Mendel’s Monohybrid Crosses
1. Here we will summarize Mendel’s first series of crosses. In these crosses, he started with purebreeding peas that always produced the same color flowers. One strain produced purple flowers,
the other white flowers. These pure-breeding plants were the Parental or P generation. The
offspring that resulted from these crosses are called the F1 generation. All the F1s had purple
flowers. Next he performed a “self cross” of the F1’s, in other words, he crossed them with each
other. Their offspring are the F2 generation. He obtained the same results (shown below) for all
the characters that he examined.
P
F1
F2
Pure-Breeding
Phenotype
purple
X
Pure-Breeding
Phenotype
Æ
Phenotype =
purple
Æ
self-cross
Phenotypes =
3 Purple & 1 white
white
Questions to ask yourself about these 2 crosses:
a. How does the phenotype (appearance) of the F1 generation prove that inheritance is not the
result of “blending” ie. an appearance that is a blend of both parent’s traits? Hint, compare to
the original parents.
b. How does the phenotype (appearance) of the F2 generation prove that inheritance is not the
result of “blending” ie. an appearance that is a blend of both parent’s traits? Hint, compare to
the original parents.
2. Mendel hypothesized that there were “Factors” (or “Particles” or “Elements”) in the plants that
determine what the phenotype would be. We now know that these factors are the Genes. He
reasoned that
o
each plant had 2 copies of these Factors for each trait;
o
one factor came from the mother and one from the father.
o
This is Mendel’s Law of Segregtion.
o
there were only 2 forms of each Factor. For example, one Factor for purple flower color and
one Factor for white flower color.
o
a plant may have two of the same Factor or one of each form.
o
He also described one Factor as Dominant and the other as Recessive depending upon which
phenotype is seen in the F1 generation.
We now use different terms to describe these Factors and their alternate forms.
GENE = the heritable factor
ALLELE = the variant forms of the gene.
Question: Which trait is Dominant in the crosses shown in the example above?
3. Now we will add the Genotypes for each individual to the chart. We will represent the dominant
allele with a capital A and the recessive allele with a lower-case a. I will put parentheses around
the gametes.
Gamete
P
Pure-Breeding
Phenotype =
purple
Genotype =
AA
F2
Gametes
Phenotypes =
Meiosis
Æ
(A)
Phenotype =
Æ
X
Pure-Breeding
Phenotype =
white
Genotype = aa
F1
Meiosis
Æ
purple
Genotype =
(a)
3 Purple & 1 white
Meiosis
Æ
(A )
(a )
Aa
Æ
Genotypes =
Selfcross
AA, aA, Aa, aa
These are all the possible
combinations of the
gametes.
Individuals that have 2 different alleles for the same character are HETEROZYGOUS.
Individuals that have 2 identical alleles are said to be HOMOZYGOUS. If both alleles are
recessive, it is described as Homozygous Recessive. If both alleles are dominant, it is
Homozygous Dominant.
Questions:
a. In which generation are there Homozygous individuals?
b. In which generation are there Heterozygous individuals?
Remember that each parent has 2 alleles for each trait but contributes only 1 allele to each gamete
formed during meiosis (due to “Segregation” in meiosis).
4. The Punnett Square is a device for representing all the potential combinations of alleles in each
generation. Use the handout titled “Mendel-Punnett” to learn how a Punnett Square works.
Use a Punnett Square to represent the self-cross of the F1 generation (see page 115 and my
handout). Each box in the Punnett square represents 1 individual in the F2 generation.
Examine the boxes that represent the F2 in your Punnett Square and answer the following
questions about their genotype.
a. How many are Homozygous Dominant ?
b. How many are Homozygous recessive?
c. How many are Heterozygous?
The “Genotype ratio” for this generation is the number of Homozygous Dominant :
Heterozygous : Homozygous recessive. In this case it is 1 : 2 : 1
Look at the boxes in the Punnett Square that represent the F2 and consider their phenotypes ie.
their physical appearance. Remember that one allele is dominant and the other is recessive.
a. How many will show the dominant form of the trait?
b. How many will show the recessive form? This is the “phenotype ratio”.
The “Phenotype ratio” is the ratio of individuals with the dominant phenotype : individuals
with the recessive phenotype. In this case it is 3:1.
5. What is the LAW OF SEGREGATION? What event in Meiosis is the physical basis for
Segregation? See my meiosis figure. Be able to explain.
Dihybrid Crosses
Mendel then investigated the inheritance patterns seen when 2 or more characteristics are considered.
He considered two alternative patterns when looking at 2 characters:
Hypothesis (a) the heritable factors are transmitted independently of each other.
Hypothesis (b) they are linked together and thus transmitted together.
I will describe a series of crosses involving 2 characters that show the different outcomes predicted by
each hypothesis.
