Let $k$ be an algebraically closed field and fix $\lambda\in k,\lambda\neq 0,1$. Consider the curve in $\mathbb P^3$ given by $yz-wx=yw-(x-z)(x-\lambda z)=0$. Choose the coordinates on the quadric $yz-wx=0$ so that $(x:y:z:w)=(s_0t_0:s_0t_1:s_1t_0:s_1t_1)$. The equation of the curve becomes $s_0s_1t_1^2-(s_0t_0-s_1t_0)(s_0t_0-\lambda s_1t_0)=0$. I've checked that this curve is nonsingular, therefore it should be an elliptic curve (also, it is stated in Hartshorne's Deformation Theory that the curve is nonsingular). But if we choose the affine chart $s_1=t_1=1$ the equation of the curve becomes $s_0-(s_0t_0-t_0)(s_0t_0-\lambda t_0)=0$ and we know that the genus of a smooth plane curve of degree $4$ is equal to $3$. What is wrong?
If you look at the closure of your affine patch in the projective plane, you'll see that there is a singularity at infinity where $t_0=0$. In other words, your affine part looks like the smooth part of a singular quartic whose smooth model is elliptic. It's just that this point at infinity does not appear when you glue your affine patches together.