I strongly encourage you to work through the case where $x$ can take two values (e.g. 0 = dominant, 1 = recessive as in your example),  as it a useful exercise for understanding the logistic model and also the MLE approach.

The model in question is supposed to be 
$$
Y_i \sim \mathrm{Bernoulli}(\mathrm{logit}(p_i)),
$$
where $p_i = a + bx_i$, $i = 1,\ldots,n$.  Suppose the observed values are $y_1, \ldots, y_n$.

Homework:

 1. What is the probability of observing this particular string of $y_i$
    (assuming they are all independent).  Hint: split according to
    whether $y_i = 0$ or $1$ and similarly for $x_i$.
 2.  What is the log of this?
 3.  What happens to the log likelihood if $a$ gets large or small (negative)?  Same question for $b$.
 4.  Conclude that you can maximise the log likelihood by differentiating in $a$ and $b$ then finding where the derivative is
    zero (separately for $a$ and $b$).

OK, for the final answer:
Write $\#_{l,m} = \#\{i : y_i = l, x_i = m\}$, so $\#_{0,1}$ is the number of observations with $y_i = 0$ and $x_i = 1$ etc.

You'll see the problem can be parameterised more easily in terms of 
$$
p = \frac{e^{a}}{1 + e^{a}}
$$
and
$$
r = \frac{e^{a+b}}{1 + e^{a+b}}.
$$

Assuming I didn't make a mistake, we find
$$
r = \frac{\#_{1,1}}{\#_{0,1} + \#_{1,1}},
$$
i.e. the fraction of cases with $x_i = 1$ that have $y_i = 1$.

In terms of $r$, we get
$$
p(\#_{0,0} + \#_{1,0} ) = \#_{1,0} + \#_{1,1} +  r(\#_{0,1} - \#_{0,0})
$$

So, $a + b = \ln(r/(1-r))$ is the log odds have $y_i = 1$ when $x_i = 1$.

Finally, $a = \ln(p/(1-p))$ and $b = \ln(r(1-p)/p(1-r))$.