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S1.A heterozygous pea plant that is tall with yellow seeds,
TtYy,
is allowed to selffertilize. What is the probability that
an offspring will be either tall with yellow seeds, tall with green seeds, or dwarf with yellow seeds?
Answer:
This problem involves three mutually exclusive events, and so we use the sum rule to solve it. First, we must
calculate the individual probabilities for the three phenotypes. The outcome of the cross can be determined using a
Punnett square.
P
Tall with yellow seeds
= 9/(9 + 3 + 3 + 1) = 9/16
P
Tall with green seeds
= 3/(9 + 3 + 3 + 1) = 3/16
P
Dwarf with yellow seeds
= 3/(9 + 3 + 3 + 1) = 3/16
Sum rule: 9/16 + 3/16 + 3/16 = 15/16 = 0.94 = 94%
We expect to get one of these three phenotypes 15/16, or 94%, of the time.
S2. As described in chapter 2, a human disease known as cystic fibrosis is
inherited as a recessive trait. A normal couple’s first child has the disease.
What is the probability that their next two children will not have the disease?
Answer:
A phenotypically normal couple has already produced an affected child. To be affected, the child must be
homozygous for the disease allele and, thus, has inherited one copy from each parent. Therefore, since the parents are
unaffected with the disease, we know that both of them must be heterozygous carriers for the recessive diseasecausing
allele. With this information, we can calculate the probability that they will produce an unaffected offspring. Using a
Punnett square, this couple should produce a ratio of 3 unaffected : 1 affected offspring.
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View Full Document The probability of a single unaffected offspring is
P
Unaffected
= 3/(3 + 1) = 3/4
To obtain the probability of getting two unaffected offspring in a row (i.e., in a specified order), we must apply the
product rule.
3/4
×
3/4
= 9/16
=
0.56 =
56%
There is a 56% chance that their next two children will be unaffected.
S3. A pea plant is heterozygous for three genes (
Tt Rr Yy
), where
T
= tall,
t
= dwarf,
R
= round seeds,
r
= wrinkled
seeds,
Y
= yellow seeds, and
y
= green seeds. If this plant is selffertilized, what are the predicted phenotypes of the
offspring and what fraction of the offspring will occur in each category?
Answer:
One could solve this problem by constructing a large Punnett square and filling in the boxes. However, in this
case, there are eight possible male gametes and eight possible female gametes:
TRY, TRy, TrY, tRY, trY, Try, tRy,
and
try.
It would become rather tiresome to construct and fill in this Punnett square, which would contain 64 boxes. As an
alternative, we can consider each gene separately and then algebraically combine them by multiplying together the
expected phenotypic outcomes for each gene. In the cross
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This note was uploaded on 04/07/2009 for the course BIO 325 taught by Professor Saxena during the Spring '08 term at University of Texas at Austin.
 Spring '08
 SAXENA
 Genetics

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