Answer:
Each one has two fatty acid chains and the glycerol backbone is bonded to a small polar group.
Explanation:
Phospholipid is a unique form of lipid. The bonding of the glycerol backbone to the polar phosphate group makes phospholipid to have dual solubility unlike general triglycerides.
The polar head is said to be hydrophillic that is <u>water loving,</u> while the two carbon chains that retained lipid features are hydrophobic <u>water hating.</u>
Therefore if a phopholipid is placed in water, in relation to its functions as component of cell membrane, it forms a bi-layer in which the water loving portion hydrophilic head points into the surrounding watery medium, while the hydrophobic layer points inwards far away from the watery medium into the internal cellular layer to form an impermeable barrier to hydrophilic (polar) substances.
This forms the basis of the phospholipd bilayer of the cell membrane. And it controls the permeability of the cell membrane to influx substances into the cells.
Photosynthesis I know that I took the test
The correct option is B.
In the ecosystem, energy flows from one trophic level to the other. The first trophic level is that of the producers which use the energy of the sun to produce their own food. Out of the energy obtained from the sun by the producers only about 3% of it is converted into food products. The second trophic level is made up of the herbivores and the omnivores which eat the plant. Only about 10% of the energy from the plant is transferred to the animals in the second trophic level. These second trophic level animals will also transfer about 10% of the energy they obtain to the animals in the third trophic level when they are eaten. Thus, it can be seen that the energy that is transferred in the ecosystem is gradually reducing from one trophic level to another.
Answer:
Statement C is the only one that is necessarily true for exons 2 and 3. It is also true for exons 7 and 8. While statements A and B could be true, they don’thave to be. Because the protein sequence is the same in segments of the mRNA that correspond to exons 1 and 10, neither choice of alternative exons (2 versus 3, or 7 versus 8) can alter the reading frame. To maintain the normal reading frame—whatever that is—the alternative exons must have a number of nucleotides that when divided by 3 (the number of nucleotides in a codon) give the same remainder. Since the sequence of the a-tropomyosin gene is known, it is possible to check to see the actual state of affairs. Exons 2 and 3 both contain the same number of nucleotides, 126, which is divisible by 3 with no remainder.
