Transcript not separate alleles at one genetic locus

MENDELIAN GENETICS
(MODIFICATION OF MENDEL RATIO)
CHAPTER 1.
(part two)
I. Epistasis
 Two or more separate genes (not separate alleles at one
genetic locus) interact to control a phenotypic character.
 If one gene locus prevents the expression of a second gene, the
first locus is epistatic to the second, and the second is
hypostatic to the first.
Example 1: H/h and AB pathway
A antigen
A allele
HH or Hh
Precursor H substance
H Substance
hh
B allele
B antigen
Epistasis example 2: Coat color in mice
Wt coat color is agouti - A (dominant to black); Nonagouti (black) coat color - a
Pigmentation expression - B (dominant to albino); No pigmentation (albino) - b
If individual is bb, then is albino regardless of allele at a locus due to gene interaction.
Agouti Markings
 A- hair made with bands of
black pigment and yellow
pigment.
 Aa hair all black.
Coat color example of epistasis, cont.
AABB (agouti) x aabb (albino)
Genotype Phenotype
AaBb x AaBb
A-BA-bb
aaBaabb
agouti
albino
black
albino
AaBb (all agouti)
F2 ratio Final phenotypic ratio
9/16
3/16
3/16
1/16
9/16
4/16
3/16
Due to gene interaction, we see a 9:3:4 F2 ratio. The b
locus is epistatic to the a locus.
Example of dominant epistasis, a 12:3:1 ratio
Inheritance of fruit color in summer squash: two loci together
control color and a dominant allele at one locus can mask
the expression of the alleles at the second locus.
A---
white
aaB- yellow
aabb green
A is dominant to a, and the a locus is
epistatic to the b locus.
Therefore, if AaBb is crossed to AaBb, the F2 is as follows:
A-BA-bb
aaBaabb
white
white
yellow
green
9/16
3/16
3/16
1/16
12/16 white
3/16 yellow
1/16 green
II. Novel phenotypes and F2 ratios
due to gene interaction
Summer squash fruit shape inheritance:
AABB - disc shape x aabb - long shape
F1 = AaBb - disc shape
F2 = A-BA-bb
aaBaabb
disc
sphere
sphere
long
9/16
3/16
3/16
1/16
9/16 disc
6/16 sphere
1/16 long
Summary of modified F2 ratios - was Mendel just wrong?
No, none of these cases has violated the principles of segregation and
independent assortment - just added complexity.
Practice Problem
In a plant, a tall variety was crossed with a dwarf variety. All F1
plants were tall. When two F1 plants were interbred, 9/16 of
the F2 were tall and 7/16 were dwarf.
Explain the inheritance of height by a) indicating the number of
gene pairs involved and b) by designating which genotypes
yield tall and which yield dwarf.
1. When studying a single character, a ratio expressed in 16
parts suggests that two gene pairs are “interacting” during
the expression of the phenotype.
2. A 9:7 ratio implies a dihybrid condition with epistasis.
3. From any dihybrid cross of double heterozygotes, we see
the following genotypic ratio:
A-BA-bb
aaBaabb
9/16
3/16
3/16
1/16
tall
dwarf
dwarf
dwarf
4. In our problem, we see a 9:7 phenotypic ratio. If the
3:3:1 groups above were lumped together, a 9:7 ratio
emerges.
Assign tall to any plant with both A-B- and dwarf to any
plant that is homozygous recessive for either or both the
recessive alleles.
III. Complementation Analysis Reveals Whether 2 Mutations
are at the Same Locus
The mutations do complement each
other, so they are in different
complementation groups.
Mutations do not complement, so
in same complememtation group.
IV. X Linkage
 Affects males and females
differently
 Example: red-green color
blindness
 Other examples in humans:


“3” or “8”??
Hemophilia
Duchenne’s muscular
distrophy
X-linkage means genes on X chromosome
 Sex chromosomes are “unlike” in many species; all others are
autosomes.
 In humans and Drosophila, males contain XY and females
contain XX.
 Y chromosome contains region of pairing homology with X:
pseudoautosomal region.
 Y contains few genes; X contains many.
Color Blindness Pedigree
Males are hemizygous
for this gene locus; only
females can be carriers.
Hemophilia in the European Royal Families
• Disorder is lethal or debilitating prior to reproductive maturation.
• X-linked recessive: heterozygous females are carriers with 50% chance
of passing trait to sons only.
• Arose from spontaneous mutation in Queen Victoria?
• Hemizygous males were affected in Russian, German, Spanish royal
families, trait not passed on in British royal family.
V. Is phenotype always due to genotype?
Penetrance vs. Expressivity
If 70% of the individuals show the
mutant phenotype, then the
penetrance is 70%.
Expressivity reflects the range of
expression of the mutant genotype.
Variable penetrance or expressivity?
Temperature effects on phenotype
Temperature can affect coat coloration.
Siamese cat fur in the extremities is
darker due to cooler temperatures. The
enzyme making darker pigment doesn’t
work well at the higher temperatures in
the rest of the body.
Temperature also affects primrose flower color and fur of
Himalayan rabbits.
Influence of chemicals on phenotype
PKU: phenotype is mental retardation due to metabolic
disorder; severity is affected by diet and whether phenylalanine
is restricted in the diet.
Phenocopies: non-hereditary phenotypic modification that
mimics a phenotype caused by a known gene mutation.
example: deafness can be caused in a developing baby
if a mother is infected with rubella during the first 12 weeks of
pregnancy; also caused by homozygous, recessive mutation