Answer:
Speciation results in biodiversity.
Explanation:
This phylogenic tree shows the changes in bacterial species from the ancestral species to the most current split in the modern classification system. This tree supports the theory that <u>speciation results in biodiversity</u>. As we know speciation is the terminal source of the latest species, in a similar way, that modification is the terminal source of genetic divergence within species (and extirpation is comparable to lack of alleles). Inequities in the movements of speciation are therefore expected to provide large scale biodiversity exemplars.
Answer:
A) The two strands of a DNA molecule are parallel and complementary.
Explanation:
DNA is a polymer of nucleotide and this nucleotide are made up of a deoxyribose sugar molecule, a nitrogenous base, and a phosphate group. In DNA genetic information of the cell is stored in the form of nitrogenous bases.
DNA is double-stranded and both the strands run anti-parallel with each other. This anti-parallel orientation provides the nitrogenous base of one strand to form hydrogen bonds with the nitrogenous base present on the opposite strand.
In DNA adenine base pairs with thymine with two hydrogen bonds and guanine base pairs with cytosine with three hydrogen bonds. So the false statement is A.
It’s B, having variation from sexual reproduction because the two parent mix genetic information so they have more of an advantage
Answer:
G - 21%
T - 29%
A - 29%
Explanation:
Nucleotide bases in DNA are complementary. Adenosine (A) binds to Thymine (T) while Cytosine (C) binds to Guanine (G). Hence the composition of A in DNA is the same as that of T; and that of C is the same as that of G.
From the information given, C is 21%
Therefore G is also 21% of the genome as C is bound to G, the therefore are the same proportion.
C and G make up 42% of the genome (that 21% + 21%).
The remaining 58% (100%-42%) is made up of A + T
Similarly the proportion of A is equal to that of T,
Hence A is 29% (half of 58%) and T is 29%.
Indirect methods like mark and recapture become a strong tool to estimate population size or density in species on which it is impossible to apply a direct methods. <em>Because of their biological and ecological characteristics, the Gypsy moth and the Green lizard populations are good examples for which mark and recapture would work well.</em>
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There are different methods to study population density. There are direct methods and indirect methods. Among these last ones, we might find the Mark-Recapture technique.
The Mark-Recapture technique assumes that
- <em>the population is closed during the sampling season, there is no mortality nor natality, </em>
- <em>marks in the individual last the whole sampling season, and they do not affect the marked individual or their behavior. </em>
- <em>marked individuals are randomly distributed in the population, and </em>
- <em>all the individuals have the same probability of being sampled.</em>
The method consists of capturing a sample of individuals belonging to the population under study. After capturing the individuals, the researcher marks and releases them again. The third step is to sample again: The researcher captures new individuals and counts how many of them are marked. These marked individuals belong to the first sample.
Indirect methods like this become a strong tool to estimate population size or density in species on which it is impossible to apply a direct method such as <em>counting individuals</em>.
For instance, if we need to estimate insects population density (<u><em>Gypsy moth population</em></u>) or reptiles population density (<u><em>Green lizard population</em></u>) because of their biological and ecological characteristics, the best way of doing it is by applying indirect methods. <em>These species characterize as small-sized, fast to escape, they can hide in small inaccessible places, they have nocturnal habits, their reproductive rate is too high, and their distribution rate is wide. </em>Among many other characteristics, their population density can not be estimated by direct methods. Mark-Recapture technique is the most suitable one.
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