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Classical Genetics
GREGOR MENDEL
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• Austrian monk
• Brought experimental and
quantitative approach to
genetics
• Bred pea plants to study
inhertance
• Why peas?
– Control mating (self- vs. crosspollination)
– Many varieties available
– Short generation time
• Gregor Mendel
– Parents donate “discrete heritable factors” (genes), to
their offspring
– These factors remain separate from one generation to
the next
• Trait – variant of an inherited characteristic
• True Breeding - when self-fertilized, only
produces offspring with
the same traits.
– Green plant only produces a green plant
Genotypes
PP X pp
_________
_____
_____
_____
P (parental) generation = true breeding plants
 F1 (first filial) generation = offspring (hybrids)
 F2 (second filial) generation = F1 offspring

_____
ALLELES: ALTERNATE
VERSIONS OF A GENE
P = purple, p = white
7 CHARACTERS IN
PEA PLANTS
Dominant vs. Recessive
(expressed) or (hidden)
MENDEL’S PRINCIPLES
1.
2.
3.
Alternate version of genes (alleles) cause
variations in inherited characteristics among
offspring.
For each character, every organism inherits
one allele from each parent.
Law of Dominance If 2 alleles are different,
the dominant allele will be fully expressed;
the recessive allele will have no noticeable
effect on offspring’s appearance.
1. Homozygous
Dominant
2. Heterozygous
3. Homozygous Recessive
 dominant
(P), recessive (p)
 homozygous = 2 same alleles (PP or pp)
 heterozygous = 2 different alleles (Pp)
 Phenotype:
expressed physical traits
 Genotype: genetic make-up
1.


