Education.com
Try
Brainzy
Try
Plus

Linkage and Chromosome Mapping Practice Problems (page 2)

based on 2 ratings
By — McGraw-Hill Professional
Updated on Apr 25, 2014

Practice 3

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.

Solution 3

Practice 4

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.

Solution 4

Solved Problems

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.

Practice 5

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."

Solution 5

Solved Problems

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.

Practice 6

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.

Solved Problems

Solution 6

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.

Solved Problems

*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.

View Full Article
Add your own comment

Ask a Question

Have questions about this article or topic? Ask
Ask
150 Characters allowed