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 418 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 418 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]
