Statistical significance test with 1x2 data and faked 2x2 contingency table Medical scientists love p-values and in a publication, I found the following patient statistics table. I stumbled over the Number of patients line. 

They divided their subject group into two subgroups TAV and BAV and I asked how they calculated the p-value, because you only have one characteristic BAV/TAV. If you look for instance at the Male row, you see that you can use a 2x2 contingency table consisting of 2 characteristics BAV/TAV and Male/Female and make e.g. a Fisher test.
What was done in the Number of patients row was to fake a 2x2 contingency table repeating and reversing the numbers:
$$\begin{array}{c|c}
347&32\\\hline
32&347
\end{array}
$$
From the viewpoint that they want to express how similar the number of patients are in the BAV and TAV group, this kind of makes sense because it gives a p-value of 1 (two tailed Fisher) when BAV and TAV have the same number of patients. However, I have never seen this and I cannot find any reference to such an approach.
Question: Can someone tell me whether it is correct what they did?
 A: The only "right" way I can think of to analyze that type of patient row is to test whether the proportions are equal.  This can be done with a one way chi-square test, which is, indeed, a 1x2 table. 
I've never seen the sort of procedure you mention - did they say in the article that that is what they did, or are you guessing/figuring out that that is what they did?
The next question is whether the two methods give the same results.  The table you  show doesn't give much, so let's test:
type <- c(TAV = 347, BAV = 32)
chisq.test(type)

and 
x <- matrix(c(347,32,32,347),ncol = 2)
chisq.test(x)

give identical p values. The first is what I suggested, the second is what you said. 
A: The p-value calculations are a little confusing.  The p-values are trying to test whether the percentages are similar for the two subgroups.  For the first row "Number of patients" the % in TAV has to be 1-% in BAV.  Therefore, one should test whether p=0.5 for one proportion:
prop.test(221, 244, p=0.5, alternative="two.sided")

For the "Male Gender" row, the % of males in TAV does not have to be 1-% of males in BAV, so we have a different test.  You can actually fill out the table:
       Male   Female  Total
TAV    221     126     347
BAV    23      9       32
Total  244     135

This can be tested using Fisher's exact test (although Barnard's exact test is superior) giving the p-value reported.
> data<-matrix(c(221,126,23,9),ncol=2,byrow=TRUE)
> fisher.test(data,alternative="two.sided")

    Fisher's Exact Test for Count Data

data:  data
p-value = 0.4418
alternative hypothesis: true odds ratio is not equal to 1
95 percent confidence interval:
 0.2710563 1.5995457
sample estimates:
odds ratio 
  0.686984 

