Download Practice genetics problems

PROBLEM 1.
Hypothetically, brown color (B) in naked mole rats is dominant to white color (b). Suppose
you ran across a brown, male, naked mole rat in class and decided to find out if he was BB
or Bb by using a testcross. You'd mate him to a white (totally recessive) female, and
examine the offspring produced. Now, if only 2-3 offspring were born and they were all
brown, you'd still be uncertain whether he was BB or Bb (for instance, even though the
odds are 50:50 that you will produce a boy or girl, there are plenty of people that produce 45 girls and never a boy and vice versa). But, if the mole rats produce 50 offspring and all are
brown, then it is likely that no hidden alleles are present and that the male is BB. But, what
if white offspring are produced? You'd know that the brown parent had a hidden little "b"
allele. So, what you need to do is perform a testcross on this brown, male, heterozygous,
naked mole rat. What are the expected genotypic and phenotypic ratios of such a cross?
PROBLEM 2.
What if you bred some snap dragons and crossed a homozygous red plant (RR) with a
homozygous white plant (rr)? In botony, "true breeding" means homozygous. In this case,
100% of the F1 individuals would be pink! This is an example of "incomplete dominance,"
where both alleles contribute to the outcome. In some cases of incomplete dominance, both
alleles might contribute equally so one allele would produce red pigment and the other
white; thus, a pink plant appears. In another case, one allele may be non-functional.
Although in many cases only a single allele is needed, perhaps in this case only one-half the
amount of needed pigment is produced and so pink is due the low amount of red pigment in
the petals. Who knows. Anyway, use a Punnett's square and set up a cross between a
homozygous red plant and a homozygous white plant. Then, take the resulting offspring
and cross these among themselves as well (i.e. F1 x F1). Then, determine the phenotypic and
genotypic ratios.
PROBLEM 3.
You know that the possession of claws (WW or Ww) is dominant to lack of claws (ww). You
also know that the presence of smelly feet (FF or Ff) is dominant to non-smelly feet (ff). You
cross a male who is clawed and has smelly feet with a female who is clawed and has nonsmelly feet. All 18 offspring produced have smelly feet, and 14 have claws and 4 are unclawed. What are the genotypes of the parents?
PROBLEM 4.
You have an individual who is totally heterozygous for 2 genes that are not linked (i.e., not
on the same chromosome). One gene is for ear size (AA or Aa being big ears whereas aa is
for small ears) and the other gene is for buggy eyes (BB and Bb for buggy eyes whereas bb
represents normal eyes). If you testcross this individual, what are the resulting genotypes
and phenotypes?
PROBLEM 5.
Now then, after you've completed the problem above, lets ignore the Punnett's square and
simply look at the 4 types of offspring from the above cross. What if the actual ratios in
your testcross were not 1:1:1:1, but were as follows. What would this represent?
PERCENTAGES
GENOTYPE
PHENOTYPE
48%
AaBb
Big ears, buggy eyes
2%
Aabb
Big ears, normal eyes
2%
aaBb
Small ears, buggy eyes
48%
aabb
Small ears, normal eyes
PROBLEM 6.
The following is a genetic linkage problem involving 4 genes. You want to determine which
of the genes are linked, and which occur on separate chromosomes. You cross two true
breeding (i.e., remember that this means that they are homozygous) plants that have the
following characteristics:
PLANT 1
PLANT 2
Red flowers
White flowers
Spiny seeds
Smooth seeds
Long pollen grains
Short pollen grains
Late blooming
Early blooming
Following the above cross, all of the offspring have red flowers, spiny seeds, long pollen
grains, and early blooming (meaning, that these traits are dominant). You then testcross the
F1 generation, which you should realize by now are totally heterozygous individuals, and
obtain the ratios below. What's going on?
49% red-spiny
25% red-long
25% red-early
25% long-early
1% red-smooth
25% red-short
25% red-late
25% long-late
1% white-spiny
25% white-long 25% white-early 25% short-early
49% white-smooth 25% white-short 25% white-late
25% short-late
PROBLEM 7.
