Mendelian Genetics and Inheritance
Names ____________________ ___________________
__________________ ___________________
1. Tall (D) is dominant to dwarf (d). Give the F2 genotypic and phenotypic ratios of a cross between purebreeding tall plant and a pure-breeding dwarf plant. (Please construct a punnett square on back of pagelook on bottom of web page).
Phenotypic Ratio ___3:1 tall to dwarf_______________
Genotypic Ratio ___1:2:1________________
2. In garden peas, one pair of alleles controls the height of the plant and a second pair of alleles controls
the flower color. The allele for Tall (D) is dominant to the allele for dwarf (d) and the allele for purple (P)
is dominant to white (p). A tall purple crossed with a tall purple produces ¾ tall purple and ¼ tall white.
What is the genotype of the parents?
P1 ___DDPp________
P2 ___DdPp or DDPp___________
(Please construct a punnett square of your answer on the back of this page).
3. There are three alleles controlling the ABO blood types. IA and IB are codominant genes so that the
combination IAIB produces the AB blood type. The third allele i (sometime written I O) is recessive to the
other two alleles. Indicate which of these parents could produce the given child:
Parents
a) A x AB
b) A x O
c) A x B
d) A x AB
e) O x O
f) B x B
g) AB x AB
Child
B
A
O
O
B
O
A
Yes or No
yes
yes
yes
no
no
yes
yes
4. Congenital deafness in humans is due to the homozygous condition of either or both of the recessive
genes d or e. Both dominant D and E are necessary for normal hearing. Gene D/d affects the middle ear,
while gene E/e affects the inner ear. It does not matter how good the normal inner ear (as indicated by E_)
is; if there is something wrong in the middle ear, the individual is unable to hear. The same applies for the
other gene. Give the phenotypic results (ratios) of the following crosses:
a) DdEE x DdEE _____3:1 hearing to non_________________________
b) DdEe x DdEe _________9 hearing to 7 non-hearing_____________________
c) ddEE x DDee _________100% all hearing_____________________
5. In humans an X-linked disorder called colobroma iridia (a fissure in the iris) is a recessive trait. A
normal couple has an afflicted daughter. The husband sues the wife for divorce on the grounds of
infidelity. Would you find in his favor? Why?
Yes. It would be impossible for a normal male to pass on the trait, therefore impossible
for the daughter to be homozygous for that trait having a normal male as her father. (See below)
6. Colorblindness is an X-linked recessive gene. Two normal-visioned parents produce a color-blind child.
a. Is this child male or female? ___male_______
b. What are the genotypes of the parents:? __XCXc_______ and _XCY_________
c. What are the chances that their next child will be a colorblind daughter? _0____
7. Now let us say just for the sake of genetics that a man (Type A blood and heterozygous for a second
trait H) and a woman (Type AB and heterozygous for H) had 16 children (just for arguments sake).
Show the types of gametes each parent can make then draw in the Punnett square.
(I am unable to make my computer put in the 4 gametes that go in a column. They would be I AH, IAh, IBH,
and IBh
IAH
IAh
I OH
I Oh
IAIAHH
IAIAHh
IAIOHH
IAIOHh
IAIAHh
IAIAhh
IAIOHh
IAIOhh
IAIBHH
IAIBHh
IBIOHH
IBIOHh
IAIBHh
IAIBhh
IBIOHh
IBIOhh
Give the genotypic and phenotypic ratios for the traits above.
Genotypic
IAIAHH-1/16
IAIAHh-2/16
IAIAhh-1/16
IAIOHH-1/16
IAIOHh-2/16
IAIOhh-1/16
IAIBHH-1/16
IAIBHh-2/16
IAIBhh-1/16
IBIOHH-1/16
IBIOHh-2/16
IBIOhh-1/16
Genotypic: 8 Type A, 4 Type AB, and 4 Type B.
Know that the above cross really produced the following: 6 were found to be type A, 3 were found to be
type AB, 3 were found to be type B and 4 were found to be type O. How can you explain this? (This is an
actual situation- Bombay phenotype).
If you look at the above punnett square, there is no type O bloods available. That means that the Hh allele
must have something to do with the expression of blood type. Since 4 were found to be type O, it would
appear that anytime the hh genotype appears, the blood type does not get expressed! Another epistasis
problem.
1. DDxdd, gametes = D or d, so F1= Dd, we then cross using gametes D and d, x D and d
D
d
D
DD
Dd
d
Dd
dd
This gives you the 3 genotypes, DD, Dd and dd at a ratio of 1:2:1. Since DD and Dd have the same
phenotype, the phenotypic ratio is 3:1.
2. This is a dihybrid cross that can be simplified. We know that D is tall and that all the offspring come
out tall, so we can assume one parent is DD, and the other could be either Dd or DD, because the two
gametes from the DD parent would cause all the offspring to be tall, even if they carried the d allele. That
really only leaves the P alleles to worry about. Since ¼ of the offspring come out with the recessive trait,
that should scream to you heterozygous for both parents (look up at the punnett square right above this
answer! That leaves us with DDPp x DDPp (or DdPp).
3. a. OK, remember if you are A blood, you can be either I AIA or IAIO, either way you donate at least 1 A
allele and either another A or an A. The other parent donates either and A or a B, so the baby would have
the following chances
A
A
A
AA
AA
B
AB
AA
A
O
A
AA
AO
B
AB
BO
So, the offspring from this relationship would only have children with A, AB or B blood types.
Do the same thing for all the rest of the questions, and they should start to make sense.
4. Remember that anytime you have a dd OR an ee, the person is non-hearing.
a. DdEE x DdEE can only make 2 gametes each DE and dE, you can write the rest out but they are
redundant. So…
DE
dE
DE
DDEE
DdEE
dE
DdEE
ddEE
This gives us our classic 3:1 ratio, where the bottom right hand corner shows the expression of our
recessive alleles. Therefore, 3 out of 4 children should be normal hearing.
b. Now, our heterozygous cross does not gives us our “classical” phenotype of 9:3:3:1 because one gene
affects the other, thereby skewing our results. So, on your classic 16 squares, you need to count all the
squares that have either dd or ee, and the rest of the stuff is irrelevant.
DE
De
dE
de
DE
DDEE
DDEe
DdEE
DdEd
De
DDEe
DDee
DdEd
Ddee
dE
DdEE
DdEe
ddEE
ddEe
de
DdEe
Ddee
ddEe
ddee
All the results in red are non-hearing because of the recessive alleles.
c. ddEE x DDee is interesting. Two non-hearing parents who’s alleles would be dE or De. This would
give us one genotype for the offspring, DdEe which would be hearing!
5. Mom is a carrier so she would have the genotype X IXi (for iris). Dad carries the gene ONLY on his X
chromosome and not his Y, and since he is normal, his genotype would be XIY. Therefore.
XI
Xi
XI
XIXI
XIXi
Y
XIY
Xi Y
So, they could have an affected son, but not a daughter. This means that the man is not the true father of
the child (bad mom).
6. This is just like the above, where mom has to be heterozygous XCXc and dad only carries one X, so he
must carry the dominant allele, so his genotype would be X CY
XC
Xc
XC
XCXC
XC Xc
Y
XCY
XC Y
So once again, only sons will express the recessive allele, and the daughters have 0% chance of expressing
the recessive allele (but they can carry it on to their sons).