[Note: "o" = centromere]
Genetics 371B Practice problems--Autumn 2000 week 6
| 1. | For each of the following inversion heterozygotes, what will be the products of meiosis, assuming a single crossover between the D and E loci? Which of the meiotic products will be viable? (A diploid is represented in each case, with the two homologs of the chromosome in question.) | |
| (a) |
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| (b) |
[Note: "o" = centromere] |
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| 2. | Explain the occasional occurrence of XX males with symptoms of Klinefelter syndrome. | |
| 3. | Which of the following women would you consider to be more at risk of having a Down syndrome baby, and why? | |
| (a) | A 22-year old woman or a 38-year old woman, neither of whose relatives has had Down syndrome babies | |
| (b) | A 38-year old woman with no family history of Down syndrome babies, or a 22-year old woman, two of whose female relatives have had Down syndrome babies | |
| 4. | Give complete genetic explanations for each of these situations: | |
| (a) | Phenotypically normal parents who have red-green colorblind XO daughters, and phenotypically normal parents who have red-green colorblind XXY sons. [Remember that this form of colorblindness is X-linked.] | |
| (b) | A pair of otherwise identical twins, one of whom is normal and the other has Down syndrome. | |
| 5. | What are the types of sperm that could (in principle) be formed by an XYY male? | |
| 6. | In pea, the genes for plant height (T = tall, t = short, recessive) and plant color (D = dark, d = light, recessive) are known to be on separate chromosomes. However, a plant breeder notices that when she crosses a particular TtDd plant with a recessive ttdd plant, the progeny consisted solely of TD and td phenotype plants in equal proportions; furthermore, there were only about half as many seeds as usual. | |
| (a) | What was unusual about this result? | |
| (b) | Suggest an explanation for the results. | |
| 7. | A partial map of some genes on one arm of the Drosophila X chromosome is shown:
A new graduate student treats normal males with X-rays and mates them to fully homozygous recessive (abdefg/abdefg) females. From the progeny, he picks out the females and mates them to fully recessive (abdefg/Y) males. The resulting F2 progeny phenotypes were:
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| (a) | What is unusual about these results? | |
| (b) | The student's advisor takes one look at the results and concludes that they can be explained by an inversion. How could she rule out deletions or translocations? | |
| (c) | Assuming that the advisor is correct, which segment of the chromosome do you think is inverted? Explain. | |
| (d) | Suggest a molecular test of your hypothesis, stating your predicted result. Assume that you can make a radioactive probe for any desired gene on the chromosome, and that you have a complete restriction map of the normal chromosome. | |
| (e) | Explain the two rare progeny classes given the hypothesis of a chromosomal inversion. | |
| 8. | For this one, again, the answer is being provided; you should provide the explanation of how to arrive at the answer.
A tall, tetraploid pea plant (genotype TTtt) is crossed to a short, tetraploid plant (tttt). Assuming that tall (T) is dominant, and that it only takes one copy of the dominant allele to give a dominant phenotype, what ratio of progeny phenotypes do you expect for this cross? Also assume that there is no aneuploidy in the progeny. (This one may take some thought--sketching out the various outcomes of meiosis may help.) The answer: Tall and short in 5:1 ratio |
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| 9. | In fruit fly, black body (b) and reduced bristle (rd) are recessive alleles of linked autosomal genes. In a strain that was b +/+ rd, it was noted that flies occasionally showed small patches of the recessive phenotype--mostly lone patches of black body and twin patches-- black body patches next to reduced bristle patches. The lone black patches and the twin patches occurred in ratio of 5:6. | |
| (a) | Explain how the patches arose, with diagrams of the chromosomes and chromatids as necessary (including the centromere). | |
| (b) | Draw a map of the chromosome showing relative distances between the centromere and each of the two genes. | |
| (c) | Closer examination of the flies revealed rare, lone patches of reduced bristle. Suggest two distinct mechanisms not involving point mutation by which these lone patches may have arisen. | |
| 10. | Another question where the answer is being provided; you provide the explanation.
Some molds (such as Aspergillus) grow as expanding disks, new growth being added at the outer edge of the colony. A consequence of this pattern of growth is that mitotic recombination, instead of giving patches as in Drosophila, causes sectoring as shown below. |
 Genes for mold color (y+ = normal, y = yellow, recessive), colony morphology (r+ = normal, r = rough edges, recessive), growth density (g+ = normal, g = sparse, recesssive), and hyphal texture (m+ = normal, m = mottled, recessive) are known to be linked in one such mold. A haploid strain that shows all four normal phenotypes is mated to a haploid strain showing all four recessive phenotypes. When the resulting diploids were grown, most colonies showed the dominant phenotypes, but some sectors of recessive phenotype were seen:
What is the arrangement of these four genes with respect to the centromere? Include the relative distances. Answer:
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| 11. |
Challenge question:
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| (a) | Assuming that all sectors resulted from mitotic recombination, how would you explain the difference in sector size? In particular, how would you explain the half-sectored (i.e., half red, half white) colonies? | |
| (b) | Counting the number of sectors gives you an idea of how many times mitotic recombination must have occurred in the population you are sampling--but it doesn't tell you the actual frequency of mitotic recombination in some chromosome interval (i.e., what percent of divisions result in recombination in that interval). How might you use the half-sectored colonies to deduce the frequency of mitotic recombination in the Centromere-Ade interval? | |
| 12. | This one was an exam question (in slightly modified form) in 1998.
A brain disorder in humans comes in two forms--reduced production of a peptide neurotransmitter (the more common, recessive form), or elevated levels of the same neurotransmitter (a rare dominant form). It is known that the only difference between patients with the reduced expression form and unaffected individuals is that in the reduced-expression patients, there is a point mutation in a gene needed for neurotransmitter synthesis. Using pedigrees of families showing the more common defect and two polymorphic loci, you construct a LOD score profile to map the gene, while your colleague uses pedigrees of the rare form to construct a separate and independent LOD score profile:
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| (a) | Draw a map of the region for unaffected individuals, showing the two polymorphic loci and the gene along with the predicted map distances. | |
| (b) | What kind of chromosomal aberration might you find in patients showing the elevated expression phenotype? (Your answer should explain the skewed LOD scores seen in these pedigrees.) | |
| (c) | Briefly suggest a molecular explanation for the overexpression phenotype based on your answer in (b) | |
Also do:
1999 Exam 2, Question 3
A certain recessive mutation (ade-) in yeast causes colonies to be red in color, while Ade+ colonies are white. Single cells of a heterozygous Ade+/ade- strain were plated on growth medium and allowed to grow into colonies. Most of the colonies were white, but some showed red sectors. The red-sectored colonies showed differences in sector size: some red sectors were small, while others were larger, and in a few instances, the red "sector" was exactly half the colony. 