Can someone explain how the map distance between the genes were calculated. You
ID: 85295 • Letter: C
Question
Can someone explain how the map distance between the genes were calculated. You are mapping three linked genes in tomato plants. The recessive alleles for the three genes (when in the homozygous condition) cause absence of anthocyanin G), respectively. Two true-breeding parents were crossed, producing trihybrid (heterozygous) Fls, which were then test-crossed. The 3000 offspring were grouped into the following phenotypic classes (a trait not mentioned in a particular phenotypic class can be considered to be dominant or wild-type for that trait): Phenotype um r Observed hairless 249 jointless, hairless 40 jointless 931 wild type for all three 270 anthocyaninless, jointless, hairless 278 anthocyaninless, hairless 941 anthocyaninless 32 anthocyaninless, jointless 259 a) What are the original parental phenotypes? b) Which of the three genes is in the middle? c) Calculate the map distance between the three genesExplanation / Answer
1. The original parental phenotype will be of higher numbers in the cross. In this case the higher number of phenotype is for Jointless - 931 and anthocyaninless, hairless -941. Hence, the parental phenotype is only this two
2. The middle genes will be anthrocyanin less since for this to occur it has to undergo double recombination and hence its frequency will be the least among the other phenotype.
3.
The four pairs of phenotype classes are:
Parental (P) class, which is most frequent, and which shows the same phase relationships as the parents
two single recombinant (I & II) classes, between the middle locus and each outside locus
one double recombinant (D) class, between the middle locus and both outside loci
As this requires two events, this is the rarest class
The two members of each recombinant pair should occur in approximately equal numbers,
because each crossover produces two products
To determine gene locus order:
Most common phenotypes are parental: no recombinants
Rarest phenotype will be the double-recombinants:
Comparing the table we have found anthocyaninless is in the middle
Hence the gene order is Jointless anthocyaninless, hairless
Frequency of recombination between Jointless and anthocyaninless will be Double recombinants + single recombinants/ Total number of offspring
Here it is = (32+40) +(249+259)/3000 = 0.19 or 19% = 19cM
For anthocyaninless hairless it will be = (32+40)+(270+278) /3000 = 620/3000 = 0.20 or 20% = 20cM
For Joint less and hairless it will be = (249+259) + (270+278)/3000 = 0.35 0r 35% = 35 cM
Phenotype Count Class Jointless 931 P Anthocyaninless, hairless 941 P Jointles, hairless 40 D anthocyaninless 32 D Hairless 249 I Wild type for all three 270 II Anthocyaninless jointless hairless 278 II Anthocyaninles, jointless 259 I