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A) Genomic Section of DNA 19.8 Kb cloned in 3 Kb vector, therefore total size of pSP1 = 22.8 Kb

A) Genomic Section of DNA 19.8 Kb cloned in 3 Kb vector, therefore total size of pSP1 = 22.8 Kb Fully digested with Hind III gives 9.7, 7.5, 2.8, 2.3 and 0.5 (sum of fragments = 22.8 Kb) therefore no double bands are present.

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A) Genomic Section of DNA 19.8 Kb cloned in 3 Kb vector, therefore total size of pSP1 = 22.8 Kb

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  1. A) Genomic Section of DNA 19.8 Kb cloned in 3 Kb vector, therefore total size of pSP1 = 22.8 Kb Fully digested with Hind III gives 9.7, 7.5, 2.8, 2.3 and 0.5 (sum of fragments = 22.8 Kb) therefore no double bands are present. pSP1 Partially digested with Hind III and sections generated cloned and insert size measured. Therefore, sections generated must be made up from Hind III fragments that lie next to one another. Layout the possible fragments that can make up the insert sizes: Table 1: Clone Insert size Lyase Kan Possible (kb) activity fragments 1 2.8 + - 2.8 or 2.3 + 0.5 2 12.5 + - 9.7 + 2.8 or 9.7 + 2.3 + 0.5 3 12.0 + - 9.7 + 2.3 4 10.3 - + 7.5 + 2.8 or 7.5 + 2.3 +0.5 5 13.1 + + 7.5 + 2.8 + 2.3 + 0.5 6 5.6 + - 2.8 + 2.3 + 0.5 Start with the most straightforward information : Clone 3, 12 Kb lyase positive must be 9.7 joined to 2.3 (9.7-2.3) Look for next simplest: Clone 1, 2.8 lyase positive 2.8 or 2.3 + 0.5. Lyase msut lie in the common 2.3 fragment and therefore 2.3 must be joined to 0.5 (2.3-0.5) The rest fall into place very easily: Clone 2, 12.5 lyase positive must be 9.7 + 2.3 + 0.5 (9.7-2.3-0.5) Clone 4: 10.3 not lyase positive but kan + must therefore not contain 2.3 therefore 7.5 is joined to 2.8 and Kan lies on 7.5 or 2.8 (7.5-2.8) Clone 6: 5.6 lyase positive, but kan negative means 7.5 has kan resistance, 2.3 joined to 0.5 joined to 2.8 (9.7-2.3-0.5-2.8) Clone 5: 13.1 lyase and kan positive (7.5-2.8-0.5-2.3) Lyase 2.3 0.5 9.7 Summary: 2.8 7.5 Kan

  2. Rational behind Question b: pHG327 3 kb Bam HI Kpn I Pst I Eco RI Eco RI Not drawn to scale 2.1 EcoR1 fragment Inserted

  3. Use guide lines or graph paper b) Scaled Map of 2.1 Kb lyase containing fragment 1,500 2,000 2,100 0 1,000 500 Mark on known vector sites Pst I Bam HI Kpn I Eco RI (10 U) - 3000, 2100 Eco RI Eco RI Pst I (10 U) - 5,100 Pst I Kpn I (10 U) - 3650, 800, 650 Either or Kpn I Kpn I Kpn I Pst 1 (10 U) + Kpn I (0.5 U) - 5100, 4300, 3650, 1450, 800, 650 (Full Pst, partial Kpn. Linarised at Pst site, then fragments generated from section lying next to each other 5100 800 4300 Kpn I 3650 650 Pst I Kpn I 1450

  4. 1,500 2,000 2,100 0 1,000 500 Mark on known vector sites Pst I Bam HI Kpn I Kpn I Kpn I Kpn I Bam HI (10 U)- 3500, 700, 300 (sum = 4500. Short by 600 therefore must be extra two 300) either Bam HI Bam HI or Bam HI or Bam HI or Bam HI (10 U) and Kpn I -3000, 500, 350, 300 (sum 4150. Short by 950, therefore must be two extra 300 and an extra 350 Kpn I Kpn I Kpn I 300 300 300 500 350 350 Bam HI Bam HI Bam HI Bam HI Bam HI Composite Kpn I Kpn I Kpn I Bam HI Bam HI Bam HI Bam HI Eco RI Eco RI Pst I Bam HI

  5. Alginate gene C) The two sized clearing zones come from the fact that the Eco RI fragments can insert into the expression vector in two orientations. Small clearing zones come from the fragment inserted into the vector so that it is not orientated to make use of the lactose operator, Eco RI 2.1 Kb fragment must have its own alg promoter to show any activity under these circumstances. Large clearing come from Alginate gene using both its own promoter and the lactose operator switch on with IPTG. Large Clearing zones Alg promoter Lactose operator Alginate gene 2.1 Kb insert Transcription Transcription in presence of IPTG Small Clearing zones Alg promoter Lactose operator 2.1 Kb insert Transcription Transcription of non-sense sequence in presence of IPTG

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