Law of Segregation: the 2 alleles
for each character separate during
gamete formation.
AaBBccDdEE (2n)  ABcdE (n)
segregated
Every gamete has an allele for every trait
LAW OF
SEGREGATI
ON
monohybrid cross punnett
square 
Ratios
Genotypic Ratio
HD : HET : hr = 1:2:1
Pheontypic Ratio
Dominant : Recessive =3:1
Testcross: determine if dominant trait expressed is
homozygous or heterozygous by crossing with
recessive (pp) – Draw both punnett squares
100% Purple
Homozygous
dominant
50% Purple
Heterozygous
Law of Independent Assortment:
(hyperlink)
 Each
pair of alleles segregates (separates)
independently during gamete formation
 Eg. color is separate from shape
 Monohybrid
cross: study 1 character
 eg. flower color
 Dihybrid cross: study 2 characters
 eg. flower color & seed shape
DIHYBRID CROSS
 P-
AABB x aabb
AB
ab
DIHYBRID CROSS
 P-
AABB x aabb
AB
ab
 F1
 What
AaBb x AaBb (heterozygous)
are the possible gametes
DIHYBRID CROSS
 P-
AABB x aabb
AB
ab
 F1
AaBb x AaBb (heterozygous)
 What
AB
are the possible gametes
Ab
aB
ab X AB
Ab
aB
ab
AB
Ab
•
•
•
•
•
F2 phenotypic ratio
AB aB Ab ab -
Hardy-Weinberg equilibrium
• Hypothetical, non-evolving population
– preserves allele frequencies
• Serves as a model (null hypothesis)
– natural populations rarely in H-W equilibrium
– useful model to measure if forces are acting on a
population
• measuring evolutionary change
G.H. Hardy
mathematician
W. Weinberg
physician
Null Hypothesis
• One states a null hypothesis by stating “nothing
is happening”
• For Evolution
– Null Hypotheis for a Hardy-Weinberg theory exampleEvolution is not occurring because the HardyWeinberg theory has been satisfied.
• If fails the equation then we have a violation in
the H.W. Theory. Therefore what is occurring?
__________
Hardy-Weinberg equilibrium
• Population of Hardy-Weinberg assumes
– Random mating
– Large Population
– Negligible mutation rate
– No gain/loss of individuals
– NO Natural Selection
Hardy-Weinberg theorem
• Counting Alleles
– assume 2 alleles = B, b
– frequency of dominant allele (B) = p
– frequency of recessive allele (b) = q
• frequencies must add to 1 (100%), so:
p+q=1
BB
Bb
bb
Hardy-Weinberg theorem
• Counting Individuals
– frequency of homozygous dominant: p x p = p2
– frequency of homozygous recessive: q x q = q2
– frequency of heterozygotes: (p x q) + (q x p) = 2pq
• frequencies of all individuals must add to 1 (100%), so:
p2 + 2pq + q2 = 1
BB
Bb
bb
H-W formulas
• Alleles:
p+q=1
B
• Individuals:
p2 + 2pq + q2 = 1
BB
BB
b
Bb
Bb
bb
bb
Using Hardy-Weinberg equation
population:
100 cats
84 black, 16 white
How many of each
genotype?
p2=.36
BB
q2 (bb): 16/100 =
.16
q (b): √.16 = 0.4
p (B): 1 - 0.4 = 0.6
2pq=.48
Bb
q2=.16
bb
What assume
Must
are the genotype
populationfrequencies?
is in H-W equilibrium!
Using Hardy-Weinberg equation
p2=.36
Assuming
H-W equilibrium
2pq=.48
q2=.16
BB
Bb
bb
p2=.20
=.74
BB
2pq=.64
2pq=.10
Bb
q2=.16
bb
Null hypothesis
Sampled data
How do you
explain the data?
ANOTHER PROBLEM
• Fraggles are mythical, mouselike
creatures that live beneath flower gardens.
• Of the 100 fraggles in a population, 91
have green hair(F) and 9 have grey hair(f).
• Assuming genetic equilibrium:
– What are the gene frequencies of F and f?
– What are the genotypic frequencies?
• q2  q  p  p2  2pq
ANSWERS TO PROBLEM
• Gene frequencies:
– F = 0.7 and f = 0.3
• Genotypic frequencies
– FF = 49% or 0.49
– Ff = 42% or 0.42
– f f = 9% or .09
EXTENDING MENDELIAN GENETICS
The relationship between genotype and phenotype
is rarely simple
Complete Dominance:
heterozygote and homozygote
for dominant allele are
indistinguishable
• Eg. YY or Yy = yellow seed
Incomplete Dominance: F1
hybrids have appearance that
is between that of 2 parents
• Eg. red x white = pink flowers

Snapdragons (flowers)
WW +RR = Pink
 F1  produces …


F2  produces …
Fig. 14-10-3
P Generation
Red
CRCR
White
CWCW
CR
Gametes
CW
Pink
CRCW
F1 Generation
Gametes 1/2 CR
1/
CW
2
Sperm
1/
2
CR
1/
2
CW
F2 Generation
1/
2
CR
Eggs
1/
2
CRCR
CRCW
CRCW
CWCW
CW

Sickle Cell anemia
SS = no sickle
 Ss = carrier
 ss = has sickle cell
 The heterozygous advantage

The intermediate expression of a trait allows for an
evolutionary advantage
 denotes a resistance to malaria


Think of a mix of good and bad cells
Codominance: phenotype of both alleles is expressed
• Eg. red hair x white hairs = roan cattle
Multiple Alleles: gene has 2+ alleles
• Eg. human ABO blood groups
•
•
•
•
•
Alleles = IA, IB, i
IA,IB = Codominant
Type A produces B antibodies and vice versa
Agglutination = blood clotting
What happens when Type A is exposed to Type B blood
• Type AB = no antibodies – universal recipients
• Type O – if exposed to A or B produces Anti-A and Anti-B
but is the universal donor
BLOOD TYPING
Phenotype
(Blood Group)
Genotype(s)
Type A
IAIA or IAi
Type B
IBIB or IBi
Type AB
IA IB
Type O
ii
BLOOD TRANSFUSIONS
Blood transfusions must match blood type
 Mixing of foreign blood  clumping  death
 Rh factor: protein found on RBC’s (Rh+ = has
protein, Rh- = no protein)

