A testcross of trihybrid Drosophila produced the following phenotypes and number

Question

A testcross of trihybrid Drosophila produced the following phenotypes and number of offspring.A table showing phenotypes and the number of offspring with each phenotype is below. A plus sign is used for wild type phenptype. a letter indicates the mutant phenotype for that gene.the order of genes given in the data table may not be the correct order(since you dont know it yet).
+++ 669
abc 653
++c 121
ab+ 139
+b+ 2280
a+c 2215
+bc 3
a++ 2
based on these data, which is correct order of these gene on the chromosome? ( please explain it for me)
a) data indicate they are noton the same chromosome
b) acb
c) bac
d)abc
e) abac
a) data

Explanation / Answer

The gene order is a c b d.

Recombination between a and c occurred at a frequency of

100%(139 + 3 + 121 + 2)/(669 + 139 + 3 + 121 + 2 + 2,280 + 653 + 2,215)

= 100%(265/6,082) = 4.36%

Recombination between b and c in cross 1 occurred at a frequency of

100%(669 + 3 + 2 + 653)/(669 + 139 + 3 + 121 + 2 + 2,280 + 653 + 2,215)

= 100%(1,327/6,082) = 21.82%

Recombination between b and c in cross 2 occurred at a frequency of

100%(8 + 14 + 153 + 141)/(8 + 441 + 90 + 376 + 14 + 153 + 64 + 141)

= 100%(316/1,287) = 24.55%

The difference between the two calculated distances between b and c is not

surprising because each set of data would not be expected to yield exactly identical

results. Also, many more offspring were analyzed in cross 1. Combined, the

distance would be

100%[(316 + 1,327)/(1,287 + 6,082)] = 22.3%

Recombination between b and d occurred at a frequency of

100%(8 + 90 + 14 + 64)/(8 + 441 + 90 + 376 + 14 + 153 + 64 + 141)

= 100%(176/1,287) = 13.68%

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