BI164
Answers to Mendelian genetics questions
Spring, 2007
1.
The father has normal vision and must therefore be hemizygous for the normal
vision allele. The mother must be a carrier and hence the source of the colorblind allele of her son. The daughter must inherit the dominant allele from her
father. She has a 50% chance of inheriting the recessive allele from her mother.
Therefore the probability that the daughter is a carrier is 1/2.
2.
The father (I-2) does not show the disease and must therefore be DY. The mother
(I-1) must be a carrier of the disease so she is Dd. The affected son (II-2) must be
dY, having inherited the d allele from his mother. The other son (II-1) does not
show the disease and must be DY. The daughter (II-3) had a 50% chance of
inheriting the d allele from her mother. Her husband(II-4), from outside the
family, does not have the disease and is therefore DY. Their son (III-1) has the
disease and must be dY. He must have inherited the d allele from his mother so
we know that II-3 must be Dd.
3.
The results from part (a) indicate the mutation for Bar eyes is an X-linked trait.
Furthermore, these results suggest that the Bar allele is dominant. The daughters
of a Bar male must receive the Bar allele from their father. Since their mothers do
not have the Bar allele, the daughters are heterozygotes and the Bar allele is
expressed. Sons must receive a Y chromosome from their father and a wildtype
allele from their mother. The hemizygous males therefore show the wildtype
eye.
We can test our hypothesis with a test cross, yielding the results in part b. The
Bar females in the F1 are heterozygotes. The males are hemizygous for the wild
type allele. Each offspring must receive the wildtype allele from the father. Half
of the offspring will receive the Bar allele from the mother and half will receive
the wildtype allele. For daughters, half will be Bar/wildtype and half will be
homozygous for wildtype, yielding a 1:1 ratio of Bar to wildtype. For sons, the
males must receive the Y chromosome from their father. Half of the sons will
receive a Bar allele from the mother and half will receive a wildtype allele, again
resulting in a 1:1 ratio of Bar to wildtype.
4.
These results suggest that incomplete dominance is the type of inheritance in
these cattle. The roan cattle are intermediate in coat color. When roan is crossed
with roan, we expect one red (homozygote) to two roan (heterozygote) to one
white (homozygote).
5.
The allele for taillessness is dominant. A cross of tailless x tailless mice produces
more tailless mice than wildtype mice.
The ratio of 2:1 brings to mind lethal genes. We know from the above information
that tailless mice do not breed true. The expected ratio of three tailless mice to one
wildtype mouse is not observed. The most likely explanation is that a homozygote
for the tailless allele is a lethal combination. The surviving tailless mice do not
have that lethal combination because they are heterozygous. The heterozygotes
outnumber the wildtype mice 2:1, as expected.
6.
The chance that a child will be a girl or a boy is independent of other offspring.
The probability of the next child being a boy is 1/2 in both cases.
7.
Calculate the probability of obtaining:
a. An Aa BB Cc zygote from a cross of Aa Bb Cc x Aa Bb Cc
Chance of Aa – 2/4
Chance of BB – 1/4
Chance of Cc - 2/4
Since these genes assort independently, the overall probability is the product
of the three individual probabilities = 1/2 x 1/4 x 1/2 = 1/16
b. An Aa BB cc zygote from a cross of aa BB Cc x AA bb CC
Chance of Aa – 1/1
Chance of BB – 0/1
Chance of cc - 0/1
Therefore, for two reasons, there is no way to realize the Aa BB cc genotype
in this cross.
c. An A B C phenotype from a cross of Aa Bb CC x Aa Bb cc
Chances of A_ - 3/4
Chances of B_ - 3/4
Chances of C_ - 1/1
Overall probability - 3/4 x 3/4 x 1 = 9/16
d. An a b c phenotype from a cross of AA BB CC x AA BB CC
There is no source of recessive alleles for any of the loci and therefore the
phenotype cannot be produced.
8. The results suggest a simple two-allele, single locus system. The results conform
to the expected 3:1 ratio of phenotypes for a monohybrid cross.
The only true-breeding purple flowers will be the homozygotes (PP). They make
up a third of the purple flowers. All of the white-flowered progeny will be true
breeding.
