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Oogenesis
• As embryo until
menopause...
• Ovaries
• Primordial germ cells (2N)
• Oogonium (2N)
• Primary oocyte (2N)
• Between birth & puberty;
prophase I of meiosis
• Puberty; FSH; completes
meiosis I
• Secondary oocyte (1N);
polar body
• Meiosis II; stimulated by
fertilization
• Ovum (1N); 2nd polar body
Spermatogenesis
• Puberty until death!
• Seminiferous tubules~
location
• Primordial germ cell (2n)~
differentiate into….
• Spermatogonium (2n)~
sperm precursor
•
•
•
•
Repeated mitosis into….
Primary spermatocyte (2n)
1st meiotic division
Secondary spermatocyte
(n)
• 2nd meiotic division
• Spermatids (n)~Sertoli
cells….
• Sperm cells (n)
Comparing Mitosis & Meiosis
Ch. 14 - Mendelian Genetics and the
Inheritance of Genetic Traits
• Modern genetics began with Gregor
Mendel’s quantitative experiments with pea
plants
Stamen
Carpel
Figure 9.2A, B
Gregor Mendel (Father of Genetics)
• Discovered the fundamentals of Genetics
in the 1860’s
• Lived in Austria and studied in Vienna
• Worked with Garden Peas (Pisum sativum)
• Gathered a huge amount of numerical data
• Discovered the frequency of how traits are
inherited
• Established basic principles of Genetics
MENDEL’S PRINCIPLES
• The science of heredity dates back to
ancient attempts at selective breeding
• Until the 20th century, however, many
biologists erroneously believed that
– characteristics acquired during lifetime could be
passed on
– characteristics of both parents blended
irreversibly in their offspring
Reason Mendel worked with
Garden Peas
•
•
•
•
Easy to grow
Many variations were available
Easy to control pollination (self vs cross)
Flower is protected from other pollen
sources
(reproductive structures are completely
enclosed by petals)
• Plastic bags can be used for extra
protection
• Mendel crossed
pea plants that
differed in certain
characteristics
and traced the
traits from
generation to
generation
• This illustration
shows his
technique for
cross-fertilization
Figure 9.2C
White
1
Removed
stamens
from purple
flower
Stamens
Carpel
PARENTS
(P)
2 Transferred
Purple
pollen from
stamens of white
flower to carpel
of purple flower
3 Pollinated carpel
matured into pod
4
OFFSPRING
(F1)
Planted
seeds
from pod
• Mendel studied
seven pea
characteristics
FLOWER
COLOR
Purple
White
Axial
Terminal
SEED
COLOR
Yellow
Green
SEED
SHAPE
Round
Wrinkled
POD
SHAPE
Inflated
Constricted
POD
COLOR
Green
Yellow
STEM
LENGTH
Tall
Dwarf
FLOWER
POSITION
• He hypothesized
that there are
alternative forms
of genes (although
he did not use that
term), the units
that determine
heredity
Figure 9.2D
Introductory Questions #4
Traits:
Flower Position:
Axial
& terminal
Seed Color:
Yellow & green
Height:
Tall
& short
1) Monohybrid cross: Two hybrid plants are tall.
How many of the offspring would you predict will
be short if there were 400 produced?
2) Dihybrid cross: Two hybrid plants with yellow
seeds and axial flowers are crossed. How many
of the offspring would you predict will have axial
flowers with green seeds if 3750 are produced?
3) Trihybrid cross: Both parents are heterozygous
for all three traits. How many will be tall with
terminal flowers and yellow seeds if 250 are
produced?
Genetic Vocabulary
•
•
•
•
•
•
Punnett square: predicts the
results of a genetic cross between
individuals of known genotype
Homozygous: pair of identical
alleles for a character
Heterozygous: two different
alleles for a gene
Phenotype: an organism’s traits
Genotype: an organism’s genetic
makeup
Testcross: breeding of a
recessive homozygote X dominate
phenotype (but unknown
genotype)
Mendelian Genetics
Character
(heritable feature, i.e., fur color)
Trait
(variant for a character, i.e., brown)
True-bred
(all offspring of same variety)
Hybridization
(crossing of 2 different true-breeds)
P generation
(parental-beginning gen.)
