Snow geese (Chen caerulescens) come in two color types, white “snows” and “blues” with dark bodies. A single gene controls coloration, where the dark (“blue”) allele (D) is dominant. A population of 30,012 geese includes 9236 dark individuals. Genetic testing reveals that 7636 of the 9236 dark individuals are heterozygous (Dd). Calculate the chi-square statistic from these data. What is the value of the chi-square statistic? A.) 588.83 B.) 13247.35 C.) 6983.45 D.) 0.0154

Snow geese (Chen caerulescens) come in two color types, white “snows” and “blues” with dark bodies. A single gene controls coloration, where the dark (“blue”) allele (D) is dominant. A population of 30,012 geese includes 9236 dark individuals. Genetic testing reveals that 7636 of the 9236 dark individuals are heterozygous (Dd). Calculate the chi-square statistic from these data. What is the value of the chi-square statistic? A.) 588.83 B.) 13247.35 C.) 6983.45 D.) 0.0154

Human Heredity: Principles and Issues (MindTap Course List)

11th Edition

ISBN:9781305251052

Author:Michael Cummings

Publisher:Michael Cummings

Chapter19: Population Genetics And Human Evolution

Section: Chapter Questions

Problem 9QP: Using the HardyWeinberg Law in Human Genetics Suppose you are monitoring the allelic and genotypic...

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Snow geese (Chen caerulescens) come in two color types, white “snows” and “blues” with dark bodies. A single gene controls coloration, where the dark (“blue”) allele (D) is dominant. A population of 30,012 geese includes 9236 dark individuals. Genetic testing reveals that 7636 of the 9236 dark individuals are heterozygous (Dd).Use the table below to determine the critical value to compare to the chi-square statistic. Is this population evolving? Degrees of freedom P = 0.05 1 3.841 2 5.991 3 7.815 Group of answer choices A.) yes B.) no C.) inconclusive
Snow geese (Chen caerulescens) come in two color types, white “snows” and “blues” with dark bodies. A single gene controls coloration, where the dark (“blue”) allele (D) is dominant. A population of 30,012 geese includes 9236 dark individuals. Genetic testing reveals that 7636 of the 9236 dark individuals are heterozygous (Dd). According to the Hardy-Weinberg equilibrium, what is the expected frequency of genotype Dd? Group of answer choices A.) 0.492 B.) 0.181 C.) 0.297 D.) 0.453
Snow geese (Chen caerulescens) come in two color types, white “snows” and “blues” with dark bodies. A single gene controls coloration, where the dark (“blue”) allele (D) is dominant. A population of 30,012 geese includes 9236 dark individuals. Genetic testing reveals that 7636 of the 9236 dark individuals are heterozygous (Dd). According to the Hardy-Weinberg equilibrium, what is the expected frequency of genotype DD? A.) 0.189 B.) 0.120 C.) 0.435 D.) 0.033
Snow geese (Chen caerulescens) come in two color types, white “snows” and “blues” with dark bodies. A single gene controls coloration, where the dark (“blue”) allele (D) is dominant. A population of 30,012 geese includes 9236 dark individuals. Genetic testing reveals that 7636 of the 9236 dark individuals are heterozygous (Dd). According to the Hardy-Weinberg equilibrium, what is the expected number of individuals of genotype Dd? Group of answer choices A.) 14752.40 B.) 13596.09 C.) 8898.56 D.) 3608.26
Snow geese (Chen caerulescens) come in two color types, white “snows” and “blues” with dark bodies. A single gene controls coloration, where the dark (“blue”) allele (D) is dominant. A population of 30,012 geese includes 9236 dark individuals. Genetic testing reveals that 7636 of the 9236 dark individuals are heterozygous (Dd). What is the actual frequency of genotype dd? Group of answer choices A.) 0.692 B.) 0.319 C.) 0.427 D.) 0.453
Snow geese (Chen caerulescens) come in two color types, white “snows” and “blues” with dark bodies. A single gene controls coloration, where the dark (“blue”) allele (D) is dominant. A population of 30,012 geese includes 9236 dark individuals. Genetic testing reveals that 7636 of the 9236 dark individuals are heterozygous (Dd). What is the actual frequency of allele D? A.) 0.819 B.) 0.181 C.) 0.435 D.) 0.347
Snow geese (Chen caerulescens) come in two color types, white “snows” and “blues” with dark bodies. A single gene controls coloration, where the dark (“blue”) allele (D) is dominant. Researchers using genetic testing are able to determine the following numbers of individuals of each genotype in another population of geese: DD = 10576, Dd = 14503, dd = 4922. -According to the Hardy-Weinberg equilibrium, what is the expected number of individuals of genotype DD? -Calculate the chi-square statistic from these data. What is the value of the chi-square statistic?
Snow geese (Chen caerulescens) come in two color types, white “snows” and “blues” with dark bodies. A single gene controls coloration, where the dark (“blue”) allele (D) is dominant. A population of 30,012 geese includes 9236 dark individuals. Genetic testing reveals that 7636 of the 9236 dark individuals are heterozygous (Dd). How many individuals are DD? A.) 9236 B.) 1600 C.) 7636 D.) 20776
The European land snail, Cepaea nemoralis, exhibits 3 phenotypes in shell color; brown, pink, and yellow. Shell color is controlled by one gene that has only two alleles. Brown (B) is dominant to yellow (b) and pink is the heterozygous condition. In one population of 520 snails, the following color phenotypes were recorded: 236 brown, 231 pink and 53 yellow. What are the actual allelic frequencies?
This is a classic data set on wing coloration in the scarlet tiger moth (Panaxia dominula). Coloration in this species had been previously shown to behave as a single-locus, two-allele system with incomplete dominance. Data for 1612 individuals are given below: White-spotted (AA) =1469 Intermediate (Aa) = 138 Little spotting (aa) =5 Calculate the following frequencies: A=, a=, AA=, Aa=, aa=
A new kind of tulip is produced that develops only purple or pink flowers. Assume that flower color is controlled by a single-gene locus and that the purple allele (C) is dominant to the pink allele (c). A random sample of 1000 tulips from a large cultivated field yields 847 purple flowers (out of which 476 are heterozygous) and 153 pink flowers. If p = frequency of C allele and q = frequency of the c allele. Calculate p and q. A. p = 0.4 q = 0.6 B. p = 0.61 q = 0.39 C. p = 0.847 q = 0.153 D. p = 0.82 q = 0.18
A total of 1000 members of a Central American population are typed for the ABO blood group. In the sample, 421 have blood type A, 168 have blood type B, 336 have blood type O, and 75 have blood type AB. Part A Use this information to determine the frequency of ABO blood group alleles in the sample. Recall that when considering genes with three alleles whose frequencies are represented by the variables p, q, and r, the sum of genotype frequencies resulting from trinomial expansion is: (p+q+r)² =p² + 2pq+q2+2pr+r²+2gr = 1