MIT Department of Biology 7.013: Introductory Biology - Spring 2004 Instructors: Professor Hazel Sive, Professor Tyler Jacks, Dr. Claudette Gardel NAME_____________________________________________________________TA__________________ SEC____ 7.013 Problem Set 3 FRIDAY March 5, 2004 Problem sets will NOT be accepted late. Question 1 Shown below is the sequence of a short eukaryotic gene. Only the 5’ to 3’ sequence of the “RNA-like” or complementary strand of the template strand is shown. 5’- GATGAGTTAT10 AATATTTCTC20 TCCAGGCATG30 GAGTATTCCG40 GTGTGCGATC50 TCCCCCTTTG60 GACCACCCTG70 GGTTGCCCTC80 TAAGCATAAT90 AGTTGGCCAT100 ACGTTTCTGT110 AATTAAAATT120 TGTTTGCCTC130 ATGT -3’ This is what’s known about this sequence… v RNA polymerase and associated proteins recognize the sequence ‘TATAAT’ and initiate transcription five nucleotides downstream of the sequence at position 18. v The intron splice sites have been determined to be ‘CUU’ (5’ splice site) and ‘AAG’ (3’ splice site). These sequences and the sequences between these sites will be spliced out as introns in the processing towards mature mRNAs. v The organism cleaves the RNA and adds poly-A tails immediately following the sequence ‘AGUUGG.’ These tails are 14 nucleotides long. RNA is produced and processed in the following order: transcription, splicing, polyadenylation, and 5’ capping. a) What is the sequence of the first 20 RNA base pairs of mRNA? Denote 5’ and 3’ ends. CUCUCCAGGCAUGGAGUAUU b) What is the sequence of the processed, cytoplasmic-localized mRNA produced from this gene? 5’cap- CUCUCCAGGCAUGGAGUAUUCCGGUGUGCGAUCUCCCCCAUAAUAGUUGGAAAAAAAAAAAAAA 3’ c) What is the sequence of the short protein encoded by this gene? Put the answer in the table below next to “original sequence”. (Use the single letter code from chart at end.) What would the protein sequence be if the following point mutations occurred? Protein Type of mutation choose from frameshift, nonsense, missense, silentOriginal sequence MEYSGVRSPP mutation of base 33 (G mutated to T) MDYSGVRSPP missense mutation of base 36 (T mutated to A) ME nonsense mutation of base 51 (T mutated to C) MEYSGVRSPP silent mutation of base 70 (G mutated to C) MEYSGVRSPP In intron (~silent)Question 2 Below is a diagram of the “central dogma.” For each step (X, Y, Z) fill in the following: a) Name of the process b) Name of the primary enzyme or enzyme complex c) Template Macromolecule – the template read by the enzyme or enzyme complex d) Monomers polymerized –to form the new macromolecule e) Initiation site – the name of the sequence in the template that directs the enzyme to begin polymerization. YZ DNAX ‡ RNA ‡ Protein X Y Z a DNA Replication Transcription Translation b DNA Polymerase RNA Polymerase ribosome c DNA DNA RNA d Deoxyribonucleoside triphosphates Ribonucleoside triphosphates Amino Acids e Origin of replication promoter AUG, start codon Question 3 You’ve gained a position as a technician in a prestigious yeast lab. Your first assignment is to learn the cellular localization of product of the gene, bio701, studied by the lab. One exciting technique involves creating a protein fusion where your protein of interest is fused with GFP, green fluorescent protein. The resulting fused protein will fluoresce green when excited at the appropriate wavelength. To make this fusion you must fuse the bio701gene with the gfpgene. Certain vectors have been made that simplify the construction by allowing you to insert your gene in frame with the gfp gene. See below. gfp ampicillin Resistance gene BamH1 Bgl II pGFP 7.5kb Promoter 2You know that the BglII site sequence in pGFP is in frame with the gfpgene (such that the AGA will encode arg (R) and the TCT will correspond to ser (S)). (see below) The bio701 coding region is about 1500 bases long. The actual sequence corresponding to proximal and distal ends of corresponding to the coding region bio701is shown below. The middle 1.4 kb are not shown. 5’ATGaccatgggcgacaagaagagcccgaccaggcc... (1.4kb of internal base pairs) ...gggcgacatgtcagcagtcaatgatgaatcttTGA 3’ The first ATG corresponds to the start codon for bio701. The capitalized TGA at the end corresponds to the stop codon of bio701. Since your goal is to make a protein fusion where the N terminus of Green Fluorescent protein is replaced by the Bio701 protein, you must engineer a DNA construct that fuses the bio701DNA in frame with the gfpgene sequence. And since you must design primers 20 nucleotides long to amplify bio701from yeast genomic DNA you cleverly decide to engineer restriction sites at the end of these primers that will allow you to cut the resulting amplified product with BamH1 and BglII, and insert it into the BamH1 BglII sites of pGFP. (You’ve checked that there are no BamHI or BglII sites within bio701.) These restriction sites are given below. The arrows represent where the enzyme will cleave the DNA. BamH1 5’-G GATCC-3’ BglII 5’-A GATCT-3’ 3’-CCTAG G-5’ 3’-TCTAG A-5’ So first you design a primer to amplify bio701in one direction with an engineered BamH1 site attached at the 5’ end such that you will be able to cut the final amplified PCR product with BamH1 enzyme to insert into the BamH1 site of pGFP. Your primer looks like… Forward primer: 5’ ggatccatgaccatgggcga 3’ Your boss quickly looks at your design and approves. Design a primer to amplify the reverse direction with an engineered BglII restriction site that will allow you to digest the PCR product and then ligate it into the BglII site in frame with the gfpgene. Reverse primer: 5’ agatctaagattcatcattg 3’ 3Assuming your primers work, fill in the blanks to outline a working cloning strategy. PCR amplify the bio701gene. Cut PCR product (or bio701) with BamH1 and BglII Cut pGFP (or vector) with BamH1 and BglII Ligate gel-purified cut fragments Transform bacteria Select for Transformants on Ampicillin plates Your desired construct should look something like this: gfp ampicillin Resistance gene BamH1 Bgl II bio701 pBioGFP 9kb After you have obtained bacterial transformants, you wish to isolate their plasmids to screen and determine that everything went correctly with the ligation. You pick 4 bacterial colonies and harvest the plasmid clones from them and perform restriction digests. Then you perform gel electrophoresis with the restriction digests. The results are as follows: BamH1 BglII BamH1 + BglII Clone
View Full Document
Unlocking...