Post on 18-Dec-2015
The Chromosomal Basis of
Inheritance
Chp. 15
Genes are located on…
CHROMOSOMES!
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 aSex-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
Y
Xw+ Xw Xw+ Xw
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 = Xw+ Xw and Xw+ Y (all wild type)
• F2 = Xw+ Xw
Xw+ Xw+ Xw+ Xw+ Xw
Y Xw+ Y Xw Y
Linked Genes
• Linked Genes = genes on same chromosome– Tend to be inherited together
black bodies and vestigial wings
Wild type
b+ b+ vg+ vg+ b b vg vg
F1 = b+ b vg+ vg
Gametes: b+ vg+ b vg
b+
vg+
b+
vg+
b
vg
b
vg
Wild type Black body & vestigial wing
b+
vg+
b
vg
Test cross of F1
If on different chromosomes (independent assortment), then
b+ b vg+ vg x b b vg vg
b+
vg
b bb
vg+vgvg
Gametes: b+vg+; b+vg; b vg+; b vg b vg
Test cross of F1
If on different chromosomes (independent assortment), then
b+ vg+ b+ vg b vg+ b vg
b vg b+ b vg+ vg b+ b vg vg b b vg+ vg b b vg vg
b+ b vg+ vg x b b vg vg
Body: Normal Normal Black Black Wing: Normal Vestigial Normal Vestigial
1 : 1 : 1 : 1
b+ b+ vg+ vg+ b b vg vg
F1 = b+ b vg+ vg
Gametes: b+ vg+ b vg
b+
vg+
b+
vg+
b
vg
b
vg
Wild type Black body & vestigial wing
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
b
vg
b
vg
Gametes: b+ vg+ or b vg b vg
Test cross of F1
If on same chromosome with NO CROSSOVER, then:
b+ vg+ b+ vg b vg+ b vg
b vg b+ b vg+ vg b+ b vg vg b b vg+ vg b b vg vg
b+ b vg+ vg x b b vg vg
Body: Normal Normal Black Black Wing: Normal Vestigial Normal Vestigial
Test cross of F1
If on same chromosome with CROSSOVER, then:
b+ b vg+ vg x b b vg vg
b+
vg+
b
vg
b
vg
b
vg
Gametes: b+ vg+ or b vg b vg
b
vg
b+
vg
b
vg+
b+ vg or b vg+
Test cross of F1
If on same chromosome with CROSSOVER, then:
b+ vg+ b+ vg b vg+ b vg
b vg b+ b vg+ vg b+ b vg vg b b vg+ vg b b vg vg
b+ b vg+ vg x b b vg vg
Body: Normal Normal Black Black Wing: Normal Vestigial Normal Vestigial
RATIO ???
Parental Types
Recombinants
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)
PRACTICE1. 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.
PRACTICE1. 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
Calculate the % recombination and the map distance between the two genes.
24 / 280 = .086 8.6% 8.6 map units
parentalrecombinants
PRACTICE2. 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)
A C D B
1020 10
PRACTICE3. 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%
m j l k
69 6
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
5
3 4
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.
1 2
5
3 4
6 7
XAY
XAY
XAXa
XAXa
XAXaXAXA
XaY
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 sex-linked 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 re-imprinted 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