Answer:
independent assortment, crossing over and random fertilization
Explanation:
There are three mechanisms by which sexual reproduction increases genetic variation in progeny. First, random fertilization is the process by which gametes randomly combine in order to increase the possibilities for genetic variation. Second, the independent assortment refers to the process by which pairs of homologous chromosomes independently separate from one another during meiosis. Finally, crossing over, also known as recombination, is the process by which two homologous chromosomes' non-sister chromatids can exchange genetic material during meiosis I (prophase I).
Answer: RNAs are not processed before translation in prokaryotes, this process only takes place in eukaryotes.
Explanation:
Messenger RNA or mRNA is a single-straded ribonucleic acid that transfers the genetic information from the DNA (deoxyribonucleic acid) molecule of the cell nucleus to a ribosome (which are the machinery responsible for protein synthesis) in the cytoplasm. mRNA determines the order in which the amino acids of a protein will be joined and acts as a template or pattern for the synthesis of that protein. To accomplish this, the DNA molecule must be transcribed into an RNA molecule, which is used for protein synthesis.
The messenger RNA obtained after transcription is known as primary transcribed RNA or precursor RNA or pre-mRNA, which in most cases is not released from the transcription complex in a fully active form, but in eukaryotes it must undergo modifications before it can perform its function (RNA processing or maturation). These modifications include:
- Elimination of fragments (splicing): In most cases, the <u>mRNA undergoes the removal of internal, non-coding sequences called introns, and the connection of exons. This does not occur in prokaryotic cells</u>, as they do not have introns in their DNA.
- Protection by CAP: <u>Addition to the 5' end of the structure called "cap" or "capping"</u>, which is a modified guanine nucleotide, 7-methylguanosine triphosphate, via a 5'-5' triphosphate linkage, instead of the usual 3',5'-phosphodiester linkage. This cap is necessary for the normal RNA translation process and to maintain its stability.
- Polyadenylation signal: <u>Addition to the 3' end of a poly-A tail, a long polyadenylate sequence, whose bases are all adenine</u>. Its addition is mediated by a sequence or polyadenylation signal (AAAAAA), located 11-30 nucleotides upstream of the original 3' end. This tail protects the mRNA from degradation, and increases its half-life in the cytosol, so that more protein can be synthesized.
The mature mRNA (in eukaryotes) is transferred to the cytosol of the cell through pores in the nuclear envelope. Once in the cytoplasm, ribosomes are coupled to the mRNA. However, in prokaryotes, ribosome binding occurs while the mRNA strand is being synthesized. After a certain amount of time, the mRNA is degraded into its component nucleotides by ribonucleases. So, the transcription and translation processes are carried out in a similar way as in eukaryotic cells but they occur simultaneously. But, the fundamental difference is that, in prokaryotes, the messenger RNA does not undergo a maturation process and, therefore, no cap or tail is added and no introns are removed. Moreover, it does not have to leave the nucleus as in eukaryotes, because in prokaryotic cells there is no defined nucleus.
So, RNAs are not processed before translation in prokaryotes, this process only takes place in eukaryotes.
It seems that you forgot the answer choices, but lucky for you, I know the answer..
Answer: creating skin tissue by combining cells from different cell lines
Removal of brain tissue via conventional surgery with a scalpel would be termed biopsy but removal of brain tissue by way of vacuum suction would be called neurosurgical vacuum suction.
The biopsy This involves a small incision and hole in the skull, the removal of small amount of brain tissue and its examination under the microscope.
The neurosurgical vacuum suction includes
In this example the marker DNA includes fragments that have 23,130, 9,416, 6,557, 4,361, 2,322, 2,027, and 564 base pairs. Appro
iris [78.8K]
Answer:
48,377
Explanation:
by adding all the fragments