F1 generation (first filial generation)
F2 generation (second filial generation)
Homologous chromosomes bear the two
alleles for each characteristic
• Alternative forms of a gene (alleles) reside
at the same locus on homologous
chromosomes
GENE LOCI
P
P
a
a
B
DOMINANT
allele
b
RECESSIVE
allele
GENOTYPE:
PP
aa
HOMOZYGOUS
for the
dominant allele
HOMOZYGOUS
for the
recessive allele
Bb
HETEROZYGOUS
Figure 9.4
Mendel’s principle of segregation describes
the inheritance of a single characteristic
• From his
experimental data,
Mendel deduced
that an organism
has two genes
(alleles) for each
inherited
characteristic
– One characteristic
comes from each
parent
Figure 9.3A
P GENERATION
(true-breeding
parents)
Purple flowers
White flowers
All plants have
purple flowers
F1
generation
Fertilization
among F1
plants
(F1 x F1)
F2
generation
3/
of plants
have purple flowers
4
1/
4 of plants
have white flowers
• A sperm or egg
carries only one
allele of each pair
– The pairs of alleles
separate when
gametes form
– This process
describes Mendel’s
law of segregation
– Alleles can be
dominant or
recessive
Figure 9.3B
GENETIC MAKEUP (ALLELES)
P PLANTS
Gametes
PP
pp
All P
All p
F1 PLANTS
(hybrids)
Gametes
All Pp
1/
2
1/
P
P
2
p
P
Eggs
Sperm
PP
F2 PLANTS
Phenotypic ratio
3 purple : 1 white
p
p
Pp
Pp
pp
Genotypic ratio
1 PP : 2 Pp : 1 pp
The principle of independent assortment
is revealed by tracking two
characteristics at once
• By looking at two characteristics at once,
Mendel found that the alleles of a pair
segregate independently of other allele pairs
during gamete formation
– This is known as the principle of independent
assortment
Remember……..
• Every Trait within a diploid organism will
have two alleles
• These alleles are separated during Meiosis
Traits: Seed color, seed shape, height
Alleles:
Y or y
R or r
T or t
Parents are diploid
Gametes are produced: only one allele is
present for each trait in each gamete
Parents Genotype
Example:
monohybrid
Dihybrid
Trihybrid
Possible Alleles
combinations
for one gamete
Yy
YyRr
YyRrTt
Y & (y)
YR (3 OTHERS)
yRt (7 OTHERS)
Mendel’s principles reflect the
rules of probability
• Inheritance follows
the rules of
probability
– The rule of
multiplication and the
rule of addition can
be used to determine
the probability of
certain events
occurring
F1 GENOTYPES
Bb female
Bb male
Formation of eggs
Formation of sperm
1/
B
1/
2
B
2
B
B
1/
b
1/
1/
2
b
B
b
1/
4
b
b
4
B
1/
2
4
b
F2 GENOTYPES
1/
4
Figure 9.7
HYPOTHESIS:
DEPENDENT ASSORTMENT
RRYY
P
GENERATION
rryy
Gametes
RRYY
ry
RY
Eggs
2
1/
2
RY
1/
2
Gametes
1/
Sperm
2
F2
GENERATION
1/
Eggs
1/
ry
1/
1/
4
4
4
Ry
ry
Actual results
contradict
hypothesis
4
RY
4
RY
1/
RRYY
RrYY
RRYy
4
RrYy
rY
1/
RrYY
rrYY
rrYy
ACTUAL
RESULTS
SUPPORT
HYPOTHESIS
1/
rY
RrYy
Figure 9.5A
ry
RY
RrYy
RY
ry
Pg. 257
rryy
RrYy
F1
GENERATION
1/
HYPOTHESIS:
INDEPENDENT ASSORTMENT
RrYy
RrYy
RRyy
Rryy
rryy
Ry
1/
RrYy
rrYy
Rryy
4
4
ry
9/
16
3/
16
3/
16
1/
16
Yellow
round
Green
round
Yellow
wrinkled
Yellow
wrinkled
• The chromosomal basis of Mendel’s
Principles
Figure 9.17
Important ratios to Remember
Cross
BB x bb
Phenotypic
4:0 (100% Dom.)
Genotypic
4:0 (all) Bb
BB x Bb
4:0 (100% Dom.)
1:1 (50% BB,Bb)
Bb x Bb
3:1 (75% Dom, 25% Rec) 1:2:1 (BB, Bb, bb)
Bb x bb
1:1(50% Dom, 50% Rec) 1:1 (50% Bb,bb)
Two traits (not linked):
AaBb x AaBb
9:3:3:1
Introductory Questions #5
1)
A Monohybrid cross and Dihybrid cross always
produces Phenotypic rations of 3:1 and 9:3:3:1. What
phenotypic ratio will produced from a trihybrid cross?
