have some golden bud and some silver bud. Not
only that but we can understand clearly what
the frequency will be. Count them!
Bb + Bb = 2Bb
bb + bb = 2bb
2 out of 4 will have golden bud. 2 out of 4
will have silver bud. Half out offspring will
have silver bud! The ration is 50:50.
The second possibility tells us a number of
things.
(1) Both parents need at least one b trait each
for the silver bud to pass on if it is a
recessive trait.
(2) If any silver bud is produced in the
offspring then the mystery parent B? must be
Bb. It can not be BB.
Remember:
Homozygous Dominant: BB = Golden Bud.
Heterozygous: Bb = Golden Bud
Homozygous Recessive: bb = Silver Bud.
So if the golden bud parent when crossed with a
silver bud parent produced only Golden Bud,
then the parent must be Homozygous Dominant for
that trait. If the parent produced any silver
bud then it must be Heterozygous.
