To determine: The percentage of warfarin-resistant rats in town Olf.
Introduction: Rodenticides are the pesticides that have a capability to kill the whole rodent species. Rodents have their own importance in nature but sometimes require some control over their population, as they could transmit disease, destroy crops, and may cause ecological damage. Warfarin is a common anticoagulant that could be used as a rodenticide to kill mice and rats.
After the discovery, warfarin has been widely used and later it resulted in the development of species of warfarin-resistant mice and rats. Another recently developed bromadiolone called as the super-warfarin (second-generation warfarin) has a greater ability to accumulate in the liver and kill the organism.
Want to see the full answer?
Check out a sample textbook solutionChapter 17 Solutions
Biology: The Unity and Diversity of Life (MindTap Course List)
- Half of the worlds population eats rice at least twice a day. Much of this rice is grown in flooded conditions, and different strains of rice are tolerant (survive) or intolerant (die) under these conditions. Rice breeders used genetic crosses to test whether tolerance to flooding is a dominant trait. Researchers used three true-breeding flood-tolerant strains, FR143, BKNFR, and Kurk, and two true-breeding flood- intolerant strains, IR42 and NB, in the crosses. Results were obtained from three sets of crosses and are reported in the Table below: Results of cross of F1 to tolerant parent: F1 plants were crossed with the tolerant parent of the cross. Number of Plants Progeny Analyzed from Intolerant Tolerant Cross Alive Dead Total 1. F2 results of cross: IR42 FR13A 187 77 264 IR42 BKNFR 192 73 265 NB Kurk 142 52 195 2. Results of cross of F1 to intolerant parent: (F1 of IR42 FR13A) IR42 14 17 31 (F1 of IR42 BKNFR) IR42 15 10 25 (F1 of NB Kurk) NB 21 35 56 3. Results of cross of F1 to tolerant parent: (F1 of IR42 FR13A) FR13A 31 0 31 (F1 of IR42 BKNFR) BKNFR 28 0 28 (F1 of NB Kurk) Kurk 40 0 40 Do the data support the hypothesis that the tolerance trait is dominant? Justify your conclusion by explaining the results from each of the three sets of crosses in terms of genotypes and phenotypic ratios. Source: T. Setter et al. 1997. Physiology and genetics of submergence tolerance in rice. Annals of Botany 79:6777.arrow_forwardUsing the HardyWeinberg Law in Human Genetics Suppose you are monitoring the allelic and genotypic frequencies of the MN blood group locus (see Question 2 for a description of the MN blood group) in a small human population. You find that for 1-year-old children, the genotypic frequencies are MM = 0.25, MN = 0.5, and NN = 0.25, whereas the genotypic frequencies for adults are MM = 0.3, MN = 0.4, and NN = 0.3. a. Compute the M and N allele frequencies for 1-year-olds and adults. b. Are the allele frequencies in equilibrium in this population? c. Are the genotypic frequencies in equilibrium?arrow_forwardAn island of the Galápagos archipelago is home to a medium ground finch that subsists mainly by eating seeds. A severe drought struck the island. During the drought, plants produced fewer seeds, and the finches soon depleted the stock of small and soft seeds, leaving only large and hard seeds that were difficult to process. In this environment, finches with deeper beaks were more likely to survive and pass their advantageous traits to their offspring by means of the principle called the 'inheritance of acquired characteristics'. Question 11 options: A) True B) Falsearrow_forward
- In a large herd of 5468 sheep, 76 animals have yellow fat, and the rest of the members of the herd have white fat. Yellow fat is inherited as a recessive trait. This herd is assumed to be in Hardy-Weinberg equilibrium. A. What are the frequencies of the white and yellow fat alleles in this population? B. Approximately how many sheep with white fat are heterozygous carriers of the yellow allele?arrow_forwardYou are studying a population of penguins in Antarctica. Your DNA analysis of this population reveals that for the feather color pattern gene, 35 individuals are homozygous dominant, 35 individuals are heterozygous, and 30 individuals are homozygous recessive. After observing this population for several years, you repeat your DNA study and find that the current generation of penguins has 15 individuals that are homozygous dominant, 10 individuals that are heterozygous, and 75 individuals that are homozygous recessive. Which of the following hypotheses for this data would be supported by this data based on your understanding of Hardy-Weinberg equilibrium? This population of penguins is maintaining Hardy-Weinberg equilibrium due to its large population size. The penguins are randomly choosing mates, which has led to Hardy-Weinberg equilibrium. The penguins are very isolated, which is preventing gene flow from affecting Hardy-Weinberg equilibrium. The recessive phenotype in…arrow_forwardThe gamma globulin of human blood serum exists in two forms, Gm(a+) and Gm(a-), specified respectively by an autosomal dominant gene Gm(a+) and its recessive allele Gm(a-). Broman et al. (1963) recorded the tabulated phenotypic frequencies in three Swedish populations. Assuming the populations were at Hardy-Weinberg equilibrium, calculate the frequency of heterozygotes in each population.arrow_forward
