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Assuming independent assortment, the possible number of differenet gamete types from 3 homologous pairs is
3
4
8
9
16
Book says 8 but i keep getting 15 from the following method.
3 homologous pairs (mom and dad)
1 2 3
M D M D M D
Gametes Formed
1) M, D
2) M, D
3) M, D
Thus there a 6 different ones so i use a combination formula
6!/2!(6-2)! = 6!/2!x4! = 6x5/2 = 15
Where did i go wrong 😕
Update i now realize that you MUST have at least 1 gamete from each chromosome. Thus _ _ _ (since there are 2 possiblities at each pair)
2 x 2 x 2 = 8 🙂
Note: This same method is used to determine the number of possible codons (a sequence of 3 nucleotides) from which each position has A,T,C or G
4 X 4 X 4 = 64.
Later guys.
3
4
8
9
16
Book says 8 but i keep getting 15 from the following method.
3 homologous pairs (mom and dad)
1 2 3
M D M D M D
Gametes Formed
1) M, D
2) M, D
3) M, D
Thus there a 6 different ones so i use a combination formula
6!/2!(6-2)! = 6!/2!x4! = 6x5/2 = 15
Where did i go wrong 😕
Update i now realize that you MUST have at least 1 gamete from each chromosome. Thus _ _ _ (since there are 2 possiblities at each pair)
2 x 2 x 2 = 8 🙂
Note: This same method is used to determine the number of possible codons (a sequence of 3 nucleotides) from which each position has A,T,C or G
4 X 4 X 4 = 64.
Later guys.
