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g100num [7]
3 years ago
13

A hypothetical bat species (species 1) lives in a city. Another hypothetical bat species (species 2) established a population in

the same city after a number of individuals escaped from a zoo. In isolation, each species prefers to roost in buildings that are three or more stories tall.
After bat species 2 becomes established in the city, you observe that bat species 1 is only found roosting at the top of short buildings that are less than 3 stories tall, while bat species 2 roots only at the top of building that are three or more stories tall.
Roosting areas in buildings of any height are the _________________ of both bat species.
A. adaptation
B. fundamental niche
C. resource partitioning
D. realized niche
E. competitive exclusion
Biology
1 answer:
nikitadnepr [17]3 years ago
3 0
<h2>B) is the correct option </h2>

Explanation:

In the given hypothetical condition roosting areas in buildings of any height are the fundamental niche of both bat species

  • Niche is the role played by an organism that how they can live in an environment including their diet and shelter
  • Fundamental niche represents all the environmental conditions where a species is able to live or can live
  • There is no ecological barriers in case of this type of niche
  • Since both bat species 1 and 2 can roost in buildings of any height this means there is no ecological barrier and heights of building represent all the conditions where bat species can live so it clearly represents fundamental niche
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Incomplete dominance is the inheritance pattern where the dominant allele did not mask the recessive allele completely and form a mix of both alleles. Here the inheritance is the incomplete inheritance. The ratio of F2 generation is 1:2:1.

Given:

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Phenotype     Observed(O)    Expected (E)    O-E      (O-E)2     (O-E)2/E

Blue                 42                   43.44               -1.44       2.0736     0.0477

cyan                86                   83.46                 2.54     6.4516     0.0773

green              39                    40.1                    -1.1       1.2100      0.0302

Total              167                     167                                               0.1552

Chi-square value = 0.155

Degrees of freedom = no. of phenotypes – 1

Df = 3-1 = 2

Critical value = 5.99

Chi-square value of 0.155 is less than the critical value of 5.99. So we accept the null hypothesis.

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