Download The Chromosomal Basis of Inheritance

The Chromosomal
Basis of
Inheritance
Chp. 15
Genes are located on…
CHROMOSOMES!
Human Genome Project
Chromosomal Basis of Mendel’s Laws…
Page
275
Thomas Hunt MORGAN – first to locate a
specific gene on a specific chromosome
Drosophila melanogaster
FEMALE
MALE
WILD TYPE
(red eyes)
MUTANT
(white eyes)
Drosophila allele symbols
• Gene symbol comes from mutant
– Ex: white eyes  w
• Wild type (normal phenotype) is dsignated
with a “+”
– Ex: normal (red) eyes  w+
• If mutant is recessive, use lower case…
• If mutant is dominant to wild type, use
upper case…
White eyed male crossed with a
wild-type female…
• All F1 had red (wild-type) eyes
• F2 had 3 wild type : 1 white
BUT…
ONLY MALES had WHITE eyes
Thus, eye color “linked” to sex
Gene for white eye
color located on the
“X” chromosome*
Symbols:
Xw+ = wild type
Xw = white eye
*Called a
Sex-Linked Gene
PRACTICE: Punnett Squares with
Sex Linked Genes
• P Generation =
wild-type female & white eyed male
Xw+ Xw+ x
Xw Y
• F1 = ?
Xw+
Xw+
Xw Xw+ Xw
Xw+ Xw
Y
Xw+ Y
Xw+ Y
PRACTICE: Punnett Squares with
Sex Linked Genes
• P Generation =
wild-type female & white eyed male
Xw+ Xw+ x
Xw Y
• F1 =
• F2 =
Xw+ Xw and Xw+ Y (all wild type)
Xw+
Y
Xw+
Xw
Xw+ Xw+ Xw+ Xw
Xw+ Y
Xw Y
Linked Genes
• Linked Genes = genes on same chromosome
– Tend to be inherited together
Wild type
black bodies and vestigial wings
Wild type
Black body &
vestigial wing
b+
b+
b
b
vg+
vg+
vg
vg
b+ b+ vg+ vg+
Gametes:
b+ vg+
b b vg vg
b vg
b+
b
vg+
vg
F1 = b+ b vg+ vg
Test cross of F1
If on different chromosomes (independent
assortment), then
b+ b vg+ vg
b+
x
b
b b vg vg
b
vg+
b
vg
Gametes: b+vg+; b+vg; b vg+; b vg
vg
b vg
vg
Test cross of F1
If on different chromosomes (independent
assortment), then
b+ b vg+ vg x b b vg vg
b+ vg+
b vg
Body:
Wing:
b+ vg
b vg+
b vg
b+ b vg+ vg b+ b vg vg b b vg+ vg b b vg vg
Normal
Normal
1
Normal
Vestigial
:
1
Black
Normal
:
1
Black
Vestigial
:
1
Wild type
Black body &
vestigial wing
b+
b+
b
b
vg+
vg+
vg
vg
b+ b+ vg+ vg+
Gametes:
b+ vg+
b b vg vg
b vg
b+
b
vg+
vg
F1 = b+ b vg+ vg
Test cross of F1
If on same chromosome with NO
CROSSOVER, then:
b+ b vg+ vg x b b vg vg
b+
b
b
b
vg+
vg
vg
vg
Gametes: b+ vg+ or b vg
b vg
Test cross of F1
If on same chromosome with NO
CROSSOVER, then:
b+ b vg+ vg x b b vg vg
b+ vg+
b vg
Body:
Wing:
b+ vg
b vg+
b vg
b+ b vg+ vg b+ b vg vg b b vg+ vg b b vg vg
Normal
Normal
Normal
Vestigial
Black
Normal
Black
Vestigial
Test cross of F1
If on same chromosome with CROSSOVER,
then:
b+ b vg+ vg x b b vg vg
b+
b
b
b
vg
vg+
vg+
vg
vg
vg
Gametes: b+ vg+ or b vg
b+ vg or b vg+
b vg
Test cross of F1
If on same chromosome with CROSSOVER,
then:
b+ b vg+ vg x b b vg vg
Recombinants
b+ vg+
b+ vg
b vg
b vg+
b vg
b+ b vg+ vg b+ b Parental
vg vg b b vg+ vg b b vg vg
Types
Body:
Wing:
Normal
Normal
Normal
Vestigial
Black
Normal
RATIO ???
Black
Vestigial
Parental Types
965 + 944 =
1909 flies
Recombinants
206 + 185 =
391 flies
% Recombinants
391 recomb. = .17 or
2300 total
17%
b
vg
17
map
units
Linkage Map: uses recombination frequencies to
map relative location of genes on chromosomes
1 map unit = 1 % recombination freq.
ex: b-vg = 17%
b-cn = 9%
cn-vg = 9.5%
• Other chromosomal maps:
– Cytogenic map – actually pinpoints genes on
physical location of chromosome (bands)
– DNA sequencing/physical map – gives order
of nucleotides for a gene and intergenic
sequences in # of b.p. (base pairs)
PRACTICE
1.
In tomatoes, round fruit shape (O) is dominant to
elongated (o), and smooth skin (S) is dominant to
fuzzy skin (s). Test crosses of F1 individuals
heterozygous for these pairs of alleles gave the
following results:
12 elongated-smooth
123 round-smooth
133 elongated-fuzzy
12 round-fuzzy
Are these genes linked?
Calculate the % recombination and the map distance
between the two genes.
PRACTICE
1.
