BS 50 Genetics and Genomics Week of Oct 10

Transcription

1 BS 50 Genetics and Genomics Week of Oct 10 Additional Practice Problems for Section 1. Two purebreeding strains of mice are crossed to produce a mouse heterozygous for dominant and recessive alleles of three linked genes (A vs. a, B vs. b, E vs. e). A series of F1 triply heterozygous mice are testcrossed and the resulting progeny were: A B E 8 A B e 342 A b E 100 A b e 62 a B E 49 a B e 94 a b E 317 a b e total progeny Justify your answer to each of the following questions: a. What are the genotypes of the purebred parents of an F 1 triple heterozygote? b. What is the map order of the three genes? (map distances are not required)

2 2. Genes A, B, and R are linked to one another. A genetic map of the locations of these genes is shown below: A R B cM cM--- cm = m.u. = 1% recombination a. You cross a pure-breeding A R strain with a pure-breeding a r strain, and then testcross the F 1 progeny. What percentage of the testcross progeny do you expect to be phenotypically A r? b. You cross a pure-breeding A R B strain with a pure-breeding a r b strain and then testcross the F 1 progeny. What percentage of the testcross progeny do you expect to be phenotypically A R B?

3 3. (18 pts) A strain of mice is homozygous for certain alleles of five genes: AA bb cc DD ee. A second strain is aa BB CC dd EE. Genes D and E are linked to each other, being 20 cm apart. All other genes assort independently. The two strains are crossed to generate F l individuals. The F l females are test crossed (that is, mated to homozygous recessive males). A) Considering only genes A, B, and C, what is the probability of obtaining an offspring with the same phenotype as the F l parents? B) Considering all the genes, what is the probability of obtaining an offspring with the same phenotype as the F l parents?

4 Problem 4 (18 pts) A true-breeding hairy-winged Drosophila female is mated with a truebreeding yellow-bodied, white-eyed male. The F 1 are all phenotypically wildtype (normalwinged, black-bodied, and red-eyed). The F 1 are then crossed among themselves and the F 2 that emerge are: Females: 757 wild-type 243 hairy-winged Males 390 wild-type 360 yellow-bodied and white-eyed 130 hairy-winged 110 hairy-winged, yellow-bodied and white-eyed 4 yellow-bodied 1 hairy-winged and yellow-bodied 3 white-eyed 2 hairy-winged and white-eyed (Descriptions list mutant phenotypes only...so if a fly is listed as just hairy-winged, it is wildtype for eye and body color, etc.) a) Does the inheritance of these traits follow Mendel s second law? Explain. b) What are the genotypes of the parents? Note linkage relationships if appropriate [Map distance(s) need not be calculated. However, define your symbols...capitalize dominant alleles and use lowercase for recessive alleles.] c) What are the genotype(s) of the F 1 females and males? [Be consistent with the nomenclature you defined in part b.]

5 Problem 5 (12 pts) A disease affecting sheep was discovered to be due to a new virus. The virus was isolated and grown on sheep cells in culture. Biochemically, it was determined the virus contained four components, which were called W, X, Y, and Z. To determine which of these components contained the genetic information of the virus, each component was radioactively labeled by growing the virus on radioactive sheep cells. Radioactively-labeled virus were then allowed to attach to nonradioactive sheep cells and after a brief period of time, the cells were centrifuged to remove and then separate the virus particles from the cells. After centrifugation, the supernatant contained the viral pieces and the pellet contained the infected cells. The following data were collected: % Radioactivity supernatant pelleted cells W X Y Z a) On the basis if these results, which of the viral components cannot be carrying genetic information? Why? Three of the components turned out to be nucleic acids, The base compositions were as follows: %A %G %C %T %U X Y Z b) What can be determined about the type of nucleic acid found in each of these three components? c) Due to experimental error, one of the nucleic acid components turned out to be sheep DNA. Which component must be the sheep DNA? Why?

6 6) The following experiment is a modification of the Meselson-Stahl experiment: generation 0: Bacteria are grown for many generations on 14 N ("light" nitrogen). generation 1: The cells are then shifted to 15 N ("heavy" nitrogen) for one generation. generation 2: Finally, the cells are then shifted back to 14 N for one generation. If these samples of DNA are centrifuged in a CsCl gradient, the DNA molecules will band according to their density. Thus, L/L, H/L, and H/H DNA will be distinguishable. What proportion of the DNA do you expect to be in each band after 0, l and 2 generations? L/L H/L H/H generation 0: generation 1: generation 2:

7 Problem 6. The following represent various stages of cell division in cells from the same individual organism. A) B) C) D) E) What stage of cell division is represented by each drawing? Choose from the following list: Mitotic Metaphase Mitotic Anaphase Mitotic Telophase Meiotic Metaphase I Meiotic Anaphase I Meiotic Telophase I Meiotic Metaphase II Meiotic Anaphase II Meiotic Telophase II a. b. c. d. e. f. What is n for this organism?