Answer a, b with explanation:
the best way to determine whether there is linkage or not we use chi-square analysis. It tells us that it is highly doubtful that the three genes assort independently. To determine linkage by simple check, look at gene-pairs. As this is a test-cross, independent assortment of genes predicts a 1:1:1:1 ratio.
Comparing shrunken and white, the frequencies are:
+ + (113 + 4)/total
s wh (116 + 2)/total
+ wh (2708 + 626)/total
s + (2538 + 601)/total
it indicates that these is no independent assortment between white and shrunken, which indicates a linkage.
Now Comparing shrunken and waxy, the frequencies are:
+ + (626 + 4)/total
s wa (601 + 2)/total
+ wa (2708 + 113)/total
s + (2538 + 116)/total
It indicates that there is no independent assortment between waxy and shrunken, which indicates a linkage.
Comparing white and waxy, the frequencies are:
+ + (2538 + 4)/total
wh wa (2708 + 2)/total
wh + (626 + 116)/total
+ wa (601 + 113)/total
This calculation shows that there is no independent assortment between white and waxy, which indicates a linkage. Because all three genes are linked therefore the strains must be + s +/wh + wa and wh s wa/wh s wa (compare most frequent, parentals, to least frequent, double crossovers, to obtain the gene order). The cross can be written as: P = + s +/wh + wa * wh s wa/wh s wa
F1 = as in problem
Crossovers between white and shrunken are: 113+116+4+2= 235
Crossovers between shrunken and waxy are: 601+626+4+2=1233
Dividing by the total number of progeny and multiplying by 100 percent yields the following map:
3.5 m.u. 18.4 m.u.
white shrunken waxy
Answer c with explanation:
Interference = 1 – (observed double crossovers/expected double crossovers)
= 1 – 6/(0.035)(0.184)(6,708)
=0.86