BLOOD TYPING PROBLEM:

A man who is heterozygous with type A blood marries
a woman who is homozygous with type B blood. What
possible blood types might their children have?
Pleiotropy: single gene has multiple phenotypic
effects (Eg. sickle cell anemia)

Phenylketonuria (PKU) metabolic disorder

Phenylalanine hydroxylase ( enzyme that converts
PA into tyorsine which is then easily metabolized)
Epistasis: one gene alters the phenotypic
expression of another gene (eg. albinism - white
fur color in mammals)
ee overrides the
expression of all
the B,b genotypes
Polygenic Inheritance: the effect of 2 or more
genes acting upon a single phenotypic character
(eg. skin color, height)
Nature and Nurture: both genetic and
environmental factors influence phenotype
Hydrangea flowers vary in shade and intensity of color
depending on acidity and aluminum content of the soil.
Genetic Disorders


Autosome – Chromosome 1-22
1) Autosomal Recessive Disorders – homozygous recessive

Cystic Fibrosis




the malformation chloride ion transport protein
Prevents the liquidification of mucus
5% are carries for CF
Tay sachs



Inability to break down lipids in the brain
Progressively worsens
Symptoms begin at before age of 2, and most die by age of 5
Sickle Cell Anemia
 PKU
 Albinism
 Cretinism


Malformation of bones
Fig. 14-16
Parents
Normal
Aa

Normal
Aa
Sperm
A
a
A
AA
Normal
Aa
Normal
(carrier)
a
Aa
Normal
(carrier)
aa
Albino
Eggs
 EXTRA
INFO
 If a recessive allele that causes a disease is rare,
then the chance of two carriers meeting and mating
is low
 Consanguineous matings (i.e., matings between
close relatives) increase the chance of mating
between two carriers of the same rare allele
 Most societies and cultures have laws or taboos
against marriages between close relatives
Copyright © 2008 Pearson Education Inc., publishing as Pearson Benjamin Cummings
Autosomal Dominant Disorders
Homozygous dominant will not survive through
embryonic stage
 Huntington’s Disease

Symptoms show around the age of 40
 Children have a 50/50 chance of inheritance if
parents have the disorder
 Flailing movements and brain deterioration


Achnodroplasia

Dwarfism
Mendelian Inheritance in Humans
Pedigree: diagram that shows the relationship
between parents/offspring across 2+ generations
Woman =
Man =
Trait expressed:
PEDIGREE ANALYSIS
GENETIC TESTING
 May
be used on a fetus to detect genetic disorders
 Amniocentesis: remove amniotic fluid around
fetus to culture for karyotype
 Chorionic villus sampling: insert narrow tube
in cervix to extract sample of placenta with fetal
cells for karyotype
Thomas Morgan
Drosphila melanogaster (fruit flies)
 Follow human sexual chromosome patterns (XX,
XY)
 Wild type – dominant phenotype
 Mutant type – recessive phenotype

Fig. 15-3
Sex Linkage

In one experiment, Morgan
mated male flies with white
eyes (mutant) with female
flies with red eyes (wild type)

The F1 generation all had
red eyes

The F2 generation showed
the 3:1 red:white eye ratio,
but only males had white
eyes
Correlating Behavior of a Gene’s Alleles with
Behavior of a Chromosome Pair

If Y held alleles for eye color
for a male then a pure breed
male for red eyes crossed with
a female white eyed would
produce red eyed males!!
XW
XR
YR
XW

All Males have white eyes

Y does not have allele for eye
color
XW
XR
Y
XW
 Morgan
determined that the white-eyed
mutant allele must be located on the X
chromosome
 Morgan’s finding supported the chromosome
theory of inheritance
 First experiment that showed gene linkage

