Answer:
Yes it is milk.
Explanation:
The main component would be the milk, the other 2 would be silk and vanilla to create the substance! Since vanilla is a main component !! It can be classified as milk !!
Dominant, <span>if a dominant allele (wrinkled) and a recessive allele (round) are together, forming a heterozygous plant, any possible combination of alleles is possible when the plant breeds, so the seed with a wrinkled allele can still produce round seeds. </span>
The correct statement is D. APICAL MERISTEMS ARE SOMETIMES CHANGED INTO FLORAL MERISTEMS.
Apical Meristems are found in the tips of the roots and in the buds of the shoots. They supply cells for the plants to grow in length.
Apical Meristems are found in herbaceous plants, woody plants, grasses, and flowering plants.
In flowering plants, shoot apical meristem develops into an inflorescence meristem which produces the floral meristem. The floral meristem is responsible for the production of the sepals, petals, stamens, and carpels of the flower.
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.
<em>Answer</em><em>:</em><em>Nitrogen fixation</em><em> </em><em>is</em><em> </em><em>the</em><em> </em><em>nitrogen</em><em> </em><em>fixing</em><em> </em><em>bacteria</em><em>.</em>