Drawing a bode plot of transfer function with unknown gain
Ok, let's do some math to explain more explicitly what has been said in some comments to your question.

Let's rewrite G as another TF multiplied by K:

$$
G(p) = K \cdot G_n(p)
$$

where

$$
G_n(p) = \dfrac{1} {p \cdot (1+0.1p) \cdot (1+0.05p)}
$$

is the normalized (with respect to K) TF. 

Let's define the logarithmic (dB) amplitude response of the system this way:

$$
A_{(dB)}(\omega) = 20 \log_{10} \left| G(j\omega) \right|
$$

We see easily that:

$$
A_{(dB)}(\omega) =
\\[1em]
= 20 \log_{10} \left| K \cdot G_n(j\omega) \right| =
\\[1em]
= 20 \log_{10} \left| K \right| 
+ 20 \log_{10} \left| G_n(j\omega) \right| =
\\[1em]
= K_{(dB)} + A_{n(dB)}(\omega)
$$

Where \$A_{n(dB)}\$ is the amplitude response relative to the normalized TF and
\$K_{(dB)}\$ is the constant K expressed in dB:

\begin{align*}
A_{n(dB)}(\omega) &amp;= 20 \log_{10} \left| G_n(j\omega) \right|
\\[1em]
K_{(dB)} &amp;=  20 \log_{10} \left| K \right| 
\end{align*}

From that you can see that the only difference in the amplitude Bode plot between the original and the normalized TF is just a vertical shift, so the corner frequencies of both plots will remain the same. 

Here is an LTspice simulation that shows practically the situation:





Of course I had to choose a value for K (100 = 40dB), but you can easily see that any change to K will just change the amount of the vertical shift.