1. The document discusses gene interactions and how they can modify Mendelian ratios by altering phenotypic expression. It provides examples of different types of epistatic interactions like complementary, duplicate, dominant, and recessive gene interactions. 2. It also covers complementation analysis, which is used to determine if two mutations that cause the same phenotype are in the same or different genes. If the mutations complement, producing wild-type offspring, they are in different genes, while failure to complement means they are in the same gene. 3. The functional consequences of mutations are described, including loss-of-function mutations like amorphic and hypomorphic, and gain-of-function mutations like hypermorphic and neomorphic.

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Gene InteractionsGene Interactions
GeneticsGenetics
Chapter 4, Part 2Chapter 4, Part 2

2. 4.3 Gene Interaction Modifies Mendelian
Ratios
• Genes work together to build the complex
structures and organ systems of plants and
animals
• The collaboration of multiple genes in the
production of a single phenotypic characteristic or
group of related characteristics is termed gene
interaction

3. • Geneticists use a variety of symbols for alleles
• Dominant alleles:
– an italic uppercase letter (D) or
– letters (Wr)
– an italic letter or group of letters with the + superscript (Wr+
)
• Recessive alleles:
– an italic lowercase letter (d) or
– an italic letter or group of letters (Wr)
Representing Alleles
S / Sm / Sm+
s / sm / Sm
Note that the mutant
usually gets the
letter name!
Note that the mutant
usually gets the
letter name!

4. Representing alleles in Drosophila
• Example: body color
– Ebony mutant phenotype is indicated by e
– Normal gray (wild-type) is indicated by e+
• e+
/e+
: gray homozygote (wild type)
• e+
/e: gray heterozygote (wild type)
• e/e: ebony homozygote (mutant)
OR
• +/+: gray homozygote (wild type)
• +/e: gray heterozygote (wild type)
• e/e: ebony homozygote (mutant)

5. Gene Interaction in Pathways
• Single-gene trait describes an inherited
variation of a gene that can produce a mutant
phenotype
• However, this is not a complete depiction of
genetic reality
• Numerous genes contribute to the normal red
eye color of Drosophila, including those
responsible for production of eye pigments or
transport proteins

6. Three Genes Involved in
Drosophila Eye Color
• The brown gene
produces an enzyme in
a pathway that
synthesizes a bright red
vermilion pigment;
mutant flies, bb, have
brown eyes
• Note: gene named after
the mutant, not the gene
it encodes for!
• The vermillion genes
produces an enzyme in
a pathway that
synthesizes a brown
pigment; mutant flies, vv,
have bright red eyes
• The white gene encodes
a transporter that carries
pigment to the eye; flies
that do not produce this
protein have white eyes

7. Three Distinct Types of Genetic Pathways
• Biosynthetic pathways are
networks of interacting genes
that produce a molecular
compound as their endpoint
• Signal transduction pathways
receive chemical signals from
outside a cell and initiate a
response inside the cell
• Developmental pathways direct
growth, development and
differentiation of body parts and
structures

8. HOW DO GENES CONTROL
BIOSYNTHETIC PATHWAYS?
HOW DO WE FIND A GENE
THAT AFFECTS A CERTAIN
PATHWAY?
Biosynthetic Pathways…

9. The One-Gene-One Enzyme Hypothesis
• George Beadle and Edward
Tatum were among the first to
investigate biosynthetic
pathways
• They studied growth variants of
the fungus, Neurospora crassa
• Their proposal, the one-gene-
one enzyme hypothesis came
out of their experiments
Red Bread Mold; http://www.biosci.missouri.edu/shiu/

10. Prototroph: an
organism that can
synthesize all of its
amino acids
Auxotroph: an
organism that has lost
the ability to
synthesize certain
substances required
for its growth and
metabolism
This experiment
looked at amino acids,
but you could look at
other synthetic
pathways!
Beadle and Tatum Experiment
CAUSING
MUTATION!

11. The Hypothesis Made a Connection
Between Genes, Proteins and Phenotypes
• Each gene produces
an enzyme
• Each enzyme has a
specific role in a
biosynthetic pathway
that produces the
phenotype
• Each mutant
phenotype due to
the loss or
malfunction of a
specific enzyme

12. Genetic Dissection to Investigate Gene
Action
• Biosynthetic pathways consist of sequential steps
• Completion of one step generates the substrate for the next
step in the pathway
• Completion of every step is necessary to produce the end
product
• Genetic dissection is an experimental approach taken to
investigate the steps of biosynthetic pathways

13. Genetic Dissections: Horowitz’s Experiments
on Met-
Mutants of Neurospora
• Horowitz’s analysis
aimed to:
• Determine the
number of
intermediate steps in
the methionine
synthesis pathway
• Determine the order
of the steps
• Identify the step
affected by each
mutation

14. Genetic Dissection: Results
of Horowitz’s Experiments
• Whether or not a mutant strain grows on a medium containing a component of the
pathway allows determination of the step at which the mutant is blocked
• Mutation of an enzyme will cause the pathway to become blocked.
• If we give an intermediate from before the block, we can’t pass the block and the mutant will not grow.
• If we give an intermediate after the block, the mutant will grow.
• The blocked step is also identified by the substance that accumulates in the
auxotroph
• Imagine one mutant:
X Give D, E or F: Will grow!Give A, B or C: Won’t grow!
What Accumulates?

