Homework 11 - KEY
BIO 2
1.
Amy and Todd have two children. Their first child, Mike, is normal. However, their second
child, Sylvia, has cystic fibrosis.
a.
Draw a pedigree of this family. Indicate the genotypes and phenotypes of each person
in the family. If you cannot determine the genotype of a person with absolute
certainty, indicate all the possibilities.
Todd
(Cc; normal)
Amy
(Cc; normal)
Mike
(CC or Cc; normal)
b.
Sylvia
(cc; diseased)
Amy and Todd want to have another child. What is the probability that the child will be
a boy with cystic fibrosis? Show your calculation.
0.5 = Probability child will get disease allele from father
0.5 = Probability child will get disease allele from mother
(0.5)(0.5) = 0.25 = Probability child will get allele from both parents
(0.5) = Probability child will be a boy.
(0.5)3 = 0.125 = Probability child will be a boy with cystic fibrosis
2.
Juanita has a brother, Carlos, with X-linked hemophilia. However, her mother and father are
both healthy.
a.
What is the probability that Juanita is a carrier (heterozygote) for the disease?
Indicate how you arrived at your answer.
Carlos must have received the mutant allele on the X chromosome he got
from his mother. Since she is healthy, she must be a carrier. If she is a
carrier, Juanita has a 50% chance of having inherited the mutant allele from
her. Therefore, there is a 50% chance that Juanita is a carrier.
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b.
What is the probability that Juanita’s first child will be affected if the father of the child
is a man who does not have the disease? Indicate how you arrived at your answer.
0.5 = Probability Juanita is a carrier
Since the father of the child is healthy and the disease is X-linked, he cannot
have the mutant allele (h)
XH
Y
XH
XHXH
XHY
Xh
XHXh
XhY
There is a 25% chance that the first child will be affected if Juanita is a
carrier. (0.5)(0.25) = 0.125 (= 1/16)
c.
Carlos gets married and his wife gets pregnant with their first child. What is the
probability that the child will be a boy with hemophilia? Assume that Carlos’ wife is not
a carrier for the disease. Use a Punnett Square to support your answer.
Xh
Y
XH
XHXh
XHY
XH
XHXh
XHY
The chance is 0%. The couple cannot have a child of either sex who has the
disease.
3.
Among unicorns, horn color is under the control of one gene with two alleles that exhibit
incomplete dominance: GG = gold, GG’ = brass, and G’G’ = silver. Horn length is under the
control of one gene with two alleles that exhibit complete dominance and recessivity, where L
= long and l = short.
a.
A pure-breeding male unicorn with a long gold horn is mated with a pure-breeding
female unicorn with a short silver horn. Use a Punnett Square to predict the
phenotype(s) and genotype(s) of their offspring.
LG
lG’
LlGG’
All offspring will be LlGG’ with long brass-colored horns.
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b.
If one of the offspring of the mating in part a is mated to another unicorn with same
genotype, what fraction of their offspring will have long gold horns? Use the
multiplication rule to arrive at your answer, not a Punnett Square.
The genotype of the long gold-horned offspring is L_GG.
From a mating between two unicorns with the genotype LlGG’, the probability
of L_ = 3/4 and the probability of GG is 1/4.
(3/4)(1/4) = 3/16
4.
Your cat, Tricky, has had a litter of 8 kittens:
2 black males
1 black female
1 tortoise-shell female
1 gray female
1 gray male
2 orange males
If you randomly pick up one of the kittens, what is the chance that you will pick up either a
black male or a tortoise-shell female? Show your calculation.
Probability of picking a black male = 2/8
Probability of picking a tortoise-shell female = 1/8
These events are not independent; if one type of kitten is picked, the other one
won’t be.
2/8 + 1/8 = 3/8
5.
Micah’s sister was born with Tay Sachs Disease, a fatal autosomal recessive disorder. Micah
and his wife, Connie, want to have a child. Assuming that Connie is a known carrier for Tay
Sachs, what is the probability that their first child will have the disease? Be very careful in
determining the probability that Micah is a carrier – this problem is trickier than it might
appear!
If Micah’s sister is affected, then his parents must both be carriers. Normally, a child
of two carriers would have a ¼ chance of being cc, ½ chance of being Cc, and ¼
chance of being CC. However, we know that Micah is not cc since he is normal.
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Therefore, cc drops out of the probability and Micah has a 2/3 chance of being a
carrier.
The probability of Connie being a carrier is 1.
The probability of two carriers having an affected child (in any pregnancy) is ¼.
(2/3)(1)(1/4) =1/6
6.
Pattern-baldness is caused by an autosomal mutation that is dominant in males and recessive
in females. Carolyn’s maternal grandmother was pattern-bald. However, neither her mother
nor father is pattern-bald. What is the probability that at least one of her two sons will be
pattern-bald? Show your calculation.
Note: In writing this question, I should have told you that Carolyn’s husband is not
pattern bald. (I did not grade this question.)
If Carolyn’s maternal grandfather was pattern-bald, then she must have been B’B’
(where B’ is the mutant allele). Carolyn’s mother must be B’B – and since the
disorder is recessive in females, she is normal. Carolyn’s father must be BB since he
is normal. Therefore, Carolyn has a ½ chance of being a carrier for the mutant allele
and a ½ chance of passing the mutation to each of her children.
She has two sons. If the sons get the mutant allele from her they will be pattern
bald. The possible ways that at least one of her sons would be affected are:
First son affected, second son normal (1/2)(1/2) = 1/4
First son normal, second son affected (1/2)(1/2) = 1/4
First son affected, second son affected (1/2)(1/2) = 1/4
These scenarios are mutually exclusive. Therefore, the probabilities should be added
to determine the overall risk.
The probability that if she is affected she will have at least one affected son is 3/4.
Since Carolyn’s chance of being affected is ½, the overall probability that she will
have at least one affected son is (1/2)(3/4) = 3/8.
7.
If Margaret and Greg are both carriers for sickle-cell anemia and decide to have four children,
what is the probability that:
a.
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All four children will have the disease?
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(1/4)4 = 1/256
b.
None of the children will have the disease?
(3/4)4 = 27/256
c.
All four children will be diseased girls?
1/256= Chance all four children will be diseased
(1/2)4 = 1/16 = Chance all four children will be girls
(1/256)(1/16) = 1/4,096
d.
Only one of the four children will have the disease?
There are four ways that only one of the four children would have the
disease (where N = normal and D = diseased)
DNNN = (1/4)(3/4)3 = 27/256
NDNN = (1/4)(3/4)3 = 27/256
NNDN = (1/4)(3/4)3 = 27/256
NNND = (1/4)(3/4)3 = 27/256
Since these are mutually exclusive, we add the probabilities together:
= 108/256 = 27/64
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