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Mendelian Genetics
DiHybrid Crosses
Independent Assortment
Vs. Linkage
The Dihybrid Cross
 As
the name suggests, a dihybrid
cross is a cross between parents that
differ from each other with regard to
two traits.
 The question that arises is the
degree to which the inheritance of
one trait influences the inheritance of
the other.
Independent Assortment
 The
name implies that the traits will
mix (assort) independently of each
other.
 The independent assortment
hypothesis assumes that the
inheritance of one trait is separate
from the other – the inheritance of
one is random with regard to the
other
Independent Assortment
Results
 If
we were to run a monohybrid
cross, we would expect a 3:1 ratio in
the F2 generation
 If we ran 2 monohybrid crosses
simultaneously, and they assorted
independently of each other, we
could assume that they would mix
and match in every combination
Independent Assortment
Results
 Given
as a probability, a 3:1 ratio is
really a ¾ chance of dominant
expression and a ¼ chance of
recessive expression
 Using the product law of probability
the likelihood of dominant expression
of both traits is 9/16, one dominant
and one recessive 3/16, and both
recessive 1/16
Results expressed as a ratio
 The
probabilities described in the
previous frame are more commonly
referred to as a 9:3:3:1 ratio
 This was Gregor Mendel’s
hypothesis. Mendel hypothesized
that traits were inherited
independently, and his hypothesis
was supported by the results of his
dihybrid cross experiments with peas
Mendel’s Dihybrid Cross
Analyzing Results
 In
the Punnett square given, the
inheritance of each allele was
consistent with the law of
segregation, alleles separated with
equal probability in gamete
formation
 Also, the R allele was as likely to be
inherited with the Y allele as with the
y allele. Seed color and seed shape
were inherited independently
Conclusions
 Mendel
performed dihybrid crosses
with several combinations of traits.
Each time he observed the predicted
9:3:3:1 ratio. His conclusions form
the Law of Independent Assortment
 A modern understanding of
Independent Assortment recognizes
the fact that genes are on
chromosomes
Chromosome Theory
 If
genes are on chromosomes, then
allele pairs are a result of pairing of
homologous chromosomes
 Homologous chromosomes separate
during meiosis, just as alleles
separate during gamete formation
Chromosome Theory
 We
remember that homologous
chromosomes synapse, forming a
tetrad, attach to a spindle fiber, align
and separate during meiosis 1
 Each tetrad is a different homologous
pair, attached to a different spindle
fiber
 The division of pair #1 has no effect
on the division of any other pair
Modern view of Mendel
 So
Independent Assortment is the
result of genes for one trait being
located on a different chromosome
than the genes for the other trait.
 But what if they aren’t? Peas
demonstrated Independent
Assortment, but will all pairs of
genes?
No. They won’t.
 Let’s
take humans for example. We
have 23 homologous pairs. We also
have way more than 23 genes. It
stands to reason that some of those
genes will be on the same
chromosome
 Genes on the same chromosome are
“linked” genes. They will be
inherited together
What would linkage look like?
 To
use Mendel’s peas as an example,
the dihybrid cross we referenced
earlier studied seed color and seed
shape
 P1: Round, Yellow x wrinkled, green
 F1: all Round, Yellow
 P2: Round, Yellow x Round, Yellow
 F2: Make a prediction
F2 results, Linked Genes
 First
we need to consider what we
started with. If the genes are linked,
they should remain in the same
combination they started in (back in
the P1 generation)
 P1: round, yellow x wrinkled, green
 So if round is linked, it’s linked to
yellow and similarly wrinkled would
be linked to green (“parental types”)
Or to look at it another way
 So
basically, 2 linked genes
travelling together behave pretty
much like a single gene
 That means linkage for a dihybrid
cross would give the same kind of
results as a monohybrid cross
 We would expect an F2 result of
3 Round, Yellow : 1 wrinkled, green
 Note, all offspring are parental types
Comparing Predictions
Independent
assortment
 9:3:3:1 ratio in the F2
 All phenotype
combinations are
represented, both
parental and
recombinant
 Genes are on different
chromosomes, so
there is no
impediment to
recombination
Linkage
 3:1 ratio in the F2
 Only the parental
combinations are
represented, no
recombinant offspring
are produced
 Genes are on the
same chromosome so
they always stay
together
Dihybrid Cross – Determine if the genes for wing
shape and eye color are Linked or Independent
P1
F1
P2
F2
=
=
=
=
Straight wings, Red eyes X Curly wings, Orange eyes
all Straight wings, Red eyes
Straight wings, Red eyes x Straight wings, Red eyes
36 straight wing, red eye
12 straight wing, orange eye
13 curly wing, red eye
4 curly wing, orange eye
Which F2 phenotypes are parental?
Which are recombinant?
P1
F1
P2
F2
=
=
=
=
Straight wings, Red eyes X Curly wings, Orange eyes
all Straight wings, Red eyes
Straight wings, Red eyes x Straight wings, Red eyes
36 straight wing, red eye
12 straight wing, orange eye
13 curly wing, red eye
4 curly wing, orange eye
Which F2 phenotypes are parental?
Which are recombinant?
P1
F1
P2
F2
=
=
=
=
Straight wings, Red eyes X Curly wings, Orange eyes
all Straight wings, Red eyes
Straight wings, Red eyes x Straight wings, Red eyes
36 straight wing, red eye
12 straight wing, orange eye
13 curly wing, red eye
4 curly wing, orange eye
9
3
3
1
Conclusion



