Two offspring with normal phenotype, III5 and II6, both gave birth to sick children, indicating that both diseases A and B are recessive inheritance. According to non-sex inheritance, III5 and II6 should be carriers with normal phenotype, but the title gives that III5 (male) does not carry pathogenic genes, and there is only one answer. The pathogenic gene is inherited by sex, accompanied by X inheritance. III5 is a dominant non-pathogenic gene.

As can be seen from the above, the genotype of III6 is XAXa(A, A cannot be superscript, please forgive me). A and A control the same trait (onychomycosis) and belong to alleles, and the segregation of alleles follows the phenomenon of gene segregation. IV 14 genotype is xaiv15, with normal phenotype (Xaxa or XAXA), V 16 is a boy, and the genotype is x (? ) y, the genotype of the disease is XaY. Where Y comes from IV 14 and X comes from IV 15, the probability of his illness (XaY) should be 0.2(IV 15 is the probability of Xa) ×1/2 (the pathogenic gamete of Xaxa1/2) = 0.

Similarly, IV 12 genotype XbY.

64 A and a, b and b are alleles, so D is excluded. B picture is not inherited by X chromosome, and it is also excluded. The A-disease gene and B-disease gene in Figure A and Figure C are linked and located on the same chromosome, regardless of the exchange situation, that is, Figure A and Figure B are not separated, and Figure A and Figure B are not separated ... If this is the case, then the sons of generation III5 and III6 are either not sick (XabY) or both are sick (XABY), and the existence of IV 12 shows that this situation is true. If it is the case in Figure C, then the offspring must suffer from only one disease, either a disease (XAbY) or b disease (Xaby). The existence of IV 1 1 and IV 13 shows that this situation is also wrong. Only when two diseases are inherited independently and located on different chromosomes can there be offspring like IV. Considering the exchange, because III5 is normal and both of them are not sick, A and B should be connected. So choose one

65 II3 genotype is (XABXab), and there is a boy X (? ) y, because x (? ) from II3, suffering from two diseases at the same time, namely XaBY, suffering from onychomycosis, namely XaBY, Xab comes from the exchange of linked genes, and the probability is lower than that of the original XabY. So the incidence of the two diseases is high.