The following is a genetic linkage problem also involving 4 genes. You want to determine
which of the genes are linked, which occur on separate chromosomes, and the distances
between the linked genes. You cross 2 true breeding (i.e. homozygous) plants that have the
following "unusual" characteristics:
PLANT 1
PLANT 2
Red flowers
White flowers
Long pollen grains
Short pollen grains
Dumb backtalk
Smart backtalk
Mean disposition
Nice disposition
All of the offspring have red flowers, long pollen grains, give smart backtalk, and have a
nice disposition (meaning, that these traits are dominant). You then testcross the F1
generation, and obtain the ratios below. How many chromosomes are involved in the
linkages, and what are the positions of the linked genes relative to one another?
45% red-long
25% red-dumb 25% long-dumb 48% red-mean 43% long-mean
5% red-short
25% red-smart
25% long smart
2% red-nice
7% long-nice
5% white-long 25% white-dumb 25% short-dumb 2% white-mean 7% short-mean
45% white-short 25% white-smart 25% short-smart 48% white-nice 43% short-nice
PROBLEM 8.
In the ABO blood system in human beings, alleles A and B are codominant and both are
dominant to the O allele. In a paternity dispute, a type AB woman claimed that one of four
men was the father of her type A child (the child would be type A with a genotype of either
be AA or AO). Which of the following men could be the father of the child on the basis of
the evidence given?
A. The Type A father?
B. The Type B father?
C. The Type O father?
D. The Type AB father?
PROBLEM 9.
A brown-eyed, long-winged fly is mated to a red-eyed, long-winged fly. The progeny are: 51
long, red ; 53 long, brown ; 18 short, red ; 16 short, brown Using solely the information
provided, what are the genotypes of the parents?
PROBLEM 10.
A strange woman has a bizzare condition known as "Cyclops" syndrome, where she has a
single eye in the middle of her forehead. The allele for the normal condition (i.e. NO
"Cyclops" syndrome) is recessive (cc). Her father is a Cyclops, as well as her mother. Her
father's mother was normal. What is the genotype of the strange woman's father?
PROBLEM 11.
In calico cats, there is an X-linked gene with 2 alleles that control fur color. BB is a black
female; B'B' is a yellow female; B'B (heterozygous) is a calico female; B' is a yellow male;
and B is a black male. You have recently taken over judge Wapner's job on the People's
Court and a woman brings in a black female cat that has given birth to 4 calico female
kittens and 2 black male kittens. You must decide which of the defendent's male cats is
guilty: the black one or the yellow one.
PROBLEM 12.
A common form of red-green color blindness in humans is caused by the presence of an Xlinked recessive allele. Given simply that, please answer the following:
A. Can two color-blind parents give birth to a normal son or daughter?
B. Can two normal parents produce a color-blind daughter?
C. Can two normal parents produce a color-blind son?
PROBLEM 13.
When studying an inheritance phenomenon, a geneticist discovers a phenotypic ratio of
9:6:1 among offspring of a given mating. Give a simple, plausible explanation of the results.
How would you test this hypothesis?
PROBLEM 14.
In an epistasis situation, PP or Pp is purple and pp is yellow. CC and Cc encode the ability
to produce color whereas cc prevents color production resulting in an albino (i.e., the C
allele either allows, or prevents, P from functioning to produce color). Given the following
parental matings, provide the ratios of the offspring that are either purple, yellow, or
albino. Remember: all offspring must have at least one big C to produce color or they will be
albino.
OFFSPRING RATIOS
PARENTAL
CROSSES
PPCC x PPCC
PPCC x ppcc
ppcc x ppCc
Ppcc x PpCc
purple
yellow
albino
PROVIDE EXPLANATIONS FOR
EACH OF YOUR ANSWERS
Monohybrid Cross:
In humans, brown eyes (B) are dominant over blue (b)*. A brown-eyed man marries a blueeyed woman and they have three children, two of whom are brown-eyed and one of whom is
blue-eyed. Draw the Punnett square that illustrates this marriage. What is the man’s genotype?
What are the genotypes of the children?
