It is important to realise in this question that the cross is not like a normal Mendelian cross, but that all previously accumulated genetic infortmation is inherited.

It would be impossible to create an offspring who was two-thirds G and one-third M. This can be explained as follows: each successive generation doubles the number of possible 'parts' that it can be made out of. For example the first generation is purely G or M, whereas the second generation has two 'parts' - it can be GG, MM or GM. Additionally, the third generation has four parts and can be GGGG, GGGM, GGMM, GMMM or MMMM. Thus, the number of 'parts' is clearly $2^n$ where n is an integer.

In order to be able to be composed on one-third M, we are essentially asking if there is a value of n such that $\frac{2^n}{3}$ is an integer. It can be seen that there is no value of n to make this true because $2^n$ generates numbers which are divisible only by the prime number 2, but by no other primes. Because 3 is a prime number, this means that the expression can never yield an integer.

We are looking to find a composition which is within 1% of $\frac{1}{3}$. which is equivalent to the range $\frac{99}{300}$ - $\frac{101}{300}$ which is $0.33 - 0.33\dot{6}$.

Using trial and error:
$\frac{3}{8} = 0.375$
$\frac{5}{16} = 0.313$
$\frac{11}{32} = 0.348$
$\frac{21}{64} = 0.328$
$\mathbf{\frac{43}{128} = 0.3359}$

Thus, 128 = 2$^7$, and so there need to be a minimum of 8 generations. Can you draw out the crossing scheme to create this final progeny?