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Welcome to IB 201!
Genetics and Evolution
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A phenotype ratio of 9:3:3:1 in the
offspring of a mating between two
individuals that are heterozygous
for two traits occurs when:
A. the genes reside on the same
chromosome
B. each gene contains two
mutations
C. the gene pairs assort
independently during meiosis
D. only recessive traits are scored
E. none of the above
Strategies for Learning Genetics
Strategy
Exam Grades
Take notes during lecture
and review before exam
Worst
Summarize lecture in your
own words after class
Intermediate
Self-questioning during
lecture
Best
Why Genetics and Evolution?
Evolution is unifying principle of biology.
“Nothing in biology makes sense except
in the light of evolution”
Theodosius Dobzhansky (geneticist)
Darwin’s Postulates
 Individuals within species are variable.
 Some of these variations are passed on to
offspring.
 In every generation, more offspring are
produced than can survive.
 Survival & reproduction of individuals are not
random. Those that survive and reproduce
are those with the most favorable variations.
They are naturally selected.
Darwin’s Success
He compiled massive amounts of
evidence that all these postulates are
true.
Showed conclusively that individuals
within species are variable, and that
some of these variations are passed on
to the offspring.
Darwin’s Problems
Because mechanisms of inheritance were not
known, two objections were raised:
1. Inheritance acts like pigment in paint
(“blending inheritance”). Any new variant
would mix with existing traits and be swamped
by them. Thus, even if new variation could be
created, it would not persist.
2. Variation within species is limited. Once
existing variation is exhausted, evolution by
natural selection will grind to a halt.
Discoveries in genetics solved both these problems.
Mendel solves one of Darwin’s
Problems
 When Origin was published, an unknown
monk was cultivating peas in central Europe.
Gregor Mendel would demonstrate that
objection 1 is invalid by showing that
inheritance is particulate, not blending.
 Darwin had Mendel’s paper but did not read
it,; he also did crosses but did not go beyond
the F1!
Morgan solves the other
 In 1900 DeVries, Von Tschermak, and Correns
published experiments similar to Mendel’s, and
attributed priority of discover to him.
 Within a few years, T.H. Morgan & colleagues
showed that new hereditary variation (mutation)
occurs in every generation.
 By beginning of the 20th century, both objections
to Darwin’s principles had been conclusively
eliminated by discoveries in genetics.
“Modern Synthesis”
 Evolution by natural selection was not fully
accepted until the hereditary mechanism was
understood.
 The “Modern Synthesis” is the synthesis of
Mendelism and Darwinism that occurred in
the early 20th century.
 The Modern Synthesis is the foundation of
modern evolutionary biology.
Modern genetics is also based on
evolutionary biology and the Modern
Synthesis
 Medical population genetics: patterns of
genetic disease and disease resistance
 Genetic epidemiology:identification and
spread of new disease vectors: Hanta virus,
AIDS, SARS
 Genome biology: Sequencing projects for
non-human organisms can tell us where did
human genes come from, which genes are
conserved and which have diverged,and how
did eukaryotic genomes come to be
organized the way they are (introns, exons,
repetitive DNA).
Course Outline
Deviations from Mendelism:
Epistasis; Unusual Modes of Inheritance
Genetic Data Analysis:
Probability & Statistics
Chromosomal Inheritance:
Chromosomal Abnormalities; Sex Determination
Mapping:
Gene and Genome Mapping
Traits Affected by Genes & Environment
Quantitative Traits
Genes in Populations
Genetic Mechanisms of Evolution; Population Genetics of Disease
and Disease resistance
Genomes and Genome Evolution
Additional Deviations from
Mendelism
Lethal Alleles
Epistasis
Unusual sex linkage
Sex influenced inheritance
Genetic Anticipation
Manx Cats
Lethal alleles
F1:
Mm
x
Mm
F1:
1 MM
2 Mm
F2:
1 Lethal: 2 Manx:
1 mm
1 Normal
F2 phenotypic ratio: 2:1 instead of 3:1
Other lethal mutations
Achondroplasia (humans)
Yellow body color (domestic mice)
Curly wings (Drosophila)
Agouti:
wild type
P:
F1:
BB CC
agouti
x
Bb Cc
agouti
bb cc
albino
Simple dominant phenotype?
