Chapter 5 Chromosome Mapping in Eukaryotes1) The term normally applied when two genes fail to assort independently is ________.A) discontinuous inheritanceB) Mendelian inheritanceC) linkageD) tetrad analysisE) dominance and/or recessivenessAnswer: CSection: 5.22) Assume that a cross is made between AaBb and aabb plants and that all the offspring are either AaBb or aabb. These results are consistent with ________.A) complete linkageB) alternation of generationsC) codominanceD) incomplete dominanceE) hemizygosityAnswer: ASection: 5.23) Assume that a cross is made between AaBb and aabb plants and that the offspring fall into approximately equal numbers of the following groups: AaBb, Aabb, aaBb, aabb. These results are consistent with ________.A) independent assortmentB) alternation of generationsC) complete linkageD) incomplete dominanceE) hemizygosityAnswer: ASection: 5.24) Assume that a cross is made between AaBb and aabb plants and that the offspring occur in the following numbers: 106 AaBb, 48 Aabb, 52 aaBb, 94 aabb. These results are consistent with ________.A) sex-linked inheritance with 30% crossing overB) linkage with 50% crossing overC) linkage with approximately 33 map units between the two gene lociD) independent assortmentE) 100% recombinationAnswer: CSection: 5.21Copyright © 2012 Pearson Education, Inc.5) Assume that regarding a particular gene, one scored 30 second-division ascospore arrangements and 70 first-division arrangements in Neurospora. What would be the map distancebetween the gene and the centromere?A) 15B) 30C) 60D) 70E) Insufficient information is provided to answer this question.Answer: ASection: 5.106) The phenomenon in which one crossover increases the likelihood of crossovers in nearby regions is called ________.A) chiasmaB) negative interferenceC) reciprocal genetic exchangeD) positive interferenceE) mitotic recombinationAnswer: DSection: 5.47) Which of the following methods is involved in determining the linkage group and genetic mapin humans?A) syntenic testing and lod score determinationB) twin spots and tetrad analysisC) tetrad analysis and bromodeoxyuridineD) zygotene and pachytene DNA synthesisE) chiasmatype and classical analysesAnswer: ASection: 5.68) The genes for mahogany eyes and ebony body are approximately 25 map units apart on chromosome 3 in Drosophila. Assume that a mahogany-eyed female was mated to an ebony-bodied male and that the resulting F1 phenotypically wild-type females were mated to mahogany, ebony males. Of 1000 offspring, what would be the expected phenotypes, and in whatnumbers would they be expected?Answer: mahogany = 375; ebony = 375; wild type = 125; mahogany-ebony = 125Section: 5.29) Assume that there are 12 map units between two loci in the mouse and that you are able to microscopically observe meiotic chromosomes in this organism. If you examined 200 primary oocytes, in how many would you expect to see a chiasma between the two loci mentioned above?Answer: 48Section: 5.22Copyright © 2012 Pearson Education, Inc.10) Draw a diagram of the chromosomal events that will ultimately result in the segregation of alleles (A and a) during meiosis II rather than meiosis I.Answer: Section: 5.211) Given that loci A and B in Drosophila are sex-linked and 20 map units apart, what phenotypic frequencies would you expect in male and female offspring resulting from the following crosses? (Assume A and B are dominant to a and b, respectively.)(a) AaBb (cis) female × ab/Y male(b) AaBb (trans) female × ab/Y male(c) aabb female × AB/Y maleAnswer: (a) AB = 40; ab = 40; Ab = 10; aB = 10 (sexes have the same phenotypes)(b) Ab = 40; aB = 40; AB = 10; ab = 10 (sexes have the same phenotypes)(c) all males = ab; all females = ABSection: 5.212) Phenotypically wild F1 female Drosophila, whose mothers had light eyes (lt) and fathers hadstraw (stw) bristles, produced the following offspring when crossed with homozygous light-strawmales:Phenotype Numberlight-straw 22wild 18light 990straw 970Total 2000Compute the map distance between the light and straw loci.Answer: 2 map unitsSection: 5.53Copyright © 2012 Pearson Education, Inc.13) Assume that the genes for tan body and bare wings are 15 map units apart on chromosome #2in Drosophila. Assume also that a tan-bodied, bare-winged female was mated to a wild-type male and that the resulting F1 phenotypically wild-type females were mated to tan-bodied, bare-winged males. Of 1000 offspring, what would be the expected phenotypes, and in what numbers would they be expected?Answer: wild type = 425; tan-bare = 425; tan = 75; bare = 75Section: 5.514) Assume that investigators crossed a strain of flies carrying the dominant eye mutation Lobe on the second chromosome with a strain homozygous for the second chromosome recessive mutations smooth abdomen and straw body. The F1 Lobe females were then backcrossed with homozygous smooth abdomen, straw body males, and the following phenotypes were observed:smooth abdomen, straw body 820Lobe 780smooth abdomen, Lobe 42straw body 58smooth abdomen 148Lobe, straw body 152(a) Give the gene order and map units between these three loci.(b) What is the coefficient of coincidence?Answer: (a) Lobe is in the middle.smooth abdomen——5——Lobe———————15————————straw body(b) zeroSection: 5.515) In Drosophila, assume that the gene for scute bristles (s) is located at map position 0.0 and that the gene for ruby eyes (r) is at position 15.0. Both genes are located on the X chromosome and are recessive to their wild-type alleles. A cross is made between scute-bristled females and ruby-eyed males. Phenotypically wild F1 females were then mated to homozygous double mutant males, and 1000 offspring were produced. Give the phenotypes and frequencies expected.Answer: scute = 425; ruby = 425; wild type = 75; scute-ruby = 75Section: 5.516) Assume that a cross is made between AaBb and aabb plants and that the offspring occur in the following numbers: 106 AaBb, 48 Aabb, 52 aaBb, 94 aabb. These results are consistent with which arrangement of genes?Answer: In the AaBb parent, the dominant alleles are on one homolog, and the recessive alleles are on the other.Section: 5.24Copyright © 2012 Pearson Education, Inc.17) In the fruit fly, Drosophila melanogaster, a spineless (no wing bristles) female fly is mated toa male that is claret (dark eyes) and hairless (no thoracic bristles). Phenotypically wild-type F1 female progeny