In each of the following cases, determine the consequences of a single crossover within the inverted region of a pair of homologous chromosomes with the gene order A B C D in one chromosome and a c b d in the other. There is a two situation of centromere for that gene. 

(A) Centromere is not included within the inversion.

(B) Centromere is included within the inversion

Following consequence are resulting are

I - The crossover chromatids consist of a dicentric and an acentric

II- One Cross over product is duplicated for the terminal region contain in A and deficient for the terminal region containing D

III- Both cross over products is monocentric

Following is correct combination for consequence?

TLS Online TPP Program

#Question id: 15617

#SCPH01 Biochemistry

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;

TLS Online TPP Program

#Question id: 15617

#SCPH06 I Botany

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;

TLS Online TPP Program

#Question id: 15617

#SCPH28 | Zoology

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;

TLS Online TPP Program

#Question id: 15618

#SCPH01 Biochemistry

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?

TLS Online TPP Program

#Question id: 15618

#SCPH06 I Botany

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?

TLS Online TPP Program

#Question id: 15618

#SCPH28 | Zoology

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?