Introduction to Genetic Analysis
11th Edition
ISBN: 9781464109485
Author: Anthony J.F. Griffiths, Susan R. Wessler, Sean B. Carroll, John Doebley
Publisher: W. H. Freeman
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Chapter 4, Problem 21.25P
Summary Introduction
To determine: The reason why part c of the problem say "if appropriate".
Introduction: Gene interference is a measure of the independence of crossovers from each other.
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Chapter 4 Solutions
Introduction to Genetic Analysis
Ch. 4 - Prob. 1PCh. 4 - Prob. 5PCh. 4 - Prob. 12PCh. 4 - Prob. 13PCh. 4 - Prob. 14PCh. 4 - Prob. 15PCh. 4 - Prob. 16PCh. 4 - Prob. 17PCh. 4 - Prob. 18PCh. 4 - Prob. 19P
Ch. 4 - Prob. 20PCh. 4 - Prob. 21PCh. 4 - Prob. 21.1PCh. 4 - Prob. 21.2PCh. 4 - Prob. 21.3PCh. 4 - Prob. 21.4PCh. 4 - Prob. 21.5PCh. 4 - Prob. 21.6PCh. 4 - Prob. 21.7PCh. 4 - Prob. 21.8PCh. 4 - Prob. 21.9PCh. 4 - Prob. 21.10PCh. 4 - Prob. 21.11PCh. 4 - Prob. 21.12PCh. 4 - Prob. 21.13PCh. 4 - Prob. 21.14PCh. 4 - Prob. 21.15PCh. 4 - Prob. 21.16PCh. 4 - Prob. 21.17PCh. 4 - Prob. 21.18PCh. 4 - Prob. 21.19PCh. 4 - Prob. 21.20PCh. 4 - Prob. 21.21PCh. 4 - Prob. 21.22PCh. 4 - Prob. 21.23PCh. 4 - Prob. 21.24PCh. 4 - Prob. 21.25PCh. 4 - Prob. 21.26PCh. 4 - Prob. 22PCh. 4 - Prob. 23PCh. 4 - Prob. 24PCh. 4 - Prob. 25PCh. 4 - Prob. 26PCh. 4 - Prob. 27PCh. 4 - Prob. 28PCh. 4 - Prob. 29PCh. 4 - Prob. 30PCh. 4 - Prob. 31PCh. 4 - Prob. 32PCh. 4 - Prob. 33PCh. 4 - Prob. 34PCh. 4 - Prob. 35PCh. 4 - Prob. 36PCh. 4 - Prob. 37PCh. 4 - Prob. 38PCh. 4 - Prob. 38.1PCh. 4 - Prob. 38.2PCh. 4 - Prob. 38.3PCh. 4 - Prob. 38.4PCh. 4 - Prob. 38.5PCh. 4 - Prob. 38.6PCh. 4 - Prob. 38.7PCh. 4 - Prob. 38.8PCh. 4 - Prob. 38.9PCh. 4 - Prob. 38.10PCh. 4 - Prob. 38.11PCh. 4 - Prob. 38.12PCh. 4 - Prob. 38.13PCh. 4 - Prob. 38.14PCh. 4 - Prob. 38.15PCh. 4 - Prob. 38.16PCh. 4 - Prob. 38.17PCh. 4 - Prob. 38.18PCh. 4 - Prob. 38.19PCh. 4 - Prob. 38.20PCh. 4 - Prob. 38.21PCh. 4 - Prob. 38.22PCh. 4 - Prob. 38.23PCh. 4 - Prob. 38.24PCh. 4 - Prob. 39PCh. 4 - Prob. 40PCh. 4 - Prob. 41PCh. 4 - Prob. 42PCh. 4 - Prob. 43PCh. 4 - Prob. 44PCh. 4 - Prob. 45PCh. 4 - Prob. 46PCh. 4 - Prob. 47PCh. 4 - Prob. 48PCh. 4 - Prob. 49PCh. 4 - Prob. 50PCh. 4 - Prob. 51PCh. 4 - Prob. 52PCh. 4 - Prob. 53PCh. 4 - Prob. 54PCh. 4 - Prob. 55PCh. 4 - Prob. 56PCh. 4 - Prob. 57PCh. 4 - Prob. 58PCh. 4 - Prob. 59PCh. 4 - Prob. 60PCh. 4 - Prob. 62PCh. 4 - Prob. 63PCh. 4 - Prob. 64PCh. 4 - Prob. 65PCh. 4 - Prob. 66PCh. 4 - Prob. 67PCh. 4 - Prob. 68PCh. 4 - Prob. 69P
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- I have seen that this was answered as C, Why is the answer C, how is that not evidence of it being genetic? Shouldnt it be none of the above? Question: Of the following, which supports the idea that alcoholism has no genetic or a low genetic component? a) Some strains of mice select alcohol over water 75% of the time, whereas others shun alcohol. b) The concordance value is 55% for MZ twins and 28% for DZ twins. c) Biological sons of alcoholic men who have been adopted have a rate of alcoholism more like that of their adoptive fathers. d) There is a 20% to 25% risk of alcoholism in the sons of alcoholic men. e) None of these.arrow_forwardPlease explain why the order is 3–1-4-2 and not another way, thank you so much. Question is in the image.arrow_forwardWhat is the answer for number 4 and how do you calculate this?arrow_forward
- 6). To investigate the genetic control of locomotive behavior in the diploid nematode, Caenorhabditis elegans, eight "wiggly" mutants were isolated, which wiggle in an uncoordinated way rather than smoothly gliding. These eight mutants were crossed together in all possible pairwise combinations, yielding the results summarized in the table above (+ means that all the progeny were wild type, and m means that all the progeny showed the wiggly mutant phenotype). Organize the eight mutations into con lementation groups, each consisting of a group of mutations affecting the same gene. 1 2 3 4 5 6 7 8 1 E N+E 2 3+1 m + m 4 5 + m m E+ + + EE 6 ++E E+ +EE m m + + + + + + + + + m 7 8 ++4 ∞+E +E m m ++ Earrow_forwardWhat is random (independent) assortment and when does it occur?arrow_forwardstion 6 of 18 Suppose that a geneticist discovers a new mutation in Drosophila melanogaster that causes the flies to shake and quiver. She calls this mutation quiver, qu, and determines that it is due to an autosomal recessive gene. She wants to determine whether the gene encoding quiver is linked to the recessive gene for vestigial wings, vg. She crosses a fly homozygous for quiver and vestigial traits with a fly homozygous for the wild-type traits, and then uses the resulting F, females in a testcross. She obtains the flies from this testcross. Phenotype Number of flies vg* qu+ 230 vg qu 224 vg qut vg* qu 97 99 Test the hypothesis that the genes quiver and vestigial assort independently by calculating the chi-squared, X², for this hypothesis. Provide the X2 to one decimal place. X2 = Does the X value support the hypothesis that the quiver and vestigial genes assort independently? Why or why not? the partial table of critical values for X2 calculations to test this hypothesis.arrow_forward
- Fill the punnet square for just cross 2 and then do the chi square by using the observed values of cross 2 given in the picture:arrow_forwardWhat's the difference between the three types of translocations? (Balanced, unbalanced, and Robertsonian)arrow_forwardn a tetraploid B /B /b /b, how many quadrivalent possible pairings are there? Draw them (see Figure 17-5)arrow_forward
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