Linkage and Chromosome Mapping Practice Problems (page 3)
Review the following concepts if needed:
- Recombination Among Linked Genes for Genetics
- Genetic Mapping for Genetics
- Linkage Estimates from F2 Data for Genetics
- Use of Genetic Maps for Genetics
- Crossover Suppression for Genetics
- Tetrad Analysis in Fungi for Genetics
- Recombination Mapping with Tetrads for Genetics
Linkage and Chromosome Mapping Practice Problems
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.
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.
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:
(a) Determine the linkage relationships in the parents and F1 trihybrids.
(b) Estimate the map distances.
(a) The parents are homozygous lines:
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
The map distances for six genes in the second linkage group of the silkworm Bombyx mori are shown in the table below. Construct a genetic map that includes all of these genes.
White eyes (w/w females; w/Y males) in Drosophila can be produced by the action of a sex-linked recessive gene. White eyes can also be produced through the interaction of two other genes: the recessive sex-linked gene v for vermilion eye color, and the autosomal recessive gene bw for brown eye color. Consider the parental cross: bw/bw, w+v+/w v (brown-eyed females) &£215; bw/bw, w v/Y (white-eyed males), where the F1 progeny consists of 70 brown-eyed and 130 white-eyed individuals. Estimate the distance between the sex-linked genes w and v.
Only the genotypes of the brown offspring are known for certain. The 70 brown offspring constitute only one-half of the offspring produced by noncrossover maternal gametes. Therefore, we estimate that 70 of the white individuals were also produced by noncrossover maternal gametes. Thus, 140 out of 200 F1 flies are estimated to be parental-type offspring = 70%. The other 30% must be crossover types. The best estimate of linkage between the white and vermilion loci would be 30 map units.
Elongate tomato fruit is produced by plants homozygous for a recessive gene o, round fruit shape is produced by the dominant allele at this locus (O). A compound inflorescence is the result of another recessive gene s, simple inflorescence is produced by the dominant allele at this locus (S). A Yellow Pear variety (with elongate fruit and simple inflorescence) is crossed to a Grape Cluster variety (with round fruit and compound inflorescence). The F1 plants are randomly crossed to produce the F2. Among 259 F2, 126 round, simple : 63 round, compound : 66 long, simple : 4 long, compound are found. Estimate the amount of recombination by the "square-root method."
Notice that the double-recessive phenotype (long, compound) occupies only 1 of the 16 frames in the gametic checkerboard. This genotype is produced by the union of two identical double-recessive gametes (o, s). If we let x = the frequency of formation of os gametes, then x2 = frequency of occurrence of the os/os genotype (long, compound phenotype) = 4/259 = 0.0154. Thus, . But x estimates only half of the crossover gametes. Therefore 2x estimates all of the crossover gametes = 2(0.124) = 0.248 or 24.8% recombination.
Several three-point testcrosses were made in maize utilizing the genes booster (B, a dominant plant color intensifier), liguleless leaf (lg1), virescent seedling (v4, yellowish-green), silkless (sk, abortive pistils), glossy seedling (gl2), and tassel-seed (ts1, pistillate terminal inflorescence). Using the information from the following testcrosses, map this region of the chromosome.
Following the procedures established in this chapter, we determine from each of the testcrosses the gene order (which gene is in the middle) and the percent crossing over in each region. Note that the results of testcrosses 1 and 2 may be combined, recognizing that the linkage relationships are different in the trihybrid parents. Likewise, the results of 3 and 4 may be combined, as well as those of testcrosses 6 and 7. The analyses of these seven testcrosses are summarized below in tabular form.
*Note that in testcross 3, only seven phenotypes appeared, whereas we expected eight. We suspect that the missing phenotype (wild type) is a double-crossover (DCO) type because DCO types are expected to be less frequent than the others. The two phenotypes with the highest numbers should derive from the parental (noncrossover gametes). Thus, the booster phenotype indicates that the three genes B, lg1+, and sk were on one parental chromosome; likewise, the other high-frequency progeny phenotype (liguleless and silkless) indicates that the three genes lg1, sk, and B+ were on the homologous chromosome, but we do not know which of these loci is in the middle. Assuming that the least frequent phenotype (booster, liguleless, silkless) is one of the double-crossover types, we can infer that the booster locus is in the middle; i.e., if parents were lg1+ , B, sk+/lg1, B+, sk the double crossovers would be + + + (wild type) and lg1, B, sk (liguleless, booster, silkless). This inference is confirmed by testcross 4 where all eight progeny phenotypes are present.
†In testcross 7, only four phenotypes appeared in the progeny (no explanation given for the missing phenotypes). One might be tempted to eliminate such bizarre results from a report, but it would be scientifically unethical to do so. Data selection or alteration would be considered fraudulent. A scientist must report all the data or give reasons for failing to do so. However, it is possible to establish the gene order from testcross 6, so that the type of progeny (noncrossovers, single crossovers, and double crossovers) can be identified unambiguously.
Suppose we are given a strain of Drosophila exhibiting an unknown abnormal genetic trait (mutation).We mate the mutant females to males from a balanced lethal strain (Cy Pm+ /Cy+ Pm, DSb+ /D+ Sb) where curly wings (Cy) and plum eye (Pm) are on chromosome 2 and dichaete wing (D) and stubble bristles (Sb) are on chromosome 3. Homozygosity for either curly, plum, dichaete, or stubble is lethal. The trait does not appear in the F1. The F1 males with curly wings and stubby bristles are then backcrossed to the original mutant females. In the progeny the mutation appears in equal association with curly and stubble. Drosophila melanogaster has a haploid number of 4 including an X, 2, 3, and 4 chromosome. (a) Determine whether the mutation is a dominant or a recessive. (b) To which linkage group (on which chromosome) does the mutation belong?
