Which of the following is NOT prokaryotic?

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 (Characters and plant species image in the attached files)

2). Score each taxon using the 0 and 1 notation about and add to this matrix (Matrix in the attached files)

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) (Tree in the attached files)

Answer:

Characters:

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:

NOTE: You will find the complete activity in the attached files.

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.

8 0

3 years ago

At different places one earth, depending on elevation, air pressure varies. What causes air pressure?

sleet_krkn [62]

Weight of air

Depending on elevation, the air pressure varies. This is because of the weight of the air that is exerting pressure on earth. At higher elevation, there is less layer of air that is exerting pressure so the atmospheric pressure is also less. At lower altitudes or at sea level, more layer of air is exerting pressure so the atmospheric pressure is also more. The sun’s energy has nothing to do with air pressure. The energy that the Earth gets from the sun is more or less constant. The same with density of nitrogen, and the earth’s magnetic field, they have nothing to do with air pressure.

7 0

4 years ago

Explain why mutations may or may not affect phenotype

Svetlanka [38]

Answer:

Explanation:

Mutations may or may not affect the phenotype because they can occur in coding or non coding areas of the DNA. If they occur in coding areas of the DNA, the protein that is coded in that part could be processed incompletely or have a different amino-acid in some part of its structure, this would affect the phenotype. If they occur in non coding areas, this would have no effect on the phenotype.

3 0

3 years ago