Transcript THE CHROMOSOMAL BASIS OF INHERITANCE

THE CHROMOSOMAL BASIS OF
INHERITANCE
CHAPTER 15
What you must know:
• How the chromosome theory of inheritance
connects the physical movement of
chromosomes in meiosis to Mendel’s laws of
inheritance.
• The unique pattern of inheritance in sexlinked genes.
• How alteration of chromosome number or
structurally altered chromosomes (deletions,
duplications, etc.) can cause genetic disorders.
• How genetic imprinting and inheritance of
mitochondrial DNA are exceptions to standard
Mendelian inheritance.
Chromosome theory of inheritance:
• Genes have specific
locations (loci) on
chromosomes
• Chromosomes
segregate and assort
independently
Chromosomes tagged to reveal a specific gene (yellow).
Figure 15.2a
P Generation
Yellow-round
seeds (YYRR)
Y
Y
R R
Green-wrinkled
seeds (yyrr)

r
y
y
r
Meiosis
Fertilization
Gametes
R Y
y
r
Figure 15.2b
All F1 plants produce
yellow-round seeds (YyRr).
F1 Generation
R
y
r
R
y
r
Y
Y
LAW OF INDEPENDENT
ASSORTMENT Alleles of
genes on nonhomologous
chromosomes assort
independently during gamete
formation.
Meiosis
LAW OF SEGREGATION
The two alleles for each
gene separate during
gamete formation.
r
R
Y
y
r
R
Metaphase I
y
Y
1
1
R
r
r
R
Y
y
Anaphase I
Y
y
r
R
2
y
Y
Y
R
R
1/
4
YR
r
1/
4
yr
y
Y
Y
Y
y
r
R
2
y
Y
Gametes
r
Metaphase
II
r
r
1/
4
Yr
y
y
R
R
1/
4
yR
Figure 15.2c
LAW OF INDEPENDENT
ASSORTMENT
LAW OF SEGREGATION
F2 Generation
3 Fertilization
recombines the
R and r alleles
at random.
An F1  F1 cross-fertilization
9
:3
:3
:1
3 Fertilization results
in the 9:3:3:1
phenotypic ratio in
the F2 generation.
Thomas Hunt Morgan
• Drosophila melanogaster – fruit fly
– Fast breeding, 4 prs. chromosomes (XX/XY)
• Sex-linked gene: located on X or Y
chromosome
– Red-eyes = wild-type; white-eyes = mutant
– Specific gene carried on specific chromosome
• 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
EXPERIMENT
:
P
Generation
F1
Generation
All offspring
had red eyes.
RESULTS
F2
Generation
• Are these results what you expect?
-Morgan determined that the white-eyed mutant allele must be
located on the X chromosome
:
Figure 15.4b
CONCLUSION
P
Generation
X
X
w
X
Y
w
w
Eggs
F1
Generation
Sperm
w
w
w
w
w
Eggs
F2
Generation
w
w
w
Sperm
w
w
w
w
w
w
Sex-linked genes
• Sex-linked gene on X or Y
• Females (XX), male (XY)
– Eggs = X, sperm = X or Y
• Fathers pass X-linked genes to daughters, but not
sons
• Only mothers can pass X-linked to their sons
• X-linked recessive disorders are much more
common in males than in females
Sex- Linked Genes
• For a recessive X-linked trait to be expressed
– A female needs two copies of the allele
(homozygous)
– A male needs only one copy of the allele
(hemizygous)
• Females can be affected or carrier
Transmission of sex-linked recessive traits
Sex-linked disorders
• Colorblindness
• Duchenne muscular dystrophy
• Hemophilia
X- Linked Genes Warm-Up
1. A white-eyed female fruit-fly is mated with a red-eyed male.
What genotypes and phenotypes do you predict for the
offspring?
X- Linked Genes Warm-Up
1. Neither Tim nor Rhoda has Duchenne muscular dystrophy (Xlinked recessive disorder), but their firstborn son has it. What is
the probability their 2nd child will have it?
X- Linked Genes Warm-Up
1. Colorblindness is a sex-linked recessive trait. A colorblind male
and a female with normal vision have a son who is colorblind.
What are the parents’ genotypes?
