A dark spore Sordaria fimicola culture is crossed with a light tan spore culture
ID: 177005 • Letter: A
Question
A dark spore Sordaria fimicola culture is crossed with a light tan spore culture on a com agar plate. Ten days later proto-perithecia are collected and crushed on a glass slide in a drop of water in order to release the many ascus structures held within them. 50 ascii are examined and the data is recorded on the color and number of the spores found in each ascus. Use the following data table ...to determine the gene to centromere distance for each group and then using the summed data from the entire class to recalculate the gene to centromere distance for the class data. Show all formulas and calculations used (2:2:2:2 = 2 dark; 2 light; 2 dark; 2 light spores in that order or the reverse order...whereas 2:4:2 = 2 dark 4 light: 2 dark in that order or the reverse order)Explanation / Answer
Step1: Count total number of Ascii showing Crossover. This can be done by summation of ascii 2:2:2:2 + ascii 2:4:2.
Step 2: % Calculate Asci Showing Crossover (Recombinant Frequency) by dividing number of Asci Showing Crossover with Total Asci multiplied by 100.
Step 3: To calculate the map distance, divide the percentage of crossover asci by 2. This is done since only half of the spores in each ascus are the result of crossing over.
Total No. of Ascii showing Crossover (2:2:2:2 + 2:4:2)
% Asci Showing Crossover (Recombinant Frequency)
Calculated Gene to Centromere Distance (cM)
Group 1
20 + 7 = 23
(23/50)*100 = 46
46/2 = 23
Group 2
13 + 4 = 17
(17/50)*100 = 34
34/2 = 17
Group 3
16 + 12 = 28
(28/50)*100 = 56
56/2 = 28
Group 4
14 + 19 = 33
(33/50)*100 = 66
66/2 = 33
Group 5
11 + 20 = 31
(31/50)*100 = 62
62/2 = 31
Group 6
16 + 13 = 29
(29/50)*100 = 58
58/2 = 29
Total
90 + 75 = 165
(165/300)*100 = 55
55/2 = 27.5
Results are :
Calculated Gene to Centromere Distance (cM)
Group 1
46/2 = 23
Group 2
34/2 = 17
Group 3
56/2 = 28
Group 4
66/2 = 33
Group 5
62/2 = 31
Group 6
58/2 = 29
Total
55/2 = 27.5
Total No. of Ascii showing Crossover (2:2:2:2 + 2:4:2)
% Asci Showing Crossover (Recombinant Frequency)
Calculated Gene to Centromere Distance (cM)
Group 1
20 + 7 = 23
(23/50)*100 = 46
46/2 = 23
Group 2
13 + 4 = 17
(17/50)*100 = 34
34/2 = 17
Group 3
16 + 12 = 28
(28/50)*100 = 56
56/2 = 28
Group 4
14 + 19 = 33
(33/50)*100 = 66
66/2 = 33
Group 5
11 + 20 = 31
(31/50)*100 = 62
62/2 = 31
Group 6
16 + 13 = 29
(29/50)*100 = 58
58/2 = 29
Total
90 + 75 = 165
(165/300)*100 = 55
55/2 = 27.5