Answer:
The percentage of caterpillars original energy available to the hawk is = 0.6%.
Explanation:
the answer above is given by; 24 kcal/4000 kcal x 100
Caterpillars are the larval stage of members of the order Lepidoptera. As with most common names, the application of the word is arbitrary, since the larvae of sawflies are commonly called caterpillars as well. Both lepidopteran and symphytan larvae have eruciform body shapes.
Hawks are a group of medium-sized diurnal birds of prey of the family Accipitridae. Hawks are widely distributed and vary greatly in size. The subfamily Accipitrinae includes goshawks, sparrowhawks, sharp-shinned hawks and others. This subfamily are mainly woodland birds with long tails and high visual acuity.
Answer:
transport of protons (H+) from low concentration in the mitochondrial matrix to high concentration in the mitochondrial intermembrane space
Explanation:
atpase pump can also be called atp synthase. this enzyme catalyses atp formation from adenosine diphosphate and phosphate. it has f1, stalk and f0 components. 3 positive hydrogen ions go through to make 1 adenosine triphosphate molecule. oxidative phosphorylation has to do with the loss of electrons. there would be electrons loss from NADH to FADH2. Cytochromes carries them through different series of transferases from I to IV and while on this positive hydrogen ions are released into mitochondrial matrix
positive hydrogen ions are moved back to lumen through adenosine triphosphate channels. a process called chemiosmosis. the pro
Answer:
1. Allele frequency of b = 0.09 (or 9%)
2. Allele frequency of B = 0.91 (0.91%)
3. Genotype frequency of BB = 0.8281 (or 82.81%)
4. Genotype frequency of Bb = 0.1638 (or 16.38%)
Explanation:
Given that:
p = the frequency of the dominant allele (represented here by B) = 0.91
q = the frequency of the recessive allele (represented here by b) = 0.09
For a population in genetic equilibrium:
p + q = 1.0 (The sum of the frequencies of both alleles is 100%.)
(p + q)^2 = 1
Therefore:
p^2 + 2pq + q^2 = 1
in which:
p^2 = frequency of BB (homozygous dominant)
2pq = frequency of Bb (heterozygous)
q^2 = frequency of bb (homozygous recessive)
p^2 = 0.91^2 = 0.8281
2pq = 2(0.91)(0.9) = 0.1638
Answer:
Mesophyll cells.
Explanation:
Transpiration is the evaporation of water at the surfaces of the spongy mesophyll cells in leaves, followed by loss of water vapour through the stomata . Transpiration produces a tension or 'pull' on the water in the xylem vessels by the leaves. Water molecules are cohesive so water is pulled up through the plant.
Answer:
Hershey & Chase concluded. after the experiment that the genetic material that is passed from virus to the bacteria is not the protein coat but DNA. It proves the DNA is the genetic material and not the protein.
After labelling the Phage DNA and protein, Hershey and Chase performed a series of experiments like infection, blending and centrifugation. Hershey and Chase gave full evidence of the DNA being a genetic material by their experiments. To perform the experiment, Hershey and Chase have taken T-2 bacteriophage (invaders of E.coli bacteria).
Explanation:
