Probability of inheriting a specific allele
Remember: The specific allele which ends up being inherited
from a parent to an offspring is random.
Since a parent has two alleles, and each allele is equally likely
to be inherited by an offspring, the probability that an offspring
will get a specific allele = 0.5.
This is equally true for both parents.
Probabilities for second filial generation
(Crossing first filial generation)
Example 1: Suppose you breed a pair of plants with pink flowers.
What are the possible genotypes and phenotypes of the offspring?
RR (Red), RW (Pink), and WW (White)
Question: What is the probability of getting these genotypes?
Phenotypes?
Answer:
P(RR) = P(R) ♂ × P(R) ♀ = 0.5×0.5 = 0.25
P(WW) = P(W) ♂ × P(W) ♀ = 0.5×0.5 = 0.25
P(RW) = P(R) ♂ × P(W) ♀ + P(W) ♂ × P(R) ♀ = 0.5×0.5 + 0.5×0.5 = 0.25 + 0.25 = 0.5
P(Red) = P(RR) = 0.25 P(White) = P(WW) = 0.25 P(Pink) = P(RW) = 0.5
Example 2: What if we did the same thing with eye color?
The genotype probabilities are exactly the same!
P(BB) = P(B) ♂ × P(B) ♀ = 0.5×0.5 = 0.25
♀ = 0.5×0.5 = 0.25
P(b) ♂ × P(B) ♀ = 0.5×0.5 + 0.5×0.5 = 0.25 + 0.25 = 0.5
But the phenotypic probabilities are different:
P(Blue) = P(bb) = 0.25
P(Brown) = P(BB) + P(Bb) = 0.25 + 0.5 = 0.75
P(bb) = P(b) ♂ × P(b)
P(Bb) = P(B) ♂ × P(b) ♀ +
Crosses Involving Two Genes
A cross that involves two independent traits is termed dihybrid cross.
Example: Suppose
Gene 1 has alleles A and a (A is dominant over a)
Gene 2 has alleles B and b (B is dominant over b)
The genes are on different chromosomes
Here is a doubly heterozygous individual:
A a
Possible gametes
AB
Ab
B b
aB
ab
Probabilities in gametogenesis
A a
Possible gametes
AB
Ab
B b
aB
ab
What is the probability of each gamete?
P(AB) = P(A) × P(B) = 0.5 × 0.5 = 0.25
P(Ab) = P(A) × P(b) = 0.5 × 0.5 = 0.25
P(aB) = P(a) × P(B) = 0.5 × 0.5 = 0.25
P(ab) = P(a) × P(b) = 0.5 × 0.5 = 0.25
Note: Each gamete
is equally probable
Crossing two double heterozygotes
AaBb × AaBb
What are the possible genotypes of the offspring?
AB
Ab
aB
ab
AB
AABB AABb AaBB
AaBb
Ab
AABb
AAbb
AaBb
Aabb
aB
AaBB
AaBb
aaBB
aaBb
ab
AaBb
Aabb
aaBb
aabb
Answer: AABB , AABb, AAbb, AaBB, AaBb, Aabb, aaBB, aaBb, aabb
What are the probabilities of an offspring being each genotype?
Probabilities of genotypes in dihybrid cross
What are the probabilities of each genotype?
P(AABB) = P(AB) ♂ × P(AB) ♀ = 0.25 × 0.25 = 0.0625
P(AABb) = P(AB) ♂ × P(Ab) ♀ + P(Ab) ♂ × P(AB) ♀ = 0.25×0.25 + 0.25×0.25 = 0.125
P(AAbb) = P(Ab) ♂ × P(Ab) ♀ = 0.25 × 0.25 = 0.0625
P(AaBb) = P(AB) ♂ × P(ab) ♀ + P(ab) ♂ × P(AB) ♀ + P(Ab) ♂ × P(aB) ♀ + P(aB) ♂
× P(Ab) ♀ = 0.25 (add them all up)
P(AaBB) = 0.125
P(Aabb) = 0.125
P(aaBB) = 0.0625
P(aaBb) = 0.125
P(aabb) = 0.0625
Probabilities of phenotypes in dihybrid cross
What are the probabilities of each phenotype?