Consider 2 characters which each have 2 traits:
Flower color = purple or white
plant height = tall or short
Use the following symbols for the genotypes:
Flower color = purple (A) or white (a)
plant height = tall (B) or short (b)
Cross pure breeding purple/tall plants with white/short plants. Their genotypes would be AABB x
aabb. The AABB plants can only make one type of gamete = (AB). The aabb plant can only make one
type of gamete = (ab). The outcome of this cross, the F1, will all be purple/tall and the genotype will
be AaBb.
Next, “self” the F1s. At this point, differences between the two hypotheses will become evident:
(a) If the heritable factors are linked together and inherited together, then there will be only 2 type of
gamete: (AB) and (ab)
Hypothesis (a) will produce 2 phenotypes in the F2. It can only produce phenotypes that are the
same as the original parents and they will be in the same ratios as we have seen in the monohybrid
crosses.
(b) If the heritable factors are transmitted independently of each other, there are 4 possible gametes:
(AB), (Ab), (aB), and (ab). These are all the possible gametes that can form if there is independent
assortment during meiosis.
We want to see all the possible offspring that can result from “self crossing” the F1 generation.
Use a Punnett Square to do this. Write in the 4 possible gametes across the top and down the left
side of the Punnett Square. The row across the top represents all the gametes that the male plants
could produce. The female gametes are listed down the left side. Next combine the male and
female gametes to produce the F2 generation. Fill each square by writing in the gametes shown at
the top of the column and the gametes that are at the left of the row. For example, the box in the
lower left corner is filled by the (AB) at the top of the left column and the (ab) to the left of the
bottom row. As a second example, the box in the lower right corner is filled with the (ab) from the
top of the right column and the (ab) from the left of the bottom row. This gives aabb. Each box
contains the genotype of one individual in the F2. The 16 boxes shown here are ALL the possible
combinations of the 4 types of gametes that the F1 generation can form. Some individuals will be
homozygous for both traits, some will be heterozygous for both traits, and some will be
homozygous for one trait but heterozygous for the other.
AB
Ab
aB
ab
AB
AABB
AABb
AaBB
AaBb
Ab
AABb
AAbb
AaBb
Aabb
Dark purple = purple flower AND tall plant.
Pale purple = purple flower AND short plant.
aB
AaBB
AaBb
aaBB
aaBb
ab
AaBb
Aabb
aaBb
aabb
Yellow = white flower AND tall plant.
Green = white flower AND short plant.
1. Count the number with the dominant phenotype for Flower color and the dominant phenotype
for plant height. They will be of genotype AABB or AABb or AaBB or AaBb (A_B_ is a shorthand way of writing this). Place a check mark in the box to help you count. These boxes are all
shaded dark purple. How many of them are there? 9
2. Count the number with the dominant phenotype for flower color and the recessive phenotype
for plant height. What is their genotype? AAbb or Aabb. These boxes are colored pale purple.
How many of them are there? 3
3. Count the number with the recessive phenotype for flower color and the dominant phenotype
for plant height. What is their genotype? aaBB or aaBb. These boxes are pale yellow. How
many of them are there? 3
4. Count the number the recessive phenotype for both flower color and plant height. What is the
genotype? aabb How many of them are there? 1
This is the outcome that Mendel found when conducting dihybrid crosses. The F2 always shows
new combinations of the 2 alleles for the 2 traits. When 2 genes that control different traits are on
different chromosomes, they will always give a 9:3:3:1 ratio of phenotypes. This is the evidence
for Mendel’s Law of Independent Assortment which asserts that the heritable factors are inherited
independently of each other.
You can prove to yourself that each trait (when considered by itself) still follows the rules that
Mendelsaw in Monohybrid crosses by doing the following:
‚ Count the number with the dominant phenotype for flower color. They will be of genotype AA
or Aa (A_ is a short-hand way of saying this). You should ignore plant height for now. These
will be all of the dark purple and pale purple boxes.
‚ Count the number with the recessive phenotype for flower color. They will be of genotype aa.
These will be the green and white boxes.
‚ What is the ratio of A_ to aa? 12:4 How does this ratio compare to the ratio that we saw in the
Monohybrid cross F2? 12:4 is the same as 3:1, the ratio seen in a monohybrid cross.
‚ Make the same counts for plant height phenotype (B_ and bb). B_ will be the dark purple and
green boxes. bb will be the pale purple and white box. What is the ratio of B_ to bb? 12:4 =
3:1 How does this ratio compare to the ratio seen in the Monohybrid crosses?
What is Mendel’s LAW OF INDEPENDENT ASSORTMENT? Briefly explain how the events
during Metaphase I can explain Independent Assortment. See my meiosis figure.
How do the results above explain why brothers never look exactly alike?