  A finite Borel measure $\mu$ in ${\mathbb R}^d$ is called a frame-spectral
measure if it admits an exponential frame (or Fourier frame) for $L^2(\mu)$. It
has been conjectured that a frame-spectral measure must be translationally
absolutely continuous, which is a criterion describing the local uniformity of
a measure on its support. In this paper, we show that if any measures $\nu$ and
$\lambda$ without atoms whose supports form a packing pair, then $\nu\ast
\lambda +\delta_t\ast\nu$ is translationally singular and it does not admit any
Fourier frame. In particular, we show that the sum of one-fourth and
one-sixteenth Cantor measure $\mu_4+\mu_{16}$ does not admit any Fourier frame.
We also interpolate the mixed-type frame-spectral measures studied by Lev and
the measure we studied. In doing so, we demonstrate a discontinuity behavior:
For any anticlockwise rotation mapping $R_{\theta}$ with $\theta\ne \pm\pi/2$,
the two-dimensional measure $\rho_{\theta} (\cdot): =
(\mu_4\times\delta_0)(\cdot)+(\delta_0\times\mu_{16})(R_{\theta}^{-1}\cdot)$,
supported on the union of $x$-axis and $y=(\cot \theta)x$, always admit a
Fourier frame. Furthermore, we can find $\{e^{2\pi i
\langle\lambda,x\rangle}\}_{\lambda\in\Lambda_{\theta}}$ such that it forms a
Fourier frame for $\rho_{\theta}$ with frame bounds independent of $\theta$.
Nonetheless, $\rho_{\pm\pi/2}$ does not admit any Fourier frame.