2) Solve this trihybrid cross with Pea plants
Traits: Seed color, seed shape, height
Male
Heterozygous all traits
Female
Heterozygous yellow and tall plant
w/ wrinkled seeds
a) How many offspring would you predict will be Tall with
wrinkled, yellow seeds?
b) How many offspring would have green seeds that are
round and tall?
IQ #5-Solution
1)
(male)
(female)
2)
YyRrTt x YyrrTt
# allele combo in gametes:
8
4
TT rr YY: ¼ x ½ x 1/4 = 1/32
TT rr Yy: ¼ x ½ x 1/2 = 1/16
Tt rr YY: ½ x ½ x ¼ = 1/16
Tt rr Yy: ½ x ½ x ½ = 1/8
= 9/32
Possible genotypes of the offspring: (a)
A. Answer
B. 3/32 will be predicted that will be tall with round green seeds
Introductory Questions #6
1)
2)
A female heterozygous for Seed shape and color is
crossed with a male that is heterozygous for seed shape
but homozygous recessive for seed color. How many
offspring would you predict (expect) to be yellow and
wrinkled if 500 were produced?
If only 50 offspring were yellow and wrinkled how can you
tell if your results were only due to chance? What
statistical test could you do in order to determine if there;’s
a significant difference between what you actually got (50)
vs. what you expected?
201 round & yellow
204 round & green
45 wrinkled and green
50 wrinkled & yellow
IQ#6: Question #2 (cont’d)
• Expected (calculated Values)
Expect. Value
wrinkled & green
Round & yellow
Round & green
Wrinkled & Yellow
62
187
187
62
Obs. value
45
201
204
50
IQ #6-Answers
•
RrYy x Rryy
How many will be yellow and wrinkled? (500)
Poss. genotypes: Yyrr = 1/2 x 1/4 = 1/8
Answer: 1/8 x 500 = 62 are expected to be yellow,
green
Do a chi-squared test
Chi-Square value: ???? (Homework)
Expected Value for each phenotype: 187,187, 62, 62
Chi-squared value:
Critical value from table (in lab)
7.82
(0.05 or 95%)
Sample Problem using Chi square
• Two hybrid Tall plants are crossed. If the F2
generation produced 787 tall plants and 277
short plants. Does this confirm Mendel’s
explanation?
• What is the expected value?
This is your null hypothesis (HO)
• Total number of plants: 1064
• 3:1 Phenotypic ratio
• Expected value should be: 798 tall and 266 short
(75%)
(25%)
Statistical Tools to Analyze results
• Chi-Square: Will tell you how much your data is
different from expected (calculated) results. It is
Non-Parametric and deals with different
catagorical groups vs. Parametric which deals
with numbers and which case you would use a
T-test instead.
Formula:  2 =  (o – e)2
e
2: what we are solving:
o: observed value
e: expected (calculated value)
When to use Chi-Squared Test
•
•
•
•
Can only be used with raw counts (not measurements)
Comparing Experimental & expected (theo.) values
Sample size must be more than 25 to be reliable
Aims to test the null hypothesis (H0)
– (H0): the hypothesis that there’s no difference between the
data sets
– Alternative hypothesis: there is a significant difference
• Compare with a critical value table (p values)
• To reject the (H0): value must be GREATER than the critical
value & favor the alternative hypothesis.
• Accepting the null means that there’s no significant difference
between the data sets.
Calculation of Chi Square Value
2 =  (O – E)2
E
2 = (787 – 798)2
(277 – 266)2 = 0.61
798
266
There are two categories and therefore the degrees of
freedom would be 2-1 = 1 .
+
• Look up the critical value for 1 degree of freedom:
3.84 (next slide-always given)
• If your value is LARGER than the critical value then you
reject the null hypothesis and assume that there is a
significant difference between the observed value and the
expected. Values are statistically different.
• 0.61 is less than 3.84 therefore we accept the null
hypothesis and accept that our values are similar enough
There’s no significant difference between the observed &
expected values. Values are not random.
Answers to IQ #5 & #6
IQ #5----1. 27:9:9:9:3:3:3:1 = 64
phenotypes
IQ #6----2.
