Fraction of hosts infected Case Study 2 Hosts, Parasites, and Time-Travel 0.8 0.6 0.4 0.2 0 Past Contemporary Future Time Figure \u00031 Horizontal axis is the time from which the parasite was taken, relative to the host's point in time. By definition, a parasite has an antagonistic relationship with the host it infects. For this reason we might expect the host to evolve strategies that resist infection, and the parasite to evolve strategies that subvert this host resistance. The end result might be a never-ending coevolutionary cycle between host and parasite, with neither party gaining the upper hand. Indeed, we might expect the ability of the parasite to infect the host to remain relatively unchanged over time despite the fact that both host and parasite are engaged in cycles of evolutionary conflict beneath this seemingly calm surface. This is an intriguing idea, but how might it be examined scientifically? Ideally we would like to hold the parasite fixed in time and see if its ability to infect the host declines as the host evolves resistance. Alternatively, we might hold the host fixed in time and see if the parasite's ability to infect the host increases as it evolves ways to subvert the host's current defenses. Another possibility would be to challenge the host with parasites from its evolutionary past. In this case we might expect the host to have the upper hand, since it will have evolved resistance to these ancestral parasites. Similarly, if we could challenge the host with parasites from its evolutionary future, then we might expect the parasite to have the upper hand, since it will have evolved a means of subverting the current host defenses. Exactly this sort of \"time-travel\" experiment has been done using a bacterium as the host and a parasite called a bacteriophage.1 To do so, researchers let the host and parasite coevolve together for several generations. During this time, they periodically took samples of both the host and the parasite and placed the samples in a freezer. After several generations they had a frozen archive of the entire temporal sequence of hosts and parasites. The power of their approach is that the host and parasite could then be resuscitated from this frozen state. This allowed the researchers to resuscitate hosts from one point in time in the sequence and then challenge them with resuscitated parasites from their past, present, and future. The results of one such experiment are shown in Figure 1. The data show that hosts are indeed better able to resist parasites from their past, but are much more susceptible to infection by those from their future. This is a compelling experiment but, by its very nature, it was conducted in a highly artificial setting. It would be interesting to somehow explore this idea in a natural host- parasite system. Incredibly, researchers have done exactly that with a species of freshwater crustacean and its parasite.2 Daphnia are freshwater crustacea that live in many lakes. They are parasitized by many different microbes, including a species of bacteria called Pasteuria ramosa. These two organisms have presumably been coevolving in lakes for many years, and the question is whether or not they too have been undergoing cycles of evolutionary conflict. Occasionally, both the host and the parasite produce dormant offspring (called propagules) that sink to the bottom of the lake. As a time passes, sediment containing these propagules accumulates at the bottom of the lake. Over many years this sediment builds up, providing a historical record of the host and parasite (see Figure 2). A sediment core can then be taken from the bottom of the lake, giving an archive of the temporal sequence of hosts and parasites over evolutionary time (see Figure 3). And again, as with the first experiment, these propagules can be resuscitated and infection experiments conducted. Figure \u00032 Sedimentation 1.\u0003 A. Buckling et al. 2002, \"Antagonistic Coevolution between a Bacterium and a Bacteriophage.\" Proceedings of the Royal Society: Series B 269 (2002): 931-36. xlvi 14036_FM_ptg01_hr_xl-xlviii.indd 46 2. \u0003E. Decaestecker et al., \"Host-Parasite 'Red Queen' Dynamics Archived in Pond Sediment.\" Nature 450 Unless otherwise noted, all content on this page is Cengage Learning. (2007): 870-73. 6/16/15 4:31 PM case study 2 | Hosts, Parasites, and Time-Travel xlvii Shallow Young Time Depth Deep Fraction of hosts infected 0.8 0.6 0.4 0.2 0 Past Contemporary Future Time Figure \u00034 Horizontal axis is the time from which the parasite was taken, relative to the host's point in time. Source: Adapted from S. Gandon et al., \"HostParasite Coevolution and Patterns of Adaptation across Time and Space,\" Journal of Evolutionary Biology 21 (2008): 1861-66. 