Complete question:
1). Determine the character states for the following six characters that are present in species OG, 5, 7, 15, 17, 18. Assign the character state found in the outgroup (OG) as a 0 and the alternative derived state a 1; use only two states per character. Example, stem width: 0 = thin, / = thick 1 <em>(Characters and plant species image in the attached files)</em>
2). Score each taxon using the 0 and 1 notation about and add to this matrix <em>(Matrix in the attached files)</em>
4). Reconstruct the phylogeny that most simply and accurately accounts for the distribution of synapomorphies among ingroup species. Given matrices as small and simple as the above example, one can build the tree from the bottom up in a series of sketches, adding clades or branches representing groups with the fewest synapomorphies near the base of the tree and those with the most at the tips. Using the unlabeled phylogeny as a starting point, draw the stepwise construction of a fully resolved phylogeny to account for all of the synapomorphies and include the synapomorphies that identify each clade (the first one, character 2, is included to get you started; this is shared by all members of the ingroup) (<em>Tree in the attached files)</em>
Answer:
- Stem width: Thick-1 // Thin-0
- Leaf edge (shape): Pointed-1 // Blunt-0
- Flower base (shape): Narrow-0 // Wide-1
- Flower orientation: Downward (Pendant)-1 // Upward (Vertical)-0
- Petal color: Purple-1 // Yellow-0
- Fruit shape (inset on card, on left): Round-0 // Elongated-1
- Complete Matrix in the attached files
- Order in the phylogeny reconstruction: OG --> 7 --> 5 --> 15 --> 17 and 18 (tree + evolutive changes in the attached files)
Explanation:
<em>NOTE: You will find the complete activity in the attached files. </em>
- The principle of maximum parsimony or maximum simplicity states that among all possible trees within a group of species, the most probable is the one that requires us to postulate the least number of evolutionary changes. So, to reconstruct a phylogeny we need to choose the tree that requires the less number of changes.
- To determine the character states for all the six characters, we assigned the number cero to all the characters expressed in the out-group. Thin steam, blunt leaf, narrow flower base, upward flower orientation, yellow petals, and rounded-seeds are all cero states. The other characters are 1.
- The above information of character state was used to fill in the matrix, specifying which character state belongs to each species according to their traits.
- The matrix was useful to reconstruct the phylogeny, to identify the autapomorphic trait, and to visualize all the clades.
<span>they both have two alleles
</span>they both express the dominant allele (aka the phenotype or physical traits of the allele). They are different in genotype though, which is their genetic make up because homozygous would be two dominant alleles and heterozygous is a combination of dominant and recessive allele. Homozygous recessive and heterozygous are completely different <span>though in genotype and phenotype.</span>
Answer:
1. metaphase I
2. telophase II
3. anaphase I
4.prophase II
5. prophase I
6. anaphase II
7. telophase I
8. metaphase II
Explanation:
1. During _metaphase I________, pairs of homologous chromosomes align in the center of the cell.2. During _telophase II___, the separated chromatids elongate and (usually) cytokinesis occurs, formingfour genetically distinct haploid daughter cells.3. During _anaphase I_______, homologous chromosomes separate by moving with the spindle microtubulestoward the poles.4. During __prophase II______, a spindle apparatus forms and individual chromosomes (each composed ofsister chromatids) begin to move toward the center of the cell.5. During _prophase I_______, chiasmata form and crossing-over occurs.6. During _anaphase II_______, sister chromatids separate and move toward the poles.7. During __telophase I______, the separated chromosomes cluster at the poles of the spindle and cytokinesis occurs, forming two daughter cells, each with a haploid set of replicated chromosomes.8. During ___metaphase II_____, individual chromosomes (each composed of sister chromatids) align in the center of the cell.
Answer:
NASA satellites help scientists study Earth and space. Satellites looking toward Earth provide information about clouds, oceans, land and ice. They also measure gases in the atmosphere, such as ozone and carbon dioxide, and the amount of energy that Earth absorbs and emits.
Explanation:
Gap 2 (G2): During the gap between DNA synthesis and mitosis, the cell will continue to grow and produce new proteins. ... Mitosis or M Phase: Cell growth and protein production stop at this stage in the cell cycle. All of the cell's energy is focused on the complex and orderly division into two similar daughter cells.