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 eyesThus, 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 F1If on different chromosomes (independent

assortment), thenb+ 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 F1If on different chromosomes (independent

assortment), then

b+ vg+ b+ vg b vg+ b vgb 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 vgb 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 vgb 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-smooth123 round-smooth

133 elongated-fuzzy12 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-smooth123 round-smooth

133 elongated-fuzzy12 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-Hon 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 translocationWhat chromosome alterations have occurred if

daughter cells have a gene sequence ofA-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