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Drosophila Genetics Homework Problems

Comparative review of: Drosophila, an online lab simulation, by Virtual Courseware for Inquiry-based Science Education,; and Virtual Genetics Laboratory II, an online lab simulation by the University of Massachusetts, Boston,

Labs demonstrating Mendelian genetics are illuminating for students but pose potential complications. Typical hands-on, in-person experimental models involve plants (which grow slowly) or fruit flies (which require strict time-frames for particular steps – something nearly impossible for high school or a commuter undergraduate campus). The advent of virtual labs allows students to struggle through mating strategies in order to draw conclusions about modes of inheritance (by simulating actual matings) without having to organize their real-life schedules around isolating Drosophila virgins.

The first virtual fruit fly lab road-tested was the Virtual Genetics Lab II version 3.2.0 (VGLII3.2.0) produced by a team at the University of Massachusetts, Boston. The program has pre-set “problems” which are individual crosses of parental fruit flies. Students must do subsequent additional crosses and analyze offspring to deduce the mode of inheritance. The problems can be worked on in a “practice” mode where the mode of inheritance can be revealed with a simple click, and there are replicas of these problems without this feature so students are forced to solve for the mode of inheritance through their own trial and error. However both practice and non-practice versions are available to anyone who downloads the software, making it difficult to use this as an assessment tool for distance learners or as homework. It would, however, be easy to delete practice versions on classroom computers to make this an excellent in-class activity. Progeny from one or more matings can be viewed in a summary chart that can be easily manipulated to show or hide sex, phenotype, etc., for fast, quantitative analysis. While a good tool overall, there are some drawbacks to the program. It requires download of an executable file as well as an appropriate Java environment; and with different versions for various operating systems, this could cause “hiccups” for distance learners. Additionally, some of the walk-throughs were written for a previous version of the program and have not yet been updated to reflect attributes of the new system. Finally, assessment is extremely limited since there are no quizzes or opportunities to record substantive conclusions at any point – such exercises would have to be developed separately by the instructor.

The second resource investigated was from Virtual Courseware for Inquiry-based Science Education (VCISE). VCISE allows a user to review applicability to K–12 science standards by state, showing that the learning objectives of the program will integrate easily into a course curriculum. Unlike the VGLII3.2.0, this interface operates entirely online (no downloads) and visually simulates all steps of an actual mating lab. One must put flies in mating jars, place and retrieve the jars from incubators, anesthetize the flies, sort them under a microscope, etc. Mimicking these actual steps (as opposed to simply receiving numerical data as in VGLII3.2.0) adds a completely different feel to the exercise. Whereas both programs allow one to keep track of mating results, VCISE allows the participant to save information about each step in a virtual lab notebook, again simulating a genuine lab operation. Finally, this program helps students build a lab report: students follow prompts to build a report online with quantitative analysis along with qualitative answers and images from their notebook. One of the best features of this software is that with minimal preparation, an instructor can set up a class code whereby students can save their work and build a report which can be monitored and graded online.

In summary, while both programs allow a student-directed approach whereby students choose fly matings and interpret the mode of inheritance based on offspring, the VCISE program is more ideally structured to visually engage students and facilitate instructor assessment of student learning.

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9.01 In sweet peas, purple flowers (P) is dominant to red (p) and elongate pollen grains (L) is dominant to round (l). A cross is made between a purple, long plant and a red, round plant. Both of these parents are known to be homozygous for both genes. When F1 plants are self-pollinated, the following progeny are obtained: 142 purple, long; 8 purple, round; 11 red, long; and 47 red, round. Use the chi-square method to test the hypothesis of independent assortment for this cross. If the hypothesis is rejected, propose a new hypothesis to explain what is happening here, and tell why your new hypothesis might better explain the observed ratios

Answer: Parents: PpLl x PpLl. If independent assortment is true, we would expect a 9:3:3:1 ratio. Testing this with chi-square:The chi-square value is 139, well above the table value for 3 df. Because independent assortment must not be true, these genes must be linked. This also looks correct because the numbers of parental (non-recombinant) offspring are both higher than the expected numbers.

9.02 In Drosophila, black body (b) and purple eyes (pr) are recessive mutations located on chromosome II. A true-breeding wild type female is mated with a true-breeding black body, purple eye male. All the offspring are wild type. One of these offspring is then test-crossed with a black body, purple eye fly. The results of this test cross indicate that there was a recombination rate of 0.06.

(a) Which four gametes were produced by the wild type fly used in the test cross?

Answer:b+ pr+ and b pr (nonrecombinant)
b+ pr and b pr+ (recombinant)

(b) What proportions of the four gametes would you expect from that parent? Keep in mind that they will not be in a 1:1:1:1 ratio because the genes are linked.

Answer: Recombinant gametes, b+ pr and b pr+, would be produced about 6% of the time when crossing over takes place. So we would expect to see about 3% of each of these. Nonrecombinant gametes, b+ pr+ and b pr, would occur the other 94% of the time (1-.06). So we would expect to see about 47% of each of these.

9.03 A trihybrid test cross has been done: Cc Dd Ee x cc dd ee. Out of the 1000 offspring that were produced, the phenotype class C- D- ee had 351 individuals. Does that suggest that the genes c, d, and e on the same chromosomes or on different chromosomes? Explain.

Answer: The genes are not on three different chromosomes. If that were true, the eight possible phenotypes would be produced in approximately equal numbers. Out of 1000 offspring, there would be about 125 (1000 x 1/8) of each type. 351 is significantly more than 125.

