Biology 212 General Genetics
Spring 2007
Lecture 6: “Gene Interactions”
Reading: Chap. 2 pp. 51-66
Lecture Outline:
1. Human Pedigrees
2. Dominance
3. Epistasis
4. Complementation
Lecture
1. Human Pedigrees


controlled breeding experiments unethical in humans
therefore analyze human pedigrees
pedigree=family tree showing phenotypes of individuals
Symbols used for pedigree analysis: Fig. 2.17
a. Pedigree for an autosomal dominant trait
autosomal=traits not on the sex chromosomes
Example:
HD=Huntington’s disease
 Affects both sexes
 Autosomal dominant trait
 Progressive nerve degeneration
 Results in severe physical and mental disability --> leads to death
 Late onset—strikes after reproductive age
Autosomal dominant trait
HD=Huntington’s
hd=normal
hdhd HDhd
Probability of ½ that child will grow up to be affected.
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Pedigree of large family with HD Fig. 2.18
b. Autosomal recessive trait




albinism
absence of pigment in skin, hair, iris of eyes
affects both sexes
usually transmitted by unaffected parents “carriers”
A=normal
A=albino
Aa
A- A-
Aa
aa
A-
Large pedigree for albinism Fig. 2.19
Also shows inbreeding
2. Dominance
a. Simple dominance
Mendel studied traits with clear dominance/recessiveness.
However for many traits heterozygote may have intermediate phenotype or may display
half as much of the gene product.
Round vs. wrinkled peas
Shows simple dominance at morphological level
Round
WW
Round
Ww
2
Wrinkled
ww
Shows incomplete dominance at microscopic or molecular level
microscope field
starch
Large round grains
Large irregular grains
Small irregular grains
b. Incomplete dominance: When the phenotype of the heterozygous type is
intermediate between the phenotypes of homozygotes.
Example: Snapdragons
P1
F1
F2
red
II
x
pink
Ii
¼ red II
½ pink Ii
¼ ivory ii
ivory
ii
x pink
Ii
1
2
1
Therefore a 1:2:1 ratio is observed in the F2 progeny, not a 3:1 ratio as is observed when
you have simple dominance.
c. Codominance: When the phenotype of the heterozygote represents the
expression of both alleles
Example: Seen in ABO blood groups
A, B, and O encode different antigens on the surface of blood cells
A antigen
Enzyme adds N-acetylgalactosamine to
carbohydrate on surface protein
Enzyme adds galactose to carbohydrate on
surface protein
No sugar group added
B antigen
O antigen
Genetic control of human ABO blood groups: Table 2.3
Genotype
IAIA
IAIO
IBIB
IBIO
IOIO
IAIB
Antigens present
A
A
B
B
Neither A nor B
A and B
3
Blood type
A
A
B
B
O
AB
Blood groups show dominance
IA is dominant to IO
IB is dominant to IO
But IA and IB are codominant, both traits are expressed
d. Expressivity and penetrance
Some genes are variably expressed
Variable expressivity=genes that are expressed to different degrees in different
organisms. Different genetic diseases may have different severity.
Penetrance=refers to the proportion of organisms whose phenotype matches their
genotype for a given trait.
3. Epistasis=Gene interaction that results in the F2 dihybrid ratio of 9:3:3:1 being
modified into some other ratio.
Generally one gene influences the expression of the other.
Example: Flower color in peas: See Fig. 2.24
P1
white
CCpp
x
white
ccPP
F1
CcPp purple
x
CcPp purple
F2:
CP
Cp
cP
cp
CP
CCPP
CCPp
CcPP
CcPp
Cp
CCPp
CCpp white
CcPp
Ccpp white
cP
CcPP
CcPp
ccPP white
ccPp white
Results: purple: white
9:7
Two genes control color. Gene 1: Colored vs. colorless
C- = CC and Cc
cc
colored flowers
colorless
4
cp
CcPp
Ccpp white
ccPp white
ccpp white
Gene 2: purple vs. white
P- = PP and Pp
Pp
purple flowers
white
The colorless/colored gene influences expression of the purple gene


Recessive cc is epistatic to PP or Pp; no color is produced
Leads to modified ratio 9:7 rather than 9:3:3:1
Other modified ratios: See Fig. 2.25
For extra practice work through more examples
text pp. 61-62
12:3:1 Ratio
Show how a P1 cross of white (aabb) and black (AABB) produces F1. Then when these
are interbred a 12:3:1 ratio (black:gray:white) is produced.
9:4:3 Ratio
Show how a cross of P1 AACC (agouti) and aacc (albino) produces an F1 AaCc (agouti).
When these are interbred, F2 are produced in ratio of 9 agouti:4 albino:3 black.
9:6:1 Ratio
Show how a cross of RrSs and RrSs pigs can produce a ratio of 9 red:6 sandy:1 white.
4. Complementation: When two recessive mutations affecting the same trait are
located in different genes.
To determine whether mutations are in the same gene or different genes, you perform a
“complementation test”.
Fig. 2.26
P1 white
Mutant strain 1
x
P1 white
Mutant strain 2
Observe F1 generation
If F1 have white flowers, then both
mutations affect same gene
NO COMPLEMENTATION
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If F1 have purple flowers, then each
mutation affects a different gene
COMPLEMENTATION
Principle of complementation: If two recessive mutations are alleles of the same gene,
then the phenotype of an organism that contains one copy of each mutation is mutant; if
they are alleles of different genes, then the phenotype of an organism that contains one
copy of each mutation is wild type.
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