiral staircase and are perpendicular to the helix axis. There are always two bases opposite each other, which are connected to one another by hydrogen bridges. A purine ring always forms a base pair with a pyrimidine ring and vice versa (i.e. they are complementary), since two opposite purine rings in a double helix would take up too much space, while two pyrimidine rings could not fill the space. The number of hydrogen bonds determines, among other things, the stability of the double helix. H-bonds are formed between an H of an OH or NH group and a lone pair of electrons of an O atom or an N atom when the groups approach each other at a distance of approx. 0.28 nm. The bond is not very strong, more than 10 times weaker than a major valence bond. Differences in stability arise because the base pairs form different numbers of H-bonds: while guanine and cytosine form three H-bonds, there are only two between adenine and thymine. This has the consequence that the stability of the double helix increases with a higher proportion of G - C pairings. 7th

8 The greatest influence on the stability of the double helix are the hydrophobic forces that act between the relatively closely stacked bases (the distance is about nm). Each base pair is rotated by approx. 36 around the axis of the helix in comparison to the neighboring base pair, so that approx. 10 base pairs result in a complete rotation of 360. The DNA described above is the most common form and is known as the B-form. However, depending on the water and salt content of the medium, the DNA can adopt a different geometry. Other characteristic structures are the A- and B-DNA, which you can find out more about in the atelier at the "Structure of nucleic acids" post. The "backbone" of the double helix is ​​formed by linking the sugars with the phosphates, which turns the periphery of the helix into a hydrophilic shell. The negative charges on the phosphate residues cause the two strands to be repelled. This is reduced or prevented by the binding of cations to the phosphates (neutralization of charges). Sugar and phosphate are linked like esters, sugar and base are linked N-glycosidically. Task: Why do adenine and guanine not form base pairs in DNA? Solution: p. 11 How do you isolate DNA? If you want to study DNA, you first have to isolate it from cells. The following steps are necessary for this: 1. Obtaining cell material 2. Disrupting the cells 3. Centrifuging off insoluble parts 4. Separating and cleaning nucleic acids from proteins 5. Collecting aqueous phase 6. Precipitating nucleic acids with alcohols 7. Dissolving nucleic acids in buffer 8. DNA and Separating RNA Note: Watch the film DNA isolation in the video corner (duration: approx. 7 min.) Details on DNA isolation can be found in the brochure. Denaturation of DNA What happens when DNA is heated? The double helix does not remain stable, but separates into the two single strands. This process is called denaturation or melting. 8th

9 The strands of DNA separate because the hydrogen bonds between the complementary base pairs are broken at high temperatures (e.g. 90 C) (the required temperature depends, among other things, on the length of the DNA and the G-C content). The physical properties of the DNA change, e.g. Viscosity, light absorption. This breaking up of the helix structure is called a melting process and the melting point Tm of the DNA is defined as the temperature at which half of the DNA is present as single strands. As we have seen before, the two possible base pairs A-T, respectively. G-C consists of a different number of hydrogen bonds (G-C three, A-T two). Thus, the melting point depends on the number of G-C base pairs, because these give the helix greater stability and the melting point Tm of a G-C-rich DNA is higher than that of an A-T-rich DNA. The following example serves to illustrate the change in the light absorption of the DNA during denaturation: The absorption of the DNA molecules of the bacterial strains Pneumococcus and Serratia is measured at 260 nm with increasing temperature. Single-stranded DNA absorbs about 1.4 times more (i.e. 40% more) than double-stranded DNA at 260 nm. The increase in absorption - this process is called the hyperchromic effect - is therefore a measure of the single-stranded DNA content of the sample examined: 9

10 The position of the melting curve can be influenced as it depends on the solvent: lowering the salt concentration, increasing the pH and the presence of certain organic solvents shift the curve to the left. On the right-hand side of the graphic, the dependence of the melting point on the G-C content of the respective bacterial DNA can be clearly seen. Tasks: Why is the melting point Tm of Pneumococcus DNA lower than that of Serratia DNA? What is the significance of a left shift in the melting curve? Solutions: p. 11 Renaturation Under suitable conditions, denatured ("melted") DNA segments can be converted back into the double-stranded form. This process is called renaturation. The speed of this process depends on various factors: - on the concentration of the DNA and the number of similar sequences - on the concentration of cations that neutralize negative charges on phosphates - on the temperature (the most favorable temperature is approx. 25 C below the Tm ) The rate of reassociation is expressed as the product of the DNA concentration Co and the renaturation time t, the so-called cot value. Practical procedure for renaturation the DNA to be examined is broken down into fragments of approx. 500 nucleotides by shear forces, so that a comparison of genomes of different sizes is possible regardless of the DNA length: heat to 100 C so that a complete separation is achieved Observe the regression of double-stranded DNA at 65 C and at a molar sodium ion concentration by making use of the hyperchromic effect. The speed of this reaction only depends on the initial concentration of the complementary sequences C. Hybridization The process of renaturation can be extended by pairing any two complementary nucleic acid sequences with one another and letting them form a double-stranded structure. If nucleic acids of different origins are involved, then one speaks of hybridization e.g. RNA with DNA or DNA with structurally related DNA. The principle of hybridization is to mix two preparations with single-stranded nucleic acid 10

11 and then determine the double-stranded fraction that was formed. This process is the basis of modern molecular biological techniques. Note: You will need knowledge of the principle of hybridization later for the post of gene cloning. Sources The texts and images in this script correspond to the page numbers from the following works: Lewin B .: "Genes - Textbook of Molecular Genetics", 2nd edition; P.65-71; Alberts B., Bray D., Lewis J., Raff M., Roberts K., Watson J.D .: "Molecular Biology of the Cell" 2nd Edition; Pages 57-63,,, 281 Karlson P., Doenecke D., Koolman J .: "Short textbook of biochemistry for physicians and natural scientists", 14. Edition, page 5, 9,, Knippers R .: "Molecular Genetics", 4th edition; Pages 5, 35-45, Note: In the reading corner there are textbooks available for in-depth study. Solutions Why do adenine and guanine not form base pairs in DNA? Adenine and guanine are both purines. They would take up too much space in the helix as a base pair. Why is the melting point Tm of Pneumococcus DNA lower than that of Serratia DNA? Serratia-DNA has a higher content of bases G and C (approx. 60% compared to 38% for Pneumococcus) and is therefore more stable. The result: more energy in the form of heat is required for denaturation. What is the significance of a left shift in the melting curve? A left shift means that the stability of the DNA is reduced. The consequence of this: Denaturation at lower temperature, the melting point Tm is thus lowered. Exercises with solutions can be found in the internet version of the studio! Note: The Molecular Biology revision course defines the material that is required in the examinations. Because of its brevity, however, it is not suitable as a primary source of information! 11

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