SAJC 2016 H2-Bio-TYS-ANS

2016 UCLES ‘A’ Level H2 Biology Mark Scheme 1 2016 ‘A’ Level H2 Biology Mark Scheme PAPER 1 (MCQ)

2016 UCLES ‘A’ Level H2 Biology Mark Scheme 2 PAPER 2 (CORE) QUESTION 1 Fig.1.1 shows the effect of increasing substrate co ncentration on the rate of an enzyme-catalysed reaction in the presence and absence of a non-competitive inhibitor. Fig. 1.1 (a) Explain why, in the reaction with the enzyme only, as substrate concentration increases : (i) the rate of reaction increases at first [2] 1 Active sites of available enzyme molecules are not fully occupied by substrate, increase in frequency of effective collisions between enzyme and substrate molecules (as substrate concentration increases) ; 2 Increase in concentration of enzyme-substrate complexes formed per unit time, increase in concentration of products formed per unit time ; 3 Substrate concentration is limiting factor at low substrate concentration, (ii) the rate of reaction becomes constant [ 2 ] 1 Active sites of available enzyme molecules are saturated with substrate molecules ; 2 Any extra substrate molecule has to wait until the E-S complex in the active site of the enzyme is released as products before the substrate can enter the active site ; 3 Enzyme concentration is limiting factor at high substrate concentration ;

2016 UCLES ‘A’ Level H2 Biology Mark Scheme 3 (b) Explain why, in Fig. 1.1, the addition of a non-competitive inhibitor causes the reaction to become constant at a lower rate. [2] 1 Non-competitive inhibitor binds to the enzyme at its allosteric site, results in change in 3D conformation of the enzyme, 3D conformation at active site altered ; 2 Substrate cannot bind to the active site, decrease in frequency of effective collisions between enzymes and substrate molecules / decrease in concentration of enzyme-substrate complexes formed per unit time, decrease in concentration of products formed per unit time ; (c) Draw, on Fig. 1.1, the approximate shape of the curve if a competitive inhibitor were added to the enzyme instead of a non-competitive inhibitor. [2] 1 (Labelled) Curve beginning at 0 ; 2 Vmax reached at higher substrate concentration ; enzyme + competitive inhibitor

2016 UCLES ‘A’ Level H2 Biology Mark Scheme 4 (d) The antibiotic penicillin irreversibly inhibits the activity of transpeptidase. Transpeptidase is a bacterial enzyme that cross-links cell wall peptides during the formation of bacterial cell walls. Fig.1.2 shows part of each of the molecular structure of a cell wall peptide and penicillin. Fig. 1.2 Suggest why the penicillin molecule is an effective inhibitor of transpeptidase.[2] 1 Penicillin has similar 3D conformation to the cell wall peptide (substrate for transpeptidase) and competes for binding to the sam e active site on transpeptidase as the cell wall peptide ; 2 Prevents cell wall peptide from binding to the active site, decreasing the concentration of the enzyme-substrate complexes formed per unit time, preventing the cross-linking of cell wall peptides at the active site ; Examiner’s comment: Candidates were able to suggest an explanation for the inhibition of the transpeptidase enzyme by penicillin. Not all considered binding of the inhibitor or the role of the active site. [Total: 10]

2016 UCLES ‘A’ Level H2 Biology Mark Scheme 5 QUESTION 2 Fig. 2.1 shows a eukaryotic transcription complex. Fig. 2.1 (a) Identify the molecules labelled A and B on Fig. 2.1. [2] A – Ribonucleic acid (RNA) B – RNA polymerase Examiner’s comment: Most candidates recognised the labelled molecules. (b) Explain how the control elements shown in Fig. 2.1 influence transcription. [4] Promoter: 1. DNA sequence upstream of a gene that contains the TATA box to which one transcription factor r e c o g n i s e s a n d binds, initiating a series of interactions between multiple transcription factors; 2. RNA polymerase II binds to the promoter of the DNA template with the aid of transcription factors, forming the transcription initiation complex, allowing the transcription process to start; Enhancer: 1. Control element that activator p r o t e i n ( a t y p e o f s p e c i fi c t r a n s c r i p t i o n factor) binds to via the proteins’ DNA-binding domain; 2. Activator proteins then binds to co-activators / mediator proteins via its activation domain w h i c h i n t u r n b i n d t o general transcription factors and RNA polymerase II via protein-protein interactions; 3. facilitate the correct positioning of transcription initiation complex on the promoter to increase the rate of transcription; Examiner’s comment: Many candidates gave full and detailed answers to this question and clearly understood the different control elements shown. (c) Describe the role of a silencer control element in transcription. [2]

2016 UCLES ‘A’ Level H2 Biology Mark Scheme 6 1. Silencer is a control element that repressor protein (a type of specific transcription factor) binds to. This turns off t r a n s c r i p t i o n e v e n i n t h e presence of activator proteins; 2. by blocking the binding of activator proteins to the control elements or to components of the transcription machinery; OR 3. repressor binds to its control element which is within an enhancer and act to turn off transcription even in the presence of activator proteins; Examiner’s comment: Most candidates fully appreciated the role of sil encer control elements in transcription and were able to provide detailed responses. (d) There are a number of differences between prokaryotic transcription and the eukaryotic transcription shown in Fig. 2.1. Describe a feature of the control of prokaryotic transcription that is not shown in Fig. 2.1. [2] 1. The control of prokaryotic transcription involves an operator (a region of DNA) that lies close to the promoter; 2. An active repressor binds t o t h e operator, preventing the RNA polymerase from accessing and transcribing the structural genes; Examiner’s comment: Most candidates were familiar with the control of transcription in prokaryotes and were able to describe a feature not shown in the diagram of eukaryotic transcription. [Total: 10]

2016 UCLES ‘A’ Level H2 Biology Mark Scheme 7 QUESTION 3 (a) Describe the changes shown in Fig. 3.1. ..[3] 1. Number of helper T cells decreases from weeks 1-3 and increases between weeks 3-9 2. Correspondingly, number of HIV viruses increases from weeks 1-3 and decreases between weeks 3-9 3. Number of helper T cells continue to decrease from year 1 to 10 while number of HIV viruses increase gradually between year 1-6, and then sharply from year 6-10 Examiner’s comments: Most candidates provided detailed descriptions of the changes. Some candidates went on to develop explanations for the changes, which was not a requirement of this question. (b) Suggest how the changes in the number of T helper cells shown in Fig. 3.1 would affect the health of an untreated HIV-infected individual over the course of the infection. ..[3] 1. As number of helper T cells decreases from weeks 1-3 and subsequently from years 1-10, the number of HIV viruses increase correspondingly; 2. Number of HIV decreases when number of T cells increase from weeks 3-9 3. [explain] When number of helper T cells decreases, there are less helper T cells to secrete cytokines to activate B cells into plasma cells which secretes antibodies / There are also less helper T cells to secrete cytokines to activate CD8 T cells into T cytotoxic cells; 4. immune system is weakened and cannot eliminate the HIV viruses effectively Examiner’s comments: Some candidates did not make fu