* Your assessment is very important for improving the workof artificial intelligence, which forms the content of this project
Download Answers to Quiz 3:
Vectors in gene therapy wikipedia , lookup
Genomic imprinting wikipedia , lookup
Neuronal ceroid lipofuscinosis wikipedia , lookup
Nutriepigenomics wikipedia , lookup
Gene therapy of the human retina wikipedia , lookup
Gene nomenclature wikipedia , lookup
Koinophilia wikipedia , lookup
Gene expression profiling wikipedia , lookup
Frameshift mutation wikipedia , lookup
Gene desert wikipedia , lookup
Epigenetics of human development wikipedia , lookup
Therapeutic gene modulation wikipedia , lookup
Segmental Duplication on the Human Y Chromosome wikipedia , lookup
Oncogenomics wikipedia , lookup
Polycomb Group Proteins and Cancer wikipedia , lookup
Genome evolution wikipedia , lookup
Saethre–Chotzen syndrome wikipedia , lookup
Skewed X-inactivation wikipedia , lookup
Designer baby wikipedia , lookup
Y chromosome wikipedia , lookup
Genome (book) wikipedia , lookup
Gene expression programming wikipedia , lookup
Artificial gene synthesis wikipedia , lookup
Point mutation wikipedia , lookup
Neocentromere wikipedia , lookup
Site-specific recombinase technology wikipedia , lookup
Answers to Quiz 3: 1. Interval 31B5- 31C1 is the only region common to all deletions, defining the region where gene x is found: 2. For these questions, the recessive mutations uncovered by the same deletion must be located near each other. The logic is to start with the smallest mutations, then use others to build the least ambiguous map: (a,b) (a,b,d) abd or dab cabd or dabc (a,b,c) cabdf (b,d,f) (d,e,f) cabdfe or efdbac Ans: (a) 3. Each individual chromosome will represent a separate linkage group, and since the species are unrelated, the number should equal the haploid gene complement of both genomes (22 + 9 = 31 linkage groups). Ans: (b). 4. The two species have four chromosomes that can pair- these are the four bivalents, but two sets of homologs come together to form a tetravalent. This is most probably due to a reciprocal translocation in one species, where the linkage groups found on two different chromosomes have exchanged genes. Ans: (a) 5. This type of cancer is (usually) due to the inappropriate expression of the c-myc gene, placed under the control of a transcriptional regulator (IgH enhancer) expressed in lymphocytes, in the immunoglobulin gene cluster. See pp. 492-494 in text. Ans: all of the above (d). Fetus Mrs. Mr. Simpson Simpson 6-8. Fetus Mr. Mrs. Simpson Simpson Fetus Mrs. Mr. Simpson Simpson The problem is with Mr. Simpson, who is heterozygous for a pericentric inversion. A crossover within the inversion loop formed between the two chromosome six homologs in meiosis one will generate a chromosome with duplications and deficiencies. 6. The chromosome was derived from the father, due to a crossover between homologs within the inversion loop. Ans: (a) 7. The chromosome in the child contains duplications and deficiencies. Ans: (d) 8. False. It is possible that a gamete could be generated that did not contain a single crossover within the inversion loop. 9. Oncogenesis can be caused by mutation. All these forms of mutation can be linked to specific types of cancers…see text. Ans: (e). 10. This experiment describes the process of bacterial conjugation. This mode of bacterial gene transfer requires cell-cell contact from a donor (F+) to a recipient (F-) cell. Ans: (a).