Question

DNA from a strain of Bacillus subtilis with genotype a⁺b⁺c⁺d⁺e⁺ is used to transform a strain with genotype a^-b^-c^-d^-e^-. Pairs of genes are checked for co-transformation and the following results are obtained:

Pair of genes              Cotransformation

and b⁺                                 no

and c⁺                                  no

and d⁺                                 yes

and e⁺                                 yes

and c⁺                                  yes

and d⁺                                 no

and e⁺                                 yes

and d⁺                                  no

and e⁺                                  yes

and e⁺                                 no

On the basis of these results, what is the order of the genes on the bacterial chromosome?

Answer 1

Assuming the full table form the given sequence: a, b, c, d and e:

a+ and b+ no

a+ and c+ no

a+ and d+ yes

a+ and e+ yes

b+ and c+ yes

b+ and d+ no

b+ and e+ yes

c+ and d+ no

c+ and e+ yes

d+ and e+ no

................................

Now, Genes that transform together (close together) :

a --- d, a --- e, b --- c, b --- e ,c --- e.

Genes that do not transform together (far away):

a --- b ,a --- c, b --- d ,c --- d, d --- e .

To find the order of gene, consider which ones transform together or close together:

b --- c, b --- e , c --- e.

Therefore, a possible order is b---c---e.

Seeing that , (a --- d) transform together and they are close and as (a --- e) transform together and they are also close. Therefore , a possible order is b---c---e---a.

It has seen that, d doesn't transform with b, c or e and d is far from b, c and e and as a doesn't transform with b or c. Therefore, a possible order is b---c---e---a---d.