2 =  (O – E)2
E
2 = (45 – 62)2 + (201-187)2 + (204 – 187)2 + (50-62)2 = 9.58
62
187
187
62
Critical Value: 4 groups = 3 degrees of freedom= 7.81
9.58 is greater than 7.81 therefore we reject the null hypothesis
What does this mean?
Accepting or Rejecting your
hypothesis?
• Accepting the Null (H0) means that:
– Test value is less than the critical value
– Values are similar enough
– there is a not SIGNIFICANT difference between the observed
and expected value (p<0.05). More than 95% confidence
– Chance alone cannot explain the differences observed.
• Rejecting the Null (H0) means that:
–
–
–
–
Test value is greater than the critical value
Values are very different from each other
Random Chance can cause the results
the observations are significantly different from the
expectations. (p>0.05). Evaluate the results. Less than 95%
confidence in the values
Solving Question #3
2 = ∑ (o – e)2
e
Degrees of Freedom
Critical Value
1
3.84
2
5.99
3
7.81
4
9.49
5
11.07
Formula:
Testcrosses
Determining an Unknown
Genotype
Geneticists use the testcross to
determine unknown genotypes
• The offspring of a testcross often reveal the
genotype of an individual when it is unknown
TESTCROSS:
GENOTYPES
B_
bb
Two possibilities for the black dog:
BB
b
OFFSPRING
Bb
B
GAMETES
Figure 9.6
or
Bb
All black
B
b
Bb
b
bb
1 black : 1 chocolate
Purebreds and Mutts — A Difference of
Heredity
These black Labrador puppies are purebred—
their parents and grandparents were black
Labs with very similar genetic make-ups
– Purebreds
often suffer
from serious
genetic defects
• The parents of these puppies were a
mixture of different breeds
– Their behavior
and
appearance is
more varied as
a result of their
diverse genetic
inheritance
• Independent assortment of two genes in
the Labrador retriever
Blind
PHENOTYPES
GENOTYPES
Black coat,
normal vision
B_N_
MATING OF HETEROZYOTES
(black, normal vision)
PHENOTYPIC RATIO
OF OFFSPRING
Figure 9.5B
9 black coat,
normal vision
Black coat,
blind (PRA)
B_nn
BbNn
3 black coat,
blind (PRA)
Blind
Chocolate coat,
normal vision
bbN_
Chocolate coat,
blind (PRA)
bbnn
BbNn
3 chocolate coat,
normal vision
1 chocolate coat,
blind (PRA)
How is Codominance
different from Incomplete
Dominance?
Non-single Gene Genetics
(pg. 260-262)
Incomplete dominance:
-neither pair of alleles are completely
expressed when both are present.
-Typically, a third phenotype is produced
Ex: snapdragons (pink flowers)
hypercholesterolemia
Codominance: Two alleles are expressed in a
heterozygote condition.
Ex: Human Blood types
Incomplete Dominance
Incomplete dominance results in
intermediate phenotypes
• When an offspring’s
phenotype—such
as flower color— is
in between the
phenotypes of its
parents, it exhibits
incomplete
dominance
P GENERATION
White
rr
Red
RR
Gametes
R
r
Pink
Rr
F1 GENERATION
1/
1/
Eggs
1/
F2 GENERATION
2
2
2
R
1/
2
r
1/
R
R
Red
RR
r
Pink
Rr
Sperm
1/
Pink
rR
White
rr
Figure 9.12A
2
2
r
• Incomplete dominance in human
hypercholesterolemia
GENOTYPES:
HH
Homozygous
for ability to make
LDL receptors
Hh
Heterozygous
hh
Homozygous
for inability to make
LDL receptors
PHENOTYPES:
LDL
LDL
receptor
Cell
Normal
Figure 9.12B
Mild disease
Severe disease
Many genes have more than two alleles
in the population
• In a population, multiple alleles often exist
for a characteristic
• The three alleles for ABO blood type in
humans is an example
Codominance
Codominance-Observed in Blood
Types
Blood Type Frequencies of different
Ethnic Groups
Pleiotrophy vs. Polygenic
Inheritance
Non-single Gene Genetics
Pleiotropy: genes with multiple phenotypic effect.
