Download Lab. Handouts Download Lab. #3 and Print out 3 files: http://www.mun.ca/biology/dinnes/B2250/B2250.html Quiz #2 Marks posted on Webpage

Mendelian Genetics Topics: -Transmission of DNA during cell division  Mitosis and Meiosis - Segregation - Sex linkage (problem: how to get a white-eyed female) - Inheritance and probability - Mendelian genetics in humans - Independent Assortment - Linkage - Gene mapping - 3 point test cross     - Tetrad Analysis (mapping in fungi) - Extensions to Mendelian Genetics - Gene mutation - Chromosome mutation - Quantitative and population genetics  

Linkage Chapter 6 - recombination - linkage maps Ch. 6 p. 148 – 165 Prob: 1-5, 7, 8, 10, 11, 14

Linkage of Genes - Many more genes than chromosomes - Some genes must be linked on the same chromosome; therefore not independent

Independent Assortment Linkage Fig 6-6 Fig 6-11 Interchromosomal Intrachromosomal

Two ways to produce dihybrid A B a b A b a B cis A B AaBb A b trans a b (dihybrid ) a B Gametes: AB P Ab ab P aB Ab R AB aB R ab P X X

Example Test Cross AaBb X aabb ab Exp. Obs. AB AaBb 25 10 R Ab Aabb 25 40 P aB aaBb 25 40 P ab aabb 25 10 R 100 100 How to distinguish: Parental high freq. Recombinant low freq.

Gametes: AB R Ab P aB P ab R Therefore dihybrid: A b (trans) a B Example (cont.) Gametes: AB R Ab P aB P ab R Therefore dihybrid: A b (trans) a B

Linkage Maps Genes close together on same chromosome: - smaller chance of crossovers between them - fewer recombinants Therefore: percentage recombination can be used to generate a linkage map

A B large # of recomb. a b C D small number of recombinants c d Linkage maps A B large # of recomb. a b C D small number of recombinants c d

Linkage maps example Testcross progeny: P AaBb 2146 R Aabb 43 Total 4513 1.4 map units 65 = 1.4 % RF 4513 A 1.4 mu B

Additivity of map distances separate maps A B A C 7 2 combine maps C A B 2 7 or Locus A C B (pl. loci) 2 5

Linkage Deviations from independent assortment Dihybrid gametes 2 parent (noncrossover) common 2 recombinant (crossover) rare % recombinants a function of distance between genes % RF = map distance

Gametes Number of Genes Number of Different Gametes monohybrid 1 (Aa) 2 dihybrid 2 (AaBb) 4 trihybrid 3 (AaBbCc) 8

AaBbCc X aabbcc ABC ABc abc AbC Abc aBC aBc abC abc Three Point Test Cross Trihybrid AaBbCc X aabbcc ABC ABc abc AbC Abc aBC aBc abC abc 8 gamete types 11

C ABC B c ABc A C AbC b c Abc a Three Point Test Cross Trihybrid Gametes C ABC B c ABc A C AbC b c Abc a 11

Three Point Test Cross Trihybrid AaBbCc 3 genes: Possibilities: 1. All unlinked 2. Two linked; one unlinked 3. Three linked - order ? A---B---C B---C---A B---A---C Trihybrid 11

Three Point Test Cross +, cv crossveinless +, ct cut wing Three recessive mutants of Drosophila: +, v vermilion eyes +, cv crossveinless +, ct cut wing   P +/+ cv/cv ct/ct X v/v +/+ +/+ 11

Three Point Test Cross P +/+ cv/cv ct/ct x v/v +/+ +/+ Gametes + cv ct v + + F1 trihybrid v/+ cv/+ ct/+ 11

Three Point Test Cross F1 v/+ cv/+ ct/+ x v/v cv/cv ct/ct v cv ct 8 gamete types one gamete type 11

8 gamete types F1 v/+ cv/+ ct/+ v + + 580 + cv ct 592 v cv + 45 + + + 94 v + ct 3 + cv + 5 1448 Parental (most frequent) Recombinant 11

