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If a population of 50 paramoecium present in a pool increases to 150 after an hour, what would be the growth rate of population?
(d)
Initial population of paramecium \(\mathrm{Pi}=50\)
After one hour population of paramecium \(\mathrm{Pf}=150\)
Growth rate after one hour \(=\mathrm{Pf}-\mathrm{Pi}\)
\(
=150-50=100
\)
What would be the percent growth or birth rate per individual per hour for the same population mentioned in the previous question (Question 2)?
(b) \(\begin{aligned} & \text { The per cent growth } \\ & \text { or birth rate per } \\ & \text { individual per hour }\end{aligned}=\frac{\text { Final population }- \text { Initial population }}{\text { Initial population }} \times 100\)
\(
\begin{aligned}
& =\frac{150-50}{50} \times 100 \\
& =200
\end{aligned}
\)
A population has more young individuals compared to the older individuals. What would be the status of the population after some years?
(c) A population has more young individuals compared to the older individuals. It will increase the status of the population after some years.
What parameters are used for tiger census in our country’s national parks and sanctuaries?
(b) Sometimes population size is indirectly estimated without actually counting them or seeing them. E.g. ; The tiger census in our National Parks and Tiger Reserves is often based on pug marks and faceal pellets.
Which of the following would necessarily decrease the density of a population in a given habitat?
(c) Mortality and emigration would necessarily decrease the density of a population in a given habitat.
A protozoan reproduces by binary fission. What will be the number of protozoans in its population after six generations?
(c)
Population after nth generations \(=2 n\)
Population after 6th generations \(=26=64\)
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
\(
\begin{aligned}
&\text { Answer. (b) }\\
&\begin{aligned}
& \frac{d N}{d t}=r N \\
& r=0.028-0.008=0.02 \\
& d t=2015-2005=10 \text { years } \\
& N=N_{2005}=14 \text { million } \\
& d N=\text { change in population in } 10 \text { years } \\
& \frac{d N}{d t}=0.02 \times 14 \\
& d N=0.02 \times 14 \times 10 \\
& \begin{aligned}
d N & =2.8 \text { million } \\
d N & =N_{2015}-N_{2005} \\
N_{2015} & =d N+N_{2005} \\
& =2.8+14 \\
& =16.8 \approx 17 \text { million }
\end{aligned}
\end{aligned}
\end{aligned}
\)
Amensalism is an association between two species where
(b)
Lichens are the associations of
(c) Lichens are the associations of fungus and algae.
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