0 Old core Deep Figure \u00033 1.0 Old The results of the second experiment are shown in Figure 4: The pattern is quite different from that in Figure 1, with hosts being able to resist parasites from their past and their future, more than those taken from a contemporary point in time. How can we understand these different patterns? Is it possible that this Daphnia- parasite system is also undergoing the same dynamic as the bacteriophage system, but that the different pattern seen in this experiment is simply due to differences in conditions? More generally, what pattern would we expect to see in the Daphnia experiment under different conditions if such coevolutionary conflict is actually occurring? To answer these questions we need a more quantitative approach. This is where mathematical modeling comes into play. Models begin by simplifying reality (recall that a model is \"a lie that makes us realize truth\"). Thus, let's begin by supposing that there are only two possible host genotypes (A and a) and two possible parasite genotypes (B and b). Suppose that parasites of type B can infect only hosts of type A, while parasites of type b can infect only hosts of type a. Although we know reality is likely more complicated than this, these simplifying assumptions capture the essential features of an antagonistic interaction between a host and its parasite. Under these assumptions we might expect parasites of type B to flourish when hosts of type A are common. But this will then give an advantage to hosts of type a, since they are resistant to type B parasites. As a result, type a hosts will then increase in frequency. Eventually, however, this will favor the spread of type b parasites, which then sets the stage for the return of type A hosts. At this point we might expect the cycle to repeat. In this case study you will construct and analyze a model of this process. As is common in modeling, the order in which different mathematical tools are used by scientists is not always the same as the order in which they are best learned. For example, when scientists worked on this question they first used techniques from Chapter 7 and then Chapter 10 to formulate the model. They then used techniques from Chapter 6 and then Chapter 2 to draw important biological conclusions.3 To fit with our learning objectives, however, this case study is developed the other way around. Following Chapter 2, in Case Study 2a, we will use given functions to draw biological conclusions about host- parasite coevolution. Following Chapter 6, in Case Study 2b, we will then begin to fill in the gaps by deriving these functions from the output of a model. Following Chapter 7, in Case Study 2c, we will then formulate this model explicitly, and following Chapter 10, in Case Study 2d, we will derive the output of the model that is used in Case Study 2b. 3. S. Gandon et al., \"Host-Parasite Coevolution and Patterns of Adaptation across Time and Space,\" Journal of Biology 21 (2008): 1861-66. Evolutionary Unless otherwise noted, all content on this page is Cengage Learning. 14036_FM_ptg01_hr_xl-xlviii.indd 47 6/16/15 4:31 PM (inc CASE STUDY 2a | Hosts, Parasites, and Time-Travel 10. Sketch the graph of an example of a function f that satisfies all of the following conditions: lim f sxd 22, lim f sxd 0, lim f sxd `, x l 2` xl` lim2 f sxd 2`, x l3 x l 23 29. The Michaelis-Menten equation for the rate v of the enzymatic reaction of the concentration [S] of a substrate S, in the case of the enzyme pepsin, is lim1 f sxd 2, x l3 11-28 Find the limit. xl ` What is lim v? What is the meaning of the limit in this fSgl ` 12x 2 1 5x 13. lim e x 3 context? 12. lim 3 22t tl` 2x 14. lim xl1 xl3 x2 2 9 x 1 2x 2 3 x2 2 9 16. lim1 2 x l 1 x 1 2x 2 3 sh 2 1d3 1 1 17. lim hl0 h t2 2 4 18. lim 3 tl2 t 2 8 sr 19. lim r l 9 sr 2 9d4 20. u4 2 1 21. lim 3 u l 1 u 1 5u 2 2 6u sx 1 6 2 x 22. lim xl 3 x 3 2 3x 2 23. lim2 lnssin xd 24. 25. lim xl ` sx 2 2 9 2x 2 6 lim v l 41 lim x l 2` 30. Prove that lim x l 0 x 2 coss1yx 2 d 0. 2 x 29 15. lim 2 x l 23 x 1 2x 2 3 2 xl | 42v 42v 31. Let (a) Evaluate each limit, if it exists. (i) lim1 f sxd (ii) lim2 f sxd xl0 | (iv) lim2 f sxd xl3 xl1 1 1 1 2 x21 x 2 3x 1 2 (v) lim1 f sxd xl3 (iii) lim f sxd xl0 (vi) lim f sxd xl3 32. Show that each function is continuous on its domain. State the domain. (a) tsxd 2 xl ` sx 2 2 9 x2 2 2 (b) hsxd xe sin x 33-34 Use the Intermediate Value Theorem to show that there is a root of the equation in the given interval. xl ` S xl0 (b) Where is f discontinuous? (c) Sketch the graph of f. 1 2 2x 2 2 x 4 5 1 x 2 3x 4 26. lim e x2x H if x , 0 s2x f sxd 3 2 x if 0 < x , 3 sx 2 3d2 if x . 