Question

Respond to this two part question that has different data for each:

A) A cross is made between an Hfr strain that is Str^Sx⁺y⁺z⁺ in genotype and an F^- strain that is Str^Rx^-y^-z^- in genotype. Interrupted-mating studies show that z⁺ enters the recipient strain last, and that the Str locus is very far away from y⁺, so it never enters the recipient strain. The y⁺ recombinants are then tested for the presence of the x⁺ and z⁺ alleles. The following data were obtained:

1	x+	y+	z+		351
2	x-	y+	z+		6
3	x+	y+	z-		25
4	x-	y+	z-		66
Total					sum of above

What is the gene order?

a) x y z Str

b) y x z Str

c) y z x Str

d) z x y Str

e) z y x Str

B) The bacterial cross Hfr  Str^Sx⁺y⁺z⁺ X F^- Str^Rx^-y^-z^- was performed, and all y⁺ exconjugants were tested for the presence of the x⁺ and z⁺ alleles. Remember that z⁺ enters the recipient strain last, and that the Str locus is very far away from y⁺, so it never enters the recipient strain.

The following data were obtained (the numbers are not the same as in the previous question):

1	x+	y+	z+		307
2	x-	y+	z+		6
3	x+	y+	z-		37
4	x-	y+	z-		93
Total					sum of above

What is the map distance between markers z and y?

Round your answer to the nearest 1/10 decimal place (xx.x); your answer should be between 10 and 40

Answer 1

A) By comparing two Hfr strains X+Y+Z+ = 351 and X-Y+Z+=6 , when X+ is present, the distance is high and when X is absent (X-) then map distance is low. So, it clearly shows that X+ will enter last.

Now, X+Y+Z+ = 351 and X+Y+Z- = 25

X-Y+Z+ = 6 and X-Y+Z- = 66

These two set clearly shows the presence and absence of X and Z highly affects the map distance. Z and X will be present at two extreme ends.

It is also given Z enter first, So the gene order will be (e) ZYX.

B) Here also the gene order will be same ZYX

The recombinant type for Z and Y will be X+Y+Z- =37 and X-Y+Z- =93

Total frequency = (307+6+37+93)=443

The map distance between Z and Y is = (37+93)/443*100=29.3 map unit.