Dihybrid Crosses
•A
cross that deals with two different genes or traits
•Example: eye color and hair color
Mendel did not only study the genetic
inheritance of a single trait, he also
studied the inheritance of two separate
traits with a single cross
 Instead of a punnett square containing 1
trait with two alleles, it now has 2 traits
with 4 alleles.

Mendel’s law of Independent Assortment

was created from his study of dihybrid
crosses; this law states that genes assort
independently; one gene doesn’t influence
the inheritance of another.
◦ All factors have an equal possibility of being
donated to the offspring.
◦AaBb

A yellow round, seed, genotype YyRr,
produces the following gametes
◦ YR,Yr, yR or yr




Y = yellow
y = green
R = round
r = wrinkled
Dihybrid Cross

To calcuate the offspring of a cross
between YyRr and YyRr we use a Punnett
Square:
Gametes
YR
Yr
yR
yr
YR
Yr
yR
yr
Fill in the chart with the offspring genotypes, keeping the “y’s” and the”r’s” together.
YR
Yr
yR
yr
YYRR
YYRr
YyRR
YyRr
YR
Yr
yR
yr
Notice Dominant alleles first, then Recessive.
YR
Yr
yR
yr
YYRR
YYRr
YyRR
YyRr
YR
Yr YYRr
YYrr
YyRr
Yyrr
YyRR
yR
YyRr
yyRR
yyRr
YyRr
yr
Yyrr
yyRr
yyrr
Dihybrid Crosses:
A dihybrid cross between two
heterozygous parents will always produce
a 9:3:3:1 phenotypic ratio.
 In the example:

◦
◦
◦
◦
9 Yellow Round
3 Yellow wrinkled
3 green Round
1 green wrinkled
Dihybrid Crosses

A generic ratio for any dihybrid
heterozygous cross:
◦
◦
◦
◦
9 Dominant Dominant
3 Dominant Recessive
3 Recessive Dominant
1 Recessive Recessive
True Breeding

A parent is said to be TRUE BREEDING if
it produces only one gamete for a specific
trait. It is homozygous.
◦ YYRR is true breeding for yellow, round seeds
◦ yyrr is true breeding for green wrinkled seeds

A cross between two true breeding
parents will always produce only one
possible offspring (why?)
True Breeding
 YYrr
x yyRR
 YYRR x yyRR
YyRr
YyRR
Example:

A homozygous individual for tongue
rolling and widows peak (both dominant)
mates with an individual who can not roll
their tongue and does not have a widows
peak. Calculate the phenotypic &
genotypic ratio of the offspring of both
the F1 and the F2 generations.
◦ Step #1 is to assign allele’s to the traits

What are the genotypes for the intial
parents?

TTHH x tthh

What gametes can each parent make?

Phenotypic and Genotypic ratios in the F1?
How do we get the F2?

Parent cross:

Next slide = Punnett square
Gametes
Gametes
TH
Th
tH
th
TH
Th
tH
th
Gametes
TH
TH TTHH
Th
tH
th
TTHh
TtHH
TtHh
Th
TTHh
TThh
TtHh
Tthh
tH
TtHH
TtHh
ttHH
ttHh
th
TtHh
Tthh
ttHh
tthh
Phenotypes – for each possible
genotype, write the phenotype
Genotypes
TTHH
 TTHh
 TtHh
 TtHH
 TThh
 Tthh
 ttHH
 ttHh
 tthh

___ / 16
___ / 16
___ / 16
___ / 16
___ / 16
___ / 16
___ / 16
___ / 16
___ / 16
Rolling, widows peak
 Rolling, widows peak
 Rolling, widows peak
 Rolling, widows pea
 Rolling, straight
 Rolling, straight
 Non-rolling, widows peak
 Non-rolling, widows peak
 Non-rolling, straight

Prove the 9:3:3:1 Ratio