Answer:
this mutation may change the open reading frame of the resulting RNA sequence and its final product, which is a protein in the case that this gene is used to synthesize a messenger RNA (mRNA) sequence
Explanation:
During the transcription, a region of DNA named 'gene' is used as template to produce an RNA molecule, typically a primary transcript of mRNA (pre-mRNA). Subsequently, this pre-mRNA suffers a process named RNA processing in order to generate a mature mRNA which is finally used to create a protein by a process called translation. If a deletion occurs during transcription, it may change the open reading frame (ORF) of the resulting mRNA when the mutation occurs in an exon of the protein-coding gene (i.e., occurs a frameshift mutation), while this deletion may not have any effect if it is localized within the introns which are removed during RNA processing. A frameshift mutation will change the amino acids that are added to the nascent polypeptide chain during translation.
the stomata are mostly found on the under surface of plant leaves
The experiment by Peter Agre who discovered the protein Aquaporin demonstrated the function of proteins that allow facilitated diffusion of water.
Explanation:
Aquaporins are water channel proteins that facilitated diffusion of water. Agre was able to prove its function when a frog’s oocytes showed increased water permeability when an unknown protein (membranous) from the RBCs was introduced into them. The oocytes without proteins did not facilitate water diffusion.
Aquaporins are membranous proteins that belong to larger intrinsic proteins but act as pores on the cellular membranes to facilitate water diffusion.
Answer:
Diet includes three materials such as lipids, protein and carbohydrates. Lipids are chemically digested by bile.
Proteins are chemically digested by duodenum.
Carbohydrates are chemically digested by Amylase in the mouth.
Lipids are converted into fatty acid, proteins are converted into amino acid and carbohydrates are converted into glucose.
Vitamins absorption of digestion food across villi and micro villi takes place.
Answer:
The correct answer is E. none of the above. The population will drops below 100 when t ≥ 38.
Explanation:
Given A= A0 e^kt. The population 10 years ago is A0, the population today is A(10), and we have to find the value of "k" and then the time when population drops below 100.
So, A(t) = 1700 e^kt ⇒ A(10) = 1700 e^k(10) ⇒ 800 = 1700 e^k(10) ⇒
800/1700 = e^k(10) ⇒ln (800/1700) = k(10) ln e ⇒ -0.754/10 = k ⇒
k = -0.0754.
Now you have all the parameters, so you can find the time at which the population drops below 100.
A(t) = 1700 e^kt ⇒ 100 = 1700 e^(-0.0754)t ⇒100/1700 = e^(-0.0754)t ⇒
ln(100/1700) = (-0.0754)t ln e ⇒ [ln(100/1700)]/(-0.0754) = t ⇒
t = 38.
So, the population will drops below 100 when t ≥ 38.
