<span>An adaptation is a trait that "evolved" for a particular condition or circumstance that allows the animal to survive. Something is adaptive if it is a useful trait but not necessarily one that occurred in response to a particular selection pressure.</span>
Answer. D
Explanation:
After the messenger RNA (mRNA) is formed, it has to go trough different steps before being translated into proteins, also known as maturation of the mRNA. One of the most important steps is splicing, a process that removes the introns (regions of the sequence that do not codify for any particular amino acid sequence or protein). In other words, the splicing process removes sequences that do not generate any protein, leaving only the exons to be translated into protein. However, the genes (within our DNA) will contain exons and introns; Therefore, the gene sequence will have more kilobases compared to mature mRNA as the introns in this macromolecule have been spliced out.
I hope this clarify your question
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Answer: D
Explanation:
The lipopolysaccharides are found in the outer membrane of Gram-negative bacteria. Lipopolysaccharide is complex molecule consisting of a lipid and chains of carbohydrates. They are also called endotoxin, they are the major component of the outer cell membrane. Sugars in the polysaccharide chains of endotoxin confer serologic specificity. They lplay a crucial role in bacteria resistance to antibiotics.
The nervous system and the integumentary system<span />
Answer:
3/4
Explanation:
If we assume simple dominance and independent assortment for each trait, we can use Mendel's Law of Segregation to predict the phenotypic proportions in the offspring of the parental cross AABBCc x AabbCc.
<h3><u>Gene A</u></h3>
AA x Aa
- F1 genotypes: 1/2 AA, 1/2 Aa
- F1 phenotypes: all A
<h3 /><h3><u>Gene B</u></h3>
BB x bb
- F1 genotypes: 1 Bb
- F1 phenotypes: all B
<h3 /><h3><u>Gene C</u></h3>
Cc x Cc
- F1 genotypes: 1/4 CC, 2/4 Cc, 1/4 cc
- F1 phenotypes: 3/4 C, 1/4 cc
We want to know the proportion of progeny with all dominant phenotype (A_B_C_). Since the genes are independent, we can multiply the probabilities of each gene to obtain the overall probability of having a ABC progeny:
<h3>1 A_ x 1 B_ x 3/4 C_ = 3/4 A_B_C_</h3>