Ex: sickle-cell anemia
combs in roosters
coat color in rabbits
Polygenic Inheritance: an additive effect of two or
more genes on a single phenotypic character
Ex: human skin pigmentation and height
A single gene may affect many
phenotypic characteristics
• A single gene may affect phenotype in many
ways
– This is called pleiotropy
– The allele for sickle-cell disease is an example
Pleiotropy – Sickle Cell anemia
Effects of Sickle Cell Anemia
Polygenic Inheritance (SG. #9)
P GENERATION
aabbcc
AABBCC
(very light) (very dark)
F1 GENERATION
Eggs
Sperm
Fraction of population
AaBbCc AaBbCc
Skin pigmentation
F2 GENERATION
Figure 9.16
Epistasis (SG. #11)
• Epistasis: a gene at one locus (chromosomal location)
affects the phenotypic expression of a gene at a second
locus. Ex: mice and Labrador coat color
Epistasis (SG #11)
• Examples: Labrador’s coat color
Albino Koala
•
Two Genes Involved:
Allele Symbol
-Pigment- Black (Dominant)
B
b
E/e
Chocolate (recessive)
-Expression or deposition of the Pigment
Black
Yellow
BBEE
BbEE
BBEe
BbEe
BBee
Bbee
Chocolate
bbEE
bbEe
Which genotype is missing and what group should it be listed under?
Epistasis
Study Guide Problems
1) White alleles are dominant to yellow
2) a. ¼ b. 1/8
c. ½
d. 1/32
3) Incomplete dominance
Human Genome & Genetic
Disorders
Information Gained by the Genome
Project (2003)
• Entire DNA (nucleus) composed of about
2.9 billion base pairs of nucleotides
• Six to Ten anonymous individuals were
used
• Estimated number of genes = under 30,000
• Only 1% to 2% of human DNA codes for a
protein or RNA
• On Chromosome 22: 545 genes have been
identified.
Genetic traits in humans can be tracked
through family pedigrees
• The inheritance of many
human traits follows
Mendel’s principles and
the rules of probability
Figure 9.8A
• Family pedigrees are used to determine
patterns of inheritance and individual
genotypes
Dd
Joshua
Lambert
Dd
Abigail
Linnell
D_?
Abigail
Lambert
D_?
John
Eddy
dd
Jonathan
Lambert
Dd
Dd
dd
D_?
Hepzibah
Daggett
Dd
Elizabeth
Eddy
Dd
Dd
Dd
dd
Female Male
Deaf
Figure 9.8B
Hearing
• A high incidence of hemophilia has plagued
the royal families of Europe
Queen
Victoria
Albert
Alice
Louis
Alexandra
Czar
Nicholas II
of Russia
Alexis
Figure 9.23B
Pedigree of Alkaptonuria
Table 9.9
• A few are caused by Dominant alleles
– Examples: Achondroplasia, Huntington’s
disease
Figure 9.9B
Human Disorders
The Family Pedigree
Recessive disorders:
-Cystic fibrosis
-Tay-Sachs
-Sickle-cell
Dominant Disorders:
-Huntington’s
-Polydactaly
Diagnosing/Testing:
-Amniocentesis
-Chorionic villus
sampling (CVS)
SEX CHROMOSOMES AND
SEX-LINKED GENES
• A human male has one X chromosome and
one Y chromosome
• A human female has two X chromosomes
• Whether a sperm cell has an X or Y
chromosome determines the sex of the
offspring
Human sex-linkage
•
•
•
•
•
SRY gene: gene on Y chromosome that triggers the development of
testes
Fathers= pass X-linked alleles to all daughters only (but not to sons)
Mothers= pass X-linked alleles to both sons & daughters
Sex-Linked Disorders: Color-blindness; Duchenne muscular
dystropy (MD); hemophilia
Sex-linked disorders affect mostly
males
• Most sex-linked human
disorders are due to
recessive alleles
– Examples: hemophilia,
red-green color blindness
– These are mostly seen in males
Figure 9.23A
– A male receives a single X-linked allele from his
mother, and will have the disorder, while a
female has to receive the allele from both
parents to be affected
Sex Linked Trait: Colorblindness
IQ #6-Answers
1) 27:9:9:9:3:3:3:1 = 64 phenotypes
2) RrYy x Rryy
How many will be yellow and wrinkled? (500)
Poss. genotypes: Yyrr = 1/2 x 1/4 = 1/8
Answer: 1/8 x 500 = 63 are expected to be yellow,
green
3) Do a chi-squared test
Chi-Square value:
10.08
Expected Value for each phenotype: 188,188, 63, 63
Chi-squared value:
Critical value from table (in lab)
7.82
(0.05 or 95%)
IQ#6: Question #3 (cont’d)
• Expected (calculated Values)
Expect. Value
wrinkled & green
Round & yellow
Round & green
Wrinkled & Yellow
63
188
188
63
Obs. value
45
201
204
50
Amniocentesis -Pg 270
• Karyotyping and biochemical tests of fetal
cells and molecules can help people make
reproductive decisions
– Fetal cells can be obtained through
amniocentesis
Amniotic
fluid
Amniotic
fluid
withdrawn
Centrifugation
Fluid
Fetal
cells
Fetus
(14-20
weeks)
Biochemical
tests
Placenta
Figure 9.10A
Uterus
Cervix
Several
weeks later
Cell culture
Karyotyping
Diagnostic Procedures to detect
Genetic Disorders in Babies
• Chorionic Villus Sampling (CVS) is another
procedure that obtains fetal cells for
karyotyping.
