Organic Macromolecules Don’t be scared Testosterone Adenosine Triphosphate

Organic Macromolecules Overview Organic Macromolecules Biopolymers Quiz A Proteins Carbohydrates Primary Structures Quiz C Quiz B Monosaccharides Secondary Structures Tertiary Structures Dissacharides Polysaccharides

Quiz A- Basic Organic Chem. Principles How many bonds do the following atoms form in a covalent compound? a) O b) C c) S d) N e) H f) P 2. Write down the formula of the functional group in each of the following compounds. a) amine b) amide c) alcohol d) carboxylic acid e) ketone f) aldehyde g) ester 3. Give the name for each of the following molecules. a) CHCH e) b) CH3CH2OH f) c) C2H6 g) d)

Quiz A- Basic Organic Chem. Principles Which of the following molecules might be unsaturated and which are definitely unsaturated? a) C3H8 b) C4H8 c) C3H4 d) CH3CHCHCH3 5 a) Name the following molecules. i) CH3CH2CH2CH2CH2CH2CH2CH2OH ii) CH3CH2CH2CH2CH2OH iii) CH3CH2OH b) Identify the 2 types of intermolecular bonding present in each of the above molecules. c) Which of the three molecules do you expect to have the highest boiling point temperature? Explain your answer.

Quiz A- Basic Organic Chem. Principles 6. Which type of reactions do you expect the following molecules to undergo? a) CH2CH2 b) CHCH c) CH3CH2CH – CH3 Cl 7. Classify the type of reaction occurring in each of the following cases. a) b) c) 8. Write down all the possible isomers for the compounds which have the following formulae. a) C3H8O a) C4H8

Quiz A- Basic Organic Chem. Principles 9. Identify the functional groups present in the following molecule.

Proteins Base Acid Most structurally complex organic macromolecules. The building blocks of proteins are amino – acids. Base Acid = side chain 20 amino acids in all 12 the body can synthesize termed non-essential amino acids. 8 that we obtain from our diet termed essential amino acids.

Proteins Non – essential amino acids L–Alanine L–Asparagine L–Aspartic Acid L–Cysteine L–Glutamic Acid L–Glutamine Glycine L–Proline L–Serine L–Tyrosine L–Arginine L–Histidine

Proteins Essential amino acids(must be in the diet) L–Isoleucine L–Leucine L–Lysine L–Methoinine L–Phenylalanine L–Threonine L–Tryptophan L–Valine

Proteins L–Alanine Glycine L–Serine

- + Proteins – Zwitter Ions All amino acids have the ability to form their corresponding ionic forms as follows. H N C O R H - N C O H R H N C O R +

Proteins Peptide Bond N C O R N C O R N C O R O N C R O Dehydration H

Ala • Ser • Gly Primary structure of Proteins Reactive ends Peptide Bonds Ala • Ser • Gly

………… ………… Try • Lys • Ala • Ser • Gly • Val • Pro Primary structure of Proteins Try • Lys • Ala • Ser • Gly • Val • Pro ………… ………… Polypeptide (Protein) : 50 – 2000 amino acids

Primary/Tertiary structure of Proteins L–Cysteine L–Alanine

Secondary structures - -helix Clockwise O H 3.6 Amino acids per turn of the helix

- + Secondary Structures - Zwitter ion formation Lys Asp Some amino acids have extra reactive groups e.g. Carboxylic acid (COOH) – glutamic acid – aspartic acid Amine (NH2) – lysine Lys + - + + Ionic bond + - + - - - H Asp

Secondary structures - -sheets Top view Secondary structures - -sheets Hydrogen bonding Side view

Tertiary structures - sheets - helix Has a 3D structure. Contains a combination of secondary structures & primary structures. Structure determined by side – chains.

Cys • Lys • Tyr Quiz B- Proteins 1. Which of the amino-acids have a hydrophobic side-chain? 2. Which of the amino-acids have an acidic side-chain? 3. Which of the amino-acids have an basic side-chain? 4. Which amino acids have a side chain which can become involved in protein hydrogen bonding? 5. Draw the zwitter – ion structure of the amino-acids glycine and serine. Draw the chemical structure of the tripeptide formed from the following sequence of amino-acids. • Lys • Tyr Cys Highlight the acidic end of the molecule and the basic end of the molecule.

Carbohydrates Monosaccharides Can be split into two groups : Aldoses & Ketoses. ‘ oses ’  “CH2O” ‘ Ald ’  Aldehyde ‘ Ket ’  Ketone ‘ C6H12O6’ Aldoses e.g. Glucose Ketoses e.g. Fructose Both Glucose & Fructose are Hex - ‘oses’. Sugars can be 3 to 7 carbons long.

Carbohydrates In aqueous solution sugars form rings e.g. glucose. 1 2 6 3 5 4 4 1 5 2 3 6

Carbohydrates Sugars polymerize via glycosidic  1–4 linkages. C O 1 2 3 4 5 C O H 1 2 3 4 5 C O H 1 2 3 4 5 H O Glycosidic linkage 2C6H12O6  C12H22O11 + H2O Dehydration

Carbohydrates       glucose + glucose maltose glucose + fructose sucrose   polysaccharide monosaccharide disaccharide   e.g. Glucose Maltose Starch (100s - 1000s of units)

Carbohydrates The resulting polysaccharide of  1–4 linakges of glucose is starch (storage polysaccharide in plants). Amylose Unbranched starch  1–4 linkages Starch Amylopectin Branched Starch 6 1–4 linkages & 1–6 linkages Branched  1–6 linkages occur every 30 linkages

Carbohydrates Sugars polymerize via glycosidic  1–4 linkages as well. 2 3 4 5 C O H 1 2 3 4 5 C O H 1 2 3 4 5 H O 1–4 Glycosidic linkage The resulting polysaccharide is cellulose (structural polysaccharide in plants).

Quiz C- Macromolecules The most abundant (by mass) class of organic macromolecules in the biological world is : A Carbohydrates B Proteins C Nucleic Acids (DNA & RNA) D Lipids D – ribose is a pentose. It can also be classified as an aldose. Write down a possible structure for D – ribose. The monosaccharide threose has 4 Os ! That is it has 4 oxygen atoms. What is the molar mass of threose?

Quiz C- Macromolecules A comprehension exercise 4. When one draws the structure of D – glucose there are 2 ways to do it : either a Fischer projection or a Haworth projection. 1 2 3 4 5 6 The Fischer projection shows glucose as a linear molecule and the numbering of the carbon atoms is shown on the structure. When it cyclizes it is the oxygen atom on carbon 5 which ends up in the ring. 1 2 3 4 5 6 This ring structure is called the Haworth projection and at equilibrium between the two forms the linear form is only 0.02% of the total glucose present.

Quiz C- Macromolecules The Haworth projection has the atoms on the RHS of the Fischer projection ‘down’ i.e. below the ring in the Haworth projection, and those on the LHS of the Fischer projection are up (above the ring). The –OH group on carbon 1 (the so called anomeric carbon) can either be up or down, (), or up, (). The  form is about twice as common as the  form. Draw the Haworth projection of the  form of D – idose, given the Fischer projection below. b) Draw the Fischer projection of D – galactose given the Haworth projection below. - D - galactose