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CHM 112 GAGE SUMMER I 1999 1. NAME _____Key____________________ DATE___________________________________ EXAM 4 Match the name of the compound in Column A with its classification in Column B. __E__1. keratin (16) A. monosaccharide B. disaccharide C. polysaccharide D. amino acid E. polypeptide F. enzyme G. triglyceride H. phospholipid I. sphingolipid J. glycolipid K. steroid L. prostaglandin __C__2. amylopectin __F__3. serine transaminase __L__4. PGB2 __D__5. threonine __H__6. lecithin __K__7. estradiol __B__8. maltose 2. Supply the most appropriate word or phrase for each of the following statements. a.____non-competetive_____ a. b.____select any three from list_____ type of enzyme inhibition where a molecule alters the active site of the enzyme from another point on the enzyme b. three functions that proteins perform ___________________________ ___________________________ c._____amino acid___________ c. small biochemical that can act as a buffer d. _____electrophoresis________ d. technique for separation of proteins e. _____fatty acid e. long, straight chain carboxylic acid f. two classes of substances that denature proteins f. ___heat, UV light ________ _______ ___heavy metals, pH change____ g. _substrate concentration, pH, tempg. (20) factor that affects enzyme rate 3. Classify each of the following molecules as specifically as possible by type (i.e. monosaccharide, phospholipid...) from the structure. (18) O OH OH H2N CH C OH H H OH O H CH 2 H C H O HO O OH H O OH HO H H OH H H NH 2 ______amino acid______________ ____________disaccharide____________ O O OH CH 3 HO H2 C C CH 3 H N CH C CH N H CH 2OH CH 3 CH 3 CH CH C NH 2 O HO CH 3 C HO O O ________steroid______________ ______________dipeptide___________ O C O O H3C C ( CH 2)16 (CH 2)17CH 3 O O C H2 O C H2 C (CH 2)18CH 3 _________wax___________________ 4. CH O C O C17H31 (CH 2)16CH 3 __________triglyceride_____________ Identify the type of glycosidic linkage in each of the following carbohydrates: H OH H H O HO OH O H OH O H HO H H H _______alpha-1, 4_______________ O HO HO H (6) H OH H OH O HO H H OH H OH H H H2C H O HO H HO H H OH OH ________beta-1,6__________________ 5. Complete the following biochemical reactions with appropriate products. O O H2N CH C + OH H2N CH C O O OH H2N CH C (15) H N CH C OH CH 2 CH 3 CH 2 CH 3 CH CH 3 CH CH 3 CH 3 CH 3 O H2C O (CH 2)16CH 3 O C CH2OH CH O O C H C (CH 2)14CH 3 O O H2O/H+ -----------> CH3 CH2CH 2 P N (CH 2)16CH 3 HO C HO C (CH 2)14CH 3 CHOH O O C CH3 O H O CH2CH 2 P CH3 O CH3 N CH3 O O CH3 HOH 2C O C C O HO CH 3 NaOH NR - non-saponifiable lipid --------------------------------> O 6. a. A polysaccharide has a molar mass of about 400 g/mol and is soluble in water while a triglyceride with the same molar mass is highly insoluble. Explain why. Lipids have large, non-polar hydrocarbon sections which are not attracted to water. Carbohydrates have multiple hydroxyl groups that form hydrogen bonds easily with water so the interactions, and thus the solubility, are greater. (5) b. Explain how insoluble lipids can be transported through the bloodstream. (5) Lipids attach to transport proteins such as serum albumin. Since proteins have many side groups that are polar, proteins are more easily dispersed/moved in a water environment. Protein acts as a carrier. c. What is the tertiary structure of a protein and how is it maintained? (5) The tertiary structure of a protein is the “folding” or compacting of the coiled or pleated structure. The tertiary structure is maintained by disulfide bridges (between amino acids with thiol groups), hydrogen bonds (between polar groups), salt bridges (between amines and carboxylic acid groups), and hydrophobic or dispersion forces (between non-polar groups). d. Explain from a chemical perspective why cellulose is indigestible while starch is even though they are of comparable masses and are made of the same monomers. (5) Cellulose is held together with beta-1,4 glycosidic linkages. Human enzymes are designed to attack mainly alpha linkages. Humans have no enzyme that will hydrolyze this linkage. e. From an energy perspective, explain how an enzyme is able to increase the rate of a chemical reaction. (5) An enzyme allows a different route or mechanism for the conversion from reactant to product. The enzymatic mechanism has a lower activation energy (energy barrier). Since less energy is necessary to overcome the barrier, the reaction will occur faster or the rate increases.