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~ P A S S F a c i l i t a t o r B I O L O G Y : G E N E T I C S E Q U E N C E S E d i b l e I n f o ~ B A S E D N A A couple of kick off questions (may find a worksheet to use but the questions may be better to get convo flowing anyhow): Does each new cell need DNA to be present? Where does the extra DNA come from? Why does DNA replicate? Does DNA replication end when the old DNA creates a completely new DNA? What is the type of replication that DNA goes through called? Why is it called semi-conservative replication? Where does translation take place In what direction do you read DNA? In which direction do you ‘build’ DNA? What type of bond exists between nitrogenous bases? Are these strong or weak bonds? Materials: coloured mini-marshmallows or other soft sweet, red liquorice ropes and toothpicks (ALTERNATIVELY: pegs, buttons, string or pipe cleaners can be used for base sequence pairing ) The idea is to use the liquorice as the phosphorous/ sugar back bone. The sugar is joined to the nitrogenous base with a toothpick and each marshmallow/soft sweet represents a base (ie. Different colour for U, C, A, G and T). Be sure the students remember at which point they must use Uracil rather than Thymine. Using a sequence of Amino Acids, the students are to use the bunch of goodies to play out: transcription from the template translation to the Amino Acid sequence Each group will be given a sequence of DNA which will be the template/noncoding strand. There are three variations and in order to determine whether they got the transcription and translation processes correct the resulting Amino Acid sequence is available for comparison. 1. TACAAATTCCACCTC -DNA template / Coding strand AUGUUUAAGGUGGAG Met –Phe –Lys –Val –Glu -mRNA -Amino Acid sequence 2. TACGGATTCGCAACT AUGCCUAAGCGUUGA Met –Pro –Lys –Arg –Ter 3. TACATAACCACGCGG AUGUAUUGGUGCGCC Met –Tyr –Trp –Cys –Ala Second Position of Codon U F i r s t C P o s i t A i o n (5’) G U C A G UUU Phe UUC Phe UUA Leu UUG Leu UCU Ser UCC Ser UCA Ser UCG Ser UAU Tyr UAC Tyr UAA Ter UAG Ter UGU Cys UGC Cys UGA Ter UGG Trp U C A G CUU Leu CUC Leu CUA Leu CUG Leu CCU Pro CCC Pro CCA Pro CCG Pro CAU His CAC His CAA Gln CAG Gln CGU Arg CGC Arg CGA Arg CGG Arg U C A G AUU Ile AUC Ile AUA Ile AUG Met ACU Thr ACC Thr ACA Thr ACG Thr AAU Asn AAC Asn AAA Lys AAG Lys AGU Ser AGC Ser AGA Arg AGG Arg U C A G GUU Val GUC Val GUA Val GUG Val GCU Ala GCC Ala GCA Ala GCG Ala GAU Asp GAC Asp GAA Glu GAG Glu GGU Gly GGC Gly GGA Gly GGG Gly U C A G T h i r d P o s i t i o n (3’) Exam Questions from 2006 BIOL1102: 46. A transcription unit that is 8000 nucleotides long may use 1800 nucleotides to make a protein consisting of 600 amino acids. This is best explained by the fact that: 1) There are termination exons near the beginning of mRNA. 2) There is redundancy and ambiguity in the genetic code. 3) Many nucleotides are needed to code for each amino acid. 4) Nucleotides break off and are lost during the transcription process. 5) Many non-coding nucleotides are present in mRNA. 47. A particular triplet of bases in the template sequence of DNA is AAA. The anticodon on the tRNA that binds the mRNA codon is: 1) TTT 2) UUA 3) UUU 4) AAA 48. Which of the following does not occur during the termination phase of translation? 1) A stop codon causes the Acceptor site to accept a release factor. 2) The newly formed polypeptide is released. 3) A tRNA with the next amino acid enters the P site. 4) The ribosome comes off the mRNA and the two ribosomal subunits separate. 5) Translation stops. 49. Suppose you have an actively growing culture of E. coli to which you add radioactively labelled guanine (G). You allow the culture to grow for 1 more generation (ie every cell undergoes one more round of replication). What do you predict for the resulting cells? 1) All cells would contain radioactive DNA. 2) Half of the cells would have radioactive DNA. 3) All four bases of the DNA would be radioactive 50. Why is the new DNA strand complementary to the 3’ to 5’ strands assembled in short segments (Okazaki fragments)? 1) The replication forks block the formation of larger strands. 2) DNA polymerases can assemble DNA only in the 3’ to 5’ direction 3) DNA polymerases can assemble DNA only in the 5’ to 3’ direction 4) It is more efficient than assembling complete new strands 5) Only short DNA sequences can extend off the RNA primers. 51. What would the effect of the translation be of a single base pair deletion in an organism’s DNA? 1) None 2) Changes only to the amino acid where it occurs. 3) Possible changes to the amino acid where it occurs as well as to all downstream amino acids from that point. 4) Poor replication of the DNA. 5) Immediate death to the cell.