In an earlier paper [\textit{W. G. Leavitt}, Stud. Sci. Math. Hung. 16, 15- 23 (1981; Zbl 0481.16002)] the author showed: if a class \({\mathbb{M}}\) of rings has property (E) the following are equivalent: (1) U\({\mathbb{M}}\) is hereditary. (2) Every \(0\neq R\in {\mathbb{M}}_ k\) has some image \(0\neq R/I\in SU{\mathbb{M}}.\) Here \(U{\mathbb{M}}=\{R| \quad every\quad 0\neq R/I\not\in {\mathbb{M}}\}\quad S{\mathbb{M}}=\{R| \quad I\not\in {\mathbb{M}}\quad for\quad every\quad ideal\quad I\quad in\quad R\}\) and \({\mathbb{M}}_ k=\{R| \quad I\) is an essential ideal of R for some \(I\in {\mathbb{M}}\}.\) In this paper it is shown that property (E) is rather independent of a regular class. An example is constructed of a regular class \({\mathbb{M}}\) with property (2) but for which U\({\mathbb{M}}\) is not hereditary. Replacing \({\mathbb{M}}_ k\) by \({\mathbb{M}}'\!_ k\), where \({\mathbb{M}}'\) is larger than \({\mathbb{M}}\), gives the result: Let \({\mathbb{M}}\) be a regular class. The following are equivalent: (1) U\({\mathbb{M}}\) is hereditary. (2) Every \(0\neq R\in {\mathbb{M}}'\!_ k\) has an image \(0\neq R/I\in SU{\mathbb{M}}.\) By an ingenious construction the author is able to construct a regular class \({\mathbb{M}}\) which does not have (E), but for which U\({\mathbb{M}}\) is hereditary. This example gives a negative answer to a question posed by the author in his cited paper.