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- . A geneticist examined the amino acid sequence of aparticular protein in a variety of E. coli mutants. Theamino acid in position 40 in the normal enzyme isglycine. The following table shows the substitutionsthe geneticist found at amino acid position 40 in sixmutant forms of the enzyme.mutant 1 cysteinemutant 2 valinemutant 3 serinemutant 4 aspartic acidmutant 5 argininemutant 6 alanineDetermine the nature of the base substitution thatmust have occurred in the DNA in each case. Whichof these mutants would be capable of recombinationwith mutant 1 to form a wild-type gene?1. You are studying an E. coli gene that specifies a protein. A part of its sequence is -Ala-Pro-Trp-Ser-Glu-Lys-Cys-His You recover two mutants for this gene that show no enzymatic activity. By isolating the mutant enzyme products, you find the following sequences: Mutant 1: Ala-Pro-Trp-Arg-Glu-Lys-Cys-His Mutant 2: Ala-Pro- What is the molecular basis for mutation 1 and for mutation 2? What is the DNA sequence that specifies this part of the protein? Wild-type amino acid sequence: NH3-Ala-Pro-Trp-Ser-Glu-Lys-Cys-His-COOH Mutant 1: NH3-Ala-Pro-Trp-Arg-Glu-Lys-Cys-His-COOH Mutant 2: NH3-Ala-Pro-13) A researcher is studying the synthesis of a specific amino acid found in Neurospora. She knows that the pathway begins with a precursor that is converted into the amino acid with two known intermediates (Substance Blue and Substance Green). She accurately predicts that this amino acid synthesis pathway is catalyzed by three enzymes, (I, II, and III). She subsequently identifies three mutants that she calls Mutant I, Mutant II, and Mutant III. With the information about the mutants 13) below place the enzymes in the order that the enzymes act in this pathway. (Note: the numbers I, II, don't necessarily indicate the order in which the enzymes appear.) Mutant I (only Enzyme I is mutated) is unable to synthesize the amino acid even if she provided the mutant with both Substance Blue and Substance Green. Mutant II (only Enzyme II is mutated) is able to synthesize the amino acid if she provided the mutar Substance Blue or Substance Green. Mutant III (only Enzyme III is mutated) is able…
- 1. Given is the 30 nucleotides in the human gene for hemoglobin (the oxygen-carrying protein in the red blood cells): 3’ TAC-CAC-GTG-GAC-TGA-GGA-CTC-CTC-TTC-AGA 5’ a. What is the amino acid sequence based on this mRNA? b. A very important mutation in human hemoglobin occurs in this DNA sequence, where the T at nucleotide 20 is replace with an A. The mutant hemoglobin is called sickle cell hemoglobin and is associated with severe anemia. What is the amino acid replacement that results in sickle-cell hemoglobin?7. If you have aligned orthologous gene sequences from an important, conserved gene; there is a different way you can determine reading frame. a. The following sequences of orthologous copies of the insulin receptor gene (INSR) from four species. The sequences are aligned but the reading frame is unknown. What is the reading frame for these sequences? (ie., Does the first complete codon begin at the first, second, or third nucleotide?) Hint: Think about what you know about the nature of mutations, natural selection and the genetic code. 1. Bear CCTGAAGAGCTACCICCGCICCCIGCGGCCGGAGGCTGAGAATAACCCCGGCCGCCCT 2. Human CCTGAAGAGCTACCICCGTICICIGCGGCCAGAGGCTGAGAATAATCCIGGCCGCCCI 3. Chicken CITGAAAAGCTACCIACGCTCICIGAGACCCGACGCIGAGAATAACCCTGGICGICCA 4. Cow CCTGAAGAGTTACCTCCGTTCCCIGCGGCCTGAGGCTGAGAATAACCCCGGCCGCCCT Answer:18) Based on your knowledge of lactose system in prokaryotes) and considering the following genotypes, complete the table below indicating a "+" where an enzyme is produced and a where no enzyme is produced. Note: Is is a mutant that cannot recognize allolactose
- . Seven E. coli mutants were isolated. The activity ofthe enzyme β-galactosidase produced by cells containing each mutation alone or in combination with othermutations was measured when the cells were grown inmedium with different carbon sources.Lactose +Glycerol Lactose GlucoseWild type 0 1000 10Mutant 1 0 10 10Mutant 2 0 10 10Mutant 3 0 0 0Mutant 4 0 0 0Mutant 5 1000 1000 10Mutant 6 1000 1000 10Mutant 7 0 1000 10F′ lac from mutant 0 1000 101/ mutant 3F′ lac from mutant 0 10 102/ mutant 3Mutants 3 + 7 0 1000 10Mutants 4 + 7 0 0 0Mutants 5 + 7 0 1000 10Mutants 6 + 7 1000 1000 10Assume that each of the seven mutations is one andonly one of the genetic lesions in the following list.Identify the type of alteration each mutation represents.a. superrepressorb. operator deletionc. nonsense (amber) suppressor tRNA gene (assumethat the suppressor tRNA is 100% efficient in suppressing amber mutations)d. defective CRP–cAMP binding sitee. nonsense (amber) mutation in the β-galactosidase genef.