Questions and assignments to Lecture

1. Population numbers of cockroaches double every month (30 d). What is their intrinsic rate of increase (per day)?

2. What is the intrinsic rate of increase in a human population if every family has 3 children at parent's age of 30 (there are no singles, no divorces, sex ratio 1:1)? What would be the numbers of human population after 100 years if initial numbers are 4 billion?

3. A new lake was created after building a dam. The number of fish censused after 2, 4, 6, 8 and 10 years since that time was 1000, 2000, 3500, 5000 and 6000. Estimate parameters of the logistic model using non-linear regression. Plot the data and the model on one graph.

4. Use Excel to simulate population dynamics with the discrete-time logistic model (Ricker's model) for 60 generations. Use K=100; r = 0.1, 0.5, 1.0, 1.5, 1.9, 2.2;

Lecture 6. Life-tables and k-values

In this lecture you will learn how to collect data for the analysis of population processes.

Analysis of population processes is as easy as balancing your personal budget! You need to estimate the increases and losses in population numbers due to different processes. If you are lucky, then the net change in population numbers will be equal to the algebraic sum of the effects of all studied processes. If you are less lucky, then some of the increases or losses in population numbers will be missing. In this case, the non-attributed mortality is considered as the effect of some unknown factor, we can call it "winter mortality" or "mystery disease". Additional research can be done later to study these unknown processes.

Ecological processes are usually specific to organisms' age or stage. Thus, they have to be recorded relative to the life-cycle stage. This information is usually called a "life-table". Two types of life-tables are generally used: (1) age-dependent and (2) stage-dependent.

Age-dependent life-tables

Age-dependent life table shows organisms' mortality (or survival) and reproduction rate (maternal frequency) as a function of age. In nature, mortality and reproduction rate may depend on numerous factors: temperature, population density, etc. When building a life-table, the effect of these factors is averaged. Only age is considered as a factor that determines mortality and reproduction.

Example. Consider a sheep population which is censused once a year immediately after breeding season:

Age, years (x)

Probability of surviving to age x

(lx)

No. of female offspring born to a mother of age x

(mx)

0

1.000

0.000

1

0.845

0.045

2

0.824

0.391

3

0.795

0.472

4

0.755

0.484

5

0.699

0.546

6

0.626

0.543

7

0.532

0.502

8

0.418

0.468

9

0.289

0.459

10

0.162

0.421

Only females are considered in this life-table. However, there is no problem to include male populations into the life table. Then, survival rates should be specified separately for males and females, and the sex ratio of offspring should be taken into account.

Survivorship curves

Survival probabilities l are often plotted against age x. These graphs are called "survivorship curves". They show, at what age death rates are high and low. The following graphs show two survivorship curves for domestic sheep (data from Caughley 1967) and for lapwings or green plovers (Vanellus vanellus) in Britain (data from Deevey 1947):

Survivorship curve is exponential (with negative growth) for the lapwing. This means that survival rate is independent of age. In the log-scale, survivorship curve becomes a straight line (see above).

Age-specific mortality is estimated using equation:

Sheep mortality generally increases with age; and the slope of survivorship curve becomes steeper to the end. Humans have a similar shape of survivorship curve.

Lapwings Density

Characteristics derived from life-tables

Net reproductive rate, R0, is the average number of female offspring born to a sheep considered at age 0. Consider N new-born sheep. Some of them will die without producing any offspring (zero offspring), others will produce one or several offspring. Ro is the average number of female offspring produced in the entire group of N sheep.

In our example, R0 = 1x0 + 0.845x0.045 + 0.824x0.391 + ... = 2.513. It means that an ewe produces 2.513 ewe lambs in average in a lifetime.

Average generation time, T, is estimated using equation:

In our example, T = (0x1x0 + 1x0.845x0.045 + 2x0.824x0.391 + ...) / 2.513 = 12.825 / 2.513 = 5.1 years. It means that the average age of mothers when they give birth to an eve lamb is 5.1 years.

Note: If organisms breed continuously, then generation time will be overestimated using this equation because all births are summed over the period between census dates which is equal to one age step. Generation time can be adjusted by subtracting half of age step.

Now it is possible to estimate approximate value of intrinsic rate of increase r using the following logic. We assume discrete generations with generation time T=5.1 years and net reproductive rate of R0 =2.513. If population size at zero time was N0, then after T years, the population will grow to NT = N0xR0. According to an exponential model,

N^Mo OtppT)

where ln is natural logarithm (logarithm with base e =2.718). We got the equation for r:

iaad

Note: This is an approximate estimation of r because we used a simplified assumption that generations are discrete. Accurate estimation of r will be discussed in the next chapter.