15. 1. Met 1: only on minimal media + methionine, indicating it is the
last step of the pathway. Need to add methionine to get past the
block
2. Met 2: need minimal + homocysteine , therefore block is at step
that produces homocysteine. This result also tell us that
homocysteine is the substrate converted to methionine in the
biosynthetic pathway.
3. Met 3: grows on minimal, homocysteine &
cystathionine. This tells us that Met 3 is
blocked at the step that produces
cystathionine and that cystathionine
precedes homocysteine
4. Met 4 grows with any supplementation of
minimal media. This tells us that Met 4 is
defective at a step that precedes the
production of cysteine.

16. More Recent Adjustments to the Hypothesis
• Hypothesis confirmed!: Each gene produces an
enzyme
• Each enzyme has a specific role in a biosynthetic
pathway that produces the phenotype
• Recent Adjustments:
• Some protein producing genes produce transport
proteins, structural or regulatory proteins, rather
than enzymes
• Some genes produce RNAs rather than proteins
• Some proteins (e.g. β-globin) must join with other
proteins to acquire a function

17.  So far…So far…
 AaBb x AaBbAaBb x AaBb
 Crossing genotypes leads to aCrossing genotypes leads to a
phenotypic ratiophenotypic ratio
 BUT, Genes do not act alone.BUT, Genes do not act alone.
Now let’s look at how genesNow let’s look at how genes
interact to alter phenotypicinteract to alter phenotypic
ratios….ratios….
Gene InteractionsGene Interactions

18. Colorless
precursor
Colorless
intermediate
Purple
pigment
Enzyme C Enzyme P
The recessive c allele
encodes an inactive
enzyme
The recessive p allele
encodes an inactive
enzyme
Epistasis: Gene Interactions
• Epistatic interactions happen when an allele of one gene
modifies or prevents the expression of alleles at another
gene.
• Epistatic interactions often arise because two (or more)
different proteins participate in a common cellular
function
– For example, an enzymatic pathway

19. No Interaction (9:3:3:1 Ratio)
• The expected 9:3:3:1 ratio is
seen in the absence of
epistasis: when the genes do
not interact to change the
expression of one another
www.integratedbreeding.net
Dihybrid cross, F2 progeny

20. • Cross involving the brown and vermillion
genes in Drosophila
• When pure-breeding brown flies (b/b;
v+/v+) are crossed to pure-breeding
vermillion flies (b+/b+; v/v), the F1 all have
wild type red eyes (b+/b; v+/v)
• When the F1 are interbred (b+/b; v+/v x b+/b; v+/v ), the F2
are:
• 9/16 b+/-; v+/-, wild type, red eyes
• 3/16 b/b; v+/-, brown eyes
• 3/16 b+/-; v/v, vermillion eyes
• 1/16 b/b;v/v, white eyes
• The results show that the
genes are not undergoing
epistatic interaction with one
another
No Interaction (9:3:3:1 Ratio)

21. Epistatic Interactions
• A minimum of two genes are
required for epistatic
interactions; these usually
participate in the same pathway
• There are six ways epistasis
could affect the predicted
9:3:3:1 dihybrid ratio

22. Gene interaction alters the classic 9:3:3:1 ratio seen in the F2
progeny of the dihybrid cross!
Gene interaction alters the classic 9:3:3:1 ratio seen in the F2
progeny of the dihybrid cross!
Epistatic Interactions

23. Complementary Gene Interaction (9:7 Ratio)
• Bateson and Punnett crossed two pure-breeding strains of white flowered sweet peas
• They found all the F1 were purple flowered; the F1 x F1 cross yielded 9/16 purple and 7/16
white flowered progeny
• They recognized that the two genes interact to produce the overall flower
color; when genes work in tandem to produce a single gene product, it
is called complementary gene interaction

24. Duplicate Gene Action (15:1 Ratio)
• The genes in a redundant system have duplicate
gene action; they encode the same product, or they
encode products that have the same effect in a
pathway or compensatory pathways

25. Dominant Gene Interaction (9:6:1 Ratio)
• Plants that have dominant allele(s) for just one of either of the genes will have
round fruit and those with only recessive alleles of both genes will have long fruit
• Dominant for either gene (A or B), equals one phenotype. (3+3 = 6)

26. Recessive epistasis (9:3:4)
• B and b for black and brown melanin (MC1R gene)
• E: controls deposition of pigment in hairs (TRYP1 gene)
• ee is epistatic
• Recessive epistatsis causes yellow coat color

27. Recessive epistasis (9:3:4)
Brown
eumelanin
Eumelanin
deposition
Black
eumelanin
Not able to
deposit!
May or may not be correct in your book!May or may not be correct in your book!

