Here is a list of knockout gene mutations that are desired to make in E. coli genome. You should take some time to identify what these genes are and what pathways they are either involved in and / or regulate. Gene lacZ fliE rpoA nagC ompA lexA Size of gene (bp) Protein 3075 315 990 1221 1041 609 Pathway

Here is a list of knockout gene mutations that are desired to make in E. coli genome. You should take some time to identify what these genes are and what pathways they are either involved in and / or regulate. Gene lacZ fliE rpoA nagC ompA lexA Size of gene (bp) Protein 3075 315 990 1221 1041 609 Pathway

Human Heredity: Principles and Issues (MindTap Course List)

11th Edition

ISBN:9781305251052

Author:Michael Cummings

Publisher:Michael Cummings

Chapter9: Gene Expression And Gene Regulation

Section: Chapter Questions

Problem 18QP: The following is a portion of a protein: met-trp-tyr-arg-gly-pro-thr-Various mutant forms of this...

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The pre-mRNA transcript and protein made by several mutant genes were examined. The results are given below. Determine where in the gene a likely mutation lies: the promoter region, exon, intron, cap on mRNA, or ribosome binding site. a. normal-length transcript, normal-length nonfunctional protein b. normal-length transcript, no protein made c. normal-length transcript, normal-length mRNA, short nonfunctional protein d. normal-length transcript, longer mRNA, shorter nonfunctional protein e. transcript never made
The following is a portion of a protein: met-trp-tyr-arg-gly-pro-thr-Various mutant forms of this protein have been recovered. Using the normal and mutant sequences, determine the DNA and mRNA sequences that code for this portion of the protein, and explain each of the mutations. a. met-trp- b. met-cys-ile-val-val-leu-gln- c. met-trp-tyr-arg-ser-pro-thr- d. met-trp-tyr-arg-gly-ala-val-ile-ser-pro-thr-
O The lac operon in E.coli encodes enzymes necessary for the breakdown of lactose. For each enzyme (lac Z and lac Y), indicate with a + or-whether or not it is made when there is no lactose or when there is lactose. B-galactosidase (lac Z) No Lactose Permease (lac Y) Lactose Lactose No Lactose Genotype PP0 Z Y/I P*O*Z•Y* I'POCZ Y*/I P* O©Z*Y° P O Z'Y/I P'OʻZ'Y* PP O ZY*/IP*O*Z*Y* IP OCZ Y /I P*O*Z•Y* IPO ZY*/I* P*O©Z*Y• I'PO*Z Y*/IP'O*Z*Y°
1. You are investigating a protein that has the amino acid sequence N ... Ala – Thr - Asn – Trp – Lys - Arg - Gly – Phe – Thr ... C within its primary structure. You found that several of the mutations affecting this protein produced shortened protein molecules that terminated within this region. In one of the mutants, the Asn became the terminal (last) amino acid. (a) What DNA single-base changes(s) would cause the protein to terminate at the Asn residue? (b) What other potential sites do you see in the DNA sequence encoding this protein where mutation of a single base pair would cause premature termination of translation? >
Help me figure out how I should interpret and read these graphs. We are looking to see whether the m6a acceptor mutant will have an impact on splicing. Just need a detail breakdown thanks. I am presenting at journal club and need help interpreting will the goals in mind. We are currently conducting experiments to understand how the splicing of circE7 relates to the splicing of linear E6*I. Goal is to determine whether the mutation impact splicing. Trying to understand whether m6A will impact splicing. Sm should impact splicing ratio. My PI stated that the results show that it inhibited linear splicing and promoted backsplicing. I need a detailed explanation for the entirety so I can understand
The most common MCAD mutation is shown below. The coding strand is shown for both the WT and mutant. The TATA box and kozak sequences are in parenthesis. What type of mutation is present? Wild-type:5’-ATGGCC[TATAT]ATGTCACTTGACTACGCAGCC[GCCACCATGG]ATATAGATAATGCGCGCATAGCATACTGAGGGTAGTAG-3’ Mutant:5’-ATGGCC[TATAT]ATGTCACTTGACTACGCAGCC[GCCACCATGG]ATATAGATAATGCGCGC AGAGCATACTGAGGGTAGTAG-3’ Answer: Is this a transition mutation? because there is an exchange of G instead of A? It kind of confuses me a little. help
e B 1_30*_SP23 - General Biology I (for majors)/11364 of € 2 A us page X F1 What conditions would we find on the gene of a prokaryote if there is low amounts of tryptophan within the cytoplasm? Select one: a. lactose would attach to the repressor removing it from the promoter promoting the transcription of lactase b. no repressor is on the promoter creating constant transcription of lactase c. tryptophan would attach to the repressor removing it from the promoter promoting the transcription of lactase O d. repressors would bind to the promoter stopping the transcription of lactase e. lactose would attach to the repressor removing it from the promoter promoting the transcription of tryptophan producing enzymes no repressor is on the promoter creating constant transcription of tryptophan producing enzymes tryptophan would attach to the repressor binding it to the promoter and stopping the transcription of tryptophan producing enzymes O f. O g. 2 Oh. tryptophan would attach to the…
Sickle cell hemoglobin DNA CACGTAGACTGAGG ACAC.. Sickle cell hemoglobin MRNA Sickle cell hemoglobin AA sequence 4. What type of mutation is this? Please explain why.
G-LO37 Identify the consequences of mutations in different regions of a gene. The image below represents two strands of DNA: the top one corresponds to a healthy individual, and the bottom one of a sibling potentially affected with a disease due to genetic mutations Mutation 1 A + с AUA ACA AUG Met ACG GUU GUC GUA GUG Val GCU GCC GCA GCG It will result in mRNA produced Mutation 2 It will result in no mRNA produced 500 AGG The protein produced will be normal 500 + GGG Ala The Select all that applies about Mutation 1 (position -6): AAGLys AGA Arg GGU GGC GGA GGG GAC Asp GAA Glu GAGJ Gly The protein produced will have a different amino acid 1235 ATT 1235 TTT 070 2070 ALL The mutation occurs in the promoter region, and this means that the mRNA cannot be produced 1535 The mRNA and protein will both be normal because the mutation occurs outside of the consensus region of the promoter G 1535 с Affected
The method of Northern blotting is used to determine the amountand size of a particular RNA transcribed in a given cell type.Alternative splicing (discussed in Chapter 14) produces mRNAsof different lengths from the same gene. The Northern blot shownhere was obtained using a DNA probe that is complementary tothe mRNA encoded by a particular gene. The mRNA in lanes 1through 4 was isolated from different cell types, and equal amountsof total cellular mRNA were added to each lane. Explain these results.
Consider the following simple regulatory pathways. Assume the full pathway is shown. A- E- B- F- C- G- D- 1 H- A You identify several null mutations (a complete deletion of the gene). For each mutant (ind with a - sign), determine whether the final product (I, J, K or L) is inducible, uninducible, or constitutive. 2 B 3 C 4 D inducible inducible constitutive uninducible constitutive inducible inducible E uninducible F G H > > >
Below is the double stranded DNA sequence of part of a hypothetical yeast genome encoding a very small gene. Transcription starts at nucleotide immediately following the promoter. The termination sequence is TATCTC. How many amino acids will this protein have? 5' TCATGAGATA GCCATGCACTA AGGCATCTGA GTTTATATCT CA 3' 3' AGTACTCTAT CGGTACGTGAT TCCGTAGACT CAAATATAGA GT 5'