Time units used for age measurement

If survival and reproduction are continuous processes without any cyclic change, then any time units may be suitable: days, weeks, months, years. Time units should provide sufficient resolution. For example, if the life span of an organism is 2 months, then taking 1 month as a unit will result in only two age intervals which is definitely not sufficient. In most cases, the number of age intervals is in the range from 10 to 50.

If survival or reproduction are cyclic (e.g., seasonal), then one cycle can be taken as a time unit. In this case it may happen that the number of age intervals will drop to 2 or 3. However, it may be not very dangerous if reproduction is limited to a short period within the year because there will be little age difference between organisms born in the same year.

If survival and reproduction are cyclic but the entire life span is less or equal to this cycle, then time units should be smaller than the cycle length. Age-dependent life-tables can be built for the entire population only if the breeding period is short and therefore organisms' development is synchronized. Otherwise, separate life-tables should be built for subpopulations that start their development at different seasons.

Determining survival of organisms till age x (lx)

For domestic animals or for populations reared in the laboratory, it is possible to observe the fate of a large group of individuals that all started life simultaneously.

Survivors can be counted at regular time intervals and l values can be easily estimated. Similar technique can be applied to non-moving organisms (plants, sedimentary animals). It is possible to mark a large number of individuals and to trace their fate.

This kind of analysis is usually impossible in populations of moving organisms. There are two principal methods to deduce l values in this case: by determining the age distribution in the population, or by determining the ages at death.

If the population is stationary (i.e., population numbers and age distribution do not change) then the number of new-born organisms x time units ago was the same as now; and survivors of that group of organisms are of age x. Thus,

where N(x) is the number of organisms of age x. Here we assumed that age can be accurately measured. In many species the number of "growth rings" in specific organs is equal to age in years. Examples of such organs are: stems of trees, scales of fishes, horns in sheep, roots of canine teeth in bears, otholits in fishes. The weight of the eye lens can be used for age measurement in some animal species. However, in many populations, measuring age is a difficult problem.

Assumption of population stationarity usually is taken a-priori if no historical data exist. In cases when age distribution has a periodic component, we can speculate that it resulted from fluctuations in population numbers. Periodic component can be filtered out using regression methods.

Consider a large number of carcasses whose age have been determined. We assume that the probability of detecting a carcass does not depend on the age of animal at death. The proportion of individuals that were at age x when they died is dx. These individuals survived to age x but did not survive to age x+1. Thus, d"= ;

The method of estimation of l from age of carcasses also requires stability of population numbers and of age structure.

Determining reproduction rates (mx)

Reproduction rate (=maternal frequency) is equal to the number of female offspring produced per one mother in age interval from x to x+1. Survival of mothers and offspring during this time interval should be included into m ; i.e., fflx is equal to the total number of offspring produced in one time interval and survived till the end of this period divided by the initial number of parent females at the beginning of the time interval.

Reproduction rates are easy to determine in laboratory reared animals or plants. In natural populations of mammals, maternal frequencies can be derived from the proportion of pregnant and/or lactating animals. In birds, maternal frequencies can be determined from the average number of chick per nest. Indirect measures of maternal frequencies should be used with caution because they may be biased.

Stage-dependent life-tables

Stage-dependent life tables are built in the cases when:

• The life-cycle is partitioned into distinct stages (e.g., eggs, larvae, pupae and adults in insects)

• Survival and reproduction depend more on organism stage rather than on calendar age

• Age distribution at particular time does not matter (e.g., there is only one generation per year)

Stage-dependent life tables are used mainly for insects and other terrestrial invertebrates.

Example. Gypsy moth (Lymantria dispar L.) life table in New England (modified from Campbell 1981)

Stage

Mortality factor

Initial no. of insects

No. of deaths

(s)

k-value [-ln(s)]

Egg

Predation, etc.

450.0

67.5

0.150

0.850

0.1625

Egg

Parasites

382.5

67.5

0.176

0.824

0.1942

Larvae I-III

Dispersion, etc.

315.0

157.5

0.500

0.500

0.6932

Larvae IV-VI

Predation, etc.

157.5

118.1

0.750

0.250

1.3857

Larvae IV-VI

Disease

39.4

7.9

0.201

0.799

0.2238

Larvae IV-VI

Parasites

31.5

7.9

0.251

0.749

0.2887

Prepupae

Desiccation, etc.

