QP Code: 3834 F.Y.B.Sc. in Biotechnology Semester II Examination (April 2018) Model Answers Chemistry I- Biorganic chemistry

Q.1. Do as directed: (Any fifteen) 15

Define the following terms:

1) Carbohydrates Carbohydrates are poly hydroxy aldehydes or ketones with a general formula Cn(H2O)n, where n  3 2) Dipeptide A dipeptide holds two interacting amino acids with one peptide bond

3) Isoelectric pH At isoelectric pH the amino acids exists as a zwitter ions and they bear no net charge.

4) Isomers Isomers are the compounds having the same empirical formula and the structure, but they differ from one another by a configuration around only one carbon atom.

5) Unsaturated fatty acid Unsaturated Fatty acids are carboxylic acids with atleast one or more double bonds in its hydrocarbon chain.

6) Pitch of the helix- is the height of one complete helix turn, measured parallel to the axis of the helix.

7) Saponification When animal fat/oil reacts with aqueous NaOH or KOH, they form soap and glycerol. Since this reaction leads to the formation of soap, it is called the Saponification.

Draw structures of the following biomolecules: 8) Purine base. Any one base

9) D- glyceraldehydes CHO-CHOH-CH2OH

10)

D-glucose

11)

Serine

12)

Acetone

Name the following: 13) An acidic amino acid- Aspartic acid, glutamic acid

14) One keto sugar- Fructose, ribulose

15) Sterol of biological significance - cholesterol

16) Fibrous protein present in hair - Keratin

17) Sugar present in DNA. -deoxyribose

18) Parent compound of phospholipids- Phosphatidic acid

19) The transport protein present in the red blood cell - Albumin

20) Type of bond present between base pair G-C – Hydrogen bond

Q2. a) Compare and contrast: unsaturated fatty acid and saturated fatty acids 8

Fatty acids are carboxylic acid with hydrocarbon side chain. General formula is R- COOH. The hydrocarbon side chain may be saturated or unsaturated in nature SFA USFA Saturated fatty acids don’t have Unsaturated fatty acids have one or unsaturation or double bonds in their more double bonds in their structure structure. Their general formula is R- COOH - SFA s is found in animal sources. USFA s is found in plant based sources. They are solids at room temperature They are liquids at room temperature, Eg: ghee , butter They are popularly called as oils Butyric acid- CH3 –CH2-CH2- COOH EG: Palmitolic acid has one double bond- C16:9 Palmitic acid- CH3 –(CH2)14- COOH Oleic acid has one double bond- C18:9 Lauric acid.- CH3 –(CH2)10- COOH Leinoleic acid has three double bonds- C18: 9,12, 15 , Aarachidonic acid has four double bond- C20:5,8,11,14 4M for each category

b)Define and classify monosaccharides. 7 1. Placement of its carbonyl group : If the carbonyl group is an aldehyde and attached to C-1, the monosaccharide is an Aldose; if the carbonyl group is a ketone attached to C-2, the monosaccharide is a Ketose. Therefore monosaccharides are called aldoses and ketoses derivatives of polyhydric alcohols. Glucose is an example of an aldose and fructose is an example of ketose.

2. Number of Carbon Atoms: Monosaccharides with three carbon atoms are called trioses, those with four are called tetroses, five are called pentoses, six are hexoses, and so on. These two systems of classification are often combined. For example, glucose is an aldohexose (a six- carbon aldehyde), ribose is an aldopentose (a five-carbon aldehyde), and fructose is a ketohexose (a six-carbon ketone).

(6M for classification and any three str )

OR c) Compare and contrast: 8 i. Maltose and Lactose (4M) Maltose Lactose

it’s the principle sugar found in malt it’s the principle sugar found in milk. Lactose is an important component used Lactose is an important component used by dairy and Brewery by dairy and pharma industry Maltose is composed of -D glucose Lactose is composed of β -D galactose linked to D glucose by means of a α linked to D glucose by means of a β 1→4 linkage 1→4 linkage.

ii. Starch and cellulose (4M)

Starch Cellulose

Homopolymer of Homopolymer of α-D glucose βD glucose

Amylose- unbranched Amylopectin- branched Storage polysaccharide found in Stuctural polysaccharide plants found in plants Bond between monomer is mostly Bond between monomer is α1→4 ( amylose) β1→4

α1→6 linkage is found in unbranched amylopectin, branched starch unit

d.)What are phospholipids? Draw structures of Lecithin and Cephalin. 7

Phospholipids are derived from phosphatidic acid. They contain glycerol backbone esterified with two fatty acids at 1,2 position and a phosphoric acid and a nitrogen base.

