Answer:
It recognizes and binds to a pair of "mismatched" nucleotides, preventing their translation.
Explanation:
Mut L protein is involved in mismatch DNA repair. MutL protein is complexed with MutS protein and the MutL-MutS complex recognizes all the mismatched base pairs present in the newly formed DNA strand. The complex can not recognize the "C-C" pairs. MutH protein joins the complex.
The MutH protein also has a site-specific endonuclease activity and cleaves the unmethylated DNA strand towards the 5' end of the guanine base in the GATC sequence to mark the strand for DNA repair. In this way, MutL protein, along with MutS and MutH proteins mark the mismatched DNA bases for repair so that they are not translated into a faulty protein.
Answer:
maltose
Explanation:
Amylase, any member of a class of enzymes that catalyze the hydrolysis (splitting of a compound by addition of a water molecule) of starch into smaller carbohydrate molecules such as maltose (a molecule composed of two glucose molecules).
The chemical reactions in the cell would not happen as fast and would require more energy to catalyze the reaction between the two reactants.
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Chemistry/ Example: Take breathing for example, when you breath you breath out carbon dioxide. The CO2 can't just leave like that and only 10% binds to hemoglobin. The rest turns into carbonic acid in your blood and its plasma. However, the acid is unstable, so it turns into bicarbonate and a dissociated proton (H). You have carbonic anhydrase that converts the two so you can breath out CO2; the carbonic acid separates into H2O and CO2. This process would take a LONG time without the enzyme-- CO2 build up, even minimal amounts it lethal.
Answer:
First, we take the owl out, the total number of mouse increases at a higher rate and the chipmunks are also decreasing in numbers. In an ideal ecosystem, both mice (lets call it "a") and chipmunks (lets call it "b") should increase since the restraining factor has been removed. But that is the opposite of what we should see. In that case, we will take the owl to be the "limiting factor" for the rodents.
Am
The rodents will have free movement and exercise their fitness over the area, competition sets in between the two species. So we see, a flourishes while b dies out. This can be viewed most predictably that a has an overall greater fitness and would easily get resources and strive readily, which influences the survival rate for 2.
Hope you got something in any realm of understanding?
