EBK CONCEPTS OF GENETICS
12th Edition
ISBN: 9780134818979
Author: Killian
Publisher: YUZU
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Chapter 25, Problem 14PDQ
(a)
Summary Introduction
To determine: The narrow-sense heritability (h2) for each trait in the given problem.
Introduction: The contribution of genotypic variance concerning the total
(b)
Summary Introduction
To determine: The trait that will respond likely to selection.
Introduction: The term heritability explains the ratio of the total phenotypic variation due to genetic components in a population. The values of heritability are used in animal and plant breeding to point the potential response of a population to artificial selection for a quantitative trait.
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Phenotypic variation in the tail length of mice has the following components: Additive genetic variance (VA) Dominance genetic variance (VD) Gene interaction variance (VI) Environmental variance (VE) Genetic–environmental interaction variance (VGE) a. What is the narrow-sense heritability of tail length? b. What is the broad-sense heritability of tail length?
Assume that human ear length is influenced by multiple genetic and environmental factors. Suppose you measure ear length in three groups of people, in which group A consists of five unrelated people, group B consists of five siblings, and group C consists of five first cousins. a. With the assumption that the environments of all three groups are similar, which group should have the highest phenotypic variance? Explain why. b. Is it realistic to assume that the environmental variance for each group is similar? Explain your answer.
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VA
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Chapter 25 Solutions
EBK CONCEPTS OF GENETICS
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- In a population of locusts, the mean wing length is 47 mm, the selection gradient on wing length is β = 0.12/mm, the phenotypic variance for wing length is P = 3.6 mm2, and the heritability of wing length is h2 = 0.27. In addition, we know that the additive genetic covariance between wing and leg length is 0.60 mm2. What is the expected evolutionary change in mean leg length due to selection on both wings and legs? Repeat these calculations to predict what will happen to wing length as a result of the selection on both wings and legs. What do you predict the average wing and leg lengths will be in the next generation?arrow_forwardEach graph below depicts reaction norms for three different genotypes reared in different environments. In which case(s) is/are VE> 0 (you can assume that graphs for parents and their offspring would be the same)? Mark all that apply Mean phenotype (a) (c) graph a graph b graph c graph d (b) (d) Environmentarrow_forwardMany researchers have estimated the heritability of human traits by comparing the correlation coefficients of monozygotic and dizygotic twins (see pp. 731–732). One of the assumptions made in using this method is that monozygotic twin pairs experience environments that are no more similar to each other than those experienced by dizygotic twin pairs. How might this assumption be violated? Give some specific examples of how the environments of two monozygotic twins might be more similar than the environments of two dizygotic twins.arrow_forward
- 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_forwardIn a study conducted on human height, heritability estimates that were initially done on children with an average age of 5 years old was 74%. Heritability estimates were again performed on the same individuals when they reached adulthood with an average of 40 years old. The result was 30%. Explain the results.arrow_forwardHotchkiss and Marmur noted that the percentage of cotransformation was higher than would be expected on a random basis. For example, the results show that 2.6% of the cells were transformed into M and 4% were transformed into S. If the M and S traits were inherited independently, the expected probability of cotransformation of M and S (M S) would be 0.026× 0.04 = 0.001, or 0.1%. However, they observed 0.41% M S cotransformants, four times more than they expected. What accounts for the relatively high frequency of cotransformation of the traits they observed?arrow_forward
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