(4M)

Eg: Phosphatidyl ethanol amine, Phosphatidyl serine, Phosphatidyl inositol, phosphatidylcholin (lecithin), and cephalin. Lecithin/ phosphatidylcholin, found in egg yolk and other membrane lipids. R= fatty acid

(4M for Lecithin / Cephalin, any one structure )

Q3. a). Explain N-terminal sequencing of proteins using Edmans and Sanger’s 8 Methods. Edman degradation, developed by Pehr Edman, is a method of sequencing amino acids in a peptide. In this method, the amino-terminal residue is labeled and cleaved from the peptide without disrupting the peptide bonds between other amino acid residues. Phenyl isothiocyanate is reacted with an uncharged N-terminal amino group, under mildly alkaline conditions, to form a cyclical phenylthiocarbamoyl derivative. Under acidic conditions, this derivative of the terminal amino acid is cleaved as a thiazolinone derivative. The thiazolinone amino acid is then selectively extracted into an organic solvent and treated with acid to form the more stable phenylthiohydantoin (PTH)- amino acid derivative that can be identified using chromatography or electrophoresis. This procedure can then be repeated again to identify the next amino acid (4M)

Sanger’s reagent was FDNB.FDNB specifically binds to the n terminal residue and forms a DNP amino acid. Chromatography is used to identify this amino acid. (4M)

b) Classify amino acids on the basis of the polarity of the R group. 7 The twenty amino acids are generally classified as 1. Nonpolar (hydrophobic, meaning they do not readily interact with water), 2. Polar(uncharged and 3. Hydrophilic, meaning they readily interact with water), or charged (either positively or negatively). (1M) The chemical composition of the R-group determines their classification. The various properties of the R-groups greatly influence how the amino acids interact with each other and their environment. (One example of each category- 6M)

OR

c) Schematically represent the titration curve of an amino acid and describe 8 the different stages. Titration helps in understanding of occurrence of different ionic forms of amino acids. It is a graphical representation of pH changes (Y axis) plotted against corresponding volume of NaOH of known concentration (X axis). a) Nature of amino acids in particular their ionisable groups determines the PI of protein. PI is affected by the presence of an additional –COOH in acidic amino acid or an – NH2 groups in a basic amino acids. At isoelectric point, the protein exists as a zwitter ion or dipolar ion. (2M) b) Draw titration curve ( 2M)

The titration curve for alanine is shown in Figure . At a pH lower than 2, both the carboxylate and amine functions are protonated, so the alanine molecule has a net positive charge. At a pH greater than 10, the amine exists as a neutral base and the carboxyl as its conjugate base, so the alanine molecule has a net negative charge. At intermediate pH's the zwitterion concentration increases, and at a characteristic pH, called the isoelectric point (pI), the negatively and positively charged molecular species are present in equal concentration.

This behavior is general for simple (difunctional) amino acids. Starting from a fully protonated state, the pKa's of the acidic functions range from 1.8 to 2.4 for -CO2H, and (+) 8.8 to 9.7 for -NH3 . The isoelectric points range from 5.5 to 6.2.

Titration curves show the neutralization of these acids by added base, and the change in pH during the titration. (4M)

d.) Explain the term conjugated Proteins. Give examples and state their 7 functions.

Many proteins contain only amino acids and no other chemical groups, and they are called simple proteins. A conjugated protein is a protein that functions in interaction with non-protein chemical group, which is attached by covalent bond or weak interactions. Conjugated proteins, upon hydrolysis yield, some other chemical component in addition to amino acids. The non-amino part of a conjugated protein is usually called its prosthetic group. (3M) Conjugated proteins are classified on the basis of the chemical nature of their prosthetic groups. Examples of conjugated proteins- 1. Glycoproteins- phosphoproteins, , flavoproteins, Glycoproteins are generally the largest and most abundant group of conjugated proteins. They range from glycoproteins in cell surface membranes that constitute the glycocalyx, to important antibodies produced by leukocytes. 2. Hemoproteins- Hemoglobin contains the prosthetic group known as heme. Each heme group contains an iron ion (Fe2+) which forms a co-ordinate bond with an oxygen molecule (O2), allowing hemoglobin to transport oxygen through the bloodstream. As each of the four protein subunits of hemoglobin possesses its own prosthetic heme group, each hemoglobin can transport four molecules of oxygen. 3. Lipoproteins-present on Cell membrane 4. Metalloproteins- Enzymes are complexed with metals- hexokinase, PFK etc 5. phytochromes/ cytochromes- take part in electron transport, 6. opsins and chromoproteins. (any four class and example 4M)

