A competent bacterial strain with genes a, b, and c is transformed by a donor bacterial fragment. Cotransformation frequencies for each gene pair are as follows: a and b 0.04% a and c 0.02% b and c 0.0064% Which conclusion can be definitively made from this data?

a) Genes b and c are farthest apart.

b) Gene order is a, b, then c.

c) Gene a is closer to c than to b.

d) Gene b is closer to c than to a.

e) Cotransformation of b and c is so frequent, they must be one gene

Question

Judah

Biology

15 January, 01:27

A competent bacterial strain with genes a, b, and c is transformed by a donor bacterial fragment. Cotransformation frequencies for each gene pair are as follows: a and b 0.04% a and c 0.02% b and c 0.0064% Which conclusion can be definitively made from this data?

a) Genes b and c are farthest apart.

b) Gene order is a, b, then c.

c) Gene a is closer to c than to b.

d) Gene b is closer to c than to a.

e) Cotransformation of b and c is so frequent, they must be one gene

+1

Answers (1)

Know the Answer?

Elaina Navarro · Answer 1

a) Genes b and c are farthest apart.

Explanation:

Transformation occurs when a competent bacteria cell takes up genetic material from the environment. Usually a donor cell donates its gene fragment which is then incorporated into the chromosome or plasmid of recipient bacterial cell.

Cotransformation occurs when two genes are taken up together by the recipient. The closer the genes lie to each other, more are the chances of them being taken up together. Contransformation frequency will be higher if two genes are close to each other. Here, cotransformation frequencies between three genes are given. Amongst them, the lowest frequency is 0.0064% which is present between gene b and c. Hence, gene b and c are the farthest apart.