Pg. 270
Fetus
(10-12
weeks)
Several hours
later
Placenta
Suction
Chorionic villi
Fetal cells
(from chorionic villi)
Karyotyping
Some
biochemical
tests
Figure 9.10B
UltraSound
(Pg. 269)
• Examination of the fetus with ultrasound is
another helpful technique
Figure 9.10C, D
Genetic testing can detect diseasecausing alleles
• Genetic testing can be of
value to those at risk of
developing a genetic
disorder or of passing it on
to offspring
• Dr. David Satcher, former U.S.
surgeon general, pioneered
screening for sickle-cell
disease
Figure 9.15B
Figure 9.15A
Introductory Questions #7
1) How is co-dominance different from Incomplete
dominance? Give an example of both.
2) Give an example of polygenic inheritance.
3) Name three autosomal disorders that are
dominant and three that are recessive.
4) Name the person who determined that genetic
traits can be linked and inherited together?
5) What does it mean when traits are sex linked?
Give an example of a human sex linked trait.
How can linked traits be separated?
6) What is a two point test cross? Why would you
use one?
Table 9.9
Methods of Detecting Genetic
Disorders
•
•
•
•
•
•
Amniocentesis
Ultrasound
CVS (Chorionic Villus Sampling)
PGD (Pre-implantation Genetic Diagnosis)
Fetuscopy
Genetic Counseling/Screening
PGD: Preimplantion Genetic
Diagnosis
• Used for Couples who are carriers of an
abnormal allele.
• IVF Procedure is used
• Eggs are fertilized, grown in culture and
tested for the disorder
• Normal embryos are implanted into the
uterus.
Table of Disorders
Name
Chromosome
Cellular effect
Overall
involvement or (#)
Phenotypic Result
_______________________________________________________________________________
Down Syndrome
Auto (47)
Many
Kleinfelter’s Syndrome
Sex (47)
Turner’s Syndrome
Sex (45)
Cri du Chat
Auto/Deletion #5
Fragile X
Auto & Sex
Phenylketonuria (PKU)
Auto rec.
Enzyme def.
Alkaptonuria
Auto rec.
Enzyme def.
Sickle Cell Anemia
Auto rec.
Hemoglobin Struct.
Cystic Fibrosis
Auto rec.
Tay Sachs
Auto rec.
Huntington’s Disorder
Auto Dom.
Achondroplasia
Auto Dom.
Albinism
Auto rec.
Color Blindness
Sex-linked
Muscular Dystrophy
Sex-linked
Hemophlia
Sex-linked
Alzheimer’s
Auto Dom.
Hypercholesterolemia
Auto Dom.
Chapter 15:
The Chromosomal Theory of
Inheritance
•
•
•
•
•
•
Gene linkage (Drosophila)
Wild-types & mutants
Gene mapping
Non-Disjunction (aneuploidy)
Barr bodies (inactive X)
Alterations of Chromosome
structure
• Genomic imprinting
Pgs. 274-291
Sex-linked genes exhibit a unique
pattern of inheritance
• All genes on the sex chromosomes are said
to be sex-linked
– In many organisms, the X chromosome carries
many genes unrelated to sex
– Fruit fly eye
color is a
sex-linked
characteristic
Figure 9.22A
Chromosomal Linkage
• Thomas Morgan
• Drosophilia melanogaster
• XX (female) vs. XY (male)
• Sex-linkage: genes located
on a sex chromosome
• Linked genes: genes located
on the same chromosome
that tend to be inherited
together
– Their inheritance pattern reflects the fact that
males have one X chromosome and females
have two
– These figures illustrate inheritance patterns for
white eye color (r) in the fruit fly, an X-linked
recessive trait
Female
XRXR
Male
Xr Y
XR
Female
XRXr
Xr
XRXr
Male
XRY
XRY
Xr
XRXR
XrXR
XRY
XrY
R = red-eye allele
r = white-eye allele
Male
XRXr
XR
XR
Y
Female
XrY
Xr
XR
Y
Xr
XRXr
Xr Xr
Y
XRY
XrY
Figure 9.22B-D
Genes on the same chromosome tend to
be inherited together
• Certain genes are linked
– They tend to be inherited together because they
reside close together on the same chromosome
How to Determine if Two Genes are
linked.