(a) If the mutation were a dominant (let us designate it M), then each member of the strain (pure line) would be of genotype MM. Since the trait does not appear in our balanced lethal stock, they must be homozygous recessive (M+M+). Crosses between these two lines would be expected to produce only heterozygous genotypes (M+M) and would be phenotypically of the mutant type. But since the mutant type did not appear in the F1, the mutation must be a recessive (now properly redesignated m). The dominant wild-type allele may now be designated m+.
(b) Let us assume that this is a sex-linked recessive mutation. The F1 males receive their single X chromosome from the mutant female (mm). Therefore, all males of the F1 should exhibit the mutant trait because males would be hemizygous for all sex-linked genes (mY). Since the mutant type did not appear in the F1, our recessive mutation could not be sex-linked.
Let us assume that our recessive mutation is on the second chromosome. The curly, stubble F1 males carry the recessive in the heterozygous condition (Cy m+ /Cy+ m, Sb/Sb+). Notice that we omit the designation of loci with which we are not concerned. When these carrier males are then backcrossed to the original mutant females (Cy+ m/Cy+ m, Sb+ /Sb+), the F2 expectations are as follows:
Note that the mutant cannot appear with curly. Therefore, our recessive mutation is not on chromosome 2.
Let us then assume that our mutant gene is on the third chromosome. When F1 carrier males (Cy/Cy+, Sb m+ /Sb+ m) are backcrossed to the original mutant females (Cy+ /Cy+, Sb+ m/Sb+ m), the F2 expectations are as follows:
Note that the mutant cannot appear with stubble. Hence, our recessive mutation is not on chromosome 3.
If the mutant is not sex-linked, not on 2 nor on 3, then obviously it must be on the fourth chromosome. Let us prove this. When F1 carrier males (Cy/Cy+, Sb/Sb+, m+/m) are backcrossed to the original mutant females (Cy+/Cy+, Sb+/Sb+, m/m), the F2 expectations are as follows:
Note that our recessive mutant occurs in equal association with curly and stubble, which satisfies the conditions of the problem. We conclude that this mutation is on the fourth chromosome.
A strain of Neurospora requiring methionine (m) was crossed to a wild-type (m+) strain with the results shown below. How far is this gene from its centromere?
Noncrossover asci are those that appear with the greatest frequency = 40 + 36 = 76 out of 100 total. The other 24/100 are crossover types. While 24% of the asci are crossover types, only half of the spores in these asci are recombinant. Therefore, the distance from the gene to the centromere is 12 map units. The origin of the crossover-type asci is as follows:
Two strains of Neurospora, one mutant for gene a, the other mutant for gene b, are crossed. Results are shown below. Determine the linkage relationships between these two genes.
Pattern 1 represents the noncrossover types showing first-division segregation (4 : 4) for both a and b. Pattern 2 shows second-division segregation (2 : 2 : 2 : 2) for a, but first-division segregation for b. Genes that show high frequency of second-division segregation are usually further from the centromere than genes with low frequency of second-division segregation. Judging by the relatively high frequency of pattern 2 these are probably single crossovers, and we suspect that a is more distal from its centromere than b.
Pattern 3, indicating first-division segregation for a and second-division segregation for b, cannot be generated by a single crossover if a and b are linked as shown above, but requires a double crossover.
Furthermore, pattern 4 could be produced from the linkage relationships as assumed above by a single crossover in region I.
Double crossovers (DCO) are expected to be much less frequent than single crossovers (SCO). Under the above assumptions, double-crossover pattern 3 is more frequent than one of the single-crossover patterns 4. This does not make sense, and thus our assumption must be wrong. The locus of a must be further from the centromere than b, but it need not be on the same side of the centromere with b. Let us place a on the other side of the centromere.
Now a single crossover in region I produces pattern 2, a single crossover in region II produces pattern 3, and a two strand double crossover (I, II) produces pattern 4. The percentage of asci are numerically acceptable under this assumption.
The distance a-centromere =1/2(SCOI + DCO) = 2(14 + 1) = 7.5 map units
The distance centromere-b = 1/2(SCOII + DCO) = 2(6 + 1) = 3.5map units
The cross (abc) × (+ + +) is made in an ascomycete with unordered tetrads. From the analysis of 100 asci, determine the linkage relationships between these three loci as completely as the data allow.
- 40 (abc)(abc)(+ + +)(+ + +)
- 42 (ab+)(ab+)(+ + c)(+ + c)
- 10 (a + c)(+ + c)(ab +)(+ b +)
- 8(a + +)(+ + +)(abc)(+ bc)
Pattern 1 is parental ditype (PD) for ab, ac, and bc. Pattern 2 is PD for ab, nonparental ditype (NPD) for ac and bc. Pattern 3 is tetratype (TT) for ab and ac, NPD for bc. Pattern 4, is TT for ab and ac, PD for bc. For each pair of markers the relative frequencies of each type of tetrad are as follows:
For the ab pair, PDs are not equivalent with NPDs. Thus, a and b must be linked. For pairs ac and bc, PDs are roughly equivalent with NPDs. Thus, c must be assorting independently on another chromosome. The recombination frequency between a and b is
The recombination frequency in percent is equal to 9 map units. Single crossovers between either b and its centromere or c and its centromere, or both, would produce TTs for bc. Since none occurred we can assume that the loci of b and c are both very near their respective centromeres.
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