2016 FRQ by topics Application topics ANS Updated

NANYANG JUNIOR COLLEGE H2 Biology Free Response Questions Application Syllabus 1 J2/2016 Free Response Questions – Application Syllabus Isolating, Cloning and Sequencing DNA 1. Describe the natural function of restriction enzymes and its role in formation of recombinant DNA molecule. [8] 1. Restriction endonucleases / restriction enzymes are naturally found in bacteria; 2. They protect the bacterial cells against foreign DNA from viruses i.e. phages; 3. They work by hydrolysis / breaking of phosphodiester bonds of the foreign DNA; 4. The bacterial DNA is protected by the addition of methyl groups ( -CH3) to adenines or cytosines; 5. Restriction enzyme has active site that recognise short specific nucleotide sequence / restriction site that are palindromic; (b) Explain the formation of recombinant DNA molecule. Instructions to students:  Revise the relevant section/s of your lecture notes, tutorials, tests and exams after looking at each question.  Memorize these relevant section/s before writing your answers. If you forget any point while writing, revise again and re-write the entire answer. Do not cheat by just looking at the notes and adding on the points to your answer – you are merely cheating your own learning!  The marks allocated for each question is a rough gauge of how many points there should be in your answer – they are not the absolute number of points. Thus, if you have more points than the marks allocated, write everything down. Highlight / underline all the key words and phrases in your answers. This set of answers can serve as your revision notes in future, if it is well done.  Pace your learning. Attempt 10 questions a day, everyday and you will finish revising and memorizing the Biology syllabus before the end of Term 3!

NANYANG JUNIOR COLLEGE H2 Biology Free Response Questions Application Syllabus 2 J2/2016 1. Recombinant DNA refers to DNA molecules containing DNA from two or more sources. 2. Restriction enzyme cleave the phosphodiester bonds in the sugar-phosphate backbone of both DNA strands at specific restriction sites creating sticky ends; 3. Gene-of-interest and plasmid is cut with the same restriction enzyme; 4. The single-stranded DNA can form hydrogen bonds with complementary sticky ends on the gene of interest by complementary base pairing; 5. DNA ligase catalyzes the formation of phosphodiester bonds between two nucleotides to produce recombinant DNA molecule OR 6. Restriction enzyme cleave the phosphodiester bonds in the sugar-phosphate backbone of both DNA strands at specific restriction sites creating blunt ends; 7. Gene-of-interest and plasmid is cut with the same restriction enzyme;

NANYANG JUNIOR COLLEGE H2 Biology Free Response Questions Application Syllabus 3 J2/2016 8. Linker DNA Nucleotides are is added to 3’ ends by terminal transferases to form complementary sticky ends / Specific linker DNA are added to blunt ends and cut with another restriction enzyme that create sticky ends; 9. The single-stranded DNA can form hydrogen bonds with comple mentary sticky ends on the gene of interest by complementary base pairing; 10. DNA ligase catalyzes the formation of phosphodiester bonds between two nucleotides to produce recombinant DNA molecule Explain how the recombinant plasmid can be “put back” into the bacteria. [2]  bacterial cells made competent (i.e able to take up DNA) by addition of calcium ions / CaCl2 solution;  brief heat shock treatment to create transient pores in bacterial cell membrane to allow entry/uptake of DNA; [Reject: electroporation – this is less common for bacteria] 2. [N2014/P3/Q5] Describe and explain the properties of plasmids that allow them to be used as DNA cloning vectors. [7] 1. Small circular DNA: Plasmids are induced to enter host cells easily by transformation; 2. Contain one or more selection marker genes (e.g. two antibiotic resistance genes or one antibiotic resistance gene + lac Z gene).: Allows for production of proteins giving a phenotype used to identify plasmids which have successfully incorporated the gene of interest and cells which had taken up the recombinant plasmid via transformation; 3. Insertional inactivation of one of the selection marker when the foreign gene is inserted is used to identify recombinant cells; 4. Contains unique restriction enzyme recognition sites: Allows cleavage and insertion of foreign genes using restriction enzymes and DNA ligase; 5. Contains own origin of replication: Allows them to replicate autonomously, independent of the bacterial chromosome and allow the binding of DNA polymerase to initiate replication of the plasmid with the gene of interest; 6. Exists in high copy number: Quantity of plasmid DNA that can be purified from each host cell is high / increases number of copy of genes; 3. [N2009/P3/Q4] Distinguish between a genomic DNA and cDNA library. [6] Features Genomic DNA Library cDNA Library 1. Starting material; Genomic DNA / complete set of genetic material Mature mRNA present in a specific cell type at a specific stage 2. Nature of genetic material; Contains all coding and non- coding sequences, including introns, regulatory and intergenic sequences Contains only coding sequence / exons 3. Types of vectors used; λ phage, BAC, YAC (spell in full) Plasmid, λ Phage

NANYANG JUNIOR COLLEGE H2 Biology Free Response Questions Application Syllabus 4 J2/2016 4. Genetic engineering tools to obtain library; Restriction enzyme to digest gDNA DNA ligase for ligation of fragments to vectors DNA fragments inserted into vectors Reverse transcriptase to reverse transcribe mRNA to ssDNA Primers to start synthesis of complementary strand DNA polymerase to synthesise complementary strand (RE, DNA ligase & vectors) 5. Purpose; Study regulatory sequences / Introns + alternative splicing / genes which expression pattern is unclear Study expression of proteins of a cell type / expression of a protein in different cell types / expression of a protein through different developmental phases 4. [N2014/P3/Q5] Eukaryotic genes cannot be expressed directly in the bacterial plasmid because of differences between prokaryotes and eukaryotes, including the presence of introns. Outline these problems and explain how they are overcome in order to allow expression of eukaryotic genes in plasmids within E.coli cells. [7]

NANYANG JUNIOR COLLEGE H2 Biology Free Response Questions Application Syllabus 5 J2/2016 Problem 1: 1. Eukaryotic genes possesses introns which prokaryotes do not have the cellular machinery for removal of introns; Prokaryotes lack spliceosome; 2. Mature mRNA is extracted and reverse transcriptase is then used to produce complementary DNA (cDNA), which does not contain intron sequence. Problem 2: 3. Most eukaryotic proteins consist of various subunits which are covalently attached via post-translational modification e.g. insulin. / @ chemical modification of protein to be functional 4. Prokaryotes are unable to carry out post-translational modifications as they lack endoplasmic reticulum and Golgi apparatus; 5. Genes of each subunit is cloned into separate bacterial plasmid and transformed into different bacteria cells; / idea of additional step needed t o chemically modified proteins after being synthesised. Problem 3: 6. Prokaryotic translation is initiated by addition of N-formyl-methionine, which is the first amino acid added as compared to methionine in eukaryotes, so the protein synthesized may be different; 7. Polypeptides synthesized has to be treated to cleave off N-formyl-methionine and bacterial amino acids as it may result in conformational changes in the protein, rendering it non - functional; @ -Adding of a prokaryotic/ bacterial promoter at the 5’ end  to allow for binding of RNA polymerase and