…HbS results from the substitution of valine forglutamic acid at the number 6 position in the b chainof human hemoglobin. HbC is the result of a change atthe same position in the β chain, but in this case lysinereplaces glutamic acid. Return to the genetic code table and determine whether single-nucleotide changes can account for these mutations. Then view and examine the R groups in the amino acidsglutamic acid, valine, and lysine. Describe the chemicaldifferences between the three amino acids. Predict how thechanges might alter the structure of the molecule and leadto altered hemoglobin function.9. one for each correct position of gene/enzyme and substrate) There is, of course, a metabolic pathway by which purrine is made. Four genes have been found which are responsible to produce purrine. These are the L, T, O, and C genes. Mutant cats were given various substances in their food, which either did help them produce purrine, or did not help. The results are summarized below. + means you get purrine production; - means you don't add Mewsic add Felinine add Clawsome add Lickinine mutation in T mutation in S mutation in A Mutation in C + Fill in the pathway including these 4 intermediates leading up to the formation of purrine, and clearly show at which steps the T, S, A, and C enzymes function. Either fill in this or rewrite it, where "substrate" refers to Mewsic, Felinine, Clawsome, or Lickinine; and gene refers to T, S, A, or C. Gene → [substrate]-- Gene Gene (substrate]--- Gene →[substrate]--- → [substrate]-- → purrine
- 1. Pro A wild-type strain of bacteria produces a protein with the amino acid proline (Pro) at one site. Treatment with nitrous acid causes a C to change to a U. Two different mutants are produced after nitrous acid treatment of the bacteria. One has a serine (Ser) in place of the proline, an the other has a leucine (Leu). Further treatment of the two mutants with nitrous acid produces new mutant strains, both with phenylalanine (Phe) at the site. Treatment of the phenylalanine carrying mutants produces no change. U Ser Leu Phe →Phe U с A บบบ UUC UUA LOU UUG CUU CUC CUA CUG- Phe GUU GUC G GUA GUG Leu AUU AUC lle AUA AUG-Met Val с UCU UCC UCA UCG CCU CCC CCA CCG ACU ACC ACA ACG GCU GCC GCA GCG Ser Pro Thr Ala A UAU Tyr Cys UAC UAA Stop UGA-Stop UAG UGG-Trp CAU His CAC CGU CGC CAA Gin CGA CAG CGG AAU Asn AAC AAA AAG Lys UGU UGC GAU GAC GAA GAG Asp Glu AGU AGC AGA AGG GGU GGC GGA GGG G b. Explain why further treatment of the phenylalanine mutants with nitrous acid produces no change. Arg…2. The diagram to the right shows the change in the structure of the C-terminal portion of each of the ẞ-subu- nits of human hemoglobin (HbA) in the oxyHb to deox- yHb or R-to-T transition. The hydrogen bonding interac- tion of the C-terminal ẞHis 146 residue with the side chain of Asp94, highlighted by the red ellipse, has been shown to be responsible for a major portion of the proton uptake associated with the Bohr effect. Treatment of HbA with the enzyme carboxypeptidase A (CPA) results in loss of the C-terminal ẞHis 146 and ẞTyr145 residues of the ẞ- subunits. (a) ( ) Draw a Hill plot [log(Y/[1-Y]) vs. log(pO2), Y = fraction of heme groups occupied by O2] to compare the values of the Hill coefficient nн and the O2-binding affinity at pH 7.4 of normal HbA before and after treat- ment of with CPA. (b) (' ) How will the plot for CPA-digested HbA change at pH 7.2? (c) 1 Hi5146 HN- ✓ Low PK B-chain. CH2 CH-NH-CO-CH-NH- CH₂ Tyr145 он HbO2 or R state Туг145 CH-CH2- OH co NH CH CH₂ His…The amino acid sequence of part of a protein has beendetermined:N . . . Gly Ala Pro Arg Lys . . . CA mutation has been induced in the gene encodingthis protein using the mutagen proflavin. The resultingutant protein can be purified and its amino acidsequence determined. The amino acid sequence of themutant protein is exactly the same as the amino acidsequence of the wild-type protein from the N terminus of the protein to the glycine in the preceding sequence. Starting with this glycine, the sequence ofamino acids is changed to the following:N . . . Gly His Gln Gly Lys . . . CUsing the amino acid sequences, one can determinethe sequence of 14 nucleotides from the wild-typegene encoding this protein. What is this sequence?