Dihybrid Crosses, Recombinants, Gene Linkage and Chi-squared Test
Assessment Statement
10.2.1
10.2.2
10.2.3
10.2.4
10.2.5
10.2.6
1.1.5
1.1.6
Calculate and predict the genotypic and phenotypic ratio of offspring of
dihybrid crosses involving unlinked autosomal genes
Distinguish between autosomes and sex chromosomes
Explain how crossing over between non-sister chromatids of a
homologous pair in prophase I can result in an exchange in alleles
Define linkage group
Explain and example of a cross between two linked genes
Identify which of the offspring are recombinants in a dihybrid cross
involving linked genes
Deduce the significance of the difference between two sets of data using
calculated values for t and the appropriate tables (Chi Squared Test)
Explain that the existence of a correlation does not establish that there is a
causal relationship between the two variables
Mendel also investigated the simultaneous inheritance of two pairs of contrasting
characters, using the same garden pea plants. He referred to this as a dihybrid cross.
To test the theory, Mendel crossed pure breeding plants from round seeds with yellow
cotyledons (seed leaves) with pure breeding plants from wrinkled seeds with green
cotyledons. All of the F1 generation were round, yellow peas. When the F1 generation
was allowed to cross, Mendel found that there was a ratio of 9 yellow round peas, 3 green
round peas, 3 yellow wrinkled peas and 1 green wrinkled pea. The result from the
monohybrid crosses was still valid. Yellow is dominant to green, and round / smooth is
dominant to wrinkled.
Mendel’s Dihybrid Cross
Further Test Crosses – Drosophilia melanogaster (Fruit Fly)
Mendel died in 1884, in relative obscurity in the scientific community. Many did not
know of his work or thought he did not understand his results. When his papers were
rediscovered in 1900, Mendel’s results came to be confirmed and then extended by many
others.
One was by Thomas Morgan in 1908. He used the Drosophilia as an organism to test
Mendelian Genetics. For his work, he was awarded the Nobel Prize in 1933.
Morgan showed:
 Mendel’s factors are linear sequences of genes on chromosomes (what is now
known as the Chromosomal Theory of Inheritance)
 discovered sex chromosome and sex linkage
 demonstrated crossing over, and the exchange of alleles between chromosomes,
resulting from chiasmata formed during meiosis
Why was the fruit fly chosen? It is quite common around rotting vegetable material and
the female lays hundred of eggs, which produce hundreds of offspring. The time from
hatching to adult takes about 10 days. They can be anaesthetized, and sorted. They have
four pairs of chromosomes. The characteristics, or phenotypes, are controlled by genes
not on the sex chromosomes. Therefore, the characteristics are controlled by the
autosomes, and are not sex-linked.
Morgan’s Test Cross
If a wild type fly is crossed with a mutant type, what will the offspring be. The wild type
has a white body and normal wings. These are dominant traits. The mutant has a red
body and vestigal wings (short). What will the F2 generation be? (The genes in both
cases are not linked)
Examples
1.
Two tomato plants were crossed. Tall plants (T) are dominant to short plants (t)
and round tomatoes (R) are dominant to oblong fruit (r). If one parent with the
genotype TTRr, is crossed with a parent homozygous recessive for both traits,
what will the F1 generation phenotypes and genotypes be?
2.
If the genotypes for the tomatoes above were TtRr, Ttrr, ttRr and ttrr, which two
pairs would a plant breeder select in order to obtain a phenotype ratio of 1:1:1:1?
3.
Frumpy plants have small flowers with green petals, which are dominant traits.
Large white flowers are recessive, and the genes are not linked. A plant with
small green flowers was crossed with one with large white flowers and the
offspring had either small green flowers or large green flowers. Determine the
genotypes of the parents. If the large green flowers were allowed to self
pollinate, determine the genotypes and phenotypes of the offspring.
Recombinants
Recombination is the reassortment of genes or characters into different combinations
from those of the parents.
When cross over occurs (previous notes), we see that the genes on some of the
chromosomes are kept the same, while others change. Recombination can be applied to
non-linked genes but is often restricted to linked genes, which we will talk about later.
We can identify the recombinants in the above two examples. What are the parental
combinations and what are the recombinants?
Autosomal Gene Linkage
Autosomes are chromosomes, which are not sex chromosomes. They are the other 22
pairs, for example, in humans.
Sex chromosomes are those chromosomes, which help in determining the sex of an
individual (XX or XY).
In our studies, we have studied that genes independently assort. In other words, they are
located on different chromosomes, and have no effect on each other.
In reality, we only have 22 autosomes or homologous pairs, along with the sex
chromosomes. There are many more genes (about 60,000!) in our genetic make-up. This
means that some must be located on the same chromosome.
When cross over occurs in Prophase I, the alleles are exchanged, between non-sister
chromatids of a homologous pair. But since some genes are on the same chromosome,
when cross over occurs, the genes go together. This is called a linkage group.
Linkage group – a group of genes whose loci are on the same chromosome.
The Reason
Linked genes were discovered in 1906, by R. C. Punnett and W. Bateson. (the square
guy!) When Mendel’s work was rediscovered, geneticists looked at Mendel’s ratios.
Some worked out to the characteristic 9:3:3:1 ratio for the F2 generation. In some cases,
using plants, the ratio for the F2 generation was 7:1:1:7! What happened?