23.6

0.7

0.030

0.970

0.0301

Pupae

Predation

22.9

4.6

0.201

0.799

0.2242

Pupae

Other

18.3

2.3

0.126

0.874

0.1343

Adults

Sex ratio

16.0

5.6

0.350

0.650

0.4308

Adult females

10.4

TOTAL

439.6

97.69

0.0231

3.7674

Specific features of stage-dependent life tables:

• There is no reference to calendar time. This is very convenient for the analysis of poikilothermous organisms.

• Gypsy moth development depends on temperature but the life table is relatively independent from weather.

• Mortality processes can be recorded individually and thus, this kind of life table has more biological information than age-dependent life tables.

K-values

K-value is just another measure of mortality. The major advantage of k-values as compared to percentages of died organisms is that k-values are additive: the k-value of a combination of independent mortality processes is equal to the sum of k-values for individual processes.

Mortality percentages are not additive. For example, if predators alone can kill 50% of the population, and diseases alone can kill 50% of the population, then the combined effect of these process will not result in 50+50 = 100% mortality. Instead, mortality will be 75%!

Survival is a probability to survive, and thus we can apply the theory of probability. In this theory, events are considered independent if the probability of the combination of two events is equal to the product of the probabilities of each individual event. In our case event is survival. If two mortality processes are present, then organism survives if it survives from each individual process. For example, an organism survives if it was simultaneously not infected by disease and not captured by a predator.

Assume that survival from one mortality source is si and survival from the second mortality source is s2. Then survival from both processes, si2, (if they are independent) is equal to the product of si and s2:

This is a "survival multiplication rule". If survival is replaced by 1 minus mortality [s=(1-d)], then this equation becomes:

For example, if mortality due to predation is 60% and mortality due to diseases is 30%, then the combination of these two death processes results in mortality of d = 1-(1-0.6)(1-0.3)=0.72 (=72%).

Varley and Gradwell (1960) suggested to measure mortality in k-value which is the negative logarithms of survival:

We use natural logarithms (with base e=2.718) instead of logarithms with base 10 used by Varley and Gradwell. The advantages of using natural logarithms will be shown below.

It is easy to show that k-values are additive:

The k-values for the entire life cycle (K) can be estimated as the sum of k-values for all mortality processes:

In the life table of the gypsy moth (see above), the sum of all k-values (K = 3.7674) was equal to the k-value of total mortality.

This graph shows the relationship between mortality and the k-value. When mortality is low, then the k-value is almost equal to mortality. This is the reason why the k-value can be considered as another measure of mortality. However, at high mortality, the k-value grows much faster than mortality. Mortality cannot exceed 1, while the k-value can be infinitely large.

The following example shows that the k-value represents mortality better than the percentage of dead organisms: One insecticide kills 99% of cockroaches and another insecticide kills 99.9% of cockroaches. The difference in percentages is very small (<1%). However the second insecticide is considerably better because the number of survivors is 10 times smaller. This difference is represented much better by k-values which are 4.60 and 6.91 for the first and second insecticides, respectively.

Key-factor analysis Varley and Gradwell (1960) developed a method for identifying most important factors "key factors" in population dynamics. If k-values are estimated for a number of years, then the dynamics of k-values over time can be compared with the dynamics of the generation K-value. The following graph shows the dynamics of k-values for the winter moth in Great Britain.

It is seen that the dynamics of winter disappearance (k1) is most resembling the dynamics of total generation K-value. The conclusion was made that winter disappearance determines the trend in population numbers (whether the population will grow or decline), and thus, it can be considered as a "key factor". There were numerous attempts to improve the method. For example, Podoler and Rogers (1975, J. Anim. Ecol, 44(1)) suggested regressing k over K.

But, this method was criticized recently because the meaning of a "key" factor was not explicitly defined (Royama 1996, Ecology). It is not clear what predictions can be made from the knowledge that factor A is a key-factor. For example, the knowledge of key-factors does not help us to develop a new strategy of pest control.

The key-factor analysis was often considered as a substitute for modeling. It seems so easy to compare time series of k-values and to find key-factors without the hard work of developing models of ecological processes. However, reliable predictions can be obtained only from models.

This critique does not mean that life-tables have no value. Life-tables are very important for gathering information about ecological processes which is necessary for building models. It is the key-factor analysis that has little sense.

