CARBOHYDRATES -compounds of C, H and O -originally thought of as hydrates of carbon e.g. glucose C 6 H 12 O 6 thought to be C(H 2 O) carbohydrates: -are poly-hydroxylated aldehydes and ketones -can cyclise -can form polymeric chains CARBOHYDRATES are synthesized by plants in photosynthesis (sun s energy converts CO 2 and water into glucose and O 2 ) when glucose is metabolised (oxidised): multistep process that forms carbon dioxide, water, and a great deal of energy 1
Monosaccharides simplest carbohydrates are monosaccharides (simple sugars) have three to seven carbon atoms in a chain, with a C=O at either the terminal carbon (C1) or C2 other carbon atoms bear OH group monosaccharides usually drawn with C=O near top of structure Physical Properties of Monosaccharides sweet tasting (relative sweetness varies) many hydroxy groups: hydrogen bonding polar high melting points insoluble in organic solvents like ether highly water soluble 2
Naming Monosaccharides -we classify monosaccharides by the type of carbonyl they contain (aldehyde or ketone): Classifying Monosaccharides and by the number of carbon atoms they contain: -six carbon aldo sugar -six carbon keto sugar -five carbon aldo sugar 3
Cyclic Forms of Monosaccharides The hydroxy and carbonyl groups of monosaccharides can react intramolecularly to give cyclic form: Disaccharides -some carbohydrates are complex e.g. table sugar (sucrose) is a disaccharide. 4
Disaccharides -some carbohydrates are complex e.g. table sugar (sucrose) is a disaccharide. Can be hydrolysed into monosaccharides e.g. H 1 sucrose 3 O + 1 glucose + 1 fructose (monosaccharide) (monosaccharide) (disaccharide) Polysaccharides -some carbohydrates are complex e.g. table sugar sucrose is a disaccharide. Can be hydrolysed into monosaccharides e.g. starch and cellulose are polymers of glucose: HO CH 2 OH OH O O OH CH 2 OH O OH 1 cellulose > 2000 glucose H 3 O + HO HO D-galactose-D-glucose Lactose (a disaccharide) Starch (a polysaccharide) 5
Sugars in Nucleic Acids two sugars, D-ribose and 2-deoxy-D-ribose form the building blocks of RNA and DNA, respectively. NUCLEIC ACIDS (Ch. 25.4) Nucleic acids are very high molar mass biopolymers that play an essential role in protein synthesis. ribonucleic acid (RNA) Deoxy-ribonucleic acid (DNA) 6
What is a nucleotide? -DNA strands are composed of nucleotides single strand of DNA one nucleotide -phosphoester bonds link sugar-base units together What is a nucleotide? -a sugar-phosphate-base unit is called a nucleotide: a nucleotide (fig. 25.18) 7
Bases in DNA -five bases are found in nucleic acids: (A) (G) (C) (U) (T) -base composition always follows pairing pattern e.g.: Clostridium bacteria: 37.2% A 37.2% T 12.8% G 12.8% C Bases in DNA Chargaff s Rules (p. 1054) amount of A = amount of T amount of G = amount of C -base composition always follows pairing pattern e.g.: Clostridium bacteria: 37.2% A 37.2% T 12.8% G 12.8% C 8
Base pair formation in DNA -adenine (A) hydrogen bonds only to thymine (T) -C only forms complementary hydrogen bonds to G Base pair formation in DNA two-stranded DNA double helix -two strands of DNA hydrogen bond to and coil around each other forming a double helix: (base pair) 9
Base pair formation in DNA -two strands of DNA hydrogen bond to and coil around each other forming a double helix: (base pair) two-stranded DNA double helix Chargaff s Rules Q: DNA from sea urchins is found to contain 18% G nucleotides. Use Chargaff s Rules to deduce the percentage abundance of C, T and A nucleotides. A: In DNA, G must pair with C and A must pair with T so: amount of G = 18% amount of G-C base pairs = 36% amount of T-A base pairs = 100% - 36% = 64% amount of T = amount of A = 32% Chargaff s Rules (p. 1054) amount of A = amount of T amount of G = amount of C 10
Chargaff s Rules Q: An 18 picomole sample of nucleotides from bacterial DNA yielded 3.9 pmol of adenine (A). How much C is expected to be found in the sample? A: In DNA, G must pair with C and A must pair with T so: amount of A = 3.9 pmol (= 21.67%) amount of T-A base pairs = 43.33% amount of G-C base pairs = 100% - 43.33% = 56.67% amount of G = amount of C = 28.33% (5.1 pmol) Chargaff s Rules (p. 1054) amount of A = amount of T amount of G = amount of C Replication of DNA -two strands of DNA are not identical but complementary -to produce copies of itself : 1. DNA uncoils into two separate strands 2. complimentary bases are attached new daughter DNA strand is produced featuring identical sequence 11
Replication of DNA Q: What DNA base sequence is complementary to (5 ) TCGGACATCGT (3 )? A: In DNA, G must pair with C and A must pair with T so: (5 ) TCGGACATCGT (3 ) (3 ) AGCCTGTAGCA (5 )...is the complementary sequence Replication of DNA Q: What DNA base sequence is complementary to (5 ) GTGACGACTGC (3 )? A: In DNA, G must pair with C and A must pair with T so: (5 ) GTGACGACTGC (3 ) (3 ) CACTGCTGACG (5 )...is the complementary sequence 12
Transcription of DNA -DNA provides template for synthesis of all cell s components -how are proteins made? -DNA is transcribed into another nucleic acid, RNA -messenger RNA (mrna) is a single stranded nucleic acid featuring same nucleotides as DNA except base thymine (T) is replaced by uracil (U) (see fig. 25.4) Transcription of DNA -to produce RNA readout of a gene, DNA uncoils -complimentary bases assemble to form mrna readout of the gene Translation of DNA -mrna sequence becomes translated in ribosomes where enzymes (proteins) are assembled from amino acid building blocks 13
Chemistry In Action: DNA Fingerprinting 14