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Biology 22 Problem Set 2 Fall 2010 Due November 23rd, 9:30 am Hand in the solutions to each of the following problems. Put a box around each of your answers and show your work for partial credit. You may consult your book, lecture notes, other students and/or your instructor for assistance. 1. Amino acids 57-62 were compared for an E. coli metabolic enzyme from a wild type strain and a strain with a double frameshift mutation having one nucleotide deletion followed by one nucleotide insertion. The results are shown below. Amino Acid Residue 57 58 59 60 61 62 Wild Type NH2HIS ILE GLU VAL GLY ASN -COOH Double NH2PRO SER LYS LEU ALA ASN -COOH Mutant a. Determine the base sequence of the mRNA for both the wild type and double mutant polypeptides. b. Using your wild type mRNA sequence, show which base was deleted and the site from which it was removed. c. Using your mutant sequence, show which base was added and where it was inserted. (You only need to give one of many possible solutions.) UUU = phenylalanine UUC = phenylalanine UUA = leucine UUG = leucine CUU = leucine CUC = leucine CUA = leucine CUG = leucine AUU = isoleucine AUC = isoleucine AUA = isoleucine AUG = methionine (start) GUU = valine GUC = valine GUA = valine GUG = valine PHE LEU ILE MET VAL UAU = tyrosine TYR UAC = tyrosine UAA = stop UAG = stop CAU = histidine HIS CAC = histidine CAA = glutamine GLN CAG = glutamine AAU = asparagine ASN AAC = asparagine AAA = lysine LYS AAG = lysine GAU = aspartic acid ASP GAC = aspartic acid GAA = glutamic acid GLU GAG = glutamic acid UCU = serine SER UCC = serine UCA = serine UCG = serine CCU = proline PRO CCC = proline CCA = proline CCG = proline ACU = threonine THR ACC = threonine ACA = threonine ACG = threonine GCU = alanine ALA GCC = alanine GCA = alanine GCG = alanine UGU = cysteine UGC = cysteine UGA = stop UGG = tryptophan CGU = arginine CGC = arginine CGA = arginine CGG = arginine AGU = serine AGC = serine AGA = arginine AGG = arginine GGU = glycine GGC = glycine GGA = glycine GGG = glycine CYS TRP ARG SER ARG GLY 2. Four auxotrophic mutant strains of E. coli (1, 2, 3, 4) are each blocked at a different step of the same metabolic pathway. This pathway involves compounds L, M, N, O and P. Each of these compounds is tested for its ability to support growth of mutant strains, with the results given below. (+= growth, __ = no growth) Compound Used as a Supplement L M N O P __ __ Mutant 1 + + + __ Mutant 2 + + + + __ __ __ __ Mutant 3 + __ __ __ Mutant 4 + + a. Give the order of the compounds in the metabolic pathway and indicate the specific step that is blocked in each mutant strain. b. Several sets of double mutations were produced. For example, double mutant 1, 2 has mutant forms for enzymes 1 and 2. Pairs of these double mutations were combined in partial diploids. Would a partial diploid consisting of each of the following pairs of double mutations grow on minimal medium (without supplementation with L, M, N, O and P)? 1. 1,2 and 2,3 2. 1,3 and 1,4 3. 2,4 and 1,3 3. Ten mutants of the rII region in bacteriophage T4 DNA were analyzed for complementation. These mutations are known to fall into TWO cistrons. The results are given in the table below where += complementation and __ = no complementation. 1 2 3 4 5 6 7 8 9 10 __ __ __ __ __ __ 1 + + + + __ __ __ __ __ __ 2 + + + __ __ __ __ __ 3 + + + __ __ __ __ __ __ __ 4 __ __ __ 5 + + + __ __ __ __ 6 + __ __ 7 + + __ __ __ 8 __ __ 9 __ 10 a. Indicate which mutations fall into each of the two cistrons. b. What type of mutation can explain the results for mutants 4 and 9? 4. In each of the following cases, determine whether B-galactosidase and Permease would be produced by the lac operon in the presence and absence of lactose as an inducer. Write + if the enzyme is produced and __ if the enzyme is not produced. B-galactosidase No Lactose lactose a. b. c. d. e. I+ P+ O+ Z+Y-/ IS P+ O+ Z- Y+ I- P+ O+ Z+ Y-/ I+ P+ Oc Z- Y+ I- P+ O+ Z+ Y-/ I- P+ O+ Z- YI- P+ O+ Z+Y-/ I+ P- O+ Z- Y+ I- P+ O+ Z+ Y-/ I+ P+ Oc Z+ Y+ Permease No Lactose lactose 5. Consider the following two human genes, each with two possible alleles. The PIP gene, coding for prolactin-inducible protein, is known to be on the short arm of chromosome 7. The location of the KEL gene, which codes for a specific red blood cell antigen, is unknown. Consider the pedigree shown below. Inheritance of alleles KEL1 and KEL2 was determined by assaying for the production of KEL1 and KEL2 antigens. The PIP1 and PIP2 alleles were studied with a Southern blot analysis of endonuclease Taq1 cut DNA where PIP1 shows a 4 kb fragment hybridizing to the probe while PIP2 shows a 6 kb fragment hybridizing to the probe. The results are shown below. a. List the genotypes for all individuals shown in the pedigree. Refer to individuals by the numbers given in the figure, eg. III-2. b. Is the KEL gene on chromosome 7? Explain your decision by giving specific examples from the genotypes you derived in part a. Genotype KEL 2, KEL 2 I 1 KEL 1, KEL 2 2 II 1 2 III 1 I-1 Size 6 kb 4 kb I-2 II-1 2 Southern Blot II-2 III-1 III-2 3 4 5 III-3 III-4 III-5