Question

The DNA from a his+, bio+ strain of bacteria (it can synthesize both histidine and biotin) is isolated and used to transform a strain with the genotype his-, bio- (it cannot synthesize histidine or biotin). Using replica plating techniques, the genotypes of 66 colonies were determined and are listed below. Determine the map distance (frequency of co-transformation) between these two genes. his+, bio+ 8 his-, bio+ 14 his+, bio- 18 his-, bio- 26

Answer 1

Based on the given data, the progeny are:

s.no	Genotype	Progeny	Type
1	his+, bio+	8	Parental
2	his-, bio-	26	Parental
3	his-, bio+	14	Recombinant
4	his+, bio-	18	Recombinant
Total = 66

Here, only two types of recombinants are possible. Thus, the distance between the “his” and “bio” is calculated based on their recombination frequencies:

The recombination frequency (RF) of parental is 34/ 66 = 0.515 x100 = 51.5%

The recombination frequency (RF) of Recombinants is 32 / 66 = 0.484 x 100 = 48.5%

Thus, the map distance are 51.5 (m.u.) and 48.5 (m.u.), but the first gene is determined by the genotype of highest recombinants obtained, thus the sequence is his” and “bio.”

Answer 2

To figure out the distances, we must identify our single crossover classes. We've got four of them: BAC, bAc, BaC and bac. Note that this is two pairs of reciprocals (BACand bac are reciprocals and bAc and BaC are also reciprocals). Again, we want to compare these to the parentals which are most like them to determine just where the crossover occurred to produce them. Comparing single crossover BAC to parental bAC, we see that it has recombined between the B and the A loci. Same for its reciprocal. Comparing the BaC single crossover to the Bac parental, we see that it has recombined between the A and the C genes, as has its reciprocal. Note also that the numbers of offspring give you hints as to which crossover classes go together.

Now we are ready to calculate distances. We have four offspring classes which have recombined between the B and the A genes. These are the single crossovers BAC (95) andbac (95) as well as the double crossovers Bac (5) and bAC (5). This gives us a total of 200 offspring who have crossed over between these two genes. There are a total of 1700 offspring, so we'd calculate this distance as (200/1700) x 100, which equals 11.8. So we've calculated a distance of 11.8 LMU between genes B and A.

There are also four offspring classes which have recombined between the A and C genes. bAc (50), BaC (50) and again, our double crossover classes BAc (5) and baC (5), for a total of 110. Calculating: (110/1700) x 100 = 6.5 LMU between genes A and C.