Genetics: Genetics Lecture 11, 9/24: Linkage Mapping

o Genotype of the organism which is producing crossover gametes needs to be heterozygous
o The cross should be constructed in a way to allow you to determine the genotype of the crossover chromosome by examining the phenotype of the offspring

o A sufficient number of offspring is necessary to allow you to understand the patterns.

o Yellow body: y, white eyes:w, eye shape: e
o All three traits on the X chromosome
o final numbers
o 80 yellow body, WT eye color, WT eye shape (SCO 2)
o 193 yellow body, white eyes, WT shape (SCO 1)
o 3 yellow body, WT eyes, echinus
o 4685 all 3 mutant traits (DCO)
o 4759 all 3 WT traits (DCO)
o 3 WT body, white eyes, WT eye shape
o 207 wild type body, WT eye color, WT echinus (SCO 1)
o 70 WT body , white eyes, echinus (SCO2)
o You should be able to tell the relative order of the genes and the mapping distance between the genes.

o The first is the non-crossover offspring. Offspring that inherit intact parental chromosome. Represent your largest class of offspring.

• In the drosphila it is the all WT and all Mutant, labeled NCO (non-cross over)

o Double crossover offspring (DCO). Offspring in which 2 crossovers have occurred. Smallest class in your population, least likely event.
o Single crossover 1: result of a single crossover between one of your gene pairs (relatively similar numbers)
o Single crossover 2: result of a single crossover between the 2nd gene pairs (relatively similar numbers)

o What are our possible arrangements? NOTE: 3 factor crossover mapping will only tell you which gene is in the middle

o perform theoretical double crossobers on all 3 of our possible alignments. Determine which one matches our actual DCO
o Our order is y w e , y+w+e+
o Determine the mapping distances between the genes
o identify all of the offspring with a cross between two genes, total number of offspring.
o Always need to add the DCO offspring to your number, these have a cross between y and w.

• 3 possible arrangements
• take the NCO and arrange them in 3 possible arrangements. Perform the double crosses and determine if it lines up with the DCO that is given. After doing this out the correct order is bn pr+ r, bn+ pr v+

• perform a single cross on the above order and you get
• bn pr v+ (195), bn+ pr+ v (200) then add the DCO numbers (86) to get 481. Divide that by total number (1109) and you get 43.4. This is the distance from bn → pr+
• do this again for the distance from pr to r. bn pr+ v+ (82), bn+ pr v (79). 161 plus 86 gives you 247/1108. This gives you a distance of 22.3 mu between pr+ and r

Genetics