Which of the following single-stranded DNA sequences is most likely to form a stem-loop structure? TAAGTACATTACCCCG ACTTCTTCTCCGCTGC GACCGTATGCACGGTC GCCCACGCCAGTAGTG
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- In the copies of each sequence below, divide the sequences into codons (triplets) by putting a slash between each group of three bases. Sequence ATCTTCCCTCCTAAACGTTCAACCGGTTCTTAATCCGCCGCCAGGGCCCCGCCCCTCAGAAGTTGGTSequence BTCAGACGTTTTTGCCCCGTAACAACTTGTTACAACATGGTCATAAACGTCAGAGATGGTCAATCTCTTAATGACTSequence CTACAAACATGTAAACACACCCTCAGTGGACCAACTCCGCAACATAAACCAAACACCGCTCGCGCCGAAAAAGATATGGThis 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.Convert the following DNA sequences to RNA sequences.Sequence # 1: CCGGTTCAGGCTTCACCACAGTGTGGAACGCGGTCGTCTCCGGCGACCSequence # 2:CTAAGGTTGCTAATCTCAGCGCTCCGCTGACCCCTCAGCAAAGGGCTTGSequence # 3:GCTCAATCTCGTCCAGCCATTGACCATCGTCGAGGGGTTTGCTCTGTTACSequence # 4:CAAAACGAAATCGAGCGCCATCTGCGGGCCCCGATTACGGACATCAGASequence # 5: TCCAACTCGGGGTCCGCATCGCTCCGCCGGCGACCGACGAAGTTCCGA
- 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' Draw the structure of hairpin loop that will be formed during the end of transcription.This 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.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’-CCGTTACCCAAAACGTTAAGATTTTCAAAAATTAAG
- Below is a sequence of DNA. 5'-ttaccgataattctctctcccctcttccatgattctgattaaagaaggcgagaacgaaactatttgttaatacc-3' Using the one letter code for Amino Acids, what is the predicted AA sequence of the shortest ORF (from N to C-terminal end)? Using the one letter code for Amino Acids, what is the predicted AA sequence of the longest ORF (from N to C-terminal end)?Assume a bacterial gene underwent a mutation, where a thymine base from an early portion of the coding sequence of the DNA is replaced with a cytosine (as illustrated below). Original sequence (coding strand): AGTTCCTACAAAATGGAGCTGTCTTGGCATGTAGTCTTT ...[Sequence continues with another 80 bases] New sequence: AGTTCCCACAAAATGGAGCTGTCTTGGCATGTAGTCTTT...[Sequence continues with another 80 bases] UAC encodes tyrosine, CAC encodes histine, per the coding table. (This question can be answered without use of the code table, but it is provided here as a resource.) What would the expected result of such a mutation be on the final protein product of the mutated gene (compared to the original, non-mutant product)? The protein will be very different from the original version, and likely non-functional. The protein will be cut short, ending after the first amino acid. There will be no protein produced at all. No change – the protein will be the same.…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 D 5’-GGCAATGGGTTTGTGCAATTCTAACAGTTTTTAATTC-3’ 3’-CCGTTACCCAAACACGTTAAGATTGTCAAAAATTAAG-5’ Group E 5’-GGCAATGGGTTTTGCAATTCTAAAAGTTTTTAATTC-3’ 3’-CCGTTACCCAAAACGTTAAGATTTTCAAAAATTAAG
- The sequence of the coding strand of a bacterial gene is given below. The positions of the first nine bases are numbered for your convenience. A missense mutation was introduced at position seven where the C was changed to a T resulting a mutant gene. 123456789 5'- ATGGCCCGACCGCAACTTTTCCGAGCTCTGGTGTCTGCGCAGTGACC-3 a. Write the template DNA (complementary strand) sequence for the wild type gene above b. Write the DNA sequence of the mutant gene (Both DNA strands) c. Write the sequence of mRNA produced from the mutant gene d. Write the sequence of the mutant protein using the codon usage table provided in the end of this document.1) What DNA base sequence is complementary to the following DNA sequence?TAGCGTGCATGGTGCTTAAC2) What RNA base sequence is complementary to the following RNA sequence?UAAUAGCUUGCUGAUAs you should recall, DNA, when not being actively transcribed, has a double helical structure. This portion of the DNA has had the two strands separated in preparation of transcribing for a needed protein. The following is one of the two complimentary strands of DNA: 3' - AACCAGTGGTATGGTGCGATGATCGATTCGAGGCTAAAATACGGATTCGTACGTAGGCACT - 5' Q: Based on written convention, i.e. the 3'-5' orientation, is this the coding strand or the template strand? ______________________________ Q: Assuming this strand extends from base #1 to #61 (going left to right), interpret the correctly transcribed mRNA and translated polypeptide for bases 24 - 47: mRNA: ___-___-___-___-___-___-___-___-___-___-___-___-___-___-___-___-___-___-___-___-___-___-___-___- polypeptide chain: ________--________--________--________--________--________--________--________