Is the statement that only one of $\mathbb{Q}$ and $\mathbb{R}$ is an injective abelian group consistent with $ZF\neg C$?

Question

Is the statement that only one of $\mathbb{Q}$ and $\mathbb{R}$ is an injective abelian group consistent with $ZF\neg C$ (assuming, of course, that $ZF$ is itself consistent)?

Note that if $\mathbb{Q}$ is injective, then it is a direct summand of $\mathbb{R}$. Conversely, if $\mathbb{R}$ is injective and $\mathbb{Q}$ is a direct summand of $\mathbb{R}$, then $\mathbb{Q}$ is also injective.

Also, if there is a $\mathbb{Q}$-basis of $\mathbb{R}$, then $\mathbb{Q}$ is a direct summand of $\mathbb{R}$. However, a direct sum of infinitely many injective abelian groups cannot be proven to be injective without AC. So, perhaps, all three of the following statements might be true:

1. $\mathbb{Q}$ is injective (hence, a direct summand of $\mathbb{R}$)
2. There is a $\mathbb{Q}$-basis of $\mathbb{R}$ (hence, again, $\mathbb{Q}$ is a direct summand of $\mathbb{R}$)
3. $\mathbb{R}$ is not injective

Without the axiom of choice, divisible abelian groups need not be injective. So, the statement "makes sense".