Question

The Hox genes are responsible for determining the anterior–posterior identity of body regions (segments) in the developing insect embryo. Different Hox genes are turned on (expressed) in different segments of the body, and in this way they determine which segments become head and which thorax, which develop legs and which antennas. One surprising thing about the Hox genes is that they usually occur in a row on the same chromosome and in the same order as the body regions that they control. For example, the fruit fly Drosophila melanogaster has eight Hox genes located on a chromosome in exactly the same order as the body regions in which they are expressed, from head to tail (see figure below; Lewis et al. 2003; Negre et al. 2005). If the eight genes were thrown randomly onto the same chromosome, what is the probability that they would line up in the same order as the segments in which they are expressed?

Answer 1

Solution

Back-up Theory

Number of ways of arranging n things among themselves = n! = n(n - 1)(n - 2) …… 3.2.1……...(1)

Values of n!can be directly obtained using Excel Function: Math & Trig FACT (Number).......... (1a)

Probability of an event E, denoted by P(E) = n/N …………………………………..……......………(2)

where n = n(E) = Number of outcomes/cases/possibilities favourable to the event E and

N = n(S) = Total number all possible outcomes/cases/possibilities.

Now to work out the solution,

If the eight genes were thrown randomly onto the same chromosome, the number of possible ways the genes would position themselves is the same as the number of ways of arranging themselves, which vide (1),

= 8!

= 8 x 7 x 6 x ……. x 2 x 1

= 40320

So, vide (2), N = 40320.

Now, of these 40320 ways the eight genes would position themselves, there is only one way they would line up in the same order as the segments in which they are expressed. So, vide (2), n = 1. Thus, the required probability, vide (2)

= 1/40320

= 0.00002480   Answer

DONE