It seems that this question missed the choices which are as follows:
<span><span>A. </span></span>Positive impact of a density-dependent limiting factor<span><span /><span>B. </span></span>Positive impact of a hurricane<span><span /><span>C. </span></span>Positive impact of beavers in an ecosystem<span><span /><span>D. </span></span>Positive impacts of dams on society
The correct answer that best describes the given characteristics above which are water storage, clean energy and flood prevention is option D. Positive impacts of dams on society.
C and D are the answers. A is incorrect because of asexual reproduction. B is incorrect due to it have the same number. E is not correct because meiosis is sexual reproduction.
D = m/v
Given:
D: .835 g/cm3
V: 34 cm3
M: ?
M= Dxv
M= .835 g/cm3 x 34 cm3
M= 28.39 g
Answer:
C: competitive exclusion
Explanation:
<em>The competitive exclusion principle states that organisms living in the same community while competing for the same resources cannot coexist at a constant population rate. </em>
Once some of the species within the community get a slight competitive edge over other species, they become dominant and this might lead to the extinction of the weaker species in the long run.
<u>In the experimental plot, the removal of sea stars provided mussel and barnacle with a competitive advantage over other species within the community (sea stars are predators of mussels and barnacles). This led to the dominance of mussel and barnacle and the eventual extinction of other species within the experimental plot as compared to the control plot.</u>
The correct answer is C.
Answer:
Explanation:
To calculate the recombination frequency, we have to know that 1% of recombinations = 1 map unit = 1cm. And that the maximum recombination frequency is always 50%.
The map unit is the distance between the pair of genes for which every 100 meiotic products, one of them results in a recombinant one.
So, en the exposed example:
- J and K are autosomal genes
- J and K are separated by 60 M.U.
- 60 M.U. means that there is 60% of recombination.
Cross) J K / j k x j k / j k
Gametes) JK Parental jk, jk, jk, jk
jk Parental
Jk Recombinant
jK Recombinant
One map unit equals 1% of recombination frequency. This means that every 100 meiotic products, one of them is a recombinant one.
1 M.U. -------------- 1% recombination
60 M.U. ------------ 60% recombination
30% Jk + 30% jK
100 M.U. - 60 M.U. = 40 M.U.
40M.U.--------------40 % Parental (Not recombinant)
20% JK + 20% jk
Punnet Square) JK jk Jk jK
jk JK/jk jk/jk Jk/jk jK/jk
J K / j k = 20%
j k / j k = 20%
J k / j k = 30%
j K / j k = 30%
