MLB 153 Tutorial 91.Pro.A wild-type strain of bacteria produces a protein with the amino acid proline (Pro) at one site. Treatment withnitrous acid causes a C to change to a U. Two different mutants are produced after nitrous acid treatment ofthe bacteria. One has a serine (Ser) in place of the proline, an the other has a leucine (Leu). Further treatmentof 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. UbSerLeuPhe—>PheCUUUUUCULAUUGCUUCUCCUACUGLauAUUAUCAUAAUG MetBEDGUUGUCGGUAGUGValCUCUUCCUCAUCGCCUcccCCACCGACUACCACAACCThursday 18 May 2023OCUGCCGCAGCGTheAUAUUACUAAUAGCAUCACCAACAGAAUAACAAAAAGTyrGinAanGAUGAC AspGAAGluGAGGvou cysUGCUGA StopUGG TrpCOUCOCCGASENACOGAGUAGAGGUGOCGGAGGGSera. Use appropriate codons to show how these mutations occurred. Assume only a single nucleotide changeoccurs with each treatment of nitrous acid.Explain why further treatment of the phenylalanine mutants with nitrous acid produces no change.Gly

Deamination is the process of removal of amino groups from the molecule. And DNA mutation is the…

1. 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. ·ច Pro. a. Ser Leu -Phe—>Phe C fuuu UUC ULA UUG CUU CUC CUA CUG AUU AUC AUA AUG GUU GUC G GUA GUG Phe Lau Leu 1 Mot 3 C UCU- UCC UCA UCG CCU CCC CCA CCG ACU ACC ACA ACG OCU GCC GCA GCG Ser Pro Thr Als A UAU UAC UAA UAG CAU CAC CAA CAG AAU AAC GAU GAC Tyr GAA GAG Stop His Gin AMA AAG Lys Aan Asp Glu UQU cy UGC UGA-Stop UGG G COU COC CGA COG AGU AGC AGA AGG GGU GOC GGA GGG b. Explain why further treatment of the phenylalanine mutants with nitrous acid produces no change. Arg Ser Arg Use appropriate codons to show how these mutations occurred. Assume only a single nucleotide change occurs with each treatment of nitrous acid.

1. 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. ·ច Pro. a. Ser Leu -Phe—>Phe C fuuu UUC ULA UUG CUU CUC CUA CUG AUU AUC AUA AUG GUU GUC G GUA GUG Phe Lau Leu 1 Mot 3 C UCU- UCC UCA UCG CCU CCC CCA CCG ACU ACC ACA ACG OCU GCC GCA GCG Ser Pro Thr Als A UAU UAC UAA UAG CAU CAC CAA CAG AAU AAC GAU GAC Tyr GAA GAG Stop His Gin AMA AAG Lys Aan Asp Glu UQU cy UGC UGA-Stop UGG G COU COC CGA COG AGU AGC AGA AGG GGU GOC GGA GGG b. Explain why further treatment of the phenylalanine mutants with nitrous acid produces no change. Arg Ser Arg Use appropriate codons to show how these mutations occurred. Assume only a single nucleotide change occurs with each treatment of nitrous acid.

Biology: The Unity and Diversity of Life (MindTap Course List)

14th Edition

ISBN:9781305073951

Author:Cecie Starr, Ralph Taggart, Christine Evers, Lisa Starr

Publisher:Cecie Starr, Ralph Taggart, Christine Evers, Lisa Starr

Chapter8: Dna Structure And Function

Section: Chapter Questions

Problem 4DAA: HersheyChase Experiments The graph shown in FIGURE 8.5 is reproduced from an original 1952...

