Answer to Genetic Worksheet: Monohybrid Crosses
1. Black fur is dominant over brown fur in guinea pigs. Two heterozygous black fur guinea pigs are
crossed. For the F1 generation (offspring) …
Solution:
Gene: Fur colour
B= black fur b= brown fur
Bb (heterozygous black fur male) x Bb (heterozygous black fur female)
B
b
B
BB
b
Bb
b
Bb
b
bb
Genotype(s) & ratios: BB : Bb : bb = 1 : 2 : 1
Phenotype(s) & ratios: Black fur: brown fur = 3 : 1
a) What are the chances of getting brown fur guinea pigs?
There is a 1 in 4 chance (or 25%) chance of getting brown fur guinea pigs.
b) What are the chances of getting black fur guinea pigs?
There is a 3 in 4 chance (or 75%) chance of getting black fur guinea pigs.
c) What are the chances of getting black homozygous guinea pigs?
There is a 1 in 4 chance (or 25%) chance of getting black homozygous guinea pigs.
d) What are the chances of getting black heterozygous guinea pigs?
There is a 1 in 2 chance (or 50%) chance of getting black heterozygous guinea pigs.
2. Black fur is dominant over brown fur in guinea pigs. A heterozygous male black fur guinea pig is
crossed with a homozygous brown fur female. For the F1 generation…
Solution:
Gene: Fur colour
B= black fur b= brown fur
Bb(heterozygous male black fur guinea pig) x bb(homozygous brown fur female)
B
b
b
Bb
b
bb
b
Bb
b
bb
Genotype(s) & ratios: Bb:bb = 1: 1
Phenotype(s) & ratios: Black fur: brown fur = 1:1
a) What are the chances of getting brown fur guinea pigs?
There is a 1 in 2 chance (or 50%) chance of getting brown fur guinea pigs.
b) What are the chances of getting black fur guinea pigs?
There is a 1 in 2 chance (or 50%) chance of getting black fur guinea pigs.
c) What are the chances of getting black homozygous guinea pigs?
There is a 0 in 4 chance (or 0%) chance of getting black homozygous guinea pigs.
d) What are the chances of getting black heterozygous guinea pigs?
There is a 1 in 2 chance (or 50%) chance of getting black heterozygous guinea pigs.
3. Woodrats have an “interesting” genetic trait of bringing home shiny objects. The trait of bringing
home shiny objects is dominant to the trait of carrying home only dull objects. Suppose two heterozygous
individuals are crossed.
Solution:
Gene: Bringing home objects
S = Shiny objects
s = dull objects
Ss (heterozygous male) x Ss (heterozygous female)
S
s
S
SS
Ss
s
Ss
ss
Genotype(s) & ratios: SS : Ss : ss = 1 : 2 : 1
Phenotype(s) & ratios: Shiny objects : dull objects = 3 : 1
a) How many of each genotype would be expected if only 4 offspring were produced?
1 offspring will be SS (homozygous shiny objects), 2 offspring would be Ss (heterozygous shiny objects) and 1
offspring will be ss (homozygous dull objects)
b) What is the chance these rats will have a child that brings home dull objects?
There is a 3 in 4 chance (or 75%) that there will be a child that brings home dull objects.
4. In dogs, wirey-hair is dominant to smooth hair. In a cross of a homozygous wirey-haired dog with a
smooth-haired dog…
Solution:
Gene: Dog hair
W = wirey
w = smooth
WW (homozygous wirey-haired) x ww (homozygous smooth-haired)
w
W
W
Ww
Ww
w
w
Ww
Ww
Genotype(s) & ratios: 100% Ww
Phenotype(s) & ratios: 100% wirey-haired
a) What will be the phenotype and their ratios of the F1 generation?
The phenotype of the F1 generation will be 100% wirey-haired.
b) What will be the genotypes and their ratios of the F1 generation?
The genotype of the F1 generation will be 100% Ww.
c) What would be the phenotype ratio of wire-haired to smooth-haired dogs in the F2 generation?
