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MENDEL AND
THE GENE IDEA
Due to the arrangement
of the flower petals
peas had always self
pollinated so they were
pure or homozygous for
every trait. True
breeding
homozygous
Testcross
SOLVING GENETICS PROBLEMS FOR DUMMIES
1. Read the problem and underline the key words.
2. Write out your key. B=black dominant;
b=white recessive
3. Write out the P cross. male first then female
4. Determine the gametes that can be produced by each
parent.
5. Draw a Punnett Square of proper size and write male
gametes along side and female gametes across the
top.
6. Fill in the Punnett Square. Always put capital letters in
the pair first.
7. Write out your F1 genotypic percentages:
F1g%=25% AA; 50% Aa; 25% aa
8. Write out your F1 phenotypic percentages:
F p%= 75% black
25% white
16.
AB
Ab
aB
ab
Ab
ab
AABb AaBb
AAbb Aabb
AaBb aaBb
Aabb aabb
Page 1
F1g%: 12.5% AABb
12.5% AAbb
25% AaBb
25% Aabb
12.5% aaBb
12.5% aabb
17.
AB
Ab
aB
ab
AB AABB AABb AaBB AaBb
Ab
AABb AAbb AaBb Aabb
aB
AaBB AaBb aaBB aaBb
ab
AaBb Aabb aaBb aabb
F1g%: 6.25% AABB
12.5% AABb
12.5% AaBB
25% AaBb
6.25% AAbb
12.5% Aabb
6.25% aaBB
12.5% aaBb
6.25% aabb
Page 3
2. P: BbNn X bbnn
1. P: Bbnn X bbNn
bN
bn
Bn BbNn Bbnn
bn bbNn
bbnn
F1g%: 25% BbNn
25% bbNn
25% Bbnn
25% bbnn
F1p%: 25% black, normal
25% brown, normal
25% black, short
25% brown, short
BN
Bn
bN
bn
bn
BbNn
Bbnn
bbNn
bbnn
F1g%: 25% BbNn
25% Bbnn
25% bbNn
25% bbnn
F1p%: 25% black, normal
25% black, short
25% brown, normal
25% brown, short
page 3 #3
F1 cross: BbNn X BbNn
P: BBNN X bbnn
bn
BN BbNn
BN
Bn
bN
bn
BN BBNN BBNn BbNN BbNn
Bn BBNn BBnn BbNn Bbnn
F1p%: 100% Black, long furred
bN BbNN BbNn bbNN bbNn
bn BbNn Bbnn bbNn bbnn
F2g%:
6.25% BBNN
12.5% BBNn
12.5% BbNN
25%
BbNn
6.25% BBnn
12.5% Bbnn
6.25% bbNN
12.5% bbNn
6.25% bbnn
page 3 #3
F1 cross: BbNn X BbNn
P: BBNN X bbnn
bn
BN BbNn
BN
Bn
bN
bn
BN BBNN BBNn BbNN BbNn
Bn BBNn BBnn BbNn Bbnn
F1p%: 100% Black, long furred
bN BbNN BbNn bbNN bbNn
bn BbNn Bbnn bbNn bbnn
F2g%:
6.25% BBNN
12.5% BBNn
12.5% BbNN
25%
BbNn
6.25% BBnn
12.5% Bbnn
6.25% bbNN
12.5% bbNn
6.25% bbnn
F2p%:
56.25% black; normal furred dominant; dominant
18.75% black; short furred
dominant; recessive
18.75% brown; normal furred recessive; dominant
6.25% brown; short furred
recessive; recessive
Page 6 #5
P: IAIBRr X IBirr
IBr
ir
IAR
IAIBRr
IAiRr
I Ar
IAIBrr
IAirr
IBR
IBIBRr
IBiRr
IBr
IBIBrr
IBirr
F1g%:12.5% of each
F1p%: 12.5% AB+
12.5% AB25% B+
25% B12.5% A+
12.5% A-
Page 6 #6
P: IAiRr X IBiRr
IBR
IBr
iR
ir
IAR IAIBRR
IAIBRr IAiRR
IAiRr
IAr IAIBRr
IAIBrr IAiRr
IAirr
iR
IBiRR
IBiRr
iiRR
iiRr
ir
IBiRr
IBirr
iiRr
iirr
F1p%: 18.75% AB+
18.75% B+
18.75% A+
18.75% O+
6.25% AB6.25% B6.25% A6.25% O-
Page 7 #1
4 capitals = black
3 capitals = dark brown
2 capitals = light brown
1 capitals = dishwater blonde
0 capitals = blonde
P: BBDD X bbdd
bd
BD BbDd
F1p%: 100% light brown
F1 cross BbDd X BbDd
F2p%: 6.25% black
25% dark brown
37.5% light brown
25% dishwater blonde
6.25% blonde
BD
Bd
bD bd
BD
4
3
3
2
Bd
3
2
2
1
bD
3
2
2
1
bd
2
1
1
0
Page 7 #2
4 capitals = black
3 capitals = dark brown
2 capitals = light brown
1 capitals = dishwater blonde
0 capitals = blonde
BD Bd
Bd 3
2
bd 2
1
F1p%: 12.5% dark brown
37.5% light brown
37.5% dishwater blonde
12.5% blonde
bD
2
1
bd
1
0
Page 7 #3
F1p%:
1.563% 12cm
9.375% 11cm
23.44% 10cm
