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一、选择题第25小题人类多指基因(T)是正常(t)的显性,白化基因(a)是正常(A)的隐性,都在常染色体上,而且都是独立遗传。一个家庭中,父亲是多指,母亲正常,他们有一个白化病和手指正常的孩子,则下一个孩子只有一种病和两种病的概率分别是: A、3/4,1/4 B、3/4,1/8 C、1/4,1/4 D、1/4,1/8 答案是B。本题的正确答案应是1/2,1/8。因此该题无正确答案(A、B、C、D四答案均错)。具体计算过程如下:根据题意,父亲的基因型必为TtAa,母亲的基因型必为ttAa,他们所生的孩子患多指的几率为1/2,患白化病的几率为1/4,同时患两
First, multiple choice questions 25th topic Human multi-fingered genes (T) are normal (t) dominant, albino genes (a) are normal (A) recessive, are on autosomes, and are all inherited independently. In a family where the father is multi-fingered and the mother is normal, they have a child with albinism and normal fingers. The probability that the next child has only one disease and two diseases is: A, 3/4, 1/4 B, 3/4, 1/8 C, 1/4, 1/4 D, 1/4, 1/8 The answer is B. The correct answer to this question should be 1/2, 1/8. Therefore, there is no correct answer to this question (A, B, C, D are all wrong answers). The specific calculation process is as follows: According to the question, the father’s genotype must be TtAa, and the mother’s genotype must be ttAa. Their children have a 1/2 chance of suffering from multiple fingers, and the chance of suffering from albinism is 1/4. Suffering from two