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BIO 356 / BEE 587 L04 - Lab 1: Unregulated and Self-Limited Growth Laura Streminsky 2/8/24 Q1 Suppose you have a pond with some lily pads in it, and suppose each lily pad replicates itself once per week. If it takes a year for half the pond to become covered with lily pads, how long will it take for the entire pond to be covered? 1 week after the half year point, or 53 weeks # half the pond is already covered and, since the lily pads double every week, we would expect to get another half once these lily pads replicate. Q2. How many grains of wheat did the emperor owe the inventor? Hint: recalling that the elements in the series 1, 2, 4, 8, … are the consecutive powers of 2, you can work out the sum with a single line of R code. Provide your code. #there are 63 squares on the board total_grains <- ( 2 ^ 64 ) - 1 total_grains ## [1] 1.844674e+19 #1.84467440737096×10^19 grains of wheat Q3. On what day will the aphid population reach 80 insects per plant? # 80=0.0098e^1.55t # 80/0.0098 = e^1.55t # ln (80/0.0098 ) / 1.55 = t #t=5.81 weeks, which is about 41 days. 41 days after March 18th is April 28th. Q4. An aphid weighs about 0.2 milligrams. Assuming a cornstalk weighs about 5kg, how does the total mass of aphids on a cornstalk compare to the weight of the plant on Week 14?

# 0.0098e^1.55(14)= 1.7297 × 10^7 aphids # (1.7297 × 10^7) (0.2 x 10^-6) = 3.46436 kilograms # Assuming exponential aphid growth and negligible cornstalk growth, the aphid mass will be over 60% the weight of the cornstalk after 14 weeks. Q5. Copy the code from the three code chunks above into a code chunk. a) Call both plots and show them in your compiled file. In which of them (linear scale or log scale) does the trend line fit the data better? Explain your answer by b) identifying the type of growth displayed in these plots, c) describing what this type of growth looks like in logarithmic scale. data = 'https://github.com/rafaeldandrea/BIO- 356/blob/master/covid_data.rds?raw=true' |> url () |> readRDS () |> as_tibble () |> mutate ( date = as.Date (date)) country = 'United Kingdom' initial_date = '2020-03-01' final_date = '2020-04-01' filtered_data = data |> filter ( location == country, date >= initial_date, date < final_date ) #plot = filtered_data |> ggplot ( aes (date, new_cases_smoothed)) + geom_point () + geom_smooth ( method = 'lm' ) + labs ( x = 'Date' , y = 'New cases' ) + theme ( aspect.ratio = 1 ) ## `geom_smooth()` using formula = 'y ~ x' #plot_log = plot + scale_y_log10 ()

initial_date = '2020-03-01' final_date = '2020-04-30' filtered_data = data |> filter ( location == country, date >= initial_date, date < final_date ) #plot = filtered_data |> ggplot ( aes (date, new_cases_smoothed)) + geom_point () + geom_smooth ( method = 'lm' ) + labs ( x = 'Date' , y = 'New cases' ) + theme ( aspect.ratio = 1 ) ## `geom_smooth()` using formula = 'y ~ x' #plot_log = plot + scale_y_log10 ()

## NULL # Exponential growth is no longer a good description of the spread of the epidemic because we see the prevalence stabilizing at a threshold. Tbhis could be due to increaed immunity from a lot of people already having the disease and developing their own antibodies, or from a significant portion of the population getting vaccinated. Q8. What would likely happen to the aphid population before Week 14 as the number of insects per plant continued to rise? # The aphid population would likely stabilize at a carrying capacity imposed by external restraints on critical resources like food and water Q9. Assuming the population is initially small and starts growing as it consumes the resources, what happens if it reaches F/c individuals? How much available resource is left at that point? What happens if the population surpasses F/c individuals? #since the amount of eaten resource is equal to cN, when N=F/c, we can determine that the amount of food eaten is = c(F/c) = F, so the population ate all of its food and cannot grow any further. Once the population surpasses F/c individuals, they will start dying very rapidly because they will starve Q10. If the current population size is much smaller than the carrying capacity, ie if Nt/K<<1, what simplification can we make to the expression for logistic growth? What do you conclude from this? # if Nt/K is negligible, we can remove it from the logistic growth equation, resulting in the equation N(t+1)-N(t)= rN(t), or dN/dt=rN. This equation means that, at a small population, we can estimate that the rate of growth will increase as the population grows (i.e growth will be exponential). Q11. How do you explain what happened biologically? # Bacteria, just like every other living organism, is constrained by finite resources. Factors like nutrients and space enforce a population threshold beyond which the bacteria's competition for resources means that many will die off. Q12. What would ultimately happen if the experiment persisted beyond 30 days?

