A pea plant heterozygous for seed shape (round R over wrinkled r) and seed colour (yellow Y over…

(a) $320 \times 9/16 = 180$. The other expected numbers are $60$, $60$ and $20$.

(b) $\chi^2 = (191 - 180)^2 / 180 + (55 - 60)^2 / 60 + (52 - 60)^2 / 60 + (22 - 20)^2 / 20 = 2.36$. This is below $7.81$, so the data fit $9 : 3 : 3 : 1$.

(c) Four classes, one fixed total: $4 - 1 = 3$ degrees of freedom.

(d) Round is $246$ and wrinkled is $74$. Expected are $240$ and $80$. So $\chi^2 = (246 - 240)^2 / 240 + (74 - 80)^2 / 80 = 0.60$ with $1$ degree of freedom. It is below $3.84$.

(e) For the two-by-two table of shape against colour, the expected cell numbers come from the row and column totals. The result is $\chi^2 = 1.69$ with $1$ degree of freedom, below $3.84$, so there is no sign that the two genes are linked.