Answer: c. The cuticle prevents water loss in a terrestrial environment.
Explanation:
The earthworms have a thin cuticle outside their epidermis. The cuticle performs different functions in the worm body. The cuticle protects the worm from predators. The thin cuticle remains moist with the slimy mucus. The cuticle helps in the exchange of gases from the external environment. The cuticle helps in absorbing the oxygen from the external environment whereas it expels the carbon dioxide to the external environment. The cuticle prevents the loss of moisture and allows the gaseous exchange which facilitates the survival of the worm in a harsh terrestrial environment.
Answer:
0.549 is the frequency of the F allele.
0.495 is the frequency of the Ff genotype.
Explanation:
FF or Ff genotypes determine freckles, ff determines lack of freckels.
In this class of 123 students, 98 have freckles (and 123-98= 25 do not).
If the class is in Hardy-Weinberg equilibrium for this trait, then the genotypic frequency of the ff genotype is:
q²= 25/123
q²=0.203
q=
q= 0.451
q is the frequency of the recessive f allele.
Given p the frequency of the dominant F allele, we know that:
p+q=1, therefore p=1-q
p=0.549 is the frequency of the F allele.
The frequency of the Ff genotype is 2pq. Therefore:
2pq=2×0.549×0.451
2pq=0.495 is the frequency of the Ff genotype.
The frequency of the dominant allele, F, in this class is 0.55. The frequency of the heterozygous genotype, Ff, is 0.495. This is calculated using Hardy-Weinberg equilibrium and observed phenotype frequencies.
To start, we need to calculate the frequency of the recessive allele, f, which is easily calculated as those who do not have freckles. From a total of 123 students, 98 have freckles, leaving 25 students with no freckles, which represents individuals who are homozygous for the recessive trait (ff). As these are the only individuals we can be sure of, we take the square root of their frequency to get the frequency of the recessive allele, q. In this case, q = sqrt(25/123) = 0.45. To find the frequency of the dominant allele, p, we subtract q from 1 (because p + q = 1), so p = 1 - q = 0.55.
Next, we'll calculate the frequency of the heterozygous genotype Ff.
Using Hardy-Weinberg equilibrium, we know this is represented by 2pq. Hence, the frequency of genotype Ff would be 2 × 0.55 × 0.45 = 0.495.
This process offers an example of applying the principles of population genetics and Hardy-Weinberg equilibrium to determine the likely genotype frequencies in a given group of individuals based on observed phenotype frequencies.
#SPJ11
a. 8n
b. 1n
c. 4n
d. 2n
e. 0n
b.what is the ploidy level at prophase of mitosis
The answer is d. 2n
Drosophila melanogaster have four pairs of chromosomes, so the normal cell would have 4n(four pairs) of chromosome. metaphase is the phase of cell division where the chromosome start to lining up in the middle. In meiosis II, the chromosome is already divided into half, so the total chromosome of the cell would be 2n(half of the 4n)
Answer:
punctuated equilibrium
Answer:
Explanation:
Chlorofluorocarbons: are nonflammable chemicals and have many uses in our life, containing atoms of carbon( C), chlorine(Cl), and fluorine (F).
Chlorofluorocarbons damage the earth's ozone layer.
CFCs was banned in 1996. before banned, they were used in aerosols, refrigerators, air conditioners and many other machines.
there continuous use may also cause global warming.
b. identical copies.
c. separated during mitosis.
d. made during the S phase.
e. All of the above.
Answer:
The correct answer is e. All of the above.
Explanation:
Sister chromatids are made during the synthesis phase of the cell cycle. In the synthesis phase, the homologous chromosomes get replicated and sister chromatids are produced so they are produced by duplication of chromosomes.
As sister chromatids are produced by replication, therefore, they are identical copies of parent chromosomes. These sister chromatids are joined to each other at centromere. They get separated during the anaphase of mitosis and moves to the opposite pole.
Therefore the right answer is e.