K-value = instantaneous mortality rate multiplied by time. A population that experience constant mortality during a specific stage (e.g., larval stage of insects) change in numbers according to the exponential model with a negative rate r. We cannot call r intrinsic rate of natural increase because this term is used for the entire life cycle, and here we discuss a particular stage in the life cycle. According to the exponential model:

Population numbers decrease and thus, Nt < N0. Survival is: s = N/N0 . Now we can estimate the k-value:

Instantaneous mortality rate, m, is equal to the negative exponential coefficient because mortality is the only ecological process considered (there is no reproduction):

Exponential coefficient r is negative (because population declines), and mortality rate, m, is positive.

We proved that if mortality rate is constant, then k-value is equal to the instantaneous mortality rate multiplied by time. This is analogs to physics: distance is equal to speed multiplied by time. Here, instantaneous mortality rate is like speed, and k-value is like distance. K-value shows the result of killing organisms with specific rate during a period of time. If the period of time when mortality occurs is short then the effect of this mortality on population is not large.

If instantaneous mortality rate changes with time, then the k-value is equal to its integral over time. In the same way, in physics, distance is the integral of instantaneous speed over time.

Example. Annual mortality rates of oak trees due to animal-caused bark damage are 0.08 in the first 10 years and 0.02 in the age interval of 10-20 years. We need to estimate total k-value (k) and total mortality (d) for the first 20 years of oak growth.

Thus, total mortality during 20 years is 63%.

Limitation of the k-value concept. All organisms are assumed to have equal dying probabilities. In nature, dying probabilities may vary because of spatial heterogeneity and individual variation (both inherited and non-inherited).

Estimation of k-values in natural populations. Estimation of k-values for individual death processes is difficult because these processes often go simultaneously. The problem is to predict what mortality could be expected if there was only one death process. In order to separate death processes it is important to know the biology of the species and its interactions with natural enemies. Below you can find several examples of separation of death processes.

Example #1. Insect parasitoids oviposit on host organisms. Parasitoid larva hatches from the egg and starts feeding on host tissue. Parasitized host can be alive for a long period. Finally, it dies and parasitoid emerges from it. Insect predators usually don't distinguish between parasitized and non-parasitized prey. If an insect was killed by a predator, then it is usually impossible to detect if this insect was parasitized before. Thus, mortality due to predation is estimated as the ratio of the number of insects numbers destroyed by predators to the total number of insects, whereas mortality due to parasitism is estimated as the ratio of the number of insects killed by parasitoids to the number of insects that survived predation. In this example, predation masks the effect of parasitism, and thus, insects killed by predators are ignored in the estimation of the rate of parasitism. The effect is the same as if predation occurred before parasitism in the life cycle. Thus, in the gypsy moth life table, predation was always considered before parasitism. Diseases also mask the effect of parasitism and thus they are considered before parasitism.

Example #2. It is often possible to distinguish between organisms destroyed by different kinds of predators. For example, small mammals and birds open sawfly cocoons in a different way. Suppose, 20% of cocoons were opened by birds, 50% were opened by mammals, and remaining 30% were alive. The question is what would be the rate of predation if birds and mammals were acting alone. We assume that sawfly cocoons have no individual variation in predator attack rate, and that cocoons destroyed by one predator cannot be attacked by another predator. First, we estimate total k-value for both predator groups: ki2 = -ln(0.3) = 1.204. Second, we subdivide the total k-value into two portions proportionally to the number of cocoons destroyed by each kind of predator. Thus, for birds ki = 1.204x20/(20+50) = 0.344, and for mammals k2 = 1.204^50/(20+50) = 0.860. The third step is to convert k-values into expected mortality if each predator was alone: for birds di = 1- exp(-0.344) = 0.291, and for mammals d2 = 1- exp(-0.860) = 0.577.

Questions and assignments to Lecture 6

1. Build a life table for an aphid population (aphids reproduce parthenogenetically). Estimate lx, dx, mx, Ro, T, and r. (See a picture of aphids!)

Age, days (x)

Number of survivals

Mean number of offsprings per parent

0

1000

0

1

900

0

2

820

0

3

750

0

4

680

0

5

620

0

6

550

1

7

500

2

8

450

5

9

400

10

10

350

12

11

300

10

12

250

8

13

200

6

14

100

3

15

50

1

16

0

0

2. Partial life-table. The European pine sawfly, Neodiprion sertifer, cocoons were collected at the beginning of August and dissected. Results of dissection of new (current year) cocoons are the following:

Healthy sawfly eonymph

144

Eaten by predators

125

Exit hole of parasitoid Drino inconspicua

15

Exit hole of parasitoid Pleolophus basizonus

78

Larvae of parasitoid Exenterus abruptorius

210

Exit hole or larvae of gregarious parasitoid Dahlbominus fuscipennis

23

Fungus disease

205

Total

800

Life-cycle information:

Excellent images of parasitoids are available from the PHERODIP homepage.

