To analyze:
Access to http://blast.ncbi.nlm.nih.gov/Blast.cgi and type in the sequence below in the link
Collect the information on the basis of steps mentioned in the question and determine the DNA sequence.
Introduction:
In bioinformatics, BLAST stands for “basic local alignment search tool” that finds the region of local similarity between the biological sequences. This algorithm compares a DNA sequence or an amino acid sequence to the sequences that are in the databases and calculates the statistical significance of the sequence matches. BLAST can be used to identify the members that belong to one gene family. There are different types of BLAST present used according to the query sequence.
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Check out a sample textbook solutionChapter 16 Solutions
Genetic Analysis: An Integrated Approach (3rd Edition)
- The following are DNA fragments containing a small gene. The top strand is the coding strand. Transcribe all 5 groups and translate. Group A 5’-GGCAATGGGTTTGTGCAATTCTAAAAGTTTTTAATTC-3’ 3’-CCGTTACCCAAACACGTTAAGATTTTCAAAAATTAAG-5’ Group B 5’-GGCAATGGGTTTGTGAAATTCTAAAAGTTTTTAATTC-3’ 3’-CCGTTACCCAAACACTTTAAGATTTTCAAAAATTAAG-5’ Group C 5’-GGCAATGGGTTTGTGCAATTCTAAGAGTTTTTAATTC-3’ 3’-CCGTTACCCAAACACGTTAAGATTCTCAAAAATTAAG-5’ Group D 5’-GGCAATGGGTTTGTGCAATTCTAACAGTTTTTAATTC-3’ 3’-CCGTTACCCAAACACGTTAAGATTGTCAAAAATTAAG-5’ Group E 5’-GGCAATGGGTTTTGCAATTCTAAAAGTTTTTAATTC-3’ 3’-CCGTTACCCAAAACGTTAAGATTTTCAAAAATTAAGarrow_forwardWhat is the DNA template of the following DNA coding: ATGGCTAACCTTGTAarrow_forwardThe X's correspond to the missing information. You need to determine what those X's represent DNA STRAND: XXX-CCC-GGG-GCG-XXX-XXX-XXX-GCC-ATA-TTA-XXX RNA STRAND: XXX-XXX-XXX-XXX-XXX-XXX-XXX-XXX-XXX-XXX-XXX PROTEIN: Met - X - X-X - Thr - Val - Glu-X-X-X - Trp To be clear, your answer is just the 3 sequences above without the X's (the highlighted yellow parts). There may be more than 1 correct answer for some parts of this assignment. Any answer your choose is fine. If you are unclear on how there could be more than 1 correct answer, review the lecture related to protein translation. Provide me with the: 1. Complete DNA sequence of 33 nucleotides 2. Complete RNA sequence of 33 nucleotides 3. 11 amino acidsarrow_forward
- 5'GGT ACG TTG GGG CTC CAT3' This sequence is transcribed and translated. Write the resulting amino acid sequence using the 3 letter code. Write the answer in a all capital letters. Leave a space between the amino acids. Do not write 5' and 3'. 5'GGT ACG TTG GGG CTC CAT3' This sequence is transcribed and translated. If the G in Bold changes to a T, then the result will be A) A nonsense mutation B) A frameshift mutation C) A silent substitution D) A missense mutation 5'GGT ACG TTG GGG CTC CAT3' This sequence is transcribed and translated. If the G in Bold changes to a A, then the result will be A) A nonsenese mutation B) A frameshift mutation C) A silent substitution D) A missense mutationarrow_forwardThe following are DNA fragments containing a small gene. The top strand is the coding strand. Transcribe all groups and translate. FIND THE POSSIBLE MUTATIONS Group D 5’-GGCAATGGGTTTGTGCAATTCTAACAGTTTTTAATTC-3’ 3’-CCGTTACCCAAACACGTTAAGATTGTCAAAAATTAAG-5’ Group E 5’-GGCAATGGGTTTTGCAATTCTAAAAGTTTTTAATTC-3’ 3’-CCGTTACCCAAAACGTTAAGATTTTCAAAAATTAAGarrow_forwardWhich of the following represents the sequence of an RNA transcript for which the coding strand (also known as non-template strand) of DNA has the sequence: GTACTGGCTAGCTGCTAGAA? Note all sequences are written 5'-3'. OA. AAGAUCGUCGAUCGGUCAUG OB. AAGATCGTCGATCGG TCATG OC. GTACTGGC TAGCTGC TAGAA OD. GUACUGGCUAGCUGCUAGAAarrow_forward
- The following are DNA fragments containing a small gene. The top strand is the coding strand. Transcribe all groups and translate. FIND THE POSSIBLE MUTATIONS Group B - MUTATION 5’-GGCAATGGGTTTGTGAAATTCTAAAAGTTTTTAATTC-3’ 3’-CCGTTACCCAAACACTTTAAGATTTTCAAAAATTAAG-5’ Group C- 5’-GGCAATGGGTTTGTGCAATTCTAAGAGTTTTTAATTC-3’ 3’-CCGTTACCCAAACACGTTAAGATTCTCAAAAATTAAG-5’arrow_forwardFigure 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.arrow_forward#1 HindII --- 5’ GTC ↓ GAC 3’ 5’ ACGACGTAGTCGACTTATTAT GTCGACCCGCCGCGTGTCGACCATCA 3’ 3’ TGCTGCATCAGCTGAATAATACAGCTGGGCGGCGCACAGCTGGTAGT 5’ Restriction enzyme: Recognition sequence: Number of pieces of DNA: Type of cut:arrow_forward
- 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)arrow_forwardThe following questions refer to this table of codons. Second Base U A G UUU UCU UAU UGU ] Cys Phe Tyr UUC UCC UAC UGC Ser UUA UCA UAA Stop UGA Stop Leu UUG UCG UAG Stop UGG Trp CUU CCU CAU CGU His CUC CCC CAC CGC Leu Pro Arg CUA CCA СА CGA Gin CUG CG CAG CGG AUU ACU AAU AGU Asn Ser AUC Ile ACC AAC AGC Thr AUA ACA AAA AGA Met or Start ] Lys Arg AGG AUG ACG AAG G GUU GCU GAU GGU Asp GAC GUC GCC GGC G Val Ala Gly GUA GCA GAA GGA Glu GUG GCG GAG GGG What amino acid sequence will be generated, based on the following mRNA codon sequence? 5' AUG- UCU- UCG- UUA- UCC- UUG 3' met-arg-glu-arg-glu-arg met-ser-ser-leu-ser-leu met-glu-arg-arg-glu-leu met-ser-leu-ser-leu-ser First Base Third Basearrow_forwardThis 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_forward
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