Download IntroductionPartII

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
P(bb) = P(b) ♂ × P(b) ♀ = 0.5×0.5 = 0.25
P(Bb) = P(B) ♂ × P(b) ♀ + 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
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?
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
AB
Father’s
gametes
Ab
aB
ab
aB
ab
AaBB
(“AB”)
AaBb
(“AB”)
aaBB
(“aB”)
aaBb
(“aB”)
AaBb
(“AB”)
Aabb
(“Ab”)
aaBb
(“aB”)
aabb
(“ab”)
Multiple offspring
What if we take our parents are breed them so they produce
100 offspring?
Each offspring is completely independent from the others. The
previous values still describe the probability of any individual
offspring having a specific genotype or phenotype.
But how many of the 100 offspring should have each
genotype? Each phenotype? What is the probability that
heterozygous parents will have 100 brown-eyed offspring? 99
brown-eyed offspring and 1 blue-eyed offspring? 98 browneyed offspring and 2 blue-eyed offspring? (etc.) How do we
calculate this?