Consider the $1$-types over $\mathbb{Q}$ as a dense linear order without endpoints. They are similar to cuts: for any $1$-type $p$ over $\mathbb{Q}$, we can define $L_p = \{a \in \mathbb{Q} \; | \; a < x \in p\}$ and $R_p = \{b \in \mathbb{Q} \; | \; x < b \in p\}$. It seems then possible to classify the $1$-types as follows:
(i) for some $a \in \mathbb{Q}$, $v = a \in p$. Then $p$ is isolated by this formula and is thus realized in $\mathbb{Q}$.
(ii) $p$ is omitted in $\mathbb{Q}$. Then there are at least three kinds of types: $L_p = \varnothing$ and $R_p = \mathbb{Q}$, $L_p = \mathbb{Q}$ and $R_p = \varnothing$, and, for $r \in \mathbb{R} \setminus \mathbb{Q}$, the type which corresponds to the cut corresponding to $r$, i.e. $L_p = \{a \in \mathbb{Q} \; | \; a<r\}$ and $R_p = \{a \in \mathbb{Q} \; | \; r<a\}$.
Now, I had thought that this exhausted the classification of $1$-types, but Marker (p. 122 of his book) claims that there are two other kinds of types as well: if $c \in \mathbb{Q}$, we also have the case $L_p = \{a \in \mathbb{Q} ; | \; a \leq c\}$ and $R_p = \{b \in \mathbb{Q} \; | \; c < b\}$ and the case $L_p = \{a \in \mathbb{Q} \; | \; a < c\}$ and $R_p = \{b \in \mathbb{Q} \; | \; c \leq b\}$. But I don't understand why these two other types are omitted in $\mathbb{Q}$. In fact, they seem to be isolated by the formula $v = c$. What am I missing?