Temperature at which the kinetic energy of the molecules lowers to reduce the chances of base paired sticky ends parting before they have been stabilized.

Wild type E. coli metabolizes the sugar lactose by expressing the enzyme ß-galactosidase. You have isolated a mutant that you call lac1–, which cannot synthesize ß-galactosidase and cannot grow on lactose (Lac–). During an condition you have a wild type (Lac+) strain carrying a Tn5 insertion known to be near several Lac genes on the E. coli chromosome. You grow P1 phage on this strain and use the resulting phage lysate to infect the lac1– strain, selecting for kanamycin resistance (Kanr). Among 100 Kanr transductants, you find that 82 are Lac– and 18 are Lac+. Express the distance between Tn5 and the lac1– mutation as a cotransduction frequency;

Wild type E. coli metabolizes the sugar lactose by expressing the enzyme ß-galactosidase. You have isolated a mutant that you call lac1-, which cannot synthesize ß-galactosidase and cannot grow on lactose (Lac-). During an condition  isolate  a second Lac– mutation, which you designate lac2-. Using P1 phage on this strain and use the resulting phage lysate to infect the lac2- strain, selecting for   Kanr   transductants. In this case, all 100   Kanr   transductants that are examined are Lac–. What does this result tell you about the relationship between the lac1- and lac2- mutations?

Wild type E. coli metabolizes the sugar lactose by expressing the enzyme ß-galactosidase. You have isolated a mutant that you call lac1-, which cannot synthesize ß-galactosidase and cannot grow on lactose (Lac-). During an condition  isolate  a second Lac– mutation, which you designate lac2-. Using P1 phage on this strain and use the resulting phage lysate to infect the lac2- strain, selecting for   Kanr   transductants. In this case, all 100   Kanr   transductants that are examined are Lac–. What does this result tell you about the relationship between the lac1- and lac2- mutations?

Wild type E. coli metabolizes the sugar lactose by expressing the enzyme ß-galactosidase. You have isolated a mutant that you call lac1-, which cannot synthesize ß-galactosidase and cannot grow on lactose (Lac-). During an condition  isolate  a second Lac– mutation, which you designate lac2-. Using P1 phage on this strain and use the resulting phage lysate to infect the lac2- strain, selecting for   Kanr   transductants. In this case, all 100   Kanr   transductants that are examined are Lac–. What does this result tell you about the relationship between the lac1- and lac2- mutations?

Wild type E. coli metabolizes the sugar lactose by expressing the enzyme ß-galactosidase. You have isolated a mutant that you call lac1–, which cannot synthesize ß-galactosidase and cannot grow on lactose (Lac–). During an condition isolate a mutation that constitutively expresses abnormally high levels of ßgalactosidase, which you designate lacc. Preliminary P1 transduction experiments indicate that lacc  is linked to the Tn5 insertion.  To map lacc  relative to lac1– you set up two reciprocal crosses. In the first cross you grow P1 on a strain that carries the Tn5 insertion and the lac1– mutation. You then use this lysate to infect a lacc mutant and select for Kanr. From 100 Kanr transductants examined, 20 are Lac–, 76 express ß-galactosidase constitutively and 4 show normal ß-galactosidase expression. In the second cross you grow P1 on a strain that carries the Tn5 insertion and the lacc mutation. You then use this lysate to infect a lac1– mutant, and select for Kanr. From 100 Kanr transductants examined, 81 are Lac– and 19  express ß-galactosidase constitutively. So what will be the correct order of  the Tn5 insertion and the lac1– and lacc  mutations. Express any measured distances as cotransduction frequencies.

Wild type E. coli metabolizes the sugar lactose by expressing the enzyme ß-galactosidase. You have isolated a mutant that you call lac1–, which cannot synthesize ß-galactosidase and cannot grow on lactose (Lac–). During an condition isolate a mutation that constitutively expresses abnormally high levels of ßgalactosidase, which you designate lacc. Preliminary P1 transduction experiments indicate that lacc  is linked to the Tn5 insertion.  To map lacc  relative to lac1– you set up two reciprocal crosses. In the first cross you grow P1 on a strain that carries the Tn5 insertion and the lac1– mutation. You then use this lysate to infect a lacc mutant and select for Kanr. From 100 Kanr transductants examined, 20 are Lac–, 76 express ß-galactosidase constitutively and 4 show normal ß-galactosidase expression. In the second cross you grow P1 on a strain that carries the Tn5 insertion and the lacc mutation. You then use this lysate to infect a lac1– mutant, and select for Kanr. From 100 Kanr transductants examined, 81 are Lac– and 19  express ß-galactosidase constitutively. So what will be the correct order of  the Tn5 insertion and the lac1– and lacc  mutations. Express any measured distances as cotransduction frequencies.