10.2Dihybrid crosses and gene linkage
10.2.1 Calculate and predict the genotypic and phenotypic ratio of offspring of dihybrid crosses involving
unlinked autosomal genes.
Terminology:
Dihybrid crosses involve two genes that control two characteristics. Unlinked genes are found
on different chromosomes and can be segregated by random assortment of meiosis following
metaphase II. Autosomal chromosomes are those chromosomes other than the XY gender
determining chromosomes. Genes on autosomal chromosomes are not sex linked. The following
is a worked example of a dihybrid cross. Other examples are included for you to calculate
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.
a. Smooth is dominant to Rough
b. Yellow is dominant to green
Meiosis:This process reduces the chromosome
number and randomly assorts the alleles for each
gene. Since the parents are homozygous (SSYY or
ssyy) they each produce only one type of gamete.
Fertilisation:Random fertilisation of the gametes
occurs.
Offspring:The offspring are heterozygous at each
gene loci (SsYy). Notice the chromosome number is
restored to the that of the parents.
New homologous pairs are produced
The offspring are crossed = F1 x F1 (F1 self)
Meiosis
Meiosis:
Independent assortment of alleles can be seen here.
All allele combinations of gametes are shown for each
parent.
Fertilization:
The grid shows all offspring genotype combinations
from random fertilization.
This ratio is called the F2 Dihybrid Ratio and results from a cross between two heterozygous
dihybrids.The ratio is a the expected ratio or prediction of the offspring ratio.
Actual numbers may deviate from this ratio as each fertilization is a random process
The ratio only sets the probability of a particular offspring phenotype arising from a dihybrid cross.
Dihybrid Calculation Problems
Write out your answers you will need them for section 8.2.2
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 feathers are
controlled by two unlinked genes. Using the allele key to the right
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.
Example2: Sweet peas (codominance/ dominance)
Sweet pea flowers colour 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.
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
Example4: Dihybrid with Sex linkage at one gene loci.
Colourblind 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.
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.
Allele Key:
Q- Long feather
q-short feather
A- dark feather color
a-light feather color
Allele key:
A1= white, A2= red
B= Tall, b= short
Allele key:
L = long hair coat
l =short hair coat
B =Black coat
b =brown coat
Allele Key:
XB= Normal vision
Xb= Colourblind
Y = Male chromosome
T= Tongue roller
t = non-tongue roller
Allele key:
C= Colour produced
c = no colour
produced
A = Banding
(agouti/back yellow
tip)
a =Non banding
(black)
10.2.2 Distinguish between autosomes and sex chromosomes
Autosomes and Sex Chromosomes
Autosomes:
Autosomes are all the chromosomes but not the X and Y, the so called sex chromosomes
Sex Chromosomes: XY chromosomes, XX in female and XY in males. gene on the chromosomes may be sex
linked
10.2.2 Explain how crossing over between non-sister chromatids of a homologous pair in prophase I can
result in an exchange of alleles
Recombinants from Cross over
.
Cross over occurs in prophase I of meiosis (see 8.1.2)
The diagram shows the loci of genes A and B.
The genes are on the same chromosome (A and B
are a linkage group)
The cross over involves the exchange of lengths of
DNA.
In doing so the chromosomes of a homologous pair
exchange their alleles.
When the chromatids separate at anaphase II this
will produce new combinations of alleles in the
gametes (recombinants).
The recombinants in this example are aB and Ab
10.2..3, 10.2.4, 10.2.5. 10.2.6
Explain linkage groups and Linkage Group Crosses. Explain an example of a cross between two linked
genes and Identify which of the offspring are recombinants in a dihybrid cross involving linked
genes.
Linkage group:
The genes A and B are
a linkage group. If this
genotype (AaBb) was
crossed with itself
would produce a 3:1
ratio not a 9:3:3:1
Linked genes are
genes on the same chromosome
genes inherited together
genes that do not show the expected Mendelian ratios as
predicted by the Laws of Independent Assortment
example: Sweat Peas (Lathyrus odoratus
)
.Allele Key:


Flower colour Purple (P) and Red (p)
Pollen grain shape, Long (L) and short (l)
A cross was made between a plants that where
heterozygotes at both gene loci (PpLl). Analysis of date:
There is a significant difference between observed data
and expected data for linkage (3:1 ratio with no cross
over)
The data do not agree with the theory )for an
unlinked cross with a chi square calculation of P <
5%
Conclusion the genes are linked but some cross
over has occurred.
This famous example comes from the work of W.
Batson who rediscovered the work of Mendel and
carried out his own experiments in the early 20
Century
In linkage crosses the format of the alleles allows us to see which alleles are inherited together and which
are recombinants.
1. Using the example of Sweet Peas the genes for Flower colour and pollen grain shape can be
shown. This shows the linkage of the P allele with the L allele with the line representing the
chromosome.
2. Therefore the heterozygote (at both loci) would be represented as follows
3. A recombinant would show the chromosome after cross over has completed. In this case
there could be the following recombinant alleles in the gamete.
or
Try the following calculations then check your answers on the end of this section
Example: Myotonic Dystrophy and AB Secretor
disfunction
Example: Sweet Peas
Allele Key:
Allele key for Myotonic Dystrophy

Normal (Md) with disease (md)
Allele Key for AB Secretor disfunction

Normal AB(Se) and Non AB secretion (se)
Show the cross of two normal heterozygotes at both
gene loci with cross over in one individual. Show
genotypes and identify recombinants. Assume Md is
linked to Se


Flower colour Purple (P) and Red (p)
Pollen grain shape, Long (L) and short (l)
A cross was made between a plants that where
heterozygotes at both gene loci (PpLl). Only one shows
cross over in gamete genotypes.
Show the possible genotypes and identify
recombinants. Assume P is linked to L
Example: Drosophila (Fruit Fly)
Normal wing (L) is linked to Grey Body (G) whilst
Vestigial wing (l) is linked to the mutant black (b) body.
A test cross of a potential heterozygote produced the
ratio of
Grey Normal (283),Grey Vestigial(1294),
Black Normal(1418) and Black Vestigial (241)
Using Chi square test determine if the genes are
linked or unlinked
8.3.4 Identification of recombinant offspring
Recombinant are those individuals that show combinations of genes(alleles) or characteristics not seen in the
parents
Check again the answers to the above crosses. The recombinants have been identified for you