Map Construction

• Need more than two genes to determine the orientation of genes on the chromosome

• If you have three genes, can orient them on the chromosome based on pair linkage

• Not only single crossovers will occur

Map Construction

• Non-crossovers (NCO), single crossovers (SCO) and double crossovers (DCO)

• Need to investigate three gene pairs• AaBbCc x aabbcc• A and B (SCO) – 20% recombinants (rf=0.20)• B and C (SCO) – 30% recombinants (rf=0.30)• DCO – A/B and B/C rf expected = 0.20 x 0.30 = 0.06 or

6%• The expected frequency of DCO is always much lower

than SCO

Single and Double Crossovers

Three-Point Mapping In Drosophila

• All three genes should be heterozygous• All phenotypes should be observed (usually

test cross is performed)• A sufficient number of progeny should be

produced• The double crossover genotype of the least

frequent classes is in the middle of the other two flanking genes.

Calculation of the Distances Between Genes

• Drosophila cross– Scute bristles– Echinus eyes– Crossveinless wings

•

Class Phenotype Genotype # observed

1 Scute, echinus, crossveinless

s e c 1158

2 Wild type + + + 14553 Scute s + + 1634 Echinus, crossveinless + e c 1305 Scute, echinus s e + 1926 Crossveinless + + c 1487 Scute, crossveinless s + c 18 Echinus + e + 1

Total 3248

Steps to Calculate the Distance Between Genes

• Step 1. Determine the number of the parental (noncrossover, NCO) types

• Step 2. Determine the phenotype and number of the single crossover products

• Step 3. Determine the phenotypes and number of the double crossover products

Steps to Calculate the Distance Between Genes

• To calculate a distance between two genes in a three-point mapping, add a number of single crossovers between the two genes plus the number of both double crossovers; divide by total number and multiply by 100

• Repeat for the second pair of genes• To verify, you may calculate the distance

between the third pair of genes

Map Construction

• sc- ec = 163 + 130 + 2 = 295/3248x 100=9%

• ec – cv =192 + 148 + 2 = 342/3248 x 100 = 11%

• sc – cv = 163 +192 +130 + 148 = 633/3248= 0.194x100=19.4%

• Map: sc----9cM---ec---11cM---cv

Another Cross

• Drosophila– Singed bristles sn– Crossveinless wings cv– Vermilion eyes v– On X-chromosome

Class phenotype genotype # observed

1 singed crossveinless vermilion S c v 32 crossveinless vermilion + c v 3923 vermilion + + v 344 crossveinless + c + 615 singed crossveinless S c + 326 singed vermilion S + v 657 singed S + + 4108 wild type + + + 3

total 1000

Map Construction

• Identify parental classes (NCO): - s + c v and s c+ v+

• Then mother’s genotype is s+ c v/s c+ v+• To determine the gene order, identify

double crossovers – less numerous – s c v and s+ c+ v+

• Suggest order: c+ v/ +s+

Map Construction

_____________________________________ • Between c and s:

34+32+3+3=72/1000x100=7.2%• Between s and v:

61+65+3+3=132/1000x100=13.2%c---7.2cM---s---13.2cM---v

Determining The Gene Sequence

• Method 1. There are only three possible orders- Determine the arrangement of alleles on each homolog of the heterozygous parent- Determine whether a double-crossover will produce the observed phenotype- If suggested order does not produce the observed phenotype, try another order

Determining The Gene Sequence

• Method 2.- Determine the arrangement of alleles on the homologs of the heterozygote parent- Determine the actual double-crossover phenotypes- Select the single allele that has been switched this one will be in the middle

Figure 5-10b Copyright © 2006 Pearson Prentice Hall, Inc.

Interference

• Interference reduces the expected number of multiple crossovers when a crossover event in one region of the chromosome inhibits a second event nearby.

• Interference is positive if fewer double-crossover events than expected occur and negative if more double-crossover events than expected occur.

• The coefficient of coincidence (C) is the observed number of DCOs divided by the expected number of DCOs

Interference• Inhibition of a crossover event in one region of

chromosome by another crossover nearby• Usually observed number of DCOs is less than

calculated from the distance between genes• Coefficient of coincidence, C:

C = Observed DCO/Expected DCO• Interference I = 1 - C

Coefficient of Coincidence

• For the problem we solved,

observed frequency

c= of double haploids = 0.006 = 0.63expected frequency 0.072 x 0.132

of double crossovers

Genetic Problem

• A woman has two dominant traits, cataract, which she inherited from her father, and polydactyly, which she inherited from her mother. Her husband has neither trait.

• If genes for these two traits are 15 cM apart on the same chromosome, what is the chance that the first child of this couple will have both traits?

Example 2. Predict the progeny phenotypes and numbers for this cross

Genetic Maps

• There are some problems with preparing genetic maps of chromosomes.

• The probability of a crossover is not uniform along

Genetic Maps

• Some regions are "hot spots" for recombination (for reasons that are not clear). Approximately 80% of genetic recombination in humans

• In humans, the frequency of recombination of loci on most chromosomes

Genetic Maps

• Chromosome maps prepared by counting phenotypes are called genetic maps.

• They have been prepared for many eukaryotes, including corn. Drosophila, the mouse , and the tomato.

• Genes that are present on the same chromosome are called syntenic.

Modern Methods of Map Construction

Mapping Using DNA Markers

• Polymorphic DNA markers instead of phenotypic traits

• Double haploids

Genetic versus Physical Maps• Chromosome mapping by counting the number of

recombinants produces a genetic map of the chromosome.

• But all the genes on the chromosome are incorporated in a single molecule of DNA.

• Genes are simply portions of the molecule (open reading frames of ORFs) encoding products that create the observed trait (phenotype)

• The rapid progress in DNA sequencing has produced complete genomes for many prokaryotes and several eukaryotes.

Genetic versus Physical Maps

• Having the complete sequence makes it possible to determine directly the order and spacing of the genes. Maps drawn in this way are called physical maps

• What is the relationship between the genetic map and the physical map of a chromosome?

Mapping in Humans

• Using pedigrees• The basic difficulty with mapping genes in

humans is that it is hard to get big enough pedigrees, with enough informative families.

Mapping Using Hybridoma Cells

Figure 5-23 Copyright © 2006 Pearson Prentice Hall, Inc.

Mapping in Humans

• Other strategies include:

Human Genome Project

• Chromosome viewer:– http://www.ornl.gov/sci/techresources/

Human_Genome/posters/chromosome/chooser.shtml