|
Dihybrid Genetics problems
1 .
Black fur in mice (B) is dominant to brown fur (b). Short tails (T) is
dominant to long tails (t).
What proportion of the
progeny of the cross BbTt x BBtt
will have black fur and long tails?
Answer: BbTt x BBtt
|
x
|
BT
|
Bt
|
bT
|
bt
|
|
Bt
|
BBTt
|
BBtt
|
BbTt
|
Bbtt
|
|
|
Black
|
Black
|
Black
|
Black
|
|
|
Short tail
|
Long
|
Short
|
Long
|
|
|
|
|
|
|
50%
Black Long
2. A couple has three children, all of whom have brown
eyes and blond hair. Both parents are
homozygous for brown eyes (BB),
and one is blond (rr) while the other is a redhead
(Rr). What
is the probability that the
next child will be a brown-eyed redhead?
P1
Brown Eye Blond x
Brown Eye Redhead
BB rr x BB Rr
|
x
|
BR
|
Br
|
|
|
|
Br
|
BBRR
|
BBrr
|
|
|
|
|
Brown eye
|
Brown Eye
|
|
|
|
|
Redhead
|
Blond
|
|
|
|
|
|
|
|
|
Assuming that RR is redhead then the
probability is 50% or P= 0.5, or a ½
3. In mice, black is dominant over tan and short tails
are dominant over long.
- Use B for
black and b for tan.
- Use S for
short and s for long.
- Write the
genotype for a heterozygous black ,
short-tailed mouse. BbSs
- Cross two of
these individuals.
- From the Punnett
Square, describe the phenotype of the
offspring.
|
x
|
BS
|
Bs
|
bS
|
Bs
|
|
BS
|
BBSS
|
BBBs
|
BbSs
|
BBSs
|
|
Bs
|
BBSs
|
BBss
|
BbSS
|
BBss
|
|
bS
|
BbSS
|
BbSs
|
bbSS
|
BbSs
|
|
bs
|
BbSs
|
Bbss
|
bbSs
|
bbss
|
9 Black Short; 3 Black Long; 3
Tan Short; 1 Tan Long
4. In certain bacteria, an oval shape is dominant over
round and thick cell walls are dominant over thin. Cross a heterozygous oval,
thick cell walled bacteria with a round, thin cell walled bacteria. Describe
the phenotype of the offspring.
Allele Key Q= oval
q= round T= thick t= thin
Oval Thickwall
(QqTt)
x Round (qqtt)
|
x
|
qt
|
|
QT
|
QqTt
|
|
Qt
|
Qqtt
|
|
qT
|
qqTt
|
|
qt
|
qqtt
|
Oval Thickwall
: Oval thin wall; round Thickwall;
round thin wall
1:1:1:1
5.
In guinea pigs, black coat colour is dominant over white, short hair is
dominant over long.. Show the
Punnett Square for a cross
between a homozygous black, short-haired guinea pig and a homozygous white,
long-haired guinea pig. What do the offspring look like?
Black (B) > White (b) Short (S)> Long (s)
Homozygous black Short tailed (
BBSS) x Homozygous white, long haired ( bbss)
All Heterozygous Black Short
6. In humans, the presence of freckles is due to a
dominant gene (F) and the non-freckled condition is
due to its recessive allele
(f). Dimpled cheeks (D) is dominant to
non-dimpled cheeks (d). Two
persons
with freckles and dimpled cheeks have two children. One has
freckles but no dimples and
one has dimples but no
freckles.
a. What are the genotypes of the parents?
Parents F_D_ x F_D_ Children F_ dd ,
ffD_
Therefore Parents must be heterozygous at each gene
loci
FfDd x
FfDd
b.What are the possible phenotypes and genotypes of the children which
they could produce?
|
x
|
FD
|
Fd
|
fD
|
fd
|
|
FD
|
FFDD
|
FFDd
|
FfDD
|
FfDd
|
|
Fd
|
FFDd
|
FFdd
|
FfDd
|
FFdd
|
|
fD
|
FfDD
|
FfDd
|
ffDd
|
ffDd
|
|
fd
|
FfDd
|
Ffdd
|
ffDd
|
ffdd
|
FFDD,
FFDd, FfDD, FfDd, FFdd, Ffdd,
ffDd, ffdd
Phenotypic
Ration: 9 Freckled Dimpled;3 Freckled Non Dimple; 3 Non-freckle Dimple; 1 Non
Freckle Non Dimple
9:3;3;1
c.
What are the chances that
they would have a child which lacks both freckles and dimples?
The probability is 9 in 16
d. A person
with freckles and dimples whose mother lacked both freckles and dimples
marries a
person with freckles but no
dimples (whose father did not have freckles or dimples). What are
the
chances that they would have a child which lacks both freckles and dimples?
FfDd X Ffdd
|
|
Fd
|
fd
|
|
FD
|
FFDd
|
FDd
|
|
Fd
|
FFdd
|
Ffdd
|
|
fD
|
FfDd
|
ffDd
|
|
fd
|
Ffdd
|
ffdd
|
Chance of no freckles or dimples is 1 in 8
|