Monohybrid Practice Problems (Answers)
1. Trait – Pod colour
P generation
Green – G
True Green – GG x
F1 generation
True yellow – gg
G
Gg
Gg
g
g
F1 Cross
Yellow – g
G
Gg
Gg
Heterozygous – Gg
x
G
g
G
GG
Gg
F2 generation
Genotypic ratio (Gg) 100%
Phenotypic Ratio (Green) 100%
Heterozygous - Gg
g
Gg
gg
Genotypic Ratio
GG : Gg: gg
1: 2 : 1
Phenotypic Ratio Green :yellow
2.a.
Trait – Comb type
P generation
Rose – R
True Rose – RR
F1 generation
F1 Cross
F2 generation
x single – rr
R
Rr
r
Rr
R
r
x
3 : 1
Single – r
Genotypic ratio (Rr) 100%
Phenotypic Ratio (Rose) 100%
Genotypic Ratio
Rr
R
RR
Rr
r
Rr
rr
RR : Rr: rr
1: 2 : 1
Phenotypic Ratio Rose : single
3 : 1
b. The only way for a Rose rooster and a Rose hen to have a single combed offspring (rr)
would be for each parent to be heterozygous (Rr) because each parent would have to
donate one “r” for the offspring to have two “rr”.
3. Trait - Eye colour
P generation
Brown – B blue – b
blue-eyed mother (bb) x brown eyed father (B _b_)
(we do know that brown-eyed father (B___) had a blue-eyed mother (bb) therefore he
must have received the “B” from his father and the “b” from his mother because his
mother only had “b” to provide...thereby giving him a Bb genotype
Genotypic Ratio
F1 generation
b
b
B
Bb
Bb
b
bb
bb
Bb :bb
1:1
Phenotypic Ratio Brown : blue
1 : 1
We know the brown-eyed man’s father had a genotype of B___ but we do not know what
the other alleles is. If he was BB then all his children would be Bb but if he was Bb then his
kids would have a 50% chance of having brown eyes and 50% chance of having blue eyes.
4. Trait – Body colour
white belt – W
uniform body colour – w
It would be easier to breed uniform body because it is a recessive trait and the only way it
will appear in the population is if it is homozygous recessive (ww) so it would be easier to
create a true breeding line of these. Belted hogs in the population may be masking the
recessive trait.
5. Trait – horniness?
Parent generation
F1 generation
Polled (hornless) – H
horned – h
Polled cow (H ___ ) x polled bull (H ____)
H
H
___
___
Genotypic Ratio HH : Hh: hh
1 : 2: 1
hh
Phenotypic Ratio hornless:horned
3:1
The only way to get offspring that are horned (hh) would be if each parent donated “h”.
Therefore each parent MUST be heterozygous (Hh). The probability of the next calf
having horns will also be 25%.
6. Trait – plant colour
Green – G
albino – g
Albino plant (gg) x heterozygous plant (Gg)
F1 generation
g
Gg
gg
G
g
Genotypic Ratio
Gg :gg
1:1
Phenotypic Ratio green : albino
1 : 1
7. Trait – Coat Colour
Solid coat – S
Parent generation
F1 generation
Spotted coat – s
spotted (ss)
s
x
spotted (ss)
s
ss
Genotypic Ratio
ss – 100%
Phenotypic Ratio – spotted coat
100%
8. Trait – red blood cell shape
Normal cell – N
sickle cell – n
For the woman’s cell’s to ‘sickle” she must be heterozygous (Nn) for the trait. Her
perspective husband’s cells do NOT sickle therefore he must be homozygous dominant
(NN).
Parent generation
F1 generation
Homozygous dominant (NN) x heterozygous (Nn)
N
n
N
NN
Nn
N
NN
Nn
Genotypic Ratio
NN :Nn
1:1
Phenotypic Ratio normal cell
100%
Future children will all have “Normal” looking blood cells but
there is a 50% chance that their kids will have normal cells that
will “sickle” under low oxygen conditions.