• Parasitoids D.inconspicua, P.basizonus and D.fuscipennis have several generations per year, whereas E.abruptorius has only 1 generation.

• D.inconspicua (Tachinidae) is an endoparasite and attacks larvae (4-5 instar). It emerges from the host immediately after host cocooning. It develops very fast and wins the competition with any other parasitoids.

• E.abruptorius is an ectoparasite, attacks host eonymphs a day prior to cocooning. Parasitoid larvae emerges inside the cocoon, eats the host and overwinters as larvae inside host cocoon. If the host was previously parasitized by D.inconspicua, then E.abruptorius dies.

• P.basizonus and D.fuscipennis attack host cocoons. They are ectoparasites. If another parasite (E.abruptorius) is already present in the cocoon, it will be eaten first. D. fuscipennis wins the competition with P.basizonus.

Estimate mortality caused by each natural enemy, convert it into k-value. Check that the sum of all k-values is equal to the total k-value for sawfly cocoons. Write results in the table, putting mortality processes in the order of their operation.

Simple example:

Healthy eggs

200

Desiccated eggs

100

Parasitized eggs

200

Total

500

Mortality process

Number of eggs in which this mortality process can be detected

Number of killed eggs

Mortality

Survival

k-value

1. Desiccation

500

100

0.2

0.8

0.223

2.Parasitism

400 (500-100)

200

0.5

0.5

0.693

Total

500

300

0.6

0.4

0.916

3. Estimate mortality in a predator-exclusion experiment. The fall webworm, Hyphantria cunea, larvae in colonies were counted at the beginning and at the end of experiment:

A. Control - without protection

B. Exclusion of large predators: prey were protected by a 1/2 inch cell hardware cloth

C. Exclusion of all predators: prey were protected by 1 mm cell mesh Estimate: mortality caused by large and small predators, convert it to k-value.

Initial larvae in 10 colonies

Larvae alive at the end

Control

3000

1400

Large predators excluded

3500

2900

All predators excluded

3200

3100

4. Estimate gypsy moth mortality due to virus (NPV). Gypsy moth larvae were collected in the forest at 7-day intervals and placed individually in the cups with diet. Incubation period of viral infection (from infection till death) is 7 days. Estimate: total mortality caused by virus and the k-value.

Collected larvae

Larvae that died in 7 days since collection

1-st week

200

23

2-nd week

200

7

3-rd week

200

5

4-th week

200

30

5-th week

200

58

6-th week

200

115

5. Estimate the rate of simultaneous mortality processes Gypsy moth (Lymantria dispar) pupae are destroyed by small mammals and by invertebrates. 300 laboratory-reared pupae were placed on tree boles. Three days later, 200 of them were damaged by small mammals and 50 were damaged by invertebrate predators (Calosoma sycophanta). Each pupa can be eaten just once. Estimate mortality caused by each predator guild if another predator guild was absent (note: use k-values!).

Lecture 7. Model of Leslie

The model of Leslie is one of the most healivy used models in population ecology. This is a discrete-time model of an age-structured population which describes development, mortality, and reproduction of organisms. The model is formulated using linear algebra. This model is mostly used to answer the following two questions:

1. What is the rate of exponential growth (intrinsic rate of increase)?

2. What is the proportion of each age class in the stable age distribution?

7.1. Model Structure

The model of Leslie (1945) describes 3 kinds of ecological processes:

1. Development (progress through the life cycle)

2. Age-specific mortality

3. Age-specific reproduction

Variables and parameters of the model:

• Nx, t = number of organisms in age x at time t (age is measured in the same units as time t). Usually, only females are considered and males are ignored because, as a rule, the number of males does not affect population growth.

• sx = survival of organisms in age interval from x to x+1.

• mx = average number of female offsprings produced by 1 female in age interval from x to x+1 (mortality of parent and/or offspring organisms is included)

There are two equations:

+1 0

Responses

  • lucio
    What is the intrinsic rate of increase in a human population if every family has answer 3 children?
    6 years ago

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