Answer: Each body segment of an earthworm contains four pairs of setae, which are used to move the worm. The setae of marine worms are found in bundles. Leeches do not have setae at all.
Sample Response
Explanation:
Diffusion across the alveoli
Answer:
Oxidize organic compounds AND use organic compounds as terminal electron acceptors. (Ans. E)
Explanation:
Anaerobes are the organism that does not require oxygen for their growth. The anaerobes are subdivided into three main classes:
1) Obligate anaerobes: These organisms use anaerobic metabolism to grow and killed in the presence of oxygen. Obligate anaerobe examples are Clostridium and Propionibacterium.
2) Facultative anaerobes: These organisms prefer to grow using aerobic metabolism processes and switch to anaerobic metabolism in the absence of oxygen. An example of facultative anaerobic bacteria is Lactobacillus.
3) Aerotolerent anaerobes: These organisms use anaerobic metabolism to grow and can survive in both conditions (oxygen and oxygen-free environment).
Anaerobic bacteria, oxidize organic compounds rather than oxygen & use organic compounds as terminal electron acceptors. Anaerobic bacteria release hydrogen sulfide gas as they decompose algae in the water. This process is also known as Anaerobic Respiration, in which formation of ATP without oxygen.
Answer:
0.0177
Explanation:
Cystic fibrosis is an autosomal recessive disease, thereby an individual must have both copies of the CFTR mutant alleles to have this disease. The Hardy-Weinberg equilibrium states that p² + 2pq + q² = 1, where p² represents the frequency of the homo-zygous dominant genotype (normal phenotype), q² represents the frequency of the homo-zygous recessive genotype (cystic fibrosis phenotype), and 2pq represents the frequency of the heterozygous genotype (individuals that carry one copy of the CFTR mutant allele). Moreover, under Hardy-Weinberg equilibrium, the sum of the dominant 'p' allele frequency and the recessive 'q' allele frequency is equal to 1. In this case, we can observe that the frequency of the homo-zygous recessive condition for cystic fibrosis (q²) is 1/3200. In consequence, the frequency of the recessive allele for cystic fibrosis can be calculated as follows:
1/3200 = q² (have two CFTR mutant alleles) >>
q = √ (1/3200) = 1/56.57 >>
- Frequency of the CFTR allele q = 1/56.57 = 0.0177
- Frequency of the dominant 'normal' allele p = 1 - q = 1 - 0.0177 = 0.9823
Answer:
The most appropriate answer would be option B.
The flaw in their reasoning is that they compared different molecules in a different set of organisms.
For example, they compared the DNA of lizard A and B whereas they compared the RNA of lizard B and C.
In addition, the mutation rates of different molecules (DNA and RNA) are different and thus, comparison of DNA in one set of organism and comparison of RNA in another set cannot be used as the basis for the conclusion.
