The inheritance of oat glume color is controlled by two pairs of alleles on different chromosomes, in which gene B controls the formation of melanin and gene Y controls the shape of yellow pigment.

(1) Because individuals with BbYy genotype contain B gene, their phenotype is black glume; The phenotype and proportion of the self-bred offspring of this individual are black glume: yellow glume: white glume = (9b-y-+3b-YY): 3bby-:1bby =12: 3:1.

(2) According to the phenotypes of black glume and yellow glume, we can infer that their genotypes are B-Y-, B-yy and BBY-; Since the offspring is now 4 Heiying: 3 Huangying: 1 Baiying, it can be inferred that: (3b-y-+1b-YY): 3bby-:1bby = 4: 3:1,so the genotypes of the two parents are bbyy.

(3)① Because the genotypes of black glume are B-Y- and B-yy, namely BbYY, BbYy, BByy, BBYy * *.

(2) Because the plants of Solanum lyratum are recessive homozygotes, if they cross with BB, all the offspring will be black glume, so there are three uncertain parental genotypes, namely BBYy, BBYY and BBYY; There are three recognizable parental genotypes, namely BbYy, BbYY and BBYY.

③ If the parent plant genotype is Bbyy, the F 1 genotype is BByy, and the ratio of black glume: yellow glume: white glume is 3: 0: 1.

(4) The genotype of Astragalus sinicus is bbYy or bbYY. In order to identify its genotype, the plant can be selfed to obtain F 1, and the color of oat glume can be counted. If the offspring are all Huang Ying, the plant genotype is bbYy; if Huang Ying: Bai Ying = 3: 1, the plant genotype is BBYY. ..

So the answer is:

(1) Black Hawk? 12 black English: 3 yellow English: 1 white English?

(2)BbYy and BbYy?

(3)①6? ②BbYy、BbYY、BbYY？ ③3:0: 1?

(4) F 1 is obtained by plant self-crossing, and the color of oat glume is counted; If all offspring are Huang Ying, the plant genotype is bbYY；; If the progeny Huang Ying: Bai Ying = 3: 1, the genotype of the plant is bbYy.