7.03 Fall 2003 Problem Set #3 Solutions Issued Friday, October 17, 2003 1. (a) We are analyzing mutagens that specifically induce G·CA·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·CA·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·CA·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 strandAnd 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·CA·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 ambersuppressors. 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) D = IS, X= OriT Hfr #1: A DXDB = C D D A transferred early Hfr #2: A DB = W = C D D C (and then D) transferred early Hfr #3: A DB = C D XDD C (then B then A) transferred early (b) label the order of IS from A D . . . DD 1 4 Hfr # Recombination Between IS #s Recombinatory Result Product Early 1 1 & 2 DX F --- 1 1 & 3 A DB = W = C D D Hfr C, D 1 1 & 4 DXDB = C F’ C, B1 2 & 3 (or 3&4*) A DXDB = C D D Hfr A 1 2 & 4 A DXD D Hfr A 2 1 & 2 or 3 & 4 A DB = W = C D D Hfr C, D 2 1 & 3 A D XDB= C D D Hfr A 2 (1&4*) DB = W = C F’ C, B 2 2 & 3 W = F --- 2 2 & 4 A DB = C DXD D Hfr C, B, A 3 2 & 3 (or 1&2*) A DB = C D XDD Hfr C, B, A 3 1 & 3 A D XDD Hfr A 3 1 & 4 B = C D XD F’ C, B 3 2 & 4 A DB = W = C DD Hfr C, D 3 3 & 4 XD F --- * denotes recombinatory
View Full Document
Unlocking...