Genetics 320, Problem Set Nine (15 points)
1 (2 points)
:
Consider a locus with four alleles (
A
1
to
A
4
) at the following frequencies:
p
1
= 0
.
1
,
p
2
= 0
.
2
,
p
3
= 0
.
3
,
p
4
= 0
.
4
. Assume random mating and all other HardyWeinberg
conditions. What is the frequency of
(a)
A
2
A
3
?
2
p
2
p
3
= 0
.
12
(b)
A
4
A
4
?
p
2
4
= 0
.
16
(c)
All heterozygotes involving allele
A
2
?
2
p
2
(1

p
2
) = 0
.
32
(d)
All heteroyzgotes?
1

(
p
2
1
+
p
2
2
+
p
2
3
+
p
2
4
) = 0
.
70
(e)
Of a mating between
A
1
A
2
and
A
3
A
3
parents.
Prob(
A
1
A
2
) =
2
p
1
p
2
= 0
.
04
, Prob(
A
3
A
3
)
=
p
2
3
= 0
.
09
, Hence, Prob(
A
1
A
2
×
A
3
A
3
mating) = Prob( father =
A
1
A
2
, mother =
A
3
A
3
) + Prob( father =
A
3
A
3
, mother =
A
1
A
2
)
= 2 (0
.
04
·
0
.
09) = 0
.
0072
.
2 (2 points)
:
Let
p
(0) = freq(
A
)
in the initial generation (which we call generation 0). As
suming HardyWeinberg conditions:
(a)
What is the frequency of
AA
homozygotes in the next generation (generation 1)?
p
(0)
2
(b)
What is the frequency of heterozygotes involving
A
in generation 1?
2
p
(0) [ 1

p
(0) ]
(b)
Given these genotype frequencies, show that
p
(1) = freq(
A
in generation 1)
=
p
(0)
.
p
(1) =
freq(
AA
in gen 1) + (1/2)freq(
Aa
in gen 1) =
p
(0)
2
+ (1
/
2)2
p
(0) [ 1

p
(0) ] =
p
(0)
2
+
p
(0)

p
(0)
2
=
p
(0)
3 (2 points)
:
This problem examines what happens when there are different allele frequencies
in the two sexes (as would occur if one mates males from one population with females from a
second). Assume an autosomal locus, where the frequency of allele
A
in the male and female
founding populations is
p
and
r
, respectively. Assuming all other HardyWeinberg conditions
hold,
(a)
What is the frequency of
AA
homozygotes in their offspring?
p
·
r
(b)
What is the frequency of
A

heterozygotes?
p
·
(1

r
) + (1

p
)
·
r
(c)
What is the frequency of allele
A
in their male offspring? In their female offspring?
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 Spring '09
 walsh
 Genetics, Population Genetics, Freq, m m

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