Number of Progenics Required to Establish the Genotype
of a Phenotype
Suppose we want to determine whether an A_ phenotype
is the AA or Aa genotype. We self-fertilize A_. How
many progeny of the A_ plant must be evaluated to determine
the genotype of this A_ plant with a specified probability
of erroneously rejecting Ho (Type I error).
Ho:plant is Aa
HA:plant is AA
a = 0.05
We can identify the genotype of the A_ phenotype by
observing a single aa offspring.
Let q = failure to observe a homozygous recessive offspring
of A_.
Example:
How many progeny must be evaluated to determine whether
A_ is in fact the Aa genotype with a
= 0.05 probability of erroneously concluding that an
Aa plant has the AA genotype?
q = probability to fail to observe homozygous recessive
offspring = 3/4.
p = probability of observing a homozygous recessive
offspring = 1/4 = probability of success.
q = 3/4 because each time we observe either an AA or
Aa progeny we fail to observe aa progeny.
The probability of n failures to observe an aa offspring
is a, when plant is actually
Aa genotype.
=
qn
log(P) = n log(q)
Because: When we fail to observe an offspring r = 0.
We want this probability to be
when Ho is true.
| n! |
(q)n-r(p)r = qn when r = 0 this equation simplifies to = qn |
| (n-r)!r! |
| n = |
log(P) |
= |
log(0.05) |
= |
-1.301 |
| log(q) |
log(0.75) |
-0.1249 |
n = 10.4 ~
10
If we want to be 99% certain of identifying a homozygous
recessive progeny when the parent is heterozygous, we
set P = 0.01.
P = qn
| n = |
log(P) |
= |
log(0.01) |
= |
-2 |
| log(q) |
log(0.75) |
-0.1249 |
= 16
