Dihybrid crosses involve two genes which control two characteristics. There are complications of these patterns as illustrated in the calculations that follow.
Unlinked genes are found on different chromosomes and can be segregated by random assortment of meiosis/ metaphase II
Autosomal are those chromosomes other than the XY gender determining chromosomes (not sex linked).
The following is a worked example of a dihybrid cross. Other examples are the set for you to calculate with answers available. The example is based on Mendel's pea crosses and begins with 'true' breeding or homozygous plants for two unlinked genes. The phenotype which is used includes the colour and the texture of the seeds of the pea plant.
Phenotypes: P1 Smooth Yellow Seeds X P2 Rough Green Seeds
The chromosomes are shown as homologous pairs but not as bivalents.
The offspring are crossed = F1 x F1 (F1 self)
Both parents have identical phenotype for the two characteristics (Smooth and Yellow) and are heterozygous at both gene loci.
Independent assortment of alleles
All allele combinations of gametes are shown for each parent.
Notice that S can be found with Y or equally likely with y alleles. This is because of random orientation in metaphase I of meiosis. Both possibilities are equally likely.
The Punnet grid shows all possible offspring genotypes from random fertilisation.
The phenotypic ratio for this cross is:
9 Smooth Yellow: 3 Smooth green: 3 Rough Yellow: 1 Rough Green
This is the classic dihybrid ratio, which indicates that both parent are heterozygous at both loci.
This ratio is called the F2 Dihybrid Ratio and results from a cross between two heterozygous dihybrids
The ratio is a prediction of the offspring ratio.
Actual numbers may deviate from this ratio as each fertilisation is a random process
The ratio only sets the probability of a particular offspring phenotype arising.
Heterozygotes(SsYy) and Homozygote dominants (SSYY) cannot be distinguish by observing their phenotypes. Mendel however developed a test cross for suspected heterozygotes.
The test population here is the heterozygote.
With the pea plants Mendel cold not be sure that the smooth and yellow seeds are heterozygote or rather it was a homozygous dominant.
The suspect heterozygote is crossed with a homozygous recessive at both gene loci. (ssyy)
Meiosis produces four gametes in the heterozygote which illustrates the random orientation of chromosomes in metaphase I
Meiosis only produces on type of gamete for the homozyote double recessive.
Random fertilisation of the gametes produces four possible genotypes for the offspring.
This test would have been performed on more than one cross so that the ratio of the offspring can be compared on a reasonable sample size.
Offspring genotypes show four different types.
The phenotype ration is 1 : 1 :1 : 1
The is the dihybrid test ration (back cross ratio).
Calculation: Mendel's pod colour and seed colour
The Pea plants (Pisum sativum) after flowering produces a pod that is either green or yellow.
The seeds inside also show a colour difference of green and yellow.
The inheritance of two characteristics (pod colour, seed colour)is being followed.
The gene for pod colour and seed colour are on different homologous pairs ( not linked).
Each gene has two alleles and shows complete dominance.
With characteristics like these it important to have a clear allele key to avoid confusion.
The cross begins with a homozygous dominant plant for both pod colour and seed colour.
P1 Green Pod, Yellow seed X P2 Yellow pod, Green seed
G- Green pod
g- Yellow Pod
Parents are Homozygous double dominant X Homozygous double recessive
Meiosis I separates the alleles. MII separates the sister chromatids
In this case since the parent is homozygous at both loci, all gametes are identical.
Therefore the only gamete produced from P1 is GY, and P2 is gy
Therefore all fertilisations will be between GY and gy
The offspring(F1) genotype therefore is GgYy
All the offspring will have green pods and yellow seeds inside.
The probability of a Green pod with yellow seed is 100%, P=1.0
All other probabilities of phenotype are P= 0.00 or O %
F1 Cross = F1 X F1
Remember that F1 seeds must be germinated and then the plant grown before flowering and the cross performed.
Both the F1 are heterozygous at both gene loci.
We would predict that Mendel's law of independent assortment would apply. That is in meiosis at metaphase I, that the pod alleles will separate in anaphase independently of the alleles for seed colour.
There would be an equal probability of the combinations of the Pod alleles with seed alleles.
Meiosis, metaphase I and the random orientation of the homologous pairs on the equatorial plane, means that there are four possible gamete genotypes, all equally possible.
