How did they go from this discrete model to this continuous one? The first equation from Barton 1979 (a paper in population genetics) is
$$\Delta p = \left(\frac{p^2+pq(1-s)}{1-2pqs}\right)-p$$
, where $1>s>0$, $1>p>0$ and $p+q=1$. The first expression on the RHS can be understood as the state of $p$ at the following time step. He directly indicates that the equivalent equation in a continuous time model is 
$$\frac{dp}{dt}≈spq(p-q)$$
Can you help me to make this "conversion" from discrete time to continuous time?
 A: Starting out with the discrete form:
$\frac{\Delta p}{\Delta t}=\frac{p^2+pq(1-s)}{1-2pqs}-p$
bring the solitary $p$ back into the fraction:
$\frac{\Delta p}{\Delta t}=\frac{p^2+pq(1-s)-p(1-2pqs)}{1-2pqs}$
Multiplying this out gives:
$\frac{\Delta p}{\Delta t}=\frac{p^2+pq-pqs-p+2p^2qs}{1-2pqs}$
make the substitution $q = (1-p)$. Then,
$\frac{\Delta p}{\Delta t}=\frac{p^2+p(1-p)-p(1-p)s-p+2p^2(1-p)s}{1-2p(1-p)s}$
Multiplying out gives:
$\frac{\Delta p}{\Delta t}=\frac{p^2+p-p^2-ps+p^2s-p+2p^2s-2p^3s}{1-2ps+2p^2s}=\frac{3p^2s-ps-2p^3s}{1-2ps+2p^2s}$
We will call this last quantity $LHS$
Now to find out how to get this quantity from the final result:
$\frac{dp}{dt}=pqs(p-q)$
Multiply this out:
$\frac{dp}{dt}=p^2qs-pq^2s$
Substitute $q=(1-p)$:
$\frac{dp}{dt}=p^2(1-p)s-p(1-p)^2s=p^2(1-p)s-p(1-2p+p^2)s$,
$\frac{dp}{dt}=p^2s-p^3s-ps+2p^2s-p^3s$
Collecting terms:
$\frac{dp}{dt}=3p^2s-ps-2p^3s=RHS$
Now compare $LHS$ and $RHS$ - they appear identical in 'numerator', but what has happened to the divisor $1-2ps+2p^2s$ ? I can only assume that they have decided to neglect the terms other than $1$ in the denominator, since they have stated $s<<1$ earlier in the paper.
