Answer:
B) transcription and translation.
Explanation:
DNA is the genetic material and stores genetic information. The genetic information stored in DNA is expressed in the form of genetic traits by two processes: transcription and translation. The template strand of DNA is copied into RNA strand by the enzyme RNA polymerase. The process is called transcription and occurs in the nucleus.
The RNA formed by the process of transcription serves to carry the genetic information to the cytoplasm. The mature RNA, mRNA leaves the nucleus in eukaryotes and joins ribosomes in the cytoplasm to serve as a template for protein synthesis. The process of protein synthesis is called translation as it translates the genetic information of mRNA into the amino acid sequence of proteins. Translation and transcription occur simultaneously in prokaryotes.
Answer:
The correct answer is - all cells arise from pre-existing cells.
Explanation:
Robert Virchow observed during the cell division in half that a single cell divide into two new cells therefore he suggested that, a famous aphorism Omnis cellula e cellula which means all cells arise from pre-existing cells. His observation is one of the major points of the cell theory.
This observation made by Robert Virchow becomes the fundamental basis of the cell theory which describes that cell divides to produce new cells. The other greatest observation made by him was that all cells not get sick or affected but a group of cells only.
My best answer for you is:
An earthquake with a shallow focus and a magnitude of 4.5
Answer:
Please see the answer below
Explanation:
a. The roman numerals represent the various generations in the pedigree.
I represents first generation
II represents the second generation
b. The half shaded square and circle representing the parents means that both of them are carriers for the trait in question. <u>Tay-Sachs disease is an autosomal recessive inheritance, assuming the trait is represented by the allele </u><u>a</u><u>, the gentoypes of the parents would be </u><u>Aa </u><u>and </u><u>Aa </u><u>respectively.</u>
c. Still on the assumption that the disease is represented by the allele a. If offpsring II4 which happens to be free of the Tay-Sachs allele marries a person that also does not carry the Tay-Sachs disease gene, the cross will be as shown below.
AA x AA = AA, AA, AA, and AA.
<em>All the children would be free from the Tay-Sach's allele. That is, they will be homozygous normal.</em>