github_address <- 'https://github.com/rafaeldandrea/BIO- 356/blob/master/WPP2019_POP_F01_1_TOTAL_POPULATION_BOTH_SEXES.xlsx?raw=true' UN_popdata_1950_to_2020 <- rio :: import (github_address, range = 'F17:BZ18' ) |> select ( - 'Parent code' ) |> pivot_longer ( - Type, names_to = 'year' , values_to = 'population' ) |> select ( - Type) |> mutate ( year = as.numeric (year), population = population / 1e6 ) UN_popdata_1950_to_2020 <- UN_popdata_1950_to_2020 %>% mutate ( population_nextyear = lead (population), growth_rate = 100 * (population_nextyear - population) / population ) plot_growth_vs_pop <- UN_popdata_1950_to_2020 |> filter (year < 2020 ) |> ggplot ( aes (population, growth_rate, color = year)) + geom_point () + labs ( x = 'Population (billions)' , y = 'Annual growth rate (%)' ) + theme ( legend.position = 'none' , aspect.ratio = 1 ) + scale_color_gradient ( low = 'blue' , high = 'red' ) linear_model <- lm (growth_rate ~ population, data = UN_popdata_1950_to_2020) plot_growth_vs_pop + geom_abline ( intercept = coef (linear_model)[ 1 ], slope = coef (linear_model)[ 2 ], color = 'darkgrey' )

coefficients <- coef (linear_model) slope <- coefficients[ 2 ] y_intercept <- coefficients[ 1 ] print ( paste ( "Slope:" , slope)) ## [1] "Slope: -0.18418319828217" print ( paste ( "Y-intercept:" , y_intercept)) ## [1] "Y-intercept: 2.52473602360298" # Slope= -0.18418319828217 # Y-intercept= 2.52473602360298 Q15. The statistical model that we are fitting to our UN data corresponds to one of the models of population growth we have discussed. Name that model, and by rearranging terms, identify how a and b relate to quantities associated with that model. # The equation model can be simplified as dN/dt=ab, which is the exponential growth rate model. a corresponds to r, the growth rate, and b corresponds to N, the current population size. When we replace a and b for r and N, we get the same equation we worked with in previous questions.The equation says that the rate of growth increases as the population increases.

tibble ( density = seq ( 0 , 30 , l = 1000 ), yield = Yield ( rho = density, kc = . 1 , kn = . 01 , knc = . 005 ) ) ## Error in Yield(rho = density, kc = 0.1, kn = 0.01, knc = 0.005): could not find function "Yield" test |> ggplot ( aes (density, yield)) + geom_line () + geom_vline ( aes ( xintercept = optimal_rho), color = 'red' ) + theme ( aspect.ratio = 1 ) ## Error in eval(expr, envir, enclos): object 'test' not found # the calculated value of optimal_rho matches the optimal density portrayed in the graph. Q20. In the model above, nutrient limitation creates a negative relationship between plant density and plant size. This provides a mechanistic explanation for which phenomenon named in lecture? Self - thinning ## Error in eval(expr, envir, enclos): object 'Self' not found Q21. Calculate the equilibrium stock given fishing effort E. Show your work. Hint: solve the dynamic equation above for equilibrium, i.e. equate the right-hand side to zero and solve for N. # dN/dt= r(1-N/K)N-qEN # At equilibrium, dN/dt= 0 # Factor out an N: 0= r- rN/k-qE # rN/k = r-qE # rN = k(r-qE) # N = k(r-qE)/r # Stock is inversely related to effort. When E=0, N is at it's maximum value. As E increases, the numerator in the above equation decreases, meaning that N decreases. Q22. Based on the plot above, what is the optimal number of fishing nets for this fishery? How many fish are caught under that fishing effort? Why does the catch decline beyond that fishing effort? Shouldn’t more nets always yield more caught fish?

# If E represents the number of fishing nets, then according to the graph the optimal number of fishing nets would be 50, which results in 250,000 fish caught. More nets doesn't always yield more fish because, as shown in the previous question, stock is inversely correlated to effort. The more fish caught, the less fish there are left in the water to catch in the future. Q23. Using your understanding of the ecological concepts we learned this week, explain why the catch drops so precipitously past the optimal fishing effort. Hint: Allee effects. # Initially, the population begins to drop because because the rate ate which fish are being taken out of the ecosystem is greater than the rate at which new fish are entering the ecosystem. As the population becomes smaller and smaller, additional factors like inbreeding and mate limitation begin to weaken the population. Also, a smaller gene pool contains less diversity, especially if there is inbreeding occurring, meaning the population of fish is more susceptible to dying from environmental factors such as disease, which could explain a very large sudden drop in population. Q24. Explain why this huge sensitivity of the catch beyond the optimal fishing effort is alarming from a sustainability standpoint. Hint: Describe what would happen if the fishery tried to arrive at the optimal fishing effort by trial-and-error, increasing the number of nets gradually until the catch no longer increases. # A trial-and-error approach until the breaking point would be incredibly difficult to recover from as it would not only obliterate a population of fish, but it would also have consequences in the many organisms that rely on the fish. Recovering from the breaking point wouldn't simply require adding new fish back into the ecosystem because by that point the ecosystem would have changed drastically enough that the fish that lived there before may not be able to thrive there anymore. Over fishing today is a major source of habitat destruction and food chain disruption in many parts of the world.