SOLUTIONS FOR GENETICS PROBLEMS - BIL 151
1.
Most humans (and vertebrates, in general) are genetically "programmed" to
produce a brown skin pigment known as melanin. Albinism--the inability to produce
melanin--is inherited in humans as an autosomal recessive trait. An individual with
genotype AA or Aa will have normally pigmented skin, whereas an individual with
genotype aa will produce no melanin. The albino's skin is very pale white, and the
(highly vascularized) irises of the eyes appear red because no melanin obscures the red
reflectance of blood. Predict the frequencies of all genotypes and phenotypes expected
to result from the cross of an albino male with a heterozygous female.
Parental genotypes: male:
aa
female:
Aa
Possible gametes of the male: a only
Possible gametes of the female: A or a
Punnett Square will yield 50% Aa (heterozygous; pigmented) and
50% aa (homozygous; albino)
a
2.
A
Aa
a
Aa
Do the same for a mating of two heterozygotes.
Possible gametes of either parent: A or a
Punnett Square will yield 25% AA (homozygous pigmented) 50%
Aa (heterozygous; pigmented) and 50% aa (homozygous; albino)
A
a
A
AA
Aa
a
Aa
aa
3.
What can you say about the genetic constitution of the parents of an albino
child?
Both parents must carry at least one albino allele for a child to
be albino.
4.
Given what you know about human A-B-o blood types, what parental cross
would yield an F1 consisting of 50% type A, 25% type AB, and 25% type B?
These are the typical ratios expected from a cross between an
individual with type AB blood and an individual with type A (Ao)
blood
A
o
5.
girl
A
AA
Ao
B
AB
Bo
As a genetic advisor to a hospital you are confronted with the following problem:
two women had babies in the hospital at about the same time on the same day.
Mrs. Cartman took home a boy (named Happy), and Mrs. Simpson took home a
(named Joy). However, Mrs. Simpson thought she recalled the delivery room
nurses commenting on her child as being a boy. In any case, she requests
immediate clarification of the situation.
The blood types of all concerned parents were determined as follows:
Mr. Cartman = B
Mr. Simpson = B
Mrs. Cartman = AB
Happy = A
Mrs. Simpson = o
Joy = o
Has there been an interchange of babies? Support your conclusion or be fired.
Mrs. Cartman can have ONLY the genotype AB. Her gametes
will be either A or B.
Mr. Cartman could be either BB or Bo.
possible outcomes:
B
B
A
AB
AB
B
BB
BB
B
o
A
AB
Ao
B
BB
Bo
So there are two
Or…
The Cartmans could not have produced a type o baby, as Joy is.
There has been no exchange of babies.
Just to be sure, let's look at the Simpsons' genotypes and
possible outcomes.
Mrs. Simpson can have ONLY the genotype oo. Her gametes
can carry ONLY the o allele.
Mr. Simpson could be either BB or Bo. Again, there are two
possible outcomes:
o
B Bo
Or…
o
Bo
oo
B
o
Joy is their baby.
(Actually, you could have avoided all this work if you'd just been
astute enough to notice that Mrs. Cartman is the only one in the
bunch with an A allele. She is the only one of the four parents
who could have provided the A allele that her son Happy
6.
The polled (hornless) condition in cattle (H) is dominant over the horned
condition (h) and is autosomal. A polled bull (Spike) is bred to three different cows
with the
following results:
Faye is horned (hh). She produces a polled calf.
Holly is horned (hh) She produces a horned calf.
Clover is polled (Hh). She produces a horned calf.
What are the genotypes of each animal?
Spike:
Holly:
Hh
hh
Faye:
Clover:
hh
Hh
What offspring would you expect from each of these matings? Punnet squares:
Spike x Faye…
h
H
Hh
h
hh
h
Hh
hh
h
h
H
Hh
Hh
h
hh
hh
H
h
H
HH
Hh
h
Hh
hh
Spike x Holly…
Spike x Clover…
7.