28. Dominant Epistasis (12:3:1 Ratio)
• In dominant epistasis, a dominant allele at one locus will
mask the phenotypic expression of the alleles at a second
locus, giving a 12:3:1 ratio
• E.g. in foxglove flowers a dominant allele at one locus
restricts the deposition of pigment to a small area of the
flower

29. Dominant Suppression (13:3 Ratio)
• In dominant suppression, a dominant allele at one locus completely
suppresses the phenotypic expression of the alleles at a second locus,
giving a 13:3 ratio
• In chickens, the C allele is responsible for pigmented feathers and the c allele for white
feathers
• The dominant allele of a second gene, I, can suppress the color producing effect of the C
allele, leading to white feathers in both C/- and c/c individuals
Dominant I suppresses dominant
pigment production
Dominant I suppresses dominant
pigment production

30. YOU MUST KNOW FOUNDATION
FIGURE 4.21 (p. 130-131!)

31. 4.4 Complementation Analysis Distinguishes
Mutations in the Same Gene from Mutations
in Different Genes
• When geneticists encounter organisms with the
same mutant phenotype, they ask two questions:
1. Do these organisms have mutations in the same
or in different genes?
2. How many genes are responsible for the
phenotypes observed?
Ex. Two botanists working with petunias both discover a white flower
mutation. One works in California and one works in the Netherlands.
Are the mutations in the same genes?
Ex. Two botanists working with petunias both discover a white flower
mutation. One works in California and one works in the Netherlands.
Are the mutations in the same genes?

32. HOW DO WE KNOW IF THE
GENETIC MUTATION IS THE
SAME?
We have two Drosophila with the same
phenotypic mutation….

33. Genetic Complementation
Analysis
• Genetic heterogeneity is when
mutations in different genes can
produce the same or very similar
mutant phenotypes
• Mating of two organisms with similar
mutant phenotypes can lead to wild
type offspring, a phenomenon called
genetic complementation
• Complementation testing is when two
pure breeding organisms with similar
mutant phenotypes are mated
• If complementation occurs, wild type
offspring are obtained and the
mutations are known to affect two
different genes
• When the mutations fail to
complement, the offspring have the
mutant phenotype and the mutations
are known to affect the same gene
Example of a complementation test.
Two strains of flies are white eyed because of two different
autosomal recessive mutations which interrupt different steps
in a single pigment-producing metabolic pathway.
A B

34. Complementation Analysis
• In complementation analysis
multiple crosses are performed
among numerous pure breeding
mutants to try to determine how
many different genes contribute
to a phenotype
• Mutations that mutually fail to
complement one another are
called a complementation
group
• Can’t get back to wild-type!
• Mutation is on the same gene!
• A complementation group in this
context refers to a gene

35. Genetic Complementation?
-Vermilion vs. White
-Apricot vs. Buff
Genetic Complementation?
-Vermilion vs. White
-Apricot vs. Buff

36. + = cross of pure-breeding mutants yield wild-type (complements)
- = cross of pure-breeding mutants yields only mutant progeny
+ = cross of pure-breeding mutants yield wild-type (complements)
- = cross of pure-breeding mutants yields only mutant progeny
-Mutations that fail to complement each other are on the same gene
-Complementation group: consist of one or mutants of a single gene

37. WHAT ARE THE FUNCTIONAL
CONSEQUENCES OF
MUTATION?
There are SO many different eye colors for Drosophila!

38. Functional Consequences of Mutation
(See Fig. 4.1)
• A wild type phenotype is produced when an organism has
two copies of the wild type allele
• Mutant alleles can be:
• Gain-of-function, in which the gene product acquires a
new function or express increased wild type activity
• Loss-of-function, in which there is a significant
decrease or complete loss of functional gene product

39. Loss-of-Function Mutations
Amorphic = no functionAmorphic = no function
Hypomorphic = less functionHypomorphic = less function

40. Dominant Negative Mutations
• Multimeric proteins, composed of two or more polypeptides
that join together to form a functional protein are particularly
subject to dominant negative mutations
• These are negative mutations due to their “spoiler” effect on
the protein as a whole

41. Gain-of-Function Mutations
Hypermorphic = more functionHypermorphic = more function
Neomorphic = new functionNeomorphic = new function

42. Questions?Questions?