Eye color was on sex chromosome
Fig. 15-4c
CONCLUSION
P
Generation
w+
X
X

w+
X
Y
w
Eggs
F1
Generation
Sperm
w+
w+
w+
w
w+
Eggs
F2
Generation
w
w+
Sperm
w+
w+
w
w
w
w+
The Chromosomal Basis of Sex
 In
humans and other mammals (XX and XY)
 Only the ends of the Y chromosome have regions
that are homologous with the X chromosome
 The SRY gene on the Y chromosome codes for the
development of testes
Fig. 15-5
X
Y
Fig. 15-6
44 +
XY
44 +
XX
Parents
22 +
22 +
or Y
X
Sperm
+
44 +
XX
or
22 +
X
Egg
44 +
XY
Zygotes (offspring)
(a) The X-Y system
22 +
XX
22 +
X
76 +
ZW
76 +
ZZ
32
(Diploid)
16
(Haploid)
(b) The X-0 system
(c) The Z-W system
(d) The haplo-diploid system
Morgan Experiment #2
Body color (b)
 Wild type = gray = b+
 Mutant = black = b

Wing shape (vg)
 Wild type = normal = vg+
 Mutant = vestigial = vg

Fig. 15-9-1
EXPERIMENT
P Generation (homozygous)
Wild type
(gray body,
normal wings)
b+ b+ vg+ vg+

Double mutant
(black body,
vestigial wings)
b b vg vg
Fig. 15-9-2
EXPERIMENT
P Generation (homozygous)
Wild type
(gray body,
normal wings)
Double mutant
(black body,
vestigial wings)

b b vg vg
b+ b+ vg+ vg+
F1 dihybrid
(wild type)
Double mutant
TESTCROSS

b+ b vg+ vg

Gametes?
b+ vg+
b b vg vg
b vg
b+ vg
b vg+
X
Eggs

Offspring
b+ b vg+ vg
b b vg vg
b vg
Sperm
b+ b vg vg
b b vg+ vg
Fig. 15-9-3
EXPERIMENT
P Generation (homozygous)
Wild type
(gray body,
normal wings)
Double mutant
(black body,
vestigial wings)

b b vg vg
b+ b+ vg+ vg+



If genes are on same
chromosome, based
on the P generation,
what are the possible
combination of
gametes ?
b+vg+ and b vg
Which offspring are
possible?
F1 dihybrid
(wild type)
Double mutant
TESTCROSS

b+ b vg+ vg
Testcross
offspring
b b vg vg
b vg
b+ vg
b vg+
Wild type
(gray-normal)
Blackvestigial
Grayvestigial
Blacknormal
b+ b vg+ vg
b b vg vg b+ b vg vg b b vg+ vg
Eggs
b+ vg+
b vg
Sperm
Fig. 15-9-4
EXPERIMENT
P Generation (homozygous)
Wild type
(gray body,
normal wings)
Double mutant
(black body,
vestigial wings)

b b vg vg
b+ b+ vg+ vg+
F1 dihybrid
(wild type)
Double mutant
TESTCROSS

b+ b vg+ vg
Testcross
offspring
b b vg vg
b vg
b+ vg
b vg+
Wild type
(gray-normal)
Blackvestigial
Grayvestigial
Blacknormal
b+ b vg+ vg
b b vg vg b+ b vg vg b b vg+ vg
Eggs
b+ vg+
b vg
Sperm
PREDICTED RATIOS
If genes are located on different chromosomes:
575
:
575
If genes are located on the same chromosome and
parental alleles are always inherited together:
1150
:
1150 :
RESULTS
965
:
944
:
:
575
:
575
0
:
0
206
:
185
Fig. 15-9-4
EXPERIMENT
P Generation (homozygous)
Wild type
(gray body,
normal wings)
Double mutant
(black body,
vestigial wings)

