In corn, a triple heterozygote was obtained carrying the mutant alleles s (shrunken), w (white aleurone), and y (waxy endosperm), all paired with their normal wild-type alleles. this triple heterozygote was testcrossed, and the progeny contained 116 shrunken, white; 4 fully wild type; 2538 shrunken; 601 shrunken, waxy; 626 white; 2708 white, waxy; 2 shrunken, white, waxy; and 113 waxy.

a. determine if any of these three loci are linked, and, if so, show map distances.

b. show the allele arrangement on the chromosomes of the triple heterozygote used in the testcross.

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

age range would be the answer