Answer:
1)
- frequencies of light-colored mice ≅ 0.74
- frequencies of dark-colored mice ≅ 0.26
2)
- frequencies of light-colored mice ≅ 0.13
- frequencies of dark-colored mice ≅ 0.87
3)
- q² = 0.74
- p² = 0.02
- 2pq = 0.24
4)
- q² = 0.13
- p² = 0.4
- 2pq = 0.46
5)
The dark-colored fur seems to have the greatest overall selective advantage
6)
Dark lava, that changed the color of the substrate, from light to dark.
7)
Because to produce dark color, animals from the different regions suffered different mutations that drove them to have almost the same dark fur color. All of the animals are inhabiting dark substrate, which means that this environmental condition is favoring the same phenotype.
8)
To see if the mice population is evolving, you need to take a sample of animals per year, through many years, and analyze if it is changing or not. If the population is evolving, you will notice a change in the allelic and genotypic frequencies over the years, favoring one genotype or the other. If the population is not evolving, the frequencies will keep equal through the years, it will not change.
Explanation:
Due to technical problems, you will find the complete explanation in the attached files.
The answer to your question is fish
<span>Everyone has cells because these are the building
blocks of life. These tiny particles that clump into groups that form tissues,
organs, and organ systems are
what makes organisms distinct from non-living things that exist on Earth. Where
there are cells, life is present, and in its absence life cannot exist as we
know of today. Cells are responsible for bringing different species of
organisms that are found in different ecosystems all over Earth. They are tiny
but in groups they are responsible for every living organisms that have existed
through time. </span>
Answer:
transcription of mRNA from DNA
small ribosomal subunit binds to mRNA
initiation complex formed with addition of large ribosomal subunit
translocation
codon recognition (non-initiating site)
peptide bond formation
ribosome reads a stop codon
polypeptide chain is released from the P site
ribosomal subunits dissociate
Explanation:
The above describes the process of translation in the ribosome. After transcription of DNA to mRNA, the mRNA is taken to the ribosome to undergo translation, here the mRNA binds to the small ribosomal subuits and to other initiation factors; binding at the mRNA binding site on the small ribosomal subunit then the Large ribosomal subunits joins in.
Translation begins (codon recognition; initiating site) at the initiation codon AUG on the mRNA with the tRNA bringing its amino acid (methionine in eukaryotes and formyl methionine in prokaryotes) forming complementary base pair between its anticodon and mRNA's AUG start codon. Then translocation occurs with the ribosome moving one codon over on the mRNA thus moving the start codon tRNA from the A site to the P site, then codon recognition occurs (non-initiating site again) which includes incoming tRNA with an anticodon that is complementary to the codon exposed in the A site binds to the mRNA.
Then peptide bond formation occurs between the amino acid carried by the tRNA in the p site and the A site. When the ribosome reads a stop codon, the process stops and the polypeptide chain produced is released and the ribosomal subunits dissociates.
That would be A.
Dicots grow flowers with petals in multiplies of 4 and 5, monocots in multiples of 3. See the attached pic:)
