Calculating and predicting genotypic and phenotypic ratios for unlinked
autosomal dihybrid crosses
Terminology:
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Dihybrid crosses involve two genes which control two characteristics.
There are complications of these patterns as illustrated in the calculations
that follow.
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Unlinked genes are found on different chromosomes and can be
segregated by random assortment of meiosis/ metaphase II
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Autosomes are those chromosomes other than the XY gender
determining chromosomes (not sex linked).
Mendel's Dihybrid Cross with Peas
Phenotypes: Smooth Yellow Seeds X Rough Green Seeds
The chromosomes are shown as homologous pairs but not as bivalents.
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Smooth is dominant to Rough
Yellow is dominant to green
Meiosis:
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Reduces the chromosome number and randomly assorts the alleles for
each gene
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As the parents are homozygous they each produce only one type of
gamete.
Fertilization:
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Random fertilization of the gametes.
Offspring:
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The offspring are heterozygous at each gene loci. Notice the
chromosome number is restored to that of the parents.
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New homologous pairs are produced
The offspring are crossed = F1 x F1 (F1 self)
Meiosis
Meiosis:
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Independent assortment of alleles
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All allele combinations of gametes are shown for each parent.
Fertilization:
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The grid shows all offspring genotype combinations from random
fertilization.
Phenotypic ratio:
9 Smooth Yellow: 3 Smooth green: 3 Rough Yellow: 1 Rough Green
Dihybrid Ratio:
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This ratio is called the F2 Dihybrid Ratio and results from a cross
between two heterozygous dihybrids
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The ratio is a prediction of the offspring ratio.
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Actual numbers may deviate from this ratio as each fertilization is a
random process
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The ratio only sets the probability of a particular offspring phenotype
arising.
Dihybrid Calculations
Example1: Capercaille (simple dominance/ recessive)
The Capercaille is a ground dwelling bird of the pine forest of Scotland. The
length and color of the primary feather is controlled by two unlinked genes.
Using the allele key below calculate the phenotypic ratio of an F2 beginning
with a cross between a hen bird that is homozygous recessive for both genes
and a cock bird that is homozygous dominant for both genes.
Allele Key:
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Q- Long feather
q-short feather
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A- dark feather color
a-light feather color
Example2: Sweet peas (codominance/ dominance)
Sweet pea flowers color is controlled by a codominant pair of alleles. The
flowers are white, pink and red. The height of the mature plant is controlled
by a gene showing dominance. Calculate the F2 phenotypic ratio for a cross
between two heterozygous plants at both gene loci.
Allele key:
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A1= white, A2= red
B= Tall, b= short
Example3: Dihybrid Test Cross
A suspect heterozygote Guinea pig for coat color and length is crossed with a
double homozygote recessive. As with a monohybrid test cross there is a
predictable dihybrid test cross ratio which is 1:1:1:1
Allele key:
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L = long hair coat
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l =short hair coat
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B =Black coat
b =brown coat
Example4: Dihybrid with Sex linkage at one gene loci.
Colorblind allele loci is found on the non-homologous region of the X
chromosome.
Tongue rolling alleles are carried on an autosomal chromosome
Calculate the phenotypic ratio of a cross between a male tongue roller
(heterozygous) with normal vision and a female who is heterozygous at both
loci.
Allele Key:
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XB= Normal vision
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Xb= Colourblind
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Y = Male chromosome
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T= Tongue roller
t = non-tongue roller
more dihybrid questions
Example5: Epistasis/ Mouse color: Here two genes interact to control one
characteristic.
Mouse coat color is controlled by two genes. The first produces the pigment
and the second controls the banding of the coat color pigment.
Mouse coat is either Black (not banded), Agouti(banded) or albino(no color or
banding). Agouti is actually a black hair that has a yellow tip and represents
the color of wild mice.
Calculate the phenotypic ration from a cross between two agouti mice who
are heterozygous at both gene loci.
Example6 Epistasis/ Siamese cats: here again two gene interact to control the
coat
Allele key:
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C= Color produced
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c = no color produced
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A = Banding (agouti/back yellow tip)
a =Non banding (black)
Statistical tests for genetic crosses (Chi Square)
Sometimes the data from an experiment are not measurements but counts (or
frequencies) of things, such as counts of different phenotypes in a genetics
cross.
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With frequency data you can’t usually calculate averages or do a t test,
but instead you do a chi-squared test.
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This compares observed counts with some expected counts and tells
you the probability (P) that there is no difference between them.
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In Excel the test is performed using the formula: =CHITEST (observed
range, expected range) .
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There are three different uses of the test depending on how the
expected data are calculated. Chart link will show you why we choose the test
below.
Sometimes the expected data can be calculated from a quantitative theory, in
which case you are testing whether you observed data agree with the theory.
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If P < 5% then the data do not agree with the theory.
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If P > 5% then the data do agree with the theory.
A good example is a genetic cross, where Mendel’s laws can be used to predict
frequencies of different phenotypes. In this example
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Excel formula are used to calculate the expected values using a 3:1
ratio of the total number of observations.
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The Chi square P is 53%, which is much greater than 5%, so the
results do indeed support Mendel’s law.
Chi square calculation on dihybrid data
In one of Mendel's dihybrid crosses, the following types and numbers of pea
plants were recorded in the F2 generation:
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Yellow round seeds (289) Yellow wrinkled seeds (122) Green round seeds (96)
Green wrinkled seeds (36)
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According to theory these should be in the ratio of 9:3:3:1. Do these
observed results agree with the expected ratio?