Download Genetic Problems

Genetic Problems
Greg Burchett
Riverside Community College
Genetic problems are word problems. You should first read the entire question carefully. The most
important thing is to understand what the problem is telling you, what the problem wants, and then decide
how you will get there.
1. Gamete production:
When you perform genetic problems, you need to first determine the gametes that the “parents” can produce.
This will determine how your punnet square is set up. Below are ten “parents”, and each of their unique
genotypes. On a piece of paper, show all of the different kinds of gametes which could be produced by the
following individuals. For example, the parent in “b” can make two different gametes, and the “j” parent can
make eight different gametes. Be sure to show them all! At the end of this handout are some selected
answers, so you can see if you are on the right track.
1.
2.
3.
4.
5.
aa
Bb
AA
AABB
CcDd
6. AABb
7. Aabb
8. aaBb
9 CCDdee
10. AaBbCc
Monohybrid cross: A monohybrid cross basically means that you are crossing two individuals, while
looking at only one trait (the term “mono” basically means one). You must be able to construct a punnet
square, and be able to determine both the genotypic and phenotypic ratios. Both of the questions below are
monohybrid crosses.
2. In pea plants, the genes for tall plants are dominant over the genes for short plants. Cross a homozygous
dominant tall pea plant with a short (homozygous recessive) pea plant. What percentage of short pea plants
will appear in the offspring of this cross?
3. In dogs, wire hair is due to a dominant gene (W), and smooth hair is due to its recessive allele (w).
A. If a homozygous wire-haired dog is mated with a smooth-haired dog, what would be the offspring’s
genotypes and phenotypes?
B. What type of offspring could be produced in the F2? Give genotypic and phenotypic ratios.
C. Two wire-haired dogs are mated, and among the offspring of their first litter is a smooth-haired pup. If
these two wire-haired dogs mate again, what is the chance that they will produce another smooth-hair pup?
What are the chances that the pup will be wire-haired?
D. A wire-haired male is mated with a smooth-haired female. The mother of the wire-haired male was
smooth-haired. What are the phenotypes and genotypes, and their ratios, that they could produce?
Dihybrid cross: where two traits are studied simultaneously. You will need to know how to arrive at the
phenotypic ratio, and how to construct a punnet square. Questions #4 and #5 are dihybrid crosses.
4. In pea plants, tall (T) is dominant over short (t) plants, and round seeds (R) is dominant over wrinkled
seeds (r).
A. Cross two heterozygous tall, round seeded plants. What types of offspring do you get?
B. From the above problem, choose any two offspring (your choice) and cross them. This will give you the
F2 generation for this problem. Remember, give both genotypic and phenotypic ratios. Your answers will
depend on your initial crossings.
5. In humans, the presence of freckles is due to the 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, one has dimples but no
freckles.
A. What are the genotypes and phenotypes of the parents?
B. What are the possible genotypes and phenotypes, and their ratios, of the children they could produce?
C. What are the chances that they would have a child which lacks both dimples and freckles.
D. A person with freckles a
nd dimples whose mother lacked both freckles and dimples marries a person with freckles but no dimples,
whose father lacked both freckles and dimples. What are the chances that they would have a child which
lacks both freckles and dimples?
Co-dominance: The genetics of ABO blood grouping in humans is an example of co-dominance. This is
where each trait inherited in any individual doesn’t express itself over another. The gene that produces the A
antigen and the gene that produces the B antigen on the red blood cells (RBC’s) are incompletely dominant
with the other, yet both are completely dominant over the third recessive allele which codes for the absence
of either the A or B antigen (type O blood).
IA The gene that produces the B antigen, and produces type A blood.
IB The gene that produces the A antigen, and produces type B blood.
I The gene that codes for the absence of A or B antigens, and produces type O blood.
Possible genotypes and phenotypes:
Type A blood
IAIA or IAi
Type B blood
IB IB or IBi
Type AB blood
IAIB
Type O blood
ii
Type B antigen
Type A antigen
No antigens
Type A and B antigens
Fill out the following table. Using the top row as a recipient patient, could they receive blood from the
column of donors on the left. In example, on the top row, if you have AB+ blood, you could not receive
AB+ blood from someone else.
A+
A+
AB+
BAB+
ABO+
O-
A-
B+
B-
AB+
AB-
NO
O+
O-
6. Cross a heterozygous Type A individual with a heterozygous Type B individual. What are the chances of
generating a Type O child from this cross?
Incomplete dominance: Color in snapdragons is a good example, where there is no complete dominant
genotype, and the offspring are like a “mixture” of both parents. In snapdragons, red flower color is
incompletely dominant over white flower color, the heterozygous plants have pink flowers. But remember,
in pink flowers, the genes for both red and white are still present. Question #7 is an incomplete dominance
problem.
