Answer:
A) Bacteria cannot carry out RNA splicing to remove introns and so produced a much larger protein.
Explanation:
Human is a eukaryote and has both introns and exons in its genes. Transcription of human genes forms a primary transcript that undergoes post-transcriptional modification.
One of the important even during the post-transcriptional modification is the removal of introns and joining the exons together to make a mature mRNA which in turn serves as the template for protein synthesis.
<em>E. coli</em> is a prokaryote and does not have the enzymatic machinery required for the splicing of introns.
Cloning of a complete human gene into the <em>E. coli</em> cells would not form the respective human protein since the bacterial cells would not be able to splice the introns from the primary transcript.
One of the factor is the Sun's elevation where the UV reaches us at a much lower frequency when the sun is more distant. Next is our ozone, with our ozone depleting as time goes by, the amount of solar energy it can diffuse is much lesser thus the heat is much more than the previous decades.
The muscle cells divide through Mitosis, a cellular division comprised of several other substeps.
The final step of the cell division is the cytokinesis by which two new cells are formed from cell with a multiple number of nucleus after the replication process. This steps follow the telophase.
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