F2: 9/16 B- C- 3/16 bb C- 3/16 B- cc 1/16 bb cc
albino
agouti
albino
black
F2 Phen. ratio: 9 agouti : 3 black : 4 albino
novel phenotype
Epistasis
Locus 1
BB
Bb
agouti agouti
bb
black
Locus 2
CC
Cc
cc
no effect no effect albino
Biochemical model
C enzym e present?
Colorless precursor 
B enzyme present?
Yes: CC or Cc some melanin
produced
Yes: BB or Bb agou ti
No: cc no melanin p roduced
No: bb black
CC or Cc: tyrosinase is produced (involved in
production of melanin)
BB or Bb: controls distribution of the pigment
Epistasis
Normal
dihybrid ratio
is altered from
9:3:3:1 to
9:3:4
C and B
gene have
an epistatic
interaction
Figure 10.18b
Crosses between pure lines produce novel colors.
Parental
generation
rrYY
X
Yellow
F1 generation
RRyy
Brown
Codominance?
R-YRed
Self-fertilization
F2 generation
R-Y-
rrY-
R-yy
rryy
Red
9/16
Yellow
3/16
Brown
3/16
Green
1/16
Figure 10.18c
Model to explain 9 : 3 : 3 : 1 pattern observed above: Two genes interact to produce pepper color.
Genotype
Color
Explanation of color
R-Y-
Red
Red pigment + no chlorophyll
rrY-
Yellow
Yellow pigment + no chlorophyll
R-yy
Brown
Red pigment + chlorophyll
rryy
Green
Yellow pigment + chlorophyll
Gene 1
Gene 2
R = Red
Y = Absence of green (no chlorophyll)
r = Yellow
y = Presence of green (+ chlorophyll)
(-) = R or r
(-) = Y or y
Practice Problem
In Labrador retrievers, coat color is controlled by two
loci each with two alleles B,b and E,e respectively.
When pure breeding Black labs with genotype BB EE
are crossed with pure breeding yellow labs of
genotype bb ee the resulting F1 offspring are black.
F1 offspring are crossed (Bb Ee x Bb Ee). Puppies
appear in the ratio:
9/16 black;
3/16 chocolate;
9/16 B- EB- E-
3/16 B- ee
B- ee
4/16=1/4 yellow.
3/16 bb E- 1/16 eebb
bb E- and bb ee
What genotypes correspond to these three
phenotypes?
Other kinds of epistasis
9/16 A-B-
3/16 A-bb
3/16 aaB-
1/16 aabb
Hint: usually given numbers, not fractions
27 agouti; 12 albino; 9 black
28 agoute; 11 albino; 4 black
Practice Problem
In the summer squash (Cucurbita pepo) fruit shape is determined
by two genes. Two different true-breeding spherical types were
crossed. The F1's were all disk, and the F2's segregated
35 disk, 25 spherical and 4 long. Explain these results.
What’s the first step?
Notice novel phenotype: disk, long.
What’s the next step?
Notice there are three F2 phenotypes. What kind of
inheritance will give three F2 phenotypes?
Expected F2 ratio?
Incomplete, codominance 1:2:1
Epistasis
Variation on 9:3:3:1
Practice Problem, cont.
In the summer squash (Cucurbita pepo) spherical fruit is
recessive to disk, True-breeding spherical types from different
geographic regions were crossed. The F1's were disk, and the
F2's segregated 35 disk, 25 spherical and 4 long. Explain
these results.
Are the phenotypic ratios closer 1:2:1 or to a
variant of 9:3:3:1 ?
If phenotypic ratios closer to a variant of 9:3:3:1,
then what variant is it?
Total # of individuals = 35 + 25 + 4 = 64
64/16 = 4
9*4 = 36
6*4 = 24
1*4 = 4
Phenotypic ratio close to 9:6:1
Practice Problem, cont.
In the summer squash (Cucurbita pepo) spherical fruit is
recessive to disk, True-breeding spherical types from different
geographic regions were crossed. The F1's were disk, and the
F2's segregated 35 disk, 25 spherical and 4 long. Explain
these results.
If phenotypic ratios are close to 9:6:1, then what are
the genotypes associated with each phenotype?
35 disk
9/16 A- B-
25 spherical
3/16 A- bb + 3/16 aa B-
4 long
1/16 aa bb
What were the genotypes of the original spherical
parents?
AA bb aaBB