This answer is flawed, removing it.

This does not give a complete answer. This provides a strategy for conditional approach.

Given that the nature of the problem is asking for $q$ and $2(p+1)-q$ being simultaneously primes, I think that the question can be approached conditionally (similar to zz7948).

A slight modification is shifting the focus to finding $p<q$ such that $p$ and $2p+2-q$ are simultaneously prime, for given prime $q$.

Let us assume the following version of prime $k$-tuple conjecture.

Conjecture 1

There is an absolute constant $C>0$ and a prime $q_0$ such that the number $N(q)$ of primes $p<q$ such that $p$ and $2p+2-q$ are simultaneously prime satisfies $$ N(q)\geq C\frac q{\log^2 q}\geq 1, \ \ \mathrm{ if } \ q\geq q_0. $$

If Conjecture 1 is true, then any prime $q>q_0$ is connected to some smaller prime.

If we can show (computation) that all primes $q\leq q_0$ are connected, then all primes will be connected.

This does not give a complete answer. This provides a strategy for conditional approach.

Given that the nature of the problem is asking for $q$ and $2(p+1)-q$ being simultaneously primes, I think that the question can be approached conditionally (similar to zz7948).

A slight modification is shifting the focus to finding $p<q$ such that $p$ and $2p+2-q$ are simultaneously prime, for given prime $q$. As suggested in zz7948's comment, we extend the edges to include all pairs $(p,q)$ with both $p$ and $2p+2-q$ are primes.

Let us assume the following version of prime $k$-tuple conjecture.

Conjecture 1

There is an absolute constant $C>0$ and a prime $q_0$ such that the number $N(q)$ of primes $p<q$ such that $p$ and $2p+2-q$ are simultaneously prime satisfies $$ N(q)\geq C\frac q{\log^2 q}\geq 1, \ \ \mathrm{ if } \ q\geq q_0. $$

If Conjecture 1 is true, then any prime $q>q_0$ is connected to some smaller prime.

If we can show (computation) that all primes $q\leq q_0$ are connected, then all primes will be connected.

Once we obtain that all primes are connected, we now start to remove the edges to include only the smallest $q$ with $q>p$ and $2p+2-q$ are both primes.

Then, the issue is, whether or not Prim's algorithm of finding spanning tree of a connected graph gives the desired graph.

This does not give a complete answer. This provides a strategy for conditional approach.

Given that the nature of the problem is asking for $q$ and $2(p+1)-q$ being simultaneously primes, I think that the question can be approached conditionally (similar to zz7948).

A slight modification is shifting the focus to finding $p<q$ such that $p$ and $2p+2-q$ are simultaneously prime, for given prime $q$.

Let us assume the following version of prime $k$-tuple conjecture.

Conjecture 1

There is an absolute constant $C>0$ and a prime $q_0$ such that the number $N(q)$ of primes $p<q$ such that $p$ and $2p+2-q$ are simultaneously prime satisfies $$ N(q)\geq C\frac q{\log^2 q}\geq 1, \ \ \mathrm{ if } \ q\geq q_0. $$

If Conjecture 1 is true, then any prime $q>q_0$ is connected to some smaller prime.

If we can show (computation) that all primes $q\leq q_0$ are connected, then all primes will be connected.

This answer is flawed, removing it.

This does not give a complete answer. This provides a strategy for conditional approach.

Given that the nature of the problem is asking for $q$ and $2(p+1)-q$ being simultaneously primes, I think that the question can be approached conditionally (similar to zz7948).

A slight modification is shifting the focus to finding $p<q$ such that $p$ and $2p+2-q$ are simultaneously prime, for given prime $q$.

Let us assume the following version of prime $k$-tuple conjecture.

Conjecture 1

There is an absolute constant $C>0$ and a prime $q_0$ such that the number $N(q)$ of primes $p<q$ such that $p$ and $2p+2-q$ are simultaneously prime satisfies $$ N(q)\geq C\frac q{\log^2 q}\geq 1, \ \ \mathrm{ if } \ q\geq q_0. $$

If Conjecture 1 is true, then any prime $q>q_0$ is connected to some smaller prime.

If we can show (computation) that all primes $q\leq q_0$ are connected, then all primes will be connected.