CROSSES
INTRODUCTION: Now that we have experience with monohybrid crosses and Mendel’s first principle, let
us turn our attention to MENDEL’S SECOND PRINCIPLE, the PRINCIPLE OF INDEPENDENT
ASSORTMENT. This principle pertains to any cross where two or more genes or pairs of alleles are being
followed at the same time. Such a cross is known as a DIHYBRID CROSS. We will now study a dihybrid
cross in the pea plant to illustrate the principles of Mendel’s second principle. Mendel’s work with peas
demonstrated that the gene form (allele) for yellow seed color (G) is dominant to that for green (g). A
cross of two heterozygous yellow-seeded plants (Gg) produces offspring which occur in a phenotypic
ratio of 3 yellow: 1 green. Now let us follow through the F2 generation a cross between a plant which is
homozygous for tallness and for green seeds and a plant which is homozygous for dwarf and for yellow
seeds. The genotypes of these P1(first parental) plants are respectively TTgg and ttGG. Since we are
following two genes at the same time (the one for height and the one for seed color), we must represent each
of them in both parents, the female parent (tall, green) and the male (dwarf, yellow).
Mendel’s first principle (the principle of segregation) tells us that when gametes are formed only one
member of a pair of genetic factors enters a sex cell. Consequently, all the eggs of the tall, green parent
(TTgg) will be type Tg. All the sperms of the male parent (ttGG) will be tG. This is so, since the
homologous chromosomes separate at meiosis, and in this cross we are following to allelic forms of two
genes. The gene for height (and its two forms) is associated with one pair of homologous chromosomes. The
gene for seed color (and its two forms) is associated with a different pair of homologous chromosomes. This
is the same as saying that the pair of alleles for height (T and t) is located on one pair of homologous
chromosomes and the pair of alleles for seed color (G and g) is located on a different pair. Every normal
gamete must have one of the gene forms (alleles) for height and also one for seed color. Due to the fact that
each parent in the P1generation is homozygous for both characteristics, each parent forms only one class of
gamete: Tg eggs and tG sperms. Since gametes of these types will unite at the next fertilization, all
members of the F1 will be dihybrids (double heterozygotes), TtGg. They are all tall and produce yellow
seeds since tallness and yellow are dominants.
Now let us see what is to be expected when we cross two of these dihybrids. A dihybrid plant of the
genotype TtGg can form four different classes of gametes. The reason for this is due to chromosome
behavior at meiosis. At meiosis, the chromosome carrying the T can move to the same pole as the
chromosome with the G. As a result, the chromosome with the t and the one with the g will travel to
the opposite poles. Consequently, this meiotic division will produce two classes of gametes: TG and tg.
However, it is equally possible at meiosis for the chromosome with T to move to the same pole as the
chromosome with g and the chromosome with t to travel to the opposite pole with G. This meiotic
division will yield two other classes of gametes: Tg and tG. Since both types of arrangements at meiosis
are equally possible, this means that the dihybrid TtGg can form 4 classes of gametes in equal amounts:
TG, Tg, tG, and tg.
Although Mendel knew nothing of chromosome behavior, it should be clear to you from a review of
meiosis why a dihybrid maybe expected to produce four different classes of gametes. There is nothing which
ties together the different pairs of homologous chromosomes. The way in which one pair arranges itself on
the spindle at meiosis is independent of the way any other pair arranges itself. As a result, when two or more
pairs of genetic factors segregate from each other before gamete formation, they segregate independently of
one another. This is precisely what Mendel stated in his principle of independent assortment.
Using the Punnett square method, we place the four types of gametes from one parent on the top of the
square and the four types from the other parent on the side. When we bring these together in all possible
combinations, we see that a cross of two dihybrids gives a phenotypic ratio of 9:3:3:1. This is the classic
dihybrid ratio of Mendel. We know today that independent assortment takes place when we are following two
genes (the pairs of alleles) associated with two different pairs of homologous chromosomes When we follow
two genes or two pairs of alleles associated with the same chromosome, however, the principle of
independent assortment may not hold. We say that two genes are LINKED when their loci are found on the
same chromosome. Since they are tied together on the same chromosome, the two genes cannot separate
independently of each other. In the following exercise we will ignore linkage and follow only those genes
which are located on different chromosomes.
