**Study on Radiation Effects on Mutation Rates in Drosophila Sperm**Mary Alexander conducted a study to examine the effects of radiation on mutation rates in the sperm of Drosophila. She exposed Drosophila larvae to either 3000 roentgens (r) or 3975 roentgens (r) of radiation. The adult males that developed from these irradiated larvae were mated with normal, un-irradiated females, and the number of resulting mutants was counted. (ML Alexander, 1954, Genetics 39:409-428)The observations from her study were as follows:| Group | Number of Offspring | Number of Mutants ||-------------------------|---------------------|-------------------|| Control | 45,504 | 2 || Irradiated (3000r) | 49,512 | 71 || Irradiated (3975r) | 51,159 | 70 |**Questions:**A. Calculate the mutation rate for all three groups.B. Based on this experiment, does irradiation cause mutations?C. Does the amount of radiation matter?D. At the mutation rate for 3000 roentgens, how many mutants would you expect if you had 1,000,000 offspring?

Given Radiation amount = 3000r and 3975r Number of offsprings Control = 45,504, mutated = 2…

11. Mary Alexander studied the effects of radiation on mutation rates in sperm of Drosophila. She irradiated Drosophila Larva with either 3000r (roentgen) or 3975r, collected adult males that developed from the irradiated larva, mated them with normal un-irradiated females and counted the number of mutants. (ML Alexander, 1954, Genetics 39:409-428) Her observations were: Number of offsprings 45,504 49,512 51,159 Number of mutants Group Control Irradiated (3000r) Irradiated (3975r) 71 70 A. Calculate the mutation rate for all 3 groups. B. On the basis of this experiment, does irradiation cause mutations? C. Does the amount of radiation matter? D. At the rate of mutation for 3000r, how many mutants would you get if you had 1,000,000 offsprings?

11. Mary Alexander studied the effects of radiation on mutation rates in sperm of Drosophila. She irradiated Drosophila Larva with either 3000r (roentgen) or 3975r, collected adult males that developed from the irradiated larva, mated them with normal un-irradiated females and counted the number of mutants. (ML Alexander, 1954, Genetics 39:409-428) Her observations were: Number of offsprings 45,504 49,512 51,159 Number of mutants Group Control Irradiated (3000r) Irradiated (3975r) 71 70 A. Calculate the mutation rate for all 3 groups. B. On the basis of this experiment, does irradiation cause mutations? C. Does the amount of radiation matter? D. At the rate of mutation for 3000r, how many mutants would you get if you had 1,000,000 offsprings?

Human Anatomy & Physiology (11th Edition)

11th Edition

ISBN:9780134580999

Author:Elaine N. Marieb, Katja N. Hoehn

Publisher:Elaine N. Marieb, Katja N. Hoehn

Chapter1: The Human Body: An Orientation

Section: Chapter Questions

Problem 1RQ: The correct sequence of levels forming the structural hierarchy is A. (a) organ, organ system,...

See similar textbooks

Similar questions

6). To investigate the genetic control of locomotive behavior in the diploid nematode, Caenorhabditis elegans, eight "wiggly" mutants were isolated, which wiggle in an uncoordinated way rather than smoothly gliding. These eight mutants were crossed together in all possible pairwise combinations, yielding the results summarized in the table above (+ means that all the progeny were wild type, and m means that all the progeny showed the wiggly mutant phenotype). Organize the eight mutations into con lementation groups, each consisting of a group of mutations affecting the same gene. 1 2 3 4 5 6 7 8 1 E N+E 2 3+1 m + m 4 5 + m m E+ + + EE 6 ++E E+ +EE m m + + + + + + + + + m 7 8 ++4 ∞+E +E m m ++ E
71. Myotonic Dystrophy, Fragile X, and Huntington disease are examples of human genetic diseases that exhibit earlier ages of onset and/or an increased severity in afflicted members of each successive generation in families in which the mutant alleles are segregating. These are both examples of genetic diseases which exhibit: a. partial penetrance. b. incomplete dominance. c. genetic imprinting. d. genetic anticipation. \
7. The following diagram shows 2 pairs of autosomal chromosomes in a karyotype of a biological woman. Based on the karyotype, a genetic counsellor determined that there is evidence that a translocation occurred between two non-homologous chromosomes. The other 21 pairs of chromosomes are NOT shown in this diagram. Assume no chromosomal aberrations occurred in the other chromosomes. As a student-researcher, the genetic counsellor asks you if you can determine whether the translocation occurred in a somatic cell, germ-line cell, or whether this can even be determined from the figure? Provide an answer and the reasoning for your answer. D
..
3
4). p53 (sometimes called TP53 for “tumor protein 53") is a human tumor suppressor gene that is mutated in the majority of human cancers (many tumor types). a. For each of the mutations described below (i-iv): is this a mutation you would expect to find when sequencing p53 alleles from tumor cells? Why or why not? i. A missense mutation encoding a hyperactive form of the protein. ii. A deletion of the gene. iii. An insertion in the promoter that increases transcription 10-fold. iv. A nonsense mutation. b. When sequencing the p53 gene in tumor cells, would you expect to find only mutant version(s) of the gene or a mix of mutant and wild type versions? c. For any of the mutations you said you would expect to find in tumor cells, would you expect tumor cells to be homozygous (same mutation on both chromosomes)? Why or why not? d. Individuals with Li-Fraumeni syndrome have a very high risk of tumors originating in various tissues due to inheritance of a loss-of-function mutant allele of…
39 The structures labelled A and B in the image below represent: СТАААТСGGT. Allele for red flowers B Enzyme A Allele for white flowers One allele sufficient to ATAAATCGGT produce half the pigment Select one alternative: O the locus for flower colour; a red flower the locus for flower colour; a white flower O non-homologous chromosomes; a red flower O homologous chromosomes; a pink flower
2) In humans, one X chromosome in females is inactivated. Why does a female with one X chromosome show Turner syndrome? What is the difference between two X chromosomes with one X inactivated and one X chromosome in a female?
Before Muller's discovery that radiation induces mutation, scientists had to work on spontaneous mutants that were found solely by phenotype differences in natural populations. Which of the features of Drosophila made it a fortuitous choice for Morgan and his colleagues? a. large number of visible phenotypes b. especially high rate of mutation c. both sexual and asexual reproduction d. having a long life cycle e. well-known biochemical pathways
Dr. P was using Drosophila to follow the inheritance patterns of two traits: body shape and wing size. Dr.P began her experiment by crossing wild type, homozygous dominant, males with double mutant, homozygous recessive, females. She then performed a testcross by mating Fi generation males with double mutant, homozygous recessive, females and observed the phenotypes of the F2 generation. Shorthand of the different alleles is annotated as follows: Wild type round body shape allele = b* Wild type large wing size =w* Mutant rectangle body shape allele = b Mutant small wing size = w A) Complete the table by indicating the genotypes for each generation of male and female flies. Male Genotype Female Genotype P generation F, generation testcross When Dr.P characterized and counted the F2 phenotypes, she determined the following: Round-large Rectangle-small 332 flies Round-small Rectangle-large 357 flies Number of flies 344 flies 360 flies B) Using the data above, would you be surprised (yes…