PUNNET SQUARES – DIHYBRID CROSSES
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T=tall, t=short, R=round, r=wrinkled
According to the Law of Independent Assortment, a plant that is hybrid for 2 traits will form FOUR
different gametes
TtRr
TR
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Tr
tR
tr
The combinations that are produced depend on how the chromosomes line up during metaphase in
meiosis I
The Punnet square for a dihybrid cross has 16 squares (vs. 4 squares for a monohybrid cross)
EXAMPLE: #1
What are the possible gametes for the following parents (B=black coat, b= white, R= regular eyes and
r= bulging eyes);
a) BBRr
b) BbRr
c) bbRR
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EXAMPLE #2:
Yellow seeds (Y) are dominant over green seeds (y). Round seeds (R) are
dominant over wrinkled seeds (r). If two plants that are heterozygous
for BOTH traits (dihybrid) are crossed, what will be the phenotype and
genotype ratios of the F1 generation?
SOLUTION:
Parents = YyRr x YyRr
YR
Yr
yR
yr
YR
YYRR
YYRr
YyRR
YyRr
Yr
YYRr
YYrr
YyRr
Yyrr
yR
YyRR
YyRr
yyRR
yyRr
yr
YyRr
Yyrr
yyRr
yyrr
Phenotype:
9 yellow round: 3 yellow wrinkled: 3 green round: 1 green wrinkled
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Genotype: 1 YYRR 2 YYRr 2 YyRR 4 YyRr (yellow round)
1 YYrr 2 Yyrr (yellow wrinkled)
1 yyRR 2 yyRr (green round)
1 yyrr (green wrinkled)
EXAMPLE #3:
A pea plant that heterozygous for the traits tall (T) and round seeds (R), is crossed with a dwarf with
wrinkled seeds. Determine the phenotype and genotype ratios of the F1 generation
SOLUTION:
Parents = TtRr x ttrr
TR
TtRr
TtRr
TtRr
TtRr
tr
tr
tr
tr
Tr
Ttrr
Ttrr
Ttrr
Ttrr
tR
ttRr
ttRr
ttRr
ttRr
tr
ttrr
ttrr
ttrr
ttrr
Phenotype:
4 tall round: 4tall wrinkled: 4dwarf round: 4dwarf wrinkled
=1:1:1:1
Genotype:
4 TtRr : 4 Ttrr : 4 ttRr : 4 ttrr
=1:1:1:1
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EXAMPLE #4:
In mice, grey coat colour (G) is dominant to white coat colour (g), and long tail (T) is dominant to short
tail (t). What fraction of F1 mice would you predict to have grey coats and short tails if a male ggTt
parent is crossed with a female GGtt parent?
SOLUTION:
Parents = ggTt x GGtt
gT
GgTt
GgTt
GgTt
GgTt
Gt
Gt
Gt
Gt
gT
GgTt
GgTt
GgTt
GgTt
gt
Ggtt
Ggtt
Ggtt
Ggtt
gt
Ggtt
Ggtt
Ggtt
Ggtt
Phenotype: 8: grey, long tail: 8 grey short tail
= 1:1
Genotype: 8 GgTt : 8 Ggtt
= 1:1
Therefore the fraction of mice with grey coats and short tails is 8/16 = 1/2.
Probability and Product Law
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Genetic ratios are probabilities
Recall: a 3:1 ratio for widow’s peak: straight hair means that there is a 1 in 4 or a 25% chance that
offspring will be born with straight hair
The product law states that the probability of two random events occurring is the product of the
individual probabilities of each event
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Ex. The probability of having a boy as the first child is 50% or ½ and the probability for having a
boy as the 2nd child is 50% or ½. Therefore the probability of having two boys in a row is ½ x ½ = ¼
or 25%
We can predict the probability of dihybrid crosses using the product law
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EXAMPLE 5: Fill in the table
Round or wrinkled: monohybrid
Round seed (Rr x Rr)
Yellow or green seed: Dihybrid
cross
Yellow seed (Yy x Yy)
Dihybrid cross probability
Round yellow seed (RrYy x RrYy)
Continuous variation and additive alleles
Continuous variation: when the product of one gene is affected by the product of another gene. The effects
can be additive of negate another product.
 When phenotype variation is not clear cut
 Explains the extensive variation in our population
 Examples: skin colour- controlled by three or more genes
Additive allele: the product of one allele comprises one part of the total phenotype
 An example of continuous variation
 Example: the colour of skin depends on which combination of the six alleles is expressed. Each
allele makes its own contribution to the pigment of our skin