Question

For fruit flies, straight wings (c+) are dominant to curved wings (cr+), red eyes (pr+) are dominant to purple eyes (pr), and long wings (vg+) are dominant to vestigial wings (vg). The parental genotypes for a series of crosses are wild-type females mated to homozygous recessive males. The F1 generation is heterozygous for all 3 traits regardless of sex. Several F1 generation testcrosses are performed and produces the following offspring:

c+ vg+ pr+
283
c vg pr
303
c+ vg pr+
1
c vg+ pr
1
c+ vg+ pr
47
c vg pr+
43
c+ vg pr
30
c vg+ pr+
31
Total
739

What is the total map distance between these 3 genes to the nearest tenth (with one number after the decimal)? Units are cM or mu (i.e. if the answer is 11.4 centimorgans, report as 11.4 cM). *Be sure to include units or the answer is incorrect.

Answer 1

To calculate the total map distance between three genes, we need to determine the recombination frequencies for each pair of genes using the given data. The recombination frequencies are then summed up to obtain the total map distance. In this case, the total map distance is 0.2328 cM.

Step-by-step explanation:

To calculate the total map distance between the three genes, we need to determine the recombination frequencies for each pair of genes. The recombination frequencies can be calculated by dividing the number of recombinant offspring by the total number of offspring. From the given data, we can determine the recombination frequencies for each pair:

• c+ and vg+: 44 recombinant offspring / 739 total offspring = 0.0596
• vg+ and pr+: 79 recombinant offspring / 739 total offspring = 0.1069
• c+ and pr+: 49 recombinant offspring / 739 total offspring = 0.0663

To calculate the total map distance, we sum up the recombination frequencies for each pair of genes:

Total map distance = 0.0596 + 0.1069 + 0.0663 = 0.2328 cM