Try to forget everything you know about Brownian motion.
You are told there is a process called Brownian bridge on $[0,1]$. It is Gaussian with mean $0$ and covariance $s(1-t)$ for $s<t$ and can be written as $B_t=W_t - tW_1$ where $(W_t)$ is a continuous process with $W_0=0$, called Brownian motion. Of course, from what you were told about the Brownian bridge, you know that $(W_t)$ is Gaussian with mean $0$ (this is not true, see remark of John Dawkins and EDIT below).
You are told $(B_t)$ is Markov with regard to its own filtration. How would you show that $(W_t)$ is Markov with regard to its own filtration ?
EDIT: In light of the remark of John Dawkins, we should indeed say that $(W_t)$ is itself Gaussian with mean $0$, otherwise, other processes than the real Brownian motion could be solution of $B_t = W_t - t W_1$.