Explanation:
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Answer: 400 k-cal
I remember learning that 4,000 k-cal would be approximately 40 k-cal for the primary consumers, which, that would be how much would be available. Now, if we would want for the primary consumers to consume more k-cal, it's obvious that there would then have to be more k-cal from the producers level of the ecosystem.
Formula of k-calories: k-cal÷20(4)= k-cal = k-cal × 2 = k-cal
As we plug in this formula into our problem above, our answer would be explained below.
40,000 ÷ 20 = 2,000 × 2 = 4,000 ÷ 100 = 400 k-cal
Answer: 400 k-cal
Answer:
Replication would not occur on either the leading or lagging strand.
Explanation:
In DNA replication in E. coli, the enzyme primase is used to attach a 5 to 10 base ribonucleotide strand complementary to the parental DNA strand. The RNA strand serves as a starting point for the DNA polymerase that replicates the DNA. If a mutation occurred in the primase gene, replication would not occur on either the leading or lagging strand.
Answer:
some of the things they may be competing for would be sunlight,nutrients in the soil, and water
Answer:tailing at 3' end protects mRNA from attack by 3' exonuclease and 5' capping is useful for recognition of mRNA
Explanation:
The mRNA formed and released from the DNA template is known as primary transcript. In mammalian system, it undergoes intensive modification to become the mature mRNA.
Post transcription processing includes removal of Introns, splicing of exons, poly-A tailing at 3'end and capping at 5' end.
The Poly-A tailing at 3' end occurs in the nucleoplasm, the 3' end is polyadenlated involves with 20-250 nucleotides long. This tail protects mRNA from attack by 3' exonuclease activity
5' end capping end also is done in the nucleus. The cap is useful in recognition of mRNA by the translating machinery.
