Give the economic importance of colchicine.

Question 1

Answer · 2026-03-23T12:39:39.615Z

Step 1: Determine the genotypes of the parents. Haemophilia is an X-linked recessive disorder. Let X^H represent the allele for normal blood clotting and X^h represent the allele for haemophilia. John has normal blood clotting, so his genotype is X^HY. Jane has normal blood clotting, but they have a haemophiliac son (X^hY). The son inherited the Y chromosome from John and the X^h chromosome from Jane. Therefore, Jane must be a carrier. Her genotype is X^HX^h. Step 2: Perform the genetic cross between John (X^HY) and Jane (X^HX^h). Possible gametes from John: X^H, Y Possible gametes from Jane: X^H, X^h Punnett Square: |c|c|c| & X^H & Y \\ X^H & X^HX^H & X^HY \\ X^h & X^HX^h & X^hY \\ Step 3: Determine the genotypes of their children. The possible genotypes for their children are X^HX^H (normal female), X^HX^h (normal female carrier), X^HY (normal male), and X^hY (haemophiliac male). They have a haemophiliac son. The genotype of the haemophiliac son is X^hY. They have two daughters. Both daughters have normal blood clotting (as only the son is specified as haemophiliac). The possible genotypes for a daughter with normal blood clotting are X^HX^H or X^HX^h. Therefore, the genotypes of the two daughters are: Daughter 1: X^HX^H or X^HX^h Daughter 2: X^HX^H or X^HX^h a) Colchicine is economically important because it is used in plant breeding to induce polyploidy, which is the multiplication of chromosome sets. This can lead to the development of new plant varieties with desirable traits such as increased fruit size, improved vigor, or sterility (e.g., seedless fruits and vegetables).

Question 2

Step 1: Determine the genotypes of the parents. Haemophilia is an X-linked recessive disorder. Let X^H represent the allele for normal blood clotting and X^h represent the allele for haemophilia. John has normal blood clotting, so his genotype is X^HY. Jane has normal blood clotting, but they have a haemophiliac son (X^hY). The son inherited the Y chromosome from John and the X^h chromosome from Jane. Therefore, Jane must be a carrier. Her genotype is X^HX^h. Step 2: Perform the genetic cross between John (X^HY) and Jane (X^HX^h). Possible gametes from John: X^H, Y Possible gametes from Jane: X^H, X^h Punnett Square: |c|c|c| & X^H & Y \\ X^H & X^HX^H & X^HY \\ X^h & X^HX^h & X^hY \\ Step 3: Determine the genotypes of their children. The possible genotypes for their children are X^HX^H (normal female), X^HX^h (normal female carrier), X^HY (normal male), and X^hY (haemophiliac male). They have a haemophiliac son. The genotype of the haemophiliac son is X^hY. They have two daughters. Both daughters have normal blood clotting (as only the son is specified as haemophiliac). The possible genotypes for a daughter with normal blood clotting are X^HX^H or X^HX^h. Therefore, the genotypes of the two daughters are: Daughter 1: X^HX^H or X^HX^h Daughter 2: X^HX^H or X^HX^h a) Colchicine is economically important because it is used in plant breeding to induce polyploidy, which is the multiplication of chromosome sets. This can lead to the development of new plant varieties with desirable traits such as increased fruit size, improved vigor, or sterility (e.g., seedless fruits and vegetables).

Give the economic importance of colchicine.

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