Dad: TtBb
Mom: ttbb
You have to use distribution for dihybrid crosses. Meaning, the first allele of each trait has a equal chance of being paired with the other allele of the other trait. So for example with Dad, I will number the traits:
T(1)t(2)B(3)b(4)
To set up the possibilities from Dad, it would be 13, 14, 23, 24: TB, Tb, tB, tb. Same idea goes for Mom, except since all alleles are the same, you only need to make one column for Mom, since if you did all 4, the other 3 would just be repeats of the 1.
Cross:
tb
TB TtBb
Tb Ttbb
tB ttBb
tb ttbb
The phenotypic ratio is 1 Tall Brown: 1 Tall Blue: 1 Short Brown: 1 Short Blue
The answer to this statement is
True
Low power 50x (5x10)
high power 250x (5x50)
If you want to calculate the microscope power of magnification you have to multiple the magnification power of the ocular with the magnification power of the objectiv.
