The Chromosomal Basis of Inheritance Chp. 15

Genes are located on… CHROMOSOMES!

Human Genome Project

Chromosomal Basis of Mendel’s Laws… Page 275

Drosophila melanogaster Thomas Hunt MORGAN – first to locate a specific gene on a specific chromosome FRUIT FLY 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 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 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

F1 = b+ b vg+ vg b+ vg+ b+ vg+ b vg b vg Black body & vestigial wing Wild type b+ vg+ b+ vg+ b vg b vg b+ b+ vg+ vg+ b b vg vg Gametes: b+ vg+ b vg b+ vg+ b vg F1 = b+ b vg+ vg

If on different chromosomes (independent assortment), then Test cross of F1 If on different chromosomes (independent assortment), then b+ b vg+ vg x b b vg vg b+ b b b vg+ vg vg vg Gametes: b+vg+; b+vg; b vg+; b vg b vg

If on different chromosomes (independent assortment), then Test cross of F1 If on different chromosomes (independent assortment), then b+ b vg+ vg x b b vg 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 Body: Normal Normal Black Black Wing: Normal Vestigial Normal Vestigial 1 : 1 : 1 : 1

F1 = b+ b vg+ vg b+ vg+ b+ vg+ b vg b vg Black body & vestigial wing Wild type b+ vg+ b+ vg+ b vg b vg b+ b+ vg+ vg+ b b vg vg Gametes: b+ vg+ b vg b+ vg+ b vg F1 = b+ b vg+ vg

If on same chromosome with NO CROSSOVER, then: 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

If on same chromosome with NO CROSSOVER, then: 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 b+ b vg+ vg b+ b vg vg b b vg+ vg b b vg vg Body: Normal Normal Black Black Wing: Normal Vestigial Normal Vestigial

If on same chromosome with CROSSOVER, then: 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+ b vg b vg b vg Gametes: b+ vg+ or b vg b vg b+ vg or b vg+

If on same chromosome with CROSSOVER, then: 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+ b vg+ vg b+ b vg vg b b vg+ vg b b vg vg Parental Types Body: Normal Normal Black Black Wing: Normal Vestigial Normal Vestigial RATIO ???

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 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 parental recombinants 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 parental recombinants

PRACTICE 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 10 10 A 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 6 6 m 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 5 3 4 6 7

PRACTICE XAY XAXa 1 2 5 3 4 6 7 XAXA XAXa XAY XaY XAXa 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 XAXa 1 2 5 3 4 6 7 XAXA XAXa XAY 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 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”

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 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?

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