Understanding Quake I Timer

Question

Here is a snippet of code from Quake I.

    LARGE_INTEGER   PerformanceFreq;
    unsigned int    lowpart, highpart;
    OSVERSIONINFO   vinfo;
if (!QueryPerformanceFrequency (&PerformanceFreq))
    Sys_Error ("No hardware timer available");

// get 32 out of the 64 time bits such that we have around 1 microsecond resolution
lowpart = (unsigned int)PerformanceFreq.LowPart;
highpart = (unsigned int)PerformanceFreq.HighPart;
lowshift = 0;

while (highpart || (lowpart > 2000000.0))
{
    lowshift++;
    lowpart >>= 1;
    lowpart |= (highpart & 1) << 31;
    highpart >>= 1;
}

The comment indicates that they are reducing the 64 bit PerformanceFreq to 32 bits and that the resolution is about 1 microsecond (the default resolution of QueryPerformanceFrequency is 1 second).

I understand what the code is doing generally, but I don't know what (lowpart > 2000000.0) is meant to be doing or why it's there. I think it's related to ensuring the resolution is about 1 microsecond.

When I run this code through the debugger the lowpart is shifted 3 times meaning that the resolution is actually 1s / 2^3 - 0.125s or 125 milliseconds and not 1 microsecond. The documentation for QueryPerformanceFrequency says that the result is hardware dependent so I don't see how a constant value 200000.0 could work unless you were making an assumption for the result based on modern hardware of the time.

Is this code wrong about the resolution or am I missing something?