Question

Consider three genes P, Q, and R, that have the following distances in map units: P-Q = 3 m.u; P-R = 7 m.u; and Q-R = 10 m.u. What is the gene order?

a) P-Q-R
b) R-Q-P
c) Q-P-R
d) R-P-Q

Answer 1

The correct order of genes based on the provided map units is P-Q-R, with gene Q being positioned between P and R but closer to P.

Step-by-step explanation:

The question involves determining the order of genes on a chromosome based on their given recombination frequencies expressed in map units. To infer the correct order of genes P, Q, and R, we look at the distances between them: P-Q = 3 m.u; P-R = 7 m.u; and Q-R = 10 m.u. If P is 7 m.u from R and only 3 m.u from Q, this places Q in between P and R, with Q being closer to P. To check this, we can add P-Q (3 m.u) and Q-R (10 m.u) distances, which should equal the P-R distance if the order is correct. Indeed, 3 m.u + 10 m.u = 13 m.u, which is greater than the given P-R distance (7 m.u), confirming Q lies between P and R but closer to P. Thus, the correct gene order is P-Q-R.