7. Cross a red-flowered plant is crossed with a white-flowered plant.
A. What are the genotypes and phenotypes of the plants in the F1 generation?
B. What genotypes, phenotypes and ratios can be produced in the F2 generation?
C. What kinds of offspring can be produced if a red-flowered plant is crossed with a pink-flowered plant?
Give both genotypic and phenotypic ratios.
D. What kinds of offspring can be produced if a pink-flowered plant is crossed with a white-flowered plant?
Give both genotypic and phenotypic ratios.
Sex-lined problems: In this example, the genes are located on the X chromosome. Thus, a female has to be
homozygous recessive (on both X chromosomes) to express a recessive condition. The male however, has to
only have one recessive trait (on his only X chromosome) in order for this same recessive condition to be
expressed. Color blindness is an example of a condition that is sex-linked. The gene for normal vision (XC)
is dominant for the recessive gene (Xc). Since a male has only one X chromosome, whichever gene he has
will be expressed. But for a woman, if she is homozygous (XC XC) or heterozygous (XC Xc), she will have
normal vision. Only if she is homozygous recessive (Xc Xc) will she be color blind.
8. Two normal visioned people marry, and together, they have a color blind son.
A. What are the possible genotypes and phenotypes of the children that they could produce?
B. What are the kinds of offspring which it would be impossible for this couple to produce?
Now you have looked at all of the different kinds of crosses that you will be expected to understand for this
class. The following problem is a situation that arises quite often in everyday life. It is your job to look at
all of the information that I give to you, and make the correct decision of how to solve the problem. Good
luck!
9. Mr. & Mrs. Truth have a daughter, Verily, born on November 31, 1999, in the Metropolitan Community
Hospital. Mr. & Mrs. Consequences have a daughter, Natural, who was also born on November 31, 1999, in
the Metropolitan Community Hospital. Mrs. Truth, having recently seen Natural Consequences, is
convinced that there had been a baby mix-up in the nursery, and that Verily is really the daughter of the
Consequences, and that Natural is really the progeny of the Truths. Blood analyses done on all individuals
of this situation reveal the following:
Mr. Truth
Mrs. Truth
Verily
Group A
Group A
Group A
Mr. Consequences
Mrs. Consequences
Natural
Group AB
Group A
Group O
Given these data, determine to whom Natural belongs to: The Truths or Consequences!
Selected Answers
1. Selected answers for gametes: (a) a, (d) AB, (f) AB and Ab, (i) CDe and Cde. You can figure out the rest.
2. In this example, both individuals only produce one type of gamete, a “T” gamete for the first, and a “t”
gamete for the second. Therefore, the only possible offspring that a cross of these two individuals can be a
heterozygous “Tt” individual, which are tall. Since short plants are “tt,” the answer is 0%.
3a. All of the offspring will be Ww, so their genotype is Ww, and their phenotype is wire-haired.
3b. Wired-haired and smooth-haired. Genotypic ratio (1:2:1), phenotypic ratio (3:1).
3c. Since the two wire-haired dogs had a smooth-haired pup, you know for a fact that both parents are Ww.
In this example, another cross would have a 25% chance of having another smooth-haired pup, and a 75%
chance of having a wire-haired pup.
3d. You know the father is wire-haired, but you do not know if he is homozygous or heterozygous. Since
his mother is smooth-haired, you now know that he is heterozygous. You can now cross him to his smoothhaired mate, and get a 1:1 genotypic and phenotypic ratio.
4a. In both these individuals, their genotype is TtRr. Each offspring therefore produces four different
gametes (TR, Tr, tR, tr). The Punnet square is therefore four gametes by four gametes (16 possible
offspring). The phenotypic ratio for this problem works out to 9:3:3:1. Remember, not all dihybrid
problems are automatically 9:3:3:1. You must completely do the problem, then give me the answer.
5. See if you can figure these out on your own. Remember, I give you all of the important information that
you need to figure out the genotypes of the parents, and from this you can figure out any of the children.
6. In this example, the Type A individual produces two gametes (IA or i), and the Type B individual
produces two gametes (IB or i). The Punnet square is therefore a two by two square. The answer is 25%.
7a. 100% are pink (RW).
7b. Cross two RW individuals, and you will get a 1:2:1 genotypic and phenotypic ratio.
7c. 1:1 genotypic and phenotypic ratio of red and pink.
7d. 1:1 genotypic and phenotypic ratio or pink and white.
8. The two gametes that the father will produce are XC and Y, and the mother will produce two as well, XC
and Xc (you know this because both are normal, yet they had a color blind son (XcY)). The answer is that
they possibly could have a color blind son, a normal son, and two daughters (one homozygous dominant, and
the other heterozygous).
9. Have fun! :-)