Identify the mutation(s) that lead to the most loss in transcriptional activity, and discuss whether those match expectations based on the consensus sequence for the Initiator. B. Hypothesize a molecular mechanism to explain how the mutations identified in Part A result in loss of transcription activity. C. Identify whether any mutations cause transcription to initiate from a different position than wild-type, and provide a brief explanation as to why that occurs for those mutant core promoters.

Identify the mutation(s) that lead to the most loss in transcriptional activity, and discuss whether those match expectations based on the consensus sequence for the Initiator. B. Hypothesize a molecular mechanism to explain how the mutations identified in Part A result in loss of transcription activity. C. Identify whether any mutations cause transcription to initiate from a different position than wild-type, and provide a brief explanation as to why that occurs for those mutant core promoters.

Biochemistry

6th Edition

ISBN:9781305577206

Author:Reginald H. Garrett, Charles M. Grisham

Publisher:Reginald H. Garrett, Charles M. Grisham

Chapter29: Transcription And The Regulation Of Gene Expression

Section: Chapter Questions

Problem 17P

See similar textbooks

Similar questions

#4 BamI --- 5’ CCTAG ↓G 3’ 5’ ACGCCTAGGACGTATTATCCTAGGTAT CCGCCGCCGT CATCA 3’ 3’ TGCGGATCCTGCATAATAGGATCCATAGGCGGCGGCAGTAGT 5’ Restriction enzyme: Recognition sequence: Number of pieces of DNA: Type of cut:
EcoRI --- 5' G - AATTC 3' 5' AGAATTCCGACGTATTAGAATTCTTAT CCGCCGCCGGAATTCT CATCA 3' 3' TCTTAAGGCTGCATAATCTTAAGAATAGGCGGCGGCCTTAAGAGTAGT 5' Number of pieces of DNA , and type of fragment .
#3 HaelII --- 5’ CC ↓ GG 3’ 5’ ACGCCGGCCGTATTAT CCGGATCCGCCG CCGGCTGTCCCGGATCA 3’ 3’ TGCGGCCGGCATAATAGGCCTAGGCGGCGGCCGACAGGGCCTAGT 5’ Restriction enzyme: Recognition sequence: Number of pieces of DNA: Type of cut:
#2 EcoRI --- 5’ G ↓AATTC 3’ 5’ ACG ACGTATTAGAATTCTTA TCCGCCGCCGGAATTCT CATCA 3’ 3’ TGC TGCATAATCTTAAGAATAGGCGGCGGCCTTAAGAGTAGT 5’ Restriction enzyme: Recognition sequence: Number of pieces of DNA: Type of cut:
Figure 1 is a bacterial gene (1-180). The first base to be transcribed is the base located at position 77. 45 5' TTGGT CTTGG TCGGA TTCCA GAGGA TGAAG TGTTG ACAGC GCATT 3' 3 AACCA GAACC AGCCT AAGGT CTCCT ACTTC ACAAC TGTCG CGTAA 5' 46 5 AATTG ACCTT GCTGT ATTAT AGCCA AGGAC AGATC TACGA GCATG 3' 3 TTAAC TGGAA CGACA TAATA TCGGT TCCTG TCTAG ATGCT CGTAC 5' 91 5 TGCGA ACCGC AAGCA TTCGT TCTCC TAGGC TACTC GATCC CGTAA 3' 3 ACGCT TGGCG TTCGT AACCA AGAGG ATCCG ATGAG CTAGG GCATT 5 77 90 110 135 136 5 TGATG TAGCT GATTC TGTTG AAAGG CTCCT TTTGG AGCCT TTTTT 3 3' ACTAC ATCGA CTAAG ACAAC TTTCC GAGGA AAACC TCGGA AAAAA 5 156 180 Figure 1. Illustrate how termination of transcription occurs in the gene above. (Hint: position from 156 to 180)
