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 F₂would segregate in which of the following ratios?

Question

Quizplus · Accepted Answer

The SsNn cactuses have two different genes segregating independently. For the S gene, the cross between Ss x Ss will result in a classic 1:2:1 ratio. Therefore, 1/4 of the offspring will be SS (sharp spined), 1/2 will be Ss (sharp spined), and 1/4 will be ss (dull spined). 
For the N gene, the cross between Nn x Nn will result in a classic 1:2:1 ratio. Therefore, 1/4 of the offspring will be NN (spineless), 1/2 will be Nn (spineless), and 1/4 will be nn (have spines).
To determine the combined ratio, we multiply the individual ratios: 
(1/4 SS) x (1/4 nn) = 1/16 SSnn 
(1/4 SS) x (1/2 Nn) = 1/8 SSNn 
(1/4 Ss) x (1/4 nn) = 1/16 Ssnn 
(1/4 Ss) x (1/2 Nn) = 1/8 SsNn 
(1/4 ss) x (1/4 nn) = 1/16 ssnn 
(1/4 ss) x (1/2 Nn) = 1/8 ssNn 
So the final ratio is: 
1 SSnn : 2 SSNn : 1 Ssnn : 2 SsNn : 1 ssnn : 2 ssNn

We can simplify this ratio by combining like terms: 
3 sharp-spined: 3 dull-spined: 4 spineless. 
Therefore, the correct answer is C.

Gene S Controls the Sharpness of Spines in a Type