SAJC 2014 H2-Bio-TYS-ANS

2014 UCLES ‘A’ Level H2 Biology Mark Scheme 1 2014 ‘A’ Level H2 Biology Mark Scheme PAPER 1 (MCQ)

2014 UCLES ‘A’ Level H2 Biology Mark Scheme 2 PAPER 2 (CORE) QUESTION 1 (a) With reference to Fig. 1.1, describe the curve between points A and B. [2] 1 Initially, rate of reaction increases gradually e.g. from 0 to 8 arbitrary unit (a.u) as temperature increases from 35 oC (point A) to 50oC. 2 Subsequently, the rate of reaction increases rapidly e.g. from 18 to 65 a.u as temperature increases from 55 oC to 65oC (point B). Examiner’s comment: The majority of candidates correctly described the change in the rate of the reaction and supported their answers by quoting figures from the graph from both axes. Candidates who considered explanation of the shape of the graph had no t recognised the significance of the instruction to ‘describe’ the curve. (b) Explain why the reaction rate changes from: (i) A to B, [3] 1 (As temperature increases,) heat increases and increases the kinetic energy of enzymes and substrates. 2 Increasing frequency of effective collision between enzyme and substrates. 3 More enzyme-substrate complexes formed per unit time, hence the rate of reaction increases. (ii) C to D. [3] 1 At the optimum temperature at point C, the rate of reaction is at its highest. As temperature increases beyond the optimum temperature, the rate of reaction decreases drastically. 2 Increased heat increases the thermal agitation of the enzyme molecule and put strain on the bonds holding the tertiary structure of the enzyme. Eventually high enough temperature leads to the disruption of weak bonds e.g hydrogen bonds, ionic bonds and hydrophobic interaction that stabilised the conformation of the enzyme and its active site. 3 The three dimensional (3D) conformation of enzyme is eventually broken permanently. Enzyme is denatured and binding of the substrate is affected. (c) Suggest how the structural features of Taq polymerase make it thermostable. [2] 1 (Taq polymerase is a protein whose structural features are influenced by) the number, types and sequence of the different amino acids in the primary structure which determines how the polypeptide chain coils and folds into tertiary structure. Tertiary structure is maintained by bonds formed by chemical interactions between the different R groups of the amino acids. 2 (To be thermostable, Taq polymerase) would have more cysteine amino acids (to form disulfide bonds) and polar or acidic / basic amino acids with polar or charged R groups, respectively (to form ionic bonds). More disulfide bonds and ionic bonds would require higher temperature to break these bonds to denature the protein. Examiner’s comment: Candidates needed to consider features of proteins that are likely to make them more thermostable. This proved challenging for many.

2014 UCLES ‘A’ Level H2 Biology Mark Scheme 3 QUESTION 2 Fig.2.1 shows the molecular structure of maltose, which is formed from two glucose molecules. (a) Give the full name of the bond holding the two glucose molecules together in the way shown. [2] 1 α(1 4) glycosidic bond ; Examiner’s comment: The majority of candidates were able to correctly give the full name of the bond shown in Fig.2.1. (b) Describe how this bond may be broken to release two molecules of glucose. [2] 1 Hydrolysis reaction with the addition of one water molecule. 2 Bond broken between –OH group on carbon atom 1 of one α-glucose molecule and carbon atom 4 of adjacent α-glucose molecule ; Examiner’s comment: Most candidates identified the type of reaction responsible for breaking the bond but fewer were able to describe how this could be brought about. A number of candidates incorrectly referred to amylase as the enzyme involved. (c) Fig. 2.2 represents the molecular structure of a type of phospholipid. (i) Describe the arrangement of phospholipids in cell membranes. [2]

2014 UCLES ‘A’ Level H2 Biology Mark Scheme 4 1 The phospholipid molecules are arranged in two rows to form a bilayer structure of the cell membranes ; 2 The phosphate head of phospholipid molecules faces outwards and interacts with the external environment and aqueous medium of the cytoplasm while the hydrocarbon chains of fatty acids face the interior of the bilayer, forming the hydrophobic core of the cell membrane. Examiner’s comment: Candidates were familiar with the arrangement of phospholipids in cell membranes and many were able to provide full descriptions. (ii) Explain how the structure of phospholipids is related to this arrangement in cell membranes. [3] 1 The phosphate head is charged (refer to Fig 2.2) while the two hydrocarbon chains of the fatty acids are non-polar ; 2 The phosphate heads interact with the water molecules in the aqueous media (e.g. through the formation of hydrogen bonds) ; 3 The non-polar hydrocarbon chains of fatty acids form hydrophobic interactions with each other ; 4 The presence of a carbon-carbon double bond causes a kink in one of t he hydrocarbon chains ; 5 This prevents the phospholipid molecules from packing too close to each other which will reduce the fluidity of the cell membrane. Examiner’s comment: Most candidates were able to relate the structure of phospholipids to their particular arrangement in cell membranes. [Total:9]

2014 UCLES ‘A’ Level H2 Biology Mark Scheme 5 QUESTION 3 (a) Describe how bacterial chromosomes differ from eukaryotic chromosomes in terms of structure and organisation. [4] Bacterial genome Eukaryotic genome Points specified in syllabus 1. Genome size / Amount of DNA Smaller genomes, 0.6 to 10Mb./ less total DNA per cell Large genomes, being less than 10 Mb – 100,000 Mb / more total DNA per cell, about 1000 times more DNA. 2. Gene length Shorter gene sequences / more compact genetic organisation Longer gene sequences / presence of more intergenic spaces 3. Chromosome structure Circular DNA molecule which is closed covalently Linear chromosomes with 2 ends 4. Packing of DNA Does not form chromatin; DNA does not package into nucleosomes Eukaryotic DNA is complexed with histones and other proteins to form chromatin; DNA is packaged into nucleosomes 5. Introns Coding sequence proceeds from start to finish without interruption by introns (no introns) Presence of introns within genes. 6. Regulatory sequences Simple regulatory sequences such as promoters More complex regulatory sequences such as enhancers and silencers Points not specified in syllabus 7. Chromosome number Single chromosome / Haploid Many chromosomes / Diploid or polyploid 8. Presence and absence of operons Two or more genes may be expressed and regulated as a unit (genomes arranged in operons). Absence of operons. 9. Repetitive sequences Few repetitive DNA sequences. Many repetitive DNA sequences 10. Coding and non- coding DNA Most of DNA are coding sequences (codes for protein, tRNA, or rRNA. Most of DNA are non-coding. 11. Origins of replication One origin of replication present Many origins of replication present 12. Presence of extrachromosomal DNA Independent small, double stranded, circular DNA called plasmids Circular, double-stranded DNA in mitochondria / chloroplasts. 13. Telomeres Absent Present (b) Describe what occurs from the end of stage 3 up to stage 4, as shown in Fig. 3.1. [4]

2014 UCLES ‘A’ Level H2 Biology Mark Scheme 6 1 One strand of the F plasmid is nicked by an (plasmid-encoded) endonuclease at the origin of transfer, and begins unwinding from another strand; 2 5’ end of the nicked strand is transferred through the sex pilus/cytoplasmic mating bridge; 3 RNA primers then bind to the nicked strand (being transferred over to the F- cell) which allows DNA polymerase to form a complementa