Q4. a) Give a brief account of types of RNA and discusses their functions. 8 Types of RNA Function(s) 1) Messenger RNA Transfer genetics information from genes to ribosome to synthesize proteins 2) Heterogeneous nuclear Serve as precursor for mRNA RNA and other RNAs 3) Transfer RNA Transfer amino acid to mRNA for protein synthesis 4) Ribosomal RNA Provides structural framework for ribosomes 5) Small nuclear RNA Involved in mRNA processing 6) Small nucleolar RNA Plays a key role in the processing of rRNA molecules 7) Small cytoplasmic RNA Involved in the selection of proteins for export. 8) Transfer-messenger Mostly present in bacteria. Add RNA short peptide tags to proteins to facilitate the degradation of incorrectly synthesized proteins.

b) Explain denaturation and renaturation of DNA in detail. 7 Denaturation of DNA is a loss of biological activity and is accompanied by cleave of hydrogen bonds holding the complementary sequences of nucleotides together. This result in a separation of the double helix into two constituent polynucleotide chains. (2M) Diagrammatic representation –(2M) Effect of pH on denaturation ( 1M ) Effect of Temperature on denaturation (1M) Renaturation- ( 1M ) The denatured DNA can reformulate hydrogen bonds between complementary single strand, making it likely to reform double helix structure again. This process is called as renaturation

OR

c) Give an account of different conformations of DNA double helix. 8 Double helical DNA exist in many different forms like, A, B, C, D, E and Z. Out of which A, B and Z are important form. ( 1 M ) Characteristics A-DNA B-DNA Z-DNA Helix sense Right- Right- Left-handed handed handed Helix- diameter 25.5⁰A 23.7⁰A 18.4⁰A Rise per base pair 2.3⁰A 3.4⁰A 3.8⁰A Base pair per turn 11 10 12 of helix Helix pitch 25.30⁰A 35.36⁰A 45.60⁰A Shape Broadest Intermediate Narrowest Any 4 points for each form -6M

d) Explain clover leaf structure of tRNA using a suitable diagram. 7 Neat labeled clover-leaf model of tRNA- 4 Marks Descriptions – each arm of tRNA- 3Marks

5. Write a short note on: (Any three) 15

1) Types of Triacylglycerol Triacylglycerol- Neutral or depot fat. In Triacylglycerol, central glycerol molecule is condensed with three fatty acids. (1M) If all the three fatty condensed are of the same type the molecule is called as simple TG and if more than one type of fatty acid is condensed, the molecule is known as mixed TG.(2M) TG is the fat reserve of the body. It can be hydrolysed and fatty acids can be beta oxidized whenever needed. Glycerol serves as a 3 carbon skeleton reserve, which undergoes gluconeogenesis in cases of hypoglycemia. ( 2M structure, 2M types, 2M function) (2M)

2) Structure and function of Starch Structure: Starch is a homopolymer of α-D glucose. Bond between monomer is mostly α1→4 and this unbranched unit is known as Amylose. α1→6 linkage is found in amylopectin.

Function: It is the storage polysaccharide found in plants. It is also important in food industry.

( 4M structure, 1M function )

3) Denaturation of proteins Disorganization of native protein structure is called denaturation. Physical agents - heat, violent shaking, X rays, UV rays Chemical agents- Acids, alkalies, heavy metals pb,Hg, salicylate, organic solvents cause denaturation. (2M) Properties: Loss of native protein structure Primary structure is retained Denatured protein is easily digested Peptide linkage is resistant to denaturing forces Higher viscosity of denatured protein Lower surface tension in denatured protein (3M)

4) Chargaff’s rule. Chargaff's rules state that DNA from any cell of all organisms should have a 1:1 ratio (base Pair Rule) of pyrimidine and purine bases and, more specifically, that the amount of guanine should be equal to cytosine and the amount of adenine should be equal to thymine. The ratio of A+T/G+C, known as dissymmetry ration varies greatly from one species of DNA to the other 5) Functions of Nucleotides i) Act as building blocks/ precursors of DNA and RNA ii) Nucleotide triphosphate especially ATP , act as universal currency of energy in biological system.iii) Adenine nucleotides are components of coenzymes of B- complex vitamins-FAD,NAD.iv) Act as metabolic regulators.v) Act as chemical messenger-cAMP