Question

Examine the genotype for an individual who is AaBbCcDc. Predict the outcomes of meiosis and the potential gamete combinations that could form from this individual select all that apply

Answer 1

Answer: 16 different gametes

1. ABCD
2. ABCd
3. ABcD
4. ABcd
5. AbCD
6. AbCd
7. AbcD
8. Abcd
9. aBCD
10. aBCd
11. aBcD
12. aBcd
13. abCD
14. abCd
15. abcD
16. abcd

Explanation:

Gamete formation involves the random and independent segregation of alleles. During anaphase 1 (meiosis), occurs the independent separation of homologous chromosomes that migrate to opposite poles of the cell. This separation generates different chromosomal combinations in the daughter cells. There are two alternatives per homologous pair.  

The number of gametes with different combinations is 2ⁿ, where  

n = number of homologous pairs.  

In the exposed example there are 4 pairs of chromosomes, so

the number of possible combinations of chromosomes = 2ⁿ = 2⁴ = 16

Individual: AaBbCcDdEe

You can make use of the fork line method to easily figure out the possible gametes.

A.......B.........C........D → ABCD

                     ........d → ABCd

          ...........c.........D → ABcD

                       ........d → ABcd

           b.........C........D → AbCD

                     ........d → AbCd

          ...........c.........D → AbcD

                       ........d → Abcd

a.......B.........C........D → aBCD

                     ........d → aBCd

          ...........c.........D → aBcD

                       ........d → aBcd

           b.........C........D → abCD

                     ........d → abCd

          ...........c.........D → abcD

                       ........d → abcd