BIOL102 F’12
Dr. Young
Name _______________________
Due _______________
Genetics Problems
1.
Calculate the genotypic and phenotypic ratios for the following crosses in peas for
height, determined by the T gene:
a.
homozygous dominant crossed with homozygous recessive
T
T
t
Tt
Tt
t
Tt
Tt
All Tt. All Tall.
b.
homozygous dominant crossed with heterozygous
T
T
T
TT
TT
t
Tt
Tt
1 TT:1 Tt. All Tall.
c.
homozygous recessive crossed with heterozygous
t
t
T
Tt
Tt
t
tt
tt
1 Tt:1 tt. 1 Tall:1 short.
d.
heterozygous crossed with heterozygous
T
t
T
TT
Tt
t
Tt
tt
1 TT:2 Tt:1 tt. 3 Tall:1 short.
2.
In cats with the Manx trait, the M allele causes a short or absent tail, whereas the m allele
confers a normal, long tail. Cats of genotype MM die as embryos. If two Manx cats mate,
what is the probability that each living kitten has a long tail?
M
m
M
MM Mm
m
Mm mm
Since MM leads to death of the embryo and not a live birth, the Manx cats must both be
Mm. Mm will be Manx cats; mm will have a long tail. Of the live births, 1/3 will have a
long tail.
3.
A man with type AB blood has children with a woman who has type O blood. What are
the chances that a child they conceive will have type A blood? B? AB? O?
IA
IB
A
i
I i
IBi
A
i
I i
IBi
There is a 50% (1/2) chance of them having an A blood type child; 50% (1/2) chance of
them having a B blood type child. There is no chance of them having an AB or O blood
type child.
4.
If a child has an AB (IAIB) blood type, the parents
a.
can have different blood types, but neither can be blood type O.
True. Since a person with type O blood has an ii genotype, this person cannot be one of
the parents.
b.
must both be AB.
Not true. One parent must supply IA (i.e., IAIA, IAIB, or IAi) and the other IB (i.e., IBIB,
IAIB, or IBi).
c.
can be any blood type.
Not true. See a. above.
d.
must be A and B, but not AB.
Not true. See b. above.
e.
must both have different blood types.
Not true. See above.
5.
If a child belonged to blood type O (need ii), he or she could not have been produced by
which set of parents?
a.
Type A (could be IAi) mother and type B (could be IBi) father
b.
Type A (could be IAi) mother and type O (ii) father
c.
Type AB (must be IAIB, no i to donate) mother and type O (ii) father (This is the
correct answer.)
d.
Type O (ii) mother and type O (ii) father
e.
a and c could not, but both b and d could produce a type O child
6.
In cocker spaniels, black coat color (B) is dominant over red (b), and solid color (S) is
dominant over spotted (s). If two dihybrids (BbSs) were crossed, the most common
phenotype would be
BS
Bs
bS
bs
BS
BBSS
BBSs
BbSS
BbSs
Bs
BBSs
BBss
BbSs
Bbss
bS
BbSS
BbSs
bbSS
bbSs
bs
BbSs
Bbss
bbSs
bbss
Since any B will give Black and any S will give Solid, the most common phenotype is
Black Solid (9/16).
7.
In cocker spaniels, black coat color (B) is dominant over red (b), and solid color (S) is
dominant over spotted (s). If a red solid male was crossed with a black solid female to
produce a red spotted puppy, the genotypes of the parents (with male genotype first)
would be
The best way to approach this is to list what we do know. The father’s genotype is bbS_
and the mother’s is B_S_. The puppy is bbss. One b allele came from dad so the other
must have come from mom. She must be Bb. One s allele came from dad so he must be
Ss. The other s allele came from mom so she must be Ss. Therefore, father’s genotype is
bbSs and mother’s genotype is BbSs.
8.
Harold works in a fish market, but the odor doesn’t bother him because he has anosmia,
an X-linked recessive lack of sense of smell. Harold’s wife, Shirley, has a normal sense
of smell. Harold’s sister, Maude, also has a normal sense of smell, as does her husband,
Phil. Maude and Phil’s daughter, Marsha, has a normal sense of smell, but their identical
twin boys, Alvin and Simon, cannot detect odors. Harold and Maude’s parents, Edgar and
Florence, can smell normally.
a.
Draw a pedigree for this family, indicating affected individuals and as much
genotypic information as can be determined.
b.
Xa
Ya
If we know that Shirley is a carrier for the anosmia allele, what is the probability
of Harold and her having a child with anosmia? What is the probability that they
would have a child that is a carrier for anosmia? Explain/show your reasoning.
Since this is an X-linked recessive disorder, Harold must be XaY. We now know
that Shirley is X Xa.
Xa
XXa
XYa
Xa
XaXa
XaY
We can see that affected ½ or 50% of the offspring are expected to have anosmia
(XaXa and XaY). We can see that their chances of having a child who is a carrier
(XXa) for anosmia is ¼ or 25%. Note that only females can be carriers for Xlinked recessive disorders.