Genetics 320, Problem Set 12
Due Wednesay 8 December 2004 at 11 am
1 (1 point)
:
Consider a population of size
N
= 10
5
and a mutation rate of
u
= 10

6
,
(a)
Assuming the Infinite alleles model, what is the probability that a random individual
is a heterozygote?
Here
4
Nu
= 4
·
10
5
·
10

6
= 0
.
4
, hence
H
=
4
Nu
1 + 4
Nu
=
0
.
4
1
.
4
= 0
.
286
(b)
Recall that the number of mutations follows a Poisson distribution (See Problem 7
of PS 1).
Assuming that the common ancestor to two random sequences occurs
2
N
generations ago, what is the probability that two randomlychosen sequences differ by
one mutation? By two mutations? By five mutations?
If the time back to the common ancestor is
2
N
generations, then the expected number of
new mutations is
2
·
(2
N
)
·
u
= 0
.
4
, and the probability of
k
mutations follows a Poisson
distribution with parameter 0.4. Hence,
Prob
(1
mutation
) = (0
.
4)
1
e

0
.
4
/
1! = 0
.
268
.
Prob
(2
mutations
) = (0
.
4)
2
e

0
.
4
/
2! = 0
.
054
.
Prob
(5
mutations
) = (0
.
4)
5
e

0
.
4
/
5! = 0
.
000057
(c)
What is the probability for this population size that two randomlychosen alleles have
there most recent common ancestor in the last 10,000 generations? 25,000 generations?
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 Fall '07
 Walsh,Weinert
 Genetics, Evolution, Mutation, common ancestor, WAA

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