8 gamete types F1 v/+ cv/+ ct/+ v + + 580 + cv ct 592 v cv + 45 + + + 94 v + ct 3 + cv + 5 1448 Parental Recombinant 268 Recombinant Parental 268 1448 = 18.5 % 11

8 gamete types F1 v/+ cv/+ ct/+ v + + 580 + cv ct 592 v cv + 45 + + + 94 v + ct 3 + cv + 5 1448 Parental Parental Recombinant 191 Recombinant 191 1448 = 13.2 % 11

8 gamete types F1 v/+ cv/+ ct/+ v + + 580 + cv ct 592 v cv + 45 + + + 94 v + ct 3 + cv + 5 1448 Parental Recombinant 93 Parental Recombinant 93 1448 = 6.4 % 11

Calculate Recombination Fraction 1. v - cv R v cv 45 + 89 R + + 40 + 94 268 / 1448 = 18.5 % 2. v - ct R + + 94 + 5 R v ct 89 + 3 191/1448 = 13.2 % 3. ct - cv R ct + 40 + 3 R + cv 45 + 5 93/1448 = 6.4 % 11

Three point test cross Observations: all 3 RF < 50 % 3 genes on same chromosome v-----cv largest distance ct in middle map v-------ct-------cv = cv-------ct-------v 13.2 + 6.4 = 19.6 > 18.5 !! Why ? 12

Three Point Test Cross P +/+ ct/ct cv/cv x v/v +/+ +/+ gametes + ct cv v + + F1 trihybrid v + + + ct cv 11

Three Point Test Cross Double crossover class rarest: v---cv P v + + v + P + ct cv + cv R v ct + v + R + + cv + cv X X X X 3 13

Three Point test cross 1. Double crossovers not counted in v--cv RF 2. Double crossovers generate P types (with respect to v--cv) 3. Double crossovers not detected as recombinants Consequence: underestimate of v----cv map distance Greater distance of genes  greater error 14

Double recombinant class: (3 + 5) x 2 = 16 268 + 16 = 284 284/1448 = 19.6 NOTE: double crossovers detected because of middle gene (ct) 15

Mapping Function Genes close together on chromosome -RF good estimate of map distance Genes far apart on chromosome - RF underestimates true map distance due to undetected multiple crossovers 17 27

Mapping Function m = avg. # crossovers per meiosis (linear with true map distance) if m = 1 (1 cross over for every meiosis) then 50 % recombinants produced Therefore: map units (mu) = m x 50 17 27

Mapping Function Mapping function: - relates RF to true map distance (better estimate for genes separated by large distances) m = -ln (1 - 2RF) mu = m x 50 Mapping function 17 27

Mapping Function m = -ln(1 - 2RF)

Mapping Function example 1. RF = 18.5 % m = 0.46 mu = 23.1 2. RF = 6.4 % m = 0.137 mu = 6.8 Summary: - short distances: use RF - long distances: use mapping function 17 27

Linkage Other Points: 1. No crossing over in male Drosophila male: AaBb A B  gametes AB, ab a b use female dihybrid: AaBb x aabb O O 16 26

Linkage 2. Linkage of genes on the X chromosome: AaBb x --Y O O Male progeny: AB Y Ab Y male progeny direct aB Y measure of female meiotic ab Y products 17 27

Fungal Genetics Fungi: important organisms in the ecosystem - decomposers - pathogens important for humans - food (Biology 4040 – Mycology) 19 29

Fun Facts About Fungi http://www.herbarium.usu.edu/fungi/funfacts/factindx.htm

Fungi

Neurospora crassa (bread mold) Morphological mutants Biochemical mutants (one gene, one enzyme)

Linkage Map Neurospora crassa Linkage group I

Fungus Life Cycle vegetative stage haploid +, - mating types brief diploid stage  meiosis n n + spores + meiosis - 2n n - n

Gamete Pool Gametes: Products of many meioses all pooled together A B a b AB AB ab ab AB ab P A B ab ab AB ab Ab AB Gamete P a b AB aB ab ab AB AB pool R a B ab AB AB ab R A b 18 28