• If results do not follow Mendel’s Law of Independent
Assortment, then the genes are probably linked
Figure 15.9-1
EXPERIMENT
P Generation (homozygous)
Wild type
(gray body, normal wings)
Double mutant
(black body,
vestigial wings)
b b vg vg
b b vg vg
Figure 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)
b b vg vg
TESTCROSS
Double mutant
b b vg vg
Figure 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
F1 dihybrid
(wild type)
Double mutant
TESTCROSS
b b vg vg
b b vg vg
Testcross
offspring
Eggs b vg
b vg
Wild type
Black(gray-normal) vestigial
b vg
b vg
Grayvestigial
Blacknormal
b vg
Sperm
b b vg vg
b b vg vg b b vg vg
b b vg vg
Figure 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
b b vg vg
Testcross
offspring
Eggs b vg
b vg
Wild type
Black(gray-normal) vestigial
b vg
b vg
Grayvestigial
Blacknormal
b vg
Sperm
b b vg vg
b b vg vg b b vg vg
b b vg vg
PREDICTED RATIOS
If genes are located on different chromosomes:
1
:
1
:
1
:
1
If genes are located on the same chromosome and
parental alleles are always inherited together:
1
:
1
:
0
:
0
965
:
944
:
206
:
185
RESULTS
Linked genes: located on same chromosome and
tend to be inherited together during cell division
Gray body, normal wings
(F1 dihybrid)
Testcross
parents
Black body, vestigial wings
(double mutant)
b vg
b vg
b vg
b vg
Replication
of chromosomes
Replication
of chromosomes
Meiosis I
b vg
b vg
b vg
b vg
b vg
b vg
b vg
b vg
b vg
Meiosis I and II
b vg
b vg
However, non parental
phenotypes were
produced!!
b vg
Meiosis II
bvg
Eggs
Recombinant
chromosomes
b vg
b vg
b vg
b vg
Sperm
Genetic Recombination: production of offspring
with new combo of genes from parents
(when does this occur during meiosis?)
• If offspring look like parents  parental types
• If different from parents  recombinants
Crossing over: explains why some linked genes
get separated during meiosis
• the further apart 2 genes on same chromosome,
the higher the probability of crossing over and
the higher the recombination frequency
Calculating recombination frequency
Linkage Map: genetic map that is based on
% of cross-over events
• Map unit: distance between genes
– 1 map unit = 1% recombination frequency
• Express relative distances and order along chromosome
• 50% recombination = far apart on same chromosome or on
2 different chromosomes
Linked Genes Warm up
1. How are linkage maps constructed? (See. Fig. 15.11
in Campbell 9th ed.)
Warm up
1. Determine the sequence of genes along a
chromosome based on the following recombination
frequencies: A-B, 8 map units; A-C, 19 map units; AD, 20 map units; B-C, 11 map units; B-D, 28 map
units.
Warm up
1. What does a frequency of recombination of 50%
indicate?
The Chromosomal Basis of Sex
• In humans and other mammals, there are two
varieties of sex chromosomes: a larger X
chromosome and a smaller Y chromosome
• Only the ends of the Y chromosome have
regions that are homologous with
corresponding regions of the X chromosome
• The SRY gene on the Y chromosome codes for
a protein that directs the development of
male anatomical features
Sex determination
varies between
animals
Human development
•
•
•
•
Y chromosome required for development of testes
Embryo gonads indifferent at 2 months
SRY gene: sex-determining region of Y
Codes for protein that regulates other genes
X-Inactivation
Barr body = inactive X chromosome; regulate gene
dosage in females during embryonic development
•
•
Cats: allele for fur
color is on X
Only female cats can
be tortoiseshell or
calico.
Nondisjunction: chromosomes fail to separate
properly in Meiosis I or Meiosis II
• A monosomic
zygote has
only one copy
of a particular
chromosome
• A trisomic
zygote has
three copies
of a particular
chromosome
Nondisjunction
• Aneuploidy: incorrect # chromosomes
– Monosomy (1 copy) or Trisomy (3 copies)
• Polyploidy: 2+ complete sets of chromosomes;
3n or 4n
– Rare in animals, frequent in plants
A tetraploid mammal. Scientists think this species may have arisen when an
ancestor doubled its chromosome # by errors in mitosis or meiosis.
