Concept explainers
To answer:
Recognition and restriction of DNA sequence by HhaI.
Introduction:
Restriction enzymes recognize and cut DNA
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EBK MICROBIOLOGY:W/DISEASES BY BODY...-
- 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.arrow_forwardA short protein segment was cleaved by two different enzymes resulting in two sets of degradation products with multiple fragments in each set. The amino acid sequence of each fragment was determined via Edman degradation. The fragments obtained from Enzyme 1 have the following sequences: LLIGVVQT RKNS SFWAA EED Meanwhile, the fragments obtained from Enzyme 2 were found to have the following sequences: QTEED SSFWAALL IGVV RKN What is the amino acid sequence of the short protein segment from which these fragments were derived? O RKNSSFWAALLIGVVQTEED O LLIGVVQTSFWAARKNSEED O SSFWAALLQTEEDIGVVRKN O IGVVRKNSSFWAALLQTEEDarrow_forwardList the sequences of RNA that would be transcribed from the following DNA template sequences. TTACACTTGCTTGAGAGTC ACTTGGGCTATGCTCATTA GGCTGCAATAGCCGTAGAT GGAATACGTCTAGCTAGCAarrow_forward
- Identify the dinucleotide CA repeat region and the score in the following sequence:TGGCACACTCACACCACACAGACAGTTAarrow_forwardA DNA strand was sequenced using the Sanger method (https://www.youtube.com/watch?v=KTstRrDTmWI). The reaction tube contained the DNA strand, fluorescently labelled dideoxynucleotide triphosphates (ddATP – yellow, ddGTP – green, ddCTP – blue, ddTTP - red), deoxynucleotide triphosphates, DNA polymerase, or its Klenow fragment. Synthesis of DNA is allowed to proceed, and the results are shown on the right: 15 14 13 12 11 10 (a) What is the sequence of the copy and the template strands? (b) If the template strand were in the 5'-3' direction, what will be the sequence of the DNA copy? Nucleotide Lengtharrow_forwardA DNA fragment with the following base sequence has some cytosine bases that are methylated (indicated by C*) and others that are unmethylated. To determine the locations of methylated and unmethylated cytosines, researchers sequenced this fragment both with and without treatment with sodium bisulfite. Give the sequence of bases that will be read with and without bisulfite treatment. —ATCGC*GTTAC*GTTGC*GTCA—arrow_forward
- Provide the sequences of the template and coding strands of a DNA double helix that was used to produce this RNA: 5'-AUUACGGUCUAU. Be sure to label the 5' and 3' ends.arrow_forwardPlease choose the right answers.arrow_forwardYou are studying a protein that contains the peptide sequence RDGSWKLVI. The part of the DNA encoding this peptide is included in the sequence shown below. 5'-CGTGACGGCTCGTGGAAGCTAGTCATC-3' 3'-GCACTGCCGAGCACCTTCGATCAGTAG-5' This sequence does not contain any BamHI restriction enzyme sites. The target sequence for the BamHI restriction nuclease is GGATCC. Your goal is to create a BamHI site on this plasmid by manipulating the DNA sequence, without changing the coding sequence of the protein. How would you do this, ie what would the new sequence be?arrow_forward
- Translate the following mRNA nucleotide sequence into an amino acid sequence, starting at the second base: 5’ - UGUCAUGCUCGUCUUGAAUCUUGUGAUGCUCGUUGGAUUAAUUGU - 3’arrow_forwardThis is part of the Escherichia coli DNA sequence that contains an inverted repeat. (Note: bottom strand is the noncoding strand). 5'-ААCGCATGAGAAAGCCCCCCGGAAGATCACСТТСCGGGGGCТТТАТАТААТТАGC-3' 3'-тTGCGTACтстттCGGGGGGCCTTCTAGTGGAAGGCCCCCGАААТАТАТТААТтCG-5' (i) Draw the structure of hairpin loop that will be formed during transcription. (ii) Illustrate how the hairpin loop structure initiates the termination of transcription.arrow_forwardBased on the following wild type DNA sequence, indicate if each of the mutations should be classified as : insertion, deletion, missense, nonsense, silent (Use the provided Genetic Code table and remember you have been given DNA sequence). Wild Type: AUGAUUCUUAAAAGU Mutant 1: AUGAUUCUUUAAAGU Mutant 2: AUGAUUCUUGAAAGU Mutant 3: AUGAUCCUUAAAAGU Mutant 4: AUGAUCCUAAAAGU Mutant 5: AUGAUCCUUAAACAGU Socond letter Key: Ala = Alanine (A) Arg Arginine (R) Asn = UUU } UAU Tyr UGU UGC Cys UGA STOP UGG Trp UCU UCC UUC Phe Ser Asparagine (N) Asp = Aspartate (D) Cys Cysteine (C) Gin = Glutamine (Q) Glu = Glutamate (E) Gly = Glycine (G) His = Histidine (H) le = Isoleucine (1) Leucine (L) Lys Lysine (K) Met = Methionine (M) Phe = Phenylalanine (F) Pro Proline (P) Ser = Serine (S) Thr Threonine (T) Trp Tryptophan (W) Tyr Tyrosine (Y) - Valine (V) UCA UCG UAA STOP UAG STOP UUA Leu UUG S CCU CC CGU CUU CUC His CGC Arg Leu Pro CAA Gin CGA CCA CCG CUA CUG CGG Leu = AGU AUU AUC } lle AUA ACU ACC ACA Ser AAC…arrow_forward
- BiochemistryBiochemistryISBN:9781305577206Author:Reginald H. Garrett, Charles M. GrishamPublisher:Cengage Learning