3 27. lim (sx 2 1 4x 1 1 2 x ) 28. lim 0.50fSg 3.0 3 10 24 1 fSg v f is continuous from the right at 3 11. lim 151 D 33. 2x 3 1 x 2 1 2 0, 2x 2 34. e x, s22, 21d s0, 1d CASE STUDY 2a Hosts, Parasites, and Time-Travel We are studying a model for the interaction between Daphnia and its parasite. Recall that there are two possible host genotypes (A and a) and two possible parasite genotypes (B and b). Parasites of type B can infect only hosts of type A, while parasites of type b can infect only hosts of type a. Here we will take equations that will be obtained in Case Studies 2b and 2d to explore the biological predictions that can be obtained from them. In Case Study 2d we will derive the functions 09631_ch02_ptg01_hr_150-154.indd 151 (1a) qstd 12 1 Mq cossct 2 q d (1b) pstd 12 1 Mp cossct 2 p d 8/4/14 2:12 PM 152 CHAPTER 2 | Limits where qstd is the predicted frequency of host genotype A at time t and pstd is the predicted frequency of parasite genotype B at time t. In these equations q , p , and c are positive constants, while Mq , Mp are positive constants that are strictly less than 12. 1. Describe, in words, how the genotype frequencies of the host and parasite change over time. Provide an explanation, in biological terms, for these dynamics. 2. How do the constants Mq and Mp affect the pattern of genotype frequencies over time? 3. The constant c is determined by the consequences of infection, in terms of reproduction, for both the host and the parasite. A large difference in reproductive success between infected versus uninfected hosts makes c large. Likewise, a large difference in reproductive success between parasites that are unable to infect a host versus those that are able to infect a host also makes c large. How does c affect the pattern of genotype frequencies over time as predicted by Equations 1? Provide an explanation for this in biological terms. 4. The constants q and p are referred to as the phase of qstd and pstd, respectively. How do these constants affect the pattern of genotype frequencies over time? 5. As you will see in Equation 3, the difference * p 2 q turns out to be important in the coevolution of the host and parasite. What does this difference represent mathematically? Explain why this quantity is a measure of the extent to which the frequency of the parasite genotype lags behind the frequency of the host genotype. Equations 1 give the genotype frequencies as functions of time. In the Daphniaparasite system described in Case Study 2 on page xlvi, these equations can also be interpreted as giving the dynamics of genotype frequencies as a function of depth in the sediment core shown in Figure 1. Depth Deep Old Shallow Time Deep Shallow W W t t1D Young Old FIGURE 1 Depth Time D Young FIGURE 2 In the experiment described in Case Study 2, researchers chose a fixed depth and extracted a layer of sediment of width W centered around this depth (see Figure 2). The contents of this layer were mixed completely, and then hosts and parasites were extracted at random from the mixture. Researchers also took deeper and shallower layers (which represent the past and the future for hosts located in the layer at ) and again completely mixed each layer. The center of these layers was a distance D from the center of the focal layer at , with D , 0 corresponding to a deeper layer and D . 0 a shallower layer (see Figure 2). The researchers then challenged hosts from the layer at with parasites from their past (that is, from the layer with D , 0), present (the layer at ), and future (the layer with D . 0). For each challenge experiment the fraction of hosts becoming infected was measured. 09631_ch02_ptg01_hr_150-154.indd 152 8/6/14 6:50 PM CASE STUDY 2a | Hosts, Parasites, and Time-Travel 153 In Case Study 2b we will show that, when a layer of sediment at location with width W is mixed completely, the frequency of type A hosts in this mixture is predicted to be (2a) qave sd 12 1 Mq cossc 2 q d 2 sin ( 12 cW ) cW Likewise, we will show that, when a layer of sediment at location with width W is mixed completely, the frequency of type B parasites in this mixture is predicted to be (2b) pave sd 12 1 Mp cossc 2 p d 2 sin ( 12 cW ) cW 6. The functions (2a) and (2b) are similar to (1a) and (1b) except that the second terms are multiplied by the quantity 2 sin ( 12 cW ) ycW. Describe how the frequency of the genotypes within a mixed layer depends on the width W of this layer. In particular, what happens as the width of the layer becomes very small (that is, when W l 0)? What happens as the width becomes very large (that is, W l `)? Provide a biological interpretation for your answers. In the experiment introduced in Case Study 2, hosts from depth were challenged with parasites from depth 1 D. This was repeated for different depths , and the overall fraction of hosts infected was measured. In Case Study 2b we will show that the predicted fraction of hosts infected from such an experiment is (3) FsDd 12 1 Mp Mq cosscD 2 * d 4 sin 2 ( 12 cW ) c 2W 2 where * p 2 q and D , 0 corresponds to parasites from a host's past and D . 