I. Linkage
     A. Genes that don't sort independently
     B. Found on same chromosome
     C. Morgan's fly results
          1. White x miniature
          2. Testcross
               a. Expected 1:1:1:1 ratio
               b. Instead had higher numbers of parental types
               c. Lower numbers of recombinants
               d. Genes don't sort independently
II. Terminology
     A. Crossover
          1. Process of reciprocal exchange of corresponding segments of homologous chromosomes
          2. OR events leading to genetic recombination
     B. Chiasma
          1. Place on chromosomes at which physical exchange occurs
     C. Parental types
          1. Offspring of a testcross which express the traits in the same combination they are expressed in
                the grandparent
     D. Nonparental types or recombinants
          1. Offspring of a testcross which express the traits in a combination different from that in the
                grandparents
     E. Genetic map
          1. Mapping genes on a chromosome based on their recombination frequencies
     F. Physical map
          1. Mapping genes on a chromosome based on their physical location
     G. Testcross
          1. Crossing an organism of unknown genotype with an individual that is homozygous recessive for
                all traits being considered
          2. The offspring of this cross reveal the genotype of the unknown parent - all previously hidden
                recessive traits are revealed in the offspring
     H. Coupling (cis)
          1. All mutant traits are located on one parent, wild type traits are located on the other parent
     I. Repulsion (trans)
          1. Mutant traits are mixed with wild type traits in each parent
     J. Recombination Frequency
          1. Results of testcross of linked genes
          2. Calculate rf: recombinants/total offspring
          3. Indicates closeness of two genes
          4. Closer they are to each other, the lower the recombination frequency
          5. Two genes far enough apart on chromosome will sort independently
     K. Map units or centimorgans
          1. Genetic distance between traits
          2. Map units = rf x 100
          3. Map unit equivalent to 1% recombination
          4. Not a physical distance
          5. Relative distance; depends on crossover rate
               a. Species dependent
               b. Sex dependent
               c. Chromosome dependent
               d. Sometimes location dependent on a chromosome
          6. For humans, actual distance of 1 centimorgan
               a. Varies from 105 to 107 base pairs
III. Two point or two factor crosses
     A. Cross heterozygote by homozygous recessive
     B. Test cross
     C. Expect 1:1:1:1 ratio
          1. Two classes have similar higher than expected frequencies
               a. Nonrecombinants or parentals
               b. Similar to parents of heterozygote
          2. Two classes have similar lower than expected frequencies
               a. Recombinants
               b. Arise from crossover occurring between genes in meiosis
     D. Indicates closeness of two genes
          1. If close together, little chance for chiasmata to form between genes
          2. If further apart, more chance for chiasmata to form between genes
IV. Mapping using data from multiple two factor crosses
     A. Multiple crossovers
          1. Less accuracy in map as distance between genes increases
          2. Once order is determined, select map distances between two closest genes
     B. Method of mapping using multiple two factor crosses
          1. Determine which genes are linked
          2. Set up pairwise crosses in all combinations
          3. Place data in ‘triangle table'
          4. Look for the largest distance between two genes.  Place these two genes on the ends of the map.
          5. If the distance between 2 genes is 50 m.u., what two possibilities exist?
          6. Starting with one of the end genes, identify which gene is the closest to it.
          7. Place this gene next to the end gene and record the distance between the two genes.
          8. For bookkeeping purposes, cross out all the map distances for the end gene.
          9. Now go to the gene just added to the map.  To which gene is it closest?
          10. Record this gene, then cross out the map distances for the previous gene.
          11. Repeat 9-10 until all the genes are mapped.
V. Mapping using three factor crosses
     A. The greater the distance between two genes, the less accurate the distance between them
          1. Possibility for multiple crossovers between the genes increases
          2. Greater accuracy in map distances if use three factor crosses
     B. Triple heterozygote x homozygous recessive
          1. In unlinked, expect 8 equal classes
          2. If totally linked, expect 2 equal classes
     C. Group reciprocal classes together
          1. Between the two classes, contain each mutant only once
          2. Occur in approximately equal numbers
     D. Largest pair (greater than 25%) parentals
     E. Lowest values are double crossovers
     F. Always determine which one gene is switched by comparing with parentals
VI. Map distances
     A. Calculated by adding classes that show recombination
     B. End-to-end distance underestimated because of double crossovers
          1. Distantly linked genes appear closer than they really are because double crossovers occur, but
                are not counted
     C. Summed short distances produce more accurate maps
          1. Fairly accurate to about 20 mu
          2. Can never detect more than 50% recombination in one cross
VII. Coefficient of coincidence
     A. Are crossovers independent of each other?
     B. If independent, double crossovers = product of frequencies of single crossovers
     C. Coefficient of coincidence = observed double crossovers/expected double crossovers
     D. Interference = 1 - coefficient of coincidence
          1. If negative, increased double crossovers
          2. If positive, inhibited double crossovers
VIII. Mapping the human genome
     A. Genetic linkage map
          1. Analysis of pedigrees
               a. Information on two different traits
               b. Can be genes, molecular markers or combination
          2. Determine parental and recombinant offspring
          3. Lod score
               a. Statistical tool to determine likelihood of linkage
               b. Lod > 3; genes linked
               c. Lod < -2; genes not linked
          4. Mapping based on recombination frequencies of traits with highest lod score
     B. Physical map
          1. Somatic cell hybridization
               a. Mouse-human hybrids
               b. Systematically lose human chromosomes until stable
               c. Screening of group of clones for trait
                    (1) Look for chromosome common to all hybrids that express trait
                    (2) Chromosome absent in all hybrids without trait
          2. Chromosomal banding patterns - low resolution mapping
          3. In situ hybridization
               a. Denature DNA, use of marker to hybridize to site
               b. FISH - fluorescent in situ hybridization
          4. Sequence-tagged sites
               a. Unique sequences that are used as markers to position genes
IX. Tetrad analysis
     A. Examination of ALL meiotic products
     B. Fungal phenotypes
          1. Colony morphology
          2. Drug resistance
          3. Nutritional requirements: mutants will not grow on minimal media
     C. Unordered spores - Saccharomyces cerevisiae
          1. Cannot determine gene-centromere distance
          2. If linked, PD >> NPD
          3. Gene-gene distance = [1/2(TT) + NPD]/total x 100
     D. Ordered spores - Neurospora
          1. Allows calculation of gene to centromere
          2. First division segregation (FDS) yields 4a then 4+
          3. second division segregation (SDS) yields 2:2:2:2 or 2a:4+2a
          4. Gene-centromere distance = 1/2(SDS)/total x 100
     E. Gene order
          1. Classify asci as parental (PD), nonparental (NPD) or tetratype (TT).  If linked, PD>>NPD
          2. Calculate gene-centromere distances
          3. If both genes are on the same side of the centromere, SDS for closer genes should usually
                produce SDS for second