b b vg vg
b+ b+ vg+ vg+
F1 dihybrid
(wild type)
Double mutant
TESTCROSS

b+ b vg+ vg
Testcross
offspring
b b vg vg
b vg
b+ vg
b vg+
Wild type
(gray-normal)
Blackvestigial
Grayvestigial
Blacknormal
b+ b vg+ vg
b b vg vg b+ b vg vg b b vg+ vg
Eggs
b+ vg+
b vg
Sperm
PREDICTED RATIOS
If genes are located on different chromosomes:
575
:
575
If genes are located on the same chromosome and
parental alleles are always inherited together:
1150
:
1150 :
RESULTS
965
:
944
:
:
575
:
575
0
:
0
206
:
185
Fig. 15-9-4
EXPERIMENT
P Generation (homozygous)
Wild type
(gray body,
normal wings)
Double mutant
(black body,
vestigial wings)

b b vg vg
b+ b+ vg+ vg+
F1 dihybrid
(wild type)
Double mutant
TESTCROSS

b+ b vg+ vg
Testcross
offspring
b b vg vg
b vg
b+ vg
b vg+
Wild type
(gray-normal)
Blackvestigial
Grayvestigial
Blacknormal
b+ b vg+ vg
b b vg vg b+ b vg vg b b vg+ vg
Eggs
b+ vg+
b vg
Sperm
PREDICTED RATIOS
If genes are located on different chromosomes:
575
:
575
If genes are located on the same chromosome and
parental alleles are always inherited together:
1150
:
1150 :
RESULTS
965
:
944
:
:
575
:
575
0
:
0
206
:
185
Fig. 15-9-4
EXPERIMENT
P Generation (homozygous)
Wild type
(gray body,
normal wings)
•Recombinant types form
through what process?
•Recombinant frequency
= ??%
Double mutant
(black body,
vestigial wings)

b b vg vg
b+ b+ vg+ vg+
F1 dihybrid
(wild type)
Double mutant
TESTCROSS

b+ b vg+ vg
Testcross
offspring
•The maximum
recombinance is 50%,
never more – at 50%
genes behave as if
unlinked
b b vg vg
b vg
b+ vg
b vg+
Wild type
(gray-normal)
Blackvestigial
Grayvestigial
Blacknormal
b+ b vg+ vg
b b vg vg b+ b vg vg b b vg+ vg
Eggs
b+ vg+
b vg
Sperm
PREDICTED RATIOS
If genes are located on different chromosomes:
575
:
575
If genes are located on the same chromosome and
parental alleles are always inherited together:
1150
:
1150 :
RESULTS
965
:
944
:
:
575
:
575
0
:
0
206
:
185
Morgan’s Experiment Animation
A
linkage map is a genetic map of a chromosome
based on recombination frequencies
 Distances between genes can be expressed as map
units; one map unit, or centimorgan, represents a
1% recombination frequency
 Map units indicate relative distance and order, not
precise locations of genes
Copyright © 2008 Pearson Education Inc., publishing as Pearson Benjamin Cummings
Fig. 15-11
RESULTS
Recombination
frequencies
9%
Chromosome
17%
b
cn = cinnabar (red eye
color)
9.5%
cn
vg

The b-vg frequency is slightly less than the sum
of b-cn and cn-vg due to double crossovers
Cancels out some of the b-vg recombination frequency
Sex Linkage and Sex-Linked Disorders
Usually, the ends of sex chromosomes are
homologous – this allows for pairing
 1990 discovery of SRY gene – sex determining
region on the Y


Codes for a protein that regulates many genes
including the gene that turns embryonic gonads into
testes


The absence results in formation of ovaries
X-chromosome shows codominance in women



All females have one Xchromosome inactivated
Random which cell has which X active
Calico cats
Sex-Linked Disorders
Colorblindness
 Duchenne Muscular Dystrophy
X

Dytrophin protein missing
 Weakening of muscles and
coordination
 Do not live past 20s
Xc


Hemophilia

Inability to produce 1 or more
blood clotting factors
Y
Xc
Genomic Imprinting
Expression of these genes is variable depending
on which parent the gene came from,
 The allele from the mother determines phenotype


Or the father’s allele determines phenotype
Mitochondrial DNA
Passed down through maternal inheritance
 If mother has a mitochondrial disorder all
offspring will have the same disorder

POP
QUIZ