Which one of the following options represents the combination of all correct statements?

• Genetic linkage and mapping of genes:
-This Question is related to genetic linkage and the mapping of genes on a chromosome based on the frequency of recombination events.
-Bacteriophages that have two mutations causing different plaque phenotypes, and the genes responsible for these mutations are 9 cM (centimorgans) apart.
-Genetic linkage refers to the tendency of genes that are physically close on a chromosome to be inherited together more frequently.
-The closer two genes are, the less likely they are to undergo recombination during gamete formation.
-Recombination is the exchange of genetic material between homologous chromosomes during meiosis.
-Based on the provided information, we have to predict the outcome of a genetic cross involving bacteriophages with specific genotypes.
-The expected proportions of different plaque types are determined by the distances between the genes and the likelihood of recombination events occurring during the mixing and infecting process.

• Explaination:
-Cross given c+ m- x c- m+. Since the genes are 9 cM apart means the the genes are linked.
-Therefore the proportion of parental types (c+ m- and c- m+) will be much larger than the recombinant types (c+ m+ and c- m-).
-Based on this we can eliminate the proportions given in A and B.
-Distance between the genes is recombination frequency percentage.
-9 cM distance means recombination frequency of 9%.
-Therefore, proportion of each recombinants (c+ m+ and cm-) = 0.09/ 2 = 0.045.
-So, the proportion of each parentals (c+ m- and c- m+) = (1-0.09)/ 2 = 0.91/2 = 0.455.
-For statement C, Total = 45+455+455+45 = 1000.
-Number of each recombinants (c+ m+ and c- m-) = 0.045 x 1000 = 45; Number of each parentals (c+ m- and c- m+) = 0.455 x 1000 = 455.
-For statement D, Total = 65+ 680+685+70 = 1500. Number of each recombinants (c+ m+ and c- m-) = 0.045 x 1500 = 67.5; Number of each parentals (c+ m- and c- m+) = 0.455 x 1500 = 682.