Hypothesis Testing Part III
Mather
Example:
Mather, K. Measurement of Linkage in Heredity.
pgs. 8-9.
Aa x aa testcross ÷ 1/2 Aa : 1/2 aa progeny
P (Aa) = 1/2, P (aa) = 1/2 for zygotes
Probability of two Aa progeny = 1/2 x 1/2 = P (Aa) x
P (Aa)
Probability of two Aa progeny = 1/4
Probability of two aa progeny = 1/2 x 1/2 = P (aa) (Paa)
Probability of two aa progeny = 1/4
Probability of one Aa and one aa progeny
= P (Aa) P (aa) + P (aa) P (Aa)
= (1/2) (1/2) + (1/2) (1/2)
= 1/4 + 1/4 = 1/2
Probability for gametes uniting to form various types
of zygotes:
Families:
Probability of various types of familes of two zygotes,
given that P(Aa) = 1/2 = P(aa):
| (1/2 Aa + 1/2 aa)2 = |
1/4 (event two Aa progeny) |
| + |
1/4 (event two aa progeny) |
| + |
1/2 (event one Aa and one aa progeny,
order ignored) |
| = |
(1/2Aa + 1/2aa)(1/2Aa + 1/2aa) |
| = |
1/2Aa x 1/2Aa + 1/2Aa x 1/2aa + 1/2aa
x 1/2Aa |
| + |
1/2aa x 1/2aa = 1/4(Aa, Aa) + 1/2(Aa,
aa) + 1/4(aa, aa) |
| |
|
Now suppose we have a family of size three. We can
use the branching method:
What is the probability of all three progeny having
the aa genotype?
P
(aa) = 1/2 first child
P (aa)
= 1/2 second child
P (aa)
= 1/2 third child
Families: The probability of the first child being
of the aa genotype is independent of the probability
of a second child’s genotype and of the third child’s
genotype.
P (three
aa children in family of size 3)
= P(aa)
x P(aa) P(aa)= 1/2 x 1/2 x 1/2 = (1/2)3
= 1/8
We can find the expected proportion (probability) of
each type of family when the family includes three progeny.
(1/2
Aa + 1/2 aa) (1/2 Aa + 1/2 aa) (1/2 AA + 1/2 aa)
will give the probability of all combinations of genotypes
for a family of size 3. We can combine the probabilities
of some combinations that differ in the order of birth,
but produce families of the same type. For example,
we can combine families that have the order Aa, aa,
aa with families that have the order aa, Aa, aa and
the order aa, aa, Aa. The above families all have one
Aa and two aa progeny. To get all possible combinations
we use:
(1/2
Aa + 1/2 aa) (1/2 Aa + 1/2 aa) (1/2 Aa + 1/2 aa)
= (1/2 Aa + 1/2 aa)3
