Problem Set 13 Answers
1.a. Mitotic recombinations occur twice as frequent as the appearance of mosaic patches; i.e. half of mitotic recombinations result in mosaics.
b. If mitotic recombination occurs between the centromere and y, then twin spots can be produced. If it occurs between y and sn, then a single spot marked with the Sn phenotype can be made. Much more rare a double event can be made, which could give rise to a Y clone.
Twin Spot Mosaic:
* y + * y + * y + Y * y + WT
* y + * + sn * y + * + sn
X à è Twin Spot or Wild-type
* + sn * y + * + sn Sn * + sn WT
* + sn * + sn * + sn * y +
Single Spot Mosaic:
* y + * y + * y + WT * y + WT
* y + * y sn * + + * + sn
X à è Single Spot or Wild-type
* + sn * + + * y sn Sn * y sn WT
* + sn * + sn * + sn * + +
c. The frequency of each type of mosaic will depend on the distance between the gene pair and the centromere and the distance between the two genes. Twin spot mosaics arise out of recombinations between the gene pair and the centromere; single spot mosaics arise from recombinations between the two genes. Therefore, the greater the distance between the gene pair and the centromere, the more likely a twin-spot clone will appear. The greater the distance between the two genes, the more likely twin-spot clones will appear.
2.a. Assuming both mutations are recessive, the trans configuration is best for generating twin spot mosaics.
b. The further the markers from the centromere, the more frequent twin spots will appear.
c. The closer together the markers are, the more frequent twin spots will appear.
d. If you were to generate a twin spot mosaic, use a pair of markers that are close together but far away from the centromere.
3. To determine whether a mutation that causes non-disjunction in meiosis has a similar effect on somatic divisions, generate an animal that is heterozygous for a specific marker (ex. white w/+) and homozygous for the mutation that causes non-disjunction. If the mutation has an effect on somatic divisions, you would expect spontaneous white (w/w) clones in the animal.
4.a. The focus of action is where the mutant gene must be expressed to see the phenotype.
b. For twin spot mosaics involving a dominant mutation and a recessive marker, use the cis configuration:
* M c * M c * M c MC * M c M
* M c * + + * M c * + +
X à è Twin Spot or M
* + + * M c * + + WT * + + M
* + + * + + * + + * M c
c. You would not expect to see twin spots in either configuration if M was not cell autonomous.
d. 50% of mitotic recombinations in the cis configuration will result in twin spot mosaics; see 4.b.
e. To optimize the number of twin spot mutations, use the c marker. “c” is the marker farthest away from the centromere that is closest to mutation “M.”
f. The answers to b-e would be unchanged if M was recessive and the marker mutations were dominant. A cis configuration would still be required for twin spot mosaics, which would arise 50% of the time that mitotic recombinations occurred.
5.a. Fate mapping shows where body parts originate from the cellular blastoderm in Drosophila. Meiotic mapping allows you to map genes via recombination.
b. Twin spot mosaics results from recombination between the centromere and a gene pair. Single spot mosaics result from recombination between the genes or when only one marker is present.
c. Mosaic analysis provides the gene’s focus of action; GFP expression studies tell where a promoter is active or where a protein is produced.
6. To determine how near the eye and antenna are on a fate mate, observe gynandromorphs with cell-autonomous markers that are expressed in the eye and antenna. The distance between the two parts on a fate map will be determined by the number of animals in which the eyes and antennae are of different sex over the total.
7.a. True, if sex determination were not cell autonomous, then gynandromorphs would not form.
b. False, the production of twin spots involving two recessive mutations requires that recombination occurs between the centromere and a (if a is the closer to the centromere).