Answer:
D
an increase in the length of the fatty acid tails.
Explanation:
As average annual temperatures decrease, we would expect to find phospholipids within the cell membranes of organisms to have both saturated
and unsaturated fatty acid tails. Saturated fatty acids tails are arranged linearly, in a way that maximizes interactions between the tails and
decreases bilayer fluidity. Unsaturated fatty acids, on the other hand, have more distance between the tails, fewer intermolecular interactions and
more membrane fluidity. Longer tails and decreased cholesterol also decrease fluidity
The other two Domains being "Bacteria" and "Eukarya" (which includes us humans). Archaebacteria<span> are characterized by having different cell call components, coenzymes and RNA Sequences compared to bacteria. ... </span>Eubacteria<span> have a rigid cell wall. However, they can be both gram positive and gram-negative.</span>
Answer:
The correct answer would be -
If the type of food available changes, then the frequency of beak also changes because the beak of the bird suited to food will survive successfully.
Explanation:
According to the theory of natural selection, an organism that is able to adapt according to the change in the environment, it helps in their survival and increases their number in the system.
It is given that the type of food available is changing, so it will lead to the change in the frequency of bird beaks in that particular area. So, if they adapt to survive under the changes in the available food type, otherwise not be able to survive and die.
So,
If the type of food available changes, then the frequency of beak also changes because the beak of the bird suited to food will survive successfully.
I think it’s the first one
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
