121 MCB Spring 2011 Problem Set 2: 1. You have taken a job in a biotech company that studies different metabolic enzymes for commercial use. Your first assignment is to study the enzymatic properties of galactokinase from S. cerevisiae, the product of the GAL1 gene. You decide on a strategy to express Gal1 protein in E. coli and purify the protein for your enzymatic studies. At your disposal is a PCR machine, a DNA oligo synthesizer, yeast genomic DNA, and an expression vector for E. coli (shown below). This expression system uses a promoter from the bacteriophage T7. Shown at the bottom is a portion of the multiple cloning site (MCS) in which you will insert your DNA.
Design oligonucleotide primers that will allow you to insert the GAL1 gene into this expression vector, using the SacI (GAGCTC) and XbaI (TCTAGA) restriction sites. The total length of each primer should be 20 nucleotides.
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Answer: What you need to know: Sequence of SacI and XbaI restriction sites (given information) SacI XbaI 5'-GAGCTC-3' 5'-TCTAGA-3'
Sequence of the 14 most 5' bases of GAL1 (given in class Lecture 2 or via the SGD web site) 5'-ATG-ACT-AAA-TCT-CA-3' (start codon underlined)
Sequence of the complement of the 14 most 3' bases of GAL1 (including a stop codon) (derived from sequence) Sense strand: 5'-TA-TAT-GAA-TTA-TAA-3'(stop codon underlined) Complement: 3'-AT-ATA-CTT-AAT-ATT-5' So: 5' Forward Primer (total length = 20 bases) This simply consists of the 5' sequences of GAL1 preceded by the SacI site: 5'- GAGCTCATGACTAAATCTCA-3' (SacI site underlined) 3' Reverse Primer (total length = 20 bases) This consists of the complementary 3' sequences preceded by the XbaI site: 5'-TCTAGATTATAATTCATATA-3' (XbaI site underlined; note this sequence is written in the 5' 3' direction)
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2. You are the president of a biotech company that isolates and sells novel restriction endonucleases. One of your employees has isolated two isoschizomers of Sau3A I (recognizes GATC), one called EcoB II and EcoB IV. EcoB II generates a two base 5' overhang and EcoB IV generates a four base 3' overhang following cleavage. You are putting together the new company catalog and you are checking to make sure that all of the information is correct for describing these new enzymes. What is an isoschizomer? From what organism were these enzymes isolated? What are the reaction products produced following cleavage with each of these enzymes? In your company catalog, can you state that your new enzymes are compatible for cloning DNA fragments into BamH I sites? Why or why not? What organism: Eco = E. coli strain B Reaction products: Eco B II: two base 5' overhang:
5' 3'
NNNNGATCNNNN3' NNNNCTAGNNNN5'
5' 3'
NNNNG3' NNNNCTA5'
5'
ATCNNNN3' 3' GNNNN5'
EcoB IV: four base 3' overhang: NNNNGATCNNNN3' 3' NNNNCTAGNNNN5' Compatability with BamHI: Neither enzyme is compatible with BamHI. While the GATC recognition sequence of both is imbedded within the BamHI sequence (GGATCC), digestion with BamHI yields a 4 base 5' overhang that cannot anneal with the reaction products of either EcoB II or EcoB IV.
5' 5'
NNNNGATC3' 3' NNNN5'
NNNN3' 3' CTAGNNNN5'
5'
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3. To prepare a genomic DNA it library, is necessary to fragment the genome so that it can be cloned into a vector. A common method is to use a restriction endonuclease. How many different genomic fragments would you expect if you digested human genomic DNA 9 with Sau3A I? (Note: the haploid human genome is ~3.2 x 10 base pairs.) How many total fragments would you get if you used EcoR I? A restriction nuclease that recognizes a four-base-pair site will cut on average 1 out of 256 fourbase-pair sites (1/4)4 . Thus, you would expect on the order of 1.3 x 107 different fragments (3.2 x 109/256) from a Sau3A digestion of human DNA. Similarly, you would expect on the order of 7.8 x 105 different fragments (3.2 x 109 /4096) from an EcoRI digestion. However, because the genome is diploid, there would be 15.6 x 105 total fragments produced per genome (2 x 7.8 x 105). These estimates are based on the assumptions that the genome is 50% GC and that the sequence is random: neither of which is true, but they won't lead you far astray. Genomic libraries are often made from fragments obtained by cleaving the DNA with Sau3A I such that only partial digestion of the DNA is achieved. Why? A set of overlapping DNA fragments will be generated by partial digestion. In class we discussed on important reason to do this, which is to help ensure that an entire yeast gene would be contained within one piece of DNA. With the genomes of higher eukaryotes this isn't possible using genomic DNA. Here, libraries constructed from sets of overlapping fragments of a genome are still valuable because they can be used to arrange cloned sequences in their original order on the chromosomes, and thus obtain the DNA sequence of a long stretch of DNA. Overlapping clones were crucial in the sequencing of the human genome. 4. In the very first round of PCR using genomic DNA, the DNA primers prime synthesis that terminates only when the cycle ends (or if a random end of DNA is encountered). Yet by the end of 20 to 30 cycles, the only visible product is defined precisely by the ends of the DNA primers. In what cycle is a double stranded fragment of the correct size first generated? It is at the end of the third cycle when the first product appears where the ends correspond fully to the DNA primers used. See the diagram in Lecture 3 for how this works. 5. The two strands of 4 phage DNA differ from each other in their GC content. Owing to this property, they can be separated in an alkaline cesium chloride gradient (the alkalinity denatures the duplex DNA). When the RNA synthesized by 4 phage is isolated from infected E. coli cells, it is found to form RNA-DNA hybrids with both strands of 4 DNA. What does this result tell you? Answer: Because RNA can hybridize to both strands, the RNA must be transcribed from both strands. This does not mean, however, that both strands are used as a template for each gene. The expectation is that different genes are transcribed in different directions along the DNA. The most direct test would be to purify a specific RNA coding for a specific protein and then hybridize it to the T4 genome. Only one strand of the genomic DNA should hybridize to the purified RNA.
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