2016 VJC H2 Biology Prelims Paper 2 Answers

VICTORIA JUNIOR COLLEGE BIOLOGY DEPARTMENT JC2 PRELIMINARY EXAMINATIONS 2016 Higher 2 BIOLOGY 9648/02 Paper 2 Core Paper Answers 14 September 2016 1 (a) (i) Identify structures P and Q and describe briefing their functions. [3]  P – Nucleolus [1/2]  Transcription of ribosomal RNA [1/2]  Site of ribosome assembly [1/2]  Q – Smooth endoplasmic reticulum [1/2]  Site of synthesis of lipids [1/2]  Detoxification of drugs and poisons [1/2]  Stores calcium ions required for contraction in muscle cells [1/2] (ii) Contrast the structure of a lysosome with structure P. [2] Lysosome P (Nucleolus)  Membrane-bound [1/2]  Not membrane-bound [1/2]  Contains hydrolytic enzymes [1/2]  Contains DNA coding for rRNA [1/2] (b) (i) Name a carbohydrate that functions as a storage molecule for T-helper cells.  Glycogen [1] (ii) Describe three structural differences between ce llulose and the carbohydrate in (bi). [3] Cellulose Glycogen  Made up of β-glucose  Made up of α-glucose  Joined by β 1,4 glycosidic bonds  Joined by α 1,4 glycosidic bonds and α 1,6 glycosidic bonds  Unbranched, straight chains  Branched brush-shaped  Alternate subunits rotated 180 o  Alternate subunits in the same orientation  Inter-chain hydrogen bonds present  No cross-linkages between adjacent chains (iii) Explain how the presence of two types of bonds in amylopectin enables it to carry out its function. [2]

1    VJC H2 Biology Paper 2 Preliminary examination 2016     α 1,4 glycosidic bonds between subunits within a branch [1/2]  α 1,6 glycosidic bonds at branch points [1/2]  form branched helical structure  compact for storage function  Both bonds can be broken enzymatically to release α glucose for respiration [1/2]  Hydrolysis of α 1,6 glycosidic bonds breaks up am ylopectin into many branches for more efficient breakdown [1/2] (c) Phosphofructokinase is an allosteric enzyme. Explain how the presence of an allosteric inhibitor affects the enzymatic activity of an allosteric enzyme. [2]  Allosteric inhibitor binds to allosteric site [1/2]  Causes the enzyme conformation to change to inactive state [1/2]  Active site not complem entary to substrate [1/2]  Prevent effective collision and formati on of enzyme-substrate complex [1/2] 2. The diagram below shows an enzyme involved in the activation of tRNA for translation in prokaryotes. (a) (i) Explain the mode of action of this enzyme [3]  amino-acyl tRNA synthetase;  has a specific active site that is  complementary to specific tRNA anticodons and a specific amino acid  Ref. to induced fit theory  catalyses the attachment of a specific am ino acid to the 3’ stem of the tRNA in the formation of the amino-acyl tRNA complex  by lowering the activation energy of the reaction  through the formation of an enzyme structure complex (ii) Explain the significance of having more than one type of the enzyme named in (ai) in the cell [2]

2    VJC H2 Biology Paper 2 Preliminary examination 2016     There are 20 different amino acids and hence 20 different amino-acy-tRNA synthethases are needed  This ensures that each of the 20 ami no acids are correctly linked to their tRNAs/ ref to specificity of enzyme for substrate  As the anticodons of the tRNA bind by complementary base pairing to the condons in the P and A site of the ribosome  When an amino acid has been linked to a tRNA, it will be incorporated into a growing polypeptide chain at a position dictated by the codon of the mRNA.  Allowing the primary stru cture of the polypeptide to be synthesised correctly according to the codons of the mRNA that is being translated (b) How does the order of nucleotides in a gene encode the information that specifies the primary struct ure of a polypeptide? Include two features of the genetic code in your answer.[3]  Transcription of the gene by RNA polymerase produces a complementary sequence of mRNA;  Three consecutive nucleoti des on mRNA make one codon;  One codon codes for one amino acid;  Although more than one codon can code for the same amino acid due to the degenerate nature of the genetic code;  The ribosome read the codons one after another with no space between codons as the genetic code is non-overlapping.  The ribosome thus joins the amino ac ids in the correct sequence as coded for by the codon sequence to form the polypeptide’s primary structure/ the codon sequences hence determine the number, type and sequence of amino acids of the polypeptide synthesized by the ribosome  As the genetic code is punctuated w here 3 codons do not code for amino acids but function as stop codons that mark the end of translation. The ribosome stops polypeptide synthesis when a stop codon is located in the ribosome A site as a release factor enters the site to release the completed polypeptide. (c) Explain how different polypeptides can be synthesised simultaneously from a single mRNA in prokaryotes. [2]  A prokaryotic mRNA is a polycistronic mRNA;  And contains the coding sequence for more than one polypeptide/ structural gene product involved in a related metabolic pathway;  Each coding sequence has its own start and stop codon;  Allows more than one ribosome to bi nd to the polycistronic mRNA and start simultaneous translation beginning at the start codon  more than one translation initiation complex can be formed at a time;  Translation of each polypeptide stops when the ribosomes read the stop codon for the coding sequence 3

3    VJC H2 Biology Paper 2 Preliminary examination 2016    (a) Using Fig 3.1, explain the mode of control of the Arg operon. [2]  Negative control of arg operon;  as the repressor (activated by arginine) is required to switch / turn off gene expression; (b) Explain why it is useful for a bacterial cell to decrease expression of the structural genes when arginine is present. [2]  Trp genes code for enzymes (involved in / necessary for) (anabolism / synthesis) of tryptophan  Decreased expression helps to conserve resources that could be diverted for other uses /preventing wastage of resources (c) Name and describe the process which can result in a population of bacteria acquiring the same allele needed to increase their likelihood of survival.  specialised transduction;;  Viral DNA integrates into a specific location;  When it excises as the cell enters the lytic cycle;  The bacterial DNA removed along with the excision of the viral DNA;  will be those that are near to the prophage on the bacterial chromosome;  DNA transferred will therefore be about the same; (d)(i) Briefly describe the role of DNAase in this experiment.  Digest naked DNA fragments (ii) How does the lack of DNAas e in the experiment result in the growth of the hybrid bacterial colonies?  Without the DNAase, the naked DNA fragments from bacteria which have died may be taken up by the other strain via transformation  DNA fragment will be small enough to cross over the filter

4    VJC H2 Biology Paper 2 Preliminary examination 2016    4. With reference to Fig. 4.1 (a) (i) describe the effect of the 2 gene mutations on the occurrence of cancer in Asians. [1]  70% of never smokers with lung cancer have mutations in EGFR gene compared to 27% of ever smokers with lung cancer;  While 4 % of never smokers with lung cancer have mutations in Ras gene compared to 20% of ever smokers with lung cancer; (ii) suggest how never smokers in Asia developed lung cancer [2]  never smokers who get lung cancer c ould have inherited a dominant mutation in the EGFR gene and experienced a loss of heterozygosity for two or more tumour suppressor genes;;  as seen from the high per centage of never smokers having the mutation in the EGFR gene compared to ever smokers suggesting th