The answer is $11$. For a construction, tile the array with $10 \times 11$ rectangles (all with the same orientation), and put all members of one species in each rectangle. Then all rows and columns have at most $11$ species. For the bound, define a species' reach to be the total number of rows and columns that contain at least one member of the species. Claim: For each species, the reach is at least $21$. Proof: Suppose otherwise, and let's say there were $r$ rows and $c$ columns containing it. Then by AM-GM, $$110 \leq rc \leq \left(\dfrac{r+c}{2}\right)^2 \leq 100$$which is clearly false. Then the sum of all reaches is at least $21 \cdot 110$, and because there are $2 \cdot 110$ columns and rows, the expected number of species in a randomly selected row/column is at least $$\dfrac{21 \cdot 110}{2 \cdot 110} = 10.5$$which implies at least one line with 11 species exists. Thus we're done. Yay!

I claim the answer is 11. To show the answer is at most 11, it suffices to construct an arrangement where every row or column has at most 11 distinct species. It is indeed possible: Tile the grid using 11 by 10 rectangles, and for each of the 110 tiles, put all of the guinea pigs of a certain species. Now, we must show that it is always possible to find a column or row with at least 11 distinct species. Assume for the sake of contradiction that there is some grid, for which every row and column has at most 10 distinct species. Let $DR_i$ be the number of distinct species on row $i$ and, $DC_j$ be the number of distinct species on column $j$. Now, consider the sum $S_1 = \sum DR_i + \sum DC_i$, which we double-count. First, we note that because $DR_i, DC_j \leq 10$, $S_1$ is at most $2 \cdot 110 \cdot 10 = 2200$. Now, for a species $s$, let $DS_s$ be the number of rows or columns which have a guinea pig of species $s$, and let $S_2 = \sum DS_s$. Observe that because each appearance in a row or column contributes 1 to $DS_s$, and at the same time contributes 1 to some $DR_i$ or $DC_j$ term, in fact, $$\sum DR_i + \sum DC_j = S_1 = S_2 = \sum DS_s$$I now claim that $DS_s \geq 21$ for all $s$. ProofLet $r$ be the number of rows with some guinea pig of species $s$, and similarly, let $c$ be the number of columns. Since a guinea pig can only be at one of these rows and columns, there are at most $rc$ guinea pigs of species $s$. This means $rc \geq 110$. Therefore, by AM-GM, $$\frac{DS_s}{2} = \frac{r + c}{2} \geq \sqrt{rc} \geq \sqrt{110}$$implying $DS_s \geq 2\sqrt{110} \approx 20.97$. But since $DS_s$ must be an integer, we have $DS_s \geq 21$, as desired. Since there are 110 species, then $S_2 \leq 21 \cdot 110 = 2310$. Therefore, $$2310 \leq S_2 = S_1 \leq 2200$$a contradiction, which finishes the proof. return 0;