Ecology Practice - Population Growth

If a problem says: calculate the population growth/size for a species in the next year experiencing exponential growth.... which equation should you use?

dN/dt = rN r = growth rate N = population size

If a problem gives you the carrying capacity..... which equation should you use?

dN/dt = rN((K-N)/K) Since K = carrying capacity, you need to use the equation that uses K!

How many squirrels die if the death rate is .12 and the population is 252? Round to the nearest whole number.

m = .12 N = 252 D = N / m D = 252 / .12 D = 72 72 squirrels die

If you are given number of deaths and births.... which equation gives you growth rate?

r = (B-D) / N B = # of births D = # of deaths N = population size

If you are given per capita birth rate and per capita death rate.... which equation gives you growth rate?

r = b - m because b = birth rate and m = death rate

If a problem give you the initial population size and a rate of birth... which equation do you use?

r = b - m dN/dt = rN When given no m (death rate) you can assume that the m = 0. So... r = b - 0 r = b

There are 252 deer in a population. There is no net immigration or emigration. If 47 deer die and 32 deer are born in one year, what is the population size at the end of the month? Round your answer to the nearest whole number.

B = 47 D = 32 N = 252 r = (B-D) / N r = (47-32) / 252 r = 15 / 252 r = .05952 dN/dt = rN dN/dt = .05952 (252) = 14.99 15 individuals are added Population size at the end of the month is 252 + 15 = 267

A hypothetical population has a carrying capacity of 1,500 individuals and r(max) is 1.0 over a year. What is the population size after 1 year with an initial population of 1600?

N = 1600 r = 1 K = 1500 dN/dt = rN ( (K-N)/K) = 1(1600) ((1500-1600)/1500) = -107 Population size = N + dN/dt = 1600 + (-107) = 1,493

A hypothetical population has a carrying capacity of 1,500 individuals and r(max) is 1.0 over a year. What is the population growth after 1 year with an initial population of 1750?

N = 1750 r = 1 K = 1500 dN/dt = rN ( (K-N)/K) = 1(1750) ((1500-1750)/1500) = -292 Growth is negative. The population loses 292 individuals.

A hypothetical population has a carrying capacity of 13,500 individuals and r(max) is .03 over a year. What is the population size after 30 years with an initial population of 2000?

N = 2000 r = .03 K = 13,500 N(t) = N (1+r)^t = 2000 (1 + .03) ^30 = 2000 (1.03) ^30 = 2000 (2.427) = 4,854.52 Growth is positive. The population is 4,855.

A hypothetical population has a carrying capacity of 13,500 individuals and r(max) is .03 over a year. What is the population growth after 1 year with an initial population of 2000?

N = 2000 r = .03 K = 13,500 dN/dt = rN((K-N)/K) = .03(2000) ((13500-2000)/13500)) = 60 (.85185) = 51 Growth is positive. The population adds 51 individuals

In a population of 400 deer, the per capita birth rate is 0.16 per year and the per capita death rate is 0.22 per year. What will be the total population in the next year?

N = 400 b = .16 m = .22 r = b - m = .16 - .22 = -.06 dN/dt = rN = -.06 (400) = -24 Growth is negative. The population will decrease by 24 individuals in the year. 400 - 24 = 376 Next year the population is 376 deer.

In a population of 400 deer, the per capita birth rate is 0.16 per year and the per capita death rate is 0.22 per year. What would be the total population in 30 years?

N = 400 b = .16 m = .22 r = b - m = .16 - .22 = -.06 dN/dt = rN = -.06 (400) = -24 N(t) = N (1+r)^t = 400 (1+ (-.06))^30 = 400 (.94)^30 = 400 (.156255) = 62.5022 Growth is negative. This equation shows us that in 30 years the population will only be 62 deer.

In a population of 750 fish, 25 die on a particular day while 12 are born. What is the per capita growth rate for the day?

N = 750 B = 12 D = 25 r = (B-D) / N r = (12-25) / 750 r = (-13) / 750 r = -.017

If a problem says: calculate the population growth/size for a species 20 years from now experiencing exponential growth.... which equation should you use?

N(t) = N (1+r)^t r = growth rate N = population size t = time interval (aka years)

How many squirrels are born if the birth rate is .06 and the population is 252? Round to the nearest whole number.

b = .06 N = 252 B = N / b B = 252 / .06 B = 36 36 squirrels are born

In a populations of 600 squirrels, the per capita birth rate in a particular period is .06 and the per capita death rate is .12. What is the per capita growth rate? What kind of growth is it?

b = .06 m = .12 N = 600 r = b - m r = .06 - .12 = -.06 r = -.06 It is negative growth because r is less than 1.

A population of bobcats has an annual per capita birth rate of 0.08 and an annual per capita death rate of 0.02. Calculate an estimate of the number of individuals added to (or lost from) a population of 1,000 individuals in one year.

b = .08 d = .02 N = 1000 r = b - m = .08 - .02 = .06 dN/dt = rN = .06 (1000) = 60 Growth is positive. 60 bobcats are added to the population.