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| | The pSB1C3 is supposed to be 2049bp, however, it appears around the marker 3000bp. | | The pSB1C3 is supposed to be 2049bp, however, it appears around the marker 3000bp. |
| | We have the pSB1C3 in K381001 sequenced and found its total length is 2944bp, and its total sequence is as below, | | We have the pSB1C3 in K381001 sequenced and found its total length is 2944bp, and its total sequence is as below, |
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| − | <html><style type="text/css">ol{margin:0;padding:0}p{margin:0}.c4{width:468pt;background-color:#ffffff;padding:90pt 72pt 90pt 72pt}.c3{text-align:justify;direction:ltr}.c2{color:#0000ff}.c1{color:#f50000}.c0{font-size:10pt}body{color:#000000;font-size:12pt;font-family:Times New Roman}h1{padding-top:12pt;line-height:1.0;text-align:left;color:#000000;font-size:16pt;font-family:Arial;font-weight:bold;padding-bottom:3pt}h2{padding-top:12pt;line-height:1.0;text-align:left;color:#000000;font-style:italic;font-size:14pt;font-family:Arial;font-weight:bold;padding-bottom:3pt}h3{padding-top:12pt;line-height:1.0;text-align:left;color:#000000;font-size:13pt;font-family:Arial;font-weight:bold;padding-bottom:3pt}h4{padding-top:12pt;line-height:1.0;text-align:left;color:#000000;font-size:14pt;font-family:Times New Roman;font-weight:bold;padding-bottom:3pt}h5{padding-top:12pt;line-height:1.0;text-align:left;color:#000000;font-style:italic;font-size:13pt;font-family:Times New Roman;font-weight:bold;padding-bottom:3pt}h6{padding-top:12pt;line-height:1.0;text-align:left;color:#000000;font-size:11pt;font-family:Times New Roman;font-weight:bold;padding-bottom:3pt}</style><div class="c4"><p class="c3"><span class="c0">GTGGAGTT</span><span class="c2 c0">CACCCTG</span><span class="c0 c1">C(lack a C)</span><span class="c2 c0">CCTTTTTCTTT</span><span class="c1 c0">A(lack a A)</span></p><p class="c3"><span class="c0 c2">AAACCGAAAAGATTACTTCGCGTTATGCAGGCTTCCTCGCTCACTGACTCGCTGCGCTCGGTCGTTCGGCTGCGGCGAGCGGTATCAGCTCACTCAAAGGCGGTAATACGGTTATCCACAGAATCAGGGGATAACGCAGGAAAGAACATGTGAGCAAAAGGCCAGCAAAAGGCCAGGAACCGTAAAAAGGCCGCGTTGCTGGCGTTTTTCCACAGGCTCCGCCCCCCTGACGAGCATCACAAAAATCGACGCTCAAGTCAGAGGTGGCGAAACCCGACAGGACTATAAAGATACCAGGCGTTTCCCCCTGGAAGCTCCCTCGTGCGCTCTCCTGTTCCGACCCTGCCGCTTACCGGATACCTGTCCGCCTTTCTCCCTTCGGGAAGCGTGGCGCTTTCTCATAGCTCACGCTGTAGGTATCTCAGTTCGGTGTAGGTCGTTCGCTCCAAGCTGGGCTGTGTGCACGAACCCCCCGTTCAGCCCGACCGCTGCGCCTTATCCGGTAACTATCGTCTTGAGTCCAACCCGGTAAGACACGACTTATCGCCACTGGCAGCAGCCACTGGTAACAGGATTAGCAGAGCGAGGTATGTAGGCGGTGCTACAGAGTTCTTGAAGTGGTGGCCTAACTACGGCTACACTAGAAGAACAGTATTTGGTATCTGCGCTCTGCTGAAGCCAGTTACCTTCGGAAAAAGAGTTGGTAGCTCTTGATCCGGCAAACAAACCACCGCTGGTAGCGGTGGTTTTTTTGTTTGCAAGCAGCAGATTACGCGCAGAAAAAAAGGATCTCAAGAAGATCCTTTGATCTTTTCTACGGGGTCTGACGCTCAGTGGAACGAAAACTCACGTTAAGGGATTTTGGTCATGAGATTATCAAAAAGGATCTTCACCTAGATCCTTTTAAATTAAAAATGAAGTTTTAAATCAATCTAAAGTATATATGAGTAAACTTGGTCTGACAGCTCGAGGCTTGGATTCTCACCAATAAAAAACGCCCGGCGGCAACCGAGCGTTCTGAACAAATCCAGATGGAGTTCTGAGGTCATTACTGGATCTATCAACAGGAGTCCAAGCGAGCTCGATATCAAATTACGCCCCGCCCTGCCACTCATCGCAGTACTGTTGTAATTCATTAAGCATTCTGCCGACATGGAAGCCATCACAAACGGCATGATGAACCTGAATCGCCAGCGGCATCAGCACCTTGTCGCCTTGCGTATAATATTTGCCCATGGTGAAAACGGGGGCGAAGAAGTTGTCCATATTGGCCACGTTTAAATCAAAACTGGTGAAACTCACCCAGGGATTGGCTGAGACGAAAAACATATTCTCAATAAACCCTTTAGGGAAATAGGCCAGGTTTTCACCGTAACACGCCACATCTTGCGAATATATGTGTAGAAACTGCCGGAAATCGTCGTGGTATTCACTCCAGAGCGATGAAAACGTTTCAGTTTGCTCATGGAAAACGGTGTAACAAGGGTGAACACTATCCCATATCACCAGCTCACCGTCTTTCATTGCCATACGAAATTCCGGATGAGCATTCATCAGGCGGGCAAGAATGTGAATAAAGGCCGGATAAAACTTGTGCTTATTTTTCTTTACGGTCTTTAAAAAGGCCGTAATATCCAGCTGAACGGTCTGGTTATAGGTACATTGAGCAACTGACTGAAATGCCTCAAAATGTTCTTTACGATGCCATTGGGATATATCAACGGTGGTATATCCAGTGATTTTTTTCTCCATTTTAGCTTCCTTAGCTCCTGAAAATCTCGATAACTCAAAAAATACGCCCGGTAGTGATCTTATTTCATTATGGTGAAAGTTGGAACCTCTTACGTGCCCGATCAACTCGAGT</span><span