Linkage and Chromosome Mapping Practice Problems

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By — McGraw-Hill Professional
Updated on Apr 25, 2014

Review the following concepts if needed:

Linkage and Chromosome Mapping Practice Problems

Practice 1

In the human pedigree below where the male parent does not appear, it is assumed that he is phenotypically normal. Both hemophilia (h) and color blindness (c) are sex-linked recessives. Insofar as possible, determine the genotypes for each individual in the pedigree.

Solved Problems

Solution 1

The linkage relationship of the males' genes on their single X chromosome is obvious from their phenotype. Thus, I1, I2, and III3 are all hemophilic with normal color vision and therefore must be hC/Y. Nonhemophilic, colorblind males II1 and II3 must be Hc/Y. Normal males II2, II6, and III1 must possess both dominant alleles HC/Y. III2 is both hemophilic and color blind and therefore must possess both recessives hc/Y. Now let us determine the female genotypes. I3 is normal but produces sons, half of which are color blind and half normal. The X chromosome contributed by I3 to her color-blind sons II1 and II3 must have been Hc; the X chromosome she contributed to her normal sons II2 and II6 must have been HC. Therefore, the genotype for I3 is HC/Hc.

Normal females II4, II5, and II7 each receive hC from their father (I2), but could have received either Hc or HC on the X chromosome they received from their mother (I3). II4 has a normal son (III1) to which she gives HC; therefore, II4 is probably hC/HC, although it is possible for II4 to be hC/Hc and produce an HC gamete by crossing over. II5, however, could not be hC/HC and produce a son with both hemophilia and color blindness (III2); therefore, II5 must be hC/Hc in order to give the crossover gamete hc to her son.

Practice 2

A kidney-bean-shaped eye is produced by a recessive gene k on the third chromosome of Drosophilia. Orange eye color, called "cardinal," is produced by the recessive gene cd on the same chromosome. Between these two loci is a third locus with a recessive allele e that produces ebony body color. Homozygous "kidney," cardinal females are mated to homozygous ebony males. The trihybrid F1 females are then testcrossed to produce the F2. Among 4000 F2 progeny are the following:

Solved Problems

(a)   Determine the linkage relationships in the parents and F1 trihybrids.

(b)   Estimate the map distances.

Solution 2

(a)   The parents are homozygous lines:

Solved Problems

The linkage relationships in the trihybrid F1 can also be determined directly from the F2. By far the most frequent F2 phenotypes are kidney, cardinal (1761) and ebony (1773), indicating that kidney and cardinal were on one chromosome in the F1 and ebony on the other.

(b)   Crossing over between the loci k and e produces the kidney, ebony (128) and cardinal (138) offspring. Double crossovers are the triple mutants (6) and wild type (8). Altogether there are 128 + 138 + 6 + 8 = 280 crossovers between k and e:

280/4000 = 0:07 or 7% crossing over = 7 map units

Crossovers between a and cd produced the single-crossover types kidney (97) and ebony, cardinal (89). Double crossovers again must be counted in this region.

97 + 89 + 6 + 8 = 200 crossovers between a and cd

200/4000 = 0:05 or 5% crossing over = 5 map units

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