Chapter 9: Dihybrid Inheritance

Biol 1408 : Chapter 9
Patterns of Inheritance
The Experiment
P generation
(true-breeding
parents)
  We previously
discussed mono hybrid crosses
×
Purple
flowers
F1 generation
White
flowers
All plants have
purple flowers
Fertilization
among F1 plants
(F1 × F1)
F2 generation
3 of plants
1 of plants
4
4
have purple flowers have white flowers
Mono Hybrid Crosses
  A cross between true
breeding plants that
differ in only one
character trait
  Examples such as
flower color, plant
height, pod color, pod
shape,…
  Outcome of F2
generation was 3:1 with
respect to phenotype
  With respect to
genotype , the outcome
is 1:2:1
1
9.5 Dihybrid Crosses
  A dihybrid cross is a mating of parental varieties
that differ in two characters.
  Mendel performed the following dihybrid cross
Parent 1 : Round seeds ( 1st character state) that were
yellow colored (2nd character state)
Parent 2 : Wrinkled seeds ( variation on 1st character state)
that were green colored (variation on 2nd
character state)
(from earlier discussion, round and yellow are dominant
traits)
© 2012 Pearson Education, Inc.
9.5 Dihybrid Crosses
  First, how do we represent an organism with two
character state ?
  A one character state with respect to color (yellow) is
represented as YY (homozygous dominant), Yy
(heterozygous), or yy (homozygous recessive).
  Two character states, such as Round and Yellow
seeds, can thus be represented as
  RR YY ( homozygous dominant for both traits)
  Rr Yy (heterozygous for both traits)
  rr yy (homozygous recessive for both traits)
© 2012 Pearson Education, Inc.
9.5 Dihybrid Crosses
  Mendel reasoned that if seed shape and seed color
were traits related to each other, they should sort out
evenly in the gametes.
rryy
P generation RRYY
Gametes RY
×
ry
  Since the parents are double homozygous, the
gametes don t immediately tell the story of what
happens.
  What will a cross between these two plant above give
us ?
© 2012 Pearson Education, Inc.
2
9.5 Dihybrid Crosses
rryy
P generation RRYY
Gametes RY
×
ry
RrYy
F1 generation
  The F1 generation is thus 100 % double heterozygous.
  And since R ( round) and Y (yellow) are dominant
characteristics, all seeds from the F1 generation plants will
yield round, yellow seeds.
9.5 Dihybrid Crosses
rryy
P generation RRYY
Gametes RY
×
ry
RrYy
F1 generation
  What if we now do a cross between F1 plants ?
  Since Mendel thought that parental traits are always related
to each other, so they should sort out in the gametes
together.
9.5 Dihybrid Crosses
rryy
P generation RRYY
Gametes RY
F1 generation
×
ry
RrYy
  In other words, if the traits were dependent on each other,
they should sort out dependently into the gametes ( called
dependent assortment)
  Thus the R and Y trait, coming from the one parents would
always travel together, and the r and y trait would do the
same as well
3
9.5 Dihybrid F1 cross assuming parental traits
are related
  The gametes would contain the parental traits if they are
related by parental origin (dependent assortment)
  The expected outcomes would thus
be 3:1 in phenotypes (3 showing
round and yellow seeds, 1 having
wrinkled green seeds)
RY
ry
RY
RRYY
RrYy
ry
RrYy
rryy
gametes
Female
Male
RrYy
RrYy
9.5 Dihybrid F1 cross assuming parental traits
are related
F1 generation
RrYy
Sperm
1
2
1
2
F2 generation
RY
1
2
ry
RY
Eggs
1
2
ry
Data did not support this hypothesis of dependent assortment
of parental traits and resulted in rejection of this hypothesis.
9.5 The law of independent assortment is
revealed by tracking two characters at once
  Mendel observed the following when performed this
dihybrid cross :
–  P generation: round yellow seeds × wrinkled green seeds
–  F1 generation: all plants with round yellow seeds
–  F2 generation:
–  9/16 had round yellow seeds
–  3/16 had wrinkled yellow seeds
–  3/16 had round green seeds
–  1/16 had wrinkled green seeds
Two new
phenotypes not
observed in either P
or F1 generation
© 2012 Pearson Education, Inc.
4
9.5 Dihybrid Crosses
rryy
P generation RRYY
Gametes RY
ry
×
RrYy
F1 generation
  The only way this could have happened if the traits
sorted out independently of each other in the
gametes
  Instead of two possibilities in the gametes, 4
possibilities can occur.
9.5 Dihybrid Crosses
rryy
P generation RRYY
Gametes RY
ry
×
F1 generation
RrYy
RY
Possible gametes
Ry
rY
ry
•  The traits thus move independent of each other into the gametes,
generating 4 possible gamete formation
•  Called independent assortment of the traits
•  A punnett square will show all the possible combinations.