0 to parasites from a host's future. 7. Sketch the graph of FsDd when * 0. Be as accurate as possible, showing where the maxima and minima occur as well as where the graph crosses the vertical axis. Construct similar sketches when * is small and positive as well as when * is small and negative. These plots depict the predicted fraction of infected hosts in the experiment as a function of the relative point in time from which the parasite was taken. FIGURE 3 Horizontal axis is the time from which the parasite was taken, relative to the host's point in time. Source: Adapted from S. Gandon et al., \"HostParasite Coevolution and Patterns of Adaptation across Time and Space,\" Journal of Evolutionary Biology 21 (2008): 1861-66. 09631_ch02_ptg01_hr_150-154.indd 153 Fraction of hosts infected 8. Suppose that cD is relatively small, meaning that the layers used in the challenge experiments are close to one another. Use your results from Problem 7 to explain how it is possible to obtain the experimental data like those shown in Figure 3. 0.8 0.6 0.4 0.2 0 Past Contemporary Future Time 8/6/14 6:52 PM 154 CHAPTER 2 | Limits In particular, what is true about the value of * in this case? Provide a biological interpretation of your answer. FIGURE 4 Horizontal axis is the time from which the parasite was taken, relative to the host's point in time. Source: Adapted from S. Gandon et al., \"HostParasite Coevolution and Patterns of Adaptation across Time and Space,\" Journal of Evolutionary Biology 21 (2008): 1861-66. 09631_ch02_ptg01_hr_150-154.indd 154 Fraction of hosts infected 9. Again suppose that cD is relatively small. Use your results from Problem 7 to explain how it is possible to obtain data like those shown in Figure 4. What is true about the value of * in this case? Provide a biological interpretation of your answer. 0.8 0.6 0.4 0.2 0 Past Contemporary Future Time 8/6/14 6:53 PM 416Chapter 6 | Applications of Integrals case study 2b Hosts, Parasites, and Time-Travel In Case Study 2c you will derive a model for the dynamics of the genotypes of Daphnia and its parasite. Recall that we are modeling a situation involving two possible host genotypes (A and a) and two possible parasite genotypes (B and b). Parasites of type B can infect only hosts of type A, while parasites of type b can infect only hosts of type a. You will then derive an explicit solution of a simplified version of the model in Case Study 2d. This will give the frequency of host genotype A and parasite genotype B as functions of time. These functions are qstd 12 1 Mq cossct 2 \u001fq d (1) pstd 12 1 Mp cossct 2 \u001fp d where qstd is the predicted frequency of host genotype A at time t and pstd is the predicted frequency of the parasite genotype B at time t. In these equations \u001fq , \u001fp , and c are positive constants, and Mq and Mp are positive constants that are strictly less than 12 (the biological significance of these constants is explored in Case Study 2a). In this part of the case study you will use Equations 1 to make predictions from the model that can be compared with data from the experiments. Recall that, in the experiment, a host from a fixed layer of the sediment core was challenged with infection by a parasite from either the same layer, a layer above the fixed layer (that is, from its future), or a layer below it (that is, from its past). We can view different depths in the sediment core as representing different points of time in the history of the Daphnia-parasite interaction (see Figure 1). In this way Equations 1 can equally be viewed as specifying the frequency of the host and parasite genotypes as functions of location in the sediment core. Increasing values of t correspond to shallower points in the core as shown in Figure 2. Shallow Young Time Depth Deep Deep Figure \u00031 09631_ch06_ptg01_hr_408-418.indd 416 Shallow 1.0 p(t) 0 q(t) Old core Depth Time (increasing t) Young Old