This looks like a test. What is your question about it?
a) okay so enzyme catalyzed reactions; i know that there is not change in the delta g standard and there is not change in the products (but what is the s referring to?)
b) and c) im confused about what the question is asking
d) false the rate depends on the transition state
A. False. Under steady state conditions, the concentrations of intermediates stay the same while the concentrations of the SUBSTRATE and PRODUCT change. Steady state conditions refer to the steady concentrations of of the ES complex not the concentration of the substrate and product.
B. I would think this is false as well. Glucose molecules in the human body are connected by alpha-1-4 linkages to other glucose molecules to form glycogen. The reason that this is so is because this allows glycogen to be more accessible to form simple sugars for cell energy requirements. Conversely, beta-1-4 linkages are broken down to form monomers that are used for the construction of fibers and other linkages that require high tensile strength; an example of this is Chitin.
C. False. Most of the carbohydrate patterns that are seen in nature are in that pattern for a particular reason, and are optically active. Think about D-Ribose and L-Ribose and how if your body was to use L over D in the formation of RNA, how this may affect the production of mRNA and proteins. Remember from organic chemistry that one enantiomer, say R, is usually interpreted by receptors in your body as one thing while the S is interpreted as something completely different by the same receptors.
D. I believe that this is false as well. Delta G only tells if their is enough energy for a reaction to occur, not what the rate can or will be. You are right to a point; lowering the energy for the transition state lowers the energy of activation, thus increasing the rate of the reaction. However, the transition state is just equal to the delta G and is just the energy requirement needed to go from reactants to products. The rate of a reaction is increased by several different things: catalysts, reactants (to a point), and heat (depending if you are breaking or forming bonds). All of the several things that I mentioned lower the energy requirements to overcome the transition state.
I hope this helps, but I am not guaranteeing you that all information provided is correct.
From your other posts, I see that you may not have taken organic chemistry yet. Most universities require that you take and pass organic chemistry before you can take biochemistry. If UCI allows you to take biochemistry before taken and passing organic chemistry is beyond stupid. I am mentioning this because you may not know what I am talking about for question E, so google R and S Carvone which should clarify what I meant in the answer that I provided for E.