See similar textbooks

Similar questions

15. Several mutants are isolated, all of which require com- pound G for growth. The compounds (A to E) in the bio- synthetic pathway to G are known, but their order in the pathway is not known. Each compound is tested for its ability to support the growth of each mutant (1 to 5). In the following table, a plus sign indicates growth and a minus sign indicates no growth. olord bos og elu Compound tested А В B CDEG Mutanf 1 a. What is the order of compounds A to E in the pathway? b. At which point in the pathway is each mutant blocked? c. Would a heterokaryon composed of double mutants 1,3 and 2,4 grow on a minimal medium? Would 1,3 and 3,4? Would 1,2 and 2,4 and 1,4? + + + + 3 45
HersheyChase Experiments The graph shown in FIGURE 8.5 is reproduced from an original 1952 publication by Hershey and Chase. Bacteriophage were labeled with radioactive tracers and allowed 10 infect bacteria. The virusbacteria mixtures were then whirled in a blender to dislodge any viral components attached to the exterior of the bacteria. Afterward, radioactivity from the tracers was measured. FIGURE 8.5 Detail of Alfred Hershey and Martha Chases 1952 publication describing their experiments with bacteriophage. Infected bacteria refers to the percentage of bacteria that survived the blender. How did the researchers know that the radioisotopes in the fluid came from outside of the bacterial cells and not from bacteria that had been broken apart by whirling in the blender?
HersheyChase Experiments The graph shown in FIGURE 8.5 is reproduced from an original 1952 publication by Hershey and Chase. Bacteriophage were labeled with radioactive tracers and allowed 10 infect bacteria. The virusbacteria mixtures were then whirled in a blender to dislodge any viral components attached to the exterior of the bacteria. Afterward, radioactivity from the tracers was measured. FIGURE 8.5 Detail of Alfred Hershey and Martha Chases 1952 publication describing their experiments with bacteriophage. Infected bacteria refers to the percentage of bacteria that survived the blender. After 4 minutes in the blender, what percentage of each isotope was extracellular?
HersheyChase Experiments The graph shown in FIGURE 8.5 is reproduced from an original 1952 publication by Hershey and Chase. Bacteriophage were labeled with radioactive tracers and allowed 10 infect bacteria. The virusbacteria mixtures were then whirled in a blender to dislodge any viral components attached to the exterior of the bacteria. Afterward, radioactivity from the tracers was measured. FIGURE 8.5 Detail of Alfred Hershey and Martha Chases 1952 publication describing their experiments with bacteriophage. Infected bacteria refers to the percentage of bacteria that survived the blender. The extracellular concentration of which isotope increased the most with blending?
HersheyChase Experiments The graph shown in FIGURE 8.5 is reproduced from an original 1952 publication by Hershey and Chase. Bacteriophage were labeled with radioactive tracers and allowed 10 infect bacteria. The virusbacteria mixtures were then whirled in a blender to dislodge any viral components attached to the exterior of the bacteria. Afterward, radioactivity from the tracers was measured. FIGURE 8.5 Detail of Alfred Hershey and Martha Chases 1952 publication describing their experiments with bacteriophage. Infected bacteria refers to the percentage of bacteria that survived the blender. Before blending what percentage of each isotope. 35S and 32P, was extracellular (outside the bacteria)?
HersheyChase Experiments The graph shown in FIGURE 8.5 is reproduced from an original 1952 publication by Hershey and Chase. Bacteriophage were labeled with radioactive tracers and allowed 10 infect bacteria. The virusbacteria mixtures were then whirled in a blender to dislodge any viral components attached to the exterior of the bacteria. Afterward, radioactivity from the tracers was measured. FIGURE 8.5 Detail of Alfred Hershey and Martha Chases 1952 publication describing their experiments with bacteriophage. Infected bacteria refers to the percentage of bacteria that survived the blender. Do these results imply that viruses inject DNA or protein into bacteria? Why or why not?
Under what conditions will the following partial diploid strain of E. coli convert chorismate to tryptophan (assume trp mutants function similarly to lac mutants)? trpR* trpP trpO* trpE* trpD* trpC* trpB* trpA*/ trpR trpP trpO° trpE trpD trpC trpB trpA ect the one BEST answer (2 oints! O never O only when tryptophan levels are low O constitutively O only when tryptophan is present O only when chorismate levels are low
. 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.…
8.Iron-uptake protein IupA is a human 302-aa, N-glycosylated, heterodimeric protein of considerable therapeutic interest. It contains 3 disulfide bridges that stabilise its structure. Explain what expression host you would use to produce the protein in recombinant form.
1_30*_SP23 - General Biology I (for majors)/1 of us page The anticodon sequence created from the following DNA: TACGGGGCTGAGATT F1 Select one: O a. Tyr-Gly-Ala-Glu-lle O b. AUGCCCCGACUCUAA c. UACGGGGCUGAGAUU O d. Met-Pro-Arg-Leu-STOP F2 # 80 F3 $ 000 000 F4 % F5 MacBook Air F6 & r F7 DII F8
31_30*_SP23 - General Biology I (for m 24 ed out of nove flag evious page A $ Transcribe the following DNA: TACGGGGCTGAGATT Select one: O a. UACGGGGCUGAGAUU O b. Tyr-Gly-Ala-Glu-lle C. Met-Pro-Arg-Leu-STOP O d. AUGCCCCGACUCUAA MacBook Air
DNA-Neoantigen Coupling FAM Peptide (Pep) Br- R.T., 12 h M «S- X Pep-sulf FAM HS ssDNA 4°C, 2 h, pH 9.6 M + FAM DNA-peptide conjugate