Genetics: Analysis and Principles
6th Edition
ISBN: 9781259616020
Author: Robert J. Brooker Professor Dr.
Publisher: McGraw-Hill Education
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Chapter 28, Problem 8CONQ
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Chapter 28 Solutions
Genetics: Analysis and Principles
Ch. 28.1 - 1. Which of the following is an example of a...Ch. 28.1 - 2. Saying that a quantitative trait follows a...Ch. 28.2 - The variance is a. a measure of the variation...Ch. 28.2 - 2. Which of the following statistics is used to...Ch. 28.3 - 1. For many quantitative traits, genotypes and...Ch. 28.4 - A QTL is a __________ where one or more genes...Ch. 28.4 - 2. To map QTLs, strains are crossed that differ...Ch. 28.5 - 1. In a population of squirrels in North Carolina,...Ch. 28.5 - Prob. 2COMQCh. 28.5 - 3. One way to estimate narrow-sense heritability...
Ch. 28.6 - 1. For selective breeding to be successful, the...Ch. 28.6 - Prob. 2COMQCh. 28 - Give several examples of quantitative traits.How...Ch. 28 - 2. At the molecular level, explain why...Ch. 28 - 3. What is a normal distribution? Discuss this...Ch. 28 - 4. Explain the difference between a continuous...Ch. 28 - What is a frequency distribution? Explain how such...Ch. 28 - 6. The variance for weight in a particular herd of...Ch. 28 - Two different varieties of potato plants produce...Ch. 28 - 8. If , would you conclude that a positive...Ch. 28 - Prob. 9CONQCh. 28 - When a correlation coefficient is statistically...Ch. 28 - 11. What is polygenic inheritance? Discuss the...Ch. 28 - What is a quantitative trait locus (QTL)? Does a...Ch. 28 - 13. Let’s suppose that weight in a species of...Ch. 28 - Prob. 14CONQCh. 28 - 15. From an agricultural point of view, discuss...Ch. 28 - Many beautiful varieties of roses have been...Ch. 28 - 17. In your own words, explain the meaning of the...Ch. 28 - 18. What is the difference between broad-sense...Ch. 28 - The heritability for egg weight in a group of...Ch. 28 - In a fairly large population of people living in a...Ch. 28 - When artificial selection is practiced over many...Ch. 28 - 22. Discuss whether a natural population of wolves...Ch. 28 - 23. With regard to heterosis, is each of...Ch. 28 - Here are data for height and weight among 10 male...Ch. 28 - 2. The abdomen length (in millimeters) was...Ch. 28 - 3. You conduct an RFLP analysis of head weight in...Ch. 28 - 5. Let’s suppose that two strains of pigs differ...Ch. 28 - Prob. 6EQCh. 28 - In a wild strain of tomato plants, the phenotypic...Ch. 28 - The average thorax length in aDrosophilapopulation...Ch. 28 - 9. In a strain of mice, the average 6-week body...Ch. 28 - Prob. 10EQCh. 28 - 11. A danger in computing heritability values from...Ch. 28 - For each of the following relationships,...Ch. 28 - An animal breeder had a herd of sheep with a mean...Ch. 28 - The trait of blood pressure in humans has a...Ch. 28 - Discuss why heritability is an important...Ch. 28 - From a biological viewpoint, speculate as to why...Ch. 28 - 3. What is heterosis? Discuss whether it is caused...
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- Corn 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_forwardCan you please find the best conclusion answer for this problem?arrow_forwardpage - Columbus State Uni X content/2622738/viewContent/50155728/View + Calculating the Value of Chi-Square Once you've defined the null hypothesis (and know what you're testing!), you can finally calculate the chi-square value based on your observed data and the expected distribution. There are many statistical software packages and free web-based applications that will automate the calculation for you, but it's important to understand the calculation to understand the meaning of the calculated value. Here, we'll walk through it using a calculation table. This is Mendel's historic data for the F2 generation of monohybrid cross for flower color. Recall that he expected to see in the F2 generation a ratio of 3 purple : 1 white. Chi-square calculation for data from a monohybrid cross with an expected 3 purple : 1 white ratio. Outcomes Observed (0) Expected (E) O-E (0-E)2 (O-E)²/E Purple White Totals 705 224 929 X²= The first step is to calculate the expected (E) values for purple and…arrow_forward
- If n ≥ 100, you may use normal distribution in all cases. please do it with excel formulas.arrow_forwardWhat is the expected genotype frequency of the heterozygous genotype under the Hardy-Weinberg equation P = 0.7?arrow_forwardDraw a graph of V0 vs. [S] (Michaelis-Menten plot) for these data. Put both sets of data on the one graph, use a different symbol of each plot and indicate which set of data is which. Make sure your graph has appropriate axis labels and units. You do not need to insert trend lines (as there is no rectangular hyperbola trendline in EXCEL).arrow_forward
- 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? it is not likely that the difference between observed and expected values is due to random chance, since p>0.05 it is impossible to conclude anything from this information it is likely that the difference between observed and expected values is due to random chance, since p>0.05 it is likely that the difference between observed and expected values is due to random chance, since p<0.05 the experiment was done incorrectly and must be repeated it is not likely that the difference between observed and expected values is due to random chance, since p<0.05arrow_forwardThe relationship between weight (in kilograms) and gender was estimated as below in a linear format. (Male was coded as 0 and Female was coded as 1.) Weight = -10.23 * Female + 66.73 (The p-value for the coefficient of Female is 0.03.) Base on the above equation, which of the following statements are true? (Select all that apply. There can be more than one answer.) Answer options: In the sample used to estimate the equation, the average weight of females is 66.23 kilograms In the sample used to estimate the equation, the average weight of females is 56.5 kilograms In the sample used to estimate the equation, the average weight of males is 66.73 kilograms Females are expected to weigh 10.23 kilograms less than malesarrow_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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