Organisms and Populations Test

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Organisms and Populations Test - 87

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Weekly Quiz Competition

Question 1
1 / -0

Which of the following equations correctly represents Verhulast Pearl logistic growth?

A
$$dN/dt = rN\dfrac({K-N}{K}) $$

B
$$dN/dt = \dfrac{rN}{K}$$

C
$$dN/dt ={N(K-N)}{K}$$

D
$$dN/dt =\dfrac{r(k-n)}{k}$$

Solution

S-shaped growth curve is also called Verhulast-Pearl logistic curve and is represented by the following equation.
$$\dfrac{dN}{dt}= rN(\dfrac{K-N}{K}) = rN(1-\dfrac{N}{K})$$
where $$\dfrac{dN}{dt} =$$ rate of change in population size, r = intrinsic rate of natural increase, N = population density, K= carrying
capacity and $$\dfrac({K-N}{K})$$ = environmental resistance.
Question 2
1 / -0

In 2005, for each of the 14 million people present in a country, 0.028 were born and 0.008 died during the year. Using exponential equation, the number of people present in 2015 is predicted as

A
25 millions

B
17 millions

C
20 millions

D
18 millions

Solution

Exponential equation gives population growth rate as
$$\dfrac{dN}{dt}=rN$$
Here, r = Rate of natural increase
N = Size of original population
$$\dfrac{dN}{dt}=$$ Increase in population size per unit time (1 year in this case)
r is the difference between birth and death rate.
r = 0.028 - 0.008 = 0.02
$$\dfrac {dN}{dt}= 0.02 x14 million = 0.28 million
over 10 years (2005 - 2015), this number will become = 2.8 miillion
Expected population in 2015 = (14+2.8) million
= 17 million (approx),
So, the correct answer is '17 millions'.
Question 3
1 / -0

In a life table, the number of individuals alive at the beginning of the 1 st year to 2 nd year interval is 800. During this interval, 200 individuals die. The death rate for this interval is

A
0.25

B
200

C
800

D
0.2.

Solution

Death rate =$$ \dfrac{Number of deaths}{Initial number of individuals} $$
= $$\dfrac {200}{800} =\dfrac{1} {4} = 0.25$$
Question 4
1 / -0

Percentage of individuals of a given age group in a given population is called as

A
age distribution

B
age density

C
age graph

D
age curve

Solution

Age distribution is percentage of individuals of a given age or age group. A population at any given time is composed of individuals of different ages. If age distribution is plotted for the population, the resulting structure is called as age pyramid.
Question 5
1 / -0

Population of an insect species increases explosively during rainy season. It disappears at the end of the season due to

A
Death of food plants at the end of rainy season

B
J-shaped growth curve.

C
Increase in predator population

D
S-shaped growth curve.

Solution

Population at the end of the season decreases because sources in the environment are not unlimited and even if sources are limited then also given habitat can support only a maximum possible number . This limit is called carrying capacity. So the population is limited due to J shaped curve. So the correct option is "J- shaped growth curve"
Question 6
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If in a population showing sigmoid growth curve the value of K - N/K is 0.95, then what is true for this population.

A
The carrying capacity has been nearly realized.

B
The carrying capacity has not been substantially realized and it can accommodate many more organisms.

C
The carrying capacity is only 50% realized.

D
Data is insufficient to comment anything.

Solution

In the given example, K represents carrying capacity or the maximum capacity that can exist and N represents the actual population present at a particular time. If in a population showing sigmoid growth curve the value of K - N/K is 0.95, then the carrying capacity of this population has not been substantially realized and it can accommodate more organisms. This population has maximum growth rate because it shows that environment can sustain more population.
Thus, the correct answer is 'The carrying capacity has not been substantially realized and it can accommodate many more organisms.'
Question 7
1 / -0

Study the following statements about Adamsia and select your option
A) It is a parasite on clown fish.
B) It is an example of commensalism along with clown fish.
C) It is called sea anemone.
D) It is called hermit crab.

A
A and B are correct

B
B and C are correct

C
C and D are correct

D
A and D are correct
Solution
- Adamsia is also called the sea anemone, and it belongs to the Phylum Cnidaria in the animal kingdom.
- Adamsia shows a symbiotic relationship with clownfish in which the Adamsia get its food with the help of clownfish, and the clownfish is being protected from its enemies due to the presence of stinging cells in the anemone. This type of symbiotic relationship is called the commensalism. Hence, both options B and C are correct.
Question 8
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Multiple Choice Questions.
A sedentary sea anemone gets attached to tho shell lining of hermit crab. The association is

A
Amensalism

B
Ectoparasitism

C
Symbiosis

D
Commensalism.

Solution

Interaction between sea anemone and hermit crab is Proto-cooperation in which both the species are benefited but this interaction is not obligatory. So this relationship is commensalism where both of them are not negatively affected.
So, the correct answer is Commensalism.
Question 9
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Read the following given examples.
(i) Plant-pollinator mutualism
(ii) African catfish in our rivers for aquaculture.
(iii) Fish species with its parasites
(iv) Nile perch and cichlid fish
Which out of these options correctly gives the examples of "Co-extinctions" as a cause of biodiversity loss?

A
(i), (ii)

B
(ii), (iii)

C
(ii), (iv)

D
(i), (iii)

Solution

Coextinction is the phenomenon in which there is loss of a host species which results in the loss of other species that depends on it. This leads to cascading effects across trophic levels. It it observed during mutualism. Loss of pollinator results in the highest number of plant coextinctions as this affects the reproduction in plants. This phenomenon is also observed parasitism. Extinction of a host, fish species results in the extinction of its parasites.
Thus, the correct answer is '(i), (iii).'
Question 10
1 / -0

Population density in India has shifted from 81 per sq. km. in 1921 to _______ per sq. km. in 2011.

A
153

B
248

C
382

D
196
Solution
- Population density refers to the number of people living per square kilometer.
- The formula for population density is $$Dp = $$$$\dfrac{N}{A}$$,
where, Dp = Population density,
N = Total population,
A = Land area covered by the population expressed in square miles or square kilometers.
The population density in India has shifted from 81 per square. kilometer. in 1921 to 382 per square kilometer in 2011.
So, the correct answer is option C.

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Re-Attempt Weekly Quiz Competition

Organisms and Populations Test - 87

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Organisms and Populations Test - 87

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Question 1

$$dN/dt = rN\dfrac({K-N}{K}) $$

$$dN/dt = \dfrac{rN}{K}$$

$$dN/dt ={N(K-N)}{K}$$

$$dN/dt =\dfrac{r(k-n)}{k}$$

Solution

Question 2

25 millions

17 millions

20 millions

18 millions

Solution

Question 3

0.25

200

800

0.2.

Solution

Question 4

age distribution

age density

age graph

age curve

Solution

Question 5

Death of food plants at the end of rainy season

J-shaped growth curve.

Increase in predator population

S-shaped growth curve.

Solution

Question 6

The carrying capacity has been nearly realized.

The carrying capacity has not been substantially realized and it can accommodate many more organisms.

The carrying capacity is only 50% realized.

Data is insufficient to comment anything.

Solution

Question 7

A and B are correct

B and C are correct

C and D are correct

A and D are correct

Solution

Question 8

Amensalism

Ectoparasitism

Symbiosis

Commensalism.

Solution

Question 9

(i), (ii)

(ii), (iii)

(ii), (iv)

(i), (iii)

Solution

Question 10

153

248

382

196

Solution

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