Figure 1 is a bacterial gene (1-180). The first base to be transcribed is the base located at position 77. 45 5 TTGGT CTTGG TCGGA TTCCA GAGGA TGAAG TGTTG ACAGC GCATT 3' 3 AACCA GAACC AGCCT AAGGT CTCCT ACTTC ACAAC TGTCG CGTAA 5 46 77 90 5' AATTG ACCTT GCTGT ATTAT AGCCA AGGAC AGATC TACGA GCATG 3' 3 TTAAC TGGAA CGACA TAATA TCGGT TCCTG TCTAG ATGCT CGTAC 5' 91 110 5 TGCGA ACCGC AAGCA TTCGT TCTCC TAGGC TACTC GATCC CGTAA 3' 3 ACGCT TGGCG TTCGT AACCA AGAGG ATCCG ATGAG CTAGG GCATT 5' 135 136 5 TGATG TAGCT GATTC TGTTG AAAGG CTCCT TTTGG AGCCT TTTTT 3 3 ACTAC ATCGA CTAAG ACAAC TTTCC GAGGA AAACC TCGGA AAAAA 5 156 180 Figure 1. (i) Identify both the hexameric sequences of the promoter region in the coding strand above.
E D ot OH ON O Replication R SORO Transcription ORNA processing Translation Gene expression 657 F Q Search % T G LOCLE Y F7 & 87 F8 U 99+ 8 F9 53879 10 F10 O 2 H J K L CV BN MO < F12
Coding DNA 5’- GTG ACT CGT TGT GCC ATT GCA GCT AAA CAC TTC GAG CCC TGT- 3’ mRNA 5’- GUG ACU CGU UGU GCC AUU GCA GCU AAA CAC UUC GAG CCC UGU- 3’ What is the polypeptide sequence?
This is part of the Escherichia coli DNA sequence that contains an inverted repeat. (Note: top strand is the coding strand). 5'-AACGCATGAGAAAGCCCCCCGGAAGATCACCTTCCGGGGGCTTTATATAATTAGC-3' 3'-TTGCGTACTCTTTCGGGGGGCCTTCTAGTGGAAGGCCCCCGAAATATATTAATCG-5' (i) Draw the structure of hairpin loop that will be formed during the end of transcription. (ii) Describe the function of the hairpin loop during transcription.
Given Sequence: 3’-TACGGTCCGGATTCGGTAGCTAGCATC-5’ Provide: Complementary Strand: Transcript Amino Acid Sequence 2.Given Sequence: 5’-GGGCATATGCCGTTTACCGGTTTGACTAAATAACCA-3’ Provide: Complementary Strand: Transcript Amino Acid Sequence 3.Given Sequence: 3’-AAC CAA TAC GTG AGG ATA CCA AGT AAC ACT CCC-5’ Provide: Complementary Strand: Direct Transcript: Transcript for Translation: Amino Acid Sequence:
40 Shown below is the antisense strand of DNA. What is the amino acid sequence that corresponds to this code? 5' AAAGCATACCGG 3' Second letter G. UUU Phe UCU UAU) Tyr UGC Cys UGU UUC UCC UAC Ser UAA Stop UGA Stop A UAG Stop UGG Trp UUA UCA UUG Leu UG CAU His CGU CGC CGA CGG CU CUU CUC CUA CUG CAC Pro Leu Arg CCA CAA) Gin CCG CAG AAU AUU AUC lle AUA AUG Met ACG ACU AGU AAC Asn AGC Ser ACC ACA Thr AAA1 AGA AAG Lys AGG Arg GAU) Asp GGC GAC Ala GAA GGU GUU GUC GUA Val GCU GCC Gly GCA GGA Select an answer and submit. For keyboard navigation, use the up/down arrow keys to select an answer. a PRO-VAL-SER-PHE PHE-ARG-MET-ALA GLY-HIS THR-LYS d LYS-ALA-TYR-ARG First letter Third letter
Give me nucleotide sequence with pairing. Like this ATC TCA TGA GCC TAG AGT ACT. CGG