BI 164
Spring, 2008
Practice questions for Mendelian genetics
Answers to these problems will be posted next week. You are encouraged to work
through the problems first before you look up the answers. Try to work on the
problems in small groups.
1.
We know that the most common form of color blindness results from an Xlinked recessive gene. A couple with normal color vision has a daughter with
normal vision and a son who is color-blind. What is the probability that the
daughter is a carrier for the color-blindness allele? In other words, what is
the probability that the daughter is heterozygous for the trait?
Answer: We know the father has the N allele and the mother must be a heterozygote,
therefore we can set up the Punnett square as below:
Alleles in Sperm
Alleles in Egg
N
Y
NN
NY
Nn
nY
N
n
The daughters are the ones without the Y allele, so the probability is 50%.
2.
In the pedigree below, the shaded symbols represent people affected with a
neurological disorder caused by an X-linked recessive allele. Let’s call the
normal allele D and the recessive allele d. What are the genotypes of every
person in this pedigree?
Answer: Because the males cannot be carriers for an X-linked trait, fill in the males first.
Any unaffected males must have genotype DY. Affected males have genotype dY. Any
unaffected female that gives rise to an affected male must have genotype Dd.
BI 164
3.
Spring, 2008
The mutation for Bar-shaped eyes in Drosophila has the following
characteristics of inheritance.
a. Bar males x wildtype (round-eyed) females produce wildtype sons and
Bar daughters
b. The Bar females from the mating in part a, when crossed with wildtype
males, yield a 1:1 ratio of Bar: wildtype sons and Bar:wildtype daughters.
What mode of inheritance do these characteristics suggest?
Answer: Part a suggests that the gene is sex linked, because the cross affects the sexes
differently. It appears to be on the X chromosome rather than the Y, because the father
passes the trait to daughters rather than to sons. In addition, the Bar allele appears to be
dominant, because daughters with only one copy have the Bar phenotype.
We can show the hypothesis for part a like this (the wildtype is recessive):
b+b+
X
bY
b+b : b+Y
(wildtype females x Bar males)
(1:1 ratio of Bar females: wildtype males)
Part b is asking us to do a test cross, as follows (again, remember that b is dominant to
b+):
Alleles in Sperm
Alleles in Egg
b+
b
b+
Y
b+b+
(wild-type daughter)
b+b
(bar daughter)
b+Y
(wild-type son)
bY
(bar son)
BI 164
4.
Spring, 2008
In Shorthorn cattle, both red coat color and white coat color are true breeding.
Crosses of red cattle x white cattle produce offspring that are uniformly
reddish brown but thickly sprinkled with white hairs. This type of coat color
is called roan. Crosses of roan x roan produce 1/4 red : 1/2 roan: 1/4 white.
What kind of genetic inheritance can explain these results?
Answer: A 1:2:1 ratio reveals that the heterozygote is distinct from either homozygote.
Therefore the red is incompletely dominant over the white (if you consider the traits
blended) or the colors are co-dominant (if both expressed).
5.
The tailless trait in the mouse results from an allele of a gene on chromosome
17. A cross of tailless x tailless mice produces tailless and wildtype offspring
in a ratio of 2 tailless: 1 wildtype. All tailless offspring from this cross when
mated with wildtype produce a 1: 1 ratio of tailless to wildtype.
a. Is the allele for the tailless trait dominant or recessive?
b. What genetic mechanism can explain the results of these crosses?
Answer: A 2:1 ratio suggests that there is an allele that is lethal when homozygous.
The first cross shows that taillessness is dominant (because the cross recovers wild-type
progeny) and works correctly when both parents are heterozygous:
t
t+
t
tt
(lethal tailless)
t+t
(tailless)
t+
t+t
(tailless)
t+t+
(wildtype)
Here is the test cross, yielding 1:1 wildtype to tailless:
t+
t+
6.
t+
t+t+
(wildtype)
t+t+
(wildtype)
t
t+t
(tailless)
t+t
(tailless)
If a couple already has a daughter, what is the probability that the next child
will be a son? If the couple already had three daughters, what is the
probability that the next child will be a boy?
Answer: Assuming that the father produces sperm bearing X and Y chromosomes in a
50:50 ratio, and that the environment for fertilization does not favor one type of sperm
over the other, the probability of producing a son is always ½.
BI 164
7.
Spring, 2008
Consider three gene pairs, Aa, Bb and Cc, each of which affects a different
character. In each case, the uppercase letter signifies the dominant allele. The
three genes are located on different chromosomes. Calculate the probability
of obtaining:
a.
b.
c.
d.
An Aa BB Cc zygote from a cross of Aa Bb Cc x Aa Bb Cc
An Aa BB cc zygote from a cross of aa BB Cc x AA bb CC
An A B C phenotype from a cross of Aa Bb CC x Aa Bb cc
An a b c phenotype from a cross of AA BB CC x AA BB CC
Answer: The probability of each multi-locus genotype (when the loci are unlinked) is
the product of the probabilities of the genotypes for each locus, as in the table below. If
you are not sure how to calculate the probabilities for each locus, you could make a
Punnett Square for each locus and each cross.