Let’s go over an example.
In sweet peas, purple flowered plants are dominant to red, and long pollen plants are
dominant to round pollen. When a homozygous purple and homozygous long plant is
crossed with a red and round pollen plant, the resulting F1 phenotype is all purple and
long, as you would expect.
What would be the expected ratio for the F2 generation, if two of the previous plants are
mated?
This was not the ratio. The results were:
7 purple long: 1 purple round: 1 red long: 7 red round
This seemed to mean that more PL and pl gametes were produced than the Pl and pL
gametes. What is the explanation?
If no cross over occurs, the phenotypes will have the parental characteristics and few
recombinants.
Example
In fruit flies (Drosophila) body colour and wing length are linked. A tan body (b+) is
dominant to a black body (b) and long wings (w+) are dominant to short wings. Cross a
homozygous tan, homozygous long winged fly with a mutant black and short wing fly.
What would be the F1 and F2 generations in terms of genotypes and phenotypes?
1.
The snowy gull has white feathers and a yellow beak, but some birds have black
feathers and an orange beak. Black and orange are recessive and the genes are
linked. A pure breeding bird with white feathers and a yellow beak was mated
with one with black feathers and orange beak. The offspring had white feathers
and yellow beak. Two of these were mated and 75% of the offspring had the
dominant phenotypes and the rest had the recessive. Explain.
2.
Transparent wings and clubbed antennae are autosomely lined dominant alleles
in the fairy fly. Grey wings and smooth antennae are the characteristics for
being homozygous recessive for both alleles. A fly with grey wings and smooth
antennae was mated with an unknown fly and all of the offspring had transparent
wings and clubbed antennae. These offspring were all mated to flies, with the
recessive traits and the total offspring were:
i.
ii.
iii.
iv.
227 transparent wings, clubbed antennae
10 transparent wings and smooth antennae
9 grey wings, smooth antennae
231 grey wings, clubbed antennae
Determine the genotypes of all the flies.
The Chi-squared Test
When you do a cross, generally you need a large amount of real life data to ensure that
your results are significant. You would then compare the ratio with the expected ratio.
The chi-squared test is used to analyze the data produced from a mono or dihybrid cross,
and determine if your results were simply by chance, or if the results are significant.
The formula is as follows:
A probability of less than 5% is considered to be significantly different than the expected
value, and therefore the original hypothesis or expectations need to be re-examined. To
put it another way, the greater the value of the chi-squared above the critical value, the
greater the difference between the observed and expected values.
Once we calculate the chi-square, we have to look at our degrees of freedom.
Degrees of freedom are the number of observations minus the minimum number
required to uniquely define the figure. In other words, if we are looking at two classes of
a phenotype, we calculate the degrees of freedom by:
n - 1
where n is the number of classes. For example, if you are looking at red eyes vs. white
eyes, the degree of freedom would be 1, as 2-1=1 or n-1.
For every degree of freedom, there is a critical value. Since in Biology, the probability is
5% or 0.05, we look at the probability (chi-square) vs. the degrees of freedom for 5%.
See handout.
Basically, if your value is bigger than the critical value for 5%, the result is by chance
and your original assumption (hypothesis) is correct. If the value is smaller, there is a
less than 5% probability that the results are due to chance, so your original assumption
may be flawed.
Confused? Let’s do an example.
Ex. In a hypothetical cross, we expect the phenotypic ratio to be 1:1. The actual results
were 45:55, instead of 50:50. Is this deviation due to chance or is it significant?
Example 2
In a genetics experiment, tall pea plants were crossed with short pea plants. The resulting
F1 was self-fertilized and F2 consisted of 787 tall plants and 277 short plants. Does this
result confirm Mendel’s explanation?
Genetics Problems
1.
List the possible gametes that would result from the segregation of the following
parental genotypes.
a. RR
b. Mm
c. AABB
d. DdSs
e. gGHhRr
2.
In a species of plants, flowers appearing at the end of the stem are dominant to
flowers appearing in the middle of the stem. Predict the phenotype for this F2
generation, for a cross involving a homozygous dominant plant with recessive
plant.
3.
In garden peas, red flowers dominate white flowers and green pod colour is
dominant to yellow pod colour.
a. Cross two heterozygous red flowered plants with green pods
b. Cross a heterozygous red flowered plant with green pods with a plant that
is completely recessive for all the traits.
c. Cross two plants that are recessive for both traits.
4.
In snapdragons, tallness (T) is dominant to dwarfness (t), while red flower (Cr)
and white flower (Cw) show incomplete dominance. A dwarf red snapdragon is
crossed with a plant homozygous for tallness and white flowers. Give the
genotypes and phenotypes of the expected F1. If the F1 flowers were crossed,
what would be the expected F2 genotypes and phenotypes.
5.
A dominant allele, A causes yellow colour in rats. The dominant allele of
another independent gene, R, produces black coat colour. When two dominants
occur together (A_R_), they interact to produce grey. Rats of the genotype (aarr)
are cream coloured. If a grey male and a yellow female, when mated, produce
offspring in the following approximate ratio, 3/8 yellow, 3/8 grey, 1/8 cream and
1/8 black, what are the genotypes of the two parents?
6.
What are the genotypes of a yellow male rat and a black female that, when
mated, produce 46 grey and 54 yellow offspring? Does a chi-square test bear
you out?