9. Dihybrid cross problems.
a. PP ss x pp SS
Flower phenotype
100% purple
Pod phenotype
100% spiny
Combined traits
100% purple flowers, spiny pods
Flower phenotype
50% purple
Pod phenotype
100% spiny
Combined traits
1/2 purple flowers, spiny pods
50% white
100% spiny
1/2 white flowers, spiny pods
Flower phenotype
75% purple
Pod phenotype
100% spiny
Combined traits
3/4 purple flowers, spiny pods
25% white
100% spiny
1/4 white flowers, spiny pods
Pod phenotype
50% spiny
50% smooth
50% spine
50% smooth
Combined traits
3/8 purple flowers, spiny pods
3/8 purple flowers, smooth pods
1/8 white flowers, spiny pods
1/8 white flowers, smooth pods
b. Pp SS x pp ss
c. Pp Ss x Pp SS
d. Pp Ss x Pp ss
Flower phenotype
75% purple
25% white
e. Pp Ss x Pp Ss
Flower phenotype
75% purple
25% white
Pod phenotype
75% spiny
25% smooth
75% spiny
25% smooth
Combined traits
9/16 purple flowers, spiny pods
3/16 purple flowers, smooth pods
3/16 white flowers, spiny pods
1/16 white flowers, smooth pods
f. Pp Ss x pp ss
Flower phenotype
50% purple
50% white
Pod phenotype
50%% spiny
50% smooth
50%% spiny
50% smooth
Combined traits
1/4 purple flowers, spiny pods
1/4 purple flowers, smooth pods
1/4 white flowers, spiny pods
1/4 white flowers, smooth pods
10.
Autosomal dominant
If we assume I-2 is heterozygous for the trait, we can find no part of the
pedigree that cannot be explained by inheritance as an autosomal dominant.
Autosomal recessive
We can reject this mechanism because II-1 and II-2 produced a son without
the trait.
X-linked dominant
If this mechanism is correct, the grandfather I-1 must be hemizygous for the
recessive allele and the grandmother I-2 must be heterozygous because their
children include both a male and female without the trait. II-2 must have
married a woman heterozygous for the trait (II-1) because they produce sons
with and without the trait. II-6 is hemizygous for the trait. His wife (II-7)
must be homozygous recessive. All of their daughters must inherit a
dominant allele from the father and the sons will lack the trait since they must
inherit their only allele from their mother.
X-linked recessive
We can reject this mechanism because female I-1 or II-1 could only have sons
with the trait. II-5, III-1 and III-4 lack the trait so we know the trait cannot be
an X-linked recessive.
In summary, we have two possible mechanisms of inheritance for this trait.
We cannot distinguish between them with the present information.
However, the fact that every daughter of a man with the trait also has the trait
indicates that X-linked dominance is the more likely mechanism. With
autosomal dominance, one would expect half of the daughters of fathers II-2
and II-6 to have the trait and half to lack it.
11. ABO blood type questions
a. An O child from the marriage of two A individuals
This result is possible if both parents are heterozygotes (IAi)
b. An O child from the marriage of an A to a B
Also possible if both parents are heterozygous (IBi and IAi)
c. An AB child from the marriage of an A to an O
Not possible. The parent with O blood cannot contribute the B allele
needed.
d. An O child from the marriage of an AB to an A
Not possible. The O child must be ii and the AB parent has no i allele to
contribute.
e. An A child from the marriage of an AB to a B
Possible if the B parent is heterozygous (IBi).
12. For each of the first cousins, one parent had a sibling with the disease. We know
that their grandparents both had to be heterozygous for the PKU gene since
neither had the disease but one of their children did have the disease. Therefore
the two siblings (II-2 and II-4) have two chances in three they have the disease
(since they are not homozygous recessive, there are three genotypes –
homozygous dominant, heterozygous with PKU gene inherited from the mother,
heterozygous with the PKU gene inherited from the father).
The chance that each first cousin (III-1 and III-2) inherited the PKU gene from
her/his parent is 1:2 by Mendel’s Law of Segregation. So, the probability that
each first cousin is a carrier for the PKU allele is 2/3 x 1/2 = 1/3
When the two cousins marry, four genotypes are possible but only one will cause
the disease (homozygous recessive). So, the overall probability is:
Probability III-1 has the PKU gene x probability that III-2 has the PKU gene x
probability of producing a homozygous recessive = 1/3 x 1/3 x 1/4 = 1/36