In tomatoes, round fruit shape (O) is dominant to
elongated (o), and smooth skin (S) is dominant to
fuzzy skin (s). Test crosses of F1 individuals
heterozygous for these pairs of alleles gave the
following results:
12 elongated-smooth
123 round-smooth
parental
recombinants
133 elongated-fuzzy
12 round-fuzzy
Calculate the % recombination and the map distance
between the two genes.
24 / 280 = .086  8.6%
8.6 map units
PRACTICE
2.
The cross-over percentages between linked genes are
given below:
A – B = 40%
C – D = 10%
B – D = 10%
B – C = 20%
A – C = 20%
What is the sequence of genes on the chromosome?
(draw a map and label distance between genes)
20
A
10
10
C
D
B
PRACTICE
3.
Recombination frequency is given below for several gene
pairs. Create a linkage map for these genes, and show
the map unit distances between loci (genes).
j, k = 12%
k, l = 6%
j, m = 9%
l, m = 15%
9
m
6
6
j
l
k
Sex Chromosomes
and sex-linked genes:
XX = female
XY = male
•Father’s gamete determines
sex of child
•Presence of a Y
chromosome (SRY genes)
allows development of
testes/male characteristics
Inheritance of sex-linked
genes
•Sex-linked gene = gene
carried on sex chromosome
(usually X)
•Females (XX) only express
recessive sex-linked
phenotypes if homozygous
recessive for the trait
•Males (XY) will express what
ever allele is present on the X
chromosome = hemizygous
PRACTICE
• What are the possible phenotypes of the
offspring from a woman who is a carrier for
a recessive sex-linked allele and a man
who is affected by the recessive disorder?
1 normal female:
1 affected female:
1 normal male:
1 affected male
PRACTICE
• Two normal color-sighted individuals produce
the following family (see pedigree). Fill in the
probably genotypes of the numbered individuals.
Solid symbols represent color blindness.
1
2
3
4
5
6
7
PRACTICE
• Two normal color-sighted individuals produce
the following family (see pedigree). Fill in the
probably genotypes of the numbered individuals.
Solid symbols represent color blindness.
XAY 1
3 XAXA
5
XAY
2 XAXa
4 XAXa
6 7
XaY XAXa
Sex-linked Disorders in Humans
• Duchenne Muscular Dystrophy
Sex-linked Disorders in Humans
• Duchenne Muscular Dystrophy
• Hemophilia
Sex-linked Disorders in Humans
• Duchenne Muscular Dystrophy
• Hemophilia
• Fragile X
Sex-linked Disorders in Humans
•
•
•
•
Duchenne Muscular Dystrophy
Hemophilia
Fragile X
(Baldness & red-green color-blindness)
X Inactivation: females have two X chromosomes,
but only need one active X
•One X condenses in each cell during
embryonic development  Barr body
•Females are a “mosaic” if heterozygous for a sexlinked trait ex: Calico cats
Chromosomal Alterations
• Aneuploidy – 1 more/less chromosome
– Due to NONDISJUNCTION: separation of
homologous chromosomes (Anaphase I) or
sister chromatids (Anaphase II) fails
Chromosomal Alterations
• Aneuploidy – 1 more/less chromosome
– Due to NONDISJUNCTION: separation of
homologous chromosomes (Anaphase I) or
sister chromatids (Anaphase II) fails
– Trisomy = 1 extra chromosome (2n + 1)
– Monosomy = 1 less chromosome (2n – 1)
HUMANS – cannot have more than 47 or less
than 45 chromosomes & NEED AT LEAST
ONE “X” to survive
Aneuploid Disorders
• Down Syndrome: Trisomy 21
• Klinefelter Syndrome: XXY
• Trisomy X: XXX
• Turner Syndrome:
Monosomy X (X0)
– Only viable monosomy in
humans!
Polyploidy
• Polyploidy = more than two complete sets
of chromosomes (nondisjunction)
– TRIPLOIDY = 3n
– Humans:
• n = haploid = 1 set = 23 chromosomes
• 2n = diploid = 2 sets = 46 chromosomes
• 3n = triploid = 3 sets = 69 chromosomes
COMMON IN PLANT KINGDOM
Activity: Polyploid Plants
Alterations of Chromosome Structure
Prader Willi & Angelman Syndrome
Cri du chat
• Deletion on
chromosome #5
CML
• Translocation (22 & 9)  “Philadelphia
Chromosome”
PRACTICE:
Two non-homologous chromosomes have
genes in the following order:
A-B-C-D-E-F-G-H-I-J & M-N-O-P-Q-R-S-T
What chromosome alterations have occurred if
daughter cells have a gene sequence of
A-B-C-O-P-Q-G-J-I-H
on the first chromosome?
PRACTICE:
Two non-homologous chromosomes have
genes in the following order:
A-B-C-D-E-F-G-H-I-J & M-N-O-P-Q-R-S-T
deletion inversion
translocation
What chromosome alterations have occurred if
daughter cells have a gene sequence of
A-B-C-O-P-Q-G-J-I-H
on the first chromosome?
Genomic Imprinting
• When it matters which parent you inherited
the allele from…
– Occurs during formation of gametes
– Methyl groups (-CH3) added to DNA and
“silence” alleles
– When offspring produce own gametes,
parental imprinting is erased & alleles reimprinted according to sex of offspring
– Ex: insulin–like growth factor 2
Genomic Imprinting
“Extranuclear Genes”
• Mitochondria (mtDNA), chloroplasts, etc..
inherited from mother through the egg