Answer:
Step-by-step explanation:
a) The late marker is c+. This is because the c+ marker is present in the genotype of the F- cells (StrR a-b- c+), but it is not present in any of the genotypes of the 1000 colonies that grew on A-deficient media. This suggests that the c+ marker is the last marker to be transferred during recombination.
b) The gene order is a-b-c+. This is because the a+ marker is present in the genotypes of 495 colonies, the b+ marker is present in the genotypes of 395 colonies, and the c+ marker is present in the genotype of 10 colonies. The order in which these markers are transferred during recombination reflects the order in which they are located on the chromosome.
c) The map distance between the a and b markers is approximately 0.1 centimorgans (cM), and the map distance between the b and c markers is approximately 0.9 cM. This can be calculated using the following formula:
map distance = (number of recombinants / total number of offspring) x 100
For the a and b markers, the number of recombinants is 95 (the number of colonies with the genotype a+b- c+) and the total number of offspring is 1000, so the map distance is:
map distance = (95 / 1000) x 100 = 0.1 cM
For the b and c markers, the number of recombinants is 190 (the number of colonies with the genotype a+b+c-) and the total number of offspring is 1000, so the map distance is:
map distance = (190 / 1000) x 100 = 0.9 cM
Therefore, the map distances between the 3 markers are 0.1 cM (a and b), 0.9 cM (b and c).