Genetics Problem Solutions
1. Diabetes has been found to be inherited (in many cases) through a recessive allele
(d). Two parents who are non-diabetic have a diabetic child. Write the genotypes
of all three persons.
Answer: Both parents are heterozygous, Dd and the child is homozygous recessive, dd.
Explanation:
What are all of the possible genotypes and corresponding phenotypes?
Genotype
DD
Dd
dd
Phenotype
Normal (non-diabetic)
Normal (non-diabetic)
Diabetic
Since diabetes is inherited as a recessive allele, in order to be afflicted with diabetes an
individual must be genotypically homozygous recessive (dd). Therefore the child in this
question is dd.
If the child is dd, then he/she MUST have received 1 recessive allele (d) from mom and
one recessive allele (d) from dad. Each parent has at least one recessive allele. Since they
are both phenotypically normal, then their phenotypes must be heterozygous (Dd).
2. In corn, there are two alleles for the gene for sugar content. High sugar content
(su) is recessive to low sugar content (Su). Give the phenotypes for each of the
following individuals: Susu, susu, SuSu.
SuSu – low sugar content
Susu – low sugar content
susu – high sugar content
3. Skin color in a fish is inherited via a single gene with four different alleles. How
many different genotypes would be possible in this system? List them.
10 possible genotypes; Gene designation = C; Alleles = CA, CB, CC, CD
All possible genotypes:
CA CA
CA CB
CA CC
CA CD
CB CB
CB CC
CB CD
CC CC
CC CD
CD C D
4. Achondroplasia, a form of dwarfism, is inherited as a dominant condition. Two
dwarfs, both of whom had one parent of normal height, marry and plan to a have
child. Their first child is born and is a dwarf. What is the probability that this
child is heterozygous?
Since achondroplasia is a dominant allele, the two dwarf parents can either be
homozygous dominant (DD) or heterozygous (Dd). However, each of these dwarfs had a
parent that had normal height, (meaning that parents’ genotype must have been dd). So,
they could have only contributed d to the dwarfs in question, therefore the dwarfs
genotype must be Dd.
Setting up a punnett square will predict the potential offspring of these two dwarfs:
D
d
D
DD
Dd
d
Dd
dd
From this cross, there is a 3:1 ratio of dwarf to normal. Of the 3 phenotypically dwarf
individuals, 2 are predicted to be heterozygous.
Therefore the answer to the question is 2/3.
5. A pair of dominant alleles influence coat color in cattle wherein homozygotes for
one allele have red coats, homozygotes for the other allele have white coats, and
heterozygotes have roan coats that are produced by an intermix of both red and
white hairs.
Suppose that you start with a herd of 20 red cows, 40 roan cows, 20 white cows, one
red bull, two roan bulls, and one white bull. How you would manage your herd to
achieve the following goals: (you can choose which bulls will mate with which cows)
a. Maximize the number of roan calves produced.
(Could you achieve 100%?)
b. Minimize the number of roan calves produced.
(Could you achieve 0%?)
c. Produce only roan and red calves.
Based upon the information given in the problem, you should be able to deduce that coat
color is based on codominance.
CRCR = red coats; CWCW = white coats; CRCW = roan coats
a. You could mate the white bull (CWCW) with red cows (CRCR) or the red bull (CRCR)
with white cows (CWCW). Either of these situations will result in 100% roan calves
(CRCW).
b. If you mate the white bull with white cows or the red bull with red cows, you will only
get white and red calves, respectively. Yes, in this way you would achieve 0% roan
calves.
c. If you mate the red bull with roan cows, the only offspring you will get is red and roan.
CR
CR
CW
CRCW
CRCW
CR
CRCR
CRCR
6. Male pattern baldness is a recessive sex-linked trait on the X chromosome. A
woman, whose father had male pattern baldness, marries a man with this trait.
What is the probability that any son born will have pattern baldness?
First set up what we know based on the information provided:
Genotype
XBXB
XBXb
XbXb
XBY
XbY
Phenotype
normal, female
normal, female
bald, female (tendency for hair loss)
normal, male
bald, male
The male in this question is XbY, the woman has at least one Xb from her father and the
homozygous recessive condition is rare, therefore she is most likely XBXb.
XB
Xb
Xb
XbXB
XbXb
Y
XBY
XbY
The probability that a son will be born with pattern baldness is 1/2
7. In humans, normal skin color (A) is dominant over albino (a). A diabetic albino
man marries a normal woman whose mother was an albino and whose father was
diabetic. What are the genotypes of the man and the woman? What proportion of
their children would be expected to be both non-diabetic and have normal color?
We are looking at two traits in this question. The man is aadd. The woman was normal
(for both traits) so she must have at least one dominant allele for each trait: A__D__
She has a mother that is albino, so her mother could have only supplied her with the
recessive allele, a. Her father was diabetic, so he could only provide her with the
recessive allele, d. Therefore her genotype, with respect to these two traits is AaDd.
ad
AD
AaDd
Ad
Aadd
aD
aaDd
ad
Aadd
¼ of their children will be expected to be non-diabetic and have normal color.
8. A person with Rh+ blood has a specific protein in his/her blood. Persons with Rhblood do not have this particular protein in their blood. Rh + is dominant to Rh-.
Also, normal insulin production dominates abnormal insulin production. If two
individuals are heterozygous for Rh+ and normal insulin production, what probable
phenotypes might their children be?
The individuals in question are heterozygous for both traits, +-Ii
+I
+i
-I
-i
+I
++II
++Ii
+-II
+-Ii
+i
++Ii
++ii
+-Ii
+-ii
-I
+-II
+-Ii
--II
--Ii
-i
+-Ii
+-ii
--Ii
--ii
9 Rh+,normal: 3 Rh+, diabetic: 3Rh-, normal: 1 Rh-diabetic
9. Crosses between long and oval shaped radishes produced 159 long and 156 oval
types. Crosses between oval and round gave 203 oval and 199 round. However,
crosses between oval and oval gave 121 long, 243 oval and 119 round. How can you
explain these results?
These results are best explained by incomplete dominance. The oval phenotype is
intermediate to either long or round and if you cross individuals that are heterozygous,
oval, the resulting offspring will produce both of the original phenotypes as well as the
incomplete phenotype. Recall the red and white snapdragons!
10. A man with group A blood marries a woman with group B blood. Their child
has group O blood. What are the genotypes of these individuals? What other
genotypes and in what frequencies would you expect in offspring from this
marriage?
Again, list what we know based on the information provided. We know, that the ABO
blood type exhibits both a multiple allele phenomenon and codominance.
Genotype
IAIA
IAIO
IBIB
IBIO
IAIB
IOIO
Phenotype
A
A
B
B
AB
O
In order for someone to have type O blood, genotypically they must be IOIO. Therefore
they must have received one O allele from one parent and the other O allele from the
other parent. In this case the father MUST be IAIO and the mother MUST be IBIO.
To determine the genotypes in the offspring of these two individuals, you would set up
the following punnett square:
IA
IO
IB
IAIB
IBIO
IO
IAIO
IOIO
We would expect ¼ IAIB: ¼ IBIO: ¼ IAIO: ¼ IOIO