Answer:
Explanation:
Le parti principali del tessuto linfatico sono il midollo osseo, la milza, la ghiandola del timo, i linfonodi, i linfonodi e le tonsille.
That would be A. tepid moist
Hope I helped!
Answer:
may be
the discovery of new evidence
Explanation:
Accepted theories may be modified or overturned as new evidence and perspective emerges. Scientists are likely to accept a new or modified theory if it explains everything the old theory did and more. The process of theory change may take time and involve controversy, but eventually the scientific explanation that is more accurate will be accepted.
Answer:
The correct answer is B) Transcription, 5' cap addition, addition of poly-A tail, exon splicing, passage through nuclear membrane.
Explanation:
The transcription process in eukaryotes takes place in the nucleus of the cell and after transcription post-transcriptional modification also takes place in the nucleus which is necessary to guide the mRNA out of the nucleus.
First, the process of transcription takes place in which DNA is transcribed to mRNA by an enzyme called RNA polymerase. After transcription post-transcriptional modifications takes place in the given order.
1. Capping: In capping process 7-methylguanosine is added by capping enzyme at 5' end of mRNA.
2. Polyadenylation: In polyadenylation, many poly-A residues are added at the 3' end of the mRNA called the poly-A tail.
3. Intron splicing: Introns are the non-coding sequence present in mRNA which are spliced out of mRNA and all exons are joined together.
After these post-transcriptional process, the mature mRNA is transported out of the nucleus through the nuclear membrane.
Answer:
(a) number of strands (n) = time (t) ÷ proportionality constant (k)
(b) The time needed for the bacterial to double its initial size is 3.36 hours.
Explanation:
(a) Let the rate (time) be represented by t and the amount (number) of strands of bacteria be represented by n
t is proportional to n, therefore, t = kn (k is the proportionality constant)
Since t = kn, then, n = t/k
(b) Initial amount of strands = 300
Amount of strands after 2 hours = 300 + (300 × 20/100) = 300 + 60 = 360
k = t/n = 2/360 = 0.0056 hour/strand
Double of the initial size is 600 (300×2 = 600)
Time (t) needed for the bacterial to double its initial size = kn = 0.0056×600 = 3.36 hours
