2018 Collated Biomolecules & Enzymes STQ MS

2018 Collated Biomolecules and Enzymes (DNA, Gene Expression) 2018 / H2 / AJC PRELIM / P2 Q11Fig. 1.1 shows the effect of pH on the activity of a protease enzyme at the optimal temperature of 37oC. Fig. 1.1(a)Draw, on Fig. 1.1, the approximate shape of the curve if the same experiment is conducted at 25oC.Similar shape to Fig. 1.1 but lower than curve in Fig. 1.1;[1](b)Explain with reasons the shape of the curve you have drawn.At lower temperature, lower kinetic energy of molecules;Less effective collisions between enzyme and substrates, less enzyme-substrate complex formation per unit time / lower rate of enzyme-substrate complex formation;[2]

(c)Using information from the graph, explain why proteases stored in vesicles with pH 7.2 cannot break down vesicular membrane proteins and suggest how these proteases can be activated through increase in pH.(At least one)With low activity of just 200 U/ml, enzyme is inactive at pH 7.2;Optimal enzyme activity of 900 U/ml (accept reasonable figure quoted, correct units quoted) is highest at pH 9;At pH 9 (accept argument at pH 7.2): charges on acidic and basic R-groups altered;Contact and catalytic residues has the correct charge to catalyse the reaction at pH 9 (accept converse);R-group interactions such as ionic and hydrogen bonds are altered;Tertiary structure / 3D conformation / configuration of enzyme is that of an active enzyme;Active site shape is complementary to shape of substrate;Maximum rate of enzyme-substrate complex formation;[max 4m](At least one)pH can be increased to optimum pH by pumping of H+ ions out of the vesiclesVesicle membrane has proton pumps;[6][Total: 9]

2018 / H2 / EJC PRELIM / P2 Q1 ( Cancer Eukaryotic Cell Structure)2The synthesis of collagen is shown in Fig. 1.1. ACProcess B(tropocollagen)Fig. 1.1(a)(i)Describe two important functions of structure A in the synthesis of collagen. ……………………………………………………………………………………………………… ……………………………………………………………………………………………………… ……………………………………………………………………………………………………… ...…………………………………………………………………………………………………[2]Structure A is the nuclear pore;

1.Allows mature mRNA to leave nucleus and enter cytosol for translation into collagen by ribosom2.Allows RNA polymerase/ ribonucleotides to enter nucleus to synthesise mRNA/tRNA durtranscription of collagen gene;3.Allows exit of mRNA to ribosomes to be translated; 4.Allows tRNA to leave nucleus for amino acid activation/ amino acid attachment for use translation during collagen synthesis;(any 2)(ii)Tropocollagen leaves the cell to be assembled to form collagen fibrils via Process B. OutProcess B. ……………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………[3]1.Process B is exocytosis;2.Secretory vesicle containing tropocollagen is transported to cell surface membrane afuses with the cell surface membrane to release tropocollagen extracellularly;3.This process requires ATP;(b)Suggest how chemical modification such as hydroxylation in organelle C results in collagen havinhigh tensile strength. …………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………[2]1.Hydroxylation is the process where hydroxyl groups are added/ attached/ joined to polypeptide chain;2.To enable formation of numerous interchain hydrogen bonds during the formationtropocollagen triple helix, giving rise to high tensile strength;

Fig. 1.2 shows two electron micrographs. One of the electron micrograph shows part of a normal cell while other electron micrograph shows part of a cancer cell. The white arrows point to an organelle within the cThe appearance of this organelle in both cell types were visibly different. The cancer cell had a higher actithan the normal cell. Organelle (indicated by white arrows) in a normal cellOrganelle (indicated by white arrows) in a cancer cell Fig. 1.2(c)(i)Describe the visible difference between the organelle indicated by the white arrows in Fig. 1between the cancer cell and the normal cell. ………………………………………………………………………………………………………………………………………………………………………………………………………[1]1.Nucleoli in the cancer cell were denser/darker/more darkly stained and larger;(ii)Account for the higher activity of the cancer cell. …………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………[2]

nthesis, which are key components of ribosomes required for translation;1.Nucleoli are the sites for rRNA synthesis which are key components of ribosomes requifor translation;2.Larger and denser nucleoli in cancer cells due to increased rRNA enable cancer cellachieve higher rates of protein synthesis to enable excessive proliferation/growth;[Total:

2018 / H2 / EJC PRELIM / P2 Q23The rate of glycolysis is regulated by the action of ATP on the enzyme phosphofructokinase (PFK), the third enzyme in the glycolysis pathway. PFK catalyzes the formation of fructose 1,6-bisphosphate from fructose-6-phosphate and ATP.A graph of PFK against F6P concentration would exhibit the ‘sigmoidal curve’ typical of allosteric enzymes.Fig. 2.1 shows the effect of substrate F6P (fructose 6-phosphate) concentration on the activity of PFK at different ATP concentrations. PFK is an allosteric enzyme with a quaternary structure that is inhibited by high levels of ATP. High levels of ADP and AMP will increase activity of the enzyme. F6PFig. 2.1(a)(i)With reference to PFK, explain the term ‘allosteric enzyme’. ……………………………………………………………………………………………………… ...…………………………………………………………………………………………………[1]5.PFK is an enzyme that has more than 1 subunits/ multimeric/ 4 subunits 6.with the allosteric site where AMP/ADP/ATP can bind to; Reject references to ‘site other than active site’

(ii)On Fig. 2.1, label the graphs which reflect low ATP and high ATP concentrations respectively. [1] F6PFig.2.1All or none;(b)Enzyme researchers have used a model of allosteric enzyme mechanism called the ‘symmetry model’ to explain the action of PFK. F6P binds with great affinity only to the active state but not to the inactive state. Binding of one F6P molecule will progressively shift PFK structure from the inactive state to the active state.A graph of PFK against F6P concentration would exhibit the ‘sigmoidal curve’ typical of allosteric enzymes. (i)Explain how the binding of a molecule like AMP to PFK can increase the activity of PFK. ……………………………………………………………………………………………………… ……………………………………………………………………………………………………… ……………………………………………………………………………………………………… ..…………………………………………………………………………………………………[2]

1.Binding of AMP to allosteric site of PFK will induce conformation change in PFK from inactive to active state;2.Conformation change will help to stabilize the active state of PFK / and will lead to the active site being made more complementary to F6P; (ii)The enzymatic mechanism in PFK is similar to that of the oxygen-binding mechanism of the transport protein, haemoglobin. Sketch the shape of the oxygen binding graph for haemoglobin at increasing concentrations of oxygen in the space below. [1]% oxygen saturation of haemoglobin / %Oxygen partial pressure / mmHgS-shaped curve starting from origin (gentle slope followed by steep curve, followed by gentle slope)(iii)Using your knowledge of the structure of haemoglobin, explain t