Download Answer Key to Problem Set 5

The Saylor Foundation’s “Answer Key to Problem Set 5”
1. What is the molecular basis of sickle cell disease? What is the mutation that
occurs and how does this result in sickle-shaped red blood cells?
Sickle cell is caused by a mutation of the HBB gene. The HBB gene is
responsible for production of beta globin, a protein subunit of hemoglobin.
Abnormal beta globin results in distorting of the red blood cell into the sickle
shape.
2. If Eric's mother does not have the sickle cell trait, what is the probability that Eric
has the trait?
Eric’s father has the trait (HbS/HbA), and Eric’s mother does not have the trait
(HbA/HbA). Therefore, Eric has a 50% chance of being a carrier for the mutated
hemoglobin gene, and there is a 0% chance that Eric has sickle cell disease.
The Punnett square below illustrates the possible outcomes of a fertilization even
between one of the mother’s eggs and one of the father’s sperm.
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Rather than writing out the entire sequence above, you can simply use the
Punnett square to predict the probable outcomes.
3. If Eric’s mother does have the sickle cell trait, what is the probability that Eric has
the trait?
Eric’s father has the trait (HbS/HbA) and if Eric’s mother has the trait (HbS/HbA),
there is a 50% chance of being a carrier for the mutated hemoglobin gene.
4. Why are certain groups more likely to carry the sickle cell trait?
In areas where malaria is prevalent, people with the sickle cell trait have a
survival benefit. The slight sickling shape permits some resistance to malaria
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parasite. It is most common in people with recent ancestry to people with
African or Mediterranean descent.
5. Sickle cell disease was once very likely to lead to mortality prior to reproductive
age. If this is the case, how has sickle cell trait remained so common? Shouldn't
natural selection have reduced its frequency in the human population?
Being a carrier for the trait (HbS/HbA) usually does not affect ones normal
behaviors or health and in some cases can actually be a benefit (see malaria
example from question 5). Therefore, being a carrier for the trait increased
fitness. Additionally, modern medicine and treatments help alleviate symptoms
and reduce episodes, allowing people to live into and past reproductive age.
6. Should the NCAA require testing of student-athletes? Why or why not?
Answers will vary. Though your answer is your opinion, it should be justified with
evidence.
7. What are the three possible genotypes of their children?
HbS/HbA
HbS/HbS
HbA/HbA
8. What are the possible phenotypes?
HbS/HbA = Carrier of trait; slight sickling to some red blood cells
HbS/HbS = Has sickle cell disease. Red blood cells are sickled.
HbA/HbA = Normal hemoglobin production and normal red blood cells
9. Is sickle cell dominant, recessive, or co-dominant? Justify your answer.
Sickle cell is co-dominant. If a person receives an allele for sickle cell (HbS) from
one parent and allele for regular hemoglobin production (HbA) from the other
parent, then they will express both traits.
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10. If Eric does have the sickle cell trait, what is the probability that their first child
would have the sickle cell trait?
If Eric has the trait (HbS/HbA) and Gina has the trait (HbS/HbA), there is a 50%
chance that their first child will be a carrier for the mutated hemoglobin gene.
See the Punnett square below, which shows the possible outcomes.
11. If Eric has the sickle cell trait, what is the probability that their second child would
have the sickle cell trait?
Each fertilization event is independent of any previous event. Therefore, if Eric
has the trait (HbS/HbA) and Gina has the trait (HbS/HbA), there is a 50% chance
their second child will be a carrier for the mutated hemoglobin gene.
12. If Eric does not have the sickle cell trait, what is the probability that their first
child would have sickle cell trait?
If Eric does not have the trait (HbA/HbA) and Gina has the trait (HbS/HbA), there
is a 50% chance their first child will be a carrier for the mutated hemoglobin
gene. See the Punnett square below for the possible outcomes.
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13. If Eric has the sickle cell trait, what is the probability that their first child would be
unaffected by sickle cell?
If Eric has the trait (HbS/HbA) and Gina has the trait (HbS/HbA), there is a 25%
chance their first child will unaffected. See the Punnett square below for possible
outcomes.
14. If Eric has the sickle cell trait, what is the probability that their first child would
have sickle cell disease?
If Eric has the trait (HbS/HbA) and Gina has the trait (HbS/HbA), there is 25%
chance that child will have sickle cell disease. See the Punnett square below for
possible outcomes.
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15. If Eric and Gina decide to have 2 children, and he has the sickle cell trait, what is
the probability that both of their children would have sickle cell disease?
If Eric has the trait (HbS/HbA) and Gina has the trait (HbS/HbA), there is a 6.25%
chance that both of their children would have sickle cell disease.
0.252 = 0.0625 = 6.25%
16. If Eric and Gina decide to have 2 children, and he has the sickle cell trait, what is
the probability that both of their children would have sickle cell trait?
If Eric has the trait (HbS/HbA) and Gina has the trait (HbS/HbA), there is a 25%
chance that both of their children would have sickle cell trait.
0.502 = 0.25 = 25%
17. If Eric and Gina decide to have 2 children, and he has the sickle cell trait, what is
the probability that at least one of their children would have sickle cell disease?
If Eric has the trait (HbS/HbA) and Gina has the trait (HbS/HbA), there is 43.75%
chance that at least one child will have the disease.
The possible phenotypes for each child are listed across the top and down the
left side of the matrix below (this is not a Punnett square). The boxes within the
matrix are the possible outcomes after having two children (e.g. both children
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with the disease, one with the disease and one carrier of the trait, etc.).
Diseased (D)
Carrier (C)
Healthy (H)
Diseased (D)
Carrier (C)
Healthy (H)
DD (6.25%)
DC (12.5%)
DH (6.25%)
CD (12.5%)
CC (25%)
CH (12.5%)
HD (6.25%)
HC (12.5%)
HH (6.25%)
18. If you were Gina's and Eric's doctor, what would you tell them in order for them
to make an informed decision regarding whether or not to have children?
I would tell Eric and Gina about the probabilities of having a child with normal
hemoglobin production, a child with the sickle cell trait, and a child with sickle cell
anemia. I would then give Eric and Gina information about the health risks
associated with sickle cell, the prognosis, the treatment options, and treatment
outcomes.
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