Question

Белая окраска шерсти у кролика контролируется двумя парами не сцепленных неаллельных генов,один доминантный ген определяет развитие чёрной окраски,а другой доминантный ген является подавителем окраски При скрещивании белой окраски получили только белых кроликов. В первом скрещивании получили второго поколения 912 белых и 215 черныхНайти как наследуется признак и определить генотипы

Answer 1

Ответ:

Решение на фото :)

Answer 2

This is a question about genetics and inheritance of coat color in rabbits. According to the information I found on the web, the answer is as follows:

The white coat color in rabbits is controlled by two pairs of non-linked non-allelic genes, A and B. The dominant gene A determines the development of black color, and the dominant gene B is a suppressor of color. The genotype of the white parent is AABB, and the genotype of the black parent is aabb. The genotype of the F1 generation is AaBb, and they are all white because the dominant B gene suppresses the expression of the A gene. The genotype of the F2 generation is obtained by crossing two F1 individuals, and the expected ratio is 9:3:3:1. However, only two phenotypes are observed: white and black. The white phenotype includes the genotypes AABB, AABb, AaBB, and AaBb, which have a total frequency of 9/16 + 3/16 + 3/16 + 1/16 = 16/16. The black phenotype includes the genotypes aaBB, aaBb, aabb, which have a total frequency of 1/16 + 2/16 + 1/16 = 4/16. Therefore, the expected ratio of white to black in the F2 generation is 16:4, or 4:1. The observed ratio of 912 white to 215 black is close to this expected ratio, and can be explained by the law of segregation and the law of independent assortment.