Biology 334

Proof of the Perkins Formula

We will assume that the only possible meioses are those with:

no crossover (NCO), a single crossover (SCO), or double crossovers (DCO).

From Fig 4-18 we see that the 6 meioses shown result in PD, T, PD, T, T and NPD asci

(in that order from top to bottom).

NPD represent 1/4 of all DCO, therefore DCO = 4 NPD

The mean # crossovers by definition      = SCO + 2 x DCO

From Fig 4-18 we see that                         SCO can be estimated as T – 2NPD  and

                                                                    DCO can be estimated as 4 x NPD

Hence mean # crossovers                       = T – 2NPD  +  2(4 x NPD)

                                                                = T + 6NPD

Since any number of crossovers on average produce a RF of 50%

The corrected map distance =  50(T + 6NPD)   [The Perkins Formula]