class="c0">TACCAATGCTTAATCAGTGAGGCACCTATCTCAGCGATCTGTCTATTTCGTTCATCCATAGTTGCCTGACTCCCCGTCGTGTAGATAACTACGATACGGGAGGGCTTACCATCTGGCCCCAGTGCTGCAATGATACCGCGAGACCCACGCTCACCGGCTCCAGATTTATCAGCAATAAACCAGCCAGCCGGAAGGGCCGAGCGCAGAAGTGGTCCTGCAACTTTATCCGCCTCCATCCAGTCTATTAATTGTTGCCGGGAAGCTAGAGTAAGTAGTTCGCCAGTTAATAGTTTGCGCAACGTTGTTGCCATTGCTACAGGCATCGTGGTGTCACGCTCGTCGTTTGGTATGGCTTCATTCAGCTCCGGTTCCCAACGATCAAGGCGAGTTACATGATCCCCCATGTTGTGCAAAAAAGCGGTTAGCTCCTTCGGTCCTCCGATCGTTGTCAGAAGTAAGTTGGCCGCAGTGTTATCACTCATGGTTATGGCAGCACTGCATAATTCTCTTACTGTCATGCCATCCGTAAGATGCTTTTCTGTGACTGGTGAGTACTCAACCAAGTCATTCTGAGAATAGTGTATGCGGCGACCGAGTTGCTCTTGCCCGGCGTCAATACGGGATAATACCGCGCCACATAGCAGAACTTTAAAAGTGCTCATCATTGGAAAACGTTCTTCGGGGCGAAAACTCTCAAGGATCTTACCGCTGTTGAGATCCAGTTCGATGTAACCCACTCGTGCACCCAACTGATCTTCAGCATCTTTTACTTTCACCAGCGTTTCTGGGTGAGCAAAAACAGGAAGGCAAAATGCCGCAAAAAAGGGAATAAGGGCGACACGGAAATGTTGAATACTCATACTCTTCCTTTTTCAATATTATTGAAGCATTTATCAGGGTTATTGTCTCATGAGCGGATACATATTTGAATGTATTTAGAAAAATAAACAAATAGGGGTTCCGCGCACATTTCCCCGAAA</span><span class="c1 c0">AGTGCCACCTGACGTCTAAGAAACCATTATTATCATGACATTAACCTATAAAAATAGGCGTATCACGAGGC</span><span class="c0">GCA</span></p></div>
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| | Through sequence analysis, we could find that the base in blue have a 99% homology with the right pSB1C3. It lacks a C and a A in the beginning. In the end, it has a 71 base homology with the right pSB1C3. The other sequence left do not accord with the right pSB1C3. | | Through sequence analysis, we could find that the base in blue have a 99% homology with the right pSB1C3. It lacks a C and a A in the beginning. In the end, it has a 71 base homology with the right pSB1C3. The other sequence left do not accord with the right pSB1C3. |
Revision as of 02:23, 6 October 2011
OUC China Experience
Applications of BBa_K594015
From the beginning, we use the linearized plasmid backbone pSB1C3 in the 2011 spring DNA distribution. But the standardization with them all end up with failures.
Then, we began to seek for a new plasmid backbone. We found the K381001 in pSB1C3. If we cut the pSB1C3 with EcoR I and Pst I, we could get pSB1C3 plasmid backbone for standardization.
User Reviews
UNIQ85e071499a3fa346-partinfo-00000000-QINU
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DonQuixia
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When we use the pSB1C3 for standardization, we found that the pSB1C3 have the right antibiotic resistance but a wrong DNA length.
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UNIQ85e071499a3fa346-partinfo-00000002-QINU
The verification
The pSB1C3 is supposed to be 2049bp, however, it appears around the marker 3000bp.
We have the pSB1C3 in K381001 sequenced and found its total length is 2944bp, and its total sequence is as below,
Through sequence analysis, we could find that the base in blue have a 99% homology with the right pSB1C3. It lacks a C and a A in the beginning. In the end, it has a 71 base homology with the right pSB1C3. The other sequence left do not accord with the right pSB1C3.