9.04 The following two recessive genes are known for corn: brittle endosperm (b) and glossy leaf (g). Two true-breeding plants with contrasting phenotypes are crossed (both plants are homozygous for both genes). The F1 offspring are all wild type. Then, an F1 plant is test-crossed with a brittle/glossy parent, and the following offspring are obtained:

(a) Explain your why these data indicate that the two genes are linked.

Answer: The numbers are not close to a 1:1:1:1 ratio which would be the case if the genes were not linked.

(b) Estimate how far apart (in map units) they are on the chromosome.

Answer: r = (16+25)/(16+238+254+25) = 41/533 = 0.08 (8 mu)

(c) What were the genotypes and phenotypes of the two parents of the F1 plant? If this is not obvious at first, start by drawing a diagram of the F1 chromosomes, then work backward to what the parents must have been like.

Answer: The non-recombinant (parental) phenotypes are glossy and brittle, so the original parents must have been glossy (g g b+ b+) x brittle (g+ g+ b b)

9.05 A geneticist working with corn has found that three genes: a, b, and c, are linked. She does as several 2-point test crosses in order to map the loci of the three genes. She determines that the rate of recombination between genes a and c is 0.1; between a and b, 0.27; and between b and c, 0.2.

(a) Determine the order of the three genes, and draw a map to show there approximate relative locations.


(b) Explain why the sum of the map distances for the two closest pairs of genes is not equal to the distance between the two genes that are farthest apart.

Answer: Since double crossovers cannot be detected in 2-point crosses, the distance between the widest spaced genes is actually greater than the calculated recombination rate obtained from the a-b test cross.

9.06 Consider a 3-point cross in which the fully heterozygous kk+ ll+ mm+ was crossed to a fully homozygous recessive kk ll mm. The table below shows the number of offspring having each of the eight possible genotypes. Construct a genetic map for these loci, and calculate the interference value.


9.07 Consider the following three Drosophila genes. Yellow versus wild-type body color is determined by the alleles y and y+, vermilion eyes versus wild-type eye color is determined by the alleles v and v+, and singed versus normal wild-type bristles by the alleles sn and sn+. All three of these genes are located on the X-chromosome (sex-linked), and in each case the wild-type allele is dominant over the mutant allele. A three-point cross was done as follows. Two flies with contrasting phenotypes were mated. Both parents were taken from purebreeding cultures, but the records of the parent flies have been lost, so you dont know what their exact phenotypes were. One of the heterozygous female offspring of this cross was then mated with a wild-type male. The male offspring of this cross appear in the following numbers:

(a) What were the phenotypes of the homozygous parents of the heterozygous female

Answer: vermillion x singed, yellow

(b) Construct a gene map for these three loci.


(c) Why were only the male offspring of this cross counted? (How would the results have been different in the females?)

Answer: Because a wild-type male is used as the tester parent, only the male offspring will tell us about crossing over. All the female offspring would be wild. Think about why this would be true.

9.08 A trihybrid Drosophila experiment is in progress. The three genes are: sepia eyes (se), black body (b), and curved wings (c). All three mutant traits are recessive to the wild type. The locus of the se gene is on chromosome III while the b and c genes are both on chromosome II, separated by 27 map units. A true-breeding female with curved wings has been mated with a true-breeding male with black body and sepia eyes. The progeny are all wild type. If a female offspring of this cross is test crossed with a black, sepia, curved male, what proportions of phenotypes would you expect? (Hint: Try the forked line method, but remember to consider that two of the genes are linked and will not sort independently)


9.09 If two loci are 10 map units apart, what proportion of the meiotic divisions will contain a single crossover in the region between these two loci, assuming that no multiple crossovers occur? Remember that crossing over occurs while each bivalent has 4 chromatids.

Answer: Approximately 2 x 10%, or 20% of the meiotic events would involve a single crossover within the bivalent. Each crossover produces 2 out of 4 chromatids that are recombinants.

9.10 Assume that a strain of Neurospora with allele D are crossed to one with allele d.

(a) If there were no crossing over between the centromere and this gene, what order of ascospores would you expect to see in the ascus? (Show both possible orders.)

Answer: DDDDdddd and ddddDDDD

(b) If there was one crossover, what orders would you expect? (Show all possible orders.)

Answer: DDddDDdd, DDddddDD, ddDDddDD, and ddDDDDdd

(c) Using chromosome diagrams, illustrate how meiosis, with or without crossing over, would account for the answers to parts a and b.


9.11 Two strains of Neurospora,a and a+ were crossed. This cross produced the following patterns of asci. On the basis of the these data, determine the distance in map units between the a/a+ locus and the centromere.

Answer: Fraction of 2nd div segregations = (19+22+18+25)/(200) = 0.42. Fraction of recombinant chromatids is one-half this, so map distance is (0.42)(1/2)(100) = 21 mu.

9.12 We have seen that tetrad analysis using Neurospora can be used to study which chromatids are involved in recombination. Assume that there are three mutant alleles e, f, and g that are linked. You mate a wild type strain with a strain containing all three mutant genes. The two homologous chromosomes in the zygote at the beginning of meiosis would look something like the diagram shown. Predict the order of spores (genotypes) in the ascus for each of the following recombination events.

(a) A double crossover with chromatids 2 and 3 between gene e and f, and with chromatids 2 and 3 between f and g.

(b) A double crossover with chromatids 2 and 3 between gene e and f, and with chromatids 1 and 4 between f and g.

(c) A double crossover: with chromatids 1 and 3 between gene e and f, and with chromatids 1 and 4 between f and g.


Delta State University > Biological Sciences >Dr. Tiftickjian >Genetics >Problem set 9 solutions