Biology 202
Problem Set 4
2/16/07
From Griffiths:
Chapter 4, problems 1, 2, 4, 5, 6, 7, 9, 12, 13, 14, 17, 21, 24, 25,
29, 30
Answers to the following questions are due in February 28
th
1. Two linked genes, (
A
) and (
B
), are separated by 18 cM.
A man with genotype
AB
/
ab
marries a woman who is
ab
/
ab.
a. What is the probability that their first child will be
Ab
/
ab
?
The recombination frequency between these gene loci is 18%.
The man is thus expected
to produce a total of 18% recombinant gametes, which includes two reciprocal products
(9%
A
/
b
+ 9%
a
/
B
).
The mother produces only one kind of gamete (
ab
).
For the first
child to be
Ab
/
ab
, he/she must inherit a specific recombinant gamete (
A
/
b
) from the father
(probability = 0.09), and any gamete (
a
/
b
) from the mother (probability = 1).
Therefore,
the final probability of the first child being
Ab
/
ab
= (0.09)(1) = 0.09 (or 9%).
b. What is the probability that their first two children are
ab
/
ab
?
The father is expected to produce a total of 18% recombinant gametes (9%
A
/
b
+ 9%
a
/
B
), and 82% nonrecombinant gametes (41%
A
/
B
+ 41%
a
/
b
).
Use the multiplicative rule
to determine the probability of the couple’s first two children being
ab
/
ab
: (0.41)(1) ×
(0.41)(1) = 0.1681 (or 16.8%).
2. You just bought two black guinea pigs from the pet store that are known to be
heterozygous (
Bb
).
You also know that black fur (
BB
) is dominant over white fur (
bb
),
and that a lethal recessive allele is located only one cM away from the recessive
b
allele.
You decide to start raising your own guinea pigs, but after mating these animals several
times you discover they only produce black progeny.
a.
How would you explain this result?
The recessive lethal allele is tightly linked to, and thus cosegregates with, the recessive
b
allele, which is lethal in the homozygous state (
bb
).
b.
If the original black guinea pigs produce an average of 10 offspring per
mating, how many matings would you have to make before you’d expect to see a
white guinea pig?
1% of the gametes will be recombinant (Bl or bL)
0.5% of the gametes will be Bl
0.5% of the gametes will be bL
white animals can have the following genotype:
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probability
bL/bL
(0.005)(0.005) = 0.000025
bl/bL
(0.495)(0.005) = 0.002475
bL/bl
(0.005)(0.495) = 0.002475
probability of white animal =
0.004975
1/0.004975 = 201 therefore on average you’d have to look at 201 progeny before seeing a
white animal so on average you’d have to do 21 matings.
c.
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 Spring '08
 KieberHogan
 Biology, molecular biology, Genetics, 1%, 16.8%, CEN, 13.4%

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