Answer:
Adding more substrate would overcome the effect of the compound
Explanation:
- Enzymes are biochemical catalysts that speed up chemical reactions. They act on specific substrate to convert them to products.
- Compounds known as inhibitors slow down the rate of enzyme activity.
- Inhibitors are classified as competitive and non-competitive inhibitors.
- Competitive inhibitors will compete with the substrate to bind the active sites on the enzyme. The effect of competitive inhibitors may be reduced by increasing the concentration of the substrate.
- The compound added by the biologist was a competitive inhibitor and therefore adding more substrate would overcome its effect on enzyme catalysis
- Non-competitive inhibitors binds the active site of the enzyme permanently and prevents the substrate from accessing the active sites.
Answer:
D. Genes received from the offsprings parents.
Explanation:
Answer:
5
Explanation:
Firstly, we convert what we have to percentage compositions.
There are two parts in the molecule, the sulphate part and the water part.
The percentage compositions is as follows:
Sulphate- (103.74)/(103.74 + 58.55) × 100% = apprx 64%
The water part = 100 - 64 = 36%
Now, we divide the percentages by the molar masses.
For the CuSO4 molar mass is 64 + 32 + 4(16) = 160g/mol
For the H2O = 2(1) + 16 = 18g/mol
Now we divide the percentages by these masses
Sulphate = 64/160 = 0.4
Water = 36/18 = 2
The ratio is thus 0.4:2 = 1:5
Hence, there are 5 water molecules.
Answer:
Water would not be able to transport nutrients -‐-‐ in plants, or in our bodies -‐-‐ nor to dissolve and transport waste products out of our bodies. ... Cohesiveness, adhesiveness, and surface tension: would decrease because without the +/-‐ polarity, water would not form hydrogen bonds between H20 molecules.
The answer for this issue is:
The chemical equation is: HBz + H2O <- - > H3O+ + Bz-
Ka = 6.4X10^-5 = [H3O+][Bz-]/[HBz]
Let x = [H3O+] = [Bz-], and [HBz] = 0.5 - x.
Accept that x is little contrasted with 0.5 M. At that point,
Ka = 6.4X10^-5 = x^2/0.5
x = [H3O+] = 5.6X10^-3 M
pH = 2.25
(x is without a doubt little contrasted with 0.5, so the presumption above was OK to make)