Perform a Two Point Test Cross:
Parents: AaBb X aabb
Possible gametes: AB, Ab, aB, ab X ab
Following Mendelian principles of independent
assortment (not linked on the same chromosome)
then:
ab
AB
Ab
aB
ab
AaBb
Aabb
aaBb
aabb
(25%)
(25%)
(25%)
(25%)
If Genes are Linked
• More Parental types should be present in the
offspring and fewer recombinants.
Parental type
ab
recombinant
recombinant Parental type
AB
Ab
aB
ab
AaBb
Aabb
aaBb
aabb
(more)
40%
(less)
10%
(less)
10%
(more)
40%
Figure 9.18
Generating Recombinants in
Drosophila
Figure 9.19C
Crossing Over Developing
Genetic Maps
Pgs. 278-281
Crossing over produces new
combinations of alleles
• This produces gametes with recombinant
chromosomes
• The fruit fly Drosophila melanogaster was
used in the first experiments to demonstrate
the effects of crossing over
Genetic Recombination (pg. 281)
•
•
•
Crossing over
Genes that DO NOT assort
independently of each other
Genetic maps
The further apart 2 genes
are, the higher the probability that
a crossover will occur between
them and therefore the higher the
recombination frequency
Linkage maps
Genetic map based on
recombination frequencies
Geneticists use crossover data to map
genes
• Crossing over is more likely to occur
between genes that are farther apart
– Recombination frequencies can be used to map
the relative positions of genes on chromosomes
Chromosome
g
c
l
17%
9%
9.5%
Figure 9.20B
Generating Recombinant Offspring
Pg. 280
• A partial genetic map of a fruit fly
chromosome
(pg. 281)
Mutant phenotypes
Short
aristae
Black
body
(g)
Long aristae
(appendages
on head)
Gray
body
(G)
Cinnabar
eyes
(c)
Red
eyes
(C)
Vestigial
wings
(l)
Brown
eyes
Normal
wings
(L)
Red
eyes
Wild-type phenotypes
Figure 9.20C
Genetic Map of Drosophila
(pg. 281)
• Alfred H. Sturtevant, seen here at a party
with T. H. Morgan and his students, used
recombination data from Morgan’s fruit fly
crosses to map genes
Figure 9.20A
Sex-Linked Patterns of
Inheritance and Non-Disjunction
Sex-Linked Patterns of Inheritance
(male)
(female)
Parents’
diploid
cells
X
Y
Male
Sperm
Egg
Offspring
(diploid)
Figure 9.21A
• Other systems of sex determination exist in
other animals and plants
– The X-O system
Grasshoppers, cockaroaches,
some insects
– The Z-W system
Birds, fish, some insects
Sex Determinant
chromosome is in the ovum
– Chromosome number
Bees & ants
Figure 9.21B-D
Abnormal numbers of sex chromosomes do
not usually affect survival
• Nondisjunction can produce gametes with
extra or missing sex chromosomes
– Unusual numbers of sex chromosomes upset
the genetic balance less than an unusual
number of autosomes
Accidents During Meiosis Can Alter
Chromosome Number
Homologous pairs
fail to separate
during meiosis I
Pg. 285
Nondisjunction
in meiosis I
Normal
meiosis II
Gametes
n+1
n+1
n–1
n–1
Number of chromosomes
Figure 8.21A
Chromosomal Errors
Nondisjunction: members of
a pair of homologous
chromosomes do not
separate properly during
meiosis I or sister
chromatids fail to separate
during meiosis II
Aneuploidy: chromosome
number is abnormal
• Monosomy~ missing
chromosome
• Trisomy~ extra chromosome
(Down syndrome)
• Polyploidy~ extra sets of
chromosomes
• Fertilization after Non-disjunction in the
mother results in a zygote with an extra
chromosome
Egg
cell
n+1
Zygote
2n + 1
Sperm
cell
n (normal)
Figure 8.21C
ALTERATIONS OF CHROMOSOME NUMBER AND
STRUCTURE
• To study human chromosomes
microscopically, researchers stain and
display them as a karyotype
