Assume that during meiosis I none of the C chromosomes disjo

Assume that during meiosis I, none of the C chromosomes disjoin at metaphase, but they separate into dyads (instead of monads) during meiosis II. Each resultant gamete participated in fertilization with a normal haploid gamete. What combinations will result? Check all that apply.

A) two copies of chromosome A , two copies of chromosome B , two copies of chromosome C

B) one copy of chromosome A , two copies of chromosome B , three copies of chromosome C

C) two copies of chromosome A , two copies of chromosome B , five copies of chromosome C

D) two copies of chromosome A , one copy of chromosome B , three copies of chromosome C

E) two copies of chromosome A , two copies of chromosome B , one copy of chromosome C

F) two copies of chromosome A , two copies of chromosome B , three copies of chromosome C

G) two copies of chromosome A , two copies of chromosome B

H) two copies of chromosome A , two copies of chromosome B , four copies of chromosome C

Solution

Answer:

E) two copies of chromosome A, two copies of chromosome B, one copy of chromosome C

F) two copies of chromosome A, two copies of chromosome B, three copies of chromosome C

Explanation:

AABBCC---due disjoin of chromosome C at metaphase of meiosis I, it results ABCC & AB after meiosis I.

Due to separation of dyads of CC chromsomes during meiosis II, it results ABCC (1) & AB(3) gametes.

ABCC + normal haploid gamete (ABC)= AABBCCC

AB + ABC = AABBC