Download PROBABILITY IN GENETICS

PROB ABIL ITY IN GE NET ICS
Independent events
In meisois, which partner chromosome migrates to a specific pole is
a random event. Thus alleles in a heterozygote always segregate to
different gametes and genes on different chromosomes should always
show independent assortment. Even gene on the same chromosome will
show random assortment unless they are "linked". Unless evidence
suggests otherwise, it is safe to assume genes are independent.
The joint probability of two independent events happening together is the
product of the individual probabilities. ie P(AB) = P(A) • P(B)
Examples of use:
gamete frequencies: given a genotype of Aa, Bb Cc dd, the probability
of an (a,b,c,d) egg or sperm is 1/2 •1/2 • 1/2 • 1 = 1/8
phenotypic and genotypic frequencies in progeny:
given the legends:
De_ normal
de/de endosperm
lethal, no seed
S_ starchy
ss sugary
D'D dwarf
D'D short
DD tall
Hs_ hairy sheath
hs/hs normal
If a De/de, Ss, D'D Hs/hs plant is self fertilized, what fraction of the
progeny will be a) De/De, Ss, D'D, hs/hs ?
Taking one gene at a time in the order listed, the answer is:
!/3 • 1/2 • 1/2 • 1/4 (the lethal de/de will not be a seed!)
b) Normal endosperm, sugary, tall, normal sheath?
1 • 1/4 • 1/4 • 1/4
If the events are not independent, then the formula changes to
P(AB) = P A • (B/A) or P(B) • P(A/B), where A/B the probability of event
A given that event
B has occurred, etc.
These are conditional
probabilities
A common use of conditional probabilities occur in pedigrees. If two
normal parents have a child with a recessive trait, we know that each
parent is heterozygous. We also know that an u naffected brother or
sister of the affected 'aa' child has a 2/3 chance of being heterozygous.
given the fact he or she is normal so can't be ‘aa'.
The probability that events A or B occur is P(A) + P(B) - P(AB)
If A and B are mutually exclusive the P(AB) = 0 so drops out.
Example: In humans the ability to taste PTC is dominant. A man and
woman are tasters but each had a nontaster parent. What is the
probability their first child will be a taster o r a son?
P(taster) = 3/4 and P(son) = 1/2 and the probability of a taster son is
3/4 • 1/2 or 3/8, so the answer is 10/8 - 3/8 = 7/8
Binomial and Multinomial Di stributions
Distributions of offspring in families generally deal with binomial or
multinomial possible outcomes.
The formula for the "general term" of a multinomial (trinomial shown)
is:
N!
P (a )r P (b )s P (c)t
r ! s! t !
where N is the total number of offspring, r is the number of one kind, s
anther, t another and a is one possible outcome, b and c and the others
(perhaps homozygous dominant, heterozygous, and homozygous
recessive).
N! = N•(N-1)•(N-2) ---•(1) ; so for example 5! = 5•4•3•2•1 = 120
0
By definition, 0! = 1 and (x) also = 1
Example problem: What fraction of families with 6 children will have 3
boys and 3 girls?
Answer:
3
3
6!/3! 3! (1/2) (1/2)
The second part of the equation shows the probability for one specific
birth order that gives the projected outcome (3 boys then 3 girls for
example) while the first part of the equation shows the number of birth
orders that give the same overall outcome of 3 boys and 3 girls.
Going back to the two heterozygous tasters, if they have 5 children,
what is the probability that 4 will be tasters and 1 a nontaster?
4
1
5!/4! 1! (3/4) (1/4)
If they have 4 children what is the probability they will all be tasters?
4
4!/4! 0! (3/4) (1/4)
0
A question that comes up in many situations is the probability of "at least
one". The easiest way to solve this question is to use the formula:
P(at least one) = 1-P(0)
For the cross Tt by tt, what is the probability that at least 1 of 6 progeny
will have the recessive phenotype? P(all dominant) = P(0 recessive) =
6
(1/2) = 1/64 or 0.0156; so P(at least one recessive) = 1-0.0156 = 0.9844
How many progeny should I examine if I want to be at least 99% certain
that the dominant parent is heterozygous?