In cats, yellow fur (B) is dominant to black fur (b). The heterozygous "tortoiseshell" or "calico" condition is exhibited only by the female, and is a result of mosaic
expression. The alleles B and b are sex-linked. What offspring (genotypes, phenotypes
and frequencies) would you expect from the cross of a black male with a yellow female?
Remember: the allele occurs only on the X chromosome.
XB
Xb XBXb
Y XBY
XB
XBXb
XBY
Offspring are expected to be 100% yellow if male, and 100%
calico if female.
8.
What kinds of offspring would you expect to result from the cross of a black male
with a tortoise-shell female?
XB
Xb XBXb
Y XBY
Xb
XbXb
XbY
Females will be 50% calico, 50% black.
Males will be 50% yellow, 50% black.
9.
How about the cross of two tortoise-shell animals?
This is highly unlikely, as the only male tortoise-shell animal
would have Klinefelter's Syndrome, and have a sex chromosome
genotype of XXY (or even more X chromosomes than two).
If such a mating were to occur, however, the colors of the
offspring would be extremely difficult to predict, as the
aneuploid sex chromosomes would migrate in various
unpredictable ways during meiosis.
Trust us. Just get all those cats neutered or spayed and don't
even try this.
10.
In guinea pigs rough hair (R) is dominant with respect to smooth hair (r), and
black hair (B) is dominant with respect to white (b). These genes are not linked (i.e., the
locus for hair color is not on the same chromosome as the locus for hair texture), nor
are they sex-linked. Cross a guinea pig heterozygous for hair texture and color, with a
white-smooth-haired guinea pig. Describe the predicted F1 in terms of phenotype and
genotype frequencies.
Heterozygous guinea pig: RrBb
White, smooth haired guinea pig: rrbb
RB
Rb
rb RrBb Rrbb
rB
rrBb
rb
rrbb
Expected phenotypes:
25% rough black
25% rough white
25% smooth black
25% smooth white
11. Over several years, a pair of guinea pigs produced the following offspring
25% Rough-black
25% Smooth-black
25% Rough-White
25% Smooth-White
Describe the phenotypes and genotypes of each parent.
They are the same guinea pigs in the first part of the question.
RrBb and rrbb.
12. Alas! Your prize potato patch is infested with greedy, potato-munching leperchauns!
Not to
be confused with the more attractive and popular "leprechaun" (Homunculus
patrickii), the "leperchaun" (Homunculus odoratus) has a strongly disagreeable
odor, and extremities that fall off messily when the organism is handled. This is an
effective defense mechanism.
At your friendly neighborhood BillyBob's Bargain Barn of Biological Control, you
purchase a pair of special, mutant ferrets--each of which is true breeding. This
strain of ferret is highly prized by potato farmers because of two simple, autosomal
traits affecting (1) coat color and (2) appetite for leperchauns.
Coat color has two alleles. Black fur (B) is dominant to the more desirable brown
fur (b), which camouflages the ferrets as they lie in ambush in the potato patch. Diet
preference in these voracious carnivores also has two alleles. Unfortunately, the
allele which drives the ferrets to chase and devour leperchauns (p) is recessive to
the gene which drives them to chase and devour your ankles (P).
Your pair consists of a true-breeding black, ankle-biting male and a truebreeding brown, leperchaun-eating female. What type of offspring do you expect to
obtain in the F1?
Recall that "true breeding" for a particular trait means that the
organism in question is homozygous for that trait.
A true-breeding black ankle biter will be BBPP
A true-breeding brown leprechaun-eater will be bbpp
The F1 generation of this crossing should be 100% BbPp
What about the F2?
