Genetic Analysis: An Integrated Approach (3rd Edition)
3rd Edition
ISBN: 9780134605173
Author: Mark F. Sanders, John L. Bowman
Publisher: PEARSON
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Textbook Question
Chapter 2, Problem 7P
If a chi-square test produces a chi-square value of 7.83 with 4 degrees of freedom,
a. In what interval range does the P value fall?
b. Is the result sufficient to reject the chance hypothesis?
c. Above what chi-square value would you reject the chance hypothesis for an experiment with 7 degrees of freedom?
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Students have asked these similar questions
You perform a chi-square test to compare observed and expected values and obtain a chi-square value of 9.4 with 3 degrees of freedom. What do you conclude?
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it is likely that the difference between observed and expected values is due to random chance, since p<0.05
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Chapter 2 Solutions
Genetic Analysis: An Integrated Approach (3rd Edition)
Ch. 2 -
1. Compare and contrast the following terms:
a....Ch. 2 - For the cross , what is the expected genotype...Ch. 2 - 37. Galactosemia is an autosomal recessive...Ch. 2 - In mice, black coat color is dominant to white...Ch. 2 - Two parents plan to have three children. What is...Ch. 2 - Consider the cross AaBbCCAABbCc. a. How many...Ch. 2 - If a chi-square test produces a chi-square value...Ch. 2 -
8. Determine whether the statements below are...Ch. 2 - In the datura plant, purple flower color is...Ch. 2 - 10. The dorsal pigment pattern of frogs can be...
Ch. 2 - 11. Black skin color is dominant to pink skin...Ch. 2 - A male mouse with brown fur color is mated to two...Ch. 2 - 13. Figure 2.12 shows the results of Mendel’s...Ch. 2 - 14. An experienced goldfish breeder receives two...Ch. 2 -
15. The accompanying pedigree shows the...Ch. 2 -
16. A geneticist crosses a pure-breeding strain...Ch. 2 - Suppose an F1 plant from Problem 16 is crossed to...Ch. 2 - 18. In pea plants, the appearance of flowers along...Ch. 2 - 19. If two six-sided dice are rolled, what is the...Ch. 2 - Experimental Insight 2.1 describes data, collected...Ch. 2 -
21. The accompanying pedigree shows the...Ch. 2 - 22. The seeds in bush bean pods are each the...Ch. 2 - List all the different gametes that are possible...Ch. 2 - Organisms with the genotypes AABbCcDd and AaBbCcDd...Ch. 2 - 25. Blue moon beans produce beans that are either...Ch. 2 - In the fruit fly Drosophila, a rudimentary wing...Ch. 2 - In pea plants, plant height, seed shape, and seed...Ch. 2 - A variety of pea plant called Blue Persian...Ch. 2 - 29. In tomato plants, the production of red fruit...Ch. 2 - A male and a female are each heterozygous for both...Ch. 2 - A woman expressing a dominant phenotype is...Ch. 2 - Two parents who are each known to be carriers of...Ch. 2 - 33. An organism having the genotype AaBbCcDdEe is...Ch. 2 - 34. A man and a woman are each heterozygous...Ch. 2 - For a single dice roll, there is a 16 chance that...Ch. 2 - You have four guinea pigs for a genetic study. One...Ch. 2 - 37. Galactosemia is an autosomal recessive...Ch. 2 - Sweet yellow tomatoes with a pear shape bring a...Ch. 2 - A cross between a spicy variety of Capsicum annum...Ch. 2 - Alkaptonuria is an infrequent autosomal recessive...Ch. 2 - 41. Humans vary in many ways from one another....Ch. 2 - 42. In chickens, the presence of feathers on the...Ch. 2 -
43. A pure-breeding fruit fly with the...Ch. 2 - 44. Situs inversus is a congenital condition in...Ch. 2 - 45. Domestic dogs evolved from ancestral grey...Ch. 2 - Alleles of the IGF-1 gene in dogs, encoding...Ch. 2 - 47. The accompanying pedigree shows a family in...Ch. 2 -
48. A pea plant that has the genotype RrGgwwdd is...Ch. 2 - Go to the OMIM website...Ch. 2 - Prob. 50PCh. 2 - Prob. 51PCh. 2 -
52. In humans, the ability to bend the thumb...
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- Example 3.5.7 Fish Vertebrae Consider the distribution of vertebrae given in Table 3.5.1. In Exam- ple 3.5.5 we found that the mean of Y is µy = 21.49. The variance of Y is VAR(Y) = of = (20 – 21.49) x Pr{Y = 20} 21) %3D + (21 – 21.49)? x Pr{Y + (22 – 21.49)? x Pr{Y = 22} + (23 – 21.49) x Pr{Y = 23} = (-1.49) x 0.03 + (-.49) x 0.51 + (0.51) x 0.40 + (1.51)² × 0.06 = 2.2201 x 0.03 + .2401 x 0.51 + .2601 x 0.40 + 2.2801 x 0.06 = 0.066603 + 0.122451 + 0.10404 + 0.136806 = 0.4299. The standard deviation of Y is ay = V0.4299 - 0.6557. %3Darrow_forwardCorn plants from a test plot are measured, and the distribution of heights at 10-cm intervals is recorded in the following table: Height (cm) Plants (no.) 100 20 110 60 120 90 130 130 140 180 150 120 160 70 170 50 180 40 Calculate (a) the mean height, (b) the variance, (c) the standard deviation, and (d) the standard error of the mean. Plot a rough graph of plant height against frequency. Do the values represent a normal distribution? Based on your calculations, how would you assess the variation within this population?arrow_forwardIf the chi-square value is 0 under the degree of freedom 1. What could be the interpretation and does it fit in the population of Hardy-Weinberg equilibrium?arrow_forward
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