The four sets of data above form a continuous map over six loci. Determine the l

Question

The four sets of data above form a continuous map over six loci. Determine the locus order and distances among loci for each set of data. Determine the gene order and the distances between all six loci. In each case, a triple dominant homozygote parent (ABD//ABD) has been crossed with a triple recessive homozygote tester (abd//abd) to produce an F_1 triple heterozygote (AaBbbd). The heterozygote was then back-crossed to the tester strain (abd//abd), and the numbers of resultant progeny counted. A ' +' symbol indicates the phenotype corresponding to the dominant allele at each locus. So, each parental cross can also be written in the form a^+b^+d^+//a^+b^+d^+ X abd//abd, or in short-hand form +++//+++ X abd//abd.

Explanation / Answer

I.By observing the double crossover (lowest number of count), the arrangement of the gene should be eni. Because in double crossovers the members of the middle pair of alleles between the chromosomes are interchanged.

Map Distance:

+i+= 135 -------- single crossover between n and i

e+n= 137 --------- single crossover between n and i

+in = 100--------- single crossover between e and n

e++ = 92 ------ single crossover between e and n

II.By observing the double crossover (lowest number of count), the arrangement of the gene should be igm.

Map Distance:

++m= 92 -------- single crossover between g and m

gi+= 88 --------- single crossover between g and m

g+m = 44--------- single crossover between g and i

+i+ = 36 ------ single crossover between g and i

III.By observing the double crossover (lowest number of count), the arrangement of the gene should be amg .

Map Distance:

a+m= 90 -------- single crossover between m and g

+g+=82 --------- single crossover between m and g

a++ = 60--------- single crossover between a and m

+gm = 52 ------ single crossover between a and m

IV.By observing the double crossover (lowest number of count), the arrangement of the gene should be gin.

Map Distance:

ig+= 176 -------- single crossover between n and i

++n= 184 --------- single crossover between n and i

+g+ = 33--------- single crossover between g and i

i+n = 27 ------ single crossover between g and i

a______m___________g_______i____________________n_______________e    ----- (genes)

  14               20                   10                           40 32 -------- distance (map units)

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