9.5 Dihybrid Cross among F1 plants with
independent sorting out of traits in gametes
RrYy
F1 generation
Sperm
1
4
1
4
RY
1
4
rY
Eggs
1
4
Ry
1
4
ry
RY
1
4
rY
1
4
Ry
1
4
ry
RRYY
RrYY
RRYy
RrYy
RrYY
rrYY
RrYy
rrYy
RRYy
RrYy
RrYy
rrYy
RRyy
Rryy
Rryy
rryy
9
16
Yellow
round
3
16
Green
round
3
16
Yellow
wrinkled
1
16
Green
wrinkled
5
9.5 The law of independent assortment is
revealed by tracking two characters at once
  Mendel observed that the F2 offspring
–  a 9:3:3:1 phenotypic ratio.
–  had new non-parental combinations of traits
–  9 showing double dominant trait, 3 heterozygous in 1st trait and
homozygous in 2nd trait, 3 homozygous in 1st trait and heterozygous in
2nd trait, 1 double recessive
  Mendel
–  suggested that the inheritance of one character has no effect on the
inheritance of another,
–  suggested that the dihybrid cross is the equivalent to two independent
monohybrid crosses ((3:1) x (3:1))
–  He called this the law of independent assortment.
© 2012 Pearson Education, Inc.
9.5 Dihybrid Cross among F1 plants with
independent sorting out of traits in gametes
Sperm
1 of RR YY
2 of RR Yy
2 of Rr YY
4 of Rr Yy
9
1 of RR yy
2 of Rr yy
3
1 of rr YY
2 of rr Yy
1 of rr yy
RrYy
F1 generation
Genotypes
3
1
1
4
1
4
RY
1
4
rY
Eggs
1
4
1
4
RY
1
4
rY
1
4
Ry
1
4
ry
RRYY
RrYY
RRYy
RrYy
RrYY
rrYY
RrYy
rrYy
RRYy
RrYy
RRyy
Rryy
RrYy
rrYy
Rryy
rryy
Ry
ry
9
16
Yellow
round
3
16
Green
round
3
16
Yellow
wrinkled
1
16
Green
wrinkled
9.5 The law of independent assortment is
revealed by tracking two characters at once
  The following figure demonstrates the law of
independent assortment as it applies to two
characters in Labrador retrievers:
–  black versus chocolate color,
–  normal vision versus progressive retinal atrophy
(blindness).
–  Black and normal vision are dominant traits
© 2012 Pearson Education, Inc.
6
  What is the result
Blind
if we cross these
two animals ?
Phenotype
Black coat,
normal vision
BBNN
Genotype
Black coat,
blind (PRA)
B_nn
Blind
Chocolate coat,
Phenotype
blind (PRA)
bbnn
Genotype
Chocolate coat,
normal vision
bbN_
F1
100 %
Double
heterozygous
Black coat,
normal vision
BBNN
Black coat,
normal vision
BbNn
  BBNN gives
gametes of BN
Blind
Chocolate coat,
  bbnn gives
normal vision
bbN_
gametes of bn
Chocolate coat,
blind (PRA)
bbnn
9.5 Dihybrid Cross among F1 Labradors with
independent sorting out of traits in gametes
BbNn
F1 generation
Genotypes
1 of BB NN
2 of BB Nn
2 of Bb NN
4 of Bb Nn
9
2 of Bb nn
1 of BB nn
3
1 of bb NN
2 of bb Nn
3
1 of bb nn
1
BN
Sperm
bN
Bn
bn
BN
BBNN
BBNn BbNN
BbNn
Bn
BBNn
BBnn
BbNn
Bbnn
9 of 16
Black
Normal
3 of 16
Black
blind
3 of 16
Brown
Normal
1 of 16
Brown
Blind
Eggs
bN
BbNN
bn
BbNn Bbnn
BbNn
bbNN
bbNn
bbNn
bbnn
7
9.5 The law of independent assortment is
revealed by tracking two characters at once
  The result of a cross between two F1 animals
Mating of double heterozygotes (black coat, normal vision)
BbNn
×
BbNn
Blind
Blind
Phenotypic
ratio of the
offspring
9
Black coat,
normal vision
3
Black coat,
blind (PRA)
1
3
Chocolate coat, Chocolate coat,
blind (PRA)
normal vision
9.5 The law of independent assortment is
revealed by tracking two characters at once
Blind
Blind
Phenotypic
ratio of the
offspring
9
Black coat,
normal vision
3
Black coat,
blind (PRA)
1
3
Chocolate coat, Chocolate coat,
blind (PRA)
normal vision
  9 have at least one copy of each dominant trait
  3 have at least one copy of the dominant allele of the 1st trait and no
dominant allele of the 2nd trait
  3 have at least one copy of the dominant allele of the 2nd trait and no
dominant allele of the 1st trait
  1 has only recessive alleles of both traits
8