Download Sex Linked Problems - Mercer Island School District

Sex Linked Problems – Answer Key
1. If a human male and female produce children, what proportion of their offspring should be males? What
proportion should be females? Illustrates using a Punnett square.
Male = 50%
Female=50%
2. What conditions are necessary in order for colorblindness to appear in women?
Both X chromosomes carry the recessive allele XcXc.
3. In humans, normal vision (XC) is dominant to colorblindness (Xc) and is sex-linked. A normal-visioned
man, whose father was colorblind, marries a colorblind woman. What are the chances that a son will be
colorblind. A daughter? Explain.
C
X Ycrossed with XcXc
There is a 100% chance that the son will be color blind. The daughter will have normal vision
but carry the recessive XCXc trait.
4. Hemophilia is due to a sex-linked recessive gene (Xh) and the normal condition to the gene (XH). A
hemophiliac man marries a woman who is not. Their first son has hemophilia. What are the chances
that their daughter, if they had one, will be hemophilic?
There is a 50% chance their daughter will be a hemophilic. She can either receive an XH or an
Xh from her mom.
5. The “Porcupine Men” who appeared in England during the 18th and 19th centuries had their whole body,
except the palms, soles, and head, covered with cylindrical bristle-like outgrowths nearly an inch long.
The condition appeared when the child was seven or eight weeks old. Judging from the accompanying
pedigree, what type of inheritance is most likely involved?
Sex linked on the Y” chromosome.
NOTE: Circles represent females and squares males. The individuals who are shaded have the trait.
6. In humans, pseudohypertrophic muscular dystrophy is a condition in which the muscles gradually waste
away, ending in death in the early teens. In some families it is dependent upon a sex-linked recessive gene.
This type occurs only in boys and has never been reported in girls. Why is it not to be expected in girls?
In order for a female to have PHT, her parents must be a woman carrying the gene and a male
with the condition, this is impossible because the male would die before passing it on.
7. Why does the sex-linked gene for pseudohypertrophic muscular dystrophy not become eliminated from
the human race since all boys showing the trait die before reaching maturity?
If a woman carries the trait, she can have children with a man without the trait, and they can
have a daughter who carries the gene, who will then pass it to the kids, and so on.
8*. In a certain family there are two boys and two girls. On of the boys develops pseudohypertrophic muscular
dystrophy, and dies at 14 years of age. The other boy and the two girls grow up and marry. What are the
chances of their offspring showing this condition?
Assuming the boy does not marry a carrier, no chance of him passing it on. The females have a
1/8 chance of having a PMD son. ½ chance being a carrier X ¼ chance having infected son = 1/8.
9. Ichthyosis hystrix gravior (a greatly thickened horney condition of the skin) is a rare human abnormality, but
in the single extensive pedigree which has been studied it occurs only in males. All the sons of each affected
father have the condition. Females are not only unaffected, but never transmit the gene for this defect. Can you
suggest a possible explanation for this curious and unusual type of inheritance?
It is carries on the Y chromosome.
10*. In humans, aniridia (a type of blindness) is due to a dominant gene. Optic atrophy (another type of
blindness) is due to a recessive sex-linked gene. A man blind from optic atrophy marries a woman blind from
aniridia. Would any of their children be expected to be blind? Which type of blindness would they have?
Yes. Aniridia.
Assuming these disorders are very rare, we can assume the “unknown” alleles in the parent’s
genotypes code for non-blindness.
Aniridia –A, non-aniridia (normal) –a
Optic atrophy – Xb, non-optic atrophy (normal) – XB
XbY aa crossed with XB XB Aa
Only Aniridia can be passes on (aa X Aa). Half the children would receive the dominant allele
(A). For Optic atrophy, only females could be carriers since the mother passes on her normal X
(XB) to her sons and the father passes the blind allele to her daughters (Xb).
11. A normal-visioned man marries a normal-visioned woman whose father was color-blind. They have two
daughters who grow up and marry. The first daughter has five sons, all normal-visioned. The second daughter
has two normal-visioned daughters and a colorblind son. Diagram the family history, including the genotypes
of all the individuals mentioned. (Put diagram on the next page.)
Blood Grouping Problems – Answer Key
1. If a man with A type blood marries a woman of O type blood and they have five children, all of blood
type A:
a. What is the most probable genotype of the man?
A
A
I I
b. What is the genotype of the woman?
o o
i i
Of the children?
I A io
2. A friend of yours has B type blood. He knows his mother has O type blood.
a. What is his blood genotype?
o
I i
B
b. What genotypes of blood might his father have?
I I , I i , IA IB
With this basic information, find probably solutions to the following genetic problems:
B
B
B
o
3. What blood types might possibly result in children of a family whose mother has B blood and whose
father has AB blood?
B o
I i or IB IB x IA IB
B, A, AB
4. Suppose a father of blood type A and a mother of type B has a child of type O. What types are possible
in their children?
A o
I i x IB io A, B, AB, O
5. Suppose a father of type B and a mother of type O have a child of type O. What are the chances that
their next child will be type O? __1/2_______ Type B? _____1/2_____ Type A?
____O_______ Type AB? ____O________.
6. Assuming you do not know the blood type of your future husband or wife, but you do know yours, what
blood type might you possibly expect your children to have? What blood types can’t they possibly
have?
If type A could be IA IA
or IA io
- could have all possibilities
If AB, no chance of type O
If O, no chance of type AB
7. Two type AB parents took home a newly born type A baby from the hospital and decided it was not
their baby because it did not seem to resemble either. They claimed another couple had their baby. The
other parents were both type A and took home a type O baby. What is their solution to this problem?
Need further tests for the type A baby. Can’t determine from blood type alone. We know that
the type 0 baby is not the two type AB parent’s child.
8. You are the judge in a case in which a type O man claimed a $50,000 inheritance after the death of type
A and type AB parents.
a. What would be your decision? Explain.
No. Could not be related because AB parent does not have a recessive allele to pass on.
IA iBx IA io
b. What if the man had type B blood? Explain.
It is possible. Would need further tests
iB io x IA io
9. From question 7, make a pedigree which represents this problem. Label each individual with their blood
type and their genotype.