– A karyotype usually shows 22 pairs of
autosomes and one pair of sex chromosomes
• Preparation of a Karyotype
Blood
culture
Packed red
And white
blood cells
Hypotonic solution
Stain
White
Blood
cells
Centrifuge
3
2
1
Fixative
Fluid
Centromere
Sister
chromatids
Pair of homologous
chromosomes
4
5
Figure 8.19
An extra copy of chromosome 21 causes
Down syndrome
• This karyotype shows three number 21
chromosomes
• An extra copy of chromosome 21 causes
Down syndrome
Figure 8.20A, B
• The chance of having a Down syndrome
child goes up with maternal age
Figure 8.20C
Autosomal Polyploid Disorders
Syndrome/Disorder
• Down’s Syndrome
• Patau Syndrome
• Edwards Syndrome
Chromsome # Affected
21
13
18
Jacobs Syndrome
Table 8.22
Changes that can occur in
Chromosome Structure
(Abnormalities)
Alterations of chromosome structure can cause
birth defects and cancer
• Chromosome breakage can lead to
rearrangements that can produce genetic
disorders or cancer
– Four types of rearrangement are:
deletion, duplication, inversion, and translocation
• Chromosomal changes in a somatic cell can
cause cancer
– A chromosomal translocation in the bone
marrow is associated with chronic myelogenous
leukemia
Chromosome 9
Chromosome 22
Reciprocal
translocation
“Philadelphia chromosome”
Activated cancer-causing gene
Figure 8.23C
Chromosomal Errors
•
•
•
•
•
Alterations of chromosomal structure:
Pg. 327
Deletion: removal of a chromosomal segment
Duplication: repeats a chromosomal segment
Inversion: segment reversal in a chromosome
Translocation: movement of a chromosomal segment to
another
Example of a Chromosomal
Deletion
• Cri Du Chat: “Cat cry” syndrome
– Effects chromosome #5
– Altered facial Features “moon face”
– Severe mental retardation
Barr Bodies
• Inactive X Chromosome
Pg. 284
• Predominant in females
• Dark Region of chromatin is visible at the edge of
the nucleus within a cell during interphase.
(Please see Figure 15.11)
• A small fraction of the genes located on this X
chromosome usually are expressed.
• Inactivation is a random event among the somatic
cells.
• Heterozygous individuals: ½ cells alleles expressed
• Ex. Calico cat & Tortoise shell (Variegation)
Calico Kitten w/Barr Bodies
Example of Variegation
Barr Bodies
Outbreeding vs. Inbreeding
• Inbreeding
-Increases homozygosity in the population.
-Increases frequency of genetic disorders
-Amplifies the homozygous phenotypes
• Outbreeding:
-Leads to better adapted offspring
-Heterozygous advantage & Hybrid Vigor become
evident and buffers out undesirable traits
Genomic Imprinting
• Def: a parental effect
on gene expression
• Identical alleles may
have different effects
on offspring, depending
on whether they arrive
in the zygote via the
ovum or via the sperm
Fragile X Syndrome
Fragile X Syndrome
• More common in Males
• Common form of
Mental Retardation
• Thinned region on tips
of chromatids
• Triplicate “CGG” repeats over
200 to 1000 times
• Normal: repeat 50 X or less
• Commonly seen in Cancer cells
• Varies in severity:
Pg. 327-328
Mild learning disabilities ADD  Mental retardaton
• A man with Klinefelter syndrome has an
extra X chromosome
Poor beard
growth
Breast
development
Underdeveloped
testes
Figure 8.22A
• A woman with Turner syndrome lacks an X
chromosome
Characteristic
facial
features
Web of
skin
Constriction
of aorta
Poor
breast
development
Underdeveloped
ovaries
Figure 8.22B
A
B
a
b
a
B
A B
a
b
Tetrad
A
b
Crossing over
Gametes
Figure 9.19A, B
Independent Assortment in Budgie
Birds