Figure \u00032 8/7/14 5:04 PM case study 2b | Hosts, Parasites, and Time-Travel 417 In the experiment introduced in Case Study 2 on page xlvii, researchers chose a fixed depth \u001f and extracted a layer of sediment of width W centered around this depth. This layer is shown in Figure 3. Depth Deep Shallow W W p(t) 1.0 q(t) 0 Figure \u00033 \u001f+D \u001f Time Old D Young After the contents of this layer were completely mixed, hosts and parasites were extracted at random from the mixture. Researchers also took deeper and shallower layers (that represent the past and the future for hosts located in the layer at \u001f) and completely mixed each. The center of these layers was a distance D from the center of the focal layer (see Figure 3). They then challenged hosts from the layer at \u001f with parasites from their past (that is, from the layer with D , 0), present (the layer at \u001f), and future (the layer with D . 0). For each challenge experiment the fraction of hosts becoming infected was measured. We can use our model to predict the fraction of hosts infected. To do so, we first need to know the predicted frequency of hosts of type A in the layer at \u001f as well as the frequency of the parasites of type B in the layer at \u001f 1 D. 1.\t\u0007\u0003\u0003Consider a focal layer at location \u001f with width W as shown in Figure 4. Depth Deep Shallow W q(t) W \u001f - -- 2 Figure \u00034 \u001f W \u001f + -- 2 Time Old Young \u0007\u0003\u0003The frequency of host type A will vary across the depth of this layer as specified by the function qstd. Show that, when this layer is completely mixed, the frequency of A in the mixture is given by qave s\u001fd 1 2 sin ( 12 cW) 1 \u001eq cossc\u001f 2 \u001dqd 2 cW \u0007\u0003\u0003Hint: You might want to use the trigonometric identity sinsx 1 yd 2 sinsx 2 yd 2 cos x sin y 09631_ch06_ptg01_hr_408-418.indd 417 8/7/14 4:54 PM 418Chapter 6 | Applications of Integrals 2.\t\u0007\u0003\u0003Consider another layer at a location a distance D from \u001f, again with thickness W, as shown in Figure 5. The frequency of parasite type B will vary across depth in this layer, as specified by the function pstd. Show that, when this layer is completely mixed, the frequency of B in the mixture is given by pave s\u001f 1 Dd 1 2 sin ( 12 cW) 1 Mp cosscs\u001f 1 Dd 2 \u001epd 2 cW Depth Deep Shallow W p(t) W \u001f+D W \u001f +D--- \u001f +D+-- 2 2 \u001f Figure \u00035 Time Old Young 3.\t\u0007\u0003\u0003Suppose that hosts from the layer at \u001f are challenged with parasites from the layer at \u001f 1 D. Use the facts that only B parasites can infect A hosts and only b parasites can infect a hosts to explain why the fraction of hosts infected in this challenge experiment is predicted to be Is\u001fd pave s\u001f 1 Ddqave s\u001fd 1 f1 2 pave s\u001f 1 Ddgf1 2 qave s\u001fdg The final step is to recognize that the experiment was actually conducted with several different, randomly chosen depths \u001f. Therefore we need to average Is\u001fd in Problem 3 over the possible depths. Because Is\u001fd is periodic, we need only average over one period of its cycle. Its average is therefore F 1 T y T 0 Is\u001fd d\u001f where T 2\u001cyc is the period of Is\u001fd. 4.\t\u0007\u0003\u0003Show that (2) FsDd 12 1 Mp Mq cosscD 2 \u001e*d 4 sin 2 ( 12 cW) c 2W 2 \u0007\u0003\u0003where \u001e* \u001ep 2 \u001eq. \u0007\u0003 Hint: You might want to use the trigonometric identity cos x cos y 12 scossx 1 yd 1 cossx 2 ydd Equation 2 was used in Case Study 2a to predict the experimental outcome expected under different conditions. 09631_ch06_ptg01_hr_408-418.indd 418 8/7/14 4:57 PM 484 CHAPTER 7 | Differential Equations (0 , h , 1). The equations become division phase is triggered by high concentrations of a molecule called MPF (maturation promoting factor). The production of this factor is stimulated by another molecule called cyclin, and MPF eventually inhibits its own production. Using M and C to denote the concentrations of these two biomolecules (in mgymL), a simple model for their interaction is dp1 c1 p1sh 2 p1d 2 m1 p1 dt dp 2 c 2 p 2 sh 2 p1 2 p 2d 2 m 2 p 2 2 c1 p1 p 2 dt Suppose that m1 m 2 3, c1 5, and c 2 30. (a) Construct the phase plane, including all nullclines, equilibria, and arrows indicating the direction of movement in the plane when 35 , h , 1. (b) Construct the phase plane, including all nullclines, equilibria, and arrows indicating the direction of movement 1 in the plane when 10 , h , 35. (c) Construct the phase plane, including all nullclines, equilibria, and arrows indicating the direction of movement 1 