Testcross:
In dogs, there is an hereditary deafness caused by a recessive gene, “d.” A kennel owner has a
male dog that she wants to use for breeding purposes if possible. The dog can hear, so the
owner knows his genotype is either DD or Dd. If the dog’s genotype is Dd, the owner does not
wish to use him for breeding so that the deafness gene will not be passed on. This can be tested
by breeding the dog to a deaf female (dd). Draw the Punnett squares to illustrate these two
possible crosses. In each case, what percentage/how many of the offspring would be expected to
be hearing? deaf? How could you tell the genotype of this male dog? Also, using Punnett
square(s), show how two hearing dogs could produce deaf offspring. If two hearing dogs were
both Dd, what kind(s) of gametes (eggs/sperm) could each produce?
Dihybrid Cross:
In humans, there is a gene that controls formation (or lack thereof) of muscles in the tongue that
allow people with those muscles to roll their tongues, while people who lack those muscles
cannot roll their tongues. The ability to roll one’s tongue is dominant over non-rolling. The
ability to taste certain substances is also genetically controlled. For example, there is a
substance called phenylthiocarbamate (PTC for short), which some people can taste (the
dominant trait), while others cannot (the recessive trait). The biological supply companies
actually sell a special kind of tissue paper impregnated with PTC so students studying genetics
can try tasting it to see if they are tasters or non-tasters. To people who are tasters, the paper
tastes very bitter, but to non-tasters, it just tastes like paper. Let’s let R represent tongue-rolling,
r represent a non-roller, T represent ability to taste PTC, and t represent non-tasting.
Suppose a woman who is both a homozygous tongue-roller and a non-PTC-taster marries a man
who is a heterozygous tongue-roller and is a PTC taster, and they have three children: a
homozygous tongue-roller who is also a PTC taster, a heterozygous tongue-roller who is also a
taster, and a heterozygous tongue-roller who is a non-taster. If these parents would have a bunch
more children so that they had 12 in all, how many of those 12 would you expect to be nontasters who are homozygous for tongue-rolling? If the first child (the homozygous tongue-roller
who is also a PTC taster) marries someone who is heterozygous for both traits, draw the Punnett
square that predicts what their children will be.
Multiple Alleles and Codominance:
#1. Some genes have more than two alleles. One of the best-known examples is the gene that is referred
to as the “ABO Blood Group,” which actually has quite a number of alleles. However, we will
discuss/consider only the three most-common of these. This gene codes for the structure of a certain
antigen on the surface of our RBCs. The three alleles we will work with are symbolized by I A, IB, and i.
However, keep in mind that a person can only have two alleles, two copies of a gene. Thus, the possible
genotypes are IAIA, IAi, IBIB, IBi, IAIB, or ii. (Sometimes, you will see these simplified as AA, AO, BB,
BO, AB, and OO, but that does make it harder to remember that these are all alleles for the same gene.)
The allele, IA, codes for, “make type A antigen,” the allele I B codes for, “make type B antigen,” and (to
simplify things somewhat) the i allele codes for, “I don’t know how to make either A or B.” Thus, both
IAIA and IAi individuals receive instructions to “make type A antigen,” and both I BIB and IBi individuals
receive instructions to “make type B antigen.” Individuals who are I AIB receive two sets of instructions:
“make type A” and “make type B,” so they have both the A and B forms of that antigen on the surface of
their RBCs. People who are ii don’t have any instructions to make either A or B, so by “default” they
make what we refer to as type O antigens. Since both I A and IB code for “make something” whereas i
codes for, “I don’t know how, therefore, both I A and IB are dominant over i. However, since both I A and
IA code for “make something,” neither of them is dominant over the other. Thus we say that I A and IB are
codominant over i.
Suppose a person with type A blood and a person with type B blood get married. What are the possible
genotypes their children could have?
#2 There is another gene that codes for another, different antigen that also occurs on the surface
of our RBCs, and technically, that gene also has multiple alleles. However, most people either
have or do not have one particular allele called the “d” allele. This gene codes for an antigen
that is called “Rh factor” because it was first discovered in Rhesus monkeys. People who have
instructions to “make d antigen” are referred to as Rh+ (the allele is often symbolized by the
letter “R”), while those who have “I don’t know how to make d antigen” instructions are called
Rh– (the allele can be symbolized by “r”). Since this is a totally separate gene than the ABO
blood group, if you’re doing a genetic cross that involves both ABO and Rh, that would be a
dihybrid cross.