Wing shape and eye color are independent.
Straight wings were not exclusively inherited with
red eyes, they appeared in the F2 generation
paired with orange eyes in statistically significant
proportions
The gene for wing shape must be located on a
different chromosome than the gene for eye color
Dihybrid Cross – Determine if the genes for eye
color and bristles are Linked or Independent
P1
F1
P2
F2
=
=
=
=
red eyes, normal bristles X brown eyes, stubble bristles
all red eye, normal bristle
red eyes, normal bristles X red eyes, normal bristles
28 red eyes, normal bristles
0 red eyes, stubble bristles
0 brown eye, normal bristles
9 brown eye, stubble bristle
Are the genes for eye color and bristles on the
same chromosome or on different chromosomes?
P1
F1
P2
F2
=
=
=
=
red eyes, normal bristles X brown eyes, stubble bristles
all red eye, normal bristle
red eyes, normal bristles X red eyes, normal bristles
28 red eyes, normal bristles
0 red eyes, stubble bristles
0 brown eye, normal bristles
9 brown eye, stubble bristle
Parental vs Recombinant
P1
F1
P2
F2
=
=
=
=
red eyes, normal bristles X brown eyes, stubble bristles
all red eye, normal bristle
red eyes, normal bristles X red eyes, normal bristles
28 red eyes, normal bristles
0 red eyes, stubble bristles
0 brown eye, normal bristles
9 brown eye, stubble bristle
3
0
0
1
Parental vs Recombinant
P1 =
F2 =
red eyes, normal bristles X brown eyes, stubble bristles
28 red eyes, normal bristles
0 red eyes, stubble bristles
0 brown eye, normal bristles
9 brown eye, stubble bristle
3
0
0
1
All the F2 offspring were parental types, no recombination occurred
Are the genes for eye color and bristles on the
same chromosome or on different chromosomes?
F2 =
28 red eye, normal bristles
0 red eyes, stubble bristles
0 brown eye, normal bristles
9 brown eye, stubble bristle
These results are what you would expect from Linked genes (genes
located on the same chromosome), but they do not take into account
the real occurrence of Crossing Over. Real life scenarios will produce
some level of genetic recombination even when genes are linked
Characteristic Data from Linked Genes
F2 =
22 red eyes, normal bristles
2 red eyes, stubble bristles
3 brown eye, normal bristles
8 brown eye, stubble bristle
In this data set, we have 30 “parental type” F2 offspring and 5
“recombinant” type offspring. This represents 14.3% recombination.
This is far too small to be a result of independent assortment, so the
genes must be linked. Recombination is the result of Crossing Over
Barbara McClintock
Barbara McClintock
won a Nobel Prize
for her research in
Cytogenetics.
 She used maize
(corn) for her
research

Why Corn?
Easy to
control
crosses
 Every
kernel of
corn is a
genetically
unique
offspring

Lab - DiHybrid Cross in Corn
 Dihybrid
= crossbreed for 2 traits
– Seed color and seed shape
 P1:
Purple Smooth x Yellow Wrinkled
 Fi: all Purple Smooth
 P2: Purple Smooth x Purple Smooth
 F2:
We will need to make alternative
predictions for the F2 offspring based on
Linkage vs. Independent Assortment
F2 Predictions - Linked
F2 Predictions - Linked
F2 Predictions –
Independent Assortment
F2 Predictions –
Independent Assortment
F2 Results
 Working
in pairs, count and record
the number of each phenotype
 Compile the data to get class totals
 Evaluate the results and determine if
the genes are linked or independent
– Purple,
– Purple,
– Yellow,
– Yellow,
Smooth
Wrinkled
Smooth
Wrinked