Tetrad Analysis Some Fungi and algae: 4 products of a single meiosis can be recovered Advantages: 1. haploid organism - no dominance 2. examine a single meiosis - test cross not needed 3. small, easy to culture 4. Tetrad Analysis - map gene to centromere 19 29

Ascus with ascospores 30

Tetrad Analysis Types of Tetrads: 1. Unordered - 4 products mixed together 2. Ordered (linear) - 4 products lined up, each haploid nucleus can be traced back through meiosis 3. Octads - mitotic division after meiosis 8 products (2 x 4) 20 31

Linear Tetrad Analysis Life Cycle: + = a+ a a a a + a + + Meiosis + Diploid Haploid + Mating: a x +  a /+ n n 2n 4 haploid products

Linear Tetrad Analysis mitosis a a a a + a + + + + + 4 haploid products 8 haploid spores (Octad)

Linear Tetrad Analysis Two types of asci: 1. no crossover----> first division segregation (MI) 2. crossover between gene and centromere-----> second division segregation (MII) 21 33

Mapping gene to centromere First Division a a a a a a + + + + No Crossover + + 22 34

First division segregation meiosis A a

Mapping gene to centromere Second division a a a + a + + a + a + + crossover 22 34

Second division segregation ** recombinant

1st and 2nd Division segregation First Division Second division a a a a a a a + a a a + + + + + a + + No Crossover a + + + + Crossover

Mapping gene to centromere a + a + + a a + + a a + + a a + a + + a + a + a 43 43 3 4 3 4 Total = 100 MI = 86 MII = 14 I II 23 35

Mapping gene to centromere MI = 86 MII = 14 14/100 = 14 % of meioses showed a crossover ½ of the crossover products recombinant RF = ½ x 14 % = 7 % a 7 m.u. 24 36

Unordered Tetrad Analysis 1. still products of a single meiosis 2. can not map gene to centromere 3. linear tetrads can be analyzed as unordered 4. map distance between linked genes a b a b a+ b+ meiosis X a+ b+ n n 2n 25 37

Three kinds of unordered tetrads: b a b a b+ a b a+ b+ a b+ a b a b+ meiosis a+b+ a+b a+b a+b+ a+b a+b+ 1. Parental Ditype 2. Nonparental Ditype PD NPD T 3. Tetratype 26 38

PD a b a+ b+ T NPD a b a b+ a+ b+ a+ b a b+ a+ b

Unlinked genes PD = NPD a b + + X a b + + meiosis a + + b a/+, b/+ PD 25 37

Unordered Tetrads Unlinked Genes: PD = NPD NPD----> all products recombinant T--------> ½ products recombinant PD-----> all parental type PD = 58 RF = ½ T + NPD T = 40 T + NPD + PD NPD = 2 RF = 0.22 22 m.u. 27 39

Tetrad Analysis Types of Tetrads: 1. Ordered (linear): map gene to centromere 2. Unordered: map genes 20 31

Linkage: Summary Recombination: generates variation (inter and intrachromosomal) Genetic maps: - genes linked on the same chromosome - location of new genes relative to genes already mapped 28 40

Linkage: Summary Hunting for genes (Human Diseases) - genetic markers: DNA variation - co-inheritance with diseases using pedigree information - recombinants used to estimate linkage - MUN Medical Genetics 28 40

Linkage vs. Association Association - test if a disease and a marker allele show correlated occurrence in a population Linkage – test if disease and a marker allele show correlated transmission within a pedigree 4,4 1,1 1,4 2,4 3,4 1,2 Linkage 3,3 1,3 3,4 1,3 3,3 2,3 1,4 2,2 2,4 4,4 Association

Genetic linkage in humans Nail- patella syndrome Rare disease inherited as a dominant

Genetic linkage in humans ABO Blood group marker: Alleles: A, B, O Genotypes: OO, AO, BO, AB Two Genes: N-P syndrome; Blood Group Evidence for linkage

Genetic linkage in humans http://www.ndsu.nodak.edu/instruct/mcclean/plsc431/linkage/linkage5.htm