Aneuploidy of Sex Chromosomes
• Nondisjunction of sex chromosomes produces
a variety of aneuploid conditions
• Klinefelter syndrome is the result of an extra
chromosome in a male, producing XXY
individuals
• Monosomy X, called Turner syndrome,
produces X0 females, who are sterile; it is the
only known viable monosomy in humans
Nondisjunction
Klinefelter Syndrome: 47XYY, 47XXY
Nondisjunction
Turner Syndrome = 45XO
Alterations of Chromosome
Structure
• Breakage of a chromosome can lead to four
types of changes in chromosome structure
– Deletion removes a chromosomal segment
– Duplication repeats a segment
– Inversion reverses orientation of a segment
within a chromosome
– Translocation moves a segment from one
chromosome to another
Chromosomal Mutations
Chromosomal Mutations
Exceptions to Mendelian
Inheritance
Genomic Imprinting
• Genomic imprinting: phenotypic effect of gene
depends on whether from M or F parent
• Methylation: silence genes by adding methyl groups
to DNA
Inheritance of Organelle Genes
• Extranuclear genes (or
cytoplasmic genes) are found in
organelles in the cytoplasm
– Mitochondria, chloroplasts,
and other plant plastids
carry small circular DNA
molecules
• Extranuclear genes are
inherited maternally because
the zygote’s cytoplasm comes
from the egg
Variegated (striped or spotted)
leaves result from mutations in
pigment genes in plastids, which
generally are inherited from the
maternal parent.
Genetic Testing
Reasons for Genetic Tests:
• Diagnostic testing (genetic disorders)
• Presymptomatic & predictive testing
• Carrier testing (before having children)
• Pharmacogenetics (medication & dosage)
• Prenatal testing
• Newborn screening
• Preimplantation testing (embryos)
Prenatal 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
Karyotyping can detect nondisjunctions.
Down Syndrome = Trisomy 21
Review Questions
1. What is the pattern of inheritance of the trait
(shaded square/circle) shown in the pedigree?
1. How many chromosomes are in a human cell that
is:
a) Diploid?
b) Triploid?
c) Monosomic?
d) Trisomic?
Warm up Questions
1. What is a Barr body?
Chi-Square Test
Chi-Square (χ2) Test
• Used to determine if there is a significant difference
between the expected and observed data
• Null hypothesis: There is NO statistically significant
difference between expected & observed data
– Any differences are due to CHANCE alone
Chi-Square (χ2) Formula
How to use the Chi-Square Test
1.
Determine null hypothesis
– All frequencies are equal –OR– Specific frequencies given already
2. Use formula to calculate χ2 value:
– n = # of categories, e = expected, o = observed
3. Find critical value using table (Use p=0.05).
– degrees of freedom (df) = n – 1
4. If χ2 < Critical Value, then ACCEPT null hypothesis. Differences in data are
due to chance alone.
If χ2 > Critical Value, REJECT the null hypothesis: Differences
in data are NOT due to chance alone!
Q1: Chi Square
•
A hetero red eyed female was crossed with a red
eyed male. The results are shown below. Red
eyes are sex-linked dominant to white, determine
the chi square value. Round to the nearest
hundredth.
Phenotype
# flies observed
Red Eyes
134
White Eyes
66
Chi Square Strategy
• Given—observed
• You have to figure out expected. Usually
to do a Punnett square to figure this out
• Plug in
+
+
Observed—134 red eyes, 66 white eyes
Chi-Square
XR
Xr
XR
XR XR
XR Xr
Y
XR Y
Xr Y
white
red
+
(134-150)2
/150
(66-50)2
+
Expected
/50
3:1 ratio
134+ 66=200
1.70666 + 5.12
150 red
50 white
6.83
Chi-Square Analysis Practice
• Two true-breeding Drosophila are crossed: a normal-winged,
red-eyed female and a miniature-winged, vermillion-eyed
male. The F1 offspring all have normal wings and red eyes.
When the F1 offspring are crossed with miniature-winged,
vermillion-eyed flies, the following offspring resulted:
– 233 normal wing, red eye
– 247 miniature wing, vermillion eye
– 7 normal wing, vermillion eye
– 13 miniature wing, red eye
• What type of conclusions can you draw from this experiment?
Explain your answer.
Sample Problem
• You buy a package of M&Ms from the factory store and
find the following: 20 brown, 20 blue, 20 orange, 20
green, and 20 yellow M&Ms.
• According to the M&M website, each package of candy
should have 13% brown, 24% blue, 20% orange, 16%
green, 13% red, and 14% yellow M&Ms.
• You realize you are missing Red M&M’s in your package!
Is this acceptable, or did something happen in the
factory during the packaging process?
• Use the Chi-Square Test to answer this question.