There are a lot of ways to build a model where DC fails. However, all of them that I'm aware of involve adding at least a messy *set* of reals (or rather, taking a forcing extension and then passing to an intermediate structure where some prescribed set becomes messy). It occurred to me recently that I don't know how to kill DC by only changing the reals; and after playing around with it for a couple days, I still don't see how to do it.

I suspect I'm missing some very obvious ideas, but I haven't been able to find it on my own. And it nags at me a bit that I don't even know these basic facts about breaking DC, hence my asking here.

Specifically, there are a few ways to phrase the question. The simplest one is:

Q1. Does ZF+$V=L(\mathbb{R})$ imply DC?

*EDIT: I didn't quite ask the question I intended here. Although I'm very pleased to have a (negative) answer to the above question, the one I had in mind when I wrote this was:*

Q1'. If $M\models$ ZFC, does $L(\mathbb{R})^M\models$ DC?

*(I am very very tired right now, and I managed to convince myself that every model of ZF+$V=L(\mathbb{R})$ is the $L(\mathbb{R})$ of some ZFC-model, which is of course completely bonkers.) The negative answer Jing Zhang gives is the $L(\mathbb{R})$ of a very non-choicey model.*

At the more ambitious end, we have:

Q2. Suppose $M\models$ ZFC and $G$ is a set of reals which is generic over $M$. Is DC necessarily true in $HOD^{M[G]}(M\cup G)$?

*I do mean "$M\cup G$" instead of "$M\cup G\cup \{G\}$" - I want to include only the individual reals in $G$. Also, note that $HOD(X)$ makes sense even when $X$ is a transitive class.*

Even Q1 seems implausible; Q2 seems downright ridiculous. But I can't cook up a counterexample at the moment.

In general, I'm interested in learning when a substructure of a forcing extension can be easily seen to have DC:

Q3. What are some properties of symmetric submodel constructions which "quickly" imply DC?

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