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EPISTASIS
v2.1/07
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Prerequisite: Dihybrid Cross
In the dihybrid cross, you dealt with two different traits found on two different pairs of
chromosomes. Setting up an epistasis problem will be much the same as a dihybrid problem.
That is, you will be dealing with two sets of genes. The difference will be in interpreting the
results (i.e. figuring out the phenotypes).
What is epistasis? Epistasis is a term used to describe the interaction of genes at different loci
(positions) on the chromosomes. Whenever a gene at one locus influences the expression of a
gene at another locus, the first gene is said to be epistatic to the other. The first set of genes
“masks” or suppressed the expression of the second set of genes.
A common epistatic trait is albinism (lack of pigment). You are all familiar with white
laboratory rats. These are albino rats. Rats may also have agouti (grey) or black coats. These are
determined by the alleles G and g. The genes for albinism are a separate set of alleles. These
genes are either A or a.
Key:
G = gray
g = black
A = color or pigment is present
a = color or pigment is absent
Because we’re talking about two different genes, the genotype written for an animal would have
four letters – just like when we did the dihybrid crosses. So a rat might be genotype GGAa for
this combination of two genes.
How do these two genes interact? First, let’s think about each gene separately. If a rat is Gg,
then it would be a gray rat. If the rat is homozygous recessive for coat color, or gg, it would be a
black rat. How about the other gene? The albino gene can be thought of as an “on/off” switch
for coat color. Black or gray coat cannot be expressed when the albino gene is homozygous (aa)
because this gene combination doesn’t let ANY pigment get expressed in the fur. For example, a
rat who is GgAa would be gray, but a rat that is Ggaa is albino (white) regardless of the
gray/black gene alleles.
What color would a rat who is GGaa be?
How about if the rat was genotype ggAa?
Try writing the genotypes for the following rats:
Gray, homozygous for both traits:
Black, homozygous for both traits:
Gray, heterozygous for both traits:
Black, heterozygous for pigment:
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Answers:
1. GGAA
2. ggAA
3. GgAa
4. ggAa
Here are two more problems for you to try:
Key:
G = gray
g = black
A = color or pigment is present
a = color or pigment is absent
Problem 1: Two black rats have albino and black offspring. Draw a Punnett Square and show
all genotypes.
 Black rats must have genotype ggAa in this case.
 Recall the gamete formation unit to form the gametes for each of these rats:
A
gA
a
ga
g
Each black rat could then form two kinds of gametes:
gA and ga.
 To find the offspring, fill in the Punnett Square:
gA
ga
gA
ga
 Check each offspring. Any that contain the A allele will have a color and any that
contain the aa pair of alleles will be albino.
1. What percent of the offspring of this cross is albino?
2. What is the genotype of the albino offspring?
3. What percent of the offspring of this cross are black?
4. Is its possible for these parents to produce any grey offspring? Why or why not?
Answers: 1. 1/4th or 25% 2. ggaa
3. 3/4th or 75%
only had the recessive alleles for coat color.
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4. No, because the parents
Problem 2: A gray rat, heterozygous for both traits, is mated with a black rat heterozygous for
pigmentation. What are the offspring? Give the phenotypes after you have drawn a Punnett
Square.
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Answer sheet: EPISTASIS
Problem 1: Mating two black rats:
gA ga
gA
ggAA ggAa
ga
ggAa ggaa
Problem 2: Mating a gray and a black rat:
Genotype of gray rat = GgAa, gametes are GA, Ga, gA, ga.
Genotype of black rat = ggAa, gametes are gA and ga
GA
Ga
gA
ga
gA
GgAA
GgAa
ggAA
ggAa
ga
GgAa
Ggaa
ggAa
ggaa
Phenotypes are:
2/8 or 1/4 albino
3/8 are gray
3/8 are black
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