RI Nov 14 H2P2 ans

Raffles Institution Nov 2014 (H2 Biology) Paper 2 (for 9744 syllabus) 2017 Nov 2014 H2 Bio Paper 2 N14P2Q1 (a) With reference to Fig 1.1, describe the curve between points A and B [2]. 1. As temperature increases from 350C (pt A) to 500C, the rate of reaction of enzyme increases gradually from 0 to 8a.u. 2. However, from 500C to 650C (pt B), there is a steeper increase in rate of reaction of enzyme from 8a.u. to 65 a.u. This increase is about 7 times that of the increase in rate from 350C to 500C . (b) Explain why the reaction rate changes from: (i) A to B [3] 1. As temperature increase from 35 0C (at A) to 65 0C (at B), there is increase in kinetic energy of the enzyme and substrate molecule. 2. which increases frequency of effective collisions* between substrate and enzyme thus rate of formation of enzyme-substrate complex* increases ; 3. increasing temperature also increases the number of molecules having sufficient energy to overcome the activation energy* barrier to form the products of reaction; (ii) C to D [3] 1. At 750C (pt C), the optimum temperature* of Taq polymerase, there is greatest number of molecular collisions and hence max rate of reaction 2. As temperature increase further to 850C (pt D), which is beyond optimum temperature, thermal agitation / vibration of enzyme molecule breaks hydrogen, ionic bonds and other weak interactions that stabilizes the 3D conformation resulting in denaturation* 3. The enzyme active site* no longer complementary in shape* and charge to the substrate and the rate of reaction decreases steeply from maximum rate of 95 a.u. to almost 0a.u. (c) Suggest how the structural features of Taq polymerase make it thermostable [2] 1. More Cysteine* residue may be present in the Taq polymerase allowing the formation of more disulphide linkages* with another cysteine residue to maintain the 3D conformation 2. Disulphide linkages are strong covalent bonds difficult to be broken by high temperature 3. Hence the 3D conformation of active side maintained \ N14P2Q2 (a) Give the full name of the bond holding the two glucose molecules together in the way shown. [2] 1. -1,4-glycosidic bond (b) Describe how this bond may be broken to release two molecules of glucose. [2] 1. Addition of water in hydrolysis* reaction 2. Enzymatic action of maltase 3. Acid hydrolysis by heating it with hydrochloric acid (either 2. or 3.)

Raffles Institution Nov 2014 (H2 Biology) Paper 2 (for 9744 syllabus) 2017 (c) (i) Describe the arrangement of phospholipids in cell membranes. [2] 1. 2 layers of phospholipid molecules in a bilayer 2. With the hydrocarbon tails facing the inside of the membrane and the phosphate heads facing the outside next to aqueous medium (ii) Explain how the structure of phospholipids is related to this arrangement in cell membranes. [3] 1. Phosphate heads are charged and hydrophilic and will form hydrogen with water 2. Hydrocarbon tails are non-polar and therefore hydrophobic and arranged away from aqueous medium 3. And forms hydrophobic interactions with the hydrocarbon tails of other phospholipid molecules to form a hydrophobic core in the bilayer structure [Total : 9] N14P2Q3 (a) Describe how bacterial chromosomes differ from eukaryotic chromosomes in terms of structure and organization. [4] Feature Prokaryotic genome Eukaryotic genome Size 105-107 base pairs 107-1011 base pairs Appearance Generally a single, circular molecule Multiple, linear molecules Molecule Double Helix DNA Association with proteins Yes – relatively less e.g. histone-like proteins Yes – large amounts e.g. histones, scaffold proteins Level of DNA packing/coiling Relatively low: DNA double helix some looping around histone-like proteins (A) Unfolded chromosome from E. coli has a diameter of 430µm. (B) DNA is folded into chromosomal looped domains by protein- DNA associations. Six domains are shown, but actual number is High: (A) DNA double helix is associated with proteins called histones. DNA molecules are negatively- charged, histone A B C D Formation of looped domains Looped domains Supercoiling Circular chromosomal DNA DNA double helix Histones Linker DNA (“string”) Nucleosome (“bead”) Nucleosomes (10-nm fiber) - Euchromatin Nucleosome Protein scaffold 30-nm fiber - Heterochromatin Looped domains (300-nm fiber) Metaphase chromosome

Raffles Institution Nov 2014 (H2 Biology) Paper 2 (for 9744 syllabus) 2017 about 50. (C) Supercoiling cause further compaction, such that it fills an area of about 1 µm. are positively-charged. DNA thus is held around histones by electrostatic interactions. Most of DNA is wound around octamers of 8 histone proteins to form nucleosomes, the 10 nm fibre. Remainder of DNA, called linker, joins adjacent nucleosomes. (B) The 10-nm fibre coils around itself to form a 30 nm chromatin fiber (or solenoid). (C) The 30-nm fibre forms loops called looped domains (a 300-nm fibre) when associated with scaffold proteins. (D) Supercoiling present. The loops further coil and fold to produce characteristic metaphase chromosome. Location Nucleoid region, not membrane- bound Nucleus (surrounded by nuclear envelope) Extrachromosomal DNA Present – plasmids (much smaller rings of DNA compared with bacteria chromosome) No plasmids (however mitochondria and chloroplast have their own DNA) Number of genes 4,500 25,000 Non-coding regions (between and within genes) Not common – typically less than 15% Common – about 98% i. introns None present (rare; only in some genes) Many present ii. promoters Present iii. enhancers/ silencers Rarely present Present iv. repeated sequences Few Many v. operons Many Few known ones (e.g. in nematodes) Origin of replication (per chromosome) One Many *Genome refers to a complete set of genetic material in a particular cellular component.

Raffles Institution Nov 2014 (H2 Biology) Paper 2 (for 9744 syllabus) 2017 (b) (i) Describe what occurs from the end of stage 3 up to stage 4, as shown in Fig. 3.1. [4] 1. After contacting a recipient cell, the sex pilus retracts, pulling the two cells together. 2. the F+ cell then forms a temporary cytoplasmic mating bridge* with the F- cell and transfers its F plasmid DNA to it. 3. one strand of the double-stranded F plasmid is nicked by a nuclease. 4. The free 3’ end of the nick is extended by DNA polymerase for the synthesis of a new complementary strand using the intact strand as the template 5. The newly synthesized strand displaces the nicked strand which is transferred concurrently, via the 5’ end, across the mating bridge into the recipient cell. 6. Upon completion of a unit length of the plasmid DNA (after 1 round), another nick occurs to release the original strand and end the replication of the newly synthesized strand; (ii) Explain the role of F plasmid. [3] 1. F factor on F plasmid, codes for proteins necessary for the formation of sex pili and subsequent cytoplasmic mating bridge, 2. allowing for conjugation to occur between bacteria. 3. This allows for bacterial genes to be transferred