P(“AB”) = P(AABB) + P(AABb) + P(AaBB) + P(AaBb) = 0.0625 + 0.125 + 0.125 +
0.25 = 0.5625
P(“Ab”) = P(AAbb) + P(Aabb) = 0.0625 + 0.125 = 0.1875
P(“aB”) = P(aaBB) + P(aaBb) = 0.0625 + 0.125 = 0.1875
P(“ab”) = P(aabb) = 0.0625
If you are familiar with it, note that these
probabilities give you the classic
9:3:3:1
ratio
Exercise
Cross AaBb father with an aaBb mother:
What are the probabilities of each genotype and phenotype?
Father’s gametes: P(AB) = 0.25; P(Ab) = 0.25; P(aB) = 0.25; P(ab) = 0.25
Mother’s gametes: P(aB) = 0.5; P(ab) = 0.5
Genotypic probabilities:
P(AaBB) = P(AB) ♂ × P(aB) ♀ = 0.25 × 0.5 = 0.125 = 1/8
P(AaBb) = P(AB) ♂ × P(ab) ♀ + P(Ab) ♂ × P(aB) ♀ = 0.25 × 0.5 + 0.25 × 0.5 = 0.25 =
1/4
P(Aabb) = P(Ab) ♂ × P(ab) ♀ = 0.25 × 0.5 = 0.125 = 1/8
P(aaBB) = P(aB) ♂ × P(aB) ♀ = 0.25 × 0.5 = 0.125 = 1/8
P(aaBb) = P(aB) ♂ × P(ab) ♀ + P(ab) ♂ × P(aB) ♀ = 0.25 × 0.5 + 0.25 × 0.5 = 0.25 = 1/4
P(aabb) = P(ab) ♂ × P(ab) ♀ = 0.25 × 0.5 = 0.125 = 1/8
Phenotypic probabilities:
P(“AB”) = P(AaBB) + P(AaBb) = 0.125 + 0.25 = 0.375 = 3/8
P(“Ab”) = P(Aabb) = 0.125 = 1/8
P(“aB”) = P(aaBB) + P(aaBb) = 0.125 + 0.25 = 0.375 = 3/8
P(“ab”) = P(aabb) = 0.125 = 1/8
As a Punnet Square
Mother’s gametes
Father’s
gametes
aB
ab
AB
AaBB
(“AB”)
AaBb
(“AB”)
Ab
AaBb
(“AB”)
Aabb
(“Ab”)
aB
aaBB
(“aB”)
aaBb
(“aB”)
ab
aaBb
(“aB”)
aabb
(“ab”)
Binomial Distribution
Consider crossing heterozygous black guinea pigs
(Bb) among themselves. We could ask the following
questions:
1. What is the probability of the first four offspring
being alternately white and black?
2. What is the probability among the four offspring of
producing three black and one white in any order?
Binomial Formula
Bernoulli Trial: Random experiments are
called Bernoulli trials if
●the same experiment is repeated several times
●there are only two possible outcomes (success
and failure) on each trial
●the repeated trials are independent
●the probability of each outcome remains the
same for each trial
Bernoulli trials can always be represented by a tree diagram.
Let the outcome 'success' be denoted by S and the outcome
'failure', by F. If P(S) = p, and P(F) = q, then p + q = 1.
The tree diagram for the experiment repeated twice is:
p
q
p
S
q
p
F
S
S
F
q
F
The binomial probability formula: the probability of getting k
successes in n trials is given by the formula:
b n ,k ; p
n!
pk qn
k! n k !
k
Multiple offspring
We now consider the case of parents producing 100 offspring?
The production of each offspring is completely independent
from that of the others.
The questions we ask here are
a. How many of the 100 offspring should have each
genotype? Each phenotype?
b. What is the probability that heterozygous parents will have
100 brown-eyed offspring? 99 brown-eyed offspring and 1
blue-eyed offspring? 98 brown-eyed offspring and 2 blueeyed offspring? (etc.)