Gene s controls the sharpness of spines in a type of cactus. cactuses with the dominant allele, s, have sharp spines, whereas homozygous recessive ss cactuses have dull spines. at the same time, a second gene, n, determines whether or not cactuses have spines. homozygous recessive nn cactuses have no spines at all.
if doubly heterozygous ssnn cactuses were allowed to self-pollinate, the f2 would segregate in which of the following ratios?
a) 3 sharp-spined: 1 spineless
b) 1 sharp-spined: 2 dull-spined: 1 spineless
c) 1 sharp-spined: 1 dull-spined: 1 spineless
d) 1 sharp-spined: 1 dull-spined
e) 9 sharp-spined: 3 dull-spined: 4 spineless

The answer is E) 9 sharp-spined:3 dull-spined:4 spineless.

It is known that the in dihybrid cross (a cross between two different traits), the phenotypic ratio between the offspring in the F2 generation is 9:3:3:1:
9 - individuals with both dominant traits (SSNN),
3 - individuals with the first dominant trait and the second recessive trait (SSnn),
3 - individuals with the first recessive trait and the first recessive trait (ssNN),
1 - individual with both recessive traits (ssnn).

According to this, there will be 9 sharp-spined cactuses (SSNN), 3 dull-spined cactuses (ssNN), and 4 spineless cactuses (3 cactuses with SSnn genotype and 1 cactus with ssnn genotype).

Need the answers if its a multiple choice question