in the plane when 0 , h , 10 . (d) From your results to parts (a), (b), and (c), determine how habitat destruction is expected to affect the coexistence of the two species. dM M C 1 CM 2 2 dt 11M dC 2M dt (a) Suppose that 2, 1, 10, and 1. Construct the phase plane, including all nullclines, equilibria, and arrows indicating the direction of movement in the plane. (b) From your answer to part (a), what is the qualitative nature of the dynamics of M predicted by this model? What does this predict about the dynamics of cell division? (c) For any equilibrium found in part (a), specify whether it is locally stable, unstable, or if the information is inconclusive. Source: Adapted from S. Nee et al., \"Dynamics of Metapopulations: Habitat Destruction and Competitive Coexistence,\" Journal of Animal Ecology 61 (1992): 37-40. 25. Cell cycle dynamics The process of cell division is periodic, with repeated growth and division phases as the cell population multiplies. It has been suggested that the Source: Adapted from R. Norel et al., \"A Model for the Adjustment of the Mitotic Clock by Cyclin and MPF Levels,\" Science 251 (1991): 1076 -78. CASE STUDY 2c Hosts, Parasites, and Time-Travel In this part of the case study you will formulate a mathematical model for the antagonistic interactions between Daphnia and its parasite using differential equations. Let's suppose that there are two possible host genotypes (A and a) and two possible parasite genotypes (B and b). Parasites of type B can infect only hosts of type A, while parasites of type b can infect only hosts of type a (see Table 1). We will derive a set of two coupled differential equations that model the dynamics of the frequency of A in the host population and B in the parasite population. Table 1 The outcome of challenges between different host and parasite genotypes. 09631_ch07_ptg01_hr_482-486.indd 484 Host A Host a Parasite B Infection occurs Infection does not occur Parasite b Infection does not occur Infection occurs 8/7/14 7:57 PM CASE STUDY 2c | Hosts, Parasites, and Time-Travel 485 A common differential equation used in biology to model the frequency dynamics of a particular genotype is BB (1) df f s1 2 f dsr1 2 r 2d dt where f is the frequency of type 1, and r 1 and r 2 are the per capita reproduction rates of the two types. For example, see Exercise 7.2.16. We will use an equation of this form for both the host and the parasite populations. Suppose the per capita reproduction rate of uninfected hosts is rq and that for infected hosts is rq 2 sq. The constant sq is assumed to satisfy the inequality 0 , sq , rq and represents the reduction in reproductive output of a host due to infection. Similarly, the per capita reproduction rate of a parasite that is able to infect a host is rp and that for one unable to infect a host is rp 2 sp (the parasite can reproduce in the absence of the host, but it does so less well). The constant sp is assumed to satisfy the inequality 0 , sp , rp and represents the reduction in reproductive output of a parasite if it is unable to infect a host. Let's use q to denote the frequency of type A individuals in the host population and p to denote the frequency of type B individuals in the parasite population. Suppose that host-parasite encounters occur at random with respect to genotype. 1. With random encounters, the average per capita reproduction rate for hosts of a given type is rB p 1 rbs1 2 pd, where rB and rb are the reproduction rates of the host when encountering a type B or type b parasite, respectively. Show that the average per capita reproduction rates of hosts of type A and a are therefore type A: rq 2 psq type a: rq 2 s1 2 pdsq 2. With random encounters, the average per capita reproduction rate for parasites of a given type is rA q 1 ras1 2 qd, where rA and ra are the reproduction rates of the parasite when encountering a type A or type a host, respectively. Show that the average per capita reproduction rates of parasites of type B and b are therefore type B: rp 2 s1 2 qdsp type b: rp 2 qsp 3. Suppose both q and p satisfy differential equations of the form given in Equation 1. Show that q and p therefore satisfy dq sq qs1 2 qds1 2 2pd dt dp sp ps1 2 pds2q 2 1d dt 4. Construct the phase plane including all nullclines, equilibria, and arrows indication the direction of movement in the plane. 5. Explain, qualitatively, how the frequencies of the two parasite genotypes are predicted to change over time. Similarly, explain how the frequencies of the two host genotypes are predicted to change over time. 09631_ch07_ptg01_hr_482-486.indd 485 8/1/14 5:51 PM