ⅱ( 1) As can be seen from the figure, ⅰ4 is not bald, while ⅱ6 is bald, and the genotype is hh, so ⅰ4 is HH; And Ⅱ 4 is a color-blind boy, the genotype is XbY, Ⅰ 4 is not color-blind and XBXBXBXBXB, so Ⅰ 4 genotype is XBXBXB ... Little baldness is dominant among boys, Ⅲ 3 is hh for boys who are not sick, XBY for boys who are not red and green color-blind, so Ⅲ 3 genotype is hhXBY. ..
(2) The first generation 1 individual is male, without alopecia, and the genotype is hh. Therefore, the gene that causes Ⅳ 2 baldness cannot come from the first generation 1 individual.
(3) First, determine the genotype, Ⅲ1is hhXBY, Ⅲ 2 is HhXBXb；; The probability of baldness in offspring is 12, and the probability of red-green color blindness in sons is 12. If ⅲ 1 and ⅲ 2 have a son, the probability of suffering from only one disease is 12×( 1? 12)+( 1? 12)× 12= 12
(4) Pay attention to the correct format when writing the genetic map of Ⅰ1and Ⅱ 2.
So the answer is:
Ⅰ Polygenic Chromosome Abnormality
ⅱ( 1)HhXBXb？ hhXBY
(2) 1?
(3) 1/2
(4)