Can I tell the interface to default the properties I don't supply to null? What would let me do this

No. You cannot provide default values for interfaces or type aliases as they are compile time only and default values need runtime support

Alternative

But values that are not specified default to undefined in JavaScript runtimes. So you can mark them as optional:

interface IX {
  a: string,
  b?: any,
  c?: AnotherType
}

And now when you create it you only need to provide a:

let x: IX = {
    a: 'abc'
};

You can provide the values as needed:

x.a = 'xyz'
x.b = 123
x.c = new AnotherType()

You can't set a default value to a property of an interface.

Use abstract class in addition to the interface (which only sets the default value and doesn't implement anything else):

    public interface IA {
        int Prop { get; }

        void F();
    }

    public abstract class ABase : IA {
        public virtual int Prop
        {
            get { return 0; }
        }

        public abstract void F();
    }

    public class A : ABase
    {
        public override void F() { }
    }

Virtuals may have defaults. The defaults in the base class are not inherited by derived classes.

Which default is used -- ie, the base class' or a derived class' -- is determined by the static type used to make the call to the function. If you call through a base class object, pointer or reference, the default denoted in the base class is used. Conversely, if you call through a derived class object, pointer or reference the defaults denoted in the derived class are used. There is an example below the Standard quotation that demonstrates this.

Some compilers may do something different, but this is what the C++03 and C++11 Standards say:

8.3.6.10:

A virtual function call (10.3) uses the default arguments in the declaration of the virtual function determined by the static type of the pointer or reference denoting the object. An overriding function in a derived class does not acquire default arguments from the function it overrides. Example:

struct A {
  virtual void f(int a = 7);
};
struct B : public A {
  void f(int a);
};
void m()
{
  B* pb = new B;
  A* pa = pb;
  pa->f(); //OK, calls pa->B::f(7)
  pb->f(); //error: wrong number of arguments for B::f()
}

Here is a sample program to demonstrate what defaults are picked up. I'm using structs here rather than classes simply for brevity -- class and struct are exactly the same in almost every way except default visibility.

#include <string>
#include <sstream>
#include <iostream>
#include <iomanip>

using std::stringstream;
using std::string;
using std::cout;
using std::endl;

struct Base { virtual string Speak(int n = 42); };
struct Der : public Base { string Speak(int n = 84); };

string Base::Speak(int n) 
{ 
    stringstream ss;
    ss << "Base " << n;
    return ss.str();
}

string Der::Speak(int n)
{
    stringstream ss;
    ss << "Der " << n;
    return ss.str();
}

int main()
{
    Base b1;
    Der d1;

    Base *pb1 = &b1, *pb2 = &d1;
    Der *pd1 = &d1;
    cout << pb1->Speak() << "\n"    // Base 42
        << pb2->Speak() << "\n"     // Der 42
        << pd1->Speak() << "\n"     // Der 84
        << endl;
}

The output of this program (on MSVC10 and GCC 4.4) is:

Base 42
Der 42
Der 84

Interfaces are reference types.

A field of an interface type will always be one pointer wide, and will default to null.

If you assign a struct that implements the interface, the struct will be boxed, and the field will contain a reference to the boxing object.

Wow, everybody is such a pessimist around here. The answer is yes.

It ain't trivial: by the end, we'll have the core function, a supporting struct, a wrapper function, and a macro around the wrapper function. In my work I have a set of macros to automate all this; once you understand the flow it'll be easy for you to do the same.

I've written this up elsewhere, so here's a detailed external link to supplement the summary here: http://modelingwithdata.org/arch/00000022.htm

We'd like to turn

double f(int i, double x)

into a function that takes defaults (i=8, x=3.14). Define a companion struct:

typedef struct {
    int i;
    double x;
} f_args;

Rename your function f_base, and define a wrapper function that sets defaults and calls the base:

double var_f(f_args in){
    int i_out = in.i ? in.i : 8;
    double x_out = in.x ? in.x : 3.14;
    return f_base(i_out, x_out);
}

Now add a macro, using C's variadic macros. This way users don't have to know they're actually populating a f_args struct and think they're doing the usual:

#define f(...) var_f((f_args){__VA_ARGS__});

OK, now all of the following would work:

f(3, 8);      //i=3, x=8
f(.i=1, 2.3); //i=1, x=2.3
f(2);         //i=2, x=3.14
f(.x=9.2);    //i=8, x=9.2

Check the rules on how compound initializers set defaults for the exact rules.

One thing that won't work: f(0), because we can't distinguish between a missing value and zero. In my experience, this is something to watch out for, but can be taken care of as the need arises---half the time your default really is zero.

I went through the trouble of writing this up because I think named arguments and defaults really do make coding in C easier and even more fun. And C is awesome for being so simple and still having enough there to make all this possible.