A bacterial strain is mutant for three genes ( a, b, and c ). When transformed w

Question

A bacterial strain is mutant for three genes (a, b, and c). When transformed with DNA carrying wild-type copies of the genes, the cotransformation frequencies for each pair of genes are as follows:

a and b 0.63%

a and c 0.10%

b and c 0.78%

What is the correct conclusion from this data?

A bacterial strain is mutant for three genes (a, b, and c). When transformed with DNA carrying wild-type copies of the genes, the cotransformation frequencies for each pair of genes are as follows:

a and b 0.63%

a and c 0.10%

b and c 0.78%

What is the correct conclusion from this data?

Genes a and c are farthest apart. Gene b is closer to a than to c. Gene order is a, c, b. Cotransformation of a and c is so rare, they must be next to each other. Genes b and c are farthest apart.

Explanation / Answer

Genes a and c are farthest apart.

Higher the cotransformation frequency indicates shorter distance between two genes.

Closes distance will be between b and c.

SInce cotransformation frequency of a and c is least hence it will have highest distance between them.

So,

Genes a and c are farthest apart.

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.