DOC PREVIEW
MIT 7 03 - Genetics

This preview shows page 1-2-3 out of 9 pages.

Save
View full document
View full document
Premium Document
Do you want full access? Go Premium and unlock all 9 pages.
Access to all documents
Download any document
Ad free experience
View full document
Premium Document
Do you want full access? Go Premium and unlock all 9 pages.
Access to all documents
Download any document
Ad free experience
View full document
Premium Document
Do you want full access? Go Premium and unlock all 9 pages.
Access to all documents
Download any document
Ad free experience
Premium Document
Do you want full access? Go Premium and unlock all 9 pages.
Access to all documents
Download any document
Ad free experience

Unformatted text preview:

Name: ____________________________________ 7.03 - Genetics - Fall 2004 Massachusetts Institute of Technology Professor Chris Kaiser Professor Gerry Fink Professor Leona Samson 1Name: ____________________________________ 1. You are studying three autosomal recessive mutations in the fruit fly Drosophila melanogaster. Flies that are homozygous for the hb– mutation are “humpbacked” (wild-type flies are straight-backed). Flies that are homozygous for the bl– mutation are “blistery-winged” (wild-type flies are smooth-winged). Flies that are homozygous for the st– mutation are “stubby-legged” (wild-type flies are long-legged). You mate flies from two true-breeding strains, and the resulting F1 flies are all are straight-backed, smooth-winged, and long-legged. F1 females are then mated to males that are humpbacked, blistery-winged, and stubby-legged. In the F2 generation, among 1000 progeny resulting from this cross, you observe the following phenotypes: Phenotype Number humpbacked, blistery-winged, and stubby-legged (26 flies) humpbacked, blistery-winged, and long-legged (455 flies) humpbacked, smooth-winged, and long-legged (24 flies) straight-backed, blistery-winged, and stubby-legged (27 flies) straight-backed, blistery-winged, and long-legged (4 flies) straight-backed, smooth-winged, and stubby-legged (442 flies) straight-backed, smooth-winged, and long-legged (22 flies) (a, 6 pts.) The male flies that were bred to the F1 generation in order to produce the F2 generation were humpbacked, blistery-winged, and stubby-legged. On each of their chromosomes, they have the alleles hb– bl– st–. Using this notation, state the genotype of each of the two true-breeding parental strains (i.e. the two strains in the P generation). Genotype of one parental strain: Genotype of the other parental strain: 2Name: ____________________________________ (b, 6 pts.) How many flies are found in the class that is the reciprocal class of the humpbacked, blistery-winged, and stubby-legged flies? (c, 7 pts.) What is the genetic distance between the hb and bl loci? (Label your answer with the proper units.) (d, 7 pts.) What is the genetic distance between the bl and st loci? (Label your answer with the proper units.) (e, 6 pts.) Draw a genetic map showing the correct order of the hb, bl, and st loci. 3Name: ____________________________________ 2. The following mouse pedigree shows the segregation of two different mutant traits. The mutant trait indicated by the dots is dominant, whereas the mutant trait indicated by the stripes is recessive. Assume 100% penetrance and no new mutations. (Squares = males, circles = females.) = expressing dominant mutant trait, caused by the “A*” allele Mouse #1Mouse#2? Mouse#4Mouse #3 = expressing recessive mutant trait, caused by the “b” allele = expressing both mutant traits (a, 6 pts.) Assuming that both mutant traits are due to linked autosomal genes that are 6 cM apart, fill in the following chart using the allele notation indicated by the key above. Blocks in the chart that cannot be filled in conclusively should be indicated as “inconclusive.” NOTE: One line of the chart is already filled in correctly for you. Number of “A*” alleles Number of “a” alleles Number of “B” alleles Number of “b” alleles Mouse #1 Mouse #2 Mouse #3 0 2 0 2 Mouse #4 4Name: ____________________________________ (b, 6 pts.) Assuming that both mutant traits are due to linked autosomal genes that are 6 cM apart, fill in the boxes with the alleles possessed by mouse #4 on each of the two homologs of this autosome that are depicted in the diagram below. Homolog inherited from mouse #1 “A*” or “a” locus “B” or “b” locus Homolog inherited from mouse #2 (c, 7 pts.) Assuming that both mutant traits are due to linked autosomal genes that are 6 cM apart, what is the probability that the mouse indicated by a question mark will show both mutant traits (the trait encoded by “A*” and the trait encoded by “b”)? 5Name: ____________________________________ (d, 8 pts.) Assuming that the recessive mutant trait is caused by a gene on an autosome and the dominant mutant trait is caused by a gene on the X chromosome, fill in the following chart using the allele notation indicated by the key above. Blocks in the chart that cannot be filled in conclusively should be indicated as “inconclusive.” Number of “A*” alleles Number of “a” alleles Number of “B” alleles Number of “b” alleles X-linked autosomal Mouse #1 Mouse #2 Mouse #3 Mouse #4 (e, 6 pts.) Assuming that the recessive mutant trait is caused by a gene on an autosome and the dominant mutant trait is caused by a gene on the X chromosome, what is the probability that the mouse indicated by a question mark will show only the recessive mutant trait assuming that the mouse is born female? 6Name: ____________________________________ 3. You are working with a mutant strain of yeast that is dark tan (wild-type yeast are white). The “dark tan” phenotype of the haploid cells you are working with is caused by two different mutations in the same strain. The two mutations are designated drk1– and drk2–. (a, 6 pts.) Mating of the drk1– drk2– double mutant to wild-type yeast produces diploids that are white. Sporulation of these diploids yields 50 tetrads. 4 of these tetrads (called “Type One”) contain four light tan spores. 37 of these tetrads (called “Type Two”) contain two dark tan spores and two white spores. 9 of these tetrads (called “Type Three”) contain one dark tan spore, two light tan spores, and one white spore. Categorize each of the tetrad types as parental ditype (PD), tetratype (TT), or nonparental ditype (NPD). (b, 6 pts.) Are the drk1– and drk2– mutations linked? If so, give the distance between them. (Label your answer with the proper units.) (c, 6 pts.) In yeast, 1 cM of genetic distance corresponds to 3,500 base pairs of physical distance. An average yeast gene is about 1,400 base pairs long, and the longest yeast gene is 14,700 base pairs. Keeping this information in mind, you select a “Type Three” tetrad from part (a) and mate the two light tan spores from that tetrad to each other. Can you deduce


View Full Document

MIT 7 03 - Genetics

Documents in this Course
Exams

Exams

22 pages

Exams

Exams

64 pages

Exam 1

Exam 1

66 pages

Exam I

Exam I

93 pages

Exam Two

Exam Two

12 pages

Exams

Exams

27 pages

Exam 1

Exam 1

41 pages

Load more
Download Genetics
Our administrator received your request to download this document. We will send you the file to your email shortly.
Loading Unlocking...
Login

Join to view Genetics and access 3M+ class-specific study document.

or
We will never post anything without your permission.
Don't have an account?
Sign Up

Join to view Genetics 2 2 and access 3M+ class-specific study document.

or

By creating an account you agree to our Privacy Policy and Terms Of Use

Already a member?