Genotype
Aa BB Cc
Aa BB cc
A_B_C_
aa bb cc
Prob. at A locus
1/2
1
3/4
0
Prob. at B locus
1/4
0
3/4
0
Prob. at C locus
1/2
0
1
0
Joint probability
1/16
0
9/16
0
8.
In the jimsonweed, purple flowers are dominant to white. When a particular
purple-flowered jimsonweed is self-fertilized, the progeny are 27 purpleflowered and 10-white flowered progeny. What proportion of the purpleflowered progeny will breed true? What proportion of the white-flowered
progeny will breed true?
Answer: The 3:1 ratio suggests that the parent is a heterozygote, so the genotypes of the
progeny are in the ratio 1:2:1. All of the white-flowered progeny will breed true,
because the white is recessive. For the purple progeny, only the homozygotes will
breed true, and they constitute 1/3 of the purple plants.
9.
Continuing with jimsonweed, spiny pods (S) are dominant to smooth (s).
Consider pod shape and flower color together. The genes are on different
chromosomes. What mix of phenotypes would you expect from the following
crosses? See if you can answer the questions using probabilities rather than
constructing Punnett squares.
Cross
PP ss x pp SS
Pp SS x pp ss
Flower phenotypes
100% purple
½ purple: ½ white
Pod phenotypes
100% spiny
100% spiny
Combined
100% purple spiny
½ purple spiny, ½ white
BI 164
Spring, 2008
spiny
¾ purple spiny, ¼ white
spiny
Pp Ss x Pp ss
3:1 purple: white
1:1 spiny: smooth See below*
Pp Ss x Pp Ss
3:1 purple: white
3:1 spiny: smooth See below**
Pp Ss x pp ss
1:1 purple: white
1:1 spiny: smooth ¼ of each phenotype
* 3/8 purple spiny, 3/8 purple smooth, 1/8 white spiny, 1/8 white smooth
** 9/16 purple spiny, 3/16 purple smooth, 3/16 white spiny, 1/16 white smooth
Pp Ss x Pp SS
10.
3:1 purple: white
100% spiny
For the following pedigree, which of the following mechanisms can explain
the pattern of inheritance of the trait shown in the affected individuals (the
dark symbols). Assume the condition is caused by a single gene. Which
mechanism do you think is the most likely?
Answer: The only mechanism to reject easily is autosomal recessive (see below).
For the others, label the pedigree with hypothetical genotypes.
a. Autosomal dominant – label the pedigree, using T for the dominant allele
and t for the recessive “wild-type” allele. All of the solid individuals must
be heterozygous, and the pedigree is consistent with this hypothesis.
b. Autosomal recessive – no, because all progeny 1-6 of generation III would
be affected.
c. X-linked dominant – label the pedigree, giving all the males a Y and the
appropriate T or t, depending on whether they show the trait or not. All
of the females must be heterozygous, and the pedigree is consistent with
this hypothesis.
d. X-linked recessive – This time, label the pedigree using tt for affected
females and TY for unaffected males, etc. The parents in the first
generation will have genotypes TY and tt. It is impossible for them to
produce an unaffected male, because all male offspring would have
genotypes tY. Therefore, this genetic mechanism is incompatible with the
pedigree.
BI 164
11.
Spring, 2008
We know that human blood type is determined by a three-allele system at a
single locus. For the following, state whether the child mentioned can
actually be produced from the marriage.
a. An O child from the marriage of two A individuals
yes, IAi x IAi - ii
b. An O child from the marriage of an A to a B
yes, IAi x IBi - ii
c. An AB child from the marriage of an A to an O
no, no source of IB allele
d. An O child from the marriage of an AB to an A
no, IA IB can’t contribute i allele
e. An A child from the marriage of an AB to a B
yes, IA IB x IBi - IAi
12.
Phenyloketonuria is a caused by a recessive autosomal allele. In the pedigree
below, the female (dark circle) in the pedigree is affected with
phenylketonuria. If person III-1 and III-2 were to marry (they are first
cousins), what is the probability that their child will be affected?
Answer: Let R be the normal allele and r the allele that causes PKU. Parents 1 and 2 in
generation I must both be carriers (Rr), because their daughter has the disease. In
generation 2, the probability that siblings 2 and 4 are carriers is 2/3. (The genotypic
ratios are 1:2:1, and we can eliminate the rr genotype from consideration). For
generation III, the probabilities of being carriers are ½ that of the previous generation,
assuming that the outside mates have genotypes RR. One half of 2/3 is 1/3.
To find the probability that their child would be affected, multiply their probabilities of
being carriers with the probability that the child would have the disease if the parents
were both carriers: 1/3 * 1/3 * ¼ = 1/36