- In 1956, Peter Buri, who at the time was at the State University of Iowa, published a study he conducted using Drosophila melanogaster. The title of his paper is called Gene frequency in small populations of mutant Drosophila. In this study, he tracked allele frequency changes of two alleles, bw and bw75. Which of the following statements about this study is correct? In this study, he demonstrated that it is impossible to tell whether the bw or bw75 allele will be extinct in any given population. This study is an example of evolution by natural selection. Both of the abovearrow_forwardAn anthropologist discovered that Papa New Guineans tribes that had given up the customs of cannibalism also had an avid aversion to grapefruit. The ability to taste the horrid bitterness of PTC present in grapefruit respectively human brain tissue is due to a dominant allele. A small tribe of 637 Papa New Guinea hens was tested by population geneticists the following results were displayed. Tasters Non-tasters Total Males 171 86 257 Females 337 43 380 Total 508 129 637 Another tribe, still being cannibals, trespassed the territory of the other aforementioned tribe. They captured killed and feasted on the brains of 50 PTC tasting males respectively 19 females as to access their abilities seen as a gift from gods. Calculate the increase/decrease in percentage units or percent of respective allele frequency displayed in the remaining tribe?arrow_forwardTay-Sachs disease is a recessive genetic disease. Individuals with this disease rarely survive past the age of four. In the general population, approximately 1 person in 300 carries the allele for this disease. However, in some populations, including the Irish Americans, the Ashkenazi Jews and the Cajuns from Louisiana, the proportion of Tay-Sachs carriers is much higher (1 in 27 to 1 in 50) than in other populations. Which evolutionary scenario can be predicted to produce a high frequency of Tay-Sachs disease in these populations? Select one: a. All three populations descend from a small number of settlers b. The Tay-Sachs allele is advantageous at the heterozygous state c. These populations experienced disruptive selection d. These populations experienced stabilizing selection e. These populations have higher than average mutation ratesarrow_forward
- This type of experiment, where members of an outbred populations are introduced to an inbred population as mating partners, is sometimes called a “genetic rescue”. How might measuring the average heterozygosity at a number of loci in the Swedish snake population allow you to test whether the introduced snakes had actually bred successfully with the inbred Swedish snakes?arrow_forwardThe ability to taste the compound PTC is controlled by a dominant allele T, while individuals homozygous for the recessive allele (t) cannot taste PTC. In a population consisting of 500 individuals, 347 are tasters and 153 are non-PTC tasters. Calculate the frequency of the T and t alleles in this population, and frequency of the genotypes. (Please train yourself to use the Hardy-Weinberg equation.) To present your answers, follow the format in the picture below.arrow_forwardIn Manx cats, the recessive allele for tail length can be lethal. A heterozygous dominant genotype results in a shortened tail, while a homozygous dominant genotype results in a normal tail. A homozygous recessive genotype is lethal. In an ideal Manx cat population exhibiting Hardy-Weinberg equilibrium, if the lethal allele has a frequency of 0.22, what percentage of the Manx cat population retains a shortened or normal tail length? Express your answer using three significant digits.arrow_forward
- Biology: The Dynamic Science (MindTap Course List)BiologyISBN:9781305389892Author:Peter J. Russell, Paul E. Hertz, Beverly McMillanPublisher:Cengage LearningHuman Heredity: Principles and Issues (MindTap Co...BiologyISBN:9781305251052Author:Michael CummingsPublisher:Cengage Learning