RI Nov 16 H2 P2

2016 H2 Biology P2 9744/02 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. As substrate concentration increases, frequency of effective collisions * between enzyme and substrate molecules increases; 2. Rate of enzyme-substrate complex * formation increases and rate of reaction increases as active sites * of enzymes are readily available and substrate concentration is limiting; (ii) the rate of reaction becomes constant. [2] 1. All active sites* of enzymes are saturated with substrate at any one point in time; 2. Concentration of substrate is no longer limiting and enzyme concentration is limiting, rate of reaction will remain constant (graph plateaus). (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. Inhibitor binds to site other than active site and changes conformation* of active site*; 2. Hence inhibitor effectively decreases availability of enzymes as it forms an enzyme-inhibitor complex; 3. Effects of inhibition cannot be overcome by increasing substrate concentration; (c) Draw, on Fig. 1.1, the approximate shape of t he curve if a competitive inhibitor were added to the enzyme instead of a non-competitive inhibitor. [2] 1. Correct shape and labelled; 2. approaches maximum rate of reaction; (d) Suggest why the penicillin molecule is an effective inhibitor of transpeptidase. [2] 1. Cell wall peptide and penicillin are similar in conformation and hence both are complementary in shape and charge to the active site; 2. Penicillin can act as a competitive inhibitor*for the active site *of transpeptidase; 3. Hence, cell wall peptides cannot bind to acti ve site and cross-links of cell wall peptides cannot form. 2 (a) Identify the molecules labelled A and B on Fig. 2.1. [2] A: messenger RNA B: RNA polymerase (R: general transcription factor as the molecule is unzipping and separating the double helix DNA) (b) Explain how the control elements shown in Fig. 2.1 influence transcription. [4] Promoter (c) Competitive inhibitor

1. serve as recognition site for the binding of general transcription factors * and RNA polymerase*; 2. to initiate transcription*; 3. has critical elements, TATA box* that determines the precise location of transcription start site; 4. has critical elements, CAAT and GC boxes * to improve efficiency of promoter by recruiting general transcription factors* and RNA polymerase* to promoter. Enhancer 5. when bound with specific transcription factors* known as activators*, promotes assembly of transcription initiation complex* 6. when bound with specific transcription factors* known as activators*, may recruit histone acetyltransferase* and chromatin remodeling complexes * to decondense chromatin (increase accessibility of promoter to general transcription factors and RNA polymerase) 7. increase frequency of transcription (c) Describe the role of a silencer control element in transcription. [2] 1. allow binding of specific transcription factors* called repressors* by preventing assembly of transcription initiation complex* at promoter 2. when bound with specific transcription factors* known as repressors*, may recruit histone deacetylase* and chromatin remodeling complexes * to condense chromatin (decrease accessibility of promoter to general transcription factors and RNA polymerase) 3. decreases the frequency of transcription; (d) Describe a feature of the control of prokaryotic transcription that is not shown in Fig. 2.1. [2] 1. Genes coding for proteins involved in sa me biochemical pathway usually clustered together on one operon. 2. If the repressor* is active, it binds to the operator* 3. and prevent the RNA polymerase * from binding to the promoter thus preventing transcription [Points 4 and 5 are no longer in the syllabus] 4. Formation of RNA polymerase holoenzyme* which is made up of core RNA polymerase and a sigma factor*; which recognizes and binds to both the -10 sequence/Pribnow box* and -35 sequences 5. The more similar the -10 and -35 sequences are to the consensus sequences, the stronger the promoter and therefore the higher frequency of transcription. 3 (a) Describe the changes shown in Fig. 3.1. [3] 1. Upon infection, the number of HIV viruses increase to a peak/maximum at week 3 and decline steeply to near 0 at week 8.5; 2. From the week 9 to the year 6 (since primary infection), viruses remain relatively constant near to 0 with 2 small peaks at year 2 and year 4; 3. From 6 to 8 years after infection, viruses increase steeply to a large number and plateau from year 8 to 10; 4. The T helper cells decrease steeply from initial infection to week 3 and rises and remain relatively constant till year 3; 5. From 4 to 9 years, the T helper cells decrease gradually to near to 0; 6. From 9 to 10 year, the T helper cells remain constant near to 0; (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. T helper cells help to activate specific naïve B cells into plasma B cells to make antibodies for antibody-mediated response;

2. the HIV infects more and more T helper cells, the levels of T helper cells lower from 4 to 9 years to near to 0, as the infected T cells are destroyed; 3. The increasing loss of T helper cells leads to impaired immune responses in the affected individual who then becomes increasingly susceptible to opportunistic diseases; (c) Why is HIV described as a retrovirus? [2] 1. A retrovirus is a RNA virus that duplicate via reverse transcription in the host cell; 2. using the RNA genome as a template, reverse transcriptase* produce DNA from its RNA genome by complementary base pairing; (d) Explain why viruses are described as obligate parasites. [2] 1. An obligate parasite is an organism that cannot live independently of its host; 2. They require the host cell to complete their life cycle and reproduce; 4 (a) Identify structures A, B, C and D, as shown on Fig. 4.1. [4] A : mRNA B : subunits of ribosomes C : transport vesicle D : membrane of the rough endoplasmic reticulum (RER) (b) Use Fig. 4.1 to suggest how the structures labelled B attach to the endoplasmic reticulum through the growing polypeptide. [2] 1. Molecules called signal recognition particle (SRP) bind to one end of growing polypeptide at ribosome; 2.The molecules bind to receptor protein on membrane of B, to bring ribosomes to it; (c) Suggest why the newly synthesised polypeptides shown in Fig. 4.1 cannot pass directly into the cytosol. [1] 1. Ribosome is attached to the membrane of RER where there is a membrane channel for the synthesized polypeptide to enter; 2. Polypeptide folds into its 3D conformation in the lumen and is too large to pass through the transient pores of the phospholipid bilayer of the membrane/ through the membrane channel; (d) Describe what happens to the newly synthesi sed polypeptides released from the rough endoplasmic reticulum. [3] 1. Polypeptides are enclosed in a transport vesicle; 2. It fuses with cis/ convex face of Golgi apparatus* and undergoes chemical modification/ eg: glycosylation; 3. Golgi apparatus targets and sorts the poly peptides and secretory vesicles are released from trans face; 4. The secretory vesicles fuse* with cell membrane and release polypeptides via exocytosis; 5 (a) Draw a genetic diagram to explain the results of the first cross in which all offspring had purple stems and green leaves. [3] Parental phenotype Purple stems, yellow leaves X White stems, green leaves Parental genotype NNgg X