Conditional Probability
Let P(B) equal the probability of event B;
P(A) equal the probability of event A;
P(B/A) equal the probability of event B, given that event
A has already occurred.
P(A ∩ B) is the probability that events A and B both occur.
P(A ∩
B) = P(B/A) P(A)
The probability of events A and B both occurring equals
the probability of event B (given that event A has already
occurred) multiplied by the probability of event A.
Example 1:
|
Brown
Pubs.
|
Grey
Pubs.
|
|
| Purple
flw. |
90
|
30
|
120
|
| White
flw. |
30
|
10
|
40
|
|
120 |
40
|
160
|
Let P(A) equal the probability that a soybean plant has
purple flowers. Let P(B) equal the probability that a
soybean plant has brown pubescence. Let P(A ∩
B) equal the probability that a soybean plant has both
purple flowers and brown pubescence.
Find the probability that a soybean plant has both
purple flowers and brown pubescence.
P(A) = 120/160 = 3/4; P(B) = 120/160 = 3/4;
P(A ∩ B)
= 90/160 = 9/16
Find the probability that a plant with purple flowers
will have brown pubescence.
Now P(B/A) is the probability that a soybean plant will
have brown pubescence, given that the plant has purple
flowers.
| P(B/A) = |
P(A ∩ B) |
= |
9/16 |
= 3/4 |
| P(A) |
3/4 |
We first must find P(B/A). P(B/A) is the probability that
a soybean plant has brown pubescence, given that the plant
is purple flowered.
|
Brown pubs.
|
Grey
pubs.
|
|
| Purple
flw. |
90
|
30
|
|
|
|
|
120
|
Assume we have a soybean population that is segregating
for flower and pubescence color. We pull out all the white
flowered plants from this population and will only have
purple flowered plants remaining. We have reduced the
sample space to only purple flowered plants.
P(B/A) = 90/120 = 3/4
Show that events A and B are independent. This
is the same as showing that the locus that controls flowering
color is not linked to the locus that determines pubescence
color.
In the special case where events A and B are independent,
P(A ∩ B)
= P(B) P(A).
We can determine whether events A and B are independent.
By definition, events A and B are independent when
P(A∩ B) = P(A) P(B). It is always true that
P(A∩ B) = P(A) P(B/A).
To test whether purple flower color and brown pubescence
are
independent we use P(A  B)
= P(A) P(B). In this case
P(A) = 3/4, P(B) = 3/4, P(A) X P(B) = (3/4)(3/4) = 9/16.
Previously, we found that the probability of a purple
flower and brown pubescence plant was equal to 9/16, so
events A and B are independent.
Now let's look at an example where two loci are not independent,
but are linked. We will show that we can not multiply
the probability of the genotype at the first locus times
the probability of the genotype at the second locus to
get the probability that the genotype has both traits
in the homozygous condition.
Example 2:
An SSR marker is co-dominant and is linked to a QTL conferring
resistance to Soybean Cyst Nematode (SCN). The marker
locus is Satt309 and let M indicate one marker band and
m indicate the second marker band. Then a Mm plant is
heterozygous for the Satt309 locus. The Satt309 locus
is located 1 cM from the rhg1 locus. The rhg1 locus is
a QTL for SCN resistance. Let Q designate the resistance
allele at the rhg1 locus and q designate the allele for
susceptibility to SCN. Then a Qq genotype is heterozygous
at the rhg1 locus. Consider an F2 population segregating
for the marker and QTL loci.
| P(M) = 1/2; P(Q) = 1/2; P(Q/m) = 1 - r; P(M∩Q) = P(M) X P(Q/M) = |
1 - r |
2 |
|
rhg1 |
MM
|
Mm
|
mm
|
|
| Satt309 |
|
|
|
|
|
| QQ |
|
245
|
5
|
0
|
250
|
| Qq |
|
5
|
490
|
5
|
500
|
| qq |
|
0
|
5
|
245
|
250
|
|
|
250
|
500
|
250
|
1000
|
Show that Satt309 and rhg1 loci are not independent.
Find the probability that marker assisted selection (MAS)
for the MM genotypes will identify genotypes that are
QQ for the QTL.
If the two loci are independent, we can multiply the separate
probabilities of each locus to get the probability that
a genotype will have a specific set of alleles.
P(QQ) = 250/1000 = 1/4
If two loci are independent, we can multiply the probability
of the genotype at locus one times the probability of
the genotype at locus two to get the joint probability
of the genotype at both loci.
P(A) P(B) =
P(A
B)
P(MM) P(QQ)
= P(MMQQ)
P(MM) P(QQ) = (1/4)(1/4) = 1/16 =0.0625
However, the P(MMQQ) = 245/1000 =0.245 which does not
equal 0.0625, therefore the Satt309 and rhg1 loci are
not independent.
Now to find the probability that an F2 plant with the
MM genotype has the QQ genotype. If we select MM plants,
we would like to know how successful MAS will be in identifying
homozygous SCN resistant plants.
| P(QQ/MM) = |
P(MM ∩ QQ) |
= |
0.245 |
= 0.98 |
| P(MM) |
0.25 |
98% of the plants identified with the MM genotype using
MAS will be homozygous resistant to SCN at the rhg1 locus.
Copyright
2000©, Ted Helms |
