ECOLOGY IMPORTANT NUMERICALS

1. The effective population size for completely monogamous species having 40 males and 10 females would be - (2015 dec)
a.42                                               c.20
b.32                                              d.10                 Given -
Ans.   NE= 4 NmNf/Nm+Nf                                    Nm-Number of males(40)
              =4 40.10/40+10                                        Nf=Number of females(10)
             =1600/50
             =32        (Ans is  b)                                   NE=Effective population size ?                               

2.  D = Σ n(n-1)/N(N-1) (Simpson’s Diversity Index)

Example:i

A lake contains 934 brown trout, 733 smallmouth bass, 34 catfish, 2003 carp, 234 steelheads, and 32 northern pikes.

Fish	Number	n(n-1)
Brown Trout	934	934(934-1) = 871422
Smallmouth Bass	733	733(732)=536556
Catfish	34	34(33)=1122
Carp	2003	2003(2002)=4010006
Steelheads	234	234(233)=54522
Northern Pike	32	32(31)=992
Total	N = 3970 N(N-1) = 15756930	Σ n(n-1) = 5474620

 

D=5474620/15756930

4. A fisheries biologist is maximizing her fishing yield by maintaining a population of lake trout at exactly 500 individuals.  Predict the initial instantaneous population growth rate if the population is stocked with an additional 600 fish.  Assume that r for the trout is 0.005 individuals/(individual*day). 

 Ans.For a populations growing according to the logistic equation, we know that the maximum population growth rate occurs at K/2, so K must be 1000 fish for this population.    

If the population is stocked with an additional 600 fish, the total size will be 1100.  From the logistic equation, the initial instantaneous growth rate will be: DN/dt  =  rN [1- (N/K)]   

 = 0.005(1100)[1-(1100/1000)] = = -0.55 fish / day

5.In a population of 10 million individuals birth rate is 19 per 1000 and death rate is 14 per 1000.Annual rise in population would be-

a.50,000                                                          c.14000

b.5000                                                             d.500000                 (2008)

Ans. a 

Birth rate =19/100*1000=190

Death rate =14/100*1000=140

Rise =b-d= 190-140=50`]

Annual rise=50*1000=50,000

7.The population size of a bird increased from 600-645 in one year.If the per capita birth rate of this population is 0.125 . what is the per capita death rate?

Ans. 0.05  b=B/N  d=D/N

  B=b*N=  12.5/100*600=75

  Death by the number in the population = 75-45=30

    d=D/N =30/600*100=5%=0.05

8.In an experimental population the birth rate is 18 per 1000 and death rate is 14 per 1000.Size of population is 10,000 at time,t,then what will be size of population at timee t+1

a.10,000                                                             c.10,040

b.10,140                                                             d.11,040

Ans.Birth rate is 18/100*1000=180

      Death rate is 14/100*1000=140

rate =b-d=180-140=40

so,the size of population at time t+1= 10000+40=10,040

9.The population density of an insect increases from 40-46 in one month .If the birth rate during that period is 0.4 .What is the death rate ?

a.0.25                                                   c.0.15

b.0.87                                                   d.0.40

Ans.c   Death rate=Final population-Initial population/Birth rate

                          =46-40/0.4

                         =6/0.4=   15% i,e 0.15

10.The population was expected to double within 50 years .Calculate the r for the population.

      tdouble =ln2/r

    r=ln2/t=ln2/50=0.0139

    
11.No. of mice =2000
     Born rate =1000
     Death rate=200
What is the Per capita growth rate over month ?
Ans.  Growth rate =rmax= b-d= 2000-200=800
Per capita growth rate =rmax/N=  800/2000=2/5=0.4 

12.A population is growing logistically with a growth rate =o.5/week ,with a carrying capacity =400.
What is the max.growth rate that this population can achieve?
Ans. K= 400
        r=o.5 
         N=k/2=400/2 =200
dN/dt=rN (1-N/K)
        =100*(200/400)

•         =50

NOTE -1.To  calculate the basic reproductive rate, R₀,  formula is 

R₀ = l_{xm x .      ( LIFE TABLE )} 2.dP/dt=rP-aCP   (prey)    PREY r =INTRINSIC GROWTH RATE OF PRAY

                                                     P=PREY DENSITY

                                                     a=PREDATION EFFICIENCY

                                                     C=PREDATION DENSITY

 dC/dt=B(aCP)-DC  predator         B=ASSIMILATION AFFICIENCY

                                                   D=DEATH RATE OF PREDATOR       (LOTKA VOLTERA MODEL )

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