Bordism groups and a short exact sequence

Let us consider a short exact sequence: $$1\to N\to G\to Q \to 1,$$ where $$N$$, $$Q$$, and $$G$$ can be continuous Lie groups in general (or finite groups).

• Suppose I have the data and the computations of the bordism group $$\Omega_{d}^{(Spin \times Q)/\mathbb{Z}_2},$$ where $${(Spin \times Q)/\mathbb{Z}_2}$$ means the modification of the $$Spin$$-structure to a new $${(Spin \times Q)/\mathbb{Z}_2}$$-structure. Here the $$Spin$$ and $$Q$$ shares a normal subgroup $$\mathbb{Z}_2$$ that was mod out, such that $$Spin/\mathbb{Z}_2= SO,$$ or more explicitly $$Spin(d)/\mathbb{Z}_2= SO(d),$$ where we omit the dimension $$d$$ through this post in the $$Spin \equiv Spin(d)$$.

• I also have the data and the computations of the bordism group $$\Omega_{d}^{(Spin \times N)/\mathbb{Z}_2},$$ where $${(Spin \times N)/\mathbb{Z}_2}$$ means the modification of the $$Spin$$-structure to a new $${(Spin \times N)/\mathbb{Z}_2}$$-structure. Here the $$Spin$$ and $$N$$ shares a normal subgroup $$\mathbb{Z}_2$$ that was mod out.

Question How can we compute $$\Omega_{d}^{(Spin \times G)/\mathbb{Z}_2},$$ where the precise extension from $$Q$$ to $$G$$ is given (thus the $$G$$ is determined and chosen), based on the previously known information of $$\Omega_{d}^{(Spin \times Q)/\mathbb{Z}_2}$$ and $$\Omega_{d}^{(Spin \times N)/\mathbb{Z}_2}$$?

Is there some simple way to decompose $$\Omega_{d}^{(Spin \times G)/\mathbb{Z}_2}$$ into $$\Omega_{d}^{(Spin \times Q)/\mathbb{Z}_2}$$, $$\Omega_{d}^{(Spin \times N)/\mathbb{Z}_2}$$ and other things?