The means that since both parents are heterozygous for both alleles then the Punnet square is a 4 X 4 grid.
Each box represents a possible fertilisation of a probability 1/16.
This does not mean that there will be 16 offspring. The table is a probability chart for the possible genotypes of each fertilisation.
Eg GGYY, there is only one such genotype in the table. The probability of this genotype in the population is 1 in 16.
In a population of 160 offspring we would predict that 10 would have the GGYY genotype. The same logic can be applied to other genotypes. However remember that each new fertilisation shares the same probabilities.
There are nine possible genotypes in the population of offspring.
Mendel could not observe genotypes, rather he could count the phenotypes. This type of F1 cross produces what is known as the
F2 Dihybrid ratio = 9: 3 : 3 : 1
Such a ratio, indicates a cross between two heterozygote, 2 genes. unlinked, complete dominance at both loci.
Other dihybrid calculations are a variation on the above core crosses. Students should obtain and calculate as many examples as possible. However always try to make the connections between the cross and the process of meiosis, random orientation and metaphase I. Do not fall into the trap of memorising the pattern of these crosses, this will fail you in the exam. Understanding is the key.
Additional calculations are available
In some species there is one homologous pair which carry genes that determine gender. THis work has already been covered in the section on sex chromosomes.
There are other genetic conditions (also covered in topic 4) associated with the so called 'sex linked" conditions.
Read the next section 10.2.4 first and understand the concept of a linkage group which is key to understanding of this particular assessment statement.
The diagram shows the loci of genes A and B.
The alleles A is linked to B, a is linked to b.
The homologous pairs are held together like this during prophase I.
The crossover point occurs above the loci for gene A and below the gene loci for A.
The position at which the exchange occurs is called the chiasma.
The recombinants will form between non-sister chromatids which are crossing over.
The homologous pairs remain attached at the chiasma until anaphase I when they are pulled apart.
After anaphase II, the chromatids are separated.
The linked genes are:
A with B
a with b
recombinant a with B
recombinant A with b
The genotype of the gametes from meiosis above would be AB and ab.
After cross over there would also be Ab and aB.
However most cells of the cell undergoing meiosis to form a gamete will not cross over between these two points.
Gametes Ab and ab will be high frequencies.
Gametes Ab and aB will be low frequency.
When cross over occurs there will be low frequency of recombinants produced.
The genotypes of an organism is AaBb. This looks like a dihybrid such as those in the sections above. The two genes would be on separate homologous pairs, however there is an alternative and this is known as linkage.
These two genes are actually on the same chromosome.
The A allele is linked to the B allele along the length of DNA.
The a allele is linked to the b allele along the length of DNA.
In the image take after S-phase, the homologous pair is aligned as it would be found in Prophase I of meiosis.
It might be predicted that during gamete formation of fertilisation that the A allele will be followed by the B allele. This would be more like carrying out two simultaneous monohybrid crosses.
Example: Sweat Peas (Lathyrus odoratus)
A cross was made between a plants that where heterozygotes at both gene loci (PpLl).
When dealing with linked genes the linkage group is drawn on the horizontal with a line representing the chromosome. Using the above cross the cross would be represented.
Parental phenotypes are the same PpLl being heterozygous at both gene loci.
If the gene for flower colour and pollen grain shape are linked then the genotypes are shown as left.
The gamete genotypes are shown to the left with the alleles for flower colour and pollen grain shape shown linked by the horizontal line.
The possible recombinants from cross over are shown for one parent only.
These recombinant gametes will be at low frequency.
The grid shows the cross with the possible fertilisations in the boxes.
There four expected genotypes.
There are four recombinant genotypes but cannot be observed from phenotypes.
There are two recombinant phenotypes which would be actually observed during the experiment.
The result of W Batson's cross showed that the result are neither dihybrid and unlinked (9:3:3:1)or linked with no linkage (3:1)
The result shows a rough 3:1 ratio of Purple/Long : Red/Short
There are also recombinant phenotypes at much lower frequency.
Recombinants are recognised by:
Unpredicted combinations of characteristics. (Red Long, Purple Short).
Low frequency of new combinations of phenotype.
Statistical significant difference from ratios expected from either dihybrid and unlinked (9:3:3:1)or linked with no linkage (3:1)