(2) In the same way, according to the first situation, the genotype of F 1 purple flower in group B may be AaBB, and the cross combination of parents is purple flower (AABB)× white flower (AABB). When the F 1 purple flower of group B and group C crosses, that is, AaBB×AaBb, then the phenotype and proportion of offspring are purple flower (3A_B_): white flower (66.
(3) F 1ABB in Group C produced F2 genotypes and proportions of 1ABB, 2aaabb, 2aaabb, 1ABB, 2aaabb, 1ABB, among which F2 selfed can produce safflower progeny, including 2aaabb, 4AaBb and/kl.
(4)① Select homozygous safflower to cross with group C white flowers, and then observe the phenotype of the offspring; ② If the offspring has both purple flowers and red flowers, it is a white-flowered hybrid plant.
So the answer is:
(1) Two (1) purple flowers × red flowers ((2), and the right one is 1).
(2) Purple flower: white flower = 3: 1 ((2 points), performance and proportion 1 minute)
(3) 1800(2 points)
(4)① Select homozygous safflower to hybridize with group C white flowers (2 points)? Observe the phenotype of offspring (1)
② There are both purple flowers and red flowers (2 points)