Ms. Johnston, Ms. Johnson, and Ms. Johnstone all entered the same hospital and gave birth to
baby girls on the same day, and all three babies were taken to the nursery to receive care, there.
Someone later claimed that the hospital mixed up the babies. As a hospital administrator, it is
your job to make sure that each pair of parents has the correct baby, so you order blood typing
to be done on all the parents and all the babies. Here are the results:
Blood Type Probable
Possible Gametes
Person
(Phenotype) Genotype
(Eggs/Sperm)
Ms. Johnston A+
Mr. Johnston B+
Ms. Johnson B–
Mr. Johnson
O+
Ms. Johnstone A+
Mr. Johnstone A–
Baby A
Baby B
Baby C
O+
AB–
B–
Question 1
For each of the diploid genotypes presented below, determine the genetic make up for all of the
possible haploid gametes.
a. Rr
b. RrYy
c. rrYy
d. RrYY
e. RrYyBb
Question 2
Use the Punnett square to determine all of the offspring genotypes (and their relative frequencies)
from the following crosses:
a. Rr x Rr
b. Rr x rr
c. RR x Rr
Question 3
In the problem above, the "R" allele is a dominant allele specifying for round seeds (in peas),
while the "r" allele is the recessive allele specifying for wrinkled seeds. Give the expected
frequencies (as percentages or ratios) for the phenotypes of the offspring resulting from each of
the crosses above.
Question 4
A brown mink crossed with a silverblue mink produced all brown offspring. When these F1 mink
were crossed among themselves they produced 47 brown animals and 15 silverblue animals (F2
generation). Determine all the genotypes and phenotypes, and their relative ratios, in the F1 and F2
generations.
Question 5
In sheep white is due to a dominant gene (W), black to its recessive allele (w). A white ewe
mated to a white ram produces a black lamb. If they produce another offspring, could it be white?
If so, what are the chances of it being white? List the genotypes of all animals mentioned in this
problem.
Question 6
In tomatoes the texture of the skin may be smooth or peach (hairy). The Ponderosa variety has
fruits with smooth texture. The red peach variety has fruits with peach texture. Crosses between
the two varieties produce all smooth fruits. Crosses between these smooth fruited F1 plants
produced 174 peach textured fruits and 520 smooth textured fruits. How are these skin textures
inherited?
Question 7
A brown mouse is mated is mated with two female black mice. When each female has produced
several litters of young, the first female has had 48 black and the second female has had 14 black
and 11 brown young. Deduce the pattern of inheritance of coat color and the genotypes of all of
the parents.
Question 1
Use the Punnett square to determine all of the offspring genotypes (and their relative frequencies)
from the following crosses (assume independent assortment):
a. RrYy x RrYy
b. RrYy x rryy
c. RrYy x Rryy
Question 2
In the problem above, the "R" allele is a dominant allele specifying for round seeds (in peas),
while the "r" allele is the recessive allele specifying for wrinkled seeds; in addition, the "Y" allele
specifies for the dominant yellow seed color trait, while "y" specifies for green seeds (recessive).
Give the expected frequencies (as percentages or ratios) for the phenotypes of the offspring
resulting from each of the crosses above.
Question 3
In turkeys a dominant gene R produces the familiar bronze color; its recessive allele r results in
red. Another dominant gene H results in normal feathers; its recessive allele h produces feathers
without webbing, so that they resemble tufts of hair. Two bronze turkeys with normal feathers
were mated, and their offspring consisted of 8 bronze with normal feathers, three bronze with
hairy feathers, two red with normal feathers, and one red with hairy feathers. What were the
genotypes of the parents?
Question 4
In horses black is dependent upon a dominant gene, B, and chestnut upon its recessive allele, b.
The trotting gait is due to a dominant allele T, and the pacing gait to its recessive allele, t. If a
homozygous black pacer is mated to a homozygous chestnut trotter, what will be the appearance
of the F1 generation?
