Assume that the coefficient of coincidence (CC) for a map distance of 24 cM = 0.

Question

Assume that the coefficient of coincidence (CC) for a map distance of 24 cM = 0.25. Consider the following 3-factor (trihybrid) testcross: a to b: 10.6 cM and b to +: 13.4 cM Calculate the number of progeny in each phenotype class (parentals, SCOs, and DCOs) resulting from this cross. Base your calculations on a total of 1000 progeny. Hint: The actual frequency of double-crossover progeny can be determined by multiplying the frequencies of recombination for each single crossover together (to calculate expected DCOs), and then multiplying that product by the coefficient of coincidence for the known map distance between the alleles (to calculate the actual, observed DCOs).

Explanation / Answer

Remember that CC =DCOobs / DCOexp

Crossover inference(i)=1-CC

The recombination frequency between genes a-b= 10.6% = 0.106

The recombination frequency between genes b-c=13.4 =0,134

Calculatted expected number of double crossover progeny= 0.106*0.134=0.0142

Expexted DCO frequency= 0.0142*1000=14.2 individuals

                                    DCOexp= 14.2

CC=DCOobs / DCOexp

DCOobs=CC*DCOexp

            =0.25*14.2=3.55 = 4

This means that of 1000 progeny you will see four double crossover phenotypes

Number ofnprogeny in single crossover a-b= 10.6-0.356=10.24=102/1000

                                                          b-c= 13.4-0.356= 13.0= 130/1000

The number of parental type progeny =1000-(4+102+130)= 764/1000=0.764

Map distance = % recombinations= SCOphenotypes +DCO phenotypes *100 / total progeny

                                                  =   0.764+4*100 / 1000

                                                  =   476.4/1000=0.476 this is the map distance between alleles

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