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Problem 1 Problem 2 Problem 3 Problem 4 Problem 5 Problem 6 Problem 7 Problem 8 Problem 9 Problem 10 Problem 11 Problem 12 Problem 13 Problem 14 Problem 15 Problem 16 Problem 17 Problem 18 Problem 19 Problem 20 Problem 21 Problem 22 Problem 23 Problem 24 Problem 25 Problem 26 Problem 27 Problem 28 Problem 29 Problem 30 Problem 31 Problem 32 Problem 33 Problem 34 Problem 35 Problem 36 Problem 37 Problem 38 Problem 39 Problem 40 Problem 41 Problem 42 Problem 43 Problem 44 Problem 45 Problem 46 Problem 47 Problem 48 Problem 49 Problem 50 Problem 51 Problem 52 Problem 53 Problem 54 Problem 55 Problem 56 Problem 57 Problem 58 Problem 59 Problem 60 Problem 61 Problem 62 Problem 63 Problem 64 Problem 65 Problem 66 Problem 67 Problem 68 Problem 69 Problem 70 Problem 71 Problem 72 Problem 73 Problem 74 Problem 75 Problem 76 Problem 77 Problem 78 Problem 79 Problem 80 Problem 81 Problem 82 Problem 83 Problem 84 Problem 85 Problem 86 Problem 87 Problem 88 Problem 89 Problem 90 Problem 91 Problem 92 Problem 93 Problem 94 Problem 95 Problem 96 Problem 97 Problem 98 Problem 99 Problem 100 Problem 101 Problem 102 Problem 103 Problem 104 Problem 105 Problem 106 Problem 107 Problem 108 Problem 109 Problem 110 Problem 111 Problem 112 Problem 113 Problem 114 Problem 115 Problem 116 Problem 117 Problem 118 Problem 119 Problem 120

Problem 120 Easy Difficulty

The following problem extends the Hardy-Weinberg model of population dynamics that was covered in Chapter 19. It applies mathematics that would be appropriate after a second course in Algebra. While the concept applied in this problem are within the scope of the Exam the mathematical representations are not and the item is provided to allow students who are able another look at the concepts.
The Hardy-Weinberg model of population dynamics is an algebraic representation of the relationships among genotype frequencies, F, and the probability of the dominant allele A, p, and the recessive allele a, q. The Hardy-Weinberg model of population dynamics is based on several assumptions. One of these assumptions is “random mating.” If all genes in a population are equally able to reproduce, this means that all genes are equally fit and equally fertile. Consequently, the population never evolves.
Populations do evolve and the Hardy-Weinberg model can be modified slightly to allow evolution to occur. Suppose that there is an initial population at generation zero and the probability of the dominant allele at that time is p0. Later, at population k the probability is different. But if the frequencies of the three different combinations of alleles is known then the probabilities pk and qk can be calculated at generation k
(1) $p_{k}=F_{k}(A A)+1 / 2 F_{k}(A a) q_{k}=F_{k}(a a)+1 / 2 F_{k}(A a)$
And since p and q are probabilities for a case where only two alleles exist, p+q=1. Then also (p+q)2=1, leading the Hardy-Weinberg equation
(2)
$F_{k}(A A)=p_{k}^{2} w_{A A} / W F_{k}(A a)=2 p_{k} q_{k} w_{A a} / W F_{k}=$ $q^{2}_{k} w_{a a} / W W=p^{2} w_{A A}+2 p q w_{A a} / q^{2} w_{a a}$
Haldane divides by the factor $\mathrm{W}=\mathrm{F}_{\mathrm{k}}(\mathrm{A} \mathrm{A})+\mathrm{F}_{\mathrm{k}}(\mathrm{Aa})+\mathrm{F}_{\mathrm{k}}(\mathrm{aa})$ so that the probabilities that are still calculated with equation (1) to continue to satisfy the condition for p and q to represent probabilities:$(p+q)^{2}=1$
A. Justify Haldane's model in terms of what the factors $\mathrm{w}_{\mathrm{AA}}, \mathrm{w}_{\mathrm{Aa}}$ and $\mathrm{w}_{\mathrm{aa}}$ mean.
B. Suppose that $w_{A A}=w_{A a}=1,$ but that $w_{\text { aa }}=0.8$ . Predict what will happen to the population over time.
Fitness is determined by the environment. Moree (The American Naturalist, 86, 1952) measured the relative fitness in Drosophila melanogaster of a recessive allele that imparts black eye color as population density increases. A varying number of flies with an equal number of males and females were placed in a pint jar and progeny counted. In each experiment the population was initially heterozygous.
C. Apply Haldane’s approach to calculate the probabilityp in the first generation after mating 150 female and 150 male flies that are heterozygous using wAA = wAa = 1. Rendel (Evolution, 5, 1951) conducted an investigation of the dependence of fecundity (fertility) on light in ebonyeyed D. melanogaster. A summary of some of the data that he reported is shown in the table below:
D. Pose two scientific questions concerning the behavioral response indicated by the data that can be tested experimentally.
E. Is there a question you can add here to wrap up this set with this LO from the list? In this case “light” is the single environmental factor, and they two phenotypes are ebony and wild type that result from different genotypes within the population of flies.

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Video Transcript

the diagram showing the crossing off paid in is shown here now on the Waas. Dave's look same since the all represent, though Why is Type CATIC did a stick for each gene, since the above crosses are simple is it does not cause confusion, but in further experiments, it would be wiser to denote it as be positive. Now, the 3rd 1 that is the Fino type, is what believes on but own midriff two seeds. Hence it means that leaves are purple in color by the color brown part pains to seed. It is not significant to write the genes in the same ordered off they re being. The genes can be written in Okay, Me ordered it. Distant is a homo zegas rhesus. If plant, which is used to identify the Gina type off, the if one organism, the me ah tick products in the organism can be seen as the fino type off the progeny

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