Ww (heterozygous wirey-haired) x Ww (heterozygous wirey-haired)
W
W
w
WW
Ww
w
w
Ww
ww
Genotype(s) & ratios: WW : Ww : ww = 1 : 2 : 1
Phenotype(s) & ratios: Wirey-haired : Smooth-haired = 3 : 1
The phenotype of the F2 generation will be 3 wirey-haired dogs to 1 smooth-haired dog (75% to 25%).
5. Assume that in some insects long-winged (LL or Ll) is dominant over short-winged (ll). If a longwinged male insect having an unknown genotype is test-crossed, the F1 generation is 50% long-winged
and 50% short-winged. Determine the genotype of the long-winged male insect and explain how you
determined this.
Solution:
Gene: Length of wings
W = long wings
w = short wings
W? (long-winged male insect) x ?? (female insect)
If the F1 generation is 50% long-winged and 50% short-winged, then one of the insects must be homozygous
recessive for short wings. This will be verified through a test cross:
W
w
w
Ww
Ww
w
W
Ww
Ww
Genotype(s) & ratios: all Ww (100%)
Phenotype(s) & ratios: all long-winged (100%)
This does not match the result. I will try another test cross, this time, a heterozygous male (Ww) and a
homozygous female (ww). Note – the male HAS to have a dominant allele.
W
w
w
Ww
Ww
w
w
ww
ww
Genotype(s) & ratios: Ww : ww = 1 : 1 (50% Ww and 50% ww)
Phenotype(s) & ratios: long-winged : short-winged = 1 : 1 (50% and 50%)
Therefore, by conducting various test-cross using different alleles, I have determined that the male is
heterozygous dominant (Ww) for long wings and the female is homozygous recessive (ww) for short wings.
6. A woman is homozygous dominant for short fingers. She marries a man who is heterozygous for short
fingers.
Solution:
Gene: Length of fingers
F = short
f = long
FF (homozygous short fingered woman) x Ff (heterozygous short-fingered male)
F
F
FF
f
Ff
w
F
FF
Ff
Genotype(s) & ratios: FF : Ff = 1 : 1
Phenotype(s) & ratios: all short-fingered (100%)
a) What is the probability that any of their children have long fingers?
There is 0% chance of their children having long fingers because they would need to be recessive homozygous
to have long-fingered. As shown in the Punnett cross, the F1 generation would only be homozygous dominant
for short fingers or heterozygous for short fingers.
b) Could any of their grandchildren potentially have long fingers? Why or why not?
Yes. This would apply if one of the F1 generation offspring had the genotype Ff (heterozygous short-fingered)
and he/she married a person who was homozygous recessive (ff) for long-fingered. The test cross below would
show the F2 generation:
Ff (heterzygous short fingered offspring) x ff (homozygous long-fingered partner)
F
f
Ff
f
ff
w
f
Ff
ff
Genotype(s) & ratios: FF : ff = 1 : 1
Phenotype(s) & ratios: short-fingered : long-fingered = 1 : 1 (50% chance each)
7. Jane and John are expecting a baby and know that they are both carriers (heterozygous) of the
recessive disease cystic fibrosis.
Solution:
Gene: Cystic fibrosis
C = does not have disease
c = has disease
(CC = normal, Cc = carrier, cc = has cystic fibrosis)
Cc (heterozygous male) x Cc (heterozygous female)
C
C
c
CC
Cc
w
c
Cc
cc
Genotype(s) & ratios: CC : Cc : cc = 1 : 2 : 1
Phenotype(s) & ratios: normal : carrier : has disease = 1 : 2 : 1
a) What is the probability that their child will have cystic fibrosis?
The probability that their child will have cystic fibrosis is 25% (1 in 4 chance)
b) What is the probability that their child will be a carrier of cystic fibrosis?
The probability that their child will be a carrier of cystic fibrosis is 50% (1 in 2 chance)
c) What is the probability that their child will NOT be a carrier of cystic fibrosis?
The probability that their child will not be a carrier of cystic fibrosis is 25% (1 in 4 chance)