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Page 1

Dihybrid Cross Worksheet
Page 1
1. Set up a punnett square using the following information:
Dominate allele for tall plants = D Recessive allele for dwarf plants = d Dominate allele for purple flowers = W Recessive allele for white flowers
= W
Cross a homozygous dominate parent
(DDWW) with a homozygous recessive parent (ddww)
DW DW
dw Adww Ad Ww AdWw
DW DW
All will
dwDd Ww Dd W
be Ad Ww
dw Dd Ww NWw
2. Using the punnett square in question #1:
a. What is the probability of producing tall plants with
purple flowers?
100% or 16/16
Possible genotype(s)? DdWw
b. What is the probability of producing dwarf plants
with white flowers? 0%
زه
Possible genotype(s)?
None
c. What is the probability of producing tall plants with
white flowers?
0%
None
Possible genotype(s)?
d. What is the probability of producing dwarf plants
with purple flowers?
Possible genotype(s)?
0%
None
dw Ad Ww Dd Ww
3. Set up a punnett square using the following
information:
•
Dominate allele for black fur in guinea pigs = B Recessive allele for white fur in guinea pigs =b Dominate allele for rough fur in guinea pigs = R
=
Recessive allele for smooth fur in guinea pigs
=r
Cross a heterozygous parent (BbRr) with a heterozygous parent (BbRr)
BR Br bR br BR BBRR BB Rr Bb RR BbRr
Br BB Rr BB rr Bb Rr Bbrr JOR BbRR BbRr bb RR bb Rr
br Bb Rr Bbrr bb Rr bb rr
Bb R r x Bb Rr
1. BR
a.
4. Using the punnett square in question #3:
What is the probability of producing guinea pigs with black, rough fur? q
Possible genotype(s)?
9/16
BBRR BBR BbRR
BbRr
b. What is the probability of producing guinea pigs
with black, smooth fur?
Possible genotype(s)?
3/16
BB rr Bbrr
C. What is the probability of producing guinea
pigs
with white, rough fur? 3/16
Possible genotype(s)?
d. What is the probability of producing guinea
pigs
bb RR bbRr
with white, smooth fur?
1/16
Possible genotype(s)?
bbrr
4 Possible 1. BR
Gametes
2. Br
2. Br
for each
3. bR
3. bR
4. br
4. br

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Page 2

RRTT x
x rrtt
5. Set up a punnett square using the following information:
Pane
Dominate allele for purple corn kernels = R Recessive allele for yellow corn kernels = r Dominate allele for starchy kernels = T
Recessive allele for sweet kernals = t Cross a homozygous dominate parent with a homozygous recessive parent
RT
RT RT
R+
RT ++
2
7. Set up a punnett square using the following
information:
Dominate allele for normal coat color in wolves
N
Recessive allele for black coat color in wolves
=
n
Dominant allele for brown eyes = B
Recessive allele for blue eyes = b
Cross a heterozygous parent with a heterozygous parent
nB
B nb
rt
ささ
rt
100%
NB
Nb
NB
Nb
nB
Rr Tz
nb
Heterozygous
6. Using the punnett square in question #5:
a. What is the probability of producing purple,
starchy
corn kernels?
100%
Possible genotype(s)? Ritz RITE
b. What is the probability of producing yellow,
starchy corn kernels? 0%
Possible genotype(s)?
None
C. What is the probability of producing purple, sweet
corn kernels?
01.
d. What is the probability of producing yellow, sweet corn kernels?
Possible genotype(s)?
None
0%.
None
Possible genotype(s)?
8. Using the punnett square in question #7:
What is the probability of producing a wolf with a normal coat color with brown eyes?
a.
Possible genotype(s)?
NNBB NnBB
9/16
NN Bb
b. What is the probability of producing a wolf with a
normal coat color with blue eyes?
C.
Possible genotype(s)?
3/16
Nnbb
What is the probability of producing a wolf with
i NNbb
3/16
a black coat with brown eyes?
Possible genotype(s)?
nn BB
nn
Bb
d. What is the probability of producing a wolf with
a black coat with blue eyes?
Possible genotype(s)?
nnbb
1/16
Nn Bb

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Page 3

#9
a)
Dominant Tall, Axial
-
Recessive. Short; Terminal
Tall + Terminal
X Short & Axial
P
TT aa
Х
tt AA
F₁
Page
3
# TEAQ
Tall + Axial
c) TEAa x TE Aa
F:
но
Tt Aa
9:3:3:1 Ratio
9
3 -
3
T_A_ Taa
- Tall & Axial (1/16) Tall + Terminal (1/16) Short & Axial (1/16) Short & Terminal (1/16
tt -
(116)
ttaa
#10
A_ curly
brown
aa
a) p)
a a x A a Litter #1 3 Aa
b) 오
aq 오
aq
2
2
1/16
2
Ga
Litter #2
4 aa
2 Аа

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Page 4

X
aaTt
Husband
# 11
Page 4
리
Woman Q Aatt
4 Possible Genotypes
A a Tt
A a tt
aa
Tt
aatt
#12
a)
HHRR x hhrr
all F. Hh Rr
F
2
Hh Rr x Hh Rr
9:3:3:1 Ratio
9 H_R_
3
H _ rr
3 h h R
1
hh rr
b) hh RR x HH rr
all F - Hh Rr
チュー
c) p:
F₁
Same
hhrr X
as
Part A
2 hh R 2 H_R_
1
Harr
Òr Must be Hh Rr
Possible genotypes (4)
P: hhrr x Hh Rr
Hh Rr
F:
Ahrr
hh Rr
hhrr