The following are DNA sequences from two homologous genes: TTGCATAGGCATACCGTATGATATCGAAAACTAGAAAAATAGGGCGATAGCTA GTATGTTATCGAAAAGTAGCAAAATAGGGCGATAGCTACCCAGACTACCGGAT The two sequences, however, do not begin and end at the same location. Try to line them up according to their homologous regions.
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The following are DNA sequences from two homologous genes: TTGCATAGGCATACCGTATGATATCGAAAACTAGAAAAATAGGGCGATAGCTA GTATGTTATCGAAAAGTAGCAAAATAGGGCGATAGCTACCCAGACTACCGGAT
The two sequences, however, do not begin and end at the same location. Try to line them up according to their homologous regions.
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- You have the following sequence reads from a genomicclone of the Homo sapiens genome:Read 1: ATGCGATCTGTGAGCCGAGTCTTTARead 2: AACAAAAATGTTGTTATTTTTATTTCAGATGRead 3: TTCAGATGCGATCTGTGAGCCGAGRead 4: TGTCTGCCATTCTTAAAAACAAAAATGTRead 5: TGTTATTTTTATTTCAGATGCGARead 6: AACAAAAATGTTGTTATTa. Use these six sequence reads to create a sequencecontig of this part of the H. sapiens genome.b. Translate the sequence contig in all possible readingframes.c. Go to the BLAST page of the National Center forBiotechnology Information, or NCBI (http://www.ncbi.nlm.nih.gov/BLAST/, Appendix B) and see if you canidentify the gene of which this sequence is a part byusing each of the reading frames as a query for protein–protein comparison (BLASTp).Below is a sequence of DNA. 5'-ttaccgataattctctctcccctcttccatgattctgattaaagaaggcgagaacgaaactatttgttaatacc-3' How many "reading frames" can be identified for this sequence? How many "open reading frames" can be identified for this sequence? What is the frame of the longest ORF?In addition to the standard base-paired helical structures, DNA can form X-shaped hairpin structures called cruciforms in which most bases are involved in Watson–Crick pairs. Such structures tend to occur at sequences with inverted repeats. Draw the cruciform structure formed by the DNA sequence TCAAGTCCACGGTGGACTTGC.
- You have sequenced the genome of the bacterium Salmonella typhimurium and find a protein that is 100 percent identical to a protein in the bacterium Escherichia coli. When you compare nucleotide sequences of the S. typhimurium and E. coli genes, you find that their nucleotide sequences are only 87 percent identical. How would you interpret the observations? Please make sure to select ALL correct answer options. Because genetic code is redundant, changes in the DNA nucleotide sequence can occur without change to its encoded protein. Due to the flexibility in the third positions of most codons, the DNA sequence can accumulate changes without affecting protein structure. Natural selection will eliminate many deleterious amino acid changes. This will reduce the rate of change in the amino acid sequence and lead to sequence conservation of the proteins. Protein sequences are expected to evolve and…TGAGGATGAAACTCACACCGGGGCGCAGTTTGGCACTTAGATTCTTGTACACGACCTAGTATAACACAGTT Compare this mutated sense sequence given below to the original one given above and identify and classify all mutations that can be found in this new DNA sequence? TGAGCATGAAACTCACACCGGGGGCAGTTTCGCACTTAGGATTCTTGTACAGGACCTAGTATAACAAGTTThe 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’-CCGTTACCCAAAACGTTAAGATTTTCAAAAATTAAG
- For the following sequence design the forward and reverse primer... explain and justify your answer. Gene of Interest: a tgaaacaaca aaaacggctt tacgcccgat tgctgacgct gttatttgcg 61 ctcatcttct tgctgcctca ttctgcagca gcggcggcaa atcttaatgg gacgctgatg 121 cagtattttg aatggtacat gcccaatgac ggccaacatt ggaagcgttt gcaaaacgac 181 tcggcatatt tggctgaaca cggtattact gccgtctgga ttcccccggc atataaggga 241 acgagccaag cggatgtggg ctacggtgct tacgaccttt atgatttagg ggagtttcat 301 caaaaaggga cggttcggac aaagtacggc acaaaaggag agctgcaatc tgcgatcaaa 361 agtcttcatt cccgcgacat taacgtttac ggggatgtgg tcatcaacca caaaggcggc 421 gctgatgcga ccgaagatgt aaccgcggtt gaagtcgatc ccgctgaccg caaccgcgta 481 atttcaggag aacacctaat taaagcctgg acacattttc attttccggg gcgcggcagc 541 acatacagcg attttaaatg gcattggtac cattttgacg gaaccgattg ggacgagtcc 601 cgaaagctga accgcatcta taagtttcaa ggaaaggctt gggattggga agtttccaat 661 gaaaacggca actatgatta tttgatgtat gccgacatcg attatgacca tcctgatgtc 721 gcagcagaaa ttaagagatg gggcacttgg tatgccaatg…The following is the base sequence of DNA that codes for first eight amino acids of the β chain of hemoglobin. The β chain of hemoglobin contains a total of 147 amino acids so this is a small part of the entire gene. DNA Template Strand: TACCACGTGGACTGAGGACTCCTC 1. What is the minimum number of DNA nucleotides in this whole gene? 3 2. What is the sequence of bases on the strand of DNA that is complementary to the template strand? 3' TACCACGTGGACTGAGGACTCCTC 5' . 5' ATGGTGCACCTGACTCCTGAGGAG 3' 3. What mRNA will be formed from the template strand of DNA? Sequence of mRNA formed from DNA template strand is shown below: 3' TACCACGTGGACTGAGGACTCCTC 5' . 5'AUGGUGCACCUGACUCCUGAGGAG 3 4. What amino acids will this mRNA code for? 5. If the 20th base in the template strand of the DNA is changed from T to A, rewrite the new template strand below. 6. When the template strand of the DNA is changed, this is referred to as a mutation. What kind of mutation is…I sequence the DNA of 3 people and see variation in my gene of interest as follows: Person 1: ATGCAACAATTTAATAAT Person 2: ATGCAACGACGACGACGACAATTTAATAAT Person 3: ATGCAACGACGACGACGACGACGACGACGACAATTTAATAAT What is the name for this kind of variation?
- the one above: Replicate this sense strand to create a double-stranded DNA helix TGAGGATGAAACTCACACCGGGGCGCAGTTTGGCACTTAGATTCTTGTACACGACCTAGTATAACACAGTT Compare this mutated sense sequence given below to the original one given above and identify and classify all mutations that can be found in this new DNA sequence? TGAGCATGAAACTCACACCGGGGGCAGTTTCGCACTTAGGATTCTTGTACAGGACCTAGTATAACAAGTT 2. Using this mutated DNA strand, express it as a polypeptide by using the correct reading frame. When you get to the stop codon – you may write an “*” to denote the stop codon. 3. How many amino acids were changed in the mutated polypeptide?The following is the base sequence of DNA that codes for first eight amino acids of the β chain of hemoglobin. The β chain of hemoglobin contains a total of 147 amino acids so this is a small part of the entire gene. DNA Template Strand: TACCACGTGGACTGAGGACTCCTC 1. What is the minimum number of DNA nucleotides in this whole gene? 2. What is the sequence of bases on the strand of DNA that is complementary to the template strand? 3. What mRNA will be formed from the template strand of DNA?If one DNA segment has the following base composition, 5'-CAGTTAGTCA-3', which of the following sequences is complementary? 5'-TGACTAACTG-3' 3'-TGACTAACTG-5' 3'-CAGTTAGTCA-5' 3'-TGACTAACTG-5'