7 03 Fall 2003 Problem Set 3 Solutions Issued Friday October 17 2003 1 a We are analyzing mutagens that specifically induce G C A T mutations in DNA Therefore we must determine the potential double stranded DNA sequences that will encode stop codons after going through this specific mutation We will start with 5 UAG3 The double stranded DNA that corresponds to 5 UAG3 is 3 ATC5 template strand 5 TAG3 coding strand We need to figure out what specific double stranded DNA sequences could have undergone a G C A T mutation to become the sequence above To do this just work backwards and change the AT base pairs in the above sequence into GC base pairs one pair at a time If you do this you will get the following double stranded DNA sequences the mutated bases are boldfaced 3 GTC5 template strand 5 CAG3 coding strand codes for Gln 3 ACC5 template strand 5 TGG3 coding strand codes for Trp These two sequences each underwent one G C A T mutation to become DNA that encodes the 5 UAG3 stop codon We have chosen to make only one mutation per three base pairs because the likelihood of a mutagen acting on two consecutive base pairs or two base pairs out of three is extremely small We can neglect those extremely rare events since this problem is asking why something generally happens Now repeat the process for the other two stop codons The double standed DNA that encodes 5 UGA3 is 3 ACT5 template strand 5 TGA3 coding strand And the DNA possible sequences that could have been mutated to become the above sequence are 3 GCT5 template strand 5 CGA3 coding strand codes for Arg 3 ACC5 template strand 5 TGG coding strand codes for Trp Lastly repeat for 5 UAA3 Its corresponding DNA is 3 ATT5 template strand 5 TAA3 coding strand And the DNA possible sequences that could have been mutated to become the above sequence are 3 GTT5 template strand 5 CAA3 coding strand codes for Gln 3 ACT5 template strand 5 TGA3 coding strand codes for UGA stop codon 3 ATC5 template strand 5 TAG3 coding strand codes for UAG stop codon Notice that for UGA and UAG the two candidate sequences for mutation both encode amino acids whereas for UAA only one of its candidate sequences encodes an amino acid The other two correspond to stop codons When G C A T mutagens are introduced they mutate random base pairs along the DNA sequence Given that there are more targets for the creation of TGA and TAG nonsense mutations these two mutations will occur more frequently than TAA mutations b The amber TAG mutation has the following mRNA codon tRNA anticodon and corresponding DNA coding for the anticodon portion of the tRNA mRNA codon 5 UAG3 tRNA anticodon 3 AUC5 DNA encoding anticodon portion 5 TAG3 template strand for tRNA transcription 3 ATC5 This DNA sequence is also the sequence that codes for the anticodon portion of amber suppressing tRNA alleles In other words we want the mutagen to change normal DNA sequences into this sequence so that we can have amber suppressing alleles To find which tRNA genes can be altered to become the DNA sequence above we can work backwards like in part a Now since we have a G C A T mutagen we can change AT pairs into GC pairs and vice versa Again we will change one base pair at a time If we do this we will get the following three DNA sequences the mutated bases are in boldface DNA encoding anticodon portion 5 CAG3 tRNA template for tRNAGln 3 GTC5 DNA encoding anticodon portion 5 TGG3 tRNA template for tRNATrp 3 ACC5 DNA encoding anticodon portion 5 TAA3 tRNA template for UAA stop tRNA 3 ATT5 If you follow the flow of genetic information you will see that these three sequences code for the anticodon portion of tRNAGln tRNATrp and the tRNA that recognizes the UAA stop codon So the gln and trp tRNA genes can be mutated to become amber suppressors The codons normally recognized by tRNAGln and tRNATrp are 5 CAG3 and 5 UGG3 respectively We gave full credit to those who ended the solution here But since the problem asked which genes could in principle be altered to become amber suppressors we can consider cases where two or three base pairs from each codon were altered by the mutagen In that case we would arrive at the following DNA sequences with the corresponding mRNA codons also listed DNA encoding anticodon portion 5 CGA3 tRNA template for tRNAArg 3 GCT5 mRNA codon 5 CGA3 DNA encoding anticodon portion 5 CGG3 tRNA template for tRNAArg 3 GCC5 mRNA codon 5 CGG3 DNA encoding anticodon portion 5 CAA3 tRNA template for tRNAGln 3 GTT5 mRNA codon 5 CAA3 DNA encoding anticodon portion 5 TGA3 tRNA template for UGA stop tRNA 3 ACT5 mRNA codon 5 UGA3 So assuming the mutagen can change two or three bases pairs in the same triplet we arrive at additional tRNA genes that can be mutated to become amber suppressors These are two tRNAArg genes and another tRNAGln 2 a Hfr 1 Hfr 2 Hfr 3 D IS X OriT A D XD B C D D A DB W C D D A D B C D XD D b label the order of IS from A Recombination C and then D transferred early C then B then A transferred early D DD 1 Hfr A transferred early 4 Recombinatory Result Product Early DX A DB W C D D D XD B C F Hfr C D F C B Between IS s 1 1 2 1 1 3 1 1 4 1 D XD B C D D D XD D A DB W C D D A D XDB C D D DB W C W A D B C D XD D A D B C D XD D A D XD D B C D XD A DB W C DD XD Hfr A 1 2 3 or 3 4 A 2 4 A Hfr A 2 1 2 or 3 4 Hfr C D 2 1 3 Hfr A 2 1 4 F C B 2 2 3 F 2 2 4 Hfr C B A 3 2 3 or 1 2 Hfr C B A 3 1 3 Hfr A 3 1 4 F C B 3 2 4 Hfr C D 3 3 4 F denotes recombinatory events between two IS sequences NOT flanking the OriT The p set asked only for events between sequences that flank the OriT Therefore these are extraneous answers 3 a The donor strain is Lac and the recipient strain is Lac Therefore in the Lac Kanr transductants lac1 was cotransduced with Tn5 So the distance between Tn5 and the lac1 mutation is 18 100 100 18 b Since none of the 100 Kanr tranductants were Lac we can conclude that Tn5 was never co transduced with lac2 This indicates that the distance between lac2 and Tn5 is at least one phage head 105 bp 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