31.25% 9cm
23.44% 8cm
9.375% 7cm
1.563% 6cm
6 capitals = 12 cm
5 capitals = 11 cm
4 capitals = 10 cm
3 capitals = 9 cm
2 capitals = 8 cm
1 capitals = 7 cm
0 capitals = 6 cm
P FFGGLL X ffggll
fgl
FGL
FfGgLl
F1p%: 100% 9 cm
F1 cross FfGgLl X FfGgLl
FGL FGl FgL
3
6
5
5
2
5
4
4
Fgl
4
3
fGL
5
4
fGl
4
3
fgL
4
3
fgl
3
2
2
1
2
5
4
5
4
3
4
4
3
4
3
2
3
4
3
4
3
2
3
3
2
3
2
1
2
1
1
0
4
4
3
3
3
2
3
3
2
2
2
1
3
3
2
2
2
1
2
2
1
1
1
0
Epistasis
The C/c gene is epistatic to
the B/b gene; fur can only
have color if the C gene is
present; so with a C a B
gives black and with a C a
b gives brown but with out
a C the mouse is white
p. 8 #1
B-black
b-brown
H-color
h-no color
P BbHh X BbHh
BH
Bh
bH
bh
BH
BBHH
BBHh
BbHH
BbHh
F1p%: 56.25% black
18.75% brown
25% white
Bh
BBHh
BBhh
BbHh
Bbhh
bH
BbHH
BbHh
bbHH
bbHh
bh
BbHh
Bbhh
bbHh
bbhh
p. 8 #2
B-black
b-brown
H-necessary for color
h-no color
F1p%: 37.5% black
37.5% brown
25% white
P BbHh X bbHh
bH
bh
BH BbHH
BbHh
Bh
BbHh
Bbhh
bH bbHH
bbHh
bh
bbhh
bbHh
#3
Both A and B are needed for colored flowers
Aabb X aaBb
aB
ab
Ab AaBb Aabb
ab
aaBb aabb
F1p%: 25% colored flowers
75% white flowers
In apples, green fruit is dependent on a dominant gene (G) and colored
fruit , (yellow or red) on the recessive gene (g). In the presence of gg
and a dominant gene R the color is red but when (R) is not present, that
is (rr) the color is yellow. A green-fruited (GGRR) is crossed with a
yellow-fruited (ggrr) tree. Give the F1 and F2 genotypic and phenotypic
%’s.
F1 cross GgRr X GgRr
gr
GR GgRr
F1p%: 100% Green fruit
F2p%: 75% green-fruited
18.75% red-fruited
6.25% yellow-fruited
GR
Gr
gR
gr
GR GGRR GGRr GgRR GgRr
Gr
gR
gr
GGRr GGrr
GgRR GgRr
GgRr Ggrr
GgRr
ggRR
ggRr
Ggrr
ggRr
ggrr
Environmental effect on phenotype the
phenotypic range for a specific genotype is
the norm of reaction
Most human traits are
multifactorial and are
influenced by many
factors
both genetic and
environmental
Recessively Inherited Disorders
Cystic fibrosis-defective chlorine channels in cell membranes
causes thicken mucus survive into 20’s
Tay-Sachs disease-dysfunctional enzyme that fails to break
down a specific lipid in the brain; symptoms begin at a few
months of age and usually death occurs by age of 2
Sickle-cell disease-hemoglobin molecules become fibrous
with slight pressure or temperature changes; rbc’s sickle and
block capillaries and cause excruciating pain; shows
pleiotropy (many other problems); heterozygotes are immune
to malaria
Phenylketonuria-defective enzyme that breaks down the
amino acid phenylalanine; thus it builds up in blood and
causes mental retardation; children can be put on a special
diet and will not be affected; laws enforce newborn
screenings because it is the only treatable disorder
pleiotropy
Dominantly Inherited Disorders
Huntington’s disease-Degenerative disease of the nervous
system; symptoms do not show up until individual is 35-45
years old
Achrondroplasia-a form of dwarfism; cartilage cells do not
divide and bones do not grow
p. 8 #5
P RRpp X rrPP
RR or Rr = rose comb
rrpp = single comb
PP or Pp = pea comb
R & P together = walnut comb
rP
Rp RrPp
F1p% 100% walnut comb
F1 cross RrPp X RrPp
RP
F2p%:
56.25% walnut comb
18.75% rose comb
18.75% pea comb
6.25% single comb
Rp
rP
rp
RP
RRPP RRPp RrPP
RrPp
Rp
RRPp RRpp
RrPp
Rrpp
rP
RrPP
RrPp
rrPP
rrPp
rp
RrPp
Rrpp
rrPp
rrpp