What happens if insert DNA is cut with two different restriction enzymes at the ends?

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 Lac– strain, called lac3–, is linked to the Tn5 insertion. From a strain carrying the Tn5 insertion and lac3– mutation you isolate an F’ that caries a region of the chromosome that includes both Tn5 and the linked Lac region. Introduce this F’ into an F– strain carrying lac1– by selecting for Kanr. These merodiploids express ß-galactosidase normally. What does this result tell you about the relationship between the lac3– and lac1- 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 Lac– strain, called lac3–, is linked to the Tn5 insertion. From a strain carrying the Tn5 insertion and lac3– mutation you isolate an F’ that caries a region of the chromosome that includes both Tn5 and the linked Lac region. Introduce this F’ into an F– strain carrying lac1– by selecting for Kanr. These merodiploids express ß-galactosidase normally. If the merodiploid were Lac–, 

a) the two mutations lie in the different gene.

b) Both mutation are complement each other

c) the two mutations lie in the same gene.

d) Both of the mutations is dominant to wild type.

what could you conclude about the relationship between the lac3– and lac1– mutations from  these statements?

9 In your study of a new bacterial species you have identified a transducing phage that you call Px. In addition you have worked out methods to make random transposon insertions into the bacterial genome. You have generated two different transposon insertion collections one with 105 random Tn5 (Kanr) insertions and the other with 105 random Tn10 (Tetr) insertions. You grow Px phage on the mixed collection of Tn5 insertions and use the resulting phage lysate to infect the mixed collection of Tn10 insertions. You select 10,000 Kanr  transductants and find that 80 of them are Tets. Use this information to estimate the total size of the bacterial genome assuming that both Tn5 and Tn10 insert randomly and that the average size of a fragment recombined into the recipient genome during Px transduction is 55 kbp. (Tn5 is about 5 kbp and Tn10 is 10 kbp.)

You have isolated two mutations in the Lac operon that cause constitutive expression of Lac genes. You designate these mutants Lac1– and Lac2–. Making use of an F' that carries the Lac operon with the LacY gene mutated, you construct strains that you test for both ß-galactosidase activity and Lac permease activity with results shown below.

     

Classify each mutation as dominant or recessive and as cis- or trans-acting, giving the experimental result that allows you to arrive at each conclusion. Finally, deduce what type of Lac mutation best fits the properties of Lac 1– and of Lac 2–.