Creation of target construct according to High-level expression of bioactive rec

Question

Creation of target construct according to High-level expression of bioactive recombinant human lysozyme in the milk of transgenic mice using a modified human lactoferrin BAC by Shen Liu

The 4.8-kb hLZ genomic sequence, starting from the ATG start codon to the TAA stop codon and flanked by two homology arms, was obtained by PCR using the BAC clone RP11-1143G9 (Genome Systems Inc., St. Louis, Mo.) as a template and using the following primers: hLF-hLZ-F (5-CTAGCTAGCAAAGCCCTGAATAAAGGGGCGCAGGGCAGGCGCAAGTGGCAGAGCCTTCGTTTGCCAAGTCGCCTCCAGACCGCAGACATGAAGGCTCTCATTGTTCTG-3) and hLF-hLZ-R (5-CTAGCTAGCAGGGGAGGCCAAGGCCCCAACACACCTGGGGAGAAGAGCTGGGGG CAGTGAATGGCTGAGGCTTTCTTGGGGAGCTGGGCCATCTTCTTCGGTTTTACACTCC ACAACCTTGAAC-3), where homology arms to the hLF gene are in bold, and NheI restriction sites are in italics. The PCR product was cloned into the pMD19-T cloning vector (Takara, Dalian, China) and confirmed by sequencing. A zeocin (Zeo) cassette flanked by two FRT sites for positive selection was blunt-end ligated into the HpaI site in intron 2 of the hLZ gene. The resulting targeting construct, named pMD19-hLZ-Zeo, was linearized with NheI, and the DNA fragments containing hLZ-Zeo were purified for recombineering.

In the Supplementary Materials and Methods under “Creation of targeting construct for recombineering” are listed two very long PCR primers called hLF-hLZ-F and hLF-hLZ-R. Describe the target that is used by these primers in their PCR and calculate the appropriate Tm one needs to use for determining an annealing temperature in the PCR [use Tm= 2(A+T) + 4(G+C) for your estimate]. Briefly explain the reasoning for your answers.

Explanation / Answer

PCR primers is 18-22 bp with 52-58 0C, annealing temperature, GC content 40-60%

5’CTAGCTAGCAAAGCCCTGAATAAAGGGGCGCAGGGCAGGCGCAAGTGGCAGAGCCTTCGTTTGCCAAGTCGCCTCCAGACCGCAGACATGAAGGCTCTCATTGTTCTG3’

Forward primer Tm = 2(A+T) + 4(G+C) = 5’CTAGCTAGCAAAGCCCTG3’ 2(5+3) + 4(4+6)= 56 0C [18bp]

GC content = (10/18) x 100 =55.56%

5’CTAGCTAGCAGGGGAGGCCAAGGCCCCAACACACCTGGGGAGAAGAGCTGGGGGCAGTGAATGGCTGAGGCTTTCTTGGGGAGCTGGGCCATCTTCTTCGGTTTTACACTCCACAACCTTGAAC3’

Backward primer Tm = 5’CACTCCACAACCTTGAAC3’ = 2(A+T) + 4(G+C) = 2(6+3) + 4(1+8) =540C [18bp]

GC content = (09/18) x 100 =50%

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