Question 5
Referring to the previous question (Question 4), what would be the genotypes and phenotypes
(and their expected ratios or percentages) of the offspring produced by a mating between an F1
generation individual and a chestnut pacer?
Question 6
Assume right-handedness (R) dominates over left-handedness (r) in humans, and that brown eyes
(B) are dominant over blue (b). A right-handed, blue-eyed man marries a right-handed, browneyed woman. One of their two children is right-handed/blue-eyed, while the other is lefthanded/brown-eyed. The man marries again, and this time the woman is right-handed and browneyed. They have 10 children, all right-handed and brown-eyed. What are the genotypes of the
husband and two wives?
Question 1
Yellow guinea pigs crossed with white ones always produce cream colored offspring. Two cream
colored guinea pigs when crossed produced yellow, cream and white offspring in the ratio of l
yellow: 2 cream: l white. How are these colors inherited?
Question 2
In humans the blood groups are produced by various combinations of three alleles IA, IB, and i.
Blood type A is caused by either IA IA or IA i; type B by IB IB or IB i; type AB by IA IB; and type O
by i i. Suppose a child is of blood type A and the mother is of type 0. What type or types may the
father belong to?
Question 3
Suppose a father of blood type A and a mother of blood type B have a child of type O. What
blood types are possible in their subsequent children?
Question 4
Suppose a father of blood type B and a mother of blood type O have a child of type O. What are
the chances that their next child will be blood type O? Type B? Type A? Type AB?
Question 5
Suppose a father and mother claim they have been given the wrong baby at the hospital. Both
parents are blood type A. The baby they have been given is blood type O. What evidence bearing
on this case does this fact have?
Question 6
A mother and father with normal color vision produce six male children, two of whom exhibit
red-green colorblindness. Their five female children exhibit normal color vision. Ignoring the fact
that these parents ought to seek some family planning advice, explain the inheritance of red-green
colorblindness in their male children.
Sample Genetics Problems
1. Brad, who has curly hair, and Sandi, who has straight hair, have a child. If Brad's father had
straight hair, what are the odds that Brad and Sandi's child will have curly hair? (C = curly hair, c
= straight hair)
2. Nathan and Ashlee both have normal vision, but Ashlee's mother is color blind, while none of
Nathan's forebears have ever exhibited color blindness. If Nathan and Ashlee have a child, what
are the odds that the child will be color blind? (N = normal vision, n = color blind)
3. Nina's brother is a hemophiliac. What are the odds that she will have a hemophiliac child,
given that her husband Martin is not a hemophiliac? (R = regular clotting, r = hemophilia; sexlinked trait)
4. George has brown hair; Emily has red hair. Both of their mothers had blond hair. What are
the odds that George and Emily's child will be red-haired? (B = brown hair, r = red hair, b =
blond hair; B > r > b, meaning that r is recessive to B but dominant over b.)
5. Charles is double-jointed, though his mother was not. Marie has normally jointed fingers.
What can be predicted about their child? (D = double jointed, d = normal joints)
6. Brown-eyed Amelia and green-eyed Ichabod have fallen in love, but before getting married,
they want to assess the odds of having a blue-eyed child, since Amelia's mother, very wealthy but
very controlling, has promised to disown her if this should occur. Both of Amelia's parents have
hazel eyes; Ichabod's mother has blue eyes. Should the couple marry? (B = brown, G = green, b
= blue; co-dominant with BG = hazel)
7. Micaiah, a type I diabetic, has married Theophanie, who is not diabetic, though her mother
was. What are the odds that their child will develop type I diabetes? (D = normal, d = type I
diabetes; sex linked trait).
8. The earth has passed through a radiation cloud, producing a spate of mutations. One of these
results in a new set of hair colors: the allele for purple hair (P) is dominant over that for green
hair (G), which in turn is dominant over that for orange hair (o). The trait is sex linked. Avigdor,
who has purple hair, marries Clarinda, who has green hair, not knowing that Clarinda's mother's
hair was a bright shade of orange. What can be predicted about the hair color of their child?