The F2 generation is the result of a dihybrid cross of two
individuals in the F1 generation:
BP
Bp
bP
bp
BP
BBPP
BBPp
BbPP
BbPp
Bp
BBPp
BBpp
BbPp
Bbpp
bP
bp
BbPP BbPp
BbPp Bbpp
bbPP bbPp
bbPp bbpp
Black ankle biter is coded in BLACK
Black leperchaun-eater is coded in BLUE
Brown ankle biter is coded in PINK
Brown leperchaun-eater is coded in GREEN
It's the typical 9:3:3:1 ratio you'd expect!
13.
If a black, ankle-biting ferret is selected from the F2 progeny above, what is the
probability that it will breed true in succeeding generations?
Of the nine black, ankle-biting weasels in the above cohort, only
ONE (BBPP) will breed true in succeeding generations. There's
a one in nine chance.
14.
Over several months, a black, ankle-biting ferret and a brown, ankle-biting ferret
yield the following phenotypic ratios in their offspring:
64
18
black, leperchaun-eating
black, leperchaun-eating
59
23
brown, ankle-biting
brown, leperchaun eating
What are the genotypes of the parents?
Since there are ANY brown, leperchaun eating offspring, you
know that the black anklebiter must be heterozygous for both
traits, and that the brown ankle-biter is heterozygous at the P
locus. Hence, the two parents have the genotypes:
BbPp and bbPp, respectively. Doing a Punnett square for these
genotypes, you get…
BP
Bp
bP
bP BBPP BbPp bbPP
bp BbPp Bbpp bbPp
bp
bbPp
bbpp
As you can see, this does reflect the percentages (out of 164 total
babies) of the offspring these two weasels produced over their
production times:
37.5% black ankle-biters (3/8)
12.5% black leperchaun-eaters (1/8)
37.5% brown ankle-biters (3/8)
12.5% brown leperchaun-eaters (1/8)
15. When investigators study human genetics, they often must examine "family
trees,"
or pedigrees. The following page shows two different pedigrees, each
following the expression of a single genetic trait. To figure out the type of
inheritance involved, consider each generation separately, and determine
phenotypic ratios in each set of offspring. If you can "work" the following
human pedigrees, you will have mastered several fundamental aspects of
Mendelian genetics!
Let us make the assumption that only four types of inheritance are involved in
the pedigrees described.
autosomal dominant
autosomal recessive
X-linked dominant
X-linked recessive
The symbols used are:

= male

= female
When the symbol is filled in, the individual expresses the trait. When the
symbol is empty, the individual does not express the trait.
For example, the figure below represents a marriage in which the female exhibits
the trait, but the male does not. The symbols on below the "parental" symbols
represent the children of this marriage. In this case, both sons exhibit the trait,
but the daughter does not.
Pedigrees are constructed by a process of elimination. Begin by writing down
all possible genotypes for each individual. Rule out each type of inheritance as it
becomes evident that, from the information given, a certain type of inheritance is
not possible. If a certain type of inheritance "fits" the pedigree, it means only that
that type of inheritance is possible in the given situation--not that you are
absolutely correct.
On the next page you will find two pedigrees, each typical of a particular type
of inheritance (autosomal dominant, autosomal recessive, X-linked dominant, Xlinked recessive, Y-linked dominant, or Y-linked recessive). Examine the
pedigree, and fill in the most likely genotypes for each individual on the family
tree. Finally, given what you have hypothesized about the family, write the type
of inheritance in the space provided below each pedigree.
Pedigree analysis: The family trees
Pedigree I.
Type of inheritance:
Autosomal dominant
If this were autosomal recessive, you would not expect individual #7 to not
express the trait. Note that matings between non-expressing individuals
always produce non-expressing offspring.
This tells you that the
expressed trait is dominant.
Pedigree II
Type of inheritance:
X-linked recessive
Note that only the sons of the affected mother #1 inherit the trait, as they receive
the X chromosome only from their mother.
Individual #4 is a carrier; one of her sons has inherited the trait. The mating
between #7 and #8 reveals that this is an X-linked recessive trait, since all the X
chromosomes in this family carry the recessive allele, and all individuals, either
homozygous (females) or hemizygous (males) will express it.