1. (10 marks) In an E. coli mapping experiment a cross is made between an Hfr st

Question

1. (10 marks) In an E. coli mapping experiment a cross is made between an Hfr strain that is a+ b+ d+ in genotype and an F- strain that is a- b- d- in genotype. Interrupted-mating studies show that b+ enters the recipient strain last. You select for b+ recombinants and obtain 400 that are then tested for the presence of the a+ and d+ alleles. The following data were obtained:

                     a+ b+ d+     326

                     a-   b+ d+        2

                     a+ b+ d-       14

                     a-   b+ d-       58

                                        400

A. (2 points) why select for the late entering gene in this experiment?

B. (2 points) What is the gene order (show your reasoning)?

C. (4 points) What are the frequencies of recombination between each pair of genes? Show calculations.

D. (2 points) Draw a map, showing map distances (in map units) between genes.

Explanation / Answer

B. The genotype with the maximum frequency is the parental genotype and the one with the minimum frequency is the double crossover genotype. In this case 'a+ b+ d+' is the parental genotype with highest frequency 326 and 'a- b+ d+' is the double crossover genotype with minimum frequency 2. Since, during double crossover the allele changes in the middle changes it's position with its non parallel allele and since only the allele d has changed in position (d+ in the parent allele is switched with d- in the double crossover), so the gene order is 'a d b'.

D. a-d distance is calculated by {100x[(2+14)]} / 400 = 4 cM.
d-b distance is calculated by {100x[(2+58)]} / 400 = 15 cM
The genetic map is given as follows:

a d b
|___|__________|
4cM 15cM

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