Re: McNemar Test for genetic data


John Uebersax wrote:
Bill H. wrote:

and in the case of the TDT, the pairing is the parents

Could you please help me understand how this works:

1. One looks only at parents, and only those of diseased offspring?
2. a and b denote, say, a normal and abnormal form of some gene?
3. Mother may be: aa, ab or bb
Father may be: aa, ab or bb
4. You have a 2 x 2 frequency table, but what are the rows/columns?

Father
? ?
+-------+--------+
M | | |
o ? | f1 | f2 |
t | | |
h +-------+--------+
e | | |
r ? | f3 | f4 |
| | |
+-------+--------+

I'm sorry I can't ask the question better, but I just don't understand
this.

I could understand it if it were a test of whether one is equally
likely to inherit the bad gene from either parent--but "transmission
disequilibrium" sounds more like an inter-generational thing.

p.s. I did search Google but couldn't find a good explanation of the
TDT.

--
John Uebersax PhD

Sorry, I was wrong when I said the pair was the parents plural, haven't
done this for awhile! It looks like the pair is actually ONE parent
and the rows and columns are "transmitted allele" and "non-transmitted
allele" so in fact the TDT does test for association of disease with
tranmission of the high risk allele from a parent. But I was correct
that only heterozygous parents contribute information to the test. For
example, see the example in the pdf linked below. Each of the five
affected offspring received the "d" allele from the heterozygous parent
but none received the "D" allele. So the off diagonal counts for d
transmitted/D not transmitted = 5 and the off diagonal cell for D
transmitted/ d not transmitted = 0, which results in a McNemars
chi-square of (5-0)^2/ (5+0) = 5 on 1 df.

http://www.genetics.ucla.edu/courses/iges2003/10b-TDT_Horvath.pdf

.

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