Answer all questions on a clean sheet of paper. Be sure to show all your work.
1. In the pea plant, W, represents the allele for round seeds and w stands for the contrasting gene form
(allele) for wrinkled seeds. G represents the allele for tallness and g the contrasting factor (allele) for
dwarf. Show the different types of gametes which can be produced by each of the following plants: WWGG,
WwGG, wwgg, Wwgg, WwGg, wwGg.
2. give the phenotypic ratios which are to be expected in the next generation in each one of the following
crosses:
(a) Wwgg x WWgg (b) WWgg x WwGg (c) a cross of two dihybrids (d) a test cross of a dihybrid.
3. In guinea pigs, short hair (L) is dominant to long hair (1). Black fur (A) is dominant to white or albino (a).
A female from a strain which is purebred for black fur and short hair is mated to a male from a strain
purebred for the albino condition and for long hair.
(a) What will be the phenotypes of the F I generation?
(b) If members of the F 1 generation are mated among themselves, what proportion of their offspring can be
expected to be homozygous for both traits? Give the genotypes and phenotypes of these homozygous animals.
4. In the human, the allele for normal skin pigmentation (A) is dominant to the albino trait (a), the absence
of pigment. A certain type of migraine headache (M) behaves as a dominant to the normal state (m), the
absence of headache. Write the genotypes of the following three persons:
(a) A phenotypically normal person whose mother was an albino with migraine.
(b) A person suffering from migraine who has normal skin pigmentation.
(c) An albino person who does not have headaches. However her normally pigmented parents both suffer
from migraine.

5. A certain woman is heterozygous in respect to both the skin pigmentation and headache characteristics.
She marries a man who is heterozygous at the skin pigmentation locus and who does not have headaches.
(a) Give the genotypes of the man and the woman and the different types of gametes each can produce.
(b) What phenotypes are to be expected among their children and in what ratio?
6. In the human, free ear lobes (F) are dominant to attached lobes (f). The alleles for normal (S) and for sickle
cell hemoglobin (s) show lack of dominance. The cells of the healthy heterozygote show some sickling, but
the person homozygous for sickle cell hemoglobin suffers from severe anemia. Give the genotypes of the
following two persons:
(a) A healthy woman with attached ear lobes whose red blood cells show some sickling.
(b) A man with free ear lobes whose mother also has free ear lobes but whose father has attached ones. This
man’s red blood cells show some sickling.
7. The man and woman described above marry and have children. What kinds of offspring are to be expected
and in what ratios?
8. Two persons are heterozygous for both characteristics discussed in the above two questions. They marry
and plan to have children. Answer the following:
(a) What is the chance that they will have at any birth a child with sickle cell anemia and who has free ear
lobes?
(b) What is the chance at any birth for a child whose cells show no sickling and who has attached ear lobes?
9. Using the information on migraine headache given in question #4 as well as the information in the above
three questions, write the genotypes of the following 3 persons:
(a) A man with migraine and attached ear lobes. His mother did not have migraine and had free ear lobes.
Both the man and his brother show some red blood cells which undergo sickling.
(b) A person with sickle cell anemia, who has free ear lobes, and who does not have headaches.
(c) a man who does not have sickle cell anemia and who has attached ear lobes and migraine headaches.
His child has sickle cell anemia, free ear lobes, and no headaches.
10. In certain breeds of chickens, the allele B is responsible for black feathers whereas the contrasting
allele b produces feathers that are white (except for flecks of black). Birds heterozygous for this pair of
alleles have blue feathers. Another pair of alleles influences the shape of the feathers. F produces straight
feathers whereas the contrasting gene form f’ results in the frizzled condition (twisted, brittle feathers). A
bird heterozygous for this pair of alleles is mildly frizzled. Give the phenotypic ratios to be expected from
each of the following crosses:
(a) A black, frizzled hen x a white straight rooster
(b) A white, frizzled hen x a blue, straight rooster
(c) A blue, mildly frizzled hen x a white, frizzled rooster
(d) A blue, mildly frizzled hen x a blue, mildly frizzled rooster.