Answer:
The correct answers are option A. "tethering proteins to the cell cortex", B. "using barriers such as tight junctions", C. "tethering proteins to the extracellular matrix", D. "forming a covalent linkage with membrane lipids", E. "tethering proteins to the surface of another cell"
Explanation:
According to the fluid-mosaic model, the components of cell membranes are in constant movement forming a barrier to avoid unwanted exterior component internalization and to avoid the loss of precious internal components. This constant movement could cause that proteins move across the plasma membrane. But, this is avoided by several mechanisms including:
A. Tethering proteins to the cell cortex. The cell cortex is a rigid structure made of actin and actomyosin. Proteins found in the plasma membrane are tethered to this structure to restrict their movement.
B. Using barriers such as tight junctions. Tight junctions are barriers found in epithelia made of claudin and occludin proteins. These barriers are impenetrable, which avoid the movement of proteins in the cell membrane.
C. Tethering proteins to the extracellular matrix. The extracellular matrix is made of several proteins and macromolecules that provide a structural and biochemical support to cells that are nearby. Proteins could be tethered to this rigid structure as well.
D. Forming a covalent linkage with membrane lipids. The proteins in the cell membrane that form a covalent linkage with membrane lipids are known as lipid-anchored proteins, or lipid-linked proteins.
E. Tethering proteins to the surface of another cell. When cell-cell communication take place it is possible that proteins in the cell membrane got tethered to the surface of the other cell.
The answer is B. Meiosis involves two cycle of cell division.
Yes all animals humans and bugs have hearts
Answer:
Frequency of B allele is 0.6681
Explanation:
If p represents the frequency of dominant allele and q represents the frequency of recessive allele, according to Hardy-Weinberg equilibrium:
p + q = 1
p² + 2pq + q² = 1
where p² = frequency of homozygous dominant genotype
q² = frequency of homozygous recessive genotype
2pq = frequency of heterozygous genotype
Given that number of recessive chestnut horse = 28
Total horses = 226 + 28 = 254
frequency of b² genotype = 28/254 = 0.1102
frequency of recessive b allele = √0.1102 = 0.3319
So, frequency of B allele =
1 - 0.3319 = 0.6681
Hence frequency of B allele is 0.6681
