Neurospora of genotype a + c are crossed with Neurospora of genotype + b +. The
ID: 165414 • Letter: N
Question
Neurospora of genotype a + c are crossed with Neurospora of genotype + b +. The following tetrads arc obtained (note that the genotype of the four spore pairs in an ascus are listed, rather than listing all eight spores). a. In how many cells has meiosis occurred to yield these data? b. Are genes a and b linked? If so calculate the distance between them in map units. c. Are genes b and c linked? If so calculate the distance between them in map units. d. Are genes a and c linked? If so calculate the distance between them in map units.Explanation / Answer
Answer:
a. The number of meiosis represented here is the total number of asci = 334.
b. First the type of asci present are to be designated. This is to be done for each pair of genes: PD (P), NPD (n), and T . When PD = NPD, then the genes are not linked.
A - b: PD = 137 and NPD = 141 + 2 = 143.
So, PD = NPD, therefore A - b are not linked.
c. B - c: PD = 137 + 26 + 2 = 165 and NPD = 141 + 25 =166
PD = NPD, therefore not linked.
d. A - c: PD = 137 + 141 = 278 and NPD = 3
PD > NPD, therefore linked
Calculation of map distances:
Mu = [(PD + 1/2T)/total ] x 100 = [2 + 1/2*(26 +25)/334] x 100 = 8.2 cM
RF = (second division segregation) * 1/2 / total
For a + and a + = [(26 + 25)x 0.5]/334 = 0.076 = RF, m.u = 7.6
For b + and b + = (3 x 0.5)/334 = 0.004, RF = 0.4 mu.
For + c and + c = 0, on top of centromere
So, the genetic map is:
c a +
(CEN)---------------------------------------